Antenna and Wave Propagation by Yadava

Antenna and Wave Propagation R.L. YADAVA Professor Department of Electronics and Communication Engineering Galgotias College of Engineering and Technology Greater Noida, Uttar Pradesh

New Delhi-110001 2011

ANTENNA AND WAVE PROPAGATION R.L. Yadava © 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-4291-0 The export rights of this book are vested solely with the publisher. Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Mohan Makhijani at Rekha Printers Private Limited, New Delhi-110020.

To My Parents and Children

Contents

Preface ....................................................................................................................................... xv Chapter 1

INTRODUCTION .......................................................................................... 1–12 Definition of Antenna 1 Radiation from an Antenna Historical View 3 Antenna Polarization 4 Typical Applications 5 Impedance Matching 6

2

VSWR and Reflected Power 6 Antenna Resonance 6 Bandwidth 7 Directivity, Gain and Beam Width

7

Radiation Patterns 8 Pattern Terminology 8 Types of Propagation 9 Surface Wave Propagation 9 Space Propagation 9 Troposphere Propagation 9 Ionospheric Propagation 10

Objective Type Questions Exercises 11 References 12 Chapter 2

10

ELECTROMAGNETIC WAVES AND RADIATION .......................... 13–55 Introduction 13 EM Wave Spectrum and Its Applications 14 EM Fields and Maxwell’s Equations 16 Poynting Vector and Velocity of EM Waves v

17

vi

Contents

Wave Polarization

17 18

Mathematical Interpretation of Polarization

19

Velocity of Propagation

Plane Wave and Uniform Plane Wave

20

20

Propagation of EM Waves in Different Mediums Power Flow of a Uniform Plane Wave 26 Incidence of Uniform Plane Wave 27 Oblique Incidence 29 Snell’s Law 30 Total Transmission 32 Total Reflection 33

Solved Examples 35 Objective Type Questions Exercises 55 References 55 Chapter 3

50

ANTENNA FUNDAMENTALS AND PARAMETERS ...................... 56–113 Introduction

56

Types of Antenna 56 Applications of Antenna

58

Isotropic Radiator and Radiation Field Far-field Region

Antenna Parameters

58

59

61

Input Impedance and VSWR 61 Equivalent Circuits of Tx Antenna 63 Equivalent Circuit of Rx Antenna 65 Bandwidth 66 Impedance Bandwidth 66 Pattern Bandwidth 67 Radiation Bandwidth 68 FBR 68 Radiation Resistance 69 Directivity 69 Antenna Gain and Efficiency 73 Radiation Pattern 74 HPBW, FNBW, Side Lobe Level and Antenna Resolution 75 Radiation Intensity, Beam Efficiency and Solid Angle 78

80

Effective Aperture and Effective Height Effective Aperture

80

Antenna Noise Temperature, Noise Figure and SNR Antenna Noise Temperature 84 Effective Noise Temperature and Noise Figure Signal to Noise Ratio (SNR) 86

Antenna Coupling 87 Antenna Polarization and Parameters

84

86

90

Antenna Polarization 90 Field Polarization in Terms of Two Circularly Polarized Components

91

Contents Polarization Vector and Polarization Ratio 92 Polarization Loss Factor and Polarization Efficiency

vii

92

Reciprocity Theorem 93 Solved Examples 95 Objective Type Questions 109 Exercises 111 References 113 Chapter 4

ANTENNA ARRAY ................................................................................ 114–180 Introduction

114 115

Design Considerations and Design Approach

115

Array Configurations

116 116 117 117

Broadside Array End-fire Array Collinear Array Parasitic Arrays

Principle of Pattern Multiplication 123 Array with n-isotropic Point Sources of Equal Amplitude and Linear Spacing 124 125 128

Broadside Array End-fire Array

Electronic Phased Array 129 Effect of Earth on Vertical Patterns Comparison of Methods

130

131

132

Dolph–Tchebyscheff or Chebyshev Array

Tchebyscheff Polynomial 132 Dolph Pattern Method of Obtaining Optimum Pattern Using Tchebyscheff Polynomial 134

Calculation of Dolph–Tchebyscheff Amplitude Distribution Advantages of Dolph–Tchebyscheff Distribution 136 Beam Width between First Nulls of Chebyshev Polynomial Patterns

Stacked/Rectangular Area Broadband Array Directivity

138

Super Directive Receiving Array

140

SNR and Directive Gain 140 Sensitivity Factor 141 Radiation Efficiency of Super Directive Array

Adaptive Array

142

Weighting of Signals 143 Adaptive Antenna in Cellular Systems

Binomial Array

147

Disadvantages of Binomial Array Mutual Coupling between Arrays

Solved Examples 150 Objective Type Questions Exercises 177 References 180

175

137

148 149

145

142

135 136

viii

Contents

Chapter 5

LINEAR WIRE ANTENNAS ................................................................ 181–234 Introduction

181 181

Small Dipole and Radiation Mechanism

Hertzian Dipole 183 Half-wave Dipole Antenna

186

Radiation Resistance and Input Resistance Parameters of a Dipole Antenna 190

190

Monopole Antenna 193 Folded Dipole Antenna 194 Theoretical Analysis 195 Input Impedance of Folded Dipole Antenna 197 Applications of Folded Dipole Antenna 197

198

Harmonics Antenna

199

Parametric Specifications

V-dipole Antenna

202 204

Design Parameters

206

Sleeve Wire Antenna

Sleeve Monopole 207 Design Specifications and Experimental Results Sleeve Dipole Antenna 210 Open-sleeve Dipole Antenna 211 Performance Characteristics 212

Beverage Antenna or Wave Antenna Principle of Operation

Rhombic Antenna

209

215

216

217

Advantages 220 Disadvantages 220


229

LOOP ANTENNAS ................................................................................. 235–274 Introduction

235

Historical View

235

237

Principle of Operation Radiation Fields

239

Induced EMF of Loop Antenna 242 Radiation Pattern of Loop Antenna 243 Large Loop Antenna 245 Loop Antenna Parameters 249 Maximum Effective Area and Gain

250

Multi-turn Loop Antenna 251 Impedance of a Loop Antenna 254 Resistive Part of Impedance

255

Impedance of Loop Antenna in Conducting Medium

256

Contents

ix

Ferrite Rod Antenna 259 Applications of Loop Antenna 261 Solved Examples 261 Objective Type Questions 270 Exercises 271 References 273 Chapter 7

METAL-PLATE LENS ANTENNAS ................................................... 275–302 Introduction

275

Metal-Plate Lens Antennas 276 Lens Antenna Design 277

Types of Lens Antenna

279

E-plane Lens Antenna 279 H-plane Metal-Plate Lens Antenna

282

Antenna Analysis and Design 283 Distribution of Illustration on the Aperture of the Lens Luneburg Lens Antenna 287 288

Maxwell’s Fisheye Lens Applications 289

Rotman Lens Antenna

290

Applications 291 Tolerances in Lens Antenna


286

292

300

PARABOLIC REFLECTOR ANTENNAS ......................................... 303–326 Introduction 303 Feeding Systems 306 Basic Requirements of Reflector Feeding Systems

306

Parabolic Reflector 309 Field Distribution on an Aperture of Parabolic Reflector Parabolic Reflector Antenna Parameters 314

312

316

Polarization Loss Efficiency

Parabolic Cylindrical Antenna 317 Multiple-reflector Antenna 318 Advantages of Dual Reflector


320

323

YAGI ANTENNA .................................................................................... 327–353 Introduction 327 Principle of Operation

328

x

Contents

Design Parameters Length of Elements

328 328

Design of Yagi–Uda Antenna 332 Hansen–Woodyard End Fire Array 335 Analysis of Yagi Antenna 338 System of Linear Equation Far-field Pattern 341

340

Circular Polarization from the Yagi–Uda Array Yagi–Uda Loop Antennas 343 Effects of Array Parameters 346 Effect of the Reflector 346 Effect of the Directors 346 Effect of Exciter and Wire Cross-section

342

346

Advantages and Applications 347 Solved Examples 351 Objective Type Questions 350 Exercises 351 References 352 Chapter 10

LOG-PERIODIC ANTENNA ............................................................... 354–382 Introduction 354 Mathematical Analysis and Design Parameters

355

Feed Techniques 358 Effect of Truncation on Efficiency and Radiation Pattern

Performance Characteristics and Design of LPDA Power Gain 362 Input Impedance and VSWR Design of LPDA 364 E-plane Pattern 365

363

Types of Log-periodic Antennas Log-periodic Log-periodic Log-periodic Log-periodic

359

362

367

Toothed Planar Antenna 367 Toothed Wedge Antenna 369 Toothed Trapezoid Antenna 370 Yagi–Uda Array Antenna 371

Log-periodic Yagi–Uda Array Design and Performance Characteristics Applications of Log-periodic Antenna 374 Solved Examples 376 Objective Type Questions 379 Exercises 381 References 382 Chapter 11

374

HORN AND CONE ANTENNAS ....................................................... 383–419 Introduction

383

Waveguide and Horn Antennas 383 Various Configurations of Horn Antennas

Horn Antenna Parameters

387

384

Contents

H-plane Sectoral Horn Antenna E-plane Sectoral Horn Antenna Pyramidal Horn Antenna 394

390 392 396

Polarization of Pyramidal Horn

Bi-conical Antennas

399 400

Radiation Pattern of Bi-conical Antennas

405

Broadband Slotted Cone Antenna

407

VSWR, HPBW and Polarization

408

Advantages and Applications of Horn Antenna Solved Examples 409 Objective Type Questions 415 Exercises 418 References 419 Chapter 12

xi

HELICAL ANTENNAS ........................................................................ 420–443 Introduction

420

Reflector Model of Helix

421

Parameters of Helix Antenna 423 Types of Helix Antenna 423 Axial Mode Helical Antenna Normal Mode Helical Antenna

423 428

Helical Antenna with Different Finite Ground Size Hemispherical Helix 432 Applications of Helical Antenna 434 Helical Antenna for G.P.S. Applications


430

435

440

MICROSTRIP ANTENNA .................................................................. 444–476 Introduction

444

Basic Configurations of Microstrip Antenna

Fringing Fields and Mechanism of Radiation

445

446

Advantages of Microstrip Antenna 447 Disadvantages of Microstrip Antenna 447 Applications of Microstrip Antenna 447

Feeding Techniques of Microstrip Antenna

447

Probe Feed 448 Microstrip Line Feed 448 Aperture Coupled Feed 449 Proximity Coupled Feed 450

Bandwidth Enhancement and Higher Order Modes Rectangular Patch Antenna 454 Radiated Fields 454 Design Parameters 455

452

xii

Contents Radiation Resistance Radiation Conductance Directivity 457

455 456

Circular Patch Antenna 457 Circularly Polarized Microstrip Antenna Other Parameters 463 Double Tuning 463 Coupling 463 Selection of Substrate Material Laminate Composite for MSA Photonic Band Gap Antennas Mobile Antennas 465 Integrated Antennas 465


461

464 464 464

472

SURFACE WAVE PROPAGATION ................................................. 477–511 Introduction 477 Historical View 478 Characteristics of EM Waves 479 Applications of EM Waves 480 Fundamental Equation of Wave Propagation 481 Electric Field Intensity at Finite Distance from Tx Antenna Modes of Wave Propagation 484 485

Surface Wave Propagation Surface Wave Tilt 486

Plane Earth Reflection

483

487

Reflection Coefficient for Vertical and Horizontal Polarizations

Refraction and Reflection of Waves

488

489

Refraction 490 Reflection 490 Phase Difference between the Direct and the Ground Reflected Waves

Field Strength at Finite Distance due to Ground Wave Field Strength due to Vertically Polarized Wave 494 Field Strength due to Horizontally Polarized Waves 496 Relation between A, p and b 496

Multi-hop Transmission

498

Effect of Ground 499 Effects of Polarization 499

Solved Examples 500 Objective Type Questions Exercises 509 References 511

507

493

492

Contents

Chapter 15

xiii

TROPOSPHERIC AND SPACE WAVE PROPAGATIONS .......... 512–547 Introduction 512 Troposphere 514 Troposphere Wave Propagation 514 Relation between the Radius of Curvature and Change of Dielectric Constant 518 Troposphere Scattering and Troposcattering Propagation 522 523

Transmission Loss

524

M-Curves and Duct Propagation 526

Duct Propagation Loss Diffraction 527

Space Wave Propagation 529 Space Wave Propagation Parameters Solved Examples 535 Objective Type Questions Exercises 545 References 547 Chapter 16

531

531

Line of Sight (LOS distance) Field Strength 533

542

IONOSPHERIC PROPAGATION ...................................................... 548–586 Introduction

548

Historical Views

549

Structure of Ionosphere

549

Propagation Effect as a Function of Frequency

Measures of Ionosphere Propagation

554

555

The Critical Frequency 555 The Angle of Incidence 555 Maximum Useable Frequency 557 Calculation of MUF 558 Lowest Useable Frequency 561 Optimum Working Frequency 561 Skip Distance 562 Relation between Angle of Incidence and Skip Distance Virtual Height 563

Refractive Index of the Ionosphere Effect of the Earth Magnetic Field Gyrofrequency 569 Gyromagnetic Field (GMF)

562

566 568

570

Regular and Irregular Variations in the Ionosphere Eleven-year Sunspot Cycle and 27-day Sunspot Cycle Sudden Ionospheric Disturbances (SID) 571 Ionospheric Storms 571

Fading 572 Attenuation Factor for Ionospheric Wave Propagation

570 570

574

xiv

Contents

Energy Loss due to Collision in Ionosphere Solved Examples 576 Objective Type Questions 583 Exercises 584 References 586

575

Chapter 17 ANTENNA MEASUREMENTS ........................................................... 587–638 Introduction

587 591

Anechoic Chamber

Radiation Pattern Measurement Concept of Near- and Far-fields Far-field Measurements 599 Near-field Measurements 599

596 598

Measurement of Reflectivity 600 Beam Width and Directivity Measurements Measurement of Radiation Efficiency 601

601

Wheeler Method 602 Q-Method 603

Polarization Pattern Measurement

604

Polarization Pattern Method 605 General Method of Polarization Measurement

Gain Measurement

Absolute Gain Method 612 Gain Transfer (Gain Comparison) Method Gain Measurement of CP Antenna 618

Impedance Measurement

617

618

Mutual Impedance between Dipole Antennas

Current Measurement

608

611

620

621

Measurement of Current Distribution for Antenna on a Finite Conducting Earth 622 Design Specification 622

Phase Measurement 623 Measurements of Noise Figure and Noise Temperature Solved Examples 629 Objective Type Questions 633 Exercises 636 References 637

625

Appendices ....................................................................................................................... 639–675 Glossary ........................................................................................................................... 677–785 Review Questions ............................................................................................................ 687–698 Question Bank with Solutions ...................................................................................... 699–726 Index ................................................................................................................................ 727–732

Preface

The present book is organized into two parts. The first part of the book is devoted to the study of various types of antennas and the second part to the phenomenon of wave propagations. Each chapter provides an in-depth understanding of all the important aspects of the particular topic. In addition, each chapter contains sufficient number of solved examples, exercises and many related references. The division of the book into seventeen chapters is as follows. Chapter 1 provides a historical introduction to the subject and enumerates its applications. Chapter 2 deals with the concept of electromagnetic (EM) wave radiation and wave propagation in different mediums. Chapter 3 is devoted to the fundamental properties and parameters of antennas. Chapter 4 focuses on the antenna array and its synthesis, whereas linear wire antennas and loop antennas are covered in Chapters 5 and 6 respectively. Each type of antenna has been described in detail, particularly the adaptive array, and the beverage and ferrite-rod antennas. Loop and metal-plate lens antennas have been described in Chapter 7, and Chapter 8 describes the reflector antenna in full length. Yagi–Uda and log-periodic antennas have been described in Chapters 9 and 10 respectively. Horn/Cone and helical antennas along with concepts of design and applications are explained in Chapters 11 and 12. Microstrip antennas and their various types and applications have been covered in Chapter 13. Chapters 14, 15 and 16 deal with surface, space and ionospheric wave propagations and the related parameters. Chapter 17 is fully devoted to the measurement of antenna properties, such as gain, efficiency, radiation pattern, noise-figure and noise-temperature, etc. Just after Chapter 17, there is a question bank which contains numerical problems and review questions pertaining to the entire text. So far as the appendices are concerned, Appendix A discusses the scalar and vector potentials. Appendix B provides a list of antenna’s expressions. Appendix C discusses the relative permittivity (er) and relative permeability (mr) of materials. Appendix D deals with the requirement of antennas for different applications. Appendix E throws light on frequency spectrum for various communications systems. Poincare’s sphere and Stokes parameters have also found place in Appendix F. Appendix G discusses the dB, dBm and dBi technology and their relations. The important terms are listed in the Glossary. xv

xvi

Preface

I thank the Almighty for providing me such a wonderful opportunity to be the author of such a standard book. I feel immense pleasure in offering my heartiest and humble thanks to honourable Prof. B.R. Vishwakarma, Banaras Hindu University, for his sagacious guidance and motivation, without which the book could not have seen the light of day. I express my grateful thanks to all the professors, senior faculty and scientists of the department of ECE, IT, BHU for their motivation and encouragement during the writing of the book. The support and environment provided by the Management of Galgoitas Educational Institutions (GEI) is very much appreciated. Also, the pleasant support of senior professors, HODs and Registrar of Galgotias College of Engineering and Technology deserves special thanks. The constant motivation and help of colleagues and technical staff of the Department of Electronics and Communication Engineering are sincerely acknowledged. Also, it is a great pleasure and honour for me to be associated with PHI Learning. I express my sincere gratitude to the entire team, including the production department for maintaining a high degree of precision and accuracy in the production of the book. I am indebted to all the reviewers for their valuable suggestions and comments, which have helped me in adding more student-friendly material in the book. My sincere thanks go to Prof. R.K. Yadava, Er. Surendra Prasad and Mr. Rakesh Kumar for their encouragement and silence support. Their untiring efforts in searching the materials and ardent exercises towards word processing are deeply appreciated. I also express my deep sense of gratitude to my colleagues and well wishers: Prof. Vikram Singh, Prof. Krishnraj and Prof. Yashpal Singh, Prof. B.K. Kanoujiya, Prof. V.K. Pandey and Prof. Neetha Awasthi for their unstinted cooperation in early completion of the book. My thanks are also due to Prof. R.P. Yadava, Prof. S.K. Koul, Prof. Ashok De, Prof. D.R. Bhasker, Dr. Sushrut Das and Prof. D.C. Dhubkariya for their comments and suggestions on some topics of the book. I am grateful to my parents who have inculcated in me good values and culture. My wife, Om Lalitha, expressed her solidarity with me by allowing me to work for long hours. My sincere thanks go to my children Om Vikash, Saichethana and Om Prakash as without their wholehearted cooperation and sacrifices, this book would not have come in the present shape. Last but not the least, I wish to avail myself of this opportunity to express a sense of gratitude and love to my beloved research scholars and students for their helpful support during the preparation of this book. Though every effort has been made to produce an error-free text. However, I shall feel much obliged to readers who can provide valuable suggestions/feedback to enhance the quality of the book. R.L. YADAVA

C H A P T E R

1

Introduction

DEFINITION OF ANTENNA An antenna is defined as a transitional device between free space and wave-guiding structures. An antenna may also be considered a transducer used in matching the guiding device to the surrounding medium or vice versa. The antennas used to transmit and receive radio energy are called transmitting and receiving antennas respectively. There are many structures that can radiate and receive the electromagnetic (EM) energy, but not all can serve the purpose efficiently. Most antennas are resonant devices, which operate efficiently over a relatively narrow frequency band. Antennas forward voltage and current from the transmission line and EM fields from the wave guide to launch the EM waves (radiation) into the space/medium. A complete description of radiation of EM waves through an antenna is shown in Fig. 1.1.

Transmitter E-lines

Receiver EM waves P0

Source

P1

Tx line

Tx line

FIG. 1.1

Radiation mechanism of an antenna.

An antenna must be tuned (matched) to the same frequency band as the radio system to which it is connected. If not so, reception and/or transmission of energy will be less effective. Though there are varieties of antenna such as dipoles, folded dipoles, helix, reflectors 1

2

Antenna and Wave Propagation

and patch antennas, all of them function according to the same principle—the principle of electromagnetism. Besides radiating and transmitting signals, antenna optimizes the radiation in some direction and suppresses it in the other directions, that is, antenna also acts as a directional device in addition to a probing device. In communication systems, antennas are needed for two purposes—efficient radiation and matching wave impedance in order to minimize reflections [1]. As far as their importance is concern, antennas are used at many places—homes, vehicles, radars, ships, satellites and in mobile phones as well; this is because of their compact structure at microwave frequencies. Antennas are employed in different systems in different ways. In some cases, the operational performance of the system is defined around the directional properties of the antenna. Antennas link us with the entire world. It can be said that they are the backbone of communication technology.

RADIATION FROM AN ANTENNA In order to describe the radiation of waves from an antenna, let us refer to Fig. 1.1, where a two-conductor transmission line is connected with voltage source at one end and an antenna at second end at Tx side and vice versa at Rx side. Applying the voltage across transmission line creates electric force lines which are tangential to the electric field at each point and their strength varies with the electric field intensity. The force lines have a tendency to act on the free electrons associated with each conductor and force them to move. By this tendency, the movements of charges generate current, that in turn creates magnetic field intensity. Associated with magnetic field intensity are magnetic lines of force, which are tangential to the magnetic fields. Since electric field lines start at positive charge and end at negative charge, they can also start at positive end and end at infinity or vice versa. In other words, they form closed loops neither starting nor ending at any charge. Magnetic field lines always form closed loops encircling current-carrying conductors. Therefore, it is convenient to introduce magnetic charges and magnetic currents to a region involving electric and magnetic sources. Since the existence of electric field lines between two conductors represents distribution of charges, so if the source voltage is sinusoidal the electric fields between the conductors need also to be sinusoidal with period equal to that of the applied source. The relative magnitude of the electric field intensity will be indicated by bunching of the force lines, with the arrows showing the relative directions (positive or negative). This continuous creation of timevarying electric and magnetic fields between conductors forms EM waves, which travel along the transmission line and finally enter the antenna. Now, suppose that the antenna is not there; then free space waves can be formed by connecting the open ends of the electric lines (dashed lines). The waves are also periodic, but a constant phase point P0 moves outwardly with the speed of light c and travels a distance of l/2 to point P1 in the time duration of onehalf of a period. It is observed that near the antenna the constant phase point P1 moves with a speed greater than c and approaches c at point far away from the antenna. These waves are detached from the antenna and transformed into the free space waves by forming closed loops [1]. Once EM waves are created by an electric disturbance, they travel inside

Introduction

3

the transmission line and then into the antenna, and are finally radiated as free space waves even if the disturbance is ceased. If the disturbance is continued, EM waves will exist continuously and follow in their travel behind the others. This phenomenon is termed radiation. At the receiving end the electric field lines of radiating waves strike as plane waves and are finally transformed into guided waves and received at the receiver.

HISTORICAL VIEW In general, an antenna means a metallic device (a rod or wire) used to radiate or receive radio energy. However, scientifically speaking, antennas are a group of conductors that transmit an electromagnetic field in response to EMF signals. In olden days, antennas were called aerials; in Japan it is still known as middle sky wire. The history of antennas began with the design of a loop antenna in the year 1887 by Hertz [1]. He proposed a complete radio system integrated with an end-loaded dipole transmitting antenna and a resonant square loop antenna as receiver. In addition, he also exercised testing with a parabolic reflector antenna. It was not until 1901 that Marconi managed to design an array of 50 copper wires in the form of a fan connected to the ground through a spark transmitter. Since then several antennas have been designed; patch antennas and fractal antennas are also among them. The big difference in today’s antennas is that decades ago they were mostly huge and heavy, while modern antennas are usually of small size, low profile and light weight. The greatest advancement of antenna was accomplished during the Second World War with the invention of high frequency antenna in the form of reflection apertures and arrays. Apart from designing and experimentation, numerical methods were also introduced to analyze complex antenna systems in the 1960s. The first mathematical formulation for radiation mechanism for many antennas was given by the scientists of Bell Laboratories. Their achievement bridged the gap between theory and experiments, and provided better understanding of the antenna to the people of the world. In subsequent years, various models such as moment methods, finite difference and finite element models for low frequencies, and geometrical and physical theories of diffraction for high frequencies, were also introduced to analyze various antennas. The antenna as a boundary value problem was treated long back by Abraham. However, the earliest treatments of the cylindrical centre-driven antenna as a boundary value problem are the Hallen and L.V. King models. The Hallen method leads to integral equations and approximate solutions of which give the current distribution. Using Hallen method the input impedance of the antenna can be determined as the ratio of voltage to current at the terminals, if the terminal is connected with a voltage source [2]. With the advent of high frequency antennas, the concept of radio wave and wireless communications were developed around 1920. In 1924, Professor Hidetsugu Yagi and Shintaro Uda designed and constructed a sensitive and highly-directional antenna, very important for radar, television, and amateur radio. The antenna was named Yagi–Uda antenna based on their names. A United States patent was issued to Yagi in 1932 and assigned to RCA. Such antennas were available for television by the late 1930s, but were more widespread for receiving purposes in the 1940s and 1950s. The

4


theory and design of linear wire antenna was introduced by R.W.P. King and his colleagues at Harvard University. A new antenna known as helical antenna was introduced by J.D. Kraus through a research paper entitled “The helical beam antenna”, which was published in the April 1947 issue of Journal of Electronics. The design and working principle of slot antenna and array was described by Babinet and extended by Henry Brooker. The widespread interest in antennas is reflected by the large number of books written and research papers being published on the subject [3, 4]. In recent years, antennas such as patch antenna, fractal antenna, PIFA antenna and many more have played a vital role in modern communication and related technologies.

ANTENNA POLARIZATION The polarization of an antenna is defined as the orientation of the wave (electric field) being transmitted or received during propagation with negligible loss. So all the characteristics of wave polarization can directly be correlated to antenna polarization. The polarization characteristics of an antenna are usually found to be constant over its main beam, and the polarization on the main beam peak is preferably used to describe the polarization of the antenna. However, the radiation from side lobes differs in polarization greatly from the main beam polarization. In case the direction of polarization is not specified, it is taken to be the direction of maximum gain of the antenna. However, in general, polarization of radiation varies with the direction from the centre of the radiator. A transmitting antenna is matched to a receiving antenna if its polarization, axial ratio, sense and major axis rotation are identical. Polarizations are classified as linear, elliptical and circular polarizations. An antenna is said to be vertically polarized (linear) if its electric field is perpendicular to the earth’s surface. The vertical polarization is mostly dominant in surface wave propagation, where wave propagation takes place along the earth’s curvature. On the other hand, antennas having their electric field lines along the earth’s surface are termed horizontally polarized antennas. Most communication systems use vertical and circular polarizations. A circularly polarized antenna radiates energy in both the horizontal and vertical planes and all planes in between. Circular polarization of an antenna is characterized by a factor known as axial ratio (AR)—which is the ratio of fields in the vertical and horizontal planes. The values of axial ratio are found to be 1 £ AR £ ¥. If the values of axial ratio lie between 0 and 2 dB, the antenna is said to be circularly polarized. If the axial ratio is greater than 2 dB, the polarization is often referred to as elliptical, whereas AR is infinite for linear polarizations. Polarization may also be classified as co-polarization and cross-polarization. The polarization in the plane of radiation is termed co-polarization; however cross-polarization is observed orthogonal to the co-polarization plane. With reference to directions of rotation of the field vector, polarization may also be classified as clockwise polarization (RHP) and counterclockwise polarization (LHP). In a linearly polarized system, a misalignment of polarization of 45° will degrade the signal by up to 3 dB and if misaligned by 90° the attenuation can be 20 dB or more. Likewise, in a circular polarized system, the antennas must have the same sense. For example, in the line-of-sight (LOS) propagation, it is very

Introduction

5

important that the antennas (Tx and Rx) should have identical polarizations. If not, an additional loss of 20 dB or more may be incurred. The main advantage of polarization matching between antennas (Tx and Rx) is the maximum transfer of energy with minimum loss.

TYPICAL APPLICATIONS Depending on specifications, antennas have applications in various fields—communications, sensing, policing, medical, etc. It is important to have the right antennas for each device. Circular polarization is most often used on satellite communications. This is particularly desired since the polarization of a linear polarized radio wave may be rotated as the signal passes through any anomalies (such as Faraday rotation) in the ionosphere. Furthermore, due to the position of the Earth with respect to the satellite, geometric differences may vary, especially if the satellite appears to move with respect to the fixed Earth-bound station. Circular polarization maintains signal strength which remains constant despite these anomalies. Circularly polarized antennas are normally more costly than linear polarized types, since true circular polarization is difficult to achieve. An example of true circularly polarized antenna is the helix. They are mainly used in mobile stations. The performances of these antennas greatly affect the performance of the mobile station. Helical antennas are also used in portable communication devices, such as cellular phones, because of their wideband characteristics and having the advantage of requiring small space. However, the most common circularly polarized antenna uses crossed Yagi’s for ‘near-circular’ or elliptical polarization [3]. Applications of Yagi’s would include any system usually less than 1000 MHz—in the HF and SHF portions of the spectrum, where antenna gain and directivity are factors. Avionic systems and VHF radars were very suitable technical areas for Yagi–Uda antenna applications. Vertically polarized antenna is often preferred whenever there is a need to transmit radio waves in all directions, such as in widely distributed mobile units. It also works well in the hill areas. As a result, nowadays most two-way wave communications in the frequency of greater than 30 MHz use vertical polarization. Horizontal polarization was originally chosen because there is an advantage in not having TV reception interfered with by vertically polarized stations, and hence it is used mainly to broadcast television in the USA. Since man-made radio noise is predominantly vertically polarized, the use of horizontal polarization would provide some discrimination against interference from noise. In the early days of FM radio in the 88–108 MHz frequency range, the radio stations broadcasted horizontal polarization. However, in the 1960s FM radios became popular in automobiles which used vertical polarized receiving whip antennas. That why the FCC has modified its standard and permits FM stations to broadcast elliptical or right hand polarization to improve reception to vertical receiving antennas as long as the horizontal component was dominating. There are various types of whip antenna; l/4 whip antenna, l/8 whip antenna and l/16 whip antenna; each has its own importance and applications. Naturally, mobile communication needs antennas. The right antenna improves quality of Tx and Rx, reduces power consumption, lasts longer and improves marketability of the device. In particular, stubby helical antennas are used in the frequency range of 800 MHz to 2 GHz and patch antennas for GPS devices.

6


IMPEDANCE MATCHING There is a term called impedance matching which we come across in various areas of electronics, particularly EM wave and antennas. In the field of antennas it is a sort of impedance adjustment between the source, connector (Tx) and antenna to avoid loss of energy. The source always has its own internal resistance; so wherever a load is connected to its output terminal, some of the output power is dissipated as heat, unless we maintain the source temperature close to absolute zero (–273°C). A survey on how the power varies as the load resistance is varied shows that the amount of power reaches maximum when the load resistance is the same as the source resistance and it is altered in all other cases (higher or lower). The idea of adjusting the load and source resistances/impedances is termed impedance matching. For efficient transfer of energy, the impedance of the source, the antenna and the transmission line must be the same. Sources typically are designed for 50 W impedance, and the coaxial cables (transmission lines) used with them should also have a 50 W impedance. However even efficient antenna configurations often have impedance other than 50 W. Therefore some sort of impedance matching circuit is then required to transform the antenna impedance to 50 W. Radial/Larsen antennas come with the necessary impedance matching circuitry as part of the antenna. We must use low-loss components in the matching circuits to provide the maximum transfer of energy between the transmission line and the antenna. The most common matching network is a l/4 transformer matching, which is nothing but a l/4 wavelength long transmission line of characteristic impedance Z 0 R A , where Z0 is characteristic impedance of transmission line and RA is antenna input resistance. The other matching network/devices are lumped elements, stub turners and baluns.

VSWR and Reflected Power The VSWR stands for voltage standing wave ratio and indicates the quality of the impedance matching. The values of VSWR lie between 1 and ¥; the values 1 and ¥ correspond to proper and improper matchings respectively. A high VSWR indicates that the signal is reflected prior to being radiated by the antenna. Another disadvantage of high VSWR is that very high voltages will be generated at certain points along a transmission line, which are called hot spots and may cause arcing. A VSWR of 2.0:1 is often considered good and leads to 89% power transmission. Most commercial antennas, however, are specified to be 1.5:1 or less over some bandwidth. Based on a 100 W radio, a 1.5:1 VSWR equates to a forward power of 96 W and a reflected power of 4 W, i.e. the reflected power is 4.2% of the forward power.

Antenna Resonance By definition an antenna is a form of tuned circuit consisting of resistance, capacitance, and inductance; as a result, an antenna has a resonant frequency, that is, the frequency at which the inductive and capacitive reactances cancel each other out; therefore at this frequency,

Introduction

7

an antenna is purely resistive and a combination of the loss resistance and the radiation resistance. The capacitance and inductance of an antenna are determined by the physical properties, dimensions and the surrounding environment.

Bandwidth The bandwidth (BW) of an antenna is usually defined as the frequency range within the performance of the antenna with respect to certain characteristics. It is expressed as the percentage of the difference between upper and lower frequency to the centre frequency and f − fO BW = u × 100% . The bandwidth of an antenna can also be defined in terms of f0 VSWR − 1 , where Q is a radiation patterns or VSWR/reflected power as follows: BW = Q VSWR quality factor of the antenna.

Directivity, Gain and Beam Width Directivity is the ability of an antenna to focus energy in a particular direction when transmitting or to receive energy better from a particular direction when receiving. Gain is the practical value of the directivity. The relation between gain and directivity includes a new parameter, which describes the efficiency of the antenna; G = hDmax. The simplest method of measuring gain is to compare the antenna under test with a known standard antenna. This method is known as gain transfer technique. At lower frequencies, it is convenient to use a l/2-dipole as the standard. At higher frequencies, it is common to use a calibrated gain horn as a gain standard with gain typically expressed in dBi. Beam width describes the angular aperture where the most important part of the power is radiated. In general, we referred 3 dB beam width which represents the aperture (in deg) on radiation pattern, in which about 92% of the energy is radiated in major lobe. Beam width of an antenna is measured in terms of two parameters: HPBW and FNBW. Decibels (dB) are the accepted parameter to describe the parameters of an antenna. The beauty of dB is that they may be added and subtracted. A decibel relationship (for power) is calculated using the following formula dB = 10 log Power. Table 1.1 tabulates various such relationships. TABLE 1.1

Relationships between decibels and power

Power gain

Power loss

3 dB = 2 ´ Power 6 dB = 4 ´ Power 10 dB = 10 ´ Power 20 dB = 100 ´ Power

–3 dB = 1/2 Power –6 dB = 1/4 Power –10 dB = 1/10 Power –20 dB = 1/100 Power

8


RADIATION PATTERNS Radiation or antenna pattern describes the relative strength of the radiated field at various orientations from the antenna at a constant distance. The radiation pattern is a ‘reception pattern’ as well, since it also describes the receiving properties of the antenna. Based on radiated fields from the antenna, there are two types of radiation pattern: near-field and farfield patterns. The term near-field refers to the field pattern existing close to the antenna, and far-field to the field pattern at large distances. The far-field is also called the radiation field and varies as per (1/r). The near-field is also called the induction field (although it also has a radiation component) and varies as (1/r2), where r is distance away from the antennas. For pattern measurement it is important to choose a distance sufficiently large to be in the farfield, well out of the near-field. The lowest permissible distance varies as per the dimensions of the antenna in relation to the wavelength, as r ≥

2D 2

. The intensities of near- and far-

M0 fields are found to be equal at a distance almost equal to l /2p from the transmitting antenna. Where D is larger aperture of an antenna and l0 is operating wavelength.

PATTERN TERMINOLOGY The various terms related to antenna pattern which describe radiation characteristics are: (a) Directional antenna: It is highly directive antenna, which radiates/ receives power very efficiently and effectively in the particular directions than in the rest directions. It is usually applied to the antennas whose maximum directivity is greater than that of a half-wave dipole. (b) Omni-directional antenna: Antenna which has a non-directional pattern in a given plane, and a directional pattern in any orthogonal plane. That is, omni-directional pattern is a special kind of directional pattern. (c) Isotropic pattern: It is a pattern of antenna having equal radiation in all directions. This is an ideal concept generally it is achievable only in light sources. It is usually referred for expressing the directive properties of practical antennas. Isotropic pattern is represented by a sphere whose centre coincides with the location of the isotropic radiator. Other patterns of the antenna pattern are pencil beam pattern, fan beam pattern and shaped beam pattern. (d) Principal pattern: These are the radiation patterns of lineally polarized antennas. They are measured particularly in E-plane and H-plane of radiation area. E- and H-plane patterns contain electric and magnetic field vectors as well as the corresponding direction of maximum radiation respectively. (e) Lobes: Lobes are portion of radiation pattern carrying definite amount of energy. Lobes are classified as a major lobes, minor/side lobes and back lobes. Except major lobes, all represent radiation in undesired directions. However in radar systems, side lobes are also important to minimize false target indications.

Introduction

9

(i) Major lobe/main lobe: The radiation lobe contains maximum radiation. In some cases, there may be more than one main lobe. (ii) Minor lobes: All lobes, other than major lobes in a plane are termed minor lobes. (iii) Back lobes: It appears in the plane opposite to main lobe, i.e., 180° w.r.t. main beam. (f) Pattern beam width: It is the angular width on the major lobe of radiation. There are two types of beam width: HPBW and FNBW. HPBW is the angle between two points at 3 dB down from the top on main lobes, whereas FNBW is the angle between two vectors drawn at the origin and tangent to the main beam. It is very often found that FNBW » 2 HPBW.

TYPES OF PROPAGATION Basically, there are four categories of radio-wave propagation [5].

Surface Wave Propagation In this propagation waves travel along curvature of the earth and modified by the ground or terrain over they travel. Signals heard on the medium wave band (upto 2 MHz) during the day. The surface wave suffers ground attenuation to the same factor as the free space. These ground losses are caused by the ohmic resistive losses in the conductive earth. Surface wave attenuation increases as frequency increases. Surface wave propagation is also affected by the heights as well as the distance of/between Tx and Rx antennas and the terrain, and the weather conditions along the transmission paths.

Space Propagation In this case the radio waves travel in free space or away from the object which influenced the way they travel. This type of radio wave propagation is encountered with signals travelling to and from satellites. Waves reach at Rx either directly or after reflection from the ground. Space propagation is of practical importance in the frequency band > 30 MHz. This is because at such frequencies both ionosphere and surface wave propagations are failed. The reason behind that at 30 MHz, ionosphere wave length becomes too short to be reflected, and groundwaves propagate close to the antenna and hence lost the energy before reaching to the destination. At VHF, UHF and microwave frequencies, the space wave propagation are limited to the so-called LOS (line-of-sight) distance.

Troposphere Propagation The signals are influenced by the variation of refractive index in the troposphere (approximately 15 km above the earth). This mode of propagation is often the means by which signals at VHF, UHF and microwave frequencies (in the range of GHz) are heard over extended

10


distances. The normal refraction is the main mechanism for the troposphere propagation phenomenon. Duct propagation is a special type of propagation that occurs due to temperature inversion in the troposphere. M-curves show the variation of modified index of refraction with height, and they are useful to predict, at least roughly, the transmission path that is usually expected.

Ionospheric Propagation In this mode of propagation, signals reach the destination after reflection from the ionized region in the upper atmosphere called ionosphere between 110 and 400 km from the ground under suitable conditions. There are various parameters that characterize the ionospheric propagation: the critical frequency, MUF, skip-distance as well as virtual height. Ionospheric propagation is also affected by the earth’s magnetic fields. Fading is the main factor that degrades the propagation and severe fading reduces the field strength of radio waves from 10 dB to 20 dB. Ionospheric propagation is of practical importance at medium and high frequencies (2 to 30 MHz) and useful for very long distance communication. Extremely long distance, i.e., round-the-globe communication is also possible with multiple reflections of waves by Ionospheric propagation [6]. Further details about the above topics will be provided in subsequent chapters.

OBJECTIVE TYPE QUESTIONS 1. An (a) (b) (c) (d)

antenna may be considered as a transducer that Provides matching between guiding devices and the surrounding medium Provides matching between guiding devices and other antennas Provides matching between guiding devices and waveguide None of these

2. The basic principle behind the functioning of antennas is the (a) Principle of energy conversion (b) Poynting theorem (c) Babinet’s theorem (d) Principle of electromagnetism 3. Which of these is not the purpose of use of an antenna? (a) Efficient radiation (b) Impedance matching (c) Increasing the velocity of radiation (d) Both (b) and (c) 4. History of antenna begun with (a) Maxwell (c) Marconi

(b) Hertz (d) Yagi–Uda

Introduction

11

5. The first antenna used in a communication system was (a) Patch antenna (b) Dipole antenna (c) Square loop antenna (d) Parabolic antenna 6. The polarization of (a) The orientation (b) The orientation (c) The orientation (d) None of these

an antenna is defined as of the electric field of the magnetic field of both the fields

7. Which of these is incorrect? (a) Polarization characteristics of the antenna remain constant over its main beam. (b) Polarization characteristics of the antenna remain constant over its minor beam (c) Polarization on main beam peak describes the polarization of the antenna. (d) None of these 8. A transmitting antenna is matched to a receiving antenna if (a) Polarization, axial ratio and sense are identical. (b) Polarization, axial ratio, sense and minor axis rotation are identical. (c) Polarization, axial ratio, sense and major axis rotation are identical. (d) None of these 9. Which of the following is a whip antenna? (a) l/4 whip antenna (b) l/8 whip antenna (c) Both (a) and (b) (d) None of these 10. Which of the following takes place up to 2 MHz and is affected by the height and distance between Tx and Rx? (a) Ionosphere propagation (b) Surface wave propagation (c) Both (a) and (b) (d) None of these

Answers 1. (a) 6. (a)

2. (d) 7. (b)

3. (c) 8. (c)

4. (b) 9. (c)

5. (c) 10. (b)

EXERCISES 1. An antenna is also variously termed a transitional device, directional device, resonant device and transducer. Explain each term with a suitable example. 2. Why does an antenna need a communication system? 3. With the help of a neat diagram, describe the radiation mechanism of an antenna. 4. What is the general classification of antenna based on frequency of operation? 5. What is axial ratio? Describe its advantages for different types of polarization.

12


6. List out advantages of circularly polarized antennas. 7. What are the physical meanings of ± dB, dBi, dBd, and dBm? 8. What do you mean by impedance matching in an antenna system? What should be the range of VSWR for good impedance matching? 9. Define bandwidth and write its different expressions. Also define one radian and one steradian with regard to beam width of an antenna. 10. Define radiation pattern, HPBW, gain and directivity of an antenna. 11. What is the basis of classification of different types of wave propagation? 12. Define MUF and skip-distance for ionospheric wave propagation. 13. “Microwave communication is limited to around 50 km”. Justify this statement. 14. What are the advantages of LOS communication?

REFERENCES [1] Kraus, J.D., “Antennas since Hertz and Marconi”, IEEE Trans., Antennas and Propagate, Vol. AP-33, No. 2, pp. 131–137, Feb. 1985. [2] Elliot, R.S., Antenna Theory and Design, Prentice-Hall of India, New Delhi, 1981. [3] Stutzman, S., “Bibliography for antennas”, IEEE Trans., Antennas and Propagate Magazine, Vol. 32, pp. 54–57, August 1990. [4] Lo, Y.T. and S.W. Lee, Antenna Handbook: Theory, Applications and Design, Van Nostrand Rein, New York, 1988. [5] www.rfip.eu/propagation [6] JPL’s Wireless Communication Reference, websites.

C H A P T E R

2

Electromagnetic Waves and Radiation

INTRODUCTION The existence of electromagnetic (EM) waves started with Maxwell’s equations, which he presented to the British Royal Society in 1864, in his paper entitled “A Dynamic Theory of the Electromagnetic Fields”. He had predicted theoretically the existence of electric and magnetic Fields associated with electromagnetic wave propagation. This was confirmed by Heinrich Hertz in 1893 when he conducted an experiment on a dipole fed parabolic antenna and found that it sends a signal by wave motion to a similar receiving antenna kept at a finite distance. It was the first strong support to the theoretical conclusion drawn by Maxwell for electromagnetic fields. In the same year, William Thomson proposed the waveguide theory for propagation of EM waves. Later, in 1897, Lodge described the mode properties of wave propagation first in free space and then in a hollow metallic tube named as waveguide. In 1898, J.C. Bose developed the horn antenna, which is still useful for high frequency EM wave propagation [1,2]. The properties of EM waves in a medium are characterized by electrical parameters; namely permittivity (e), permeability (m), conductivity (s) and characteristic impedance (h), and also the presence of boundary between media. The values of these parameters for free space are: e0 = 8.854e–12, m0 = 4pe–7, s = e–14 and h0 = 120p. The properties of EM waves are associated with oscillating electric and magnetic fields. These fields are perpendicular to each other as well as perpendicular to the direction of propagation. The direction of propagation is generally taken to be along the z axis. The vector in this direction is known as Poynting vector or propagation vector. There are various modes of propagation of EM waves such as TE mode (Ez = 0), TM (H z = 0) and TEM mode (Ez = 0, H z = 0). In free space, at sufficient distance from source, the wave propagates in TEM mode, where both E and H fields are perpendicular to the direction of propagation [3,4]. The field configuration of TEM wave propagation is shown in Fig. 2.1. The region close to radiating sources is most likely to carry high intensity of fields having both longitudinal and transverse components with respect to direction of propagation. 13

14

Antenna and Wave Propagation x

E H

O

z

Direction of propagation

y

FIG. 2.1

Field configuration of TEM wave propagation.

In general, these locations are characterized by complicated field structure, including reactive (stored) and real (propagated) energies, irregular phase structures and undefined polarizations.

EM WAVE SPECTRUM AND ITS APPLICATIONS The electromagnetic or EM wave spectrum (see Fig. 2.2) is a continuum of all electromagnetic waves arranged according to frequency/wavelength. This spectrum includes visible, ultraviolet and infrared, microwave, radio and gamma waves. The sun, earth and other bodies radiate electromagnetic energy of varying wavelengths. All electromagnetic energy passes through space at the speed of light i.e., 3 ´ 108 ms–1 in the form of sinusoidal waves. The spectrum of waves is divided into sections based on wavelength. The shortest waves are gamma rays, which have wavelengths of 10–6 microns or less. The longest waves

FIG. 2.2

EM wave spectrum.


15

are radio waves, which have wavelengths of many kilometres. Visible light has a particular band of electromagnetic radiation that can be seen and sensed by the human eye. This energy consists of the narrow portion of the spectrum, from 0.4 micron (blue) to 0.7 micron (red). The infrared range starts at the end of the red spectrum with wavelengths greater than 0.7 micron. 1. Radio waves: Radio waves have wavelengths that range from less than a centimetre to tens or even hundreds of metres. FM radio waves are shorter than AM radio waves. For example, an FM radio station at 100 on the radio dials (100 MHz), would have a wavelength of about three metres. An AM station at 750 on the dials (750 kHz) uses a wavelength of about 400 metres. They are used to transmit radio and television signals. Radio waves can also be used to create images. Radio waves with wavelengths of a few centimetres can be transmitted from a satellite or airplane antenna. The reflected waves can be used to form an image of the ground in complete darkness or through clouds. 2. Microwaves: Microwaves’ wavelengths range from approximately 1 mm to 30 cm. In a microwave oven, the radio waves generated are tuned to frequencies that can be absorbed by the food. The food absorbs the energy and gets warmer. The dish holding the food does not absorb a significant amount of energy and stays much cooler. Microwaves are emitted from the Earth, from objects such as cars and planes, and also from the atmosphere. These microwaves can be detected to give information, such as the temperature of the object that emitted the microwaves. They are used in various communications, media and medical laboratories for testing. 3. Infrared: Infrared is the region of the electromagnetic spectrum that extends from the visible region to about one millimetre (in wavelength). Infrared waves include thermal radiation. Infrared radiation can be measured using electronic detectors and has applications in medicine and in finding heat leaks from houses. Infrared images obtained by sensors in satellites and airplanes can yield important information on the health of crops and can help us see forest fires even when they are enveloped in an opaque curtain of smoke. The rainbow of colours we know as visible light is the portion of the electromagnetic spectrum with wavelengths between 400 and 700 nanometres. It is the part of the electromagnetic spectrum that we see, and coincides with the wavelength of greatest intensity of sunlight. Visible waves have great utility for the remote sensing of vegetation as well as for the identification of various objects by their different visible colours. 4. Ultraviolet: Ultraviolet radiation has a range of wavelengths from 400 ´ 10–7 of a metre to about 10–8 of a metre. Sunlight contains ultraviolet waves which can burn your skin. Most of these are blocked by ozone in the Earth’s upper atmosphere. A small dose of ultraviolet radiation is beneficial to human beings, but larger doses cause skin cancer and cataracts. Ultraviolet wavelengths are used extensively in astronomical observatories. Some remote sensing observations of the Earth are also concerned with the measurement of ozone. 5. X-rays: X-ray’s wavelength ranges from 10–10 to 10–12 of a metre. They are high energy waves which have great penetrating power and used extensively in medical applications as well as in inspecting welds. X-ray images of the sun can yield important clues to solar flares and other changes on our sun that can affect space weather.

16


6. Gamma rays: Gamma rays have wavelengths of less than about 10–12 of a metre. They are more penetrating than X-rays. Gamma rays are generated by radioactive atoms and in nuclear explosions, and are used in many medical applications. Images of our universe taken in gamma rays have yielded important information on the life and death of stars, and other violent processes in the universe.

EM FIELDS AND MAXWELLS EQUATIONS Let us consider a medium in which waves travel without loss of energy, i.e. wave amplitude is constant with distance. The electric and magnetic fields for such waves are defined as E = E0 cos(w t – kz) xˆ

(2.1a)

H = H0 cos(w t – kz) yˆ

(2.1b)

where E0 and H 0 are field amplitudes, w = 2p, angular frequency, t is time, k is wave number, z is distance along z-axis and xˆ and yˆ are the unit vectors along positive x and y directions. The wave number represents the rate of change of phase of field with distance that is phase of the wave changes by kr radians over distance r metres. The existence of EM waves can be predicted as the direct consequence of Maxwell’s equations. Maxwell’s equations specify the relationship between the variations of electric and magnetic fields (E, H) in time and space within medium. The electric field is generated either by time varying magnetic field (¶B/¶t) or by the free space. The H field is generated either by a time varying electric field (¶E/¶t) or by current distribution and measured in A/m. Mathematical forms as well as physical significance of Maxwell’s equations are summarized as follows:

∇ × H = J + Jc = T E + F

∂E ∂t

(2.2a)

A magnetic field is produced by a time varying electric field or by a current.

∇ × E= −

∂B ∂t

= −N

∂H ∂t

(2.2b)

An electric field is produced by time varying magnetic field. Ñ.D = r

(2.2c)

Electric field lines may either start or end on charges or continuous. Ñ.B = 0

(2.2d)

Magnetic field lines are continuous, and Ñ.J =

∂S ∂t

(2.2e)

where e, m, s and r are the electrical parameters of the medium. The permittivity e and permeability m are normally expressed relative to the values in free space; m = m0mr, e = e0er.


17

Here er and mr are known as relative permittivity and relative permeability respectively and each have value 1 for free space. Any EM wave consisting of E and H fields satisfy Maxwell’s equations, provided the ratio of the field amplitudes is constant for a given medium. Ex Hy

=

E H

=

E0 H0

=

N =I F

(2.3)

where h is known as wave/characteristic/intrinsic impedance of medium and measured in W. Since for the free space er = 1, mr = 1, the intrinsic impedance h becomes

I = I0 =

Nr N0 N0 4Q e − 7 = = = 377 : Fr F 0 F0 8.854 e − 12

(2.4)

Poynting Vector and Velocity of EM Waves Poynting vector is also known as power density of EM waves; it represents the magnitude and direction of the power flow carried by the waves per unit square metre of area parallel to the plane perpendicular to the direction of propagation. It is vector quantity and measured in W/m2. The instantaneous value of Poynting vector P = E ´ H*

(2.5)

where H* is complex conjugate of H. Usually, only the time average of the power flow over one period is of concern, that is, P = 1/2 EH zˆ . Since the ratio of E and H is constant and equal to h P=

1 2

IH 2 =

1 2I

E2

(2.6)

For the wave propagating in particular plane (plane wave), the direction of energy flow is in the direction of propagation. Thus, the Poynting vector offers a useful, easy and coordinatefree way to specify the direction of propagation as well as determining the direction of the fields if the direction of propagation in known. This can be particularly valuable where incident, reflected and transmitted waves are being examined [1].

WAVE POLARIZATION The orientations of the electric field vector of a plane wave relative to the direction of propagation define the polarization of wave. If an electric field vector of an EM wave is parallel to x-axis, wave is said to be linearly x-polarized. This wave could be generated by a straight wire antenna parallel to x-axis. Similarly y-polarized waves can also be defined and generated. If two plane waves of equal amplitude and orthogonally polarized are combined with 90° phase difference, the resulting wave is circularly polarized and the electric field

18


vector describes a circle centred on propagation vector. The field vector will rotate by 360° for every wavelength travelled. Circular polarization is generated as either RHCP or LHCP. The right hand circularly polarized wave describes a wave with E field vector rotating clockwise when looking in the direction of propagation. In most cases, the components of EM waves are not equal in amplitude or are at a phase angle other than 90° (see Fig. 2.3). y

y E1 E02

E02

E E01

E0

E0

E0

O

z

y

O

x

–E0

E

z

(a) Linear polarization

E

x

E01

O

–E0

z

(b) Circular polarization

FIG. 2.3

x

–E0 (c) Elliptical polarization

Polarization of EM waves.

The resultant polarization is elliptical polarization. Here the electric field vector still rotates at same rate but varies in amplitude with time. In case of elliptical polarization, the waves are characterized by the ratio between the maximum and minimum values of the instantaneous electric fields, called axial ratio (AR), given by AR =

Emax

(2.7)

Emin

Mathematical Interpretation of Polarization Assume that a TEM wave is propagating in the z-direction; in general an E-field vector can be decomposed in two orthogonal components at any time on the z-constant plane.

wt Ey = E02 sin(w t + q)

Ex = E01 sin Hence the resultant field is

E = (E x2 + E y2 )1/2 or

2 2 E 2 = E01 sin X t + E02 sin (X t + R )

⇒

2 E01 sin X t

E2

+

2 E02 sin (Xt + R )

E2

=1

(2.8)

19


which represents an ellipse with its semi-major and minor axes inclined to the x and y axes. That is electric field vector E constantly changes both its magnitudes and direction describing an ellipse. Such waves are known as elliptically polarized plane wave. There are different cases; Case I When q = 0 with Ex = E01 sin w t, and Ey = E02 sin w t, at any point on the z-constant plane, wave is said to be linearly polarized with sinusoidal time variation.

E = (Ex2 + Ey2 )1/2 = E0 sin X t

Hence

(2.9a)

Here tip of E describes a plane surface and is constant in direction. Case II

When q = p/2 with E01 = E02 = E0, the above equation reduces to a circle

E x2 + E y2 = E02

(2.9b)

Here tip of E describes a circle of radius E0. These waves are called circularly polarized plane wave. If d = 90° and E1 = E2 = 2 E , the wave is linearly polarized but in a plane at an angle of 45° w.r.t. the x-axis.

VELOCITY OF PROPAGATION Phase velocity This is a measure of how fast a signal travels along a line/in a medium. A radio signal (all EM waves) travels in free space at the speed of light (c) » 2.998 ´ 108 ms–1. A signal in a Tx line travels at much lesser than this speed and however in twisted pair cable the velocity of propagation may be between 40% and 75% of velocity in free space. The velocity of a point of constant phase on the EM waves in any medium is known as phase velocity (vp) and equal to vp =

X k

=

1

NF

(2.10)

It is frequency dependent and often stated either as a percentage of c or as time to distance; when time to distance figure is used, it may also be called propagation delay and will be expressed as ms/km. The velocity of propagation is also known as velocity factor and it is used in communication media such as data cables—category 5 cables, plenum data and ethernet/fast ethernet. Plenum data cables typically have vp ranging from 42% to 72% of speed of light [2]. Group velocity The group velocity of a wave is the velocity with which the variation in the shape of the waves amplitude (known as the modulation or envelope of the wave) propagates through space. The group velocity is of special importance in the propagation of modulated waves,

20


pulses and transmission through wave-guides. The phase and group velocity are related by vpvg = c2.

Plane Wave and Uniform Plane Wave When EM waves propagate with phase remains constant over a set of planes, are called plane waves. The magnitude of wave’s fields are constant in the xy-plane, and surface of constant phase (a wave front) forms a plane parallel to the xy-plane, hence the term plane wave. The oscillating electric field produces a magnetic field, which itself oscillates to recreates an electric filed and so on, accordance with Maxwell’s curl equations. This interplay between the two fields stores energy and hence carries power. Variation/modulation of the properties of the wave (amplitude, frequency or phase) then allows information to be carried in the wave between its source and destination, which is the central aim of a wireless communication system. In particular, the EM waves which electric field is independent of y and z axes and function of x and t (time), known as uniform plane wave, and have special importance in propagation. A uniform plane wave has the following properties: (i) (ii) (iii) (iv) (v)

These waves are TEM waves. E and H fields are always in time phase. The magnitude of the two fields is always constant. The stored energies are equally divided between E and H fields. The power transmitted by the two fields is in the direction of propagation.

PROPAGATION OF EM WAVES IN DIFFERENT MEDIUMS Before discuss wave propagation in various media, let us consider the criteria of these mediums in terms of electric parameters (m, e and s). In electromagnetic, the materials are classified as conductor, dielectric and lossy dielectric. We can explain this classification using Maxell’s equation for time varying fields. Ñ´ H = or

sE

+ j w E = Jc + JD

Ñ ´ H = jwe (1 +

s /jwe)E

(2.11)

The term s/we is therefore just the ratio of conduction to displacement current densities; it may be considered a mark line to divide the different materials/media. For good conductor, (s/jwe) is much greater than unity over entire radio frequency, whereas it is much lesser than unity for a good dielectric. For a lossy dielectric, e = e ¢ – je ² and s << we. The study of EM wave propagation began with the investigation of uniform plane wave, which perhaps represent the simplex form of wave propagation. The propagation of waves is described by the standard equation, called wave equation, which can be derived from Maxwell’s equations. Let us consider a linear, isotopic, and homogeneous medium. The net free charge in the region is zero (r = 0) and the existing currents in the region are

21


sE).

conduction currents, i.e. (j = to (see [3]).

Maxwell’s equations, in this particular case, are reduced

∇ × H = TE + F ∇ × E=N

∂E

(2.12a)

∂t

∂H

(2.12b)

∂t

ÑH = 0

(2.12c)

ÑE = 0

(2.12d)

Taking the curl of (2.12b) and substituting it into (2.12a) yields

∇ × ∇ × E = − NT

∂E

− NF

∂t

∂2 E ∂t 2

(2.13)

Similarly, taking the curl of (2.12a) and substituting it into (2.12b) yields

∇ × ∇ × H = NT

∂H

− NF

∂t

∂2 H ∂t 2

(2.14)

We know the vector identity Ñ ´ Ñ ´ A = Ñ(Ñ . A) – Ñ2A

(2.15)

Solving (2.13) using (2.12d) and (2.15) yields

∇ 2 E = NT

∂E ∂t

+ NF

∂2 E

(2.16)

∂t 2

This is known as electric wave equation for a medium. If electric field is in phasor form, i.e. E = Eeiwt, then the electric wave Eq. (2.16) is reduced to (2.17) Ñ2E = jwm(s + jwe)E = g 2E Similarly, we get the magnetic wave equation Ñ2H = jwm(s + jwe)H =

g 2H

where g is equal to [jwm(s + jwe)] and known as propagation constant. Since complex quantity, its square root will also be a complex quantity. 1/2

Hence, let

g = [jwm(s + jwe)]1/2 = a + jb

(2.18)

g2

is a

(2.19)

Separating real and imaginary parts from Eq. (2.19), we get ⎡ ⎛ NF B = X⎢ ⎜ 1+ ⎢ 2 ⎜⎜ ⎣⎢ ⎝

⎛T ⎞ ⎜ ⎟ ⎝ XF ⎠

2

⎞⎤ − 1 ⎟⎥ ⎟⎟ ⎥ ⎠ ⎦⎥

1/2

(2.20a)

22


⎡ ⎛ NF C =X⎢ ⎜ 1+ ⎢ 2 ⎜⎜ ⎢⎣ ⎝

⎞⎤ ⎛T ⎞ ⎟⎥ + 1 ⎜ ⎟ ⎟⎟ ⎥ ⎝ XF ⎠ ⎠ ⎥⎦

1/2

2

(2.20b)

where a is attenuation constant, measures the rate of decrease of amplitude of wave per unit length. It is measured in Np/m. b is known as phase constant and is a measure of phase shift in radians per unit length.

T ⎞ ⎛ ∇ × H = jXF ⎜ 1 + ⎟ E = jXF *E jXF ⎠ ⎝

Again

(say)

where e* = e (1 + s /jwe) is known as complex permittivity of any conducting medium of finite conductivity. The intrinsic impedance is hence defined as

I=

N N = = F* F (1 + T /jXF )

jXN

(2.21)

jXN + T

Basically there are three types of wave propagation: (a) Wave propagation in good conductor As, for a good conductor s/jwe >> 1, Hence, the propagation constant

H So, if

g

=

a

=

jXN (jXF + T ) =

g

reduces to

jXF ⎞ ⎛ jXNT ⎜1 + = (jXNT ) = (XNT ) 45° T ⎟⎠ ⎝

+ jb, we get

XNT

B =C =

(2.22)

2

The phase velocity

vp =

X C

2X

≈

(2.23)

NT

The characteristic impedance

I=

=

jXN /T

1 + jXN /T

=

⎛ XN ⎞ ⎜ ⎟ (1 + j ) ⎝ T ⎠

jXN

T

=

⎛ XN ⎞ ⎜ ⎟ 45° ⎝ T ⎠

(2.24)


h

So, if

23

is equal to Rs + jXs

XN

Rs = X s =

2T

where Rs and Xs are respectively known as surface conductor resistance and reactance. For a good conductor, the depth of penetration

E=

1

2

=

B

XNT

and hence ⎛ 1 ⎞ Rs = X s = ⎜ ⎟ ⎝ TE ⎠

(2.25)

(b) Wave propagation in good dielectric For a good dielectric, s/jwe << 1. In this condition, the propagation constant

H

= B + jC = jX

NF

T⎤ ⎡ ⎢1 − j XF ⎥⎦ ⎣

1/2

where

T ⎤ ⎡ ⎢1 − j XF ⎥⎦ ⎣

1/2

=1 −

T j 2XF

1 ⎛T ⎞ +j ⎜ ⎟ + ... 8 ⎝ XF ⎠

⎡T = ⎢ ⎢2 ⎣

⎛ N + jX NF ⎜ 1 + ⎜ F ⎝

2

Hence

H = X NF

2 ⎡ T 1⎛ T ⎞ ⎤ ⎢1 − j +j ⎜ ⎟ ⎥ 2XF 8 ⎝ XF ⎠ ⎥ ⎢⎣ ⎦

1/2

1⎛ T ⎞ ⎜ ⎟ 8 ⎝ XF ⎠

2

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(2.26a)

Separating real and imaginary parts yields

B=

T 2

2 ⎛ N 1⎛ T ⎞ ⎞ ⎜ and C = X NF 1 + ⎜ ⎟ ⎟ ⎜ F 8 ⎝ XF ⎠ ⎟ ⎝ ⎠

(2.26b)

The phase velocity

vp =

X C

1 2 ⎛ 1 T NF ⎜ 1 + ⎛⎜ ⎞⎟ ⎜ 8 ⎝ XF ⎠ ⎝

⎞ ⎟ ⎟ ⎠

=

1

NF

2 ⎛ 1⎛ T ⎞ ⎞ ⎜1 − ⎜ ⎟ ⎟ ⎜ 8 ⎝ XF ⎠ ⎟ ⎝ ⎠

(2.26c)

24


The characteristic impedance

I=

N N⎛ T ⎞ = ⎜1 + j ⎟ F (1 + T /jXF ) F⎝ 2XF ⎠

(2.26d)

where N /F is the characteristic impedance of the dielectric when s = 0; hence it is clear that the main effect of a small amount of loss is to add a small reactive component to the intrinsic impedance. (c) Wave propagation in lossy dielectric In a lossy dielectric, we know that e = e¢ – je ² and propagation constant g is equal to

H

= jX

T ⎞ ⎛ NF ⎜ 1 − j ⎟ XF ⎠ ⎝

1/2

= jX

⎛

NF ⎜ 1 ⎝

s << we, − j

that is,

T ⎞ ⎟ =B 2XF ⎠

s ¹ 0 [3]. Hence the

+ j C (say)

(2.27a)

therefore,

B=

T

N /F and C = X NF

2

(2.27b)

The intrinsic impedance of the dielectric is given by

I=

jXN

jXF + T

=

N F

T⎞ ⎛ ⎜1 − j XF ⎟⎠ ⎝

−1/2

=

N F

T ⎞ ⎛ ⎜1 + j ⎟ 2 XF ⎠ ⎝

(2.27c)

The phase velocity vp =

X C

=

X X NF

=

1

NF

(2.27d)

(d) Wave propagation in lossless dielectric In case of a lossless dielectric medium, e = e¢, e ² = 0 and s = 0. Hence the attenuation constant a = 0 and the intrinsic impedance, phase velocity and propagation constant are given by

I=

N 1 , vp = and H = j C F NF

(2.28)

where, b = X NF . Here s, m and e are together known as constitutive parameters of the medium. Due to attenuation of waves, the field strength of both E and H diminishes exponentially as the wave travels through the medium, as shown in Fig. 2.4. The distance to which the wave travels, before its field strength reduces to e–1 = 36.8% of its original value, is its skin-depth (d) and the phenomenon is known as skin-effect. Skin-depth (d) and attenuation constant (a)


25

1 0.9 0.8

E(x)

Value of E-field

0.7 0.6 0.5 0.4 0.3 0.2 0.1

0

1

2

3

4

5

6

7

8

9

10

x Distance along propagation vector (skin-depth, m)

FIG. 2.4

Attenuation of EM waves.

is related as follows: dÿ = 1/a. Thus the amplitude of the electric field strength at a point z, compared with its value at z = 0, is given by E(z) = E(0)e–z/d. If we deal with wave propagation in the dielectric medium, there is wave-attenuation if the conductivity of the medium is not zero. A quantity known a loss-tangent (tan d), which is the phase of the complex dielectric constant, is normally used as measure of the medium attenuation. It is equal to s/we, i.e., smaller the loss-tangent, lesser is the attenuation and better is dielectric. We know that the first Maxwell’s equation for any material yields Ñ ´ H = jwe (1 – js /we)E The loss properties of the materials may be treated in a similar fashion as conductive loss, by replacing s ® we ² and e ® e¢, i.e. the loss-tangent reduces to tan d = e²/e¢. Skin-effect occurs, generally at very high frequencies (in range of GHz), but we can include the effect in similar fashion as for conductive material even if a dielectric material has non-zero conductivity. Ñ ´ H = (s + jwe*)E

(2.29a)

where e * is complex permittivity and equal to e¢ – je². Ñ ´ H = (s +

we²)E + jwe¢ E

(2.29b)

26


Hence, the loss tangent of any medium can be defined as

tan E =

T + XF ′′ XF ′

(2.29c)

where (s + we ²) is considered to be effective conductivity and e¢ to be the effective permittivity of the material having conductivity (s) and permittivity (e).

POWER FLOW OF A UNIFORM PLANE WAVE If both the fields E and H of a wave are time varying, i.e. E = E0ejw t and H = H0ejw t Þ E0 jXt e , then the average power density of the wave is H =

I

Pav =

1 2

Re ( E × H *) =

E ⎛ ⎞ Re ⎜ E0 e jXt × 0 e jXt × xˆ yˆ ⎟ 2 I ⎝ ⎠

1

⎡ E02 ⎤ 1 ⎡1⎤ = Re ⎢ ⎥ zˆ = E02 Re ⎢ ⎥ 2 2 ⎣ I* ⎦ ⎢⎣ I* ⎥⎦ 1

(2.30a)

Similarly

Pav =

1 2

H 02 Re (I*)

(2.30b)

These are the expressions for average power densities in terms of E and H fields for any medium. Next let us consider the Pav for particular mediums. (a) Good conductor In the case of good conductor

I = (1 + j)

XQ 2T

XN /2T ⎛ 1 ⎞ Re (I) Re ⎜ = = ⎟= 2 2XN /2T |I | ⎝ I* ⎠ Pav =

1 2

T 2XN

T 2XN

| E0 |2

(2.31a)

Similarly Pav =

1 | H 0 |2

2 Re (I)

=

1 2

| H 0 |2

2XN

T

(2.31b)


27

(b) For lossless dielectric We know that

N F

I= Hence

Pav =

1 E02 2

I

=

1

H 02I

2

(2.32a)

(c) Good dielectric and lossy dielectric medium We know that

I=

N⎛ T ⎞ ⎜1 + j ⎟ F⎝ 2XF ⎠

which is a complex number; however E and H are not in time phase; consequently Pav has to be determined using Pav =

| E0 |2 2

| H 0 |2 ⎛1⎞ Re ⎜ ⎟ ≈ Re (I) 2 ⎝I ⎠

(2.32b)

INCIDENCE OF UNIFORM PLANE WAVE Basically there are two types of incidence of wave: normal incidence and oblique incidence. 1. Normal incidence of uniform plane waves on plane boundaries: Let us consider a uniform plane wave incident normally from left to right on a boundary between two media (Fig. 2.5). Electrical parameters of medium (1) are e1, h1, m1 and s1 and those of medium (2) are e2, h2, m2 and s2. If (Ei, Hi), (Er, Hr) and (Et , Ht) are the electric and magnetic field intensities of incident, reflected and transmitted waves, then they can be written in their phasor forms as follows: Ei = Ei e −H 1 z aˆ x = Ei e −(B1 + jC1 )z aˆ x H i = Ei e −H 1z aˆ y =

Ei

I1

e −(B1 + j C1 )z aˆ y

(2.33a) (2.33b)

2. Reflected wave Er = Er eH 1z aˆ x = Er e(B1 + jC1 )z aˆ x Hr = − H r eH 1z aˆ y = −

Er

I1

e − (B1 + jC1 )z aˆ y

(2.34a) (2.34b)

28


FIG. 2.5

Vector representation of incident, reflected and transmitted waves.

where g1 is propagation constant for medium 1 and equal to a1 + jb1 with a1 and b1 as attenuation and phase constants and

jXN1

Iˆ1 = I < R i/r =

jXF1 + T1

(2.34c)

3. Transmitted waves Et = Et e−H 2 z aˆ x = Et e −(B 2 + j C2 )z aˆ x H t = − H t e−H 2 z aˆ y =

Et

I2

e − (B 2 + j C2 )z aˆ y

(2.35a) (2.35b)

where g2 is propagation constant for medium 2 and equal to a2 + jb2 with a2 and b2 are attenuation and phase constants and

Iˆ2 = I < Rt =

jXN2

jXF 2 + T 2

(2.35c)

As the second medium is infinite in extent there will not be backward travelling component of transmitted wave, thus there will be no reflection at its right most surface. At the boundary z = 0, the boundary condition reveals that the tangential components of the total electric and magnetic fields be continuous. Since Ei, Er and Et are defined in the x direction Ei + Er = Et

at z = 0

(2.36a)

Similarly, if Hi, H r and H t are defined in the y direction Hi + Hr = Ht

at z = 0

(2.36b)


29

Substituting the phasor forms of the fields into (2.36a), we get

Ei

Er

−

Ii

Ei + Er Ei − Er

I1

=

=

Et

I2

=

Et + Er

I2

I2 Er I − I1 or = 2 Ei I1 I2 + I1

(2.37a)

is known as reflection coefficient and represented by G. Also Er

+1=

Ei

Er + Ei Ei

=

I2 − I1 +1 I2 + I1 2I2

I2 + I1

⇒

Et Ei

=

2I2

I2 + I1

(2.37b)

is known as transmission coefficient and represented by T. Again, from (2.37b) it is clear that Er Ei

+

Ei Ei

=T ⇒ 1+*=T

(2.38)

The coefficients G and T will be real only if both the mediums are lossless, that is, s1 = 0 and s2 = 0. Otherwise, they will be, in general, complex quantities i.e., G = G < qr and T = T < qt. From (2.37) it is also clear that the magnitude of T may exceed unity, but the magnitude of G will always be £ 1. If region 2 is perfect conductor, then h2 = 0 and we get G = –1 and T = 0; so there is no signal transmitted in region 2. Just as in transmission, we can define wave impedance, which describes the ratio of electric field to magnetic field at any point in space. This gives the familiar impedance transformation formula Z in ( − l) = I1

I2 + jI1 tan kl I1 + jI2 tan kl

where, k = wave-vector (bn). We can then treat the situation just as in a quarter wave transformer and multiple dielectric boundaries just as we do for transmission lines.

Oblique Incidence When a uniform wave incidence at any angle from the normal (say qi) then as resultant, i.e., reflected as well as transmitted waves also makes certain angles (say, qr and qt). This case is referred to as oblique incidence. Basically there two cases of oblique incidence: (a) parallel polarization and (b) perpendicular polarization. To define these types of polarization, first we have to specify the plane of incidence. If the electric field (E) is polarized in the

30


plane of the page, with H is perpendicular to page pointing outward. Then plane of page may be referred as plane of incidence and uniform plane is said to have parallel polarization (P-polarized). As the magnetic field H is perpendicular to plane of incidence (transverse magnetic or TM), it is also called TM polarization. In another case, H is parallel to the plane of incidence (or H lies in plane of incidence) and E is perpendicular to the plane of incidence. This is perpendicular polarization (S-polarized). Due to similar region, it is also called TE polarization. Reflection and transmission coefficients differ for these two polarizations. But reflection and transmission angles are independent of polarization.

Snells Law Consider any two rays of an incident wave (as shown in Fig. 2.6), say, rays 1 and 2 travel from medium 1 to medium 2. One part of wave is transmitted at an angle qt in the second medium, whereas another part is reflected in medium 1 itself at an angle qr.

FIG. 2.6

Reflection of EM wave at oblique incidence.

From the diagram, it is clear that the incident ray 2 travels the distance CB, whereas the reflected ray 1 travels the distance AE and the transmitted ray 1 travels the distance A to D. If v1 and v2 are the velocities of the waves in corresponding mediums, then (see [2]) CB

=

AD

v1

(2.39a)

v2

Now from geometry, CB = AB sin qi and AD = AB sin qt. Hence sin Ri sin Rt

As the velocity of uniform plane wave is

=

1

NF

v1 v2

(2.39b)


sin Ri sin R t

F2 F1

=

31

(2.39c)

From the properties of uniform plane wave, the ratio of electric and magnetic field vectors are constant and known as intrinsic impedance of the medium, i.e.,

E H

=

N =I F

Hence sin R i sin R t

=

I1 I2

(2.40)

In view of optics, if n1 and n2 are refractive indexes of medium 1 and medium 2, then sin R i sin R t

=

n1 n2

The final expression shows that

sin Ri sin R t

F 2 I2 v n = = 1 = 1 v2 n2 F1 I1

=

(2.41)

Further, AE = CB, then sin qi = sin qt which is known as Snell’s Law. For parallel polarized wave, the reflection and transmission coefficients are given as

*P =

I2 cos Rt − I1 cos Ri I2 cos Rt + I1 cos Ri

=

F1 cos Rt F1 cos Rt

− +

F 2 cos Ri F2 cos Ri

(2.42a)

and TP =

I2

2I2 cos Rt

cos R t + I1 cos Ri

=

F1 cos Rt F1 cos Rt + F 2 cos Ri 2

(2.42b)

The corresponding wave impedance in mediums 1 and 2 is given by

I1P

I2P

=

=

E1 (x ) H 1 (y)

= I1 cos Ri z =0

E1 ( x )

= I2 cos Ri =

1

H (y)

(2.43a)

z =0

I1 I2

cos Rt

(2.43b)

For the perpendicularly polarized incident waves, the reflection and transmission coefficients are given by

32


*S =

I2 sec Rt − I1 sec Ri I2 sec Rt + I1 sec Ri

=

F1 sec Rt F1 sec Rt

=

F1 sec Rt F1 sec Rt + F 2 sec Ri

− +

F 2 sec Ri F 2 sec Rii

(2.44a)

and TS =

I2

2I2 sec Rt

sec Rt + I1 sec Ri

2

(2.44b)

The corresponding wave impedance in mediums 1 and 2 is given by

I1S I2S

E1 (x )

=

E1 (y)

= I1 sec Ri z =0

E1 (x )

=

= I2 sec Ri =

1

E (y)

(2.45a)

z =0

I1 I2

cos Rt

(2.45b)

Reflection coefficients for both the polarizations could be defined in term of wave impedance as

*P =

I2P − I1P I − I1S and * S = 2S I2P + I1P I2S + I1S

(2.45c)

Total Transmission If the angle of incident is varied, there will be an angle where | G | = 0, i.e. there will be no reflection, and total transmission of parallel polarized wave occurs at the interface. For this case, Eq. (2.42a) can be written as

*P =

F1 cos R t −

F 2 (1 − sin 2R i )

F1 cos R t +

F 2 (1 − sin R i ) 2

(2.46a)

From Snell’s Law sin 2R t =

F1 F2

sin 2R i

Hence, Eq. (2.46a) can be written as F2

*P =

F2 F2 F1

cos Ri − cos R i +

F2 F1 F2 F1

− sin 2Ri

(2.46b) − sin 2Ri


33

Therefore | G | = 0 only when qi = qb (say), where 1/ 2

R i = Rb = sin

−1

⎡ F2 ⎤ ⎢ ⎥ ⎣ F 2 + F1 ⎦

1/ 2

⎡ F2 ⎤ = cos−1 ⎢ ⎥ ⎣ F 2 + F1 ⎦

1/ 2

⎡ F2 ⎤ = tan −1 ⎢ ⎥ ⎣ F1 ⎦

(2.46c)

where qb is called Brewster angle. Any arbitrary wave incident at qb will be reflected back with E-polarization parallel to the interface, and the other component of E is totally transmitted. The angle qb is 45°, provided e1 = e2. We could also find a similar angle for perpendicular polarization such that T = 0, but this would require mediums of different permeability and identical permittivity, something which does not often occur in nature qi = qb. However, if the incident wave is not arbitrarily polarized, there will be some reflection, but the reflected rays will be entirely of perpendicular polarization. Reflection coefficient G S never becomes zero, as long as the two media are different (h1 ¹ h2). Hence, we can say that Brewster angle does not exist for perpendicular polarization.

Total Reflection The incident wave is reflected back in same medium for | G P | = 1. If both media are lossless non-magnetic dielectrics, the quantity under square root (2.46b) will be positive and GP will be real provided (e2 > e1), i.e., the wave is incident from rare medium to dense medium. If, however, the wave is incident from more dense medium onto less dense medium (e2 >> e1) and if sin 2R i ≥

F1 F2

then GP becomes complex and | GP | = 1. In this particular case, incident wave is totally reflected back into denser medium. Therefore, there exist an incident angle qc (say), for which | GP | = 1 < 0°, called critical angle and given by (see [4]) R c = sin

−1

⎛ F2 ⎞ ⎜ ⎟ ⎜ F1 ⎟ ⎝ ⎠

(2.47)

For the entire incident angle greater than qc, | G P | = 1. Then qc = 90° when e2 = e1. Physically there is still some field penetration into region 2, but the field strength decays exponentially away from the boundary. This is called evanescent field and propagates along the boundary. Additional parameters, P-polarized and S-polarized power reflectivity, are expressed as P-polarized power reflectivity RP =

tan 2 (R1 − R 2 ) tan 2 (R1 + R 2 )

34


S-polarized power reflectivity RS =

sin 2 (R1 − R 2 ) sin 2 (R1 + R2 )

The above critical angles’ field reflectivity are in the form of pure phase shifts. The power reflectivity is 100%. The P- and S-polarized phase shifts differ, and functions of incidence n angle and the relative refractive index n = 2 . The formula for relative phase shift is n1

⎡ cos R sin 2 R − n2 ⎤ 1 1 ⎥ E = GS − GP = 2 tan −1 ⎢ ⎢ ⎥ sin 2 R1 ⎣ ⎦

h2P

For the total transmission, GP = 0 implies that Using Snell’s Law of refraction sin R i sin R t

Þ

Þ

⎡ ⎤ ⎛n ⎞ I2 ⎢1 − ⎜ 1 ⎟ sin 2Ri ⎥ ⎝ n2 ⎠ ⎣ ⎦

I0 ⎡

⎤ ⎛ n1 ⎞ 2 ⎢1 − ⎜ ⎟ sin Ri ⎥ n2 ⎣ ⎝ n2 ⎠ ⎦

h1P

=

Þ

h2

(2.48a)

cos qt = h1 cos qi:

n2

=

n1

1/2

= I1 [1 − sin 2Ri ]1/2

1/2

=

I0 n1

[1 − sin 2Ri ]1/2

from (2.41)

⎛ n1 ⎞ ⎡ 2 ⎤ ⎢1 − ⎜ n ⎟ sin R i ⎥ ⎝ 2⎠ ⎛ n2 ⎞ ⎣ ⎦ ⎜ ⎟ = 2 n [1 sin ] R − ⎝ 1⎠ i 2

Þ which gives

sin R1 =

n2

= sin R B

n12 + n22

(2.48b)

where qB is called the Brewster angle of total transmission. From Snell’s Law

sin R2 = or transmitted angle = 90° – q1.

n1 n12

+

(2.48c)

n22


35

SOLVED EXAMPLES Example 2.1 An uniform plane wave of frequency 2 GHz travelling in a large block of Teflon (er = 2.1, mr = 1 and s = 0). Determine the values of vp and h. Solution:

c

We know that vp =

=

Nr F r

3 × 108 2.1

= 2.07 × 108 m/s

I = I Nr /F r = 377 2.1 = 260 : Example 2.2 A wave of frequency 1 MHz travels in a large block of copper (s = 5.7 ´ 107, er = 1 and mr = 1). Determine the values of vp, h, l, a and b. Also find the phase shift between the electric and magnetic fields and the distance that the wave must travel to be attenuated by a factor 100 (40 dB). Solution:

We know

g = [jwm(jwe + s)1/2 = [j2p ´ 106 ´ 4p ´ 107 ´ (5.7 ´ 107 + j2p ´ 106 ´ 8.854 ´ 10–12)]1/2 = 2.14 ´ 104 < 45° = 1.513 ´ 104 + j1.513 ´ 104 Hence

ÿ

a

= 1.513 ´ 104 Np/m

b = 1.513 ´ 104 rad/m

and

The wave velocity vp =

X C

=

2 Q × 10 6 1.513 × 10 4

= 415.3 m/s

The wavelength in copper at the given frequency is

M= The intrinsic impedance =

jXN

2Q

C

=

2Q 1.513 × 10

4

= 415.3 × 10 −4 m/s

h

⎡ ⎤ j 2Q × 10 6 × 4Q × 10 −7 = ⎢ 7 6 −12 ⎥ ⎢⎣ (5.7 × 10 + j 2Q × 10 × 8.854 × 10 ⎥⎦

1/2

jXF + T

= 3.689 × 10 −4 < 45°

So, it is confirmed that the intrinsic impedance of a conductor should be very small. The angle of intrinsic impedance is 45°. As the field attenuated by a factor 100, i.e., e–ad = 1/100, where d is distance travelled for the given attenuation

B d = ln 100

or d =

4.605 1.513 × 10

−4

= 3.0437 × 10 −4 m

36


Example 2.3 A copper conductor of finite length supports a uniform wave to propagate at kHz. Determine the values of a, b, g and vp. Also find the value of ratio of s/we. Solution:

We know that for a good conductor

B = C = NXT /2 =

2Q × 50 × 10 3 × 4Q × 10 −7 × 5.8 × 10 7

= 11.437 × 10 6 = 3.38 × 103

I=

XN T

= 2Q × 50 × 103 × 4Q ×

10 −7 5.8

× 10 −7

= 67 × 10 −5 < 45° = 8.245 × 10 −5 < 45°

M=

2Q

C

=2 ×

3.14 3.38

= 1.85 mm

v p = 2 X /NT = 2 × 2Q × 50 × 103 /4Q × 10 −7 × 5.8 × 10 7 = 92 m/s The ratio of

T 5.8 × 10 7 = = 2.08 × 10 3 XF 2Q × 50 × 103 × 8.854 × 10 −12 Example 2.4 A linearly polarized waves of frequency 1 GHz propagating in +z direction in a medium specified with er = 3.0, m = 1.0 and s = 0.02 S/m. If the electric field magnitude at z = 0 is 2 V/m, determine (a) the wave impedance, (b) |H| at z = 0, (c) average power available in an 0.6 m2 area perpendicular to direction of propagation at z = 0, (d) time taken to travel 20 cm, and (e) distance travelled at which field strength drops to 1/20. Solution:

The ratio of

T XF

=

0.02 2Q × 1 × 10 × 3 × 8.854 × 10 −12 9

= 0.06 << 1

i.e., the medium can be considered as good dielectric (a) Wave impedance Z=

N 377 = z0 1/Fr = = 218 : F 3


37

(b) As

E H

=Z ⇒ H =

E Z

=

2

= 9.2 mA/m −1

377

(c) Total available power P = SA = EH ´

A

= 2 ´ 9.2 ´ 10–3 ´

2

0.6 2

= 5.5 mW

(d) The time taken is ratio of distance to phase velocity

t=

d v

=

d c

Fr =

0.2 3 3 × 108

= 1.2 ns

(e) As E z = E0 e− z/E

2 ⎛E ⎞ ⎛ 1 ⎞ ⇒ z = − E ln ⎜ z ⎟ = − E ln ⎜ ⎟ = − T ⎝ 20 ⎠ ⎝ E0 ⎠ = −

⎛ 1 ⎞ ln ⎜ ⎟ N ⎝ 20 ⎠ F

2 ⎛ 1 ⎞ − 2.99 = 2.743 m ln ⎜ ⎟ = − T Z ⎝ 20 ⎠ 0.01 × 218 2

Example 2.5 A dielectric material has er = 18 and tan d = 10–3 at a frequency of 200 MHz. Find the conductivity of material and distance over which the field strength drops to (1/c) of its value at the surface. Solution:

We know that

tan E =

T ⇒ T = XF 0F r tan E XF

= 2p ´ 200 ´ 106 ´ 8.854 ´ 10–12 ´ 18 ´ 10–3 = 2 ´ 10–4 S/m Since the loss tangent of given material is very small, the material may be regarded as a lossless dielectric, and the corresponding attenuation constant (a) can be used.

B=

T 2

N T 1 2 × 10 −4 1 I = = 377 = 8.885 × 10 −3 Np/m F 2 Fr 2 18

Hence the distance travelled at which the field strength goes down to the (1/e), i.e., Skin-depth (d) =

1

B

=

103 8.885

= 1.125 × 10 2 m

38


Example 2.6 At point inside a lossy medium electric field intensity of a wave is 10 V/m at frequency 300 MHz. Find the power density of the wave at this point and at distance 1 cm in the direction of propagation, if er = 8

Solution:

and

s

= 100 S/m.

We know that the intrinsic impedance of the medium is jXN ⎞ I = ⎛⎜ ⎟ ⎝ jXN + T ⎠

1/2

⎛ j 2 × Q × 3 × 108 × 4Q × 10 −7 =⎜ ⎜ 100 + j 2 × Q × 3 × 108 × 8.854 × 10 −12 ⎝ = (3.44 + j 3.44) :

1/2

⎞ ⎟ ⎟ ⎠

The power density of the wave is Pav =

ÿÿh

As

1

I*

|E0 |2 2

⎛ 1 ⎞ Re ⎜ ⎟ ⎝ I* ⎠

= 3.44 + j3.44 =

and

hence

h*

= 3.44 – j3.44

3.44 + j 3.44 (3.44)2 + 3.44 2

⎛ 1 ⎞ Re ⎜ ⎟ = 0.145 ⎝ I* ⎠

Hence Pav =

(10)2 × 0.145 2

= 7.25 W/m 2

Now since the conductivity of the medium is not zero, there will be attenuation and can be calculated from the propagation constant H = B + jC =

=

jXN (jXF 0 F r + T )

j × 2Q × 3 × 4Q × 10 −7 (100 + j × 2Q × 3 × 8 × 8.854 × 10 −12 )

= 3.43 + j 344.36

Hence

a

= 3.43 Np/m.


39

Power at a distance of 2 cm is Pave–2 ´ d ´ a = Pave–2 ´ 10

–2 ´ 343.9

= Pave–6.878

= 7.25 e–6.878 = 7.47 mW/m2 Example 2.7 Polystyrene has dielectric constant of 2.7. If a uniform plane wave is incident at angle of 35° onto it, calculate the angle of transmission and the Brewster angle. Solution:

From Snell’s law

sin Ri sin R t

sin R t =

or

F2

=

F1

1 2.7

sin 35°

qt = 20.02° ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ qB

= tan–1 ( 2.7) = 58.67°

Example 2.8 A linearly polarized plane wave incident from water onto the water–air interface at an angle 45°. Show that wave is totally reflected back. (Hint: er = 81; mr = 1 and s = 0.) Solution:

We know that R c = sin

−1

F1 F2

= sin −1

1 81

= 6.38°

which is less than the incident angle; thus the incident wave is totally reflected back. Example 2.9 A parallel polarized wave is incident from air onto (a) distilled water (er = 81), (b) flint glass (er = 10.5) and (c) paraffin (er = 2.05). Determine the Brewster angle for each case. Solution:

We know that R B = tan

−1

⎛ F2 ⎞ ⎜ ⎟ ⎜ F1 ⎟ ⎝ ⎠

Hence,

R B = tan −1 ( 81) = 83.7° R B = tan −1 ( 10.5) = 72.85° R B = tan −1 ( 2.05) = 55.06°

40


Example 2.10 A uniform plane wave incident from air onto a glass at an angle of 30° from normal. Determine the percentage of incident power that is reflected and transmitted for (a) P-polarized and (b) S-polarized wave if glass has refractive index 1.45. Solution: ⎡ sin 30° ⎤ sin Rt = sin −1 ⎢ ⎥ ⇒ Rt = 20.02° ⎣ 1.45 ⎦

(a)

For parallel polarization ÿ

h1P

=

h1

cos q1 = 377 ´ cos 30° = 326 W

I1

I2P = I2 cos R t = *P =

I2

cos 20.2° =

377 1.45

× cos 20.2° = 244 :

I2P − I1P 244 − 326 = = − 0.144 244 + 326 I2P + I1P

The percentage of power reflected is Pr

% = | * P |2 = 0.021 = 2.1%

Pinc

Hence, the transmitted power is 97.9%. (b)

For perpendicular polarization

I1S = I2S = *S =

I1 cos R i

I2

=

I2 cos R t

377° cos 30° =

= 435 : 377°

1.45 × cos 20.2°

= 277 :

I2S − I1S 277 − 435 = = − 2.22 277 + 435 I2S + I1S

The percentage of power reflected is Pr Pinc

% = | * S |2 = 0.049 = 4.9%

The transmitted power is 95.1%.


41

Example 2.11 Show that for (a) Parallel polarization tan (R1 − R 2 ) tan (R1 + R 2 )

(b) Perpendicular polarization sin (R 2 − R1 ) sin (R 2 + R1 )

Solution:

We know from Snell’s Law F1 sin R1 =

(a)

*P = =

(b)

*S =

F 2 cos R1 −

F 1 cos R 2

F 2 cos R1 +

F 1 cos R2

sin 2R1 − sin 2 R 2 sin 2R1 + sin 2R 2

=

F 2 sin R 2

sin R1 cos R1 − sin R 2 cos R 2 sin R1 cos R1 + sin R2 cos R2

sin (R1 − R 2 ) cos (R1 + R 2 )

=

sin (R1 − R 2 ) cos (R1 − R 2 )

F 1 cos R1 −

F 2 cos R2

F 1 cos R1 +

F 2 cos R 2

=

=

tan (R1 − R 2 ) tan (R1 + R 2 )

sin R2 cos R1 − sin R1 cos R 2 sin R 2 cos R1 + sin R1 cos R2

=

sin (R2 − R1 ) sin (R 2 + R1 )

Example 2.12 A light ray is incident from air on to glass at the Brewster angle. Determine the incident and transmitted angles, and also the reflection coefficient for S-polarized waves. Solution:

Since the glass has refractive index = 1.45, the incident angle is

n2

R i = R B = sin −1

R t = sin −1

n12 + n22 n1

n12

+

n22

= sin −1

1.45 1 + 1.452

1.45

= sin −1

1 + 1.452

= 55.4°

= 34.6°

i.e. the sum of the incident and transmitted angles at the Brewster condition is always 90°, i.e., sin q2 = cos qB.

I1S = I1S =

I1 cos R i

=

I1 cos R i

=

I1 I2 cos R t

377° cos 55.4° =

= 664 :

377° 1.45 × cos 34.6°

= 315.75 :

42


*S =

I2S − I1S 316 − 665 = = − 0.355 316 + 665 I2S + I1S

Example 2.13 Find the rate of energy flow per unit area of a uniform plane wave travelling with a velocity (v) = 1/NF . Solution:

Total energy of an EM wave, due to its electric and magnetic fields is given by

1 2

(F E 2 + N H 2 )

For a wave travelling with velocity (v) the rate of flow of energy per unit area would be

1 Pˆ = (F E 2 + N H 2 ) vˆ 2 Property of plane wave says that E H

N F

=

Hence the above equation can be written as ⎞ 1⎛ N F Pˆ = ⎜ F EH + N eH ⎟ vˆ ⎟ F N 2 ⎜⎝ ⎠

⇒

1 2

(

NF EH +

)

NF EH vˆ

G G ⎛ EH ⎞ ⇒ ⎜ ⎟ vˆ = E × H ⎝ v ⎠ G

G

P= E × H

So, Example 2.14

An EM wave propagating in a certain medium is described by E = 25 sin (2p ´ 106t – 6x) az V/m

(a) Determine the direction of wave propagation. (b) Compute the period T, the wavelength l and velocity u. Solution:

E = 25 sin (2p ´ 106t – 6x) az V/m

Given

(a) Direction is along az. (b) w = 2p ´ 106, b = 6 Time period T =

2Q

X

=

2Q 2Q × 10

6

= 10 −6 μs.


43

If u is the velocity of EM wave, the phase of a point is constant on the wave, i.e.

(X t −

X

which gives

−

Cu = 0

C x) = k

⇒ u=

X C

=

X ⎛⎜

dt ⎞ ⎟ − ⎝ dt ⎠

⇒

2 Q × 106 6

C ⎛⎜

dx ⎞ ⎟ =0 ⎝ dt ⎠

= 1.047 × 10 6 m/s

Wavelength l = Velocity ´ Time period = 1.047 ´ 106 ´ (2p/w) = 1.047 ´ 10–6 ´ (1 ´ 10–6) = 1.047 m Example 2.15 Given that propagation constant g 2 = jwm(s + jwe) and g = a + jb, derive the equation for a and b. Solution:

Re(g 2) = b 2 – a 2 = w2me

(i)

and

Im (H 2 ) = C 2 + B 2 = XN T 2 + X 2 F 2

(ii)

Adding (i) and (ii)

2C 2 = X 2 NF + XN T 2 + X 2F 2

C2

C

Þ

2

⎛ X 2 NF =⎜ + ⎜ 2 ⎝

=

X2 2

C =X

⎡ ⎢ NF + ⎢ ⎣

⎡ ⎢ NF ⎢ 2 ⎢⎣

XN T 2 + X 2F 2 ⎞⎟ ⎟ ⎠

2

NXF X

X2

⎛ ⎜ NF + 2 ⎝

=

N 2 2 2⎞ T +X F ⎟ X ⎠

2⎤ ⎛T ⎞ ⎥ 1+ ⎜ ⎟ ⎝ XF ⎠ ⎥ ⎦

2 ⎧ ⎫⎤ ⎛T ⎞ ⎪ ⎪⎥ ⎨ 1+ ⎜ ⎟ + 1⎬ ⎥ ⎝ XF ⎠ ⎪⎩ ⎪⎭ ⎥ ⎦

1/2

Similarly, by subtracting (i) and (ii), we get ⎡ NF B =X⎢ ⎢ 2 ⎣⎢

⎧ ⎪ ⎨ 1+ ⎪⎩

⎛T ⎞ ⎜ ⎟ ⎝ XF ⎠

2

⎫⎤ ⎪ − 1⎬ ⎥ ⎥ ⎪⎭ ⎥ ⎦

1/2

Example 2.16 A lossy material has m = 5m0 and e0 = 2e0. If at 5 MHz the phase constant is 10 rad/m, calculate (a) the loss-tangent, (b) the conductivity of the material, (c) the complex permittivity and (d) the intrinsic impedance.

44


Solution: (a) Loss-tangent tan R =

F

T

=

F0

XF

⇒

2F 0

=2

F0

(b) Conductivity

T = XF

× tan R = 2 × 2Q × f ×

F =4

× 3.14 × 5 × 10 6 × 8.854 × 10 −12

= 1.11 ´ 10–3 S/m (c) Complex permittivity

Fc = F ′

− jF ′′ ⇒

Fc = F

⎛T ⎞ − j⎜ ⎟ ⎝X ⎠

⎛ ⎞ 1.11 × 10 −3 ⎟ ⎜ 2 × 3.14 × 5 × 10 6 ⎟ ⎝ ⎠

F c = 2F 0 − j ⎜

⎛ ⎞ 1.11 × 10 −3 = 2 × 8.854 × 10 −12 − j ⎜ ⎟ ⎜ 2 × 3.14 × 5 × 10 6 ⎟ ⎝ ⎠ –12 = (17.7) – j(35.4) ´ 10

Hence, attenuation constant ⎡ NF B =X⎢ ⎢ 2 ⎣

⎛T ⎞ 1+ ⎜ ⎟ ⎝ XF ⎠

2

⎤ − 1⎥ ⎥ ⎦

⎡ 5 × N0 × 2 × F 0 2 ⎣

B =X ⎢

B= =

X

⎡5 5 − 1⎤ ⎦ c ⎣

1/2

⎤ 1 + (2)2 − 1⎥ ⎦

1/2

1/2

2 × 3.14 × 5 × 106 3 × 108

× 3.2 = 33.5 × 10 −2 ⇒

(d) Intrinsic impedance

I=

N /F 2 ⎡ ⎛T ⎞ ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ XF ⎠ ⎥⎦

1/4

=

5N0 /2F 0 2 ⎤ 1/4

⎡1 + (2) ⎣ ⎦

=

5N0 /2F 0 ⎡1 + (2)2 ⎤ ⎣ ⎦

1/4

B = 0.335


=

5/2 N0 /F 0

=

( 5)1/2

45

× 377 = 398.7 :

1.5811 1.495

G

Example 2.17 In free space (z £ 0), a plane wave with Hi = 10 cos(108t – b z) ax (in mA/ m) is incident normally on a lossless medium (e = 2e0, m = 8m0) in region z ³ 0. Determine the Hr and Er components of the reflected wave as well as Ht and Et of the transmitted wave. Solution: For free space, b1 = w/c = 108/(3 ´ 108) = 1/3 and For the lossless dielectric medium,

C2 = X NF = X N0 F 0 ×

h1

=

2 × 8 = 4X /c = 4/3

I2 = N /F = N0 /F 0 × 8/2 = 2I0 Since

G

H i = 10 cos(108t − C z) ax

We expect that G

Ei = Ei 0 cos (10 8 t − C z ) aEi G

G

G

G

aEi = aHi × aHj = aˆ x × aˆ z = − ay

where

Ei 0 = I1 H i 0 = 10 I0

Hence Now Thus Since E = hH

Similarly

Ei = –10h0 cos (108t – b1z) aˆ y mV/m Er 0 Ei 0

=*=

I2 − I1 1 E = ⇒ Er 0 = i 0 3 I2 + I1 3

Er = Er 0 = − 3.33 I0 cos (108 t + 0.33z) aˆ y mV/m ⎛ 10 ⎞ Hr = − ⎜ ⎟ cos(108 t + 0.333 z) aˆ x mA/m ⎝ 3⎠ 4Ei 0 ⎛ Et 0 ⎞ ⎜ ⎟ = U = 1 + * = 4/3 or Et 0 = 3 ⎝ Ei 0 ⎠

Thus

Et = Et 0 cos(108 t − C2 z ) aˆ Et

where Hence

aˆ Et = aˆ E1 = − aˆ y ⎛ 40 ⎞ Et = − ⎜ ⎟ ⎝ 3 ⎠

I0

4z ⎞ ⎛ cos ⎜108t − ⎟ aˆ y mV/m 3 ⎠ ⎝

h0

= 120p.

46


From which we obtain

I=

Et Ht

and H t =

Et

2I0

4z ⎞ ⎛ 20 ⎞ ⎛ Ht = ⎜ ⎟ cos ⎜ 108 − ⎟ aˆ x mA/m 3 ⎠ ⎝ 3 ⎠ ⎝ Example 2.18 Show that 1 + G = t, if an electromagnetic wave is normally incident from one medium (s1, e1, m1, h1) to second medium (s2, e2, m2, h2). Solution: We have considered uniform plane waves travelling in unbounded, homogeneous, isotropic media. When a plane wave from one medium meets a different medium, it is partly reflected and partially transmitted. The portion of the incident wave that is reflected or transmitted depends on the constitutive parameters (e, m, s) of the two media involved. Incident wave Ei, Hi is travelling (+) az in the medium. If we suppress the time factor e jwt and assume that Eis (z) = Ei 0 e−H 1z aˆ z

(i)

His (z) = Hi 0 e−H 1z aˆ y ⎛E ⎞ His (z) = ⎜ i 0 ⎟ e −H 1z aˆ y ⎝ I1 ⎠

(ii)

Ei 0 , H i 0 = Amplitudes of incident wave

H is(z), Eis(z) = Incident wave in medium 1 Reflected wave G

G

G

G

Ers (z) = Er 0 eH 2 z az

(iii)

Hrs (z) = Hr 0 eH 2 z ay or

G

⎛E Hrs (z ) = − ⎜⎜ r 0 ⎝ I1 G

⎞ H 2z ⎟⎟ e ay ⎠

(iv)

Transmitted wave Ets (z) = Et 0 e−H 2 z aˆ x

Hts (z ) = Ht 0 e −H 2 z aˆ y

(v)


⎛E ⎞ Hts (z ) = ⎜ t 0 ⎟ e−H 2 z aˆ y ⎝ I2 ⎠

47 (vi)

where g1 and g2 are propagation constants. Boundary condition Since waves are transverse at z = 0, the tangential components of E and H fields must be continuous, i.e., Ei(0) + Er(0) = Et (0)

(vii)

Hi(0) + H r(0) = Ht(0) Et 0 ⎛1⎞ ⎜ ⎟ ( Ei 0 − Er 0 ) = I2 ⎝ I1 ⎠

or

(viii)

From Eqs. (vii) and (viii), we get ⎛I ⎞ Et 0 ⎜ 1 ⎟ = Ei 0 − Er 0 ⎝ I2 ⎠

⎛I Et 0 = ⎜ 2 ⎝ I1 ⎛ Et 0 ⎜ 1 + ⎝ Et 0 =

Er 0 =

and

⎞ ⎛ I2 ⎟ Ei 0 − ⎜ ⎠ ⎝ I1

I2 ⎞ I1 ⎟⎠

⎛I = 2⎜ 2 ⎝ I1

2I2

I1 + I2

⎞ ⎟ Er 0 ⎠

⎞ ⎟ Ei 0 ⎠

Ei 0

(I2 −

I1 ) (I2 + I1 )

Ei 0

(ix)

(x)

Reflection coefficient ⎛ E ⎞ I − I1 * = ⎜ r0 ⎟ = 2 I2 + I1 ⎝ Ei 0 ⎠

(xi)

Transmission coefficient 2I2 ⎛ Et 0 ⎞ ⎟= ⎝ Ei 0 ⎠ I1 + I2

U = ⎜

From Eq. (xi) 1+* =

(I2 −

I1 + I1 + I2 ) I1 + I2

⎛ 2I2 ⎞ =⎜ ⎟ =U ⎝ I1 + I2 ⎠

(xii)

48


1+ G = t

or Hence, proved. Example 2.19 Solution:

Derive the wave equation for infinite, isotropic and homogeneous media.

The Maxwell equations for the given medium are G

G

G

∇ × H = TE + F

∂E

(i)

∂t

G

G

∇ × E = −N

∂H

G

(ii)

∂t

∇.H =0 Ñ.E = 0

(iii) (iv)

Taking the curl of Eq. (ii) yields G

⎡ ∂( ∇ × H ) ⎤ ∇ × ∇ × E = −N ⎢ ⎥ ∂t ⎢⎣ ⎥⎦ G

(v)

Putting the value of Eq. (i) in Eq. (v) gives G

G

∂ ⎡ G ∂E ⎤ ∂E ∇ × ∇ × E = − N ⎢T E + F − ⎥ = − NT ∂t ⎣⎢ ∂t ⎦⎥ ∂t G

G

G

G

NF

∂2 E ∂t 2

(vi)

Similarly taking the curl of (i) and substituting (ii) yields G

∇ × ∇ × H = − NT

G

∂H

− NF

∂t

∂2 H

(vii)

∂t 2

Ñ ´ Ñ ´ A = Ñ(ÑA) – Ñ2 A

Using vector identity

Since Ñ(ÑA) is zero, the Eqs. (vi) and (vii) reduces to G

∇ E = NT 2

G

∂E ∂t

G

+ NF

∂2 E ∂t 2

G

and ∇ H = NT 2

G

∂H ∂t

G

+ NF

∂2 H ∂t 2

which are known as electric and magnetic wave equations respectively. Example 2.20 For the given two mediums a linearly polarized wave is incident from medium 1 (s = 0, e1, m1) to medium 2 (s = 0, e2, m2). Prove that for total reflection of wave, the incident angle must be R i ≥ in mediums 1 and 2 respectively.

N2 F 2 u1 = , where u1 and u2 are the velocities of wave u2 N1F1


Solution:

Applying Snell’s Law, n1 sin qi = n2 sin qt

which gives

⎛n ⎞ sin Ri = ⎜ 2 ⎟ sin Rt ⎝ n1 ⎠ For total reflection of light, qi > 90°; hence ⎛n ⎞ sin Ri = ⎜ 2 ⎟ sin 90° ⎝ n1 ⎠

or

⎛n ⎞ sin R i = ⎜ 2 ⎟ ⎝ n1 ⎠

From reflection n2 = c N2 F 2 =

n1 = c N1F1 =

c u2 c u1

Putting the value of reflective index in above equation gives sin Ri =

sin R i =

if qi is small

or Hence, proved.

c N2F 2 c N1F1

N2 F 2 u1 = u2 N1F1

sin R i = R i =

Ri >

N2 F 2 u1 = N1F1 u2

N2 F 2 u1 = u2 N1F1

49

50


OBJECTIVE TYPE QUESTIONS Group A:

Electromagnetic Wave and Radiation

1. Electromagnetic wave can be characterized by (a) Permittivity (e) and conductivity (s) (b) Permeability (m) and characteristic impedance (h) (c) Presence of boundary between media (d) All the above is correct. 2. Assuming the Poynting vector along z-axis, choose the correct option. (a) In TE mode of propagation, Ez ¹ 0 (b) In TM mode of propagation, Hz = 0 (c) TEM mode of propagation, Ez ¹ 0, H z = 0 (d) Only (a) and (c) are correct. 3. The shortest wave is (a) Gamma rays (c) Visible light

(b) Radio waves (d) Infrared light

4. The wavelengths of microwave lies between (a) 0.1 mm and 30 cm (b) 1 mm and 30 cm (c) 1 cm and 300 cm (d) 1 m and 100 m 5. Choose the incorrect statement: (a) X-rays can be produced by excitation and ejection of core atomic electron. (b) Lien-energy gamma ray is produced by creation of particle–antiparticle pairs. (c) Gamma ray is produced by energetic ejection of core electron in heavy metals. (d) Ultraviolet light can be produced by excitation of molecule and atomic valence proton as well as neutron. 6. Given waves and their cause of production Statement 1: Radio collective oscillation of charge carrier in bulk material-plasma oscillation Statement 2: Microwave: Plasma oscillation, molecular production. Statement 3: Near Infrared: Molecular vibration, plasma oscillation. Choose the correct option: (a) Statement 1 is correct but 2 and 3 are wrong (b) Statement 2 is correct but 1 and 3 are wrong (c) All the statements are correct (d) None of the above is a correct statement 7. Choose the correct option (More than one may be correct.) (a) Microwave is absorbed by molecules that have a dipole moment in liquids. (b) In microwave oven, H2O, HCl as well as BF3 will heat up. (c) Microwave is super high frequency (SHF) and extremely high frequency (EHF) wave. (d) Microwave is produced with klystron and magnetron tubes and with solid state diodes such as Gunn and IMPATT devices.


51

8. Choose the correct option: (a) Low intensity microwave radiation is used in Wi-Fi. (b) High intensity microwave radiation is used in Wi-Fi. (c) Both (a) and (b) are correct. (d) None of these. 9. Maxwell–Faraday’s equation is G

∂B

G

(a) ∇ × E = −

∂t (c) Both (a) and (b) are correct

(b)

∫∫

B . ds = 0

(d) None of these.

10. Choose the incorrect option: (a) Gauss law is ∇ . E = (b) Gauss law is

S F0

G

w ∫∫

K

E . dA =

∂v

(c) If Ñ . B ¹ 0 and

∂v F0

, where ¶v is the total enclosed charge

w∫∫ B . dA ≠ 0 , then monopole exist in Universe. ∂v

(d) (a) and (b) are correct but c is incorrect. 11. Ampere’s circuit law states that G

K

G

G

(a) ∇ × H = J +

(c)

∫

∂D

(b)

∂t

∫

G

G

∫

H . dl =

S

L

⎛ G ∂D ⎞ ⎜⎜ J + ⎟ ds ∂t ⎟⎠ ⎝

G

G

H . dl =

S

∫ L

⎛ G ∂D ⎞ ⎜⎜ J + ⎟ ds ∂t ⎟⎠ ⎝

(d) None of these

12. Choose the correct option about Poynting vector: (a) Poynting vector represents electron flux of an electromagnetic field. (b) Energy flux of an electromagnetic field. (c) Both (a) and (b) are correct. (d) None of these 13. Choose the correct one: G

G

G

G

G

(a) p = E × H

G

(c) p = H x × Ex

G

G

G

G

G

(b)

p=H × E

(d)

p = H y × Ey

G

14. The time-averaged magnitude of the Poynting vector is (more than one option may be correct). (a) =

1 2 N0 c

G

E02 , S = Poynting vector

52


(b) =

F0 c

2

G

E02 , S = Poynting vector

2 (c) (a) is correct but (b) is wrong (d) None of these.

15. Choose the correct option:

G G (a) Linear momentum of electromagnetic field is given by p = 2 , S = Poynting C vector. (b) Radiation pressure is given by Prad = C (c) Both (a) and (b) are correct (d) None of these.

16. The concept of displacement current was a major contribution attributed to (a) Faraday (b) Lenz (c) Maxwell (d) Hertz (e) Your Professor 17. Identify which of the following expressions are not Maxwell’s equations for a timevarying field: (a) ∇ . J +

∂S

=0

∂F

(b) ∇ . D = Sv

G

−∂B (c) ∇ . E = ∂t

(e)

v∫

(d)

v∫

Hdl =

∫

(T E + F ∂E/∂t ) ds

B ds = 0

18. Choose the incorrect option: (a) Plane wave is a constant frequency wave. (b) Wave front is surface of constant phase. (c) Wave fronts are infinite parallel planes of constant amplitude normal to the phase velocity vector. (d) Wave fronts are infinite parallel phases of constant amplitude parallel to the phase velocity vector. 19. Choose the incorrect option about uniform plane wave: (a) Phone waves are TEM waves (b) E and H fields are always in time phase (c) The magnitude of the two fields is always variable. (d) The stored energy is equally divided between the E and H fields. 20. Choose the correct option. (Multiple options may be correct.) G G (a) In static EM field E and B are independent. (b) In dynamic EM field, the two s fields are interdependent. (c) Both (a) and (b) are incorrect. (d) None of these.


Group B:

53

EM Waves in Mediums

1. Which of these is not a correct form of the wave Ex = cos (w t – (a) cos (b z – w t) (b) sin (b z – w t + p /2) 2Q ⎞ ⎛ 2Q t − z (c) cos ⎜ M ⎟⎠ ⎝ T (e) cos b (z – w t)

b z)?

(d) Re(e j(w t–b z))

2. Which of the following statement is not true for the waves in general? (a) The phenomenon may be a function of time only. (b) The phenomenon may be sinusoidal or co-sinusoidal. (c) The phenomenon must be a function of time and space (d) For practical reasons it must be finite in extent. 3. If the electric field component of a wave in free space is given by E = 10 cos (107t + kz) aˆ y V/m, which of the following is incorrect? (a) The wave is transverse along x-axis (b) The wavelength,ÿ l = 188.5 m (c) The wave amplitude = 10 V/m (d) The wave number, k = 0.33 rad/m (e) The wave attenuates as it travels. 4. Select the correct option: (a) Electromagnetic wave travels faster in conductors than in dielectrics (b) In a good conductor E and H are in time phase (c) The Poynting vector physically denotes the power density leaving or entering a given volume in a time-varying field. (d) None of these. 5. Choose the incorrect option: (a) 1 + G = t, tÿ = Transmission, G = Reflection coefficient. (b) Both G and t are dimensionless and may be complex. (c) 0 £ G £ ¥. 2I2

(d) U =

I1 + I2

where

h1 G

and

6. In the wave equation ∇ E = NF 2

of the wave? G 2

are intrinsic impedances of the medium.

G

∂2 E ∂2 t

G

+ NT

∂E ∂t

, which term is responsible for attenuation G

(b) NF

(a) ∇ E G

(c) NT

h2

∂E ∂t

∂2 E ∂2t

(d) All of the three.

54


7. Two statements are given below. Read the statement and give the correct option. Statement 1: The Poynting vector represents the particular case of an energy flux vector for electromagnetic energy. Statement 2: The Poynting vector represents any type of movement in space as well as its density, and so the energy flux vector can be defined for other types of energy as well, for example, mechanical energy. (a) (b) (c) (d)

Statement (1) is correct but (2) is wrong. Statement (1) is wrong but (2) is correct. Statement (1) and Statement (2) both are correct. Both are wrong statements.

Answers Group 1. 6. 11. 16.

A (d) (c) (a) (c)

2. 7. 12. 17.

(b) (a, c, d) (b) (a, c)

3. 8. 13. 18.

(a) (a) (a) (d)

4. 9. 14. 19.

(b) (c) (a, c) (c)

Group B 1. (b) Explanation:

Q⎞ ⎛ E x = sin ⎜ C z − X t − ⎟ = cos R 2⎠ ⎝ Q⎞ Q⎞ Q⎞ ⎛ ⎛ ⎛ sin ⎜ C z − X t − ⎟ = sin ⎜ v − ⎟ = − sin ⎜ v + ⎟ 2⎠ 2⎠ 2⎠ ⎝ ⎝ ⎝ cos C (z − ut ) = cos(C z − C ut ) cos (–1) (w t –

C =X /u

bz) = cos(w t – b z)

2. (c) must be a function of both time and space. 3. (a) Given E = 10 cos(107t + kz) aˆ y E = A cos(107t + kz) aˆ y (b) Transverse in –z direction (– Ez) (c) Amplitude, A = 10 V/m (d)

w = 107, C =

X

=k

or

c c is free space velocity.

k =

10 7 3 × 108

= 0.033 rad/m

5. 10. 15. 20.

(d) (d) (c) (a, b)


55

(e) Wavelength l = Distance travel in 2p ´ radius. l = Velocity ´ Time period 2 ⎛ 2Q ⎞ 8 =c × ⎜ = 188.5 m ⎟ = 3 × 10 × 3.14 × 10 7 ⎝X ⎠

4. (c) Explanation: (a) Velocity of dielectric > velocity of conductor (b) E and H not in time phase (c) Correct 5. (c) 0 £ | G | £ 1 6. (c)

EXERCISES 1. A uniform plane wave incident from air onto glass at an angle of 60° from the horizon. Determine the amount of power reflected and transmitted for (i) P-polarization (ii) S-polarization. Refractive index of glass is 1.52 and total incident power is 60 mW. 2. A material has a relative permittivity of 2.8. If a wave is incident at an angle of 30° onto it from air, determine the angle of transmission and Brewster angle. 3. A plane wave of frequency 4 GHz is incident normally air onto a half-space of dielectric having s = 0, mr = 1 and er = j3. Find the dB value of reflected power. 4. Light is incident from air onto metal plate of refractive index 1.52 at Brewster angle. Determine incident and transmitted angles and also find the reflection coefficient for the perpendicular polarization.

REFERENCES [1] Maxwell, James, C., A Dynamic Theory of the Electromagnetic Fields, Scientific papers reprinted by Devers, NY, 1952. [2] Jordan, E.C. and K.G. Balmin, Electromagnetic Waves and Radiating Structures, 2nd ed., Prentice-Hall of India, New Delhi, 1995. [3] Das, A. and S.K. Das, Microwave Engineering, Tata McGraw-Hill, New Delhi, 2000. [4] Paul, C.R., et al., Introduction to Electromagnetic Fields, McGraw-Hill, USA.

C H A P T E R

3

Antenna Fundamentals and Parameters

INTRODUCTION Communications between human beings began first by sound through voice using devices such as drums. The visual methods such as signal flags and smoke signals were also used for this purpose. Later, for all the long distance communication, the electromagnetic waves, outside the visible region, have been employed, through the use of radio. The antenna used in this communication is termed radio antenna, which is an essential component in the radio communication system [1]. By the definition of an antenna, information can be transferred between different locations without any intervening structures. Antennas are popular in broadcast situations where one transmit terminal can serve unlimited number of receivers, which can be mobile phones (or car radio sets). The group of frequencies of the electromagnetic (EM) waves carrying this information form EM spectrum which is one of humankind’s greatest natural sources and the antenna has been instrumental in harnessing this source. Most antennas are reciprocal devices and show identical behaviour on transmit/receive signals. Antennas have directional characteristics; EM power density is radiated from a transmitting antenna with intensity that varies with the angle around the antenna.

Types of Antenna There are various configurations of antennas (Fig. 3.1); broadly they can be classified as given in Table 3.1. In other words, based on their performance and function of frequency, the commonly used antennas (across the radio spectrum) are divided into four groups: · · · ·

Electrically small antennas Resonant antennas Broadband antennas Aperture antennas 56


Dipole

Loop

Patch

Slot

Spiral

Helix

Yagi–Uda

Horn

Notch

FIG. 3.1

57

Various antenna configurations.

TABLE 3.1

Various types of antenna

Shape and geometry

Gain

Beam shapes

Bandwidth

Wire antennas—dipole, loop, helix antennas

High gain—Dish antennas

Omnidirectional— Dipole antennas

Wideband—log, spiral and helix antennas

Aperture antennas— Horn, slot antennas

Medium gain—Horn antennas

Pencil beam— dish, array antennas

Wideband—log, spiral and helix antennas

Printed antennas— patch, spiral antennas

Low gain—Dipole, loop, Fan beam—oneslot and patch antennas dimensional array

Narrow band—Patch and slot antennas.

Electrically small antennas are much less than operating wavelength. They are simple in structure and their properties are not sensitive to construction specifications. These antennas are used for VHF frequencies applications. The vertical monopole used for AM reception on cars is a best example. It is about 0.002l long and has nearly omnidirectional pattern in horizontal plane. Resonant antennas are often used where simple structure with good input impedance over a narrow band of frequencies is needed. It has a broad main beam and low or moderate gain. The l/2 dipole, Yagi antenna and patch antenna are the popular examples. There are many applications that require antenna which could operates over a wide frequency range, such antenna is known as broadband antenna. They have acceptable performance over a 2:1 bandwidth ratio of upper to lower operating frequency. Examples are log-periodic dipole, Yagi and spiral antennas. Since only portion of a broadband antenna is responsible for radiation at a given frequency, the gain is low. But it may be an advantage to have gain that is nearly constant, although low [2, 3].

58


Aperture antenna is usually several wavelengths long in one or more dimensions. The pattern usually has a narrow main beam leading to high gain. Bandwidth is moderate as large as 2:1. Most useful aperture antenna is a horn antenna; it acts as a “funnel” directing waves into the connected wave-guide.

Applications of Antenna Non-communication applications of antenna include remote sensing and industrial applications. Remote sensing systems are either active (radar) or passive (radiometry). They receive scattered energy or inherent emission from the object respectively. The received signals are proposed to infer information about the object or scenes. Industrial applications are mainly cooking and drying using high frequency waves. There are several other applications of antenna. For example, mobile communication involving aircraft, spacecraft, ships or land vehicle requires antennas. Non-broadcast radio applications such as municipal radio and amateur radio also require antennas. Personal communications devices such as pagers and cellular phones are common places where antennas are being used.

ISOTROPIC RADIATOR AND RADIATION FIELD An isotropic radiator is a theoretical point antenna that cannot be realized in practice. It radiates energy equally well in all directions, as shown in Fig. 3.2(a). The radiated energy will have a spherical wave front with its power spread uniformly over the surface of a sphere.

FIG. 3.2(a)

Isotropic radiator.

If the source transmitting power is Pt, the power density Pd at a distance R from the source can be calculated using Pd =

Pt 4Q R 2

W/m 2

(3.1)


59

Although the isotropic antenna is not practical, it is commonly used as a reference with which to compare other antennas. For a large value of R, the transmitted wave can be G approximated by uniform plane. The electric field ( E ) is ^r to direction of propagation and G G G G H is normal to E and nˆ is unit rector along the direction of propagation. Both E and H lie in the constant-phase-plane and the wave is a TEM wave [see Fig. 3.2(b)].

FIG. 3.2(b) As time-average power density is Pd =

Plane wave. 1 E2

2 I0 distance R from isotropic antennas as follows [4]:

E=

60 Pd R

, one can find the electric field at a

= 2 Erms

(3.2)

where E is peak magnitude of electric field and Erms is its rms value.

Far-field Region Generally, it is assumed that the antenna operated properly in the far-field region and radiation pattern are measured in this region only. In the far-field region, the transmitted wave of the transmitting antenna resembles a spherical wave from the point source that only locally resembles a uniform plane wave. To derive the far-field criterion for the distance R, consider the maximum antenna dimension to be D as shown in Fig. 3.3. We have ⎛D⎞ R 2 = ( R − 'l)2 + ⎜ ⎟ ⎝2⎠

2

⎛D⎞ = ⎡⎣ R2 + ('l)2 − 2R'l ⎤⎦ + ⎜ ⎟ ⎝2⎠

2

(3.3)

60


FIG. 3.3

Calculation of far-field region criterion.

For R >> Dl, Eq. (3.3) becomes 2RDl » D2/4. Therefore R=

D2 8'l

(3.4)

If we let, Dl = l0/16, which is equivalent to 22.5° phase error, be the criterion for far-field operation, we have Rfar-field =

2D 2

M0

(3.5)

Thus, the condition for far-field operation could be given by R ≥

2D 2

M0

(3.6)

where l0 is the free space wavelength and D is the maximum aperture/dimension of the antenna. The far-field conditions are summarized as follows: R>

2D 2

M0

(3.7a)

R >> D

(3.7b)

r >> l

(3.7c)

The condition R >> D is associated with the approximation R = r for use in the magnitude ⎛ 2Q r ⎞ dependence. The condition r >> l follows from br = ⎜ ⎟ >> 1. Equation (3.5) is a ⎝ M ⎠ sufficient condition for the antenna operating in the UHF region and above. However, at lower frequencies, where antenna is small compared to the wavelength, the condition of farfield region is given by Eq. (3.6), provided that all conditions in Eq. (3.7) are satisfied. In far-field region, the radiation pattern of an antenna is independent from the distance [5].


61

ANTENNA PARAMETERS Input Impedance and VSWR The input impedance is the one port impedance appearing into the antenna. It is impedance presented by the antenna to the receiver or transmitter connected to it. If the input impedance of an antenna (Zin) is not equal to the characteristic impedance (Z0) of Tx, line used to feed the antenna, impedance mismatching occurs. In this situation, total voltage and current present across the Tx line are expressed as the sum of two travelling waves moving opposite directions as on infinite Tx line. The waves travelling from left to right are regarded as incident waves having voltage and current V0 and I0, and waves travelling opposite to these as reflected waves, having voltage and current V1 and I1 (say) (Fig. 3.4). As a result of these, two travelling waves along the Tx line, third wave, are generated known as standing wave and described by voltage standing wave ratio (VSWR). The voltage V at any point on the line is equal to the sum of the voltages V1 and V0, i.e., V = V0 + V1. The ratio of reflected to incident voltages is constant, and termed as reflection coefficient (r), i.e.

Sv =

V1

(3.8)

V0

It can also be described in terms of impedances as:

Sv =

Z A − Z0

(3.9)

Z A + Z0

Hence, the input impendence of an antenna (ZA) can be expressed in terms of rv as follows: Z A = Z0

1 + Sv

(3.10)

1 − Sv

For maximum power transfer or perfect matching the input impedance of the antenna should be equal to Z0. However, VSWR is given by VSWR =

Vmax Vmin

=

I max

(3.11)

I miin

where (Vmax, Imax) and (Vmin, Imin) are the maximum and minimum values of voltage and current respectively. Hence, equation (3.11) becomes,

VSWR =

V0 + V1 V0 − V1

1+ =

V1 V0

V 1 − 1 V0

=

1 + Sv 1 − Sv

(3.12)

62

Antenna and Wave Propagation Total wave Vmax

Vmin

I Z0 Source

Zi

V

Antenna

Incident wave Reflected wave

FIG. 3.4

Impedance matching antenna as load.

and the return loss in the process of energy transfer is given by RL = 20 log |r| dB

(3.13)

That is optimum VSWR appears when |r| = 0 and corresponding value of VSWR = 1. This means all the power is being transmitted to the antenna and there is no reflection (no power loss). On the other hand, maximum VSWR occurs, when |r| = 1 and VSWR = ¥. This is case when all the power reflected back and there is no power transmission to the antenna. Typically, VSWR less than 2 is acceptable for most of the applications. The power reflected from the antenna is |r|2 times the power available from the source. The power coupled to the antenna is (1 – |r|2) times the power available from the source. Transmission coefficient for voltage *v =

2Z A Z 0 + Zi

= 1 + Sv

(3.14a)

= − Sv

(3.14b)

= 1 + Si

(3.14c)

Reflection coefficient for current

Si =

Z 0 − Z in Z 0 + Z in

Transmission coefficient for the current *i =

2 Z0 Z 0 + Zi

The equivalent circuit models of Tx and Rx antennas are shown in Figs. 3.5 and 3.6 respectively.


63

Antenna

Source Guide

A XS

RA

RS

Rr B VS

XA

(a) Thevenin’s equivalent

IS

GS

BS

BA

Gl

Gr

(b) Norton’s equivalent

FIG. 3.5

Equivalent Circuits of Tx Antenna In general, antenna impedance is expressed by ZA = RA + jXA where RA is the antenna resistance and XA is the antenna reactance. Antenna resistance RA is sum of the two resistances; radiation resistance (Rr) and loss resistance (Rl), i.e. RA = Rr + Rl. The radiation resistance (Rr) relates the radiated power to the current (or voltage) at the antenna terminals by Rr = 2Pr /|I|2 W. In this model it is considered that the source is connected directly to the antenna. If not so, i.e. if there is guiding structure (Tx line) between the source and the antenna, the Zs = Rs + jXs represent the equivalent impedance of the source/generator, transferred to the input terminals of the antenna. Existence of Tx line may cause significant loss of power. The maximum power transformation from source to antenna happens only when conjugate matching of impedance is occurred, i.e. RA = R r + Rl = R s

and

XA = – Xs

64


ZL load

Antenna

A RA XL

Rr

RL

VA XA (a) Thevenin’s equivalent

A GL

BL

BA

Gl

Gr

IA

(b) Norton’s equivalent

FIG. 3.6 Using circuit’s theory, the following expressions are derived in the condition of matched impedances: (i) The power delivered to the antenna

PA =

Vs

2

(3.15a)

8(Rr + Rl )

(ii) Radiated power Pr =

Rr (Rr + Rl )2

Vs

2

(3.15b)

8

(iii) Power dissipated as heat in antenna Pl =

Rl

Vs

(Rr + Rl )2

8

2

(3.15c)


65

(iv) Power dissipated as heat in antenna Vs

Ps = PA =

2

(3.15d)

8(Rr + Rl )2

Equivalent Circuit of Rx Antenna The incident space wave induces voltage VA (say) at the antenna terminals, when antenna is open circuited. Similar to transmitting antenna, the following expressions for delivered power are found in condition of conjugate matched impedances: (i) The power delivered to the load is

Pl =

VA

2

8 RL

VA

=

2

(3.16a)

8 RA

(ii) The power dissipated as heat in the antenna is Pa =

VA

2

Rl

(3.16b)

R A2

8

(iii) The re-radiated power is

Pr =

VA Rr 8

(3.16c)

R A2

(iv) Total received power captured is Pc =

VA2 4(Rr + Rl )

=

VA

2

4 RA

(3.16d)

Conjugate impedance matching is compulsory between the antenna and the load (receiver) to achieve maximum power delivery; RL = RA = Rl + Rr and XL = –XA. When conjugate matched impendence is achieved, only half of the captured power Pc is delivered to the load (receiver) and another half is still dissipated by the antenna as antenna losses. Antenna losses take place in terms of heat dissipation Pl and scattered power Pr. Even in case of lossless antenna, only half of the power is delivered to the load (though there is conjugate impedance matching), the other half being scattered back into space. The antenna impedance (ZA) is related to the radiated power (Pr), the dissipated power (Pl) and stored reactive energies as follows [4]:

ZA =

Pr + Pd + 2jX (Wm − We ) 0.5 I 0 I 0*

(3.17)

66


where I0 is the current at the antenna terminals, We and Wm are the average electric and magnetic energies stored in the near-field region. When these energies are equal, a condition of matching occurs, where the reactive of antenna impedance (XA) vanishes. The antenna impedance is frequency dependent thus it is matched to its source and load in a certain frequency band. The input impedance of the antenna also depends on many other factors including its geometry, feed techniques and its proximity to surrounding objects.

Bandwidth It is found that most of the antennas operate around there resonant point (frequency), i.e. there is only a limited frequency range over which they can operate efficiently. This is because outside particular frequency range (BW), the levels of reactance rise high and deteriorate satisfactory operation of the antenna. The bandwidth of antenna is usually defined as the frequency range within the performance of the antenna, with respect to a certain characteristic, conforms to a specified standard; particularly the antenna gain and (FBR) ratio hold up. The standards may be higher gain than some acceptable value or atleast FBR or the value of SWR closer to unity. It is expressed as the percentage of the difference between upper and lower frequencies to the centre frequency. As we have seen in case of impedance of the antenna, antenna characteristics are affected in different ways as frequency changes, there is unique definition of the bandwidth. For the antenna of relatively small dimensions (< half wavelength) and low frequency, the bandwidth is generally measured by impedance variation, because pattern performance is less sensitive to the frequency, i.e. the pattern changes less rapidly with the frequency. The most commonly used bandwidths are pattern bandwidth and impedance bandwidth. In addition, there is another bandwidth known as radiation bandwidth.

Impedance Bandwidth It is the range of frequencies over which the input impedance conforms to a perfect matching and hence maximum power deliberation. This standard is commonly to be VSWR £ 2 (or |r £ 0.5|) and translate to a power reflection of about 11%. Certain applications may require a more stringent specification, such as a VSWR of £ 1.5 (Fig. 3.7(a)). The operating bandwidth of the antenna could be smaller than the impedance bandwidth, since other parameters; Gain, FBR, VSWR, beam width, radiation patterns and polarization are also functions of frequency and may deteriorate over the impedance bandwidth. Impedance bandwidth also termed the fractional bandwidth of an antenna and it is a measure of how wideband the antenna is. For the narrow band antennas [5], fractional bandwidth is defined as FBW =

fu − f l f0

× 100%

(3.18)

where fu and fl = Upper and lower frequencies ⎛ f + fl ⎞ f0 = Design/centre frequency ≈ ⎜ u ⎟ or ⎝ 2 ⎠

fu . f l .


67

Frequency fl

fu

VSWR = 2

Return loss

FIG. 3.7(a)

Bandwidth determination of antenna.

The fractional bandwidth varies between 0 and 2, and is often quoted as a percentage (between 0% and 200%). The higher the percentage the wider the bandwidth. Wideband antennas typically have a fractional bandwidth of 20% or more. Antennas with a FBW of greater than 50% are referred to as ultra-wideband antennas. From Eq. (3.18) it is clear that, higher the difference between fu and fl, wider the bandwidth. However, the development of frequency-independent antenna led to unlimited bandwidth where upper and lower frequencies limits are specified independently and, in this case, FBW = fu/f0. In terms of quality factor Q and VSWR, the bandwidth of an antenna is also defined as

BW =

where Q = 2Q

fc Q

and BW =

VSWR − 1 Q VSWR

(3.19)

Total energy stored by the antenna Energy dissipated per cycle

i.e. the lower value of Q of the antenna leads to higher BW and vice versa.

Pattern Bandwidth Pattern bandwidth is the frequency range over which the pattern characteristics vary within the acceptable limits. A broadband antenna generally has a relatively low level of pattern variation over the bandwidth. However, design techniques used to optimize impedance bandwidth can degrade pattern bandwidth. The major factors associated with antennas are their resonance (centre frequencies) and BW over which they operate. Naturally, they are very important features for the operation of antenna and as such they are specifications of an antenna. Whether the antenna is used for broadcasting, Wide Local Area Netowrk (WLAN), cellular and mobile communications, satellites/radars or any other applications, the performance of the antenna is paramount, and the resonant frequency and bandwidth are of great importance.

68


Radiation Bandwidth This is another feature of antenna that changes with its operating frequency. In the case of antenna beam, it is particularly noticeable. In particular the front-to-back ratio (FBR) will fall off rapidly outside a given bandwidth, and so will the gain. In an antenna such as Yagi, this is caused by a reduction in the currents in the parasitic elements as the frequency of operation is moved away from the resonance. In such a case, the radiation pattern B/W is defined as the frequency range over which the gain of the main lobes is within 1 dB of its maximum values. In general, for many beam antennas, especially high gain it is found that the impedance BW is wider than the radiation BW, although both the bandwidths are interrelated in many respects. Similarly polarization bandwidth can also be defined as frequency range in which polarization of antenna is under acceptable limits (0 dB < AR < 3 dB).

FBR FBR stands for forward to backward radiation ratio and defined as the ratio of power radiated in desired direction to the power radiated in opposite direction (Fig. 3.7(b)). That is

FBR =

Power radiated in forward direction Power radiated in backward direction

FBR is mostly observed in array/Yagi antenna. The FBR changes with frequency of operation and its value decreases with increasing the spacing between elements. The values of FBR also depend upon tuning condition as well as electrical length of parasitic elements. The gain of a directional antenna is inversely coupled to the front-to-back ratio—as one goes up, the other goes down. Therefore higher value of FBR is obtained at the cost of gain of antenna; because diverting more radiation in opposite direction reduces the gain of antenna.

Backward radiation

FIG. 3.7(b)

Forward radiation

Forward and backward radiations.

69


Radiation Resistance Radiation resistance describes the relation between the total radiated energy from a transmitting antenna and the current flowing in it. It is fictitious parameter and represented by Rr. Radiation resistance acts as a load for the transmitter or for the radio-frequency Tx line connecting the transmitter and antenna. Antenna is a radiating device which emits radiation in the form of EM waves in the free-space, provided it is excited with proper input power. Since there is physical contact as well as impedance differences between feed and antenna, there is power dissipation. As a result, not all power supplied to the antenna is transformed into EM radiation, but some of it is lost in the heating of antenna wire. That is the total energy supplied to antenna is used in radiation as well as heating the antenna. Hence associated with each one, there are resistances—radiation resistance (Rr) and ohmic resistance (Rl). Therefore the total energy given to the antenna is sum of radiation energy and power dissipation, i.e., P = Pr + Pl = I2Rr + I2Rl = I2[(R)r + Rl] = I2R where R = Rr + Rl. The radiation resistance of a radiator and depends upon (i) (ii) (iii) (iv)

Geometry of the antenna and hence point of measurement of resistance Orientation of antenna as well as number of objects present around Length and width of antenna conductor A luminous discharge round the surface of antenna due to ionosphere of air

Presence of ground significantly affects the radiation resistance, because the EM waves radiated from it are reflected from ground which induced current in the antenna while flowing through it. The magnitude of induced current as well as its phase depends upon the height of antenna from ground. If the height is such that the induced current is in phase with antenna current, then the total current is larger and this results in a series of variation in the free space value of radiation resistance. In general since reflected waves are weaker than incident waves, fluctuation in radiation resistance decreases as height increases. Radiation resistance of wire antenna and rod-tubing antenna are found between 55 and 65 W; however for half-wave dipole its value is 72 W.

Directivity Directivity of an antenna in a given direction is the ratio of the radiation intensity in that direction and the radiation intensity averaged over all directions. The radiation intensity averaged over all directions is equal to the total power radiated by the antenna divided by 4p. If direction is not specified, then the direction of maximum radiation is implied. The directivity Dmax is defined as the value of the directive gain in the direction of its maximum values. The directive gain D(q, f) over power density radiated by an isotropic radiator fed by the same account of power is given by D(R , G ) =

P (R , G ) PT /4Q

(3.20)

70


Dmax = D0 =

and

where P(R , G ) =

1

Pmax (R , G ) PT /4Q

= 4Q

Pmax (R , G ) PT

(3.21)

G G Re [ E × H ∗]

2 PT = Total radiated power.

Directivity is a dimensionless quantity. The maximum directivity is always ³ 1. Directivity of an isotropic source This is given by

P(q, f) = P0 = constant PT = 4 p P 0 D(R , G ) = 4Q

P (R , G ) PT

=

4Q × P0 4Q × P0

=1

D0 = 1 Thus by definition, the directivity of an isotropic radiator is one, and that of other antenna will always be greater than one. Thus, directivity serves as a figure of merit relating the directional properties of an antenna w.r.t. those of an isotropic radiator. Partial directivity The partial directivity of an antenna is specified for a given polarization of the wave. It is defined as that part of the radiation intensity, which corresponds to given polarization, divided by the total radiation intensity averaged over all directions. The total directivity is the vector sum of partial directivities for any two orthogonal polarizations. D0 = Dq + D f where

DR = 4Q DG = 4Q

(3.22)

PR (PT )R + (PT )G PG (PT )R + (PT )G

(3.23)

in which Pq and Pfÿ are the radiation intensities in desired directions containing q and f field components respectively. (PT)q and (PT)f are total radiated power in all directions of q and f field components respectively.


71

Directivity in terms of relative radiation intensity 3 (q, f) As per definition, directivity can also be expressed by D0 =

[P (R , G )]max

(3.24)

[ P(R , G )]av

where power density average over a sphere is given by

[P(R , G )]av =

1 4Q

2Q

Q

0

0

∫ ∫

[P(R , G )] sin R dR dG

(3.25)

Therefore from Eq. (3.24), the directivity

D0 =

P(R , G )max 1 4Q

Q

2Q

0

0

∫ ∫

where P (R , G ) =

P((R , G )sin R dR dG )

P ((R , G ) P (R , G ) max

The integration

1 Q

2Q

0

0

∫ ∫

(3.26)

P ((R , G )sin R dR dG )

.

Q

2Q

0

0

∫ ∫

= 4Q

P (R , G ) sin R dR dG = : A is termed beam solid angle of an

antenna. The beam solid angle of an antenna is the solid angle through which all the power of radiation would flow provided its radiation intensity is constant and equal to the maximum radiation intensity Pmax for all angles with WA. Hence, the relation between the maximum directivity and the solid beam angle is found to be D = 4p/WA. The complexity of the determination of the directivity D0 depends upon the power pattern P (R , G ) to be integrated over a spherical surface. But in most practical antenna cases it is not available in closed analytical form. Hence, in practice, simpler formulas, based on the two orthogonal plane half power beam widths (HPBW) of the pattern, are often used for fast and approximate calculation of D0. Kraus formula For antennas with narrow major lobes and with negligible minor lobes, the beam solid angle WA is approximately equal to the product of the HPBWs in two orthogonal E- and H-planes, i.e., WA = qE ´ qH, where qE and qH are in radians. Hence 4Q

when qE and qH are in radians

(3.27a)

4Q × 56.94 4.1 × 10 4 when qE and qH are in degrees Dk = = R E′ R H′ R E′ R H′

(3.27b)

Dk =

RE × RH

72


Directivity is a parameter of antenna that is most of time, calculated in term measured power density. There is another approach to calculate the directivity of antenna with the help of E-plane and H-plane patterns. In experimental work, we measure power pattern in two principal planes, E-plane and H-plane. These patterns correspond to [Eq (q, 0)]2 in the E-plane pattern and [Ef (0, p/2)]2 in the H-plane pattern. In terms of these two patterns, the directivity of antenna can be expressed as 1

=

D

where D1 =

ER 1 2

∫

Q 0

1⎛ 1 1 ⎞ + ⎜ ⎟ D2 ⎠ 2 ⎝ D1

2 max

ER (R , 0)

2

which corresponds to directivity of antenna with a sin R dR

rotationally rationally symmetrical pattern ER

D2 =

EG

(3.28)

2 max

.

2 max

which corresponds to directivity of antenna with a 1 2 EG (0, Q /2) sin R dR 2 0 2 rotationally symmetrical pattern EG (0, Q /2) .

∫

Q

The arithmetic mean directivity formula (3.28) is quite accurate for the arrays made of half-wave-dipoles, however for uniform arrays of short dipoles operated in broadside and end fire configurations, the formula is found to be exact. However, for narrow beam antennas, D1 and D2 can be estimated in terms of HPBW of E-plane and H-plane patterns as follows:

D1 ≈ 16

ln 2

R E2

and

D2 ≈ 16

ln 2

R H2

Hence, resultant directivity of antenna will be

DT =

32 ln 2

R E2 + R H2

(3.29)

Equation (3.29) is known as Tai and Pereira formula for antenna directivity [6]. For qE = qH, Eq. (3.29) gives D = 4.62 whereas Eq. (3.27) gives D = 5.09. All the equations involved in getting DT and DK are obtained by considering the asymptotic expression for the directivity of an antenna with a rotationally symmetrical power pattern of the form U(q) = cosmq for (p/2 ³ q ³ 0) and U(q) = 0, for (q > p > p/2) with very large value of m. However the U max M2 and the maximum effective area is Aem = . maximum directivity is Dmax = U av 4Q


73

Antenna Gain and Efficiency Similar to directivity, antenna gain is also dimensionless quantity. The gain of an antenna is the directivity multiplied by the aperture or illumination efficiency of the antenna to radiate the energy presented at its terminal, i.e. G = hDmax

(3.30)

where h is the illumination efficiency. The gain of an antenna also defined as the ratio of the radiation intensity P in a given direction and radiation intensity that would be obtained if the power fed to the antenna is radiated isotropically (in all directions). That is G(R , G ) = 4Q

P (R , G ) Pin

(3.31)

Gain is calculated via the input power Pin, which is measurable quantity unlike directivity, which is calculated via the radiated power Pr. Since power radiated by the antenna is always less than the power fed to the antenna system, i.e., Pr £ Pin, D ³ G, unless the antenna has integrated active devices. When antenna has no losses, i.e. Pin » Pr, G(q, f) = D(q, f). However, as per IEEE standards, the gain does not include losses arising due to impedance and polarization mismatching. Therefore gain takes into account only the direct and conduction losses of the antenna system itself. Partial gains with respect to given-field polarizations are determined in the similar way as the partial directivities of antenna. In general, the narrower the beam width, the higher the gain of antenna. In experimentation, the gain of antenna, also called forward gain, is defined as the ratio of the signal transmitted by an antenna in the maximum direction to that of a standard or reference antenna. Basically there are two types of reference antenna used. One is dipole antenna, which is easily available and considered as basis antenna for many other antennas. In this case gain often expressed in dBd, i.e. gain expressed in decibels over a dipole. The second one is isotropic radiator. In this case gain specified in dBi, i.e. gain in decibels over an isotropic source. It is possible to relate two gains as dipole has a gain of 2.1 dB over an isotropic source, i.e. 2.1 dBi. In other words, the gain over an isotropic source will be 2.1 dB higher than those relative to a dipole. So, when choosing and looking for gain specifications, be sure to check whether the gain is relative to a dipole or an isotropic source. The directive gain (Gg) is a measure of the extent to which the total radiated power is concentrated in one direction. It defined as the ratio of the power density in particular direction, at a given distance by antenna under test (AUT), to the power density of an isotropic antenna, provided both are radiating the same total power. That is, it is a quantity that varies with the change in directions and solely depends on the distribution of radiated power in space. It is free from input power as well as antenna losses. However, power gain (Gp) is defined as the ratio of the radiated power density by actual antenna, to that of an isotropic antenna at same distance provided both are given same input power. That is, both define gain of an antenna; directive gain after radiation and power gain before radiation of waves. The basic difference is that the directive gain considers radiated power whereas

74


power gain involves input power to the antenna, i.e. power gain also takes into account the losses occurs in the antenna. They may be related as Gp = h G g where h is termed efficiency of antenna and its value lies between 1 and 0. In case there are no losses in the antenna, Gp = Gg. The efficiency of antenna represents the fraction of total energy supplied to it, which is converted into free space EM waves. In general, it is defined as the power radiated from antenna to total power supplied to the antenna, i.e. Antenna efficiency (h ) =

Radiated power Total input power

=

Prad Pin

=

Prad Prad + Ploss

(3.32a)

where Prad = actual power radiated Pin = power coupled to the antenna Ploss = power lost in the antenna (conductor dielectric losses) As power is proportional to resistance (i.e. P0 = I2R), antenna efficiency can also be expressed as

I=

Rr Rr + Rl

(3.32b)

where Rr and Rl are the radiation and loss resistances respectively. The total efficiency of the antenna (hT) represents the total loss of energy at the input terminals of the antenna and within the antenna structure. It includes all mismatch losses as well as the dielectric and conduction losses; hence it can be expressed as follows:

h T = h p h r h c h d = h ph r h

(3.33)

In which, h = h c hd and called radiation antenna efficiency, which is used to relate gain and directivity. The conduction and dielectric efficiencies hc and hd are measured experimentally. Here the subscripts p, r, c and d represent reflection, polarization, conduction and dielectric respectively. The reflection efficiency can be calculated in terms of reflection coefficient at the input terminal as follows:

h r = 1 – | r| 2

(3.34)

If polarization losses are negligible or zero, the total efficiency (h t) relates to the radiation efficiency (h) by

hT = h(1 – |r|2)

(3.35)

where rÿ is the volatage reflection coefficient at the input terminals of the antenna.

Radiation Pattern Radiation pattern generally describes the normalized field/power values with respect to the maximum values. The radiation pattern of antenna is the representation (or trace) of the


75

radiation properties of the antenna as a function of space coordinates (angles and distances). The radiation pattern is measured at electrically large distance where spatial (angular) distribution of the radiated power does not depend on the distance. At the large distances, the power density drops off as (r–2) in any direction [7]. The variation of power density with angular position can be plotted by the radiation pattern. Generally, the field intensities (E and H) or received/transmitted powers are measured at a constant distance from the antenna and plotted, and they are referred as field pattern and power pattern. The power pattern and field pattern are the same when they are computed and plotted in dB. However, both the patterns are related to each other field pattern is square root of power pattern, i.e. E (R , G )

I

≡ I H (R , G ) ∝

P(R , G )

(3.36)

In the view of properties of EM waves, the antenna has E plane and H plane patterns each with co- and cross-polarizations. The E plane pattern refers to the plane containing the electric field vector (Eq) and the direction of maximum radiation with Ef as cross-polarization components. Similarly, the H plane pattern contains the magnetic field vector (Hq) and the direction of maximum radiation with Hf as cross-polarization components. The radiation pattern of antenna is a 3-D plot and hence the co-ordinate system used for the same is the spherical co-ordinate (r, q, f), with antenna to be located at the origin [see Fig. 3.8(a)].

FIG. 3.8(a)

3D Antenna pattern co-ordinate system.

HPBW, FNBW, Side Lobe Level and Antenna Resolution In direction of the maximum beam, the angle between the two directions in which the radiation intensity is half of that of maximum value is termed the half power BW (HPBW). In other words, the half power BW is the range in degrees in which the radiation falls to

76


one half of its maximum values (or 3 dB down). Usually it referred to as 3 dB beam width and it increases as side lobes decrease and vice versa. It is a very important figure of merit, used to describe the resolution capabilities of an antenna. The half-power beam width is approximately equal to FNBW. Two- and three-dimensional antenna radiation patterns along with HPBW, FNBW and all lobe levels are shown in Figs. 3.8(b) and (c) respectively. In particular, the radiation pattern of a half-wave dipole antenna is shown in Fig. 3.8(d). The side lobes are power radiation peaks in addition to the main lobe. They are given as the number of decibels below the main lobe peak. FNBW refers to first null beam width. In practice the test antenna is used as receiver and Tx antenna is placed in the far-field region of the test antenna and vice versa; this is because radiated fields are plane waves in the vicinity of the Tx antenna.

FIG. 3.8(b)

2D Antenna pattern characteristics. z

First null beam width (FNBW)

Major lobe

Half power beam width (HPBW)

Minor lobes

Side lobe y

Back lobe Minor lobes x

FIG. 3.8(c)

3D Radiation pattern characteristics.


77

0 dB

10 20 30

Antenna

FIG. 3.8(d)

Typical radiation pattern of a half-wave dipole antenna.

A convenient equation for predicting the HPBW of an antenna is also given by [5]. HPBW = K

M0 DA

(3.37)

where DA is the aperture dimension in the plane of pattern. K is constant; one can use K = 70°. So, if the length of antenna is 5 cm, the beam width at 20 GHz, in the plane of length, will be 70°. Side lobe level (SLL) is defined as the ratio of the pattern value of a side lobe peak to the pattern value of the main lobe. The largest side lobe level for the whole pattern is the maximum (relative) side lobe level. Mathematically, it is given by (SLL)dB = 20 log

F (SLL) F (max)

(3.38)

where |F(max)| is the maximum value of the pattern magnitude and |F(SLL)| is the pattern value of the maximum of the highest side lobe magnitude. For a normalized pattern, F(max) is 1. Resolution of an antenna is equal to half of the first null beam width, i.e. FNBW/2. That is, if any antenna has FNBW = 3°, so it has resolution of 1.5°. We also know that half power beam width is approximately equal to half of FNBW, i.e. HPBW = FNBW/2 and product of HPBW in two orthogonal planes of the antenna pattern is referred as solid beam angle or antenna beam area, i.e. ⎛ FNBW ⎞ ⎛ FNBW ⎞ : A = (R E , R H ) HPBW = ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠E ⎝ 2 ⎠H

which reveals that the total number of transmitters (N, say) of radiation distributed uniformly over sky, which an antenna could resolve, is approximately proportional to 4p/WA. Hence D = N = 4p/WA. That is ideally the number of transmitter/point sources that an antenna can resolve is numerically equal to the directivity of the antenna.

78


Radiation Intensity, Beam Efficiency and Solid Angle Radiation intensity in a given direction is defined as the power per unit solid angle radiated by the antenna in that direction, i.e.

dPr

U=

d:

(3.39)

W/Sr

However, the radiation power density P is the Poynting vector magnitude of the far-field, i.e.

dPr

P=

W/m 2

ds

(3.40)

There is radiation between radiation intensity and radiation power density U = r2 ´ P

(3.41)

Radiation intensity is useful in calculating gain and directivity of an antenna. It is measured in W/Sr. The far-field magnitude depends on r as r–1 and hence the power density of the farfield depends on r (distance from the source) as r–2. Thus the radiation intensity U is independent from the distance (r) and depends only on the direction (q, f), i.e., angular positions. In the far-field region, the radial field component vanishes, and the remaining transverse component of the electric and magnetic fields are in phase and have magnitude related by |E| = h |H|. That is why the far-field Poynting vector has only a radial component and it is a real number corresponding to the radiation density Prad = P =

2

1 E 2

I

=

1 2

IH

2

(3.42)

Therefore, the radiation intensity can be expressed in terms of the fields as U (R , G ) =

r2 2I

E

2

=

1 2

r 2I H

2

If Eq and Ef are the field components of E along q and f directions E Then, Eq. (3.43) reduces to U (R , G ) =

or

r2 2I

U (R , G ) =

1 2I

(3.43) 2

= ER2 (r, R , G ) + EG2 (r, R , G ) .

ER2 (r, R , G ) + EG2 (r , R , G )

(3.44a)

ER2p (R , G ) + EG2p (R , G )

(3.44b)

where h is the intrinsic impedance of medium. Equations (3.44a and b) lead to a useful relation between the field pattern and power pattern.


79

Radiation intensity and radiation pattern of an isotropic radiator These are given by Pr

P(r , R , G ) =

U (R , G ) = r 2 P =

4Q r 2

Pr

= constant

4Q

U (R , G ) = 1

Þ

(3.45)

where U (R , G ) is normalized radiation intensity. From Eq. (3.45), it is clear that the normalized radiation pattern of an isotropic radiator is simply a sphere of unit radius. Beam efficiency is defined as the ratio of the power radiated in a cone of angle 2q1 (say) and the total radiated power. The angle 2q1 can be generally any angle, but usually this is the first null beam width, i.e.

Beam efficiency =

PR 1 PT

2Q

R1

0 2

0

0

0

∫ ∫ U (R , G ) sin R dR dG = Q Q ∫ ∫ U (R , G) sin R dR dG

(3.46)

Equation (3.46) defines beam efficiency, provided the antenna has its major lobe directed along the z-axis (q = 0). If q1 is the angle where the FNBW’s minima occurs in two orthogonal planes, then the beam efficiency will represent only the part of the total radiated power channeled through the main beam. Antennas of very high beam efficiency are needed in radar, radiometry and astronomy. One steradian is the solid angle with its vertex at the centre of a sphere of radius r, which is subtended by a spherical surface of area equal to that of a square with each side of length r. It is defined as Solid angle (W) =

S: Sr r2

(3.47)

where SW is angular area. As infinitesimal area ds on a surface of a sphere of radius r in a spherical co-ordinate is given by ds = r2 sin q dq df (in m2). Therefore, and

d: =

ds r2

= sin R dR dG (Sr)

ds = r2 dW

(3.48)

80


EFFECTIVE APERTURE AND EFFECTIVE HEIGHT Effective Aperture It is the ratio of the available power at the terminals of the antenna to the power flux density of a plane wave incident upon the antenna which is polarization matched to the antenna, i.e. Ae =

PA

(3.49)

Wi

where PA = power delivered to the load from the antenna Wi = power flux density of the incident wave In case a specific direction is chosen, the direction of maximum radiation intensity is implied. For aperture types of antenna, the effective area is smaller than the physical aperture area. Aperture antennas (i.e., parabolic reflector and dish) with constant amplitude and phase distribution across the aperture have the maximum effective area, which is practically equal to the geometrical area. The effective aperture of the wire antenna is much larger than the surface area of the wire itself. Using the Thevenin equivalent of a receiving antenna, the effective aperture can be expressed in terms of the antenna impedance as 2

Ae =

I A RL /2 Wi

2

=

VA RL 2Wi (Rr + Rl + RL )2 + ( X A + X L )2

(3.50)

Under the condition of conjugate matched impedance, i.e. XA = –XL and RA = Rl = RL. Equation (3.50) reduces to

Ae =

VA

2

8Wi (RA = Rl + Rr )

i.e.

Ae =

VA

2

if Rl = 0

8Wi Rr

(3.51)

The effective area, beam solid angle and gain of an antenna is linked by the expressions AeW A = l 2

and

G=

4Q Ae

M2

Hence, maximum effective area (Aem) can be given by Aem =

DM 2 4Q

=

M2 = 0.077 M 2 4Q

i.e., all the lossless antenna must have Ae ³ 0.0077l2.

(3.52)


81

The aperture efficiency of an antenna is defined as ratio of the effective aperture area and its physical area, i.e., I p =

Ae Ap

, where Ae effective area and Ap physical area of antenna.

We also know that even in case of lossless antenna only half of power delivered to the load (received), although there is conjugate matched impedance. Another half power dissipated back or scattered into space. Power delivered to the load is product of antenna effective area and incident power density. Therefore to account for another half power density, scattered and dissipated power, we must define scattered loss and capture areas, such as effective area. Scattered equivalent area It is an area which produces scattered/re-radiated power when multiplied with the incident power density, i.e.

Ps = As × Wi ⇒ As =

2

Ps

=

Wi

I A Rr 2Wi

m2

(3.53)

In case of conjugate matched impedance 2

As =

2

VA Rr 8Wi (Rr + Rl )

2

VA Rr

=

8Wi RA2

m2

(3.54)

where Rr + Rl = RA Hence, scattered power 2

Ps = AsWi =

VA Rr 8 RA2

W

(3.55)

Loss area The loss area is defined as the equivalent area which leads to power dissipated from the antenna under multification with incident power density. Therefore, similar to Eq. (3.54), we get 2

Al =

VA Rl 8Wi RA2

m2

(3.56)

W

(3.57)

2

and power loss

Pl =

VA Rl 8 RA2

Capture area The area which when multiply with the incident wave power density produces the total power intercepted by the antenna.

82

Antenna and Wave Propagation 2

PT

Ac =

=

Wi

I A ( Rr + Rl + RL ) 2Wi

(3.58)

In case of conjugate matching 2

Ac =

=

2

VA (Rr + Rl + RL )

=

8 Wi (Rr + Rl )2 VA

VA (RA + RL ) 8 Wi (RA )2

2

when RA = RL

4Wi RA2

(3.59)

On multiplication by the incident power density, Eq. (3.59) leads to total power captured by the antenna. In general, capture area is the sum of effective area, loss area and scattered area. A c = Ae + Al + As

(3.60)

Under conjugate matching for the antennas

1

Ae = Al + As =

2

Ac

(3.61a)

If conjugate matching is received for a lossless antenna, then

Ae = Al =

1 2

Ac

(3.61b)

The effective aperture has more general application to all types of antenna. Effective height The effective height of an antenna is another parameter, useful for transmitting tower type of antennas. It may be defined as the ratio of induced voltage to the incident field across the antenna,

he =

i.e.

V E

(3.62)

m

In other words, the effective height of a transmitting antenna is the physical height (length) multiplied by the ratio of the average current to the peak current, i.e. [8] he =

where he h Iav I0

= = = =

effective height (m) physical height (m) average current (A) peak current (A)

1 I

∫

h 0

I (z ) dz =

I av I0

h m

(3.63)


83

Effective height can also be expressed as a vector quantity

V = he E = he E cos R

(3.64)

where q = angle between polarization angles of antenna and wave (deg) he = effective height (m) E = field intensity of incident wave (V/m) The effective height of an antenna is also linked with radiation resistance (Rr), efficiency (h ) and effective aperture (Ae). We know that the power delivered to the load is P=

which is also equal to

1 V2 4 Rr

P = SAe =

he2 E

=

(3.65)

4 Rr

E2

I0

Ae

(3.66)

m

(3.67)

Comparing Eqs. (3.65) and (3.66) yields

Rr Ae

he = 2

I

There are some additional parameters related to scattered loss and captured areas, namely effectiveness ratio, scattered ratio and absorption ratio. They are defined as follows: Effectiveness ratio (a) Effectiveness ratio (a) =

A (Ae ) max

=

Effective area Maximum effective area

Values of a lies between 0 and 1, a = 1 implies a perfectly matched antenna of 100% efficiency. Scattered ratio (b ) The ratio of scattered area to effective area is known as scattered ratio (b); b = As/Ae and its value lies between 0 and ¥. Absorption ratio (g) The ratio of maximum effective aperture to the physical aperture, H = values lie between 0 and ¥; (Ae ) max =

VA

2

8Wi Rr

.

(Ae ) max Ap

, and its

84


ANTENNA NOISE TEMPERATURE, NOISE FIGURE AND SNR Antenna Noise Temperature In addition to collecting signals from the space, antenna also collects external noise from source such as ground, sky and other obstacles present. Every object with a physical temperature above absolute zero (0 K = –272°C) radiates the energy and amount of radiated energy is represented by an equivalent temperature (TB), which is also known as brightness temperature and expressed as (see [8]). TB(q, f) = xmTp = (1 – |r|2)Tp

(3.68)

where xm = emissivity and 0 £ xm £ 1 Tp = physical temperature (0 K) ÿ ÿ r = reflection coefficient of surface wave for the polarization of wave ÿ

Antenna noise is generally not a major component field of total system noise unless very low noise converters (LNCs) are being used. But, if 1 dB (say) noise figure is being used, we must attention to the antenna noise. For a lossless antenna, antenna temperature is strong related to the temperature of distant region of space coupled to the antenna through its radiation resistance. In this regard, the receiving antenna may be considered a remote sensing temperature measuring device. Actually the equivalent temperature caused by the surrounding sources is intercepted by the antenna and appears at the antenna terminals as an antenna temperature (TA). As far as temperature of antenna, receiver and surrounding (in a measurement set-up) is concerned there are three cases. (i) If there is no loss If in case there is neither mismatch nor guiding structure (Tx line) losses between antenna and the receiver, then the total noise power transferred to the receiver is given by Pr = kTADf where Pr = k = TA = Df =

(3.69)

total antenna noise power (W) Boltzmann’s constant = 1.38 ´ 10–23 J/K antenna temperature (0 K) bandwidth (Hz)

Therefore the noise power transfer per unit available bandwidth will be P¢r = kTA

W/Hz

(3.70)

In particular, if an antenna has effective area (Ae) and its beam is directed as a source of radiation, which produces a flux density (S) at the antenna, then total power received from the source will be Pr = S Ae Df

(3.71)

Equating (3.69) and (3.71), the power density per unit bandwidth/flux density from the source at the antenna is


S=

kTA

TA =

or

Ae

SAe

=

k

85

2

E Ae

1 2I0

k

(3.72)

where E = electric field intensity (V/m) h0 = 377 W (intrinsic impedance of free space) ÿ

In view of received power from any source at antenna (50% power only received at antenna), then the total actual flux density/Poynting vector is twice of present flux density (S). Hence

TA =

2 SAe k

2

=

1 Et Ae

I0

k

=

I0 Ht 2 Ae k

(3.73)

Once the antenna temperature is known, the source temperature (Ts) can also be determined in terms of TA and solid beam angle as follows: Ts =

: S TA :A

when (Ws < WA)

(3.74)

where WS = angular size of source (deg) WA = antenna solid beam angle (deg) (ii) If there is transmission loss If Tx line used between antenna and receiver is lossy, then the antenna temperature seen at the receiver must have significant modification to include the lines losses. As a result, we need to modify the antenna noise power at the receiver input. If l is length of Tx line, T0 is constant physical temperature and a is uniform attenuation, then the effective temperature at the receiver terminals is given by Te = TAe–2a l + T0(1 – e–2a l)

(3.75)

and total effective power received (Pr)e = kTeDf

(3.76)

(iii) If receiver has own temperature Tr If the receiver has its own centre temperature Tr (due to thermal noise), the system noise power at the receiver terminals is given by Ps = k(Te + Tr)Df = kTs Df where Ts Te Tr Ps

= = = =

system noise temperature = (Tr + Te) effective antenna noise temperature receiver noise temperature system noise power (W)

(3.77)

86


Effective Noise Temperature and Noise Figure The noise introduced by a network may also be expressed as effective noise temperature (Te). Effective noise temperature defined as that fictional temperature at the input of the network, which would account for additional noise introduced by the network itself at the output. Effective noise temperature (Te) is related to noise figure (F) as follows (see [9]): F=

Te = (F – 1)T0

or

Te + T0 T0

T ⎞ ⎛ F = ⎜1 + e ⎟ T0 ⎠ ⎝

and

(3.78)

The noise figure in dB is FdB = 10 log F, where T0 = 290 K = (273 + 17) °F. A complete set-up of receiver arrangement for system noise power measurement, along with TA, TB, Te, and Tr, is shown in Fig. 3.9. Te

TA

TB

Ts = Te + Tr Tx line Length, l

Receiver

Antenna

FIG. 3.9

Set-up of receiver arrangement.

The devices with no gain (i.e., attenuators) have a noise figure equal to their attenuation, L (absolute value, not in dB), provided their physical temperature equals T0. Thus, for an attenuator at a physical temperature Tphy, the effective noise temperature is Te = (L – 1)Tphy. Thus giving a noise figure

F =1+

(L − 1)Tphy T0

Signal to Noise Ratio (SNR) The SNR is defined as the ratio of power received to system noise power, i.e. SNR =

Pr PN


87

in which the system noise power PN (in watts) is related with the system noise temperature Ts as follows: PN = kTsDf i.e.

SNR =

Pr

(3.79)

kTs 'f

In general, both signal and noise powers must be measured at the same/equivalent point in a system and within the same bandwidth. However, if the signal and the noise are measured across the same impedances, then the SNR can also be calculated in terms of amplitudes.

SNR =

Þ

⎛A ⎞ =⎜ S ⎟ PN ⎝ AN ⎠ Pr

2

⎛P ⎞ ⎛A ⎞ SNR(dB) = 10 log ⎜ r ⎟ = 20 log ⎜ S ⎟ ⎝ PN ⎠ ⎝ AN ⎠

where AS and AN are the rms values of amplitudes. From Friis equation, the received power by an antenna is ⎛ M ⎞ Pr = Pt Gt Gr ⎜ ⎟ ⎝ 4Q R ⎠

2

Therefore PG G ⎛ M ⎞ SNR = t t r ⎜ ⎟ kTs 'f ⎝ 4 Q R ⎠

2

We also know that maximum directive gain and effective area of an antenna are related as Gt =

4Q Aet

M

2

and Gr =

4 Q Aer

M2

Hence the value of SNR reduces to

SNR =

Pt Aet Aer (M R)2 kTs 'f

(3.80)

ANTENNA COUPLING Like other parameters, it also important to discuss the interaction (coupling) between two antennas paced at finite distance in space for the purposes of Tx and Rx of EM waves. In order to describe the coupling between two antennas let us consider Fig. 3.10(a). Using the relations of voltages and currents at the terminals of each antenna as a pair of coupled equations

88


FIG. 3.10

Equivalent circuit model of coupling between two antennas. V1 = Z11I1 + Z12I2

(3.81a)

V2 = Z21I1 + Z22I2

(3.81b)

If the medium of transmission between antennas is merely free space (which is linear and isotropic), then Z12 = Z21, representing the general form of the reciprocity theorem, which states that: If I1 is a current applied to the terminals of Antenna (1) and the terminal of Antenna (2) is opened (i.e. I2 = 0), then a voltage V2 will appear at the terminals of Antenna (2) or vice versa at terminals (2) and (1), i.e. I1 = 0 and V1 appears at Antenna (1).


89

Then V2 I1

V1

=

I2

I 2 =0

(3.82) I1 = 0

i.e. the ratio of each driving currents to its resulting open circuit voltages are the same. Again from Eq. (3.81a), we have Z12 =

V1 I2

(3.83a) I1 = 0

and from Eq. (3.81b) Z 21 =

which implies that

V2 I1

(3.83b) I2 = 0

Z12 = Z21 = Zm (say)

Thus, Eq. (3.81a and 3.81b) becomes V1 = Z11I1 + ZmI2

(3.84a)

V2 = ZmI1 + Z22I2

(3.84b)

These equations may be represented as a lumped two-port equivalent circuit [Fig. 3.10(b)]. Equation (3.84) and [Fig. 3.10(b)] are an exact representation of coupling between two antennas. If Antenna 1 is driven by source with phase source voltage Vs of impedance Zs and Antenna 2 is terminated in a load impedance ZL, then input impedance (Z1) to Antenna 1, [see Fig. 3.10(c)], becomes Z1 = (Z11 − Z m ) + ⎡⎣ Z m {(Z 22 − Z m ) + Z L }⎤⎦

(3.85a)

and equivalent source impedance (Z2) is equal to Z 2 = (Z 22 − Z m ) + ⎡⎣ Z m {(Z11 − Z m ) + Z s }⎤⎦

(3.85b)

The open-circuit voltage source (Voc21) is given by Voc21 = I1Zm The power delivered by the source to Antenna 1 becomes

PT =

1 2

2

I1 R1

(3.86a)

and similarly, power delivered to the matched load will be

PR =

I1 Z m 8R2

2

(3.86b)

90


provided the load is matched to the receiving antenna, i.e., ZL = ZB (impedance across BB¢), where R1 and R2 are resistive parts of impedances Z1 and Z2. Therefore, from Eq. (3.86), we get 2

Zm ⎛ PR ⎞ ⎜ ⎟= ⎝ PT ⎠ 4R1 R2

(3.87)

where Zm is known as mutual impedance between the antennas. Suppose we modify the arrangement of elements shown in Fig. 3.10(c), and place the voltage source Vs in the terminated circuit of Antenna 2, retaining impedances Zs and ZL at their original positions [as shown in Fig. 3.10(d)]. The open-circuited voltage source Voc12 is given by (3.88) Voc12 = I2 . Zm In this situation, it is found that Z ¢1 = Zs and Z ¢2 = ZL, where Z ¢1 and Z ¢2 are impedances at Antenna 1 and Antenna 2 in case of re-arrangement of elements [see Fig. 3.10(d)]. So, we can conclude that the input impedance of an antenna when it is used for transmission is equal to the equivalent source impedance when it is used for reception, provided the terminal impedances for each antenna remain unchanged. If the distance between antennas is large, Zm will be small compared to Z11 and Z22, and Z1 = Z 1¢ = Z11 and Z2 = Z ¢2 = Z22 will be independent from the source and load impedances, Zs and ZL. Equation (3.87) indicates that the amount of power coupled inversely depends on R1 and R2 and directly on Zm. The coupling impedance (Zm) is related to angular position (q, f) in the same manner, whether antenna is used as transmitter or receiver. Hence, by interchanging R1 and R2, when the roles of the antennas are reversed, we can measure the Tx and Rx patterns of the antennas and obtain the desire results.

ANTENNA POLARIZATION AND PARAMETERS Antenna Polarization In general, polarization of EM waves is a time-harmonic field characteristic. In the case of antennas, we are concerned with polarization of fields in a plane normal to the direction of propagation. This is because in the far-field zone, the longitudinal field components are negligible, i.e., the field is a quasi-TEM field. The type of polarization, where the field vector at a given point traces an ellipse (termed polarization ellipse) as a function of time, is known as elliptical polarization. This is the most general type of polarization obtained for any phase difference dL between the fields and the field ratio (Ey/Ex). Linear and circular polarizations are the special cases of elliptical polarization. The circular polarization may be classified as RHCP and LHCP. Like the circular polarization, elliptical polarization can be RHEP and LHEP, depending on the relation between the direction of propagation and the direction of rotation. With reference to Fig. 3.11, the parameters of the polarization ellipse are given by


FIG. 3.11

Polarization ellipse at tilt angle along with field.

)

1/2

)

1/2

(

⎤ E x4 + E y4 + 2 E x2 E y2 cos(2E L ) ⎥ ⎦

(

⎤ E x4 + E y4 + 2 E x2 E y2 cos(2E L ) ⎥ ⎦

⎡1 Major axis OA = ⎢ E x2 + E y2 + ⎣2 ⎡1 2 2 Minor axis OB = ⎢ E x + E y − 2 ⎣

and tilt angle

91

U =

⎡ 2E x E y ⎤ arc tan ⎢ 2 cos E L ⎥ 2 2 ⎣⎢ E x + E y ⎦⎥

1

(3.89a)

(3.89b)

(3.90)

Field Polarization in Terms of Two Circularly Polarized Components The representation of a complex vector in terms of CP components is comparatively less easy, but more useful in calculation of the polarization parameters. Here the total field is represented as the superposition of two CP waves: left and right handed CP. E = EL(x – iy) + ER(x + iy) or

= x(ER + EL) + iy(ER – EL)

(3.91a)

Equation (3.91a) represents the relation between the linear components and circular components of the field polarization with x and y as the unit vectors. If dL is the relative phase difference between ER and EL, then dL = fL – fR, where fL and fR are phase angles of LHCP and RHCP waves. Then (Eq. 3.91a) can be re-written as E = ER(x + iy) + ELejdL(x – iy)

(3.91b)

92


Polarization Vector and Polarization Ratio The polarization vector Sˆ L is a normalized phasor of the electric field vector and it is a ⊗ complex-number vector of unit magnitude, i.e., Sˆ L Sˆ L = 1. Mathematically it is given by

E

Sˆ L =

ET

=x

Ex ET

+y

Ey ET

e jE L

with ET = E x2 + E y2

(3.92)

In particular cases the polarization vector [i.e., Eq. (3.92)] reduces to

Sˆ LL = x

Sˆ LC = as

Ex ET

+y

Ex Ex 2

Ey ET

(linear polarization)

( x ± jy) =

1 2

(3.93a)

( x ± jy) (circular polarization)

(3.93b)

E x = E x and ET = 2 E x = 2 E y

The polarization ratio is the ratio of the phasors of the two orthogonal polarization components. It is a complex number and mathematically it is given by rL = rL eE L =

Ey Ex

=

E y e jE L Ex

or

rL =

EV

(3.94)

EH

However, in the case of circular components, the polarization ratio is defined as rc = rcejdc = ER/EL. It is also known as circular polarization ratio, where t = dc/2 is termed the tilt angle. The circular polarization ratio rc is of particular interest, since the axial ratio (AR) of rc + 1 the polarization ellipse can be expressed as AR = . rc − 1

Polarization Loss Factor and Polarization Efficiency In general, the polarization of the receiving antenna differs from the polarization of the incident wave; the condition is termed polarization mismatch. A parameter representing the loss of EM power due to polarization mismatch is known as polarization loss factor (PLF) and given by (see [10]). PLF = |ri ra|2

(3.95)

where ri and ra are the polarization vectors of incident wave and receiving antenna respectively. The polarization efficiency has the same physical sense as the PLF. The value of polarization loss factor lies between 0 and 1 (i.e. 0 £ PLF £ 1). If PLF = 1, then antenna is polarizationmatched and there is no polarization power loss. If PLF = 0, then the antenna is incapable of receiving the signal.


93

RECIPROCITY THEOREM There are several forms of the reciprocity theorem; however, the most common one is: If an emf is applied to the terminals of Antenna 1 (Tx) and current is measured at the terminals of Antenna 2 (Rx), then an equal current (in both amplitude and phase) would be observed at the terminals of Antenna 1 (Tx) in case the same emf is applied to the terminals of antenna 2 (Rx). This theorem was originally proposed by Rayleigh in 1929. There are some considerations involved in the reciprocity theorem: (i) The values of emf should have the same frequency, and the media between Tx and Rx need to be linear, passive and isotropic. (ii) Generator and ammeter need to have either zero or equal impedance. Reciprocity theorem will be more powerful if impedances of generator and ammeter are large (Zs = Zm = ¥); in such cases, the generator becomes constant source and ammeter becomes an infinite-impedance voltmeter. (iii) There should be polarization matching between Tx and Rx; it is necessary because antennas (Tx/Rx) need to transmit and receive the same field components; hence there is total power radiation. In case an antenna is used as a probe to measure the radiation fields of AUT of different polarization, even though radiation patterns are the same. For example, if Tx antenna is circularly polarized and probe antenna is linearly polarized and used to measure q and f-components of radiation field one by one, then the sum of two components can be a circularly polarized pattern in either case of Tx/Rx mode. In order to prove reciprocity theorem, let us refer to Figs. 3.12 and 3.13, where an antenna is used as transmitter and receiver respectively. Since any 4-terminal network can be reduced to an equivalent T-section, in both the cases antennas (Tx and Rx) are represented by T-networks. I01 I2

V1

Z1

Z2 Zm

V1 Mesh 1

FIG. 3.12

Antenna 1 as Tx and its T-network.

From Figure 3.12 V1 I2

= Z12 = Z 21 =

V2 I1

I2 Mesh 2

94


Then from Mesh 1, the current I01 is I 01 =

V1 V1 (Z 2 + Z m ) = Z2 Zm Z1 Z m + Z1 Z 2 + Z 2 Z m Z1 + Z2 + Zm

(3.96)

Again from Mesh 2, (Z 2 + Z m )I 2 − Z m I 01 = 0

Therefore current across the ammeter will be ⎛ I Z ⎞ I 2 = ⎜ 01 m ⎟ ⎝ (Z 2 + Z m ) ⎠

Substituting the value of I01 from Eq. (3.96) yields Zm V1 (Z 2 + Z m ) V1 Z m ⎛ ⎞ I2 = ⎜ = ⎟ × Z1 Z m + Z1 Z 2 + Z 2 Z m Z1 Z m + Z1 Z 2 + Z 2 Z m ⎝ (Z 2 + Z m ) ⎠

(3.97)

In case Antenna 2 is used as transmitter (i.e., the location of source and ammeter is interchanged as in Fig. 3.13), then from Mesh 1, the current I02 is I 02 =

V2 V2 (Z1 + Z m ) = Z1 Z m Z1 Z m + Z1 Z 2 + Z 2 Z m Z2 + Z1 + Z m I02 I1

V2

Z2

Z1

Zm

V2 Mesh 1

FIG. 3.13

Antenna 2 as Tx and its T-network.

Again from Mesh 2, (Z1 + Z m )I1 − Z m I 01 = 0

Therefore the current across the ammeter ⎛ I Z ⎞ I1 = ⎜ 02 m ⎟ ⎝ (Z 2 + Z m ) ⎠

I1 Mesh 2


95

Substituting the value of I02 from Eq. (3.97) yields I1 =

V2 Z m Z1 Z m + Z1 Z 2 + Z 2 Z m

Therefore from Eqs. (3.97) and (3.98), it is clear that I1 = I2 Provided V1 = V2. Thus the radiation pattern of Tx and Rx are the same in the two cases. Using reciprocity theorem, it can also be proved that the effective lengths, antenna impedances and directivities of Tx and Rx are also equal. Power flow in antennas will also be the same, provided there is proper impedance matching in the set-up. However, there are certain limitations in the theorem; it is true only for separate antennas and not for two points on the antennas. It is valid for radiation pattern and not for current distribution in Tx and Rx. Reciprocity theorem fails when wave propagation between antennas is affected by the earth’s magnetic fields and also when communication takes place through the ionosphere.

SOLVED EXAMPLES Example 3.1 The distance between two horn antennas situated in free space is 200 m. The antennas are identical. The dimensions are 12 ´ 6 cm2 and have directive gains of 15 dB in the direction of transmission. Determine (a) whether the receiving antenna is in the far-field of the transmitter, (b) the received power, and (c) the electric field intensity at the receiving antenna. Assume that the transmitting power is 5 W at a frequency of 2 GHz. Solution:

The far-field distance will be df =

2d 2

M

=

2 × 12 2 10

= 28.8 cm

which is less than 100 m; hence both the antennas are in the far-field region of each other. The received power is (PR)dB = (PT)dB + (GT)dB + (GR)dB – 20 log fHz – 20 log Dm + 148 = 6.98 + 15 + 15 – 20 log (2 ´ 109) – 20 log (200) + 148 = 26.98 + 148 – 20 (log 2 + 9 + log 2 + 2) = 184.98 – 20 (9 + 0.477 + 0.2010 + 2) = –50.58 dB = 8.75 mW The magnitude of the electric field near the receiving antenna is obtained as

E =

60 PT GT d

=

60 × 5 × 31.62 200

= 0.487 V/m

96


Example 3.2 Calculate the maximum effective aperture of the Hertzian dipole for an incident linearly polarized uniform wave. Solution: Let a dipole of input impedance Zin = Rrad + jX be terminated along its length in an impedance ZL, such that Zin = Rrad – jX, i.e., the dipole is supposed to be lossless. For maximum response, let us suppose that incident wave is striking in broadside (q = 90°) to the antenna, i.e., the axis of dipole is parallel to the electric field of the incident wave E, then the average power density in the incident wave is

Sav = and open-circuit voltage

1 E 2

2

I0

|Voc| = |E| dl

Since there is perfect load match, maximum power transfer occurs and the power received is

PR =

Voc

2

8 Rrad

2

=

E (dl)2 8 Rrad 2

E (M 0 ) 2 ⎛ dl ⎞ ≈ 80 ⎜ ⎟ ; hence we obtained PR = , and 640 Q 2 ⎝M⎠ 2

Therefore, for a half-wave dipole Rrad

thus the maximum effective aperture is Aem =

PR

= 1.5

M02 , as h0 = 120p, which shows that 4Q

Sav the effective aperture does not necessarily depend on the physical aperture of antenna. We

know that for Hertzian dipoles, Aem depends on frequency through l0 as Aem maximum effective aperture of antenna reflects its directivity through

M02 = Dmax . Thus 4Q

M02 , and Dmax = 1.5 in 4Q

this case, when incident angle q is 90°. In general, if wave is incident at an angle q, then induced open-circuit voltage in the antenna will be |Voc| = |E| dl sin q, as (Et = E sin q), where Et is component of E along the dipole axis. Thus, the maximum effective aperture in this direction is ⎡ ⎤ M02 M02 R A = 1.5 sin = D (R , G ) ⎥ ⎢ em 4Q 4Q ⎣⎢ ⎦⎥

where D(q, f) = 1.5 sin q. Example 3.3 A 1 m long dipole antenna is fed by a 150 MHz source having resistance of 50 W and an open-circuited voltage of 100 V. If the radius of antenna wire is 4.06 ´ 10–4 m and the wire is made up of copper, determine the total time average power radiated and the power dissipated in the antenna.


FIG. 3.14(a)

Half-dipole fed by source VA.

FIG. 3.14(b)

97

Equivalent l/2 fed by source Vs.

Solution: We know that radiation resistance and reactance of dipole antenna are Rr = 73 W, Xin = 42.5 W and skin-depth is defined as

ÿÿÿd =

=

2

XNT

=

1

Q f NT 1

3.14 × 150 × 10 × 4 × 3.14 × 10 6

−7

× 5.8 × 10

7

= 5.4 × 10 −6 m

That is, the radius of wire is larger than the skin-depth at the given frequency. Hence, high frequency approximation for wire resistance can be used for calculation. Rw =

Rs 2Q rw

=

1 2 × 3.14 × 4.06 × 10

−6

× 5.8 × 10 7 × 5.4 × 10 −6

= 01.25 :/m

The ohmic power loss of the wires is given by

Pl =

Rw 2

∫

l/2 − l/2

2

I ( z) dz

Hence, for half-wave dipole ( Pl )dipole = Rw

l

Im

2

2

2

Therefore, the net ohmic resistance of the dipole is

Rl =

Rw l 2

= 0.63 :

So, the total input impedance to the antenna as seen by the source is Z A = Rl + Rr + jX in = 0.63 + 73 + j 42.5 = (73.63 + j 42.5) :

98


Then the antenna input current is IA =

Vs (Rs + Z A )

100 < 0 o

=

(50 + 73.63 + j 42.5)

= 0.765 < − 18.97o A

Therefore, the average radiated and dissipated powers are

Prad =

Ploss =

and efficiency

1

2

I A Rrad =

2 1

2

IA

2

I% =

1

Rloss =

Rrad Rrad + Rloss

2

0.765

2 1

=

2

0.765

2

× 73 = 21.36 W

× 0.63 = 184 mW

73 73 + 0.63

= 99.10%

This efficiency can be improved further by cancelling reactance Xin at the antenna terminal (i.e., by –Xin), a capacitor whose reactance is equal to –Xin at 150 MHz, i.e.,

−

1

XC

= − X in = − 42.5 :

⇒ C = 25 pF

Under this condition, antenna current becomes IA =

Prad =

Ploss =

VS Rl + Rr

1 2 1 2

IA

IA

2

2

=

100 < 0 o 123.63

Rrad =

Rloss =

1 2 1 2

= 0.809 < 0 o

0.809

2

2

0.809

× 73 = 23.89 W

× 0.63 = 206.8 mW

Example 3.4 Find the maximum effective aperture and directivity of the Hertizian dipole at the frequency of 20 MHz, if an uniform linearly polarized wave incident at an angle 45°, and also find Voc if E = 10 V/m and length is 5 cm. Solution:

(i) We know that for a dipole antenna D(q, f) = 1.5 sin q = 1.5 sin 45o = 1.5 ×

1 2

= 1.061


(ii) Hence Aem = where

99

M02 Dmax 4Q

M0 =

300 f MHz

=

300 30

Aem =

= 10 m 10 2 4 × 3.14

× 1.061 = 8.45 m

(iii) Voc = Edl sinR = 10 × 5 × 10 −2 × sin 45o = 10 × 5 × 10 −2 ×

1 2

= 0.25 V

Example 3.5 Determine the maximum effective aperture and directivity of a short dipole, supposed to be operated at f = 450 GHz. Solution: 300

M0 =

f MHz

=

300 450

= 67 cm

Maximum effective aperture can be found only when antenna losses are zero, i.e., RL = Rr + R l Þ RL = Rr is radiation resistance. We also know that =

VA

2

8 Wi Rr

where VA is induced voltage in the dipole and equals |E|dl. Also

Therefore

Wi =

Aem =

1 E

2

2 I0

⎛ dl ⎞ and Rr = 80 Q ⎜ ⎟ ⎝M⎠

(E dl)2 ⎛ E2 8 × ⎜ ⎜ 2 I0 ⎝

2 ⎞ 2 ⎛ dl ⎞ ⎟ × 80 Q ⎜ ⎟ ⎟ ⎝ M0 ⎠ ⎠

2

2

=

I0 M02 3 M02 = 8Q 320 Q 2

Example 3.6 In a 20 km microwave communication link, two identical antennas are operating at frequency 20 GHz with a power gain of 40 dB. The transmitted power is 1.5 W. Find the received power if there are no losses.

100


Solution:

As antennas are identical, Gp = Gr = G = 40 dB ⎛ 40 ⎞ G = antilog ⎜ ⎟ = 10 4 ⎝ 10 ⎠

Therefore,

M=

300 20 × 10

3

= 1.5 × 10 −2 m , Wt = 1.5 W and r = 3 ´ 104 m ⎛ M ⎞ = Gt Gr ⎜ ⎟ Wt ⎝ 4Q r ⎠

Wr

We know that

2

2 ⎛ ⎞ 1.5 × 10 −2 ⎛ M ⎞ 4 2 = G Wt ⎜ ⎟ ⎟ = (10 ) 1.5 ⎜⎜ 4 ⎟ Wt ⎝ 4Q r ⎠ ⎝ 4 × 3.14 × 3 × 10 ⎠

Wr

2

2

=

1.5 × 2.25 (37.68 × 10)

=

3.375 × 10 −8 1419.78

= 2.377 × 10 −7 = 23.77 μW

Example 3.7 An antenna with effective temperature 25 K is fed into a microwave amplifier that has an effective noise temperature of 30 K. Find the available noise power for a noise bandwidth of 5 MHz. Te = 25 K,

Solution: Hence,

Tr = 30 K and Df = 5 ´ 106 MHz

Ps = k(Te + Tr) Df = 1.38 ´ 10–23 (25 + 30) ´ 5 ´ 106 = 3.795 ´ 10–15 W

and therefore the power per unit bandwidth = 7.59 ´ 10–22 W/Hz. Example 3.8 The radial component of the radiated power density of an infinitesimal linear dipole of length l << l is given by

Pav =

ar Pm sin 2R r

2

W/m 2

where Pm = peak value of power density ÿ ÿ q = spherical co-ordinate ar = radial unit ÿÿ

Find directivity and then effective aperture of the antenna at l = 1.5 m. Solution:

Given that

Pr =

Pm sin 2R r2

, as Pav = ar Pr

Therefore the radiation density U = r2Pr = Pm sin2q.


101

Hence maximum radiation intensity Umax = Pm, as q = 90° The total power radiated can be obtained by UT =

2Q

Q

0

0

∫∫ U d : = ∫ ∫

= Pm [2Q ]

∫

Q 0

Pm sin 2 R sin R dR dG

sin 3 R dR = 4Q Pm

2 3

=

8Q 3

Pm

Hence the directivity, D=

Ae =

4QU max Prad

=

4 × Q × Pm × 3 8Q

= 1.5

M2 1.5 × 1.5 × 1.5 = 0.27 m 2 D= 4Q 4 × 3.14

Example 3.9 Calculate the effective height of a half-wave dipole antenna operating at l = 1.55 m if maximum effective area Aem = 0.15 l2. Solution:

We know that he = 2

Rr Aem

I

=2

73 × 0.15 M 2 377

= 0.3174 M = 0.3174 × 1.55 = 0.5 m

Example 3.10 The electric field intensity of wave radiated by an antenna is defined by Em = sin q sin f, where q is the angle measured from the z-axis and f is the angle measured from the x-axis. E has value only for 0 £ q £ p and 0 £ f £ p and zero elsewhere (i.e., the pattern is unidirectional with maxima in y-direction). Find the exact and approximate directivities and decibel difference. Also, find the average power density and radiation resistance if the antenna current is 4.5 A. Solution:

We know that

Pav = EN =

Þ

2Q

Q

P(R , G ) sin R

0

0

4Q

∫ ∫

E (R , G ) Emax (R , G )

dR dG

in which Emax = (q, f) = sin 90° ´ sin 90° = 1

EN = E(q, f) = sin q sin f

102


P(R , G ) =

Pav =

Pmax (R , G ) =

E2

E2

=

I0

377

2Q

Q

sin 3R sin G

0

0

4Q × 120 Q

∫ ∫

(sin 90 sin 90)2 120 Q

=

dR dG =

2Q /3 4 × 3.14 × 120Q

= 1.38 × 10 −3 W/Sr

1 120 Q

Normalized power densities 4Q

D1 =

2Q

Q

0

0

∫ ∫ 41000

D2 =

R ER H

=

sin 3R sin G dR dG 41000 90° × 90°

=

4Q 2Q /3

=6

= 5.1

⎛ 6.0 ⎞ ∇D = D1 − D2 = 10 log ⎜ ⎟ = 0.7 dB ⎝ 5.1 ⎠

R=

∫ =

P: I

2

Q 0

sin 3 R

∫

2Q

sin G dR dG

0

120Q × (4.5)

2

=

2Q /3 120 × Q × (4.5)2

= 2.743 : Example 3.11 If an antenna radiates over a half space above a perfect conducting ground plane such that E = 60 ´ 10–2 V/m at a distance of 2 km from the antenna. Calculate the radiated power and the radiation resistance if the antenna terminal current is 60 mA. Solution:

We know that the power density is given by S=

E2

I0

=

(60 × 10 −2 )2 377

= 9.54 × 10 −4 W/m 2

Pr = 2Q r 2 S = 2 × 3.14 × (2 × 10 3 )2 × 9.54 × 10 − 4 = 2.39 × 10 4 :

Pt = Rr =

1 2

I 02 Rr ⇒ Rr =

2 × 2.39 × 10 4 (70 × 10 −3 )2

2 Pr I 02 = 97.55 × 10 2 k:


Example 3.12

103

(a) The electric field of a linearly polarized EM wave is G E i = xˆ Em (x, y)e− j C z

It is incident upon a linearly polarized antenna whose polarization is given by G Ea = (xˆ + yˆ ) E (r, R , K ) Find the PLF. (b) An elliptically polarized wave travelling in positive z-direction in a medium of relative permittivity 4.5. Find the average power per unit area covered by the waves Ex = 3 sin(w t – bx)

Solution:

2

1

PLF = xˆ

(a)

Ey = 7 sin(w t – bx – 60°)

and

(xˆ + yˆ )

2

=

1 2

Hence, PLF (dB) = 10 log 0.5 = –3. (b)

We know that average power per unit area is equal to the average Poynting vector. Sav =

1 E2 2 I

=

1 E2 F r I0

2

E 2 = E12 + E22 = 9 + 49 = 58

Sav = Example 3.13

1 58 × 2

4.5

377

= 16.32 × 10 −3 W/m −2

Why is test antenna preferred for use in receivers?

Solution: In practice the AUT is used as receiver and Tx antenna is placed in the far-field region of the test antenna and vice versa. This is because radiated fields are plane waves in the vicinity of test antenna. Example 3.14 Show that the SNR for a communication link at 1 W transmitter and isotropic antenna is given by SNR =

Solution:

M2 kTS 'f (4Q R)2

We know that SNR =

PG t t Gr ⎛ M ⎞ ⎜ ⎟ kTS 'f ⎝ 4Q R ⎠

2

104


1 × 1 × 1 ⎛ M ⎞ SNR = ⎜ ⎟ kTS 'f ⎝ 4Q R ⎠ SNR =

2

M2 (4Q R)2 kTS 'f

This is because Pt = 1 W and directive gain of isotropic antennas are 1. Example 3.15 Find SNR of a communication link operating with 50 MHz bandwidth over a distance of 1500 km if the parabolic dish antennas of 1 m diameter operating at 3 GHz. The transmitter power is 10 W and the receiver system temperature is 200 K. Solution:

The power gain of parabolic circular antenna is 2

2

⎛D⎞ ⎛ 1 ⎞ Gt = Gr = 6.289 ⎜ ⎟ = 6.289 ⎜ ⎟ = 628.9 ⎝M⎠ ⎝ 0.1 ⎠ SNR =

=

PG t t Gr ⎛ M ⎞ ⎜ ⎟ kTS 'f ⎝ 4Q R ⎠

2

10 × 628.9 × 628.9 1.38 × 10 −23 × 200 × 50 × 10 6

0.1 ⎛ ⎞ ⎜ 2 ⎟ ⎝ 4 × 3.14 × 1500 × 10 ⎠

2

= 8.211 ´ 102 = 29.2 dB Example 3.16 A satellite transmitter produces an effective radiated power (ERP) at an earth station of 35 dB over 1 W isotropic. Determine SNR under the following specifications: (i) Rx antenna effective area = 3.53 m2, temperature = 25 K (ii) Receiver temperature = 75 K and bandwidth = 20 MHz (iii) Distance between earth and satellite is 26000 km Solution: We know that

ERP = PtGt

Since

SNR =

=

Pt Gt Gr ⎛ M ⎞ ⎜ ⎟ kTS 'f ⎝ 4 Q R ⎠

2

⎛ M 2 Gr ⎜ 4 Q R 2 kTS ' ⎜⎝ 4 Q ERP

⎞ ⎟ ⎟ ⎠


or

⎛ ERP Aer SNR = ⎜⎜ 2 ⎝ 4Q R kTS 'f

105

⎞ ⎟⎟ ⎠

M 2Gr where Aer = 4Q Since the receiver system consists of an antenna and a receiver, hence Tsys = 75 + 25 = 100 K. ERP = 35 dB = 3163 3162 × 3.53 ⎛ ⎞ SNR = ⎜ −22 2 12 6 ⎟ × 100 × 26 × 10 × 20 × 10 ⎠ ⎝ 4 × 3.14 × 1.38 × 10 ⎛ 11161.86 ⎞ SNR = ⎜ ⎟ = 16.6 = 12.2 dB ⎝ 672.899 ⎠

Example 3.17 An ideal omni-directional antenna has constant radiation in the horizontal plane (q), and would fall rapidly to zero outside that plane, in such a way that the pattern follows: F(0) = 1 = 0

for 60° £ q £ 120° elsewhere

Find its directivity. Solution:

The solid beam angle :A =

∫∫

(F (R , G )2 ) d : =

∫

2Q 0

∫

120 60

12 sin R dR dG

= 2Q { − cos (R )}120 60 = 2Q (0.5 + 0.5) = 2Q

Therefore, directivity D=

Example 3.18 any case?

4Q 2Q

= 2 or 3.010 dB

Describe the difference between directivity and gain. Are they the same in

Solution: Directivity of an antenna would be the same as gain, provided all the input power given appeared as radiated power, i.e. Pin = Pr. However, gain reflects the fact that practical antenna cannot function as ideal antenna. Some of the power always lost in the antenna and surroundings. This fact introduces a new parameter called efficiency, those values lie between 0 and 1.

I= which also gives

Pr Pin

G = hD

106


Example 3.19

What are the different units for gain measurement?

Solution: There are three units for gain measurement: dB, dBi and dBd. Of these, dB is absolute gain, dBi is gain relative to isotropic reference antennas, and dBd is gain relative to half-wave reference antennas. The dBi and dBd gains are related as follows: dBd = dBi – 2.15. For example, 6.1 dB of any antenna can also be expressed as 6.1 dB = 6.1 dBi = 3.95 dBd. Example 3.20

Describe ideal dipole and short dipole antennas.

Solution: There are the following differences and similarities between these antennas: (i) The physical length (Dz) of both the dipoles are the same (Dz << l). (ii) Both the dipoles have equal directivity (= 1.5), because pattern shape completely determines directivity. (iii) The ideal dipole has uniform current distribution (rectangular in shape); the current on the wire smoothly goes zero at the ends, whereas in short dipole current distribution is approximately triangular in shape and then goes to zero at the ends (see Fig. 3.15). (iv) Since the length of both the dipoles satisfies (Dz << l), the pattern of short dipole will also be sin q, the same as the radiation pattern of ideal dipole. (v) The radiation resistance of short dipole is one-fourth of that of ideal dipole, because area is one-half of that of ideal dipole. (vi) The ohmic resistance of short dipole is one-third of that of ideal dipole. (vii) Since the radiation resistance decreases more relatively to the ohmic resistance, the efficiency is lower for the short dipole than it is for an ideal dipole for the same length.

FIG. 3.15

Current distribution along (a) ideal dipole and (b) short dipole.

Example 3.21 Radiation resistance of a short dipole is one-fourth of that of an ideal dipole. Why? Solution: In short dipole, its triangular current distribution leads to an equivalent length, that is, one-half of its physical length. This is because the equivalent length is proportional to the area under the current versus distance curves (see Fig. 3.15). In turn, radiated fields are proportional to this equivalent length. Since the radiation resistance is proportional to the far-field squared, the radiation resistance will also be proportional to the equivalent length squared. Since the area of triangular shape current distribution is one-half of that uniform current distribution on the ideal dipole, so the radiation resistance is one-fourth of that of the ideal dipole and it is equal to ⎛ 'z ⎞ R = 20Q ⎜ ⎟ ⎝ M ⎠ 2

2


Example 3.22

Find the radiation resistance of a 1.575-m long short dipole operating at

⎛ 10 6 S ⎞ 1 MHz, if the diameter of the wire ⎜ T = 2 × ⎟ is 1.59 cm. ⎜ m ⎟⎠ ⎝

Solution: 2

⎛ 1.575 ⎞ −2 Rr = 20Q 2 ⎜ ⎟ = 5.45 × 10 : 200 ⎝ ⎠ 2

NX ⎡ 4Q × 10 −7 × 2Q × 10 6 ⎤ −2 RS = = ⎢ ⎥ = 1.40 × 10 : 6 2T 2.2 × 10 ⎢⎣ ⎥⎦ ⎡ ⎤ 1.575 × 1.4 × 10 −2 RL = = ⎢ = 7.26 × 10 −2 : −2 ⎥ 2.2 Q a ⎢⎣ 2 × 2 × 3.14 × 1.59 × 10 ⎥⎦ LRS

Therefore, the efficiency

I= Example 3.23

107

Rr Rr + RL

=

5.45 × 10 −2 5.45 × 10 −2 + 7.26 × 10 −2

= 6.7%

Derive the expression for an antenna following the pattern: ⎧ ⎛ ⎧Q ⎪ 1, for ⎜ ⎨ − F= ⎨ ⎝⎩2 ⎪ ⎩ 0, elsewhere

⎫

B⎬

≤

⎭

R

⎧Q ⎫⎞ ≤ ⎨ + B ⎬⎟ ⎩2 ⎭⎠

Solution: The solid beam angle :A =

∫∫

[F (R , G )2 ] d : =

∫

2Q 0

Q /2+B

∫Q

/2 −B

12 sin R dR dG

⎛ ⎧Q ⎫ ⎧Q ⎫⎞ B = 2Q {− cos (R )}QQ /2+ + B ⎬ − cos ⎨ + B ⎬ ⎟ /2 −B = − 2Q ⎜ cos ⎨ ⎩2 ⎭ ⎩2 ⎭⎠ ⎝ = –2p (–2 sin a) = 4p sin a Therefore, directivity D=

4Q

4 Q sin B

= cosec B

108


Example 3.24

Describe the importance of impedance matching in the antenna system.

Solution: Antenna impedance is important for the transfer of power, whether from a transmitter to an antenna or from an antenna to a receiver. In order to maximize the power at the Rx antenna, the receiver impedance should be conjugate to the antenna impedance (equal resistance, equal magnitude and opposite sign). In general, receivers have real value impedance, typically 50 W, so it is necessary to turn out the antenna reactance using a matching network. Example 3.25

What are disadvantages of impedance matching in antenna systems?

Solution: There are two disadvantages of impedance matching in antenna systems: (i) Ohmic losses in the network components, such as tuning, reduces antenna efficiency; (ii) Matching network provides a matching only over a narrow band of frequencies, which reduces the functional bandwidth of the antenna. Example 3.26

(a) Define plane waves; (b) Sketch linear and circular polarizations.

Solution: (a) The phase-front (surface of constant phase) of wave radiated by a finitesized antenna becomes nearly planar over small observation regions. This wave is referred as a plane wave and its E and H lie in a plane. (b) If the electric field vector of a wave moves back and forth along a line, the wave is said to be linearly polarized, whereas if the E-vector remains constant in length and rotate around the axis, the wave is said to be circularly polarized. Rotation at radian frequency w, in one of two directions, is termed sense of rotation (Fig. 3.16).

FIG. 3.16

Various types of polarization.


109

OBJECTIVE TYPE QUESTIONS 1. The horn antenna was developed by (a) J.C. Bose (1900, India) (b) Maxwell (1800, USA) (c) J.D. Kraus (1789, UK) (d) None of these 2. Directivity of antenna is determined by (a) Gain (b) Impedance pattern (c) Radiation pattern (d) Efficiency 3. Which is not a modern antenna? (a) Dipole antenna (c) Reflector antenna

(b) Slot antenna (d) Lens antenna

4. Power loss during power transformation through antenna at distance R is proportional to (R–2), whereas power loss during power transformation through a Tx line at distance R is proportional to (a) (R–2) (b) (R–2) –a R (d) (e–aR)2 (c) e 5. Directive antennas are useful in point to point communication, whereas omni-directional antennas are useful in (a) Broadside situation (b) Microwave ovens (c) As a feed in reflector (d) None of these 6. Double polarized antennas enable the doubling of communication capacity by carrying separate information on orthogonal polarizations over the same physical link and at (a) Different frequency (b) Same frequency (c) Both of these (d) None of these 7. A type of antenna acts as a funnel and directs the waves into the connecting waveguide. This is the (a) Horn antenna (b) Slot antenna (c) Reflector (d) Lens antenna Select the properties which are not possessed by the given antennas in Questions 8–10: 8. Electrically small antennas (a) High directivity (c) High input reactance

(b) Low input impedance (d) Low radiation efficiency

9. Resonant antennas (a) Moderate gain (c) Narrow bandwidth

(b) Real input impedance (d) Finite reactance

10. Broadband antennas (a) Wide BW (c) Constant gain

(b) Real input impedance (d) High reactance

110


11. Quasi-fields vary with the distance as (a) 1/r (b) 1/r2 (c) r2 (d) None of these 12. The distance at which far- and near-fields are equal is (l/2p) is termed (a) Radian sphere (b) Radian distance (c) Equi-distance (d) None of these 13. Ohmic losses on an antenna are due to (a) Antenna temperature (b) Noise figure (c) Noise temperature (d) Noise sources 14. The external noise picked up by antenna is proportional to the antenna radiation resistance and it is larger than the noise arising from internal ohmic losses. (a) True (b) False (c) Not known (d) None of these 15. The reactive part of input impedance (XA) of an antenna represents power stored in the (a) Near-field (b) Far-field (c) Both (a) and (b) (d) Quasi-field 16. The short dipole has a capacitive reactance, whereas an electrically small dipole has (a) Inductive reactance (b) Capacitive reactance (c) Both (a) and (b) (d) None of these 17. Helix antenna provides complete (a) Linear polarization (c) Elliptical polarization

(b) Circular polarization (d) Parallel polarization

18. In general, the value of AR is +ve for RHCP and –ve for LHCP. It is measured in dB, which is converted by using (a) 10 log AR (b) 10 loge AR (c) Both (a) and (b) (d) 20 log AR 19. An antenna has directivity of 20 and a radiation efficiency of 90%. What is its gain (dB)? (a) 2.55 dB (b) 12 dB (c) 12.55 dB (d) 90 dB 20. A horn antenna has half-power beam width of 29° in both the principal planes. What is the directivity in dB? (a) 22 dB (b) 45 dB (c) 24 dB (d) 17 dB

Answers 1. 6. 11. 16.

(a) (b) (b) (a)

2. 7. 12. 17.

(c) (a) (a) (b)

3. 8. 13. 18.

(c) (a) (d) (d)

4. 9. 14. 19.

(d) (d) (a) (a)

5. 10. 15. 20.

(a) (a) (a) (d)


111

EXERCISES 1. Determine the same parameters, taking d = 100 m, and dimensions of antennas are 14 cm ´ 7 cm and operating frequency is 4 GHz. 2. Determine the same parameters and gain of receiving antenna (GR) if D = 100 m, f = 2 GHz, PT = 5 W, PR = 21.7 mW, GR = 15 dB and the antenna dimensions are 12 cm ´ 6 cm. 3. Determine the electric field intensity at a distance of 10 km from a dipole antenna of directive gain of 6 dB and radiating power of 20 kW. 4. A certain antenna with an efficiency of 90% has maximum radiation intensity of 0.5 W Sr–1. Determine its directivity assuming input power of 0.5 W and radiated power of 0.4 W. 5. Determine the directivity, if the radiation intensity of an antenna is defined by

⎧⎪ 4.5 sin 2R sin 2G U= ⎨ ⎪⎩ 0

0 ≤ (R , G ) ≤ Q elsewhere

6. The power pattern of an antenna is given by P (R , G ) =

2 sin 2R sin 2G

Q

within 0 £ (q) £ p and 0 £ (f) £ p and zero elsewhere. Determine the directivity and radiation resistance if the antenna terminal current is 2.5 A. 7. In a communication links the Tx and Rx antennas are separated by a distance of 200l and have directive gains of 25 and 20 dB respectively. Calculate the amount of power received, if Tx power is 1.582 W. 8. An antenna radiates in such a way that maximum radiated field strength measured at 25 km from the antenna is 14 mW. Find, it directivity and gain in dB if efficiency is 95%. 9. A radar operating at 2.5 GHz transmits power of 200 kW. Determine the signal power density at ranges 200 km and 450 km, if the effective area of the radar antenna is 9 m2 with a 20 m2 target at 200 km. Calculate the power of the reflected signal at the radar. 10. The field pattern of an antenna is defined as E(q) = 2 cos q . cos 2q for 0 £ (f) £ p/2 Determine HPBW and FNBW. 11. Estimate the directivity of an antenna with qE = 3° and qH = 1.5°. Also calculate the gain of the antenna if the efficiency of antenna is 60%.

112


12. Two spacecrafts (A and B) are at a distance of 3 ´ 103 km. Each has directivity 200 at 2 GHz. If craft A’s receiver requires 20 dB over 1 pW, what transmitter power is required on craft B to achieve this signal level? 13. A wave travelling normally out of the page is the result of two linearly polarized components, Ex = 4 cos w t and Ey = 5 cos (w t + 90°). For the resultant wave, find (a) axial ratio, (b) tilt angle (t) and (c) nature of rotation (left or right). 14. What do you understand by effective area of an antenna? Show that it can be expressed as Ae =

V2

M 2 (m2)

4 PRr

15. In a satellite communication, Tx transmits 107 W, 10 s pulse of right-hand circular radiation at 5 GHz. The antennas at Tx and Rx have the same diameter of 100 m. What will be the maximum distance at which the signal can be received with SNR = 2.5 dB? Assume antennas have efficiency of 50% and the earth station has a system temperature of 20 K and bandwidth of 0.2 Hz. 16. In a mobile communication system, a 5-m diameter antenna radiates at 800 MHz. How much power is required to establish the link with a mobile user at 10 km with SNR = 2 dB? The bandwidth is 2 Hz, Tsys = 20 K and effective area of Rx is 20% of that of Tx antenna. 17. A 2-m long dipole made up of 10.5 mm diameter aluminium is operated at 500 kHz. Compute its radiation efficiency, assuming: (i) The uniform current distribution (ii) The triangular current distribution 18. Compute the gain of an antenna which has radiation efficiency of 90% and follow the radiation pattern ⎧1 ⎪ F (R ) = ⎨ 0.707 ⎪ ⎪⎩ 0

R ≤ R ≤ 20° 20 ≤ R ≤ 120° 120 ≤ R ≤ 180°

19. (a) Show that there is 4p (Sr) in all space by integrating dW over a sphere; (b) Describe the difference between power gain and directive gain. 20. Using far-field criterion for a linear antenna with length L, find the far-field region for two antennas L = 5l and L = 8l.


113

REFERENCES [1] Ramsay, Jack, “Highlights of antenn history”, IEEE, Ant. and Prop., Soc., Newsletter, pp. 8–20, Dec. 1982. [2] Collin, R.E. and F.J. Zucker, Antenna Theory, Part I, Mc-Graw Hill, New York, 1969. [3] Van Bladel, J., “Lorentz”, IEEE, Ant. and Prop., Magazine, Vol. 22, p. 67, April 1991. [4] Navarro, J.A. and K. Chang, Integrated Active Antenna and Spatial Power Combining, John Wiley, New York, 1996. [5] Johnson, R.C., Antenna Engineering Handbook, 2nd ed., McGraw-Hill, New York, 1992. [6] Tai, C. and C. Pereira, “An approximate formula for calculating the directivity of an antenna,” IEEE, Trans., Antenna Propagate, AP, Vol. 24, No. 2, March 1976, pp. 225–226. [7] Kraus, J.D., Antennas: For All Applications, 2nd ed., Tata McGraw-Hill, New Delhi, 2007. [8] Balanis, C.A., Antenna Theory: Analysis and Design, 2nd ed., John Wiley, India, 2007. [9] Prasad, K.D., Antenna and Propagation, Satya Prakashan, India, 2006. [10] Mott, Harold, Polarization in Antennas and Radars, John Wiley & Sons, NY, 1986.

C H A P T E R

4

Antenna Array

INTRODUCTION The study of a single small antenna indicates that the radiation fields are uniformly distributed and antenna provides wide beam width, but low directivity and gain. For example, the maximum radiation of dipole antenna takes place in the direction normal to its axis and decreases slowly as one moves toward the axis of the antenna. The antennas of such radiation characteristic may be preferred in broadcast services where wide coverage is required, but not in point to point communication. Thus, to meet the demands of point to point communication, it is necessary to design the narrow beam and high directive antennas, so that the radiation can be released in the preferred direction. The simplest way to achieve this requirement is to increase the size of the antenna, because a larger-size antenna leads to more directive characteristics. But from the practical aspect, the method is inconvenient as antenna becomes bulky and it is difficult to change the size later. Another way to improve the performance of the antenna without increasing the size of the antenna is to arrange the antenna in a specific configuration, so spaced and phased that their individual contributions are maximum in desired direction and negligible in other directions. This way particularly, we get greater directive gain. This new arrangement of multi-element is referred to as an array of the antenna. The antenna involved in an array is known as element. The individual element of array may be of any form (wire, dipole, slot, aperture, etc.). Having identical element in an array is often simpler, convenient and practical, but it is not compulsory. The antenna array makes use of wave interference phenomenon, that occurs between the radiations from the different elements of the array [1]. Thus, the antenna array is one of the methods of combining the radiation from a group of radiators in such a way that the interference is constructive in the preferred direction and destructive in the remaining directions. The main function of an array is to produce highly directional radiation. The field is a vector quantity with both magnitude and phase. The total field (not power) of the array system at any point away from its centre is the vector sum of the field produced by the individual antennas. The relative phases of individual field components 114

Antenna Array

115

depend on the relative distance of the individual element and in turn depend on the direction. The main types are: Linear array:

An array in which individual elements are equally spaced along a straight line.

Uniform linear array: Elements of array are fed with a current of equal magnitude and uniform progressive phase shift along the line. The geometrical configuration of the array may be of many types: straight line, rectangular, circular, etc., but there are strict limitations. The simplest and most practical array configuration is a straight line array. In multi-element array antenna, elements are generally l/2 long dipoles. The length of element not strictly limited to l/2; it can vary by upto 5%, provided the radiating property of the element remains unaffected.

Design Considerations and Design Approach The designs of antenna array is based on the proper selection of design parameters such as the number of elements, elements spacing, excitation techniques, directivity, gain, efficiency and beam width. In design procedure, some of the parameters are specified and others are determined using certain design expressions. The specified as well as calculated parameters vary design to design. For most of the uniform arrays, the side lobe is always around –13.5 dB, and spacing as well as length of array element is usually taken as l/2. Practically, phase array design (where maximum array radiation can be directed in desired direction) is primarily based on control of the phase excitation of the elements. In this array (i) Beam width and side lobe level can also be controlled by proper amplitude excitation and tapering of the elements. (ii) The level of the minor lobes can be controlled using binomial techniques. With reference to directivity, there are four types of array configuration: edge, uniform, optimum and binomial array. From the radiation pattern of these arrays it is found that lower the side lobe level, larger the half power beam width, and vice versa. However, it is desired to have both simultaneously; a very low side lobe level and a considerable half-power beam width. In order to optimize the side lobe and HPBW, someone must look for a compromise design which could meet both the requirements. The Dolph–Tschebyscheff array design is used effectively to achieve a good compromise between side lobe level and beam width.

ARRAY CONFIGURATIONS Broadly, array antennas can be classified into four categories: (a) (b) (c) (d)

Broadside array End-fire array Collinear array Parasitic array

116


Broadside Array This is a type of array in which the number of identical elements is placed on a supporting line drawn perpendicular to their respective axes. Elements are equally spaced and fed with a current of equal magnitude and all in same phase. The advantage of this feed technique is that array fires in broad side direction (i.e. perpendicular to the line of array axis, where there are maximum radiation and small radiation in other direction). Hence the radiation pattern of broadside array is bidirectional and the array radiates equally well in either direction of maximum radiation. In Fig. 4.1 the elements are arranged in horizontal plane with spacing between elements and radiation is perpendicular to the plane of array (i.e., normal to plane of paper.) They may also be arranged in vertical and in this case radiation will be horizontal. Thus, it can be said that broadside array is a geometrical arrangement of elements in which the direction of maximum radiation is perpendicular to the array axis and to the plane containing the array element. Radiation pattern of a broad side array is shown in Fig. 4.2. The bidirectional pattern of broadside array can be converted into unidirectional by placing an identical array behind this array at distance of l/4 fed by current leading in phase by 90°. Direction of maximum radiation

Major lobe Minor lobe

Axis of array d

FIG. 4.1

Geometry of broadside array.

FIG. 4.2

Radiation pattern of broadside array.

End-fire Array This is an arrangement of elements with the principal direction of radiation along axis of array. The basic concept of array is similar to broad side array. The elements are fed by a current of equal magnitude but their phases vary progressively usually 180° along the line in such way as to make the entire array sustain unidirectional. In other words, we can say that individual elements are excited in such a way that a progressive phase different between adjacent elements becomes equal to the spacing (in l) between the elements. End-fire array may be bidirectional also, if a two elements array is fed with the currents of equal magnitude and 180° out of phase. Since phase shift between the adjacent elements is 0° or 180°, the field amplitude now adds in phase in the plane of array. The direction of maximum radiation can be changed at will by introducing the appropriate phase-shift between successive elements of the array. In fact it is possible to produce a radar beam which sweeps around the horizon, without any mechanical motion of array, by varying the phase difference between successive elements of the array electronically.

Antenna Array

117

Collinear Array In collinear array the elements are arranged co-axially, i.e., antennas are either mounted end to end in a single line or stacked over one another. The collinear array is also a broadside array and elements are fed equally in phase currents. But the radiation pattern of a collinear array has circular symmetry with its main lobe everywhere normal to the principal axis. This is reason why this array is called broadcast or amni-directional arrays. Simple collinear array consists of two elements; however, this array can also have more than two elements (Fig. 4.3). The performance characteristic of array does not depend directly on the number of elements in the array. For example, the power gain for collinear array of 2, 3, and 4 elements are respectively 2 dB, 3.2 dB and 4.4 dB respectively. The power gain of 4.4 dB obtained by this array is comparatively lower than the gain obtained by other arrays or devices. The collinear array provides maximum gain when spacing between elements is of the order of 0.3l to 0.5l; but this much spacing results in constructional and feeding difficulties. The elements are operated with their ends are much close to each other and joined simply by insulator.

Direction of maximum radiation L

Direction of maximum radiation L

Array axis

Array axis

(a)

FIG. 4.3

(b)

(a) Vertical collinear antenna array; (b) Horizontal collinear antenna array.

Increase in the length of collinear arrays increases the directivity; however, if the number of elements in an array are more (3 or 4), in order to keep current in phase in all the elements, it is essential to connect phasing stubs between adjacent elements. A collinear array is usually mounted vertically in order to increase overall gain and directivity in the horizontal direction. Stacking of dipole antennas in the fashion of doubling their number with proper phasing produces a 3 dB increase in directive gain.

Parasitic Arrays In some way it is similar to broad side array, but only one element is fed directly from source, other element are electromagnetically coupled because of its proximity to the feed

118


element. Feed element is called driven element, while other elements are called parasitic elements. A parasitic element lengthened by 5% to driven element act as reflector and another element shorted by 5% acts as director. Reflector makes the radiation maximum in perpendicular direction toward driven element and direction helps in making maximum radiation perpendicular to next parasitic element. The simplest parasitic array has three elements: reflector, driven element and director, and is used, for example in Yagi–Uda array antenna. The phase and amplitude of the current induced in a parasitic element depends upon its tuning and the spacing between elements and driven element to which it is coupled. Variation in spacing between driven element and parasitic elements changes the relative phases and this proves to be very convenient. It helps in making the radiation pattern unidirectional. A distance of l/4 and phase difference of p/2 radian provides a unidirectional pattern. A properly designed parasitic array with spacing 0.1l to 0.15l provides a frequency bandwidth of the order of 2%, gain of the order of 8 dB and FBR of about 20 dB. It is of great practical importance, especially at higher frequencies between 150 and 100 MHz, for Yagi array used for TV reception. The simplest array configuration is array of two point sources of same polarization and separated by a finite distance. The concept of this array can also be extended to more number of elements and finally an array of isotropic point sources can be formed [2]. Based on amplitude and phase conditions of isotropic point sources, there are three types of arrays: (a) Array with equal amplitude and phases (b) Array with equal amplitude and opposite phases (c) Array with unequal amplitude and opposite phases (a) Array with equal amplitude and phases Let us consider that there are two point sources at a distance d, symmetrically situated wrt the origin (see Fig. 4.4). Let P be an observation point at distance R from the origin O. From Fig. 4.4, it is clear that ray 2 radiated from source 2 leads ray 1 due to path difference AB² involved between two rays. The path difference

AB′′ = OA′ + OB′ =

d 2

cos R +

d 2

cos R = d cos R =

d

M

cos R per wavelength

Hence, the phase difference

(Z ) = 2Q ×

d

M

cos R = C d cos

R

rad

(4.1)

If E1 = E1e− jZ /2 is field component due to source 1, then field component due to source 2 jZ /2 will be E2 = E1e (because source 2 is opposite wrt source 1).

Antenna Array

FIG. 4.4

119

Array of two point sourcescase 1.

Therefore, the total far-field at observation point P will be ⎛ e− jZ /2 + e jZ /2 ET = E0 (e − jZ /2 + e jZ /2 ) = 2E0 ⎜ ⎜ 2 ⎝

⎞ ⎟ = 2E0 cos Z /2 ⎟ ⎠

(where E1 = E2 = E0 )

⎛ C d cos R ⎞ ET = 2E0 cos ⎜ ⎟ 2 ⎝ ⎠

(4.2)

i.e., maximum value of total field ET is 2E0, when cos(y/2) = ±1. In another case, if reference point, i.e., origin O is shifted to the point source number 1, then the amplitude field pattern will remain unchanged but change in phase will occur, i.e., ET = E0 e j 0 + E0 e jZ = E0 (1 + e jZ ) ⎛ e− jZ /2 + e jZ /2 = 2 E0 e jZ /2 ⎜ ⎜ 2 ⎝

⎞ Z jZ /2 ⎫ ⎧ e ⎟ = 2 E0 e jZ /2 cos(Z /2) = 2 E0 ⎨ cos ⎬ ⎟ 2 ⎩ ⎭ ⎠

(4.3)

Hence, it is clear that phases of ET and E¢T are not the same. In order to draw the field pattern of this array, we must know the position of maxima, half-power and minima. (a) Maxima direction:

ET is maximum, i.e.,

⎛ C d cos R ⎞ ⎛ 2Q M ⎞ ⎛Q ⎞ . cos R ⎟ = cos ⎜ cos R ⎟ is maximum. ET = cos ⎜ ⎟ = cos ⎜ 2 2 ⎝ ⎠ ⎝ M ⎠ ⎝2 ⎠

Þ

⎛Q ⎞ cos ⎜ cos R ⎟ = ± 1 ⎝2 ⎠

which gives qmax = 90° or 270°.

120


(b) Half-power point directions:

Half-power occurs when 1 ⎛Q ⎞ cos ⎜ cos R ⎟ = ± 2 ⎝2 ⎠

q = 60° or 120°

Þ (c) Minima direction:

Minimum power occurs ⎛Q ⎞ cos ⎜ cos R ⎟ = 0 ⇒ R min = 0° and 180° ⎝2 ⎠

when

Hence, the pattern will be as shown in Fig. 4.5, which is similar to field pattern of broadside array and also known as broadside couplet.

FIG. 4.5

Radiation pattern array of two-point sourcescase 1.

(b) Array with equal amplitude and opposite phases Geometrical arrangement of this array is similar to previous one, except that point source 1 is out of phase (i.e. 180°) with point source 2. This means that if there is maximum field at source 1 at particular moment, the field at source 2 will be minimum at that instant and vice versa. The amplitude and phase of this array are shown in Fig. 4.6, where E1 and E2 are radiated fields due to source 1 and source 2 having phases –y/2 and y/2 with reference point O respectively. Hence, the total far-field at distance point P can be given by ET = − E1e

− jZ /2

+ E2 e

+ jZ /2

⎛ e jZ /2 − e− jZ /2 = E0 ⎜ ⎜ 2j ⎝

⎛ C d cos R ⎞ = 2E0 j sin Z /2 = 2jE0 sin ⎜ ⎟ 2 ⎝ ⎠

⎞ ⎟ × 2j ⎟ ⎠

(4.4)

Antenna Array

FIG. 4.6

121

Array of two-point sourcesCase 2.

That is, Eq. (4.4) involves the sine function instead of cos function in Eq. (4.2), and also addition operator j. Presence of j indicates that opposite phases of the point sources (1, 2) introduces a phase-shift of 90° in total field at P. For particular array with d = l/2 and term 2E0j = 1 ⎛ Q ⎞ En = sin ⎜ ⎟ R 2 cos ⎝ ⎠

(4.5)

i.e., maxima occurs at qmax = 0° and 180°, and minima occurs at qmin = 90° and 270° and half-power at q3dB = 60° and 120°. Hence, the field pattern of the array is as shown in Fig. 4.7.

FIG. 4.7

Radiation pattern array of two point sourcesCase 2.

That is, the field pattern is shifted by 90° from the horizontal in comparison to the field pattern shown in Fig. 4.5. Since this array provides maximum radiation in direction along the line joining two point sources, it can be called end-fire array; the previous one is called broadside array.

122


(c) Array of unequal amplitude and arbitrary phase It is an array in which amplitude of current radiated from both the point sources is not equal and therefore they are infinite phase difference. Let us consider source 1 as the reference for the phase and amplitude of fields due to source 1 and source 2, i.e., E1 and E2 (where E1 > E2) as shown in Fig. 4.8. Hence the total phase difference between radiations from two point sources at observation point P is

Z =

2Q

M

d cos R + B

(4.6)

where a is the phase angle between the currents I2 and I1. Current I2 leads the current I1. Hence, if aÿ = 0° and 180°, this is a special case of the previous two cases, because E1 = E2 = E 0 .

FIG. 4.8

Array of two point sourcesCase 3.

In general, the total far-field intensity at point P is given by E = E1e j 0 + E2 e jZ = E1 (1 + ce jZ )

(4.7)

where c=

E2 E1

i.e. 0 £ c £ 1

Hence, from Eq. (4.7) the magnitude of field at P is E = [ E1 {1 + c(cos Z + i sin Z )}]

or

E = E1 (1 + c cos Z )2 + (c 2 sin

Z )2 < G

E = E1 1 + c2 cos2Z + 2 c cos Z + c 2 sin 2 E = E1 [1 + c2 + 2 c cos Z < G ]1/2

Z
1/2

(4.8)

Antenna Array

123

⎛ c sin Z ⎞ in which f is phase angle at point P and it is equal to tan −1 ⎜ ⎟ . If E1 = E2, ⎝ 1 + c cos Z ⎠ c = 1 and a = 0, then Eq. (4.7) becomes Eq. (4.2), which is the case of broadside array. On the other hand, if E1 = E2, c = 1 and a = 180° then this is the case of end-fire array.

PRINCIPLE OF PATTERN MULTIPLICATION In the previous sections we have discussed the arrays of two isotropic point sources radiating field of constant magnitude (i.e., E0). In this section the concept of array is extended to nonisotropic sources. The sources identical to point source and having field patterns of definite shape and orientation. However, it is not necessary that amplitude of individual sources is equal. The simplest case of non-isotropic sources is when two short dipoles are superimposed over the two isotopic point sources separated by a finite distance. If the field pattern of each source is given by E0 = E1 = E2 = E¢ sin q Then from Eq. (4.2) the total far-field pattern at point P becomes ⎛Z ⎞ ⎛Z ⎞ ⎛Z ⎞ ET = 2E0 cos ⎜ ⎟ = 2E ′ sin R cos ⎜ ⎟ ⇒ ETn = sin R cos ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠

or

⎛Z ⎞ ETn = E (R ) × cos ⎜ ⎟ ⎝2⎠

where

Z =⎜

⎛ 2Q d ⎝ M

(4.9)

⎞ cos R + B ⎟ ⎠

Equation (4.9) shows that the field pattern of two non-isotropic point sources (short dipoles) is equal to product of patterns of individual sources and of array of point sources. The pattern of array of two isotropic point sources, i.e., cos y/2 is widely referred as an array factor. That is ET = E (Due to reference source) ´ Array factor

(4.10)

This leads to the principle of pattern multiplication for the array of identical elements. In general, the principle of pattern multiplication can be stated as follows: The resultant field of an array of non-isotropic but similar sources is the product of the fields of individual source and the field of an array of isotropic point sources, each located at the phase centre of individual source and having the relative amplitude and

124


phase. The total phase is addition of the phases of the individual source and that of isotropic point sources. The same is true for their respective patterns also. The normalized total field (i.e., ETn), given in Eq. (4.9), can re-written as E = E 1 (q ) ´ E 2 (q ) where E1(q) = sin q = Primary pattern of array ⎛ 2Q d ⎞ E2 (R ) = cos ⎜ cos R + B ⎟ = Secondary pattern of array. ⎝ M ⎠ Thus the principle of pattern multiplication is a speedy method of sketching the field pattern of complicated array. It also plays an important role in designing an array. There is no restriction on the number of elements in an array; the method is valid to any number of identical elements which need not have identical magnitudes, phase and spacing between them. However, the array factor varies with the number of elements and their arrangement, relative magnitudes, relative phases and element spacing. The array of elements having identical amplitudes, phases and spacing provides a simple array factor. The array factor does not depend on the directional characteristic of the array elements; hence it can be formulated by using pattern multiplication techniques. The proper selection of the individual radiating element and their excitation are also important for the performance of array. Once the array factor is derived using the point-source array, the total field of the actual array can be obtained using Eq. (4.10).

ARRAY WITH n-ISOTROPIC POINT SOURCES OF EQUAL AMPLITUDE AND LINEAR SPACING In this section the concept of array is extended to n-isotropic point sources, which are equally spaced by d and is fed with in-phase currents of equal amplitudes, i.e., E0 (see Fig. 4.9). Hence the total far-field pattern at the observation point P will be the vector sum of the fields of individual sources, i.e., ET = E0 (e0jZ + e2jZ + ... + e j (n +1)Z ) = E0 (1 + e2jZ + e3jZ + ... + e j (n +1)Z )

or

ET e jZ = E0 (e jZ + e2jZ + ... + enjZ )

multiplying by ejy on both sides

After subtraction we get ET (1 – ejy) = E0(1 – ejny) ⎛ 1 − e jnZ ET = E0 ⎜ jZ ⎜ ⎝1 − e

⎞ ⎟ ⎟ ⎠

(4.11)

Antenna Array

FIG. 4.9

125

Array of n-point sources.

After further modification ⎛ 1 − e jnZ /2 e+ jnZ /2 =⎜ ⎜ 1 − e jZ /2 e+ jZ /2 ⎝ = e j ( n −1)Z /2 ×

jnZ /2 ⎞ e jnZ /2 (e− jnZ /2 − e jnZ /2 ) − e− jnZ /2 ) j (n −1)Z /2 (e = e ⎟ = jZ /2 ⎟ e (e − jZ /2 − e jZ /2 ) (e jZ /2 − e jZ /2 ) ⎠

sin n Z /2 sin Z /2

⎛ n − 1⎞ If G = ⎜ ⎟ Z is the resultant phase-angle, then ⎝ 2 ⎠

⎧ sin n Z /2 ⎫ jG ET = E0 ⎨ ⎬e ⎩ sin Z /2 ⎭

(4.12)

where ejf represents the phase-term. Equation (4.12) is the total far-field pattern of a linear array of n-isotropic point sources with source 1 as reference point for phase. In case the reference point is shifted to the centre of the array from source 1, then phase angle f is eliminated and Eq. (4.12) is reduced to ⎧ sin n Z /2 ⎫ ET = E0 ⎨ ⎬ ⎩ sin Z /2 ⎭

(4.13)

Again, it is clear that Eq. (4.13) is a product of the two terms: E0, the primary pattern, and sin n Z /2

, the secondary pattern or array factor of the n-element array. There are two sin Z /2 particular cases of the n-array: broadside array and end-fire array.

Broadside Array As we know that the array is said to be broadside, if (i) maximum radiation occurs in direction perpendicular to the line of array, and (ii) array sources should be in phase, i.e.,

126


a

= 0 and y = 0 must be satisfied, i.e., q = 90°, 270°. Hence, from Eq. (4.13), it is clear that Et is maximum when sin ny/2 is maximum, i.e.,

sin nZ

or

2

nZ 2

=1

= ± (2N + 1)

Q 2

Z ≡ ± (2N + 1)

which gives

Q n

C d cos(R max ) + B = ± (2N + 1)

Q n

⎡ 1 ⎧ (2N + 1)Q ⎫⎤ ⇒ R max = cos−1 ⎢ ⎨± ⎬⎥ n ⎭⎦ ⎣ Cd ⎩ where N = 0, 1, 2, 3, ..., n and N = 0 corresponds to major lobe maxima; hence

⎡ 1 ⎧ 2Q ⎫⎤ − B ⎬⎥ (R max ) major = cos−1 ⎢ ⎨ ⎭⎦ ⎣ Cd ⎩ n

(4.14)

In either case, N ¹ 0 corresponds to minor lobe maxima, i.e.

⎡ 1 (R max ) minor = cos−1 ⎢ ⎣ Cd For broadside array,

a

= 0 and

b

⎧ (2N + 1) Q ⎫⎤ − B ⎬⎥ ⎨± n ⎩ ⎭⎦

= 2p/l; Hence

⎡ ⎧ (2N + 1) M ⎫⎤ (R max )minor = cos−1 ⎢ ⎨± ⎬⎥ 2 nd ⎭⎦ ⎣⎩

(4.15)

Similarly, Et will be minimum when

sin

or

nZ 2

= 0,

i.e.

Z = ±

2N Q n

⎡ 1 ⎛ 2N Q ⎞⎤ − B ⎟⎥ (R min )minor = cos−1 ⎢ ⎜ ± n ⎠⎦ ⎣ Cd ⎝

Antenna Array

127

Since for broadside array, a = 0, therefore ⎛ NM ⎞ (R min ) minor = cos−1 ⎜ ± ⎟ ⎝ nd ⎠

(4.16)

Beam width of major lobe (broadside array) Beam width is defined as the angle between first nulls, i.e., BW = 2 ´ qb

qb = 90 – q

qmin = (90 – qb)

or

⎛ NM ⎞ cos R min = cos (90 − R b ) = sin R b = ± ⎜ ⎟ ⎝ nd ⎠

from Eq. (4.16)

If qb is very small, ⎛ NM ⎞ ⎛ NM ⎞ sin R b ≅ R b = ± ⎜ ⎟ ⇒ Rb = ± ⎜ ⎟ ⎝ nd ⎠ ⎝ nd ⎠

Hence N = 1, which gives R b = ±

M nd

, therefore the beam width between the first nulls is

BWFN =

2M

(4.17)

nd

Here n is the total number of elements and d is spacing between the elements. Hence (n – 1)d = nd = L (total length of the array, in m) BWFN =

=

2M L

=

2 LM

2 × 57.3 LM

rad

deg =

114.6 LM

deg

Another pattern is HPBW, which is commonly half of BWFN, i.e. HPBW =

=

BWFN 2 57.3 LM

=

deg

1 LM

rad

128


End-fire Array We know that an array is said to be end-fire if the phase angle is such that it makes maximum radiation in the direction of array axis, i.e., q = 0° or 180°. Thus for an array to be end-fire, y = 0 and q = 0° or 180°, i.e., bd cos q + a = 0

bd = – a

Þ

B= −

or

2Q d

M

As we know for pattern maxima

sin

nZ 2

=1 ⇒

Z

(2N + 1)Q

= ±

n

where N = 0 corresponds to major lobe maxima.

y=0

For end-fire array

and

a = bd

C d cos(R max ) − C d = ±

Hence

cos(R max ) = ±

(2N + 1)Q

(2N + 1)Q

C nd

n +1

Beam widths of major lobes for both the arrays are shown in Fig. 4.10.

FIG. 4.10

Beam width of major lobes: (a) broadside array; (b) end-fire array. ⎡ (2N + 1)Q ⎤ (R max ) minor = cos−1 ⎢ ± + 1⎥ C nd ⎣ ⎦

(4.18)

Similarly, for pattern minima (cos R min − 1) = ±

2N Q

C nd

= ±

NM nd

1

or

⎛R ⎞ ⎛ NM ⎞ 2 sin ⎜ min ⎟ = ± ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2nd ⎠

Antenna Array

⎛ (R min )minor = 2 sin −1 ⎜ ± ⎜ ⎝

NM ⎞ ⎟ 2nd ⎟⎠

⎛ If qmin is very small, sin(qmin)minor » (qmin)minor = 2 ⎜ ± ⎜ ⎝

NM ⎞ ⎟ 2nd ⎠⎟

2N M

(R min ) minor = ±

Hence

129

(4.19)

nd

Beam width of major lobe (end-fire array) This is given by ⎛ BWFN = 2 × R min = 2 × ⎜ ± ⎜ ⎝

2N M ⎞ ⎟ nd ⎟⎠

2

2

BWFN = ± 2

where LM =

L

M

LM

rad = ± 114.6

LM

deg

(4.20)

, known as the length of the array per unit length.

ELECTRONIC PHASED ARRAY From the previous discussions, it is clear that the maximum radiation from an array can be released in the desired direction by controlling the phase excitation between the elements of array. In particular case, if the maximum radiation of an array is required to be oriented at an angle q0(0 £ q0 £ 180°), the corresponding phase excitation a between the elements will be

Z

= (C d cos R + B )R =R0 = 0

bd cos q0 + a = 0

a = – bd cos q0

(4.21)

i.e., by controlling the progressive phase difference between the array elements, the maximum radiation can be released in the desired direction, hence forming a scanning/phased array. This is basic concept of scanning/phased array operation. In order to have continuous scanning the array system must be capable of continuously varying the progressive phase between the elements. In practice it is accomplished electronically by the use of ferrite phase shifter. Hence the progressive phase change is controlled by the magnetic field within the ferrite, which in turn is controlled by the amount of current flowing through the wires wrapped around the phase shifter.

130


EFFECT OF EARTH ON VERTICAL PATTERNS In general, it is assumed that radiators are fixed far away from the earth surface; but in practice they are erected right at or within a few l off the earth surface. Under such situations, currents flow in the reflecting surface which magnitude and phase depends upon frequency, conductivity and dielectric constant of reflecting surface. These induced currents modify the radiation pattern of antenna accordingly. For the practical purposes, the resultant radiation fields are often computed on the assumption that reflecting surfaces are perfectly conducting. However, this computation is limited up to medium frequencies for the earth as reflecting surface, and radio frequency for the metallic reflector surface. The horizontal and vertical antennas located above perfect ground are shown in Fig. 4.11(a). – – – – ++ ++

–– ––

++ ++ Perfectly conducting plane

– – – – –– ––

++ ++

++ ++ (i) Vertical antenna

FIG. 4.11(a)

(ii) Horizontal antenna

Actual and image charges and current of antennas.

According to boundary conditions the ET and HN must vanish, i.e., at the surface E is normal and H is tangential. Hence the charge distribution and currents flow on conducting surface would be in such a way that boundary condition is satisfied. Therefore, the total electric and magnetic fields will not be only due to charges and currents on the antenna, but also due to these induced charges and currents. The E and H above the conducting plane can be obtained by removing this plane and replacing it by suitably located images and currents; the image charges will be mirror images of actual charges, but are opposite nature. The currents in original and image antennas will have the same direction for vertical antennas, but opposite direction for horizontal antennas. The present case can be dealt with simple ray theory, where resultant field is considered as made up of direct and reflected waves. Actual antenna and image antenna will be the sources of direct and reflected waves. The vertical component of E for the incident wave is reflected without phase reversal, whereas horizontal component will have phase reversal of 180°. The phase delay due to path difference is automatically controlled.

Antenna Array

131

FIG. 4.11(b)

ray Re flec ted

t ra y Dire c

ray Ref lect ed

Dire c

t ra y

Therefore, using image theory, it is simple to take into account the effect of earth on the radiation pattern. The earth is replaced by an image antenna, located at a distance below 2h, where h is the height of actual antenna above the earth. The field of image antenna is added to that of the actual antenna and obtain the resultant field. The shape of the vertical pattern is affected greatly, whereas horizontal pattern found remains unchanged (only the absolute value changes). The effect of the earth on the radiation pattern can also be explained using the principle of pattern multiplication of array theory [see Fig. 4.11(b)]. The vertical pattern of the antenna (or array) is multiplied by the vertical pattern of two non-directional radiations of equal amplitude and 2h spacing.

Direct and reflected rays from actual and image antennas.

In case of vertical antenna pattern there will be equal phase, whereas there will be opposite phase for the horizontal antenna. That is, vertical antenna may be treated as broadside array and horizontal antenna array as end-fire array. The resultant radiation patterns of such antennas are shown in Fig. 4.12 (a and b). Only a half-portion of the pattern is visible; the rest is hidden in both the cases.

Comparison of Methods When antennas are sloped (i.e., neither vertical nor horizontal), or when antennas are not multiples of half-wavelength long, the antenna array process fails; but image theory still be useful to obtain the resultant field. Similarly, when the finite conductivity of the earth is considered, the image theory will still be valid, whereas the array theory will no longer be adequate.

132

Antenna and Wave Propagation + O

O – (a)

+ =

× + (i) Unit pattern

(ii) Group pattern

(iii) Resultant pattern

(b)

FIG. 4.12

(a) Vertical pattern of a horizontal antenna using array theory. (b) Vertical pattern of a vertical antenna using image theory.

DOLPHTCHEBYSCHEFF OR CHEBYSHEV ARRAY Generally in the antenna design it is often desired to achieve narrowest beam width besides low side level. However these characteristics of an antenna system are so related that any attempt in the improvement of the one deteriorates other. C.L. Dolph proposed a method to minimize the beam width of the main lobe for a specific side-lobe level and vice versa. In other words, if the beam width between first nulls is specified, then the side-lobe level is minimized. Thus, Dolph array produces narrowest beam width for given side lobe-level and vice versa. Dolph’s approach indicates that reduction inside lobe can be accomplished at a cost of antenna performance in some other respect such as beam width, and gain or directivity. Dolph–Tchebyscheff current distribution is optimum for d £ l/2. It has been found that a high gain narrow beam antenna can be designed for side lobe levels of 20–30 dB in VHF and UHF bands, particularly for radar applications. A 20 dB level is considered good and 30 dB is excellent; however, it is difficult to achieve 40 dB levels.

Tchebyscheff Polynomial The Tchebyscheff polynomial is represented by letter T and defined as

and

Tm(x) = cos (m cos–1 x)

for |x| < ±1

(4.22a)

Tm(x) = cosh (m cosh–1 x)

for |x| > ±1

(4.22b)

Antenna Array

133

where m is integer and equal to 0, 1, 2, 3, … If

m = 0

Tm(x) = cos (0 cos–1x) = cos(0 d) = 1

where d = cos–1x or x = cos d. Similarly, for other values of m, we get m = 1

T1(x) = cos (1 d) = x

m = 2

T2(x) = cos (2 d) = 2cos2d – 1 = 2x2 – 1

m = 3

T3(x) = cos (3 d) = 4cos3d – 3cos d = 4x3 – 3x

m = 4

T4(x) = cos (4 d) = 2cos2 2d – 1 = 2[(2cos2 d – 1)2 – 1] = 8 cos4 d – 8 cos2 d + 1 T4(x) = 8x4 – 8x4 + 1

Further higher term can be obtained from the expression Tm+1(x) = 2xTm(x) = Tm–1(x)

(4.23)

So, for m = 5, we must put m = 4 in the above equation, i.e., T5(x) = 2xT4(x) – T3(x) = 2x[(8x4 – 8x2 + 1) – (4x3 – 3x] T5(x) = 16x5 – 20x3 + 5x Similarly, the values of T6, T7, T8, T9 and T10 are also found and summarized as follows: T0(x) = 1 T1(x) = x T2(x) = 2x2 – 1 T3(x) = 4x3 – 3x T4(x) = 8x4 – 8x2 + 1 T5(x) = 16x5 – 20x3 + 5x T6(x) = 32x6 – 48x4 + 18x2 – 1 T7(x) = 64x7 – 112x5 + 56x3 – 7x T8(x) = 128x8 – 256x6 + 160x4 – 32x2 + 1 T9(x) = 256x9 – 576x7 + 432x5 – 120x3 + 9x T10(x) = 512x10 – 1280x8 – 608x6 + 40x4 – 14x2 Hence, in general the Tchebyscheff polynomial can be expressed as Tm(x) = cos (m cos–1x) = cos md = cos (my/2) or

Tm(x) = cos (my/2)

(4.24)

134


The general characteristics found are shown in Fig. 4.13.

m Tm(x) when m is even

(x0, r)

+1 a

Tm(x) when m is odd

r

+1 –1 a

O

+1

O

–1

–1

+1 –1

x0

x0

(a) when m is even

FIG. 4.13

(x0, r)

(b) when m is odd

General characteristics of Tchebyscheff polynomial in two cases.

The side lobes arise in the region x < 1. The main lobe extends into the range x > 1. It is also seen from Tchebyscheff polynomial that the value of m and the degree of the polynomial are the same.

Dolph Pattern Method of Obtaining Optimum Pattern Using Tchebyscheff Polynomial Consider a linear array of n number of isotropic point sources; n may even or odd number. All the sources are in same phase and at uniform spacing d. The individual sources have the amplitudes A0, A1, A2, …, Ak, etc. as indicated, the amplitude distribution being symmetrical about the centre of array. The direction q = 0° is the direction perpendicular to array with origin at the centre of the array. The total field ET at a larger distance in a direction q from the even number of sources is then sum of the fields of the symmetrical pairs of sources, i.e., 3Z ⎛Z ⎞ ⎛ n − 1⎞ + ... + 2AK cos ⎜ e ET = 2A0 cos ⎜ ⎟ + 2A1 cos ⎟Z 2 ⎝2⎠ ⎝ 2 ⎠

(4.25)

where ne = 1, 2, 4, ..., 2(K+1)

Z =

2Q

M

d sin R

Each term in Eq. (4.25) represents the field due to a symmetrically disposed pair of the sources. If Therefore

ne = 2(K+1),

where K = 0, 1, 2, 3, ...

⎛ ne − 1 ⎞ ⎛ 2K + 1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

Antenna Array K = N −1

Hence

ETE = 2

∑

K =0

where N =

⎛ 2K + 1 ⎞ Z⎟ AK cos ⎜ ⎝ 2 ⎠

135 (4.26)

ne

. 2 Similarly, consider the case of a linear array of an odd number of isotropic point sources, i.e. n0 [see Fig. 4.13(b)] under the same condition. In addition the amplitude distribution is symmetrical about the centre source. The amplitude of the centre source is taken as 2A0 and sequent next amplitudes, A1, A2 and Ak. The total field ET at a larger distance in a direction q from the odd number of sources is the sum of the fields of the symmetrical pairs of sources, i.e., 3Z ⎛Z ⎞ ⎛ n −1⎞ + ... + 2AK cos ⎜ 0 ET = 2A0 + 2A1cos ⎜ ⎟ + 2A2 cos ⎟Z 2 ⎝2⎠ ⎝ 2 ⎠

(4.27)

n0 = 1, 3, 5, ..., (2K+1) Each term in Eq. (4.27) represents the field due to a symmetrically disposed pair of the sources. As

ne = 2K+1,

⎛ ne − 1 ⎞ ⎛ 2 K ⎞ ⎜ ⎟=⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

Therefore K=N

Hence

where K = 0, 1, 2, 3, ....

ETE = 2

∑

K =0

⎛ 2K AK cos ⎜ ⎝ 2

Z ⎞⎟ ⎠

where N =

ne − 1 2

(4.28)

Equation (4.28) may be recognized as a finite Fourier series of N terms. Since for K = 0, we have constant term 2A0, which represents the contribution of centre source. Similarly each higher value of K gives a higher harmonics term, which in turn represents the contribution of a pair of symmetrically disposed sources. Therefore the total field pattern is the summation of the series of terms in increasing order of Fourier series having constant term, fundamental terms and 2nd harmonic term, etc., as is being represented in case of alternating currents. However, the total field pattern of an even number of sources is also represented by a finite Fourier series, but without constant term and having odd harmonics. The coefficients A0, A1, A2, A3, … in either series are arbitrary and expressed in amplitude distribution.

CALCULATION OF DOLPHTCHEBYSCHEFF AMPLITUDE DISTRIBUTION The factor 2 that appeared in the expression of ETe (4.27) and ETo (4.28) can be omitted as we are mainly concerned with relative field pattern. If the total number of sources involved in array are n, the steps below need to be followed:

136


(a) Let the ratio of main lobe maximum to minor lobe level be specified as R, i.e.

r=

Main lobe maximum Side lobe level

i.e., r can be calculated from the side lobe level below the main lobe maximum in dB as follows: 20 log r. (b) Select Tchebyscheff polynomial of the same degree as the array polynomial, i.e., Tn–1(x0) = r and solve it for x0 using the equation

x0 =

1 ⎡ {r + r 2 − 1}1/m + {r − ⎢ ⎣ 2

r 2 − 1}1/m ⎤ ⎦⎥

where m = n–1. The above formula is used only for high degree of Techebyscheff polynomial. (i) Choose array polynomial ET from ETe = A0 z + A1 [4z 3 − 3z ] + A2 [165 − 20z 3 + 5z ] + A3 [ A3 (64z 7 − 112 z 5 + 56 z 3 − 72)] + ... ETo = A0 + A1[2z 2 − 1] + A2 [8z 4 − 8z 2 + 1] + A3 [32z 6 − 48z 4 + 18 z 2 − 1] + ...

in which z = (x/x0) (ii) Equate Tchebyscheff polynomial Tn–1(x) with array polynomial Et, i.e., Tn–1(x) = Et, and calculate the coefficients and take ratios for relative amplitudes.

Advantages of DolphTchebyscheff Distribution (i) It provides a minimum optimum beam width for a specified degree side lobe level reduction, (ii) It results in side lobes are all of the same amplitudes, unlike in uniform distribution in which side lobes near adjacent to the main lobe are largest and others progressively decrease as angle increase from main lobe. (iii) Ratio of current between centre element and end element is small, which provides ease in feeding design.

Beam Width between First Nulls of Chebyshev Polynomial Patterns ⎡ M Q ⎫⎤ ⎧1 BWFN = 2sin −1 ⎢ cos−1 ⎨ cos ⎬⎥ 2(m − 1) ⎭⎦⎥ ⎩ x0 ⎣⎢ Q d

(4.29)

Half power beam width (HPBW) and minor lobe maxima of Chebyshev polynomial patterns is given by

Antenna Array

HPBW = 2 sin −1

⎡ ⎢ ⎢ M cos −1 ⎢Q d ⎢ ⎢⎣

r ⎞ ⎫⎤ ⎛ ⎧ cosh ⎜ cosh −1 ⎟ ⎪⎥ ⎪ 2 ⎠ ⎪⎥ ⎪1 ⎝ ⎨ ⎬⎥ m − 1 ⎪ x0 ⎪⎥ ⎪⎩ ⎪⎭ ⎥ ⎦

137

(4.30)

and the angle at which null maxima occur is ⎡ M

⎧⎪ 1 ⎛ kQ ⎞ ⎫⎪⎤ cos−1 ⎨ cos ⎜ ⎟ ⎬⎥ ⎢⎣ Q d ⎝ m − 1 ⎠ ⎭⎪⎦⎥ ⎩⎪ x0

R nm = sin −1 ⎢

(4.31)

where the parameters have the usual meaning and have been defined already.

STACKED/RECTANGULAR AREA BROADBAND ARRAY As the name suggests, it is an array where elements are arranged one above the other and the array occupies a flat area of rectangular shape. It is just an extension of obtaining the field-pattern of linear array to the rectangular broadside array. The array may consist of isotropic radiators or be a continuous current sheet. If all the elements of the array operated are in phase, the array is called broadside rectangular array, i.e., maximum radiation is in the direction perpendicular to the plane of array. If a unit array is stacked upto height m, the power gain in the direction of maximum radiation over single unit is mn, provided all the elements are co-operated in phase (see Fig. 4.14), where m and n are numbers of elements along length (l) and width (b) of the array. The 3-D configuration of a rectangular array is shown in Fig. 4.15. Let the field pattern across the array in the y-direction is same for any value of z between (± b/2), while the field pattern across the array in z-direction is the same for all values of y between (± a/2). Therefore, the field pattern in x-y plane is a function of q and depends only on the y-direction (i.e., l) of the array, while the field pattern in x-z plane is function of f and depends only y n-elements

b

O

FIG. 4.14

x m-elements

Rectangular array of m ´ n isotropic radiators.

138


FIG. 4.15

A rectangular array of dimensions l ´ b in 3-D coordinate system.

on z-dimension (i.e., b) of the array. That is, array is combination of y-array of height l and z-array of height b, in addition the array also has depth in x-direction (i.e., has end-fire directivity). Therefore the resultant field-pattern of the rectangular array can be obtained by using principle of pattern multiplication for y-array and z-array. Radiation pattern of the array of dimension l ´ b is bidirectional, i.e., having forward and backward radiating beams (see Fig. 4.16). y

z

x

x

x

x

y

z

(a) x-y plane

(b) x-z plane

FIG. 4.16

Radiation pattern of rectangular array antenna.

Directivity If qhp and fhp are the half-power beam width in (x-y) and (x-z) planes respectively, the directivity of a rectangular antenna array is given by (see [3]). D=

41.287

R hp × G hp

(4.32)

However, in general the directivity of a large rectangular broadside array of height (l) and width (b) of the uniform amplitude distribution is given by

Antenna Array

4Q F (R , G ) max

D=

∫∫

139 (4.33)

F (R , G ) sin R dR dG

Subject to condition that the minor lobes are not large, where F(q, f) is a parameter termed space-power pattern and varies as the square of the space-field pattern. The space-field pattern of a large rectangular array is given as

E = E xy (R , G ) + E xz (R , G ) =

⎛ Q l sin R ⎞ sin ⎜ M ⎟⎠ ⎝ ⎛ Q l sin R ⎞ ⎜ M ⎟⎠ ⎝

×

⎛ Q b sin R ⎞ sin ⎜ ⎟ M ⎝ ⎠ ⎛ Q b sin R ⎞ ⎜ ⎟ M ⎝ ⎠

The main beam maximum is in the direction of q = f = 0; thus for larger array sin q = q, and sin f = f; therefore ⎛ Q lR ⎞ sin ⎜ ⎟ ⎝ M ⎠

E=

E2 =

⎛ Q lR ⎞ ⎜ ⎟ ⎝ M ⎠

sin 2 (R l ) (R l )

×

×

⎛ Q bG ⎞ sin ⎜ ⎟ ⎝ M ⎠ ⎛ Q bG ⎞ ⎜ ⎟ ⎝ M ⎠

sin 2 (R b ) (R b )

where ⎛ Q lR ⎞ ⎛ Q bG ⎞ ⎟ and Gb = ⎜ ⎟ ⎝ M ⎠ ⎝ M ⎠

Rl = ⎜

| E2(q, f)max |2 = 1

Hence

In the condition of maximum directivity, i.e. q at North pole, sin q = 1; therefore 4Q

D=

∫

Q /2 −Q /2

∫

Q /2

sin 2 R l

−Q /2

Rl

×

sin 2 R b

dR l dGb

(4.34)

Rb

i.e., the array is radiating only in forward direction, there is no lost of power in backward direction. Assuming the limit to ± ¥ in place of limit –p/2 to p/2, the above equation’s denominator reduces to l2/l ´ b. Therefore approximate directivity is D=

4Q

M

2

l × b=

12.56

M

2

l × b=

12.56

M2

× Area of aperture

For example, the directivity of a broadside rectangular array of height h = 20 l and width b = 12l is 34.5 dB.

140


SUPER DIRECTIVE RECEIVING ARRAY Antenna designs which result in greater directive gain than the ordinary directive gain is termed as super directive. Super directive antennas are those antennas whose directivity is very high compare to conventional antennas of the same specifications. In antenna array, super directivity is accomplished by inserting number of elements within a fixed length of array. However, it leads eventually to a very large magnitudes and rapid changes of phase in the excitation coefficients of the elements of the array and therefore adjacent elements have very large and oppositely directed currents. As a result, the ohmic losses increase and the efficiency of the array decreases very sharply. In addition, increases in reactive power relative to the radiated power and the quality factor of the array are also observed. Super directivity antenna arrays are also called super gain array, as such antennas have actual overall gains (because of very low efficiency) less than the gain of uniform array of the same length. The main limiting factor to the directivity of any array is its length. Available results show that theoretically very high directivity can be obtained from linear end-fire arrays. Dolph–Tschebyscheff arrays with element spacing less than l/2 can provide desired directivity, however, the efficiency of array system goes down. The most important parameter which characterizes the performance of receiving array antenna is the signal to noise ratio (SNR), which is proportional to the directive gains. Conventional method to design a high directive gain (hence SNR) antenna is to feed the array elements with constant amplitude and proper phase, so that radiation from each element gets added in-phase in the desired direction. But the array becomes huge when more number of elements is added in order to have high directivity. The alternative way to design an array for maximum directive gain has been proposed by E.M. Newman [4]. The array is designed in order to achieve maximum directive gain subject to a constraint on the sensitivity factor. The involved parameters are directive gain, sensitivity factor, bandwidth, efficiency, number of elements and array spacing. It is also found that the conventional receiving arrays can be substantially reduced in size without a loss in directive gain and SNR.

SNR and Directive Gain As already mentioned, SNR and directive gain are proportional; hence, if the external noise is uniformly distributed in space, the system SNR can be expressed as SNR =

where Nex Nai NR Sex D(qm,ÿ fm)

= = = = = h =

ISex × D(R m , Gm ) I N ex + N ai + N R

Power at antenna/load interface due to external noise Power at antenna/load interface due to internal noise Power at antenna/load interface due to receiver noise External signal level incident from the direction (qm, fm) Directive gain Antenna efficiency expressed by h = PL/PcL, in which

(4.35)

Antenna Array

141

PL = power delivered to the actual load (the receiver) of the antenna PcL = power delivered to the conjugate matched load assuming lossless conditions If the receiving system is external or background noise limited, then by definition

hNex >> Nai + NR Hence Eq. (4.35) reduces to Sex × D(R m , G m )

SNR =

N ex

(4.36)

i.e., SNR is proportional to the directive gain and independent from the efficiency. Therefore, the SNR could be maximized by maximizing the directive gain without effecting the efficiency so long as the system has limited background noise (i.e., hNex >> Nai + NR).

Sensitivity Factor Sensitivity factor is an important parameter to describe the performance of an array, especially in practical implementation. The sensitivity factor for an N-element array is defined by

K=

∑ | I n′ | 2 | ∑ nn=1 I n′ e − jkrn | 2

(4.37)

where I n′ = current at nth feed port rn = distance between nth feed port and observation point located in the far-field region in direction of maximum radiation. k = measure of the susceptibility of the pattern to random errors in the excitation and position of the array elements. In practice, the excitation coefficients and the positioning of the array elements which results in the desired pattern cannot be obtained as specified. A certain amount of electrical and mechanical error will always be present, i.e., in reality, if any array of N-element is built the ideal pattern is not precisely realized. This discrepancy is due to two factors; one is current difference in real antenna and ideal antenna, second one is location specification between them. If Cn be the feed port current at the nth feed port of the real array, then C n = I n′ + I n′ B n = I n′ (1 + B n )

where Cn is realized, current excitation coefficient and ( I n′ B n ) represents the error in the nth excitation coefficient and mean square value of an is denoted by 2

2

F = <| B n | >

To take into account the error associated with the location of the elements, let define s be the root mean square value of the element position error, such that

142


E2 =

k 2T 2 3

Combining the above equations yields

E 2 + F 2 = '2 =

k 2T 2 3

+ < |B n |2 >

(4.38)

where D combines measures of electrical and mechanical error. Therefore in presence of these errors the pattern of real array is treated as sum of patterns of ideal array and an error pattern.

Radiation Efficiency of Super Directive Array It is defined as the ratio of the far-field amplitude of the super directive radiation pattern in the broad side direction to the far-field amplitude that would obtained if all the elements were fed in phase. Radiation efficiency [5]

Irad =

∑1N I p ∑1N | I p |

100%

(4.39)

This relation is pre-reference to the used expression;

Irad =

| ∑1N I p |2 ∑1N | I p |2

(4.40)

Since it produces 100% efficiency for all in-phase current distribution and it also provides a more discrepancy measure of super directive array efficiency. Super directive functions become more efficient when the overall length of the array is reduced. A greater number of elements permit greater beam width reduction at the cost of radiation efficiency.

ADAPTIVE ARRAY In general the antenna element along with their transmission line feed produce a beam or beams in predetermined directions, on the other hand receiving array/antenna look in a particular direction regardless of whether any signals are arriving from this direction or not. If by processing the signals from the individual elements, an array can become active and reacts intelligently to its environment. And steering its beam towards a desired signals while simultaneously steering a null towards an undesired, interfering signal and thereby the maximum signal to noise ratio of the desired signal. This antenna is termed adaptive array antenna. Also, by appropriate sampling and dignifying the signals at the terminals of each element (of such array) and processing them with a computer, a very efficient antenna can be built up, this new antenna is termed SMART antenna.

Antenna Array

143

Earlier adaptive antennas were used as radar antenna with the side lobe elimination characteristics. The side lobe eliminator antenna consists of a conventional radar antenna where output is coupled with that of much lower gain auxiliary antennas. The gain of the auxiliary antenna is slightly greater than the gain of maximum side lobe of radar antenna. Addicting the weighted signals received by the auxiliary antenna to those received by the radar antenna permitted suppression of interfering sources located in direction other than the main beam of the radar antenna. This early use of adaptive antenna evolved to adapted array and multiple beam antennas. An antenna is an essential component of an adaptive antenna system, which is uniquely related to the disciplines of antenna design. In fact, adaptive antenna system uses antenna of various types and configurations, however, they can be classified as phased array, multiple-beam antenna and a combination of both. Each of these antenna configurations has several ports where received signals Pr appears in response to sources located in the antenna’s field of view. By characteristics, phased array have identical elements each of which has a port where the output signal is represented as M

Er = ∑ I m Fn (R m , Gm ) e jHn

(4.41)

n =1

where

Im =

Pm Gm (4Q Rm ) f 2

C2

in which Pm = power radiated by mth source Gm = gain of antenna used by mth source Rm = distance between mth source and adaptive antenna f = operating frequency Fn and Hn represent the amplitude and phase that relates Im to a signal at the antenna port. The (qm, fm) part gives angular position of mth source and measured in a suitable spherical co-ordinate system. In most of adaptive phased array antennas, Fn is identical whereas Hn is generally different for all elements of the array. For signals at the output port of multibeam antenna (MBA), the Hn are nearly equal and Fn differ; however; this fundamental difference between phased array and an MBA results in the inherently larger bandwidth of the MBA.

Weighting of Signals Weighting of signals received at the port of an adaptive antenna determines directional response of antenna to incident signals, weights attenuate and alter the phase of received signals and designed to be either frequency-independent or adaptively varied as a function of frequency. Some adaptive antenna operates entirely at the received frequency and use RF weights; however, many others have mixer amplifier at each antenna port and the weights operate at lower IF (see Fig. 4.17). All the antenna systems have frequency dependent

144


FIG. 4.17

Receiver with adaptive antenna array.

response to incident signals. Whenever frequency independent weights are used, suppression of undesired signals varies with frequency. Because of scenario-dependent, this inherent performance characteristic cannot be succinctly and accurately described. However, it is found that for frequency independent weights the cancellation parameters C (suppression of an interfering signal) of an adaptive array antenna is limited as follows (see [6]):

⎡ ⎛ 2Q ⎞ ⎤ C ≤ 20 log ⎢ K ⎜ ⎟ DW sin (R m /2) ⎥ ⎣ ⎝ c ⎠ ⎦

(4.42)

where

qm D W c

= = = =

maximum angle subtended by antenna’s field of view maximum dimension of antenna aperture nulling bandwidth velocity of light in free space

The K is constant depends on the array configuration and particular scenario of interfering and desired sources and its values ranges from +5 to +15. As we know that adaptive antennas use both phased-array and multi-beam antennas, so it is necessary to be aware of the fundamental difference between these two antennas in regard of adaptive antenna. Phased-array antennas are focused to receive signals from a particular direction by adjusting phases of array elements. The differential time delay tm associated with signals arriving at the ports of the array elements is created by inserting delay tm in the range of ⎛ pM ⎞ 0 ≤ Um ≤ U − ⎜ ⎟ ⎝ c ⎠

where p is constant. That is, the array is perfectly focused at the design frequency fc and its performance deteriorates as the operating frequency altered from the fc.

Antenna Array

145

The parameter p is constant for a signal source located adjacent to the bore sight direction of the array and array does not focused effectively even for a large bandwidth (Df = fc – f). According to rule of thumb relating Df to antenna aperture D and q1, the fractional bandwidth is given by

'f =

where F =

D M

sin R

2FM fc

(4.43)

D sin R

'f 2 f0

and known as differential path delay. In case e is less than 0.1 (a path-length error = l/10), and qÿ = 10° and D = 120l. Equation (4.43) gives 'f fc

%=

2 × 0.1 × M 120 M sin 10°

≈ 1.0%

The available result shows that for e = 0.1 the interference signal suppressed » 20 dB. Halving or doubling Df changes the signal suppressed to 26 dB or 14 dB respectively. That is signal suppression varies approximately as (Df)2. As far as MBA is concerned, most MBA uses lens or parabolic reflector antennas and focused over wide frequency band. The side lobes and receiving patterns shape changes with frequency and alter the phase of received signals significantly. Reports indicate that this effect of varying frequency does not degrade related adaptive antenna performance as much as that of an equivalent planar, because each beam of MBA performs like a phase array with its receiving beam in the bore sight direction. As the side lobes of an MBA do not dominate the determination of the weight applied to beam port, MBA with an aperture of D = 120l can suppress interfering signals more than 20 dB for 'f fc

= 5%

If the expected Df/fc meets or exceeds system requirements, the phased-array may be the best choices, however, if the estimated Df/fc is less than required, an MBA may be the best choice.

Adaptive Antenna in Cellular Systems The major limiting factor on the capacity of a cellular mobile system is interfering from cochannel mobile in neighbouring cells. Adaptive antenna technology can be used to overcome this intelligent combination of the signals at multiple antenna elements at the base station. As we have discussed, two types of antenna phased array and multiple beam antenna.

146


In cellular mobile system a set of antennas (phased array) is arranged in space and the output of each element is multiplied by a complex weight and combined by summing as shown in Fig. 4.18, where Y is resultant out of new radiation pattern. In arrangement shown in Fig. 4.18 the radiation pattern of individual elements are summed with phase and amplitude depending upon both the weights applied and their positions in space providing a new and combined pattern. If the weights are allowed to vary in time the array becomes an adaptive array, and it is exploited to improve the overall performance of mobile communication system by choosing the weights so as to optimize some measure of the system performance. The reason behind using a mobile adaptive antenna system is to improve the performance of the system in the effect of the noise and interference. If a base station in a cellular system uses an adaptive array to direct its radiation patterns towards a mobile phone which is in communication, there are the following advantages: (a) The transmitted power for a particular signal quality can be reduced in both up-link and down-link directions. In other words, the cell radius and thereby the number of base stations required to cover a given area can be increased. (b) Since the mobile transmit power is reduced, its battery life can be extended. (c) Channel delay spread is reduced because off-axis scatters are no longer illuminated. (d) Depending on the direction of the mobile the probability of base-stations causing interference to co-channel mobile in surrounding cells is reduced. (e) In the same way, the probability of mobile causing interference to co-channel base station is also reduced.

FIG. 4.18

Four-element phased array antenna.

Hence the applications of adaptive antenna to mobile system have significant advantages in terms of coverage, capacity and quality. Currently few operational mobile systems actually use adaptive antennas in standards operation; however, it is expected that in the next few years, such antennas will form a standard feature of virtually all systems.

Antenna Array

147

BINOMIAL ARRAY In order to increase the directivity of an array its total length need to be increased. In this approach, number of minor lobes appears which are undesired for narrow beam applications. In has been found that number of minor lobes in the resultant pattern increases whenever spacing between elements is greater than l/2. As per the demand of modern communication where narrow beam (no minor lobes) is preferred, it is the greatest need to design an array of only main lobes. The ratio of power density of main lobe to power density of the longest minor lobe is termed side lobe ratio. A particular technique used to reduce side lobe level is called tapering. Since currents/amplitude in the sources of a linear array is non-uniform, it is found that minor lobes can be eliminated if the centre element radiates more strongly than the other sources. Therefore tapering need to be done from centre to end radiators of same specifications. The principle of tapering are primarily intended to broadside array but it is also applicable to end-fire array. Binomial array is a common example of tapering scheme and it is an array of n-isotropic sources of non-equal amplitudes. Using principle of pattern multiplication, John Stone first proposed the binomial array in 1929 [2, 3], where amplitude of the radiating sources are arranged according to the binomial expansion. That is, if minor lobes appearing in the array need to be eliminated, the radiating sources must have current amplitudes proportional to the coefficient of binomial series, i.e. (1 + x )n = 1 + (n − 1)x +

(n − 1)(n − 2) !2

x2 +

(n − 1) (n − 2) (n − 3) !3

x 3 ± ... (4.44)

where n is the number of radiating sources in the array. For an array of total length (nl/2), the relative current in the nth element from the one end is given by n!

=

r !(n − r )!

where r = 0, 1, 2, 3, and the above relation is equivalent to what is known as Pascal’s triangle. For example, the relative amplitudes for the array of 1 to 10 radiating sources are as follows: No. of sources n n n n n n n n n n

= = = = = = = = = =

1 2 3 4 5 6 7 8 9 10

Pascal’s triangle 1 1 1 1 1 1 1 1 1

1 9

7 8 36

2 3

4 5

6 28 84

1 3

6 10

15 21

1

10 20

35

1 4

1 5

1

15 35

56 70 56 126 126

6 21 28 84

1 7

1 8 36

1 9

1

148


Since in binomial array the elements spacing is less than or equal to the half-wave length, the HPBW of the array is given by HPBW =

10.6 n − 1

=

1.06 2L

=

0.75

M

(4.45a)

LM

and directivity D0 = 1.77 n = 1.77 1 + 2LM

(4.45b)

Using principle of multiplication, the resultant radiation pattern of an n-source binomial array is given by ⎛Q ⎞ En = cosn −1 ⎜ cos R ⎟ ⎝2 ⎠

In particular, if identical array of two point sources is superimposed one above other, then three effective sources with amplitude ratio 1:2:1 results. Similarly, in case three such elements are superimposed in same fashion, then an array of four sources is obtained whose current amplitudes are in the ratio of 1:3:3:1. The far-field pattern can be found by substituting n = 3 and 4 in the above expression and they take shape as shown in Fig. 4.19(a) and (b). It has also been noticed that binomial array offers single beam radiation at the cost of directivity, the directivity of binomial array is greater than that of uniform array for the same length of the array. In other words, in uniform array secondary lobes appear, but principle lobes are narrower than that of the binomial array.

Disadvantages of Binomial Array These are: (a) The side lobes are eliminated but the directivity of array reduced. (b) As the length of array increases, larger current amplitude ratios are required.

FIG. 4.19(a)

Radiation pattern of 2-element array with amplitude ratio 1:2:1.

149

Antenna Array

FIG. 4.19(b)

Radiation pattern of 3-element array with amplitude ratio 1:3:3:1.

Mutual Coupling between Arrays The mutual coupling between array elements can be described by taking an array of two elements. For example, two dipoles of lengths L1 and L2 such that first dipole is fed by voltage V1 and second one is coupled with the first one, as shown in Fig. 4.20(a). (The second one is passive.) If I1 and I2 are the currents in the terminals, then from the network theory: Z11I1 + Z12I2 = V1

(4.46a)

Z21I1 + Z22I2 = 0

(4.46b)

where Z11 and Z22 are the self-impedances of elements (1) and (2), and Z12 ~ Z21 are the mutual impedances between the elements. In case the lengths of both the dipoles are equal in length (L1, = L2 = L), then self-impedances is also equal. Therefore (I1 + I2) [Z11 + Z12] = V1 So, if I1 = I2, we get I1 =

V1 Z11 + Z12

(4.47)

For the present case (thin half-wave length dipoles), the self-impedance values are found to be Z11 = 73.1 + j42.5 W The variations of mutual impedance between two similar half-wave dipoles with normalized distance d/l are shown in Fig. 4.20(b). In the limiting case of the separation distance d ® 0, the mutual impedance approaches self-impedance, which is to be expected.

150


FIG. 4.20

(a) Arrangement of two dipoles; (b) Mutual impedance vs d/l of dipoles.

SOLVED EXAMPLES Example 4.1 Find and plot the array factor for three identical antenna arrays, which individually consists of two isotropic elements. Elements of array are separated by 5, 10 and 20 cm and each element is excited in phase and fed by a signal of 1.5 GHz. Solution:

The operating wavelength

M=

3 × 108 1.5 × 10 9

= 20 cm

Therefore normalized separations between elements are: l/4, l/2, l. The corresponding phase difference is zero, i.e., d = 0. Therefore the array factor will be ⎛ C d cos R + E ⎞ ⎛ 2Q d cos R ⎞ ⎛ Q d cos R ⎞ F (R , G ) = cos ⎜ ⎟ = cos ⎜ ⎟ = cos ⎜ ⎟ M 2 2M ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(a) d =

M

⎛Q ⎞ , F (R , G ) = cos ⎜ cos R ⎟ ; therefore 4 ⎝4 ⎠

q = 0, F (R , G ) =

1 2

= 0.707

q = 30°, F(q, f) = 0.77

151

Antenna Array

q = 45°, F(q, f) = 0.852 q = 60°, F(q, f) = 0.92 q = 90°, F(q, f) = 1 (b) d =

q q q q q

= = = = =

M

⎛Q ⎞ , F (R , G ) = cos ⎜ cos R ⎟ ; therefore 2 ⎝2 ⎠ 0, F(q, f) = 0 30°, F(q, f) = 0.22 45°, F(q , f) = 0.444 60°, F(q , f) = 0.707 90°, F(q, f) = 1

(c) d = l, F(q, f) = cos(p cos q); therefore

q q q q q

= = = = =

0, F(q, f) = 30°, F(q , f) 45°, F(q , f) 60°, F(q, f) 90°, F(q, f)

1 = = = =

–0.99 –0.606 0 1

Radiation patterns of three-element arrays are shown in Fig. 4.21. 120

90 1

60

0.5

150 180

30 0

210

330 240

FIG. 4.21

120

270 (a)

300

90 1

60

0.5

150

120 30

180

90 1

60

0.5

150

0 180

210

330 240

270 (b)

300

30 0

210

330 240

270 (c)

300

Radiation patterns of three-element arrays: (a) d = l/4, (b) d = l/2, (c) d = l.

Example 4.2 Draw the radiation pattern of 4-isotropic sources of equal amplitudes and phases in broadside and end-fire arrays. Solution:

Broadside array:

Given that n = 4 and d = l/2, then Eq. (4.15) becomes

152


(2N + 1) ⎤ ⎡ (2N + 1)M ) ⎤ −1 ⎡ (R max ) min = cos−1 ⎢ ± ⎥ = cos ⎢ ± ⎥ 2 nd ⎦ 4 ⎣ ⎣ ⎦ ⎛ 3⎞ = cos−1 ⎜ ± ⎟ = ± 0.75 ⎝ 4⎠

which gives

(qmax)minor = ± 41.4°

or

for N = 1

138.6°

That is there will be 4 minor lobes (maxima) adjacent to major lobes. No major lobes occur for other values of N, i.e., 2, 3, …, because cos (qmax)min ® 1, which violates the rule of cosine function. The major lobe occurs at q = 90° and 270°. Again from Eq. (4.16), N = 1. ⎡ 1M ⎤ −1 ⎡ 1 ⎤ (R min ) minor = cos−1 ⎢ ± ⎥ = cos ⎢ ± ⎥ ⎣ 4 M /2 ⎦ ⎣ 2⎦

which gives (qmin)minor = ± 60° or 120° Also, N = 2. Therefore, ⎡ 2M ⎤ −1 (R min ) minor = cos−1 ⎢ ± ⎥ = cos [ ± 1] M 4 /2 ⎣ ⎦

(qmin)minor = 0° or 180° That minor lobes (minima) occurs at 0°, ±60°, ±120°, 180°; i.e., 6 lobes. No other minor lobe minima occur because for N = 3, the values of cosine become greater than one (Fig. 4.22a).

FIG. 4.22(a)

Field pattern in broadside case.

Antenna Array

End-fire array: ⎡ (2N + 1) M ⎤ −1 ⎡ (2N + 1) ⎤ (R max ) 1 = cos−1 ⎢ ± ⎥ = cos ⎢ ± ⎥ C nd ⎦ 4 ⎦ ⎣ ⎣

n = 4,

M 2

, B = − Q, N = 1

⎛ 3 ⎞ ⎛ 7⎞ ⎛ 1⎞ = cos−1 ⎜ ± + 1 ⎟ = cos−1 ⎜ ⎟ or cos−1 ⎜ ± ⎟ ⎝ 4 ⎠ ⎝4⎠ ⎝ 4⎠

in which cos–1(7/4) does not exist whereas cos–1(1/4) gives qmax = 75.5°. Similarly for N = 2, ⎛ 5 ⎞ ⎛9⎞ ⎛ 1⎞ cos−1 ⎜ ± + 1 ⎟ = cos−1 ⎜ ⎟ or cos−1 ⎜ − ⎟ ⎝ 4 ⎠ ⎝4⎠ ⎝ 4⎠

Again cos–1(9/4) does not exist whereas cos–1(–1/4) gives qmax = –75.5° ⎛ ⎜ ⎝

R min = 2sin −1 ⎜ ±

n = 4,

M 2

NM ⎞ ⎟ 2 nd ⎟⎠

,N =1

⎛ (R min )1 = 2 sin −1 ⎜ ± ⎜ ⎝

⎞ ⎟ = 2 sin −1 2.4 M /2 ⎟⎠ 1M

⎛ ⎜± ⎜ ⎝

1⎞ ⎟ = 2 × ± 30° = ± 60° 4 ⎟⎠

Similarly ⎛ (R min )2 = 2sin −1 ⎜ ± ⎜ ⎝

1⎞ ⎟ = 2 × ± 45° = ± 90° 2 ⎟⎠

⎛ (R min )3 = 2 sin −1 ⎜ ± ⎜ ⎝

3⎞ ⎟ = 2 × ± 60° = ± 120° 4 ⎟⎠

(qmin)4 = 2 sin–1(±1) = 2 ´ ±90° = ± 180° N = 5 is not possible as its values exceed one [Fig. 4.22(b)].

153

154


FIG. 4.22(b) Example 4.3

Field pattern in end-fire case.

Find the FNBW of 18l long antenna arrays.

Solution: (FNBW) broad =

114.6° L

(FNBW)end-fire = 114°

M= 2M 18 M

114.6° 18 M

=

114° 3

= 6.37°

= 38.2°

Example 4.4 Find and plot the radiation pattern of two parallel thin half-wave length electric dipoles separated by

d= Solution: (a)

M 2

, M and

3M 2

The array factor for this case is given by d=

M

⎛Q ⎞ , F (R , G ) = cos ⎜ cos R ⎟ ; therefore 2 ⎝2 ⎠

q = 0, F(q, f) = 0 qÿ = 45, F(q, f) = 0.444 q = 90, F(q, f) = 1 (b)

d = l, F(q, f) = cos(p cos q); therefore

q = 0, F(q, f) = –1 q = 45, F(q, f) = –0.606 q = 90, F(q, f) = 1. (c)

d=

3M

⎛ 3Q ⎞ , F (R , G ) = cos ⎜ cos R ⎟ ; therefore 2 2 ⎝ ⎠

q = 0, F(q, f) = 0

155

Antenna Array

qÿ = 45, F(q, f) = –0.982 q = 90, F(q, f) = 1. Radiation patterns of three-element array are shown in Fig. 4.23. 120

90 1

60

0.5

150 180

120 30 0

210

330 240

270

300

60

0.5

150

120 30

90 1

60

0.5

150

0 180

180 210

330 240

(a)

FIG. 4.23

90 1

270

300

(b)

30 0

210

330 240

270

300

(c)

Radiation patterns of three-element array: (a) d = l/2, (b) d = l, (c) d = 3l/2.

Example 4.5 Using the concept of principle of pattern multiplication, find the radiation pattern of the four-element array separated at l/2 as shown in Fig. 4.24(a).

FIG. 4.24(a) Solution: To solve this problem, we have to consider the case of binomial array. Let us consider that we have a linear array that consists of three elements which are physically placed away d = l/2 and each element is excited in phase (d = 0), the excitation of the centre element is twice as large as that of the outer two elements [see Fig. 4.24(b)].

FIG. 4.24(b)

Example 4.5.

156


The choice of this distribution of excitation amplitudes is based on the fact that 1:2:1 are the leading terms of a binomial series. Corresponding array which could be generalized to include more elements is called a binomial array. As the excitation at the centre element is twice that of the outer two elements, it can be assumed that this three-element array is equivalent to two-element array that are away by a distance d = l/2 from each other. If so, equation ⎛ NZ ⎞ sin ⎜ ⎟ ⎝ 2 ⎠ F (R , G ) = ⎛Z ⎞ N sin ⎜ ⎟ ⎝2⎠

can be used for N = 2, where it is interpreted to be the radiation pattern of this new element, i.e., F (R , G ) =

sin Z

⎛Z ⎞ ⎛ C d cos R ⎞ ⎛ Q cos R ⎞ = cos ⎜ ⎟ = cos ⎜ ⎟ = cos ⎜ ⎟ 2 ⎛Z ⎞ ⎝2⎠ ⎝ ⎠ ⎝ 2 ⎠ 2 sin ⎜ ⎟ ⎝2⎠

i.e., the array factor of these elements is the same as the radiation pattern of one of the elements. Therefore from pattern multiplication principle, the magnitude of the far-field radiated electric field from this structure can be given by ⎛ Q cos R ⎞ F (R , G ) = cos2 ⎜ ⎟ ⎝ 2 ⎠

Hence in general, for an array of n-elements: ⎛ Q cos R ⎞ F (R , G ) = cosn −1 ⎜ ⎟ ⎝ 2 ⎠

Therefore, in given question, the array could be replaced by an array of two elements containing three sub-elements (1:2:1), each and new array will have the individual excitation (1:3:3:1), and ⎛ Q cos R ⎞ 2 ⎛ Q cos R ⎞ 3 F (R , G ) = cos ⎜ ⎟ cos ⎜ ⎟ = cos 2 2 ⎝ ⎠ ⎝ ⎠

Three patterns are possible: ⎛ Q cos R ⎞ (a) The element pattern: cos ⎜ ⎟ ⎝ 2 ⎠

⎛ Q cos R ⎞ ⎜ ⎟ ⎝ 2 ⎠

157

Antenna Array

⎛ Q cos R ⎞ (b) Array factor: cos2 ⎜ ⎟ ⎝ 2 ⎠ ⎛ Q cos R ⎞ (c) The array pattern: cos3 ⎜ ⎟ ⎝ 2 ⎠

The radiation patterns are shown in Fig. 4.24(c). 120

90 1

60

0.5

150

30 0

180 210

FIG. 4.24(c)

270 (a)

60

0.5

150

120 30

90 1

60

0.5

150

0 180

180

330 240

90 1

120

210

330 240

300

270 (b)

0

210

300

30

330 240

270 (c)

300

Radiation pattern of 4-element array separated at a distance d = l/2.

Example 4.6 Show that the directivity for a broadside array of two identical isotropic in-phase point sources separated at distance d is given by D(R , G ) =

Solution:

2 sin C d 1+ Cd

As we know that the directivity of an array is given by

D=

4 Q E 2 (R , G ) max

∫ ∫

2Q

Q

0

0

2Q

Q

0

0


or D=

4 Q F (R , G ) max

∫ ∫


⎛ sin 2Z /2 ⎞ 2 2 ⎛Z ⎞ 2 ⎛Z ⎞ where F (R , G ) = E 2 (R , G ) = ⎜ ⎟ = 2 cos ⎜ ⎟ = 4 cos ⎜ ⎟ ⎝ sin Z /2 ⎠ ⎝2⎠ ⎝2⎠

158


Taking N = 2, in which y = bd cos q + a = bd cos q as a = 0 for broadside array. F(q, f) = 4 cos (0.5 bd cos q) = F(q, f)max = 4 Let

I=

2Q

∫

∫

2Q 0

Q 0

∫

Q

E 2 (R , G ) sin R dR dG = 2Q

0

⎡ 2(1 + cos Z ) sin R dR = 4Q ⎢ ⎣

∫

Q 0

∫

Q 0

⎛Z ⎞ 4 cos2 ⎜ ⎟ sin R dR ⎝2⎠

sin R dR +

∫

Q 0

⎤ cos Z sin R dR ⎥ ⎦

= 4 Q [2 + I1 (say)]

where I1 =

If

∫

Q 0

cos Z sin R dR =

∫

Q 0

cos (C d cos R ) sin R dR

x = C d cos R ⇒ dx = − C d sin R dR or sin R dR = −

dx

Cd

and the corresponding limit varies between bd and –bd. Therefore I1 =

∫

Cd − Cd

1 2 sin C d ⎛ dx ⎞ cos x ⎜ = [ − sin x ]C− Cd d = ⎟= Cd ⎝ Cd ⎠ Cd

Hence,

D=

4Q 4 2 = 2sin C d ⎤ sin C d ⎤ ⎡ ⎡ 4 Q ⎢2 + 1+ ⎥ ⎢ Cd ⎦ ⎣ C d ⎥⎦ ⎣

Similarly, we can find the directivity of end-fire array as

D=

2 sin 2 C d ⎤ ⎡ ⎢1 + 2 C d ⎥ ⎣ ⎦

Hint: In case of end-fire array, a = –bd and y = (bd cos q –ÿ bd). Take x = (bd cos q –ÿ bd), dx = – bd sin q –ÿ dq. The limit varies between 0 and –2bd. Example 4.7 A uniform array consists of 18 isotropic point sources, each separated at distance of (l/4). If the phase difference d = –90°. Calculate (i) HPBW, (ii) solid beam angle, (iii) beam efficiency and directivity, and (iv) effective aperture. Also find the improved directivity using Hansen–Woodyard uniform array approach as well as change in directivity value.

Antenna Array

Solution:

159

(i) HPBW of end-fire is given by 57.3° L/2 M

where the length of array

L = (n − 1)d = (18 − 1) × 57.3°

HPBW =

17 M × 2M 4

=

M 4

57.3 1.457

=

17 M 4

= 39.32°

(ii) Directivity

4L

D=

=

M

4 17 M

M

4

= 17 = 12.30 dB

(iii) Beam solid angle

:=

4 × Q

=

D

4 × 3.14 17

= 0.74 Sr

(iv) Effective aperture Ae =

D × M2 4Q

=

17 × M 2 12.5

= 1.353 M 2 = 1.353 × 3.52 = 16.57 cm 2

where l is 3.5 cm. We also know that

⎡ ⎛ d ⎞⎤ ⎡ ⎛ M ⎞⎤ Dim = 1.789 ⎢ 4 n ⎜ ⎟ ⎥ = 1.789 ⎢ 4 × 18 ⎜ ⎟ ⎥ = 32.02 = 15.07 dB ⎝ 4 M ⎠⎦ ⎣ ⎝ M ⎠⎦ ⎣ Therefore change in directivity DD = 15.07 – 12.30 = 2.77 dB Example 4.8

Find the current in one of dipole in an antenna array, if

(a) No mutual coupling exists between the dipoles. (b) Mutual coupling exists between the dipoles. Assume that the first dipole of array is fed by voltage 100 V and the dipoles are separated by d = l/2. Also, determine the amplitude and the phase changes.

160


Solution:

(a) In case of no coupling, Z12 = 0; hence I1 =

V1 Z11

=

100 73.1 + j 42.5

= 1.183 exp ( − j 30.70°)

(b) For the case where coupling is existing, we obtain [from Fig. 4.20(b)], Z12 = (–12.5 – j29.9) W Hence from Eq. (4.47), we get I1 = 1.2 e–j21.8°A The amplitude of the input current changes slightly while the changes in phase is more significant. Example 4.9 Design an 8-element broadside array of isotropic sources having l/2 spacing between elements. The pattern is to be optimized with a side lobe –25 dB down the minor lobe maximum. Solution:

(i) Side lobe below main lobe is maximum in dB = – 20 log (r) Þ 25 – 20 log (r)

⇒ log r =

25 20

= 1.25 ⇒ r = 17.78 = 18

(ii) n = 8; Tchebyscheff polynomial of degree (n – 1) = 8 – 1 = 7. From the relation T7(x0) = r 64 x07 − 112 x05 + 56 x03 − 7x0 = 18

where x0 can be found from the equation x0 =

=

=

x0 =

1 ⎧⎡ ⎨ r+ 2 ⎩ ⎢⎣ 1 ⎧⎡ ⎨ 18 + 2 ⎩ ⎣⎢

1 2

(r 2 − 1) ⎤ ⎥⎦

1/m

+ ⎡r − ⎢⎣ 1/7

(182 − 1) ⎤ ⎦⎥

(r 2 − 1) ⎤ ⎥⎦

+ ⎡18 − ⎣⎢

⎬ ⎭

(182 − 1) ⎤ ⎦⎥

{[18 + 17.97]1/7 + [18 − 17.97]1/7} =

1 ⎛1 1 ⎞ ⎜ log 35.97 + log 0.03 ⎟ 2 ⎝7 7 ⎠

1/m ⎫

1 2

1/7 ⎫

⎬ ⎭

{[35.97]1/7 + [0.03]1/7}

Antenna Array

161

1 ⎛ 1.556 2.4771 ⎞ 1 + ⎜ ⎟ = (0.222 + 1.7824) 2 ⎝ 7 7 ⎠ 2

=

1

=

2 1

x0 =

2

= (anti-log 0.222 + anti-log 1.7824)

= (1.6680 + 0.6060) = 1.137 = 1.14

The array of 8 elements can be shown as follows:

a3

a2

a1

a0

a0

a1

a2

a3

Each has spacing l/2

Therefore the total E due to this array, i.e., E8 can be obtained as E8 = A0 z + A1 (4z 2 − 3z ) + A2 (16z 5 − 20z 3 − 5z ) + A3 (64 z 7 − 112z 5 + 56z 3 − 7z )

T7(x) = E8 Þ 64 z

7

⎡ ⎛ x ⎞2 ⎛ x ⎞ ⎛ x ⎞⎤ − 112z + 56z − 7z = A0 ⎜ ⎟ + A1 ⎢ 4 ⎜ ⎟ − 3 ⎜ ⎟ ⎥ ⎢ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎥⎦ ⎣ 5

3

⎡ ⎛ x ⎞ + A2 ⎢16 ⎜ ⎟ ⎢ ⎝ x0 ⎠ ⎣

5

⎛ x ⎞ − 20 ⎜ ⎟ ⎝ x0 ⎠

3

⎛ x − 5⎜ ⎝ x0

⎞⎤ ⎟⎥ ⎠ ⎥⎦

5 3 ⎡ ⎛ x ⎞7 ⎛ x ⎞ ⎛ x ⎞ ⎛ x + A3 ⎢64 ⎜ ⎟ − 112 ⎜ ⎟ + 56 ⎜ ⎟ − 7 ⎜ ⎢ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎝ x0 ⎣

Equating coefficient similar terms ⎛ x ⎞ 64x 7 = 64 A3 ⎜ ⎟ ⇒ A3 = (x0 ) 7 = (1.14) 7 ⎝ x0 ⎠

log A3 = 7 log 1.14 Þ A3 = 2.502 5 ⎡ ⎛ x ⎞5 ⎤ ⎛ x ⎞ − 112 x = A2 ⎢16 ⎜ ⎟ − 112 ⎜ ⎟ A3 ⎥ ⎢ ⎝ x0 ⎠ ⎥ ⎝ x0 ⎠ ⎣ ⎦ 5

⎞⎤ ⎟⎥ ⎠ ⎥⎦

162


− 112 x05 = [16 A2 − 112 A3 ] 16 A2 = − 112 [x05 − A3 ] = A2 = 4.039 ⎛ x ⎞ 56 x = 4A1 ⎜ ⎟ ⎝ x0 ⎠ 3

A1 = 1 4

1 4

3

3

3

⎛ x ⎞ ⎛ x ⎞ − 20 A2 ⎜ ⎟ + 56 ⎜ ⎟ A3 ⎝ x0 ⎠ ⎝ x0 ⎠

[56(x0 )3 + 20 A2 − 56 A3 ]

[56(1.14)3 + 20 ´ 4.039 – 56 ´ 2.502] = 5.9085

⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ 7x = − A0 ⎜ ⎟ − 3 A1 ⎜ ⎟ + 5 A2 ⎜ ⎟ − 7 A2 ⎜ ⎟ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎝ x0 ⎠ ⎝ x0 ⎠

A0 = 3A1 – 5A2 + 7A3 – 7x0 = 3 ´ 5.9085 – 5 ´ 4.039 + 7 ´ 2.502 – 7 ´ 1.14 = 7.0645 Thus it is clear that in array centre elements have maximum current amplitude while last ones have minimum current amplitudes. And relative amplitudes will be 1.0 : 1.614 : 2.362 : 2.823 : 2.823 : 2.362 : 1.614 : 1.0 Thus tapering of current amplitudes start symmetrically on either sides of the centre source, hence Dalph–Tchebyscheff gives optimum pattern. We know that

⎡M ⎧⎪ 1 ⎛ cosh −1 (r/ 2) ⎞ ⎫⎪⎤ HPBW = 2 sin −1 ⎢ cos−1 ⎨ cosh ⎜ ⎟ ⎥ ⎜ m − 1 ⎟ ⎬⎥ x ⎢Q d ⎪ ⎝ ⎠ ⎪⎭⎦ ⎩ ⎣ Let ⎛ cosh −1 (r/ 2) ⎞ ⎧⎪⎛ cosh −1 (18 × 0.707) ⎞ ⎫⎪ 1 y=⎜ ⎟ = ⎨⎜ ⎟ ⎬ = cosh −1 (12.73) ⎜ ⎟ ⎜ ⎟ − 1 7 m ⎠ ⎪⎭ 7 ⎝ ⎠ ⎪⎩⎝

= Þ

1⎡ 1 log (12.73 + 12.732 − 1) ⎤ = log (25.416) ⎢ ⎥ ⎣ ⎦ 7 7

y = 0.2007

Hence cosh(y) = 1 +

y2 12

+

y4 14

± ... = 1 + 0.0204 = 1.020

Antenna Array

163

⎡ M ⎛ 1.020 ⎞ ⎤ ⎤ −1 ⎡ 2 HPBW = 2 sin −1 ⎢ cos−1 ⎜ ⎟ ⎥ = 2 sin ⎢ × 26.5°⎥ = 34.26° ⎝ 1.14 ⎠ ⎦ ⎣Q ⎦ ⎣ Q × M /2 The angle at first null maximum occurs ⎡ M

⎛ kQ ⎞ ⎪⎫ ⎤ ⎪⎧ 1 cos−1 ⎨ cos ⎜ ⎟⎬⎥ ⎪⎩ x0 ⎢⎣ Q d ⎝ m − 1 ⎠ ⎪⎭ ⎥⎦

R nm = sin −1 ⎢

⎡2 = sin −1 ⎢ cos−1 ⎢⎣ Q

⎧⎪ 1 ⎛ 1 × Q ⎞ ⎫⎪ ⎤ cos ⎜ ⎨ ⎟ ⎬ ⎥ = 24.82° ⎝ 7 ⎠ ⎪⎭ ⎥⎦ ⎩⎪1.14

Example 4.10 Calculate the directivity of a broadside stacked antenna of height 10.5 m and length 21 m in dB, if operating frequency f = 3.5 GHz. Solution:

We know that D=

12.56

M

h × l=

2

12.56

M

2

10.5 × 21 =

2769.48

M2

where

M= Þ Example 4.11

D=

c f

=

30 3.5

2769.48 8.572

= 8.57 cm

= 37.695 = 15.763 dB

Describe the directivity of Dolph–Tchebyscheff array.

Solution: Directivity of large DT array with side lobes in the range of –20 dB to – 60 dB, is generally defined in term of a factor called beam broaden factor (f). Where the beam broaden factor is given by [2]. ⎡2 ⎤ f = 1 + 0.636 ⎢ cosh { (cosh −1 Rv )2 − Q 2 }⎥ ⎣ Rv ⎦

where Rv is the major to minor lobe voltage ratio. The directivity relates f as follows:

D0 =

2 Rv2 1 + (Rv2 − 1) × f ×

M (L + d )

2

164


where (L + d) is the array length. The beam width of DT array can also be given in terms of D0 as follows: BW3dB =

101.5 D0

in degree. That is, the product of directivity and 3 dB beam width is approximately equal to 100. This is similar to the product of the gain and beam width for electronic amplifier. The above expression can also be considered for most of linear broadside array (see [7]). Example 4.12 Calculate the directivity of DT array antenna, if the pattern is to be optimum at side lobe –20 dB down to the minor lobe and length of array is 4l. Solution: From the question, Rv = 20 The total length of the array is (L + d) = 4l Hence the beam broaden factor ⎡2 ⎤ f = 1 + 0.636 ⎢ cosh { (cosh −1 20)2 − Q 2 }⎥ 20 ⎣ ⎦

2

2

⎡1 ⎤ ⎡1 ⎤ = 1 + 0.636 ⎢ cosh { (3.69)2 − 3.142 }⎥ = 1 + 0.636 ⎢ × 3.52 ⎥ ⎣10 ⎦ ⎣10 ⎦

2

f = 1 + 0.079 = 1.079 Therefore the directivity D0 =

2 × 20 2

M 1 + (22 − 1) × 1.079 × 4M 2

=

800 108.63

= 7.364

= 8.67 dB

Therefore the beam width of the array will be

R3dB = Example 4.13

101.5 D0

=

101.5 7.364

= 13.78°

Show that the directivity of an ordinary end-fire array can be expressed as

D=

n ⎛ M ⎞ n =1 ⎡ n − k ⎤ ⎛ 4Q kd ⎞ 1+ ⎜ sin ⎜ ⎟ ∑ k =1 ⎢ ⎟ ⎥ ⎝ 2nd ⎠ ⎣ k ⎦ ⎝ M ⎠

Antenna Array

Solution: area

165

If the spacing between array elements is uniform and constant (d, say), the beam

:A =

1 n2

∫

−2 Q

Q

∫

0

⎡ ⎛ nZ ⎞ ⎤ ⎢ sin ⎜ 2 ⎟ ⎥ ⎠⎥ ⎢ ⎝ ⎢ ⎛Z ⎞ ⎥ ⎢ sin ⎜ ⎟ ⎥ ⎣⎢ ⎝ 2 ⎠ ⎦⎥

0

2

sin R dR dG

where q is angle form the array axis. Since array pattern is not function of f, hence above equation reduces to 2Q

:A =

n

∫

2

⎡ ⎛ nZ ⎞ ⎤ ⎢ sin ⎜ 2 ⎟ ⎥ ⎠⎥ ⎢ ⎝ ⎢ ⎛Z ⎞ ⎥ ⎢ sin ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦

Q 0

2

sin R dR

in which

Z 2

=

Qd 2

(cos R − 1) ⇒

= sin R dR =

dZ 2

=

Qd M

sin R dR

M dZ Qd 2

Hence above equation reduces to

:A =

2M 2

n d

∫

Q 0

⎡ ⎛ nZ ⎞ ⎤ ⎢ sin ⎜ 2 ⎟ ⎥ ⎠⎥ ⎢ ⎝ ⎢ ⎛Z ⎞ ⎥ ⎢ sin ⎜ ⎟ ⎥ ⎣⎢ ⎝ 2 ⎠ ⎦⎥

2

dZ 2

Further, let

Z 2

=R ⇒

Z 2

= 0, if R = 0 and

Z 2

=

2Q d

M

, if R = Q

Also ⎛ nZ ⎞ ⎤ ⎡ ⎢ sin ⎜ 2 ⎟ ⎥ ⎝ ⎠⎥ ⎢ ⎢ ⎛Z ⎞ ⎥ ⎢ sin ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦

2 n =1 ⎛ 2kZ ⎞ = n + ∑ 2(n = k ) cos ⎜ ⎟ k =1 ⎝ 2 ⎠

166


Therefore the value of WA reduces to

:A =

2M 2

n d

∫

2Q d/M 0

n =1 ⎡ ⎢ n + ∑ 2(n = k ) cos k =1 ⎣

⎛ 2kZ ⎜ ⎝ 2

n =1 2(n − k ) 2M ⎡ nZ ⎛ 2kZ = 2 ⎢ + ∑ sin ⎜ 2k n d ⎣ 2 k =1 ⎝ 2

=

⎞ ⎤ dZ ⎟⎥ ⎠⎦ 2

2Q d/ M

⎞⎤ ⎟⎥ ⎠⎦ 0

n =1 (n − k ) 2M ⎡ 2Q nd ⎛ 2Q d ⎞ ⎤ ∑ + sin ⎜ 2k ⎢ ⎟⎥ 2 k M ⎠⎦ n d ⎣ M k =1 ⎝

The directivity D=

Þ

Example 4.14

D=

4Q :A

=

2n2 dQ /M 2 Q nd ⎡n − k ⎤ ⎛ 2Q d ⎞ + ∑ nk =1 sin ⎜ 2 k =1 ⎢ ⎥ M M ⎟⎠ ⎣ k ⎦ ⎝

n M (n − k ) ⎛ 2Q d ⎞ ∑ nk =1 1+ sin ⎜ 2k =1 M ⎟⎠ 2 Q nd 2 ⎝

Find HPBW and directivity of a 17-element binomial array.

Solution: HPBW =

1.06 17 − 1

=

1.06 4

= 0.253 rad = 15.2°

and directivity D0 = 1.77 17 = 7.3 = 8.64 dB

Example 4.15 Using the principle of pattern multiplication, describe radiation characteristics of binomial array antenna. Solution: We know that relative far-field pattern of two-point sources of same amplitude and phase is given by ⎛Q ⎞ En = cos ⎜ cos R ⎟ ⎝2 ⎠

If another identical array is superimposed on this array, the resultant relative far-field pattern ⎛Q ⎞ (using pattern multification) is En = cos2 ⎜ cos R ⎟ . This arrangement doubles the current ⎝2 ⎠

Antenna Array

167

amplitude of the array at centre than that at the edges. In other words the array has the three effective sources with current amplitudes ratio 1:2:1. Similarly if same procedure is repeated for an array of three sources array, then an array of four effective sources with current amplitude in ratio 1:3:3:1 is obtained. Both the above arrays have no minor lobes in radiation pattern. The total far-field pattern of array will be as shown in Fig. 4.19(a) and (b). From the above examples, it is clear that current amplitudes of the array are according to the binomial coefficients. Therefore we can design binomial array of n-sources without minor lobes for any desired directivity using principle of pattern multiplication and superimposing sources on the others. That is, the current amplitudes should correspond to the coefficient of binomial expansion. The far-field pattern of n-sources binomial array therefore will be ⎛Q ⎞ En = cos−1 ⎜ cos R ⎟ 2 ⎝ ⎠

Example 4.16 Derive the relation between length of n-elements array and its directivity. Show that directivity of end-fire array is twice that of broadside array. Solution:

We know that the directivity is defined as D=

Maximum radiation intensity of AUT Radiation intensity of isotropic antenna

=

Gmax G0

In general, the radiation intensity is defined as ⎡ sin z ⎤ G (R ) = ⎢ ⎥ ⎣ z ⎦

2

where

z=

nC d 2

cos R

i.e. f(q) = fmax =1 at q = 90°

In particular, for broadband array

G (R ) = G0 =

1 nC d

∫

∞ −∞

⎡ sin z ⎤ ⎢ ⎥ ⎣ z ⎦

2

dz =

Q nC d

Therefore Db =

G max nC d d L⎞ d ⎛ ⎛L⎞ d = = 2n = 2 ⎜ 1 + ⎟ ≈ 2⎜ ⎟ G0 Q M d⎠ M ⎝ ⎝d⎠ M

Db = 2

L

M

where L is the length of the array and

168


L = (n – 1)d Similarly, for end-fire array

G (R ) = G0 =

1

∫

nC d

∞ −0

⎡ sin z ⎤ ⎢ ⎥ ⎣ z ⎦

2

dz =

Q 2 nC d

Therefore De =

Gmax 2 nC d d L⎞ d ⎛ ⎛L⎞ d ≈ 4⎜ ⎟ = = 4 n = 4 ⎜1 + ⎟ G0 Q M d⎠ M ⎝ ⎝d⎠ M

=4

L

M

Therefore De = 2Db. Example 4.17

Show that the maximum of minor lobe of the n-array factor: (AF) n =

1 sin [nZ /2] n sin [Z /2]

is 13.46 dB down from the maximum at the major lobe in a linear array of uniform amplitude, phase and spacing. Solution:

We know that the maximum of first minor lobe occurs when ny/2 = 1. Therefore,

n

Z

n

Z

2

= ± (2n + 1)

Q 2

3Q ⎛ C d cos R + B ⎞ = n⎜ ⎟= ± 2 2 2 ⎝ ⎠

(AF) n =

⎛ Z⎞ sin ⎜ n ⎟ ⎝ 2⎠ n

Z

2

=

Q⎞ ⎛ sin ⎜ ± 3 ⎟ 2⎠ ⎝ ±3

Q

=

2Q 3

= 0.212

2

AFn = 20 log10 (0.212) = –13.46 dB i.e., maximum of first minor lobe of the array factor is 13.46 dB down from the maximum at major lobe. Example 4.18 Find the percentage change in directivity of a 12-element array in the cases of broadside and end-fire array configurations.

Antenna Array

169

Solution: n = 12 Db = 2 n Db = 4 n

d

M d

M

=

=

2 × 12 × M 2M 4 × 12 × M 2M

= 12 = 10.8 dB

= 24 = 13.8 dB

DD = Db – De = 13.8 – 10.8 = 3.002 dB % change = 3.002 ´ 100/13.8 = 21.75% w.r.t. De % change = 3.002 ´ 100/10.8 = 27.85% w.r.t. Db Example 4.19 Three isotropic point sources with spacing l/4 between them are placed along the x-axis. The excitation coefficient of each outside element is unity while that of centre element is 2. Find the resultant array factor and show that (i) No nulls exist on the pattern between 0° £ q £ 180°. (ii) Only one maximum exists at q = 90° on the pattern 0° £ÿ q £ 180°. Solution: As per question, arrangement is shown in Fig. 4.25. If E1, E2 and E3 are the amplitudes of corresponding radiation field, then the resultant field ET = E1 + E2 + E3. ET = E0

exp ( − jkr1 ) r1

+ 2E0

exp ( − jkr ) r

+ E0

exp ( − jkr2 ) r2

in which for the far-field region, r1 = r2 = r for amplitude variation, whereas for phasedistribution (from Fig. 4.25) r1 = r – d cos q r2 = r + d cos q

FIG. 4.25

3-element array placed along x-axis.

170


Therefore

ET = E0 = E0

exp( − jkr ) r exp( − jkr ) r

[2 + e jkd cos R + e− jkd cos R ]

[2 + 2 cos(x )]

where x = kd cos q

= 2E 0

exp ( − jkr ) r

[1 + cos(x )]

exp( − jkr )

ET = 2E0

r

2 cos2 (x/2)

Therefore ⎛ kd cos R ⎞ (AF)n = [1 + cos(kd cos R )] = 2 cos2 ⎜ ⎟ 2 ⎝ ⎠

If

k=

2Q

and d =

M

M 4

Hence

kd cos R 2

=

2Q 2M

,

M Q cos R = cos R 4 4

⎛Q ⎞ (AF) n = 2 cos2 ⎜ cos R ⎟ ⎝4 ⎠

(i) If qn are the positions of null on the pattern, then ⎛Q ⎞ (AF) n = 0 = 2 cos2 ⎜ cos R ⎟ ⎝4 ⎠

Þ or

Q 4

cos R n = cos−1 (0) =

nQ 2

ÿ ÿ ÿ qn = cos–1 (2n)

where, n = ± 1 ± 3 ± 5, ... Since for none of the values of n, the value of qn satisfies the cosine function, hence no null exists.

Antenna Array

171

(ii) Similarly, maximum value occurs at angle qm, if ⎛Q ⎞ 2 cos2 ⎜ cos R m ⎟ = maximum ⎝4 ⎠ ⎛Q ⎞ cos ⎜ cos R m ⎟ = ± 1 ⎝4 ⎠ ⎛Q ⎞ m = 0, ±1, ±2, ±3 = ⎜ cos R m ⎟ = cos−1 ( ± 1) = mQ 4 ⎝ ⎠ qm = cos–1(4m) So, if m = 0, q0 = cos–1(4.0) = 90°

No other value of m satisfies the cosine function, therefore only one maximum exists, on the radiation pattern 0° £ q £ 180°. Example 4.20 Design a 4-element ordinary end-fire array of isotropic sources, positioned along the x axis such that spacing between elements are d and its only one maximum occurs at q0 = 0°. Assuming d = l/2, find (a) (b) (c) (d)

Progressive phase excitation between elements Angle where nulls occur Angle where maximum of array factor occurs FNBW and directivity (dB).

Solution: (a) B = − C d =

2Q M

M

2

= Q = 180°

M ⎞ M ⎞ N⎞ ⎛ −1 ⎛ −1 ⎛ (b) R n = cos−1 ⎜ 1 − N ⎟ = cos ⎜ 1 − N ⎟ = cos ⎜ 1 − ⎟ nd ⎠ 4M /2 ⎠ 2⎠ ⎝ ⎝ ⎝ where, N = 1, 2, 3, … Therefore, for N = 1 N = 2 N = 3

q1 = cos–1(1/2) = 60° q1 = cos–1(0) = 90° q1 = cos–1(–1/2) = 120°

(c) qm = cos–1(1 – ml/nd) = cos–1(1 – 2m) where m = 0, 1, 2, …

172


Therefore, for m = 0, q0 = cos–1(1) = 0° m = 1, q0 = cos–1(–1) = 180° (d) FNBW = 2 cos–1(1 – l/nd) = 2 cos–1 (1 – l/(4 ´ 0.5l)) = 2 cos–1(1/2) = 2 ´ 60° = 120° D = 4n(d/l) = 4 ´ 4(l/2l) = 8 = 9.03 dB Example 4.21 Show that in order to have no minor lobe for a uniform array of n-elements the spacing and progressive phase shift between elements must be: (a) D = l/n; b = 0 for a broadside array (b) D = l/2n; bÿ = ± kd for an end-fire array Solution:

We know that for an array of n elements: (AF)n =

Z =

where

1 sin[nZ /2] n sin[Z /2]

C d cos R 2

C d cos R

+B =

2

as a = 0, for broadside array. (a)

In order to have no minor lobes, first null should occur at q = 0° or 180°, thus (AF) n =

nC d 2 (b)

1 sin [nZ /2] n sin [Z /2]

=Q ⇒ d =

=0

M n

Similarly, for end-fire array

Z =

C d cos R + C d 2

=

Cd 2

(cos R + 1)

⎧ Cd ⎫ sin n ⎨ (cos R + 1) ⎬ 2 ⎩ ⎭ (AF) n = ⎧Cd ⎫ (cos R + 1) ⎬ ⎨ ⎩ 2 ⎭

So, if ⎧ Cd ⎫ sin n ⎨ (cos R + 1) ⎬ = 0 ⎩ 2 ⎭

Antenna Array

173

nC d (cos R + 1) = Q 2

Þ

2nC d =Q 2 which gives d = l/2n. Example 4.22 In order to suppress grating lobes from a linear scanning array the maximum spacing between elements is as follows: d=

M 1 + cos R m

where q m is deviation of maximum radiation. What will be maximum spacing between element without lobes at f = 3.0 GHz, when the array is designed to scan a maximum angle of 30°. Solution:

l = 10 cm d=

M 1 + cos R m

, R m = scan angle

at qm = 30°

d=

M 1 + cos 30°

= 0.5359 M = 5.4 cm

Example 4.23 Show that a three-element binomial array with a spacing of d £ l/2 between the elements does not have a side lobe. Solution: The excitation coefficient of a thee-element binomial array will be 1:2:1. The normalized array factor of the array will be: ⎛ C d cos R ⎞ AF3 = cos2 ⎜ ⎟ 2 ⎝ ⎠

Hence it is clear that its maxima occur at q = 90°. In order to have no side lobe, the phase factor ⎛R ⎞ ⎝2⎠

C d cos ⎜ ⎟ at R = 0° or 180° must be equal to or less than p/2. Thus,

174


C d cos R 2

≤

Q

at q = 0°, 180°

2

2pd/2lÿ £ p/2 2Q 2M

d ≤

d ≤

Þ

Q 2

Q 2

Example 4.24 The total length of DT array is 4l. For a 30 dB side lobe level design and spacing of (l/2) between the elements along the array axis, find (i) number of elements, (ii) directivity and (iii) half-power beam width. Solution:

(i)

N = 2n + 1 = 9

as n = 4

R0 = 30 dB Þ R0 = 101.5 = 31.6222 dB ⎡2 ⎤ F = 1 + 0.636 ⎢ 662 cosh (cosh) −1 (31.662)2 − Q 2 ⎥ ⎣ 31 ⎦

(ii)

⎡ 2 ⎤ = 1 + 0.636 ⎢ (7.54) ⎥ 31.662 ⎣ ⎦

2

2

= 1 + 0.144 = 1.144 Hence

D=

=

2 R02

M ⎤ ⎡ 2 ⎢1 + (R0 − 1) f L + d ⎥ ⎣ ⎦ 2 (31.662)2 1 ⎤ ⎡ 2 ⎢1 + ((31.662) − 1) × 1.144 × 4.5 ⎥ ⎣ ⎦

=

2004.96 255.559

= 7.844

= 8.945 dB

⎡ M ⎞ M ⎞⎤ ⎛ −1 ⎛ (iii) HPBW = f ⎢cos−1 ⎜ cos R 0 − 0.443 ⎟ − cos ⎜ cos R + 0.443 ⎟⎥ L +d⎠ L + d ⎠⎦ ⎝ ⎝ ⎣

Antenna Array

175

⎡ M ⎞ 1 ⎞⎤ ⎛ −1 ⎛ = 1.144 ⎢cos−1 ⎜ − 0.443 ⎟ − cos ⎜ 0.443 ⎟ ⎥ at R = 90° L +d⎠ 4.5 ⎠ ⎦ ⎝ ⎝ ⎣ = 1.44(95.65 – 84.35] = 12.93° Using direct formula, BW3dB = 101.5/D = 101.5/7.844 = 12.93°, which is the same as above.

OBJECTIVE TYPE QUESTIONS 1. The essential condition for an array to be linear is that (a) Elements should be of equal length (b) Elements should be fed with equal current (c) Elements should be equally spaced (d) None of these. 2. The total field produced by an array is (a) Vector sum of fields produced by individual elements (b) Sum of fields produced by individual elements (c) Sum of vector fields produced by first three elements (d) None of these. 3. The essential condition for an array to be uniform linear is (a) Elements should be of equal length (b) Element should be fed with a current of equal magnitude and uniform progressive phase shift. (c) Elements should be equally spaced and opposite phased (d) None of these. 4. In multi-elements array, variation in electrical length from l/2 within ________ does not affect the radiating properties. (a) 10% (b) 5% (c) 15% (d) 1.3% 5. Bi-directional properties of broadside array can be converted into unidirectional if an identical array exciting by current leading in phase 90o is placed (a) Behind the array at l/2 distance (b) Front of the array at l/2 distance (c) Behind the array at l/4 distance (d) None of these. 6. The variation between lengths of reflector, driven element and director of 3-element parasitic array is (a) 10% (b) 5% (c) 5% (d) 1.3%

176


7. In practice parasitic arrays are successfully used in frequency range of 100 MHz– 1000 MHz. This antenna also known as: (a) Yagi–Uda antenna (b) Log antenna (c) Folded dipole antenna (d) None of these. 8. A broadside couplet is formed if two isotropic radiators operates in phase, whereas end-fire couplet is formed if two equal radiators are operated (a) In-phase quadrature at a distance of l/2 (b) In-phase quadrature at a distance of l (c) In-phase quadrature at a distance of l/4 (d) None of these. 9. Which statement is incorrect for collinear array? (a) Its other name is broadcast array (b) Gain is maximum when d = 0.3 l to 0.5 l. (c) Two elements collinear array known as collinear couplet. (d) Power gain does not increase with number of elements after four (04). 10. The directivity of a 10-element uniform linear end-fire array with separation of l/4 is (a) 10 dB (b) 20 dB (c) 0 dB (d) 2 dB 11. The directivity of a collinear array increases with (a) Increase in the length of the array (b) Decrease in the length of the array (c) Increase in the size of elements (d) None of these. 12. The FNBW for 20-element broadside array with separation l/4 is (a) 12.92° (b) 20.5° (c) 10.58° (d) 22.92° 13. This antenna array has the largest beam width: (a) Edge array (b) Uniform array (c) Optimum array (d) Binomial array 14. In electronic phased array, the direction of maximum radiation is controlled by (a) Controlling the progressive phase difference between the elements (b) Controlling the current amplitudes of the elements (c) Tapering the array (d) None of these. 15. Tapering is a technique in which ________ of the array is controlled. (a) FNBW (b) HPBW (c) Side lobe (d) Spillover 16. Main constraint with DT array is that it increases beam width at the cost of (a) Radiation pattern (b) Efficiency (c) Directivity (d) None of these

Antenna Array

17. Broadside rectangular array is a (a) Broadside array (c) Parasitic array

177

(b) End-fire array (d) None of these

18. The directivity of a broadside rectangular array of height h = 20l and width b = 12l is (a) 40.5 dB (b) 50 dB (c) 5 dB (d) None of these 19. Super directivity is accomplished by inserting a number of elements within a fixed length of array (a) True (b) False (c) Partially true (d) None of these 20. In mobile communication adaptive antenna is used mainly to reduce (a) Size of handset (b) Co-channel interference (c) SNR (d) Scattering power loss

Answers 1. 6. 11. 16.

(c) (c) (a) (c)

2. 7. 12. 17.

(a) (a) (d) (a)

3. 8. 13. 18.

(b) (c) (d) (d)

4. 9. 14. 19.

(c) (c) (a) (a)

5. 10. 15. 20.

(c) (a) (c) (b)

EXERCISES 1. What are the advantages of array antenna? Describing principle of pattern multiplication and sketch the radiation pattern of a three-element array separated at l/2. 2. Show that the directivity for an end-fire array of two identical isotropic in phase point sources separated at distance d is given by D (R , G ) =

2 sin 2C d 1+ 2 Cd

3. Design a 5-element broadside array of isotropic sources separated at l/2. The pattern is to be optimized with a side lobe –15 dB down the minor lobe maximum. 4. Calculate the directivity and effective area of DT array antenna operating at l = 3 cm, if the pattern is to be optimized a side lobe –15 dB down to the minor lobe. 5. Find HPBW, directivity, effective area and gain of a 15-element binomial array. Assume that array efficiency is 92%.

178


6. A uniform array consists of 20 isotropic point sources such that elements are l/2 away and fed in same phase. Assuming operating wavelength of 2.5 cm, find FNBW, WA, directivity and effective area. 7. Find the change in directivity of a 10-element array antenna when they are arranged in (a) broadside and (b) end-fire configurations. 8. A broadside array consisting of several l/2-long isotropic radiators is used to have a directive gain of 30 dB. Estimate array specifications. What will be these values for end-fire array? 9. Derive an expression for beam width of a uniform linear array between first nulls and also determine its value for n = 5 to 8 elements, if the spacing between elements is l/4. 10. What do you mean by electronic phased array antenna? Find the beam width of primary lobes of 4 and 10 elements in case of (a) broadside array, (b) end-fire array. Comment on the array with reference to its directivity. 11. Find the FNBW and HPBW for end-fire and broadside linear array consisting of 20 Hertzian dipoles with element spacing l/4 and l/2 respectively. 12. Distinguish between end-fire, broadside, parasitic and collinear arrays. Show that array of two isotropic sources fed with equal amplitudes and opposite phases acts as an end-fire array. 13. Find the location of the first nulls on the either side of the centre beam for a linear array of 60° in phase elements spaced at l/2 and fed with equal amplitude current. 14. What are the advantages of Dolph–Tchebyscheff array antenna? Write the expression for its design parameters. 15. Derive the expression of directivity for the length of n-elements broadside array. 16. Design broadside and end-fire array to be used for 30 MHz communication where 30 dB directive gains are needed for proper communication. 17. Describe stacked array antenna. Show that the directivity of a broadside rectangular stacked array is D=

12.56 A

M2

where A is the area of array. 18. Describe the principle of operation of super directive antenna. Show that its SNR is proportional to the directive gain and independent from the efficiency. 19. A 4-element broadside array, where elements are spaced at l/2, is operating at 250 MHz such that each element carries current in the same phase and of 0.5 A amplitude. Find its HPBW, FNBW and power radiating from the array.

Antenna Array

179

20. Describe the design procedure of a binomial array with suitable example. Also mention its disadvantages. Find the directivity and gain of a 5-element binomial array. 21. Sketch the radiation pattern of 3-element isotropic point array in end-fire and broadside array configurations. 22. What is tapering? Find the HPBW and directivity of a 19-element binomial array? 23. Find the half power bandwidth of 10-element binomial array with a spacing of l/2 between the elements. Also find the change in maximum directivity if the value obtained, assuming the array factor is 7.32 dB. 24. Describe the effect of earth on the radiation pattern of the antenna. Mention the merit of image principle in comparison to simple array method. Also sketch the radiation pattern of vertical and horizontal antennas, if they are at height h which is a multiple of l/2. 25. Design a two-element array of isotropic point sources positioned along x-axis such that spacing between elements are l/4 and its only one maxima occurs at q = 0°. Assuming end-fire condition, find: (i) Array factor of the array; (ii) Relative phase excitation of each element. 26. Design an ordinary end-fire uniform linear array with only one maximum so that its directivity is 20 dBi. If the length of array is much greater than the spacing (l/4). Find: (a) Total length of array (l) (b) HPBW (degree) and progressive phase shift. (c) Amplitude level (compared to maximum of the major lobe) of first minor lobe (dB). 27. Find the beam width and directivity of a 10-element uniform array of isotropic sources placed along the x-axis, if the spacing between the elements is l/4 and the maximum is directed at 45° from its axis. 28. Design a five-element –50 dB side lobe level DT array of isotropic elements, if spacing between elements are l/4. Also find: (a) array factor; (b) directivity and (c) HPBW. 29. Describe the stacked antenna array. Estimate the directivity of a broadside stacked array of height 10 m and length 20 m, at operating frequency f = 3.2 GHz. 30. Write short notes on the following: (a) SMART antenna (b) Mutual coupling between array antennas (c) Sensitive factor of super directive array.

180


REFERENCES [1] Prasad, K.D., Antenna and Wave Propagation, Satya Prakasan, New Delhi, 1996. [2] Balanis, C.A., Antennas, John Wiley & Sons, N.Y., 2001. [3] Kraus, J.D., Antennas for All Applications, 3rd ed., Tata McGraw-Hill, New Delhi, 2003. [4] Newman, E.H., et al., “Super directive receiving arrays,” IEEE Trans. Antennas and Propagate, AP. 26, No. 5, pp. 629–635, Sept. 1978. [5] M.M. Dagwood and A.P. Anderson, “Design of super directive array with high radiation efficiency”, IEEE Trans. Antennas and Propagate, AP. 26, No. 6, pp. 819–823, Nov. 1978. [6] Meghan, J.T. et al., “Wideband adaptive antenna nulling using tapped delay lines”, Tech Note 1979–45, Lincoln Laboratory, MIT, Lexington, Mass, June 26, 1979. [7] R.S. Elliott, “Beam width and directivity of large scanning arrays,” Parts I and II, Microwave Journal, pp. 53–60, Dec. 1963.

C H A P T E R

5

Linear Wire Antennas

INTRODUCTION In previous chapters, we have paid attention on the fundamental parameters of antenna, feeding techniques, impedance matching as well as the design of antenna and antenna arrays. This chapter is going to be the first, which will deal with antenna directly, specially wire antennas. As far as wire antenna, in general, is concerned, it is the simplest, cheapest and most prevalent and versatile antenna, and is useful for many applications. Wire antennas can be constructed from either solid wire or tubular conductors. They are resonant antennas, i.e. input reactance is zero at resonance. Examples of wire antenna are: long wire antenna, folded dipole antenna, V-dipole, rhombic, and Yagi–Uda antennas. Loop antennas are special form of wire antennas. In order to obtain complete and accurate solution for wire antennas, current on it must be solved, subject to boundary condition, which results the tangential component of E is zero along the wire. This approach gives rise to integral equation, whose solution is complex. In addition, this approach restricted to few wire antennas only. However the latest numerical methods, full wave analysis, finite difference time domain and finite elements method are simple and suitable for all kinds of wire antennas. The moment method is a rather simple and conceptual approach to solve and analyze the properties of small wire antennas.

Small Dipole and Radiation Mechanism As the name suggests, dipole means two opposite charges at a finite distance. So, before considering the study of any particular dipole (i.e. l/4, l/2 or full wave dipole), let us first understand the mechanism by which electric lines of force are generated and then detached from a dipole antenna to form the free-space waves. In order to give better physical interpretation of detachment of the field lines, we can assume a dipole carrying sinusoidal current distribution, which is a good approximation and verified experimentally. If so the current must, of course, be zero at the ends. We have been using effectively the current distribution, which found on 181

182


an open-circuited parallel wire lines. If such transmission lines are bent out to form the wire antennas, the current distribution essentially unchanged. But practically it is not strictly true; it may be only good approximation for the thin wire antennas, whose diameter is £ 0.01l [1]. In order to give a better description of radiation from a dipole antenna, let us consider a centre fed small dipole of two opposite charges having maximum separation between charges l0 (Instantaneous separation l), oscillating up and down in harmonic motion which focusing attention of electric field [2]. At the initial time t = 0, the charges are at distance l0 and undergo maximum acceleration say (v¢) as they reverse their direction, therefore instantaneous current I is zero. Later at a time T/8, the charges start moving toward each other and, they pass the midpoint at T/4 period (Fig. 5.1(a)–(c)). As soon this happens the field lines detach and new field lines of opposite nature are generated, producing maximum current ‘I’ and zero charge acceleration. As time further progresses (i.e., 3T/8 period), charges start moving opposite direction, creating additional (opposite) field lines (see Fig. 5.1(d)), and finally at time-period T/2 charges arrived at their original position (i.e., maximum separation l0, current I is again zero and acceleration is maximum) completing one cycle. That is, two equal charges of opposite nature oscillating up and down in harmonic motion generated electric field lines. If this process continues, several field lines move radically outward, detached from the antenna and form electromagnetic free space waves, which in turn called radiation. Electric field lines producing radiation from a l/2 dipole is shown in Fig. 5.2.

FIG. 5.1

Electric field lines and its detachment from dipole antenna.


FIG. 5.2

183

Electric field lines moving from half wave-dipole antenna.

The light is EM waves and both travel by the wave disturbance of the same speed. — Maxwell (1873)

HERTZIAN DIPOLE In general, an infinitesimal current carrying element termed as Hertzian dipole. Although practically it is not possible to have such a dipole, but it is interesting to examine the properties of these dipoles. Hertzian dipole is very useful to calculate the field of a large wire antenna. This can be done by considering a long wire antenna as a combination of a large number of Hertzian dipoles connected in series. Let us consider a Hertzian dipole of length dl(<
FIG. 5.3

Hertzian dipole in 3-D co-ordinate system.

184


As the diameter of dipole is very small compared to its length, it is omitted in analysis. If the dipole carries a uniform constant current I = I0 cos w t, the retarded magnetic vector potential A at a far-field point P due to dipole is given by

A=

N[I ] dl aˆ z 4Q r

(5.1)

where aˆ z is a unit vector as dipole is situated along the z-axis. with [I] = retarded current and given by I = I0 cos w(t – r/v) I = I cos(wt – br) = Re [Iej(wt–br)] 0

in which

C

=

X 2Q v M

and v is velocity of waves =

1

NF

(5.2)

.

[] is added to indicate that it is retarded current. (t – r/v) is retarded time as the phase of the wave at point P is retarded w.r.t. the phase of the current in the element by an angle (wr/v). Equation (5.2) implies that the disturbance at time t at distant point P (at distant r from original) from the element is caused by a retarded current [I] that occurred at an earlier time (t – r/v). The time-difference (r/c) is the time needed by the disturbance to travel the distance (r) at the velocity of EM wave, i.e., velocity of light c. The corresponding retarded current density [J] can be defined as J = J e j(wt–br) A/m2 m

Since potential vector A is acting in z-direction, it will have only z-component, i.e., Az. Thus, we may write A in the phasor form as follows: Az =

N I 0 dl − jCr e 4Qr

Transforming this vector from Cartesian to spherical co-ordinate using ⎡ Ar ⎤ ⎡ sin R cos G ⎢ ⎥ ⎢ ⎢ AR ⎥ = ⎢ cos R cos G ⎢ ⎥ ⎢ ⎣⎢ AG ⎦⎥ ⎣⎢ − sin R

sin R sin G cos R sin G cos G

cos R ⎤ ⎥ − sin R ⎥ ⎥ 0 ⎦⎥

⎡ Ax ⎤ ⎢ ⎥ ⎢ Ay ⎥ ⎢ ⎥ ⎣⎢ Az ⎦⎥

and considering Ax = 0, Ay = 0 and Az = Az gives We know that

Ar = Az cos q

Aq = –Az sin q and Af = 0 G (by definition) B = NH = ∇ × A

We obtained components of magnetic field (H) as follows: Hr = Hq = 0

(5.3a)


HG =

1⎤ ⎡ jC sin R ⎢ + 2 ⎥ e− jC r 4Q r ⎦ ⎣ r

I 0 dl

Similarly by using Maxwell’s equation, ∇ × H = F

∂E ∂t

185 (5.3b)

= jFX E .

We find electric field components and they are:

and

Er =

II 0 dl cos R 2Q

ER =

⎡ jC II 0 dl 1 j ⎤ − jCr sin R ⎢ + 2 − ⎥ e 2Q Cr3 ⎦ r ⎣ r

⎡1 ⎢ 2 − ⎣r

j ⎤ − jC r ⎥ e Cr 3 ⎦

Ef = 0

(5.4a)

(5.4b) (5.4c)

where I = N /F and defined as characteristic impedance of medium surrounding the dipole. The close observation of expressions for magnetic and electric fields given in Eqs. (5.3) and (5.4) reveals that there are terms varying as 1/r3, 1/r2 and 1/r and (1/r3). The terms vary with (1/r3), known as electrostatic or simply electric field. Since it corresponds to the field of an electric dipole and dominates over other terms in a region very close to Hertzian dipole/ conductor. Both equations have components varying as 1/r2 and they are known as near field or induction field, and it is predictable from the Biot–Savart’s law. Induction field is predominant at point near to dipole only and represents the energy stored in the magnetic field surrounding the current element. This energy is alternatively stored in the field and returned to the source current element during each half-cycle. As well as radiation is concerned this field is of little importance and omitted in radiation field calculations. The term varies as 1/r is known as radiation/distant or far-field, because it is the only term that remains up to the far-zone. This field accounts for the radiation of EM waves from the conductor/dipole. The radiation fields are of great significance at large distance from conductor/dipole. Since the components of the magnetic field are produced by the alternating electric fields and the electric field components arise from the alternating magnetic field and vice versa; the flow of current establishes the local induction fields, whereas the radiation fields exist as sequence of the changing induction field. Just close to the conductor the magnetic field is in phase with the current; whereas the electric field varies in phase with the change on either side of the conductor. In far-field region, the magnetic and electric fields thus have a phase-difference of p/2 radians and are at the right angles to each other in space, i.e., Eq and Hf are in phase in the far-field region. Here we are concerned mainly with far-field varying as 1/r, i.e., HG =

and

ER =

I 0 dl 4Q

sin R (j C e− j Cr )

II 0 dl sin R (j C e− j Cr ) 4Q

(5.5a)

(5.5b)

186


That is, Eq /Hf = h, which is same as in the case of plane wave radiation. The time-average power density, thus obtained

Pav =

1 2

Re ( ER × HG ) =

1 2

Re ( ER HG* ar ) =

1 2

I |HG |2 ar

Therefore, the average radiated power is Prad =

∫

=I

Pav ds =

I 02 C 2 dl 2 32 Q 2

∫

2Q

∫

0

2Q

∫

(I 0 C dl sin R )2

Q

32 Q r

2 2

0

Q 0

I r 2 sin R dR dG

I 2Q 4 ⎛ dl ⎞ sin R dR = I 0 ⎜ ⎟ 3 ⎝M⎠

2

3

as C 2 =

4Q 2

M2

If medium surrounding the dipole is air, h = 120p; then 2

Prad =

I 02 Q × 120 Q ⎡ dl ⎤ 2 2 2 ⎢ ⎥ = 40 Q [dlM ] I 0 M 3 ⎣ ⎦

This power could be considered as the power dissipated in a fictitious resistance Rrad by current I, which is equal to I0 cos w t. That is 2 Prad = I rms Rrad =

1 2

I 02 Rrad

2

⎛ dl ⎞ 2 × 40 Q ⎜ ⎟ I 02 2 ⎝M⎠ 2 ⎛ dl ⎞ = = 80 Q ⎜ ⎟ I 02 ⎝M⎠ 2

Rrad

(5.6)

Thus, the antenna requires large amount of radiation resistance to radiate sufficient amount of power. Since, Hertzian dipole is very small, the radiation resistance Rrad is very small and no feeding lines of such low impedance are available to match the dipole. Also, we have to consider uniform current that current be non-zero at the end-points of the dipole, which is practically impossible because the surrounding medium is not conducting. Hence these are limitations with the Hertzian dipole.

HALF-WAVE DIPOLE ANTENNA As name indicates, it is dipole antenna of length of half-wave length, i.e. l/2, where lÿ is operating wavelength. Half-wave dipole can be considered as the series combination of large numbers of Hertzian dipoles and fields due to l/2 dipole antenna can be found easily by integrating the field of infinitesimal dipole under specified limit, i.e. ±l/4. The basic half-


187

wave dipole antenna fed with two-wire transmission line and its current distribution is shown in Fig. 5.4(a), whereas Fig. 5.4(b) shows the geometry of field calculation due to dipole, where an infinitesimal length dl = dz of the dipole is located at distance z from the centre. The magnetic vector potential at observation point P due to dl = dz carrying current I = I0 cos wt is given by dAz =

N I 0 cos C z dz − jCr1 e 4 Q r1

(5.7)

where r1 = distance of element from the point of observation P and r – z cos q .

FIG. 5.4(a) Half-wave dipole and current distribution. FIG. 5.4(b) Geometry of field calculation. As well as the distance is concerned, we can replace r1 by r, i.e. r1 » r as they are very − jCr large, but not the phasor terms e 1 and e–jbr, as there is a significant difference between terms br1 and br. So, let us replace r1 = r – z cos q in Eq. (5.7), and find out the total magnetic vector potential as follows: Az =

N I0 4Qr

∫

M /4 − M /4

e − jC (r − z cos R ) cos C z dz =

N I 0 − jCr e 2Qr

∫

M /4 − M /4

e j C z cos R cos C z dz

After simplification, this gives − jC r

Az =

N I0 e 2 QC r

⎛Q ⎞ cos ⎜ cos R ⎟ ⎝2 ⎠ 2 sin R

for L = l/2

(5.8)

Similar to Hertzian dipole case (Section 5.2) Using B = mH = Ñ ´ A and Maxwell’s equation Ñ ´ H = jweE The expressions for magnetic and electric fields come as follows:

HG =

jI 0 e

− jCr

2Qr

⎛Q ⎞ cos ⎜ cos R ⎟ ⎝2 ⎠ sin R

(5.9a)

188


and

jI I 0 e

HR =

− jC r

⎛Q ⎞ cos ⎜ cos R ⎟ ⎝2 ⎠

2Qr

(5.9b)

sin R

It is also seen that expressions of Eq and Hf indicate that they are in time phase and orthogonal. Also, the ratio of Eq and Hf is constant and equal to h. In addition, both Eq and Hf contains two terms: first term is constant for particular distance r, whereas second term shows the variation of field with angle q, i.e., it determines the pattern of antenna. So it may be termed as pattern factor and ⎛Q ⎞ cos ⎜ cos R ⎟ 2 ⎝ ⎠ F (R ) = sin R

(5.10)

which is termed normalized electric field pattern of a half-wave dipole antenna. Radiation patterns of a centre fed dipole antenna of different lengths along with half-power beam width are shown in Fig. 5.5(a)–(d), which indicates that HPBW decreases as length of dipole increases. With the similar procedure to Hertzian dipole, the time-average power density are found to be

Prad =

I 8Q r

I 02 2 2

⎛Q ⎞ cos2 ⎜ cos R ⎟ 2 ⎝ ⎠ G ar sin 2R

(5.11)

Hence the time-average radiated power can be determined as follows:

Prad =

=

∫ ∫

Pav ds

2Q 0

∫

Q 0

⎛Q ⎞ I I 02 cos2 ⎜ sin R ⎟ 2 ⎝ 8 Q 2 r 2 sin 2R

⎠

× r 2 sin 2 R dR dG

After simplification, which gives: Prad = 36.56 I 02 and since Prad =

1 2

2 I rms × Rrad , therefore Rrad is defined as Prad =

2 Prad I 02

= 73 W

(Neglecting ohmic loss, that is Pdis and hence Rl = 0.) In order to compare the radiation resistance of these two dipoles let us assume that the length of Hertzian dipole is l/2, we found Rrad = 2 W, which is very much less than 73 W. That is, there is significant increase in the radiation resistance of the half-wave dipole antenna over a Hertzian dipole antenna. Therefore, l/2 dipole antenna is capable of delivering greater amounts of power to space than the Hertzian dipole. As the total input impedance of


FIG. 5.5

189

Radiation patterns of dipole antenna of different lengths (L = l/2 to 3/2 l).

an antenna is represented by Zin = Rin + jXin (where Rin is resistance and Xin reactance), the input impedance of a l/2-dipole antenna can be expressed as Zin = Rin + jXin = Rrad + jXin, where Rrad = Rin for a lossless dipole antenna and reactance Xin is to be 42.5 W for a dipole of length l/2; however, Xin approaches zero if the length of dipole reduced slightly. Typically, for l = 0.485l, Xin is zero and Zin = 73 W. That why, in practice, dipole antenna designed maintaining its length lesser than the actual length. Approximate parameters (length to diameter ratio, thickness and corresponding resonant length) of dipole (for Zin = 73 W and Xin = 0) are listed in Table 5.1. TABLE 5.1 L/D

% Shortening

5000 50 10

2 5 9

Resonant length of dipole for different (L/D) Dipole thickness class Very thin Thin Thick

Resonant length 0.49 l 0.475 l 0.455 l

190


Radiation Resistance and Input Resistance In general, the input impedance Zin of a device is defined as the ratio of the voltage to current at a pair of terminals or the ratio of approximate components of the electric to magnetic fields at a point. The real part of Zin is referred as input resistance Rin and Zin » Rin for a lossless antenna (i.e. Xin is zero) and resulting into the radiation of real power. The radiation resistance Rrad is referred to the maximum current for which some lengths (l = l/4, 3l/4, ..., etc.) does not occur at the input terminals of the device. In order to refer radiation resistance to the input terminal of the antenna, antenna must be lossless Rl = 0, and then the power at the input terminals is equated to the power at the maximum current. There are several formulas to relate/compare the input resistance of the dipoles (Table 5.2). TABLE 5.2 Expressions of input resistance for different range of dipole length (l) l = Length of Dipole 0 < l <

M 4

M 2

Input resistance Rin(W)

M

⎛l⎞ 20 Q ⎜ ⎟ ⎝M⎠

4

< l <

2

2

⎛ Ql ⎞ 24.7 ⎜ ⎟ ⎝M⎠

M 2

2.4

⎛ Ql ⎞ 11.14 ⎜ ⎟ ⎝M⎠

< l < l

4.17

Parameters of a Dipole Antenna Bandwidth As dipoles are resonant type radiating device, therefore, their bandwidth is low. The available data shows that for standard VSWR less than 2.0:1, the bandwidths are 16% and 8% respectively ⎛ L ⎞ ⎛ L ⎞ for ⎜ = 50 ⎟ and ⎜ = 2500 ⎟ at design frequency 300 MHz [3]. In general, the bandwidth ⎝ 2a ⎠ ⎝ 2a ⎠ is directly proportional to thickness of dipole; thicker the dipole wider its bandwidth. The minimum VSWR for thicker dipole occurs at a lower frequency than for the thinner one.

Radiation intensity We know that the radiation intensity U is the power per unit Sr, i.e., it can be expressed in watt per unit solid angle (W Sr–1). That is U=

P 4Q

(5.12)


191

From the definition of Poynting vector, the power radiated per unit surface area can be given as Sr = r2 ×

or

1 2

P 4Q r

2

⇒ Sr × r 2 =

I | HG |2max = r 2 ×

1 2I

P 4Q

=U

| ER |2max = U m

(5.13)

where Um is maximum radiation intensity, i.e. U = Um at q = p/2. Therefore substituting the value of Eq from Eq. (5.9b) gives ER (Q /2) =

and hence

Um =

I 2 I 02 4Q 2

I I 02 120 2 = I 0 = 4.8 I 02 2 8 Q 8Q

(5.14)

Directivity The directivity of a l/2 dipole antenna is defined as D = Hence from Eq. (5.14) D = Hence

4Q × 120I 02 36.56 I 02

8Q

=

120 73.12

4Q U m P

.

= 1.64

Ddb = 2.15 dB

(5.15)

Therefore the directivity of l/2-dipole, i.e. 1.64 is only slightly greater than the directivity of an ideal dipole which is 1.5. That is, the directivity of short dipole (1.5) increases to 1.64, as the length of dipole increases to a half-wavelength. Further increase in the length of dipole increase the directivity of dipole and a full wavelength dipole has directivity of 2.41. Increase in directivity continues till dipole length approaches 1.25l, and further increase in the length of dipole deteriorates the value of directivity. Maximum effective area As usual maximum effective area of l/2 dipole antenna is defined Ae =

M2 M 2 × 1.64 D0 = = 0.13l2 4Q 12.56

(5.16)

Radiation efficiency In the previous section, we have found that the power radiated by a l/2-dipole antenna is Pr = 1/2 I 02 Rr , in which power dissipation was neglected. Therefore taking into account ohmic resistance of dipole, we can define the power delivered to the dipole as follows:

192


Pr =

1 2

I02 ( Rr + Rl )

(5.17)

where Rl is ohmic resistance of the dipole corresponding to the maximum current. In general, the radiation efficiency is the ratio of power radiated to the total power supplied to the antenna at a given frequency [4], therefore Pr

I=

1 2 I 0 Rr 2

=

Pd

1 2 I 0 (Rr + Rl ) 2

Rr

=

(Rr + Rl )

(5.18)

The power dissipated by the dipole can be given as

Pdis =

1 2

Rl I 02

(5.19)

If I(z) is assumed to symmetrical current distribution of dipole, Pdis may also be defined as [5], Pdis =

=

∫

M /4 − M /4

I z2 Rs dz =

Rs I 02 2

∫

M /4 0

2 sin C z dz

⎛M⎞ Rs ⎜ ⎟ 2 ⎝4⎠

I 02

(5.20)

Comparing Eqs. (5.19) and (5.20), we get

M

Rl = Rs

4

where Rs is surface resistance [6] and equals (=)

1

XN . 2T

2Q a Therefore radiation efficiency h can be given as (see [7]). 1

I= 1+

1 8Q a

XN 1 2T 73.12

where a = radius of dipole conductor. Since for l/2 dipole, Rr = 73.12 W, the radiation efficiency h reduces to 1

I= 1+

M a

Nf 1 1 × T 8 Q 73.12


I=

or

1

M ⎛Nf ⎞ 1+ ⎜ ⎟ a ⎝ T ⎠

193 (5.21)

1/2

9.6 × 10

−4

The radiation efficiency for a half-wave dipole of circular cross-section has been computed using above two efficiency Eqs. (5.18) and (5.21). The variation of radiation efficiency of copper dipole (s = 5.8 ´ 107 mho/m) in free space as function of d/l as well as frequency has been plotted. Calculations were repeated using the conductivity of silver instead of copper and found that efficiency differs from those of copper. The same was done for aluminium half-wave dipole too and concluded that the results are useful in determining the radiation efficiency only when ohmic losses are neglected [7].

MONOPOLE ANTENNA Monopole antenna is one of the most widely used antennas throughout the RF spectrum ranging from VHF to UHF. The simple structure of monopole coupled with unique properties such as a pure vertical polarization and horizontal omnidirectional coverage, hence attract its extensive possible uses in a variety of applications. Therefore, monopole antenna is an attractive option for broadband communications. The available literature indicates that the impedance bandwidth of a simple thin-wire monopole can be increased by introducing specific changes in geometry such as folding wire, loading and thickening. Compared to simple wire monopole, modified monopoles, such as conical or rotationally symmetric monopoles are bulkier. In fact, it is extremely wide bandwidth antenna but its radiation pattern deteriorates at the higher operating frequency. Amman and Chen [8] have proposed an alternative widerband planar monopole with considerable volume reduction to replace the wire monopole for broad impedance bandwidth purposes. Monopole antenna is half in the length of half-wave dipole antenna and generally fed by a co-axial cable connected to its base. The monopole antenna is perpendicular to the plane, which is usually assumed to be infinite and perfect conducting. Using the image theory we can replace monopole antenna as a half-wave dipole antenna as shown in Fig. 5.6.

FIG. 5.6

Monopole antenna above ground plane.

194


Therefore the field produced by monopole antenna in the region above the ground plane with its image is same as the field due to l/2 wave dipole, i.e.,

HG =

jI 0 e

− jCr

2Qr

− jCr

ER =

I jI 0 e 2Q r


(5.22a)


(5.22b)

However, monopole radiates only over the hemispherical surface above the ground plane, i.e., 0 £ q £ p. That is, radiated power is half of power radiated from the dipole with the same amount of current. Thus for l/4 monopole antenna the radiated power will be Prad = 18.28 I 02 and therefore Rrad = 73/2 = 36.5 W and hence input impedance is given by Zin = (36.5 + j21.25) W

(5.23)

In fact, the radiation resistance is a function of height above the ground and it may vary between 60 W and 90 W. The main difference is that l/4 monopole radiates only in hemisphere surface whereas l/4 dipole radiates more or less in all directions. The l/2 dipole and l/4 monopole antenna also called Hertz antenna and Marconi antenna respectively.

FOLDED DIPOLE ANTENNA A folded dipole antenna (see Fig. 5.7) is a modified l/2 dipole with an additional wire connecting its two ends. Folded dipole antenna is an extremely practical wire antenna and also called ultra closed spaced array. It consists of two parallel, closely spaced l/2 dipoles join together at the outer ends forming a narrow wire loop (d << L and d << l). The antenna is fed at centre of one dipole, i.e., the dipoles have the same voltage at their ends. As well as radiation fields are concerned it is same as the l/2 dipole antenna, but input impedance differs and equal to 300 W. d

L Feed

L

= Feed

Feed

Tx mode

FIG. 5.7

L

+

Antenna mode

folded dipole antenna and its current distribution.


195

Folded dipole antenna differs from the conventional dipole mainly in two respects: directivity and wider bandwidth. The directivity of folded dipole antenna is bi-directional but because of the distribution of current in the parts of the folded dipole antenna the input impedance becomes higher, however the radiation patterns of both are equal. The folded dipole antenna does not accept power at any even harmonics (i.e., 2nd, 4th, …, etc.) of the fundamental frequency, however it works with low value of VSWR on odd harmonics (i.e. 3rd, 5th, …). This is because current distribution of l/2 and 3l/2 antennas is almost similar. That is, if any folded dipole antenna functions at 20 MHz, it will also function at 60, 100 MHz frequencies. Since a simple l/2-dipole antenna carries 73 W radiation resistance, so it is inconvenient to match it with a feed line of characteristic impedance (Z 0 ) 300 W or so. However a folded dipole antenna offers terminal resistance nearly 300 W and found suitable for such impedance matching.

Theoretical Analysis Folded dipole antenna is basically an unbalanced Tx line with unequal currents which radiates because of its unbalanced condition. A folded dipole antenna operation may be analyzed by considering its current to be composed of two distinct modes, namely Tx line mode and antenna mode. A model composed of these modes has been referred as a transmission line model [9]. This model accurately calculates the input impedance of folded dipole antenna provided the parallel wires are electrically close so that the usual transmission line equations apply. Using transmission line model the input impedance of folded dipole antenna can be expressed as:

Z in =

4 ZT Z D Z T + 2Z D

⎛ ⎞ ⎜ ⎟ 1 ⎟ = 4Z D ⎜ 1 ZD ⎟ ⎜ ⎜1 + 2 Z ⎟ T ⎠ ⎝

(5.24)

L⎞ ⎛ where ZT = jZ 0 tan ⎜ C × ⎟ . So if the length of folded dipole antenna, i.e., L = l/2, 2⎠ ⎝

then

M⎞ ⎛ 2Q ZT = jZ 0 tan ⎜ × ⎟ =∞ 4⎠ ⎝ M

Hence

Zin = 4ZD

(5.25)

i.e. the half-wave folded dipole antenna offers a four-fold increase in input impedance over its l/2 dipole version. Since input impedance of a resonant antenna l/2-dipole antenna is 73 W, therefore, Zinf = 4 ´ 73 = 292 » 300 W (i.e. very close to impedance of common twin lead transmission). Similar to half-wave dipole, l/2-folded dipole antenna also has real input impedance at resonance. Equation (5.24) can be used to predict the input impedance of folded dipole antenna of any length, provided the impedances of Tx line mode ZT and

196


corresponding linear dipole are known. The equivalent radius for a folded dipole antenna is given by (see [8]). ln (ae ) = ln (a) +

⎛d⎞ ln ⎜ ⎟ 2 ⎝a⎠ 1

(5.26)

where d is spacing between dipoles. Equation (5.26) can be sampled as ae ae (ad)1/2. Therefore the necessity of using the equivalent radius (ae) depends upon both the wire radius (a) as well as spacing between them (d). The above method is limited to a spacing of 0.01l between dipoles, but in practice many folded dipole antennas in the range of VHF, exceed this separation limit substantially as result accuracy degraded. For example, a folded dipole antenna manufactured using 19 mm aluminium tubing has a separation of 100 mm or 1/20th wavelength at 150 MHz [10]. Therefore, in general, the method fails for separations d > 0.01l, because general transmission line equations do not apply therein. The Tx line equation depends on the characteristic impedance Z0 as follows: Vs = VR cosh rl + IR Z0 sinh rl I s = I R cos H l +

VR Z0

sin H l

(5.27a) (5.27b)

where Z0 is the characteristic impedance and is given by ⎛d⎞ Z 0 = 276 log10 ⎜ ⎟ : ⎝a⎠

(5.28)

i.e., Z0 is function of the separation d and the radius a of the two-wire line. A.R. Clark has proposed the extension of the use of this method to a folded dipole antenna with spacing even greater than or equal to one-sixth of a wavelength or so. The physical length of the folded dipole antenna is increased by an extension factor 0.39 time the inter-element spacing and the NEC2 method of moment’s program is used for this purpose. Clark’s extension method is based on extended length theory of wire antennas proposed by Austin and Fourier [11]. They have shown that the resonant length of a bent monopole is constant for bending up to 90° for various configurations. Experimentally it is found that the wire with a bend less than 90° behaves electrically like longer straight wires. Bends such as ones at the tips of a folded dipoles, can hence be treated by extending the physical length of the antenna by a factor dependent on the inter-element spacing (d). Therefore the accuracy of Thiele Tx line mode can improve if the overall length (l) is increased by an amount dependent on the additional wire-length at the tips of folded dipole. The extended length relates the original lengths, separation between them and extended factor as follows: lext = Iorig + ad (5.29)

a = Unknown factor to be determined for different antennas (Using NEC2 method of moments the common value of a for folded dipole antenna is found to be 0.39.)


197

The value of a was originally found for a folded dipole antenna of length 1 metre manufactured using a 19 mm (a = 9.5 mm) aluminium tube. That is extension factor (a) can provide consistently accurate results for wide range of wire thicknesses. Comparison of three methods, NEC2, Thiele and Clarke methods is shown in Figures 2, 3 and 4 of [11] (for d = 30 mm, 100 mm and 200 mm, length l = 1 metre and radius a = 1 mm). From these plots it is concluded that the improved method is much closer to NEC2 than Thiele method as the separation increases.

Input Impedance of Folded Dipole Antenna In general the input impedance of a folded dipole antenna is given by Zin = n2 ´ 73 provided all the wires carry equal currents. Where n is number of l/2 dipoles having radiation resistance 73 Q. However for folded dipole antenna of unequal radii of the two dipoles the input impedance is modified to Z in

a ⎞ ⎛ = 73 ⎜ 1 + 2 ⎟ a1 ⎠ ⎝

2

(5.30)

Since input impedance depends not only on radius of wires, but also on the separation between them, Uda and Mushiake proposed another formula for calculating the input impedance in terms of d as: 2

d ⎤ ⎡ log ⎥ ⎢ a1 ⎥ = 73 × Zr Z in = 73 ⎢1 + d ⎥ ⎢ log ⎢ a2 ⎥⎦ ⎣

(5.31)

2

log d/a1 ⎤ ⎡ where Zr = ⎢1 + ⎥ is termed impedance transformation ratio/impedance set-up ratio. log d/a2 ⎦ ⎣ Equation (5.31) is well suited when matching is done with low impedances, e.g. directive arrays using parasitic elements because the radiation resistance of these arrays is quite low.

Applications of Folded Dipole Antenna The folded dipole antenna is very useful as an FM broadcast band receiving antenna particular as elements in Yagi-folded antenna, which is mostly used in television. In this Yagi antenna, the driven element is folded dipole antenna and remaining elements simple l/2 dipole antennas. It is constructed by cutting a piece of 300 W twin lead transmission line of length about l/2. The ends are soldered together maintaining overall length slightly less than l/2 at the desired operating frequency.

198


HARMONICS ANTENNA So far we have discussed wire antennas of finite length; less than or equal to l/2. We are also aware that an antenna resonates as long as it length is integral multiple of l/2. Therefore we can design an antenna of length even more than one half wavelengths long by combining many l/2 elements in series. The antenna designed in this fashion is termed as harmonics antenna and provides better directivity. The higher the number of l/2 elements, greater its directivity. So harmonics antenna is a long wire antenna of improved directivity over other single wire antennas. It radiates a horizontally polarized wave at low angles from 17° to 24° relative to the earth surface. The currents in adjacent half-wave section must be out of phase and hence a feeding system cannot be used that disturb this condition. Examples of long wire antenna (i.e. harmonics antennas) are V-antenna, rhombic and beverage antennas. On the basis of terminating to its characteristic impedance, the long wire antenna is classified as resonant (un-terminated) and non-resonant (terminated with Z0) LW antennas. In resonant long wire antenna because of mismatching of impedances standing waves exist along its length and the pattern becomes bidirectional corresponding to incident and reflected waves. However in case of terminated antenna all the incident waves absorbed in terminating impedance and there is no existence of reflected waves. Radiation pattern of antenna is unidirectional and uniform current and voltage exist along the axis of antenna. The resonant and non-resonant LW antennas and their directional radiation patterns are shown in Figs. 5.8 and 5.9 respectively.

FIG. 5.8

FIG. 5.9

(a) Resonant long-wire antenna; (b) Bi-directional pattern.

(a) Non-resonant long-wire antenna; (b) Unidirectional pattern.


199

The angle of radiation with reference to antenna axis depends on number of half wavelength (l/2), i.e., n. For example, direction of maximum radiation from a long-wire antenna of 16 elements (i.e. 8l long) w.r.t. antenna axis is found to be 17.5°. Large wire antennas (resonant and non-resonant) are used for transmission and reception in many communicating systems in frequency range from 30 kHz to 30 MHz. Irrespective of theoretical complexity LW antenna is useful because of its simple structure and low cost. They provide a simple and effective method of obtaining directional pattern as well as power gain. These properties of LW antennas are utilized when they are used as a circuit element in wire array antennas, V antennas and rhombic antennas.

Parametric Specifications (i) Physical length: The total physical length of LW antennas are given in terms of number of dipoles connected continuously in the antenna. That is for n-number of half wave dipoles. The length of antenna is L=

492(n − 0.05) f MHz

(5.32)

ft

(ii) Field strength: The field strength of the long-wire antenna made with odd and even number of l/2-dipoles are given by

E (r , R , G ) =

E (r , R , G ) =

where Irms I0 E r

60 I rms r

60 I rms

= = = =

r

⎞⎤ ⎡ ⎛ nQ ⎢ sin ⎜ 2 cos R ⎟ ⎥ ⎠ ⎥ 30 2 I 0 ⎢ ⎝ = ⎢ ⎥ sin R r ⎢ ⎥ ⎢⎣ ⎥⎦

⎛ nQ ⎞⎤ ⎡ ⎢ sin ⎜ 2 cos R ⎟ ⎥ ⎝ ⎠⎥ ⎢ for n even (5.33a) ⎢ ⎥ sin R ⎢ ⎥ ⎢⎣ ⎥⎦

⎛ nQ ⎞⎤ ⎡ ⎢ cos ⎜ 2 cos R ⎟ ⎥ ⎝ ⎠ ⎥ 30 2 I 0 ⎢ = ⎢ ⎥ sin R r ⎢ ⎥ ⎣⎢ ⎦⎥

⎛ nQ ⎞⎤ ⎡ ⎢ cos ⎜ 2 cos R ⎟ ⎥ ⎝ ⎠⎥ ⎢ for n odd (5.33b) ⎢ ⎥ sin R ⎢ ⎥ ⎣⎢ ⎦⎥

rms current at maximum current point (in A) peak amplitude of current electric field strength (V/m) distance between observation point and antenna

In case a long wire antenna is terminated to its characteristic impedance (i.e. nonresonant), the number of l/2 dipoles (i.e. n = odd or even) do not affect the pattern of the antenna.

200


And field strength at any point P(r, q, f) is given by ⎡ 60 I rmssin R ⎤ ⎡ IQ ⎤ (1 − cos R ) ⎥ E (r, R , G ) = ⎢ ⎥ sin ⎢ ⎣ r (1 − cos R ) ⎦ ⎣ 2 ⎦ ⎡ 30 2 I 0 ⎤ sin R ⎛ QL ⎞ = ⎢ (1 − cos R ) ⎟ ⎥ sin ⎜ r (1 − cos R ) ⎥⎦ ⎢⎣ ⎝ M ⎠

(5.34)

where L = hl/2 is the total length of antenna. (iii) Radiation resistance: The radiation resistance Rr of an n-element LW antenna in terms of radiation resistance and its number is given by Rrw = Rrd + 69 log10(n) = 73.13 + 69 log10(n)

(5.35)

where 73.13 W is radiation resistance of a l/2-dipole antenna. (iv) Angle of maximum radiation and maximum electric field intensity: The angle of maximum radiation, i.e., angle between maximum lobe and axis of antenna is given by

cos R m =

n −1 n

(5.36)

Therefore corresponding maximum electric fields (Eqs. 5.33a and b] are reduced to Emax =

30 2 I 0 n ⎡ ⎛ (n − 1)Q ⎞ ⎤ ⎢cos ⎜ ⎟ ⎥ for n is odd 2 r 2n − 1 ⎣ ⎝ ⎠⎦

(5.37a)

Emax =

30 2 I 0 n ⎡ ⎛ (n − 1)Q ⎞ ⎤ ⎢sin ⎜ ⎟ ⎥ for n is even 2 r 2n − 1 ⎣ ⎝ ⎠⎦

(5.37b)

Similarly, the maximum field for a terminated antenna (Eq. 5.34) is modified to

Emax =

or

Emax =

30 2(2n − 1) I o r 30 2(2n − 1) I 0 r

⎛ Q ⎞ 30 2(2 n − 1) I 0 sin ⎜ ⎟ = r ⎝2⎠

(5.38)

⎛ QL ⎞ sin ⎜ ⎟ ⎝ nM ⎠

(5.39)

(v) Input impedance: Using transmission line model a wire antenna can be treated as lossless parallel transmission line, i.e. a = 0 and hence g = jb. Therefore, impedance at the input terminals of resonant wire antenna can be expressed as


⎡ R + jZ 0 tan (C l) ⎤ Z in = Z 0 ⎢ L ⎥ ⎣ Z 0 + jRL tan(C l) ⎦

201 (5.40)

where l = length of wire antenna Z0 = characteristic impedance of wire antenna b = phase constant = 2p/l l = operating wavelength RL = resonant terminal load resistance. (When a wire antenna with diameter d is adjusted at height h above the earth surface, RL is given by ⎛ 4h ⎞ RL = 138 log10 ⎜ ⎟ ⎝ d ⎠

(5.41)

and its value lies between 200 W and 300 W to achieve proper impedance matching.) (vi) Maximum directivity: The maximum directivity of a LW antenna in terms of its radiation resistance is given by

Dm =

120 Rrw sin 2R max

(5.42)

where qmax is direction of maximum radiation. The voltage and current distributions along a LW antenna operating at various harmonics frequency can be obtained using Eqs. (5.33) to (5.39). For example, the voltage and current distributions of a LW antenna excited at its 1st, 2nd, 3rd and 4th harmonics are obtained as shown in Fig. 5.9(c).

FIG. 5.9(c)

Voltage and current distribution when operated at various harmonics.

202


From Fig. 5.9(c) it is clear that a long-wire antenna can be used for harmonically related frequencies (i.e. l/2, l, 3l/2, 2l, …) and therefore multi-band operation can be achieved using long-wire antenna(s). In general, long-wire antennas are found very useful in the MF (300 KHz–3 MHz) and HF (3 MHz–30 MHz) ranges. The performance characteristics of LW antenna can still be enhanced by using them in arrays/specific configurations.

V-DIPOLE ANTENNA As already mentioned, the main drawback of a conventional linear dipole is that its directivity begins to diminish due to an increasing side lobe level, when dipole length goes much beyond a wavelength. It has been reported that this undesirable performance characteristic of a conventional linear dipole can be eliminated by tilting the lengthened arm in particular shape in which V shape tilting is very common. The V shape tilted dipole antenna offers enhanced directivity and gain. The resonant V antenna is one of the cheapest antennas for providing a low angle beam for a fixed frequency operation in HF band. It is an extension of long wire antennas in such a way that two LW antennas (generally l/2 or multiple) are jointed together forming V shape and fed at apex. The terminated and non-terminated V-antennas are shown in Fig. 5.10, where both the wires are fed 180° out of phase with each other. The directivity and gain of these antennas can be increased further by lengthening the arms. The ground V-antenna offers nearly twice the gain of a single large wire antenna. The apex angle of V-antenna is also important and it varies between 36° and 72° for arm length 8l to 2l long. If a V-antenna has to be operated over a wider frequency range, the apex angle must be made the average between the optimum values for the highest and lowest frequencies in terms of the number of l/2 dipoles in each arm. In general, for a long-wire antenna, the main lobes of the pattern is observed at angle b = 36° w.r.t. the antenna axis [Fig. 5.10(b)]. If inclined angle a of V-antenna is twice of b, i.e. (2 ´ 36 = 72°) a bi-directional pattern is obtained, which is sum of pattern’s of individual arms. However bi-directional pattern of V-antenna can be converted into unidirectional by terminating antenna arms in its characteristic impedance. As a result, wires do not

(a)

FIG. 5.10

(b)

(a) Resonant V-antenna; (b) Terminated V-antenna.


203

carry reflected waves, hence all the back radiations stand nearly cancelled and antenna functions like travelling wave-antenna. In order to obtain alignment at elevation angles greater than zero, the inclined angle a of antenna would be somewhat less than 2b. The performance of V-antenna can be increased further by stacking them in array form in such a way that another V-antenna is placed at an odd multiple of l/4 from back of the first. In order to achieve end-fire action both the antenna is to be excited with a phase difference of 90°. In this way, directional pattern of antenna narrowed in horizontal plane in broadside direction and power gain increases approximately equal to number of section times used in the array. That is power gain of two elements V-antenna array is doubled, similarly triplet and so on for further increase in the number of elements in the array (see Fig. 5.11). Alternative way to minimize the back radiation of V-antenna with termination is to use V-element of considerable thickness, length and angle (b). The reflected waves on such elements are very small compared to travelling waves (outgoing waves) as a result antenna behaves as travelling wave antenna. A V-antenna of two elements, each of radius ( l/20) and length 1.25 l/4, with an inclined angle b = 90° are found to provide highly uni-directional pattern (see Fig. 5.12). Input impedance of a V-antenna is less than that of a linear dipole of the same length l. The major drawbacks of V-antenna are its stronger side-lobes and narrow BW. The inverted V-antenna is a modified form of V-antenna (see Fig. 5.13). If a V-antenna is constructed above a perfectly conducting plane, the image of it would carry the currents oppositely directed to those in antenna wire.

FIG. 5.11

V-element array antennas.

The main disadvantage of inverted V-antenna is that it has undesirable minor lobes due to other portion of the radiating lobes. Therefore, these lobes emit horizontally polarized waves in some other direction and hence inverted V-antenna may also receive some horizontally polarized waves from these waves. The inverted V-antenna is one of the travelling wave antennas used in HF band. It is effectively used for reception of waves up to 60 GHz [12].

204


FIG. 5.12

V-antenna of 1.25l long legs with radius of l/20.

FIG. 5.13

Inverted V-antenna.

Design Parameters The close observation indicates that V-antenna may be visualized as an open-circuited flared transmission line of length l and inclined angle a [Fig. 5.10(a)]. Therefore the directivity of a V-antenna in the normal direction can be increased (i.e., side lobes reduced) over that of corresponding linear dipole by choosing proper inclined angle a. In general, the larger the value of arm length l the smaller the angle a must be. Using method of moments and computer codes as well as concept of current distribution [13], the improved formulas for the optimum angle a and corresponding directivity has been achieved. The inclined angle (a) for which the directivity is greatest in the direction of the bi-sector of aÿ is given by (see [14])

and

3

B

⎛h⎞ ⎛h⎞ = − 149.3 ⎜ ⎟ + 603.4 ⎜ ⎟ ⎝M⎠ ⎝M⎠

B

⎛h⎞ = 13.39 ⎜ ⎟ ⎝M⎠

2

2

⎛h⎞ − 809.5 ⎜ ⎟ + 443.6 ⎝M⎠

⎛h⎞ − 78.27 ⎜ ⎟ + 169.77 ⎝M⎠

for 1.5 ≤

h

M

for 0.5 ≤

≤ 3.0

h

M

≤ 1.5

(5.43a)

(5.43b)


205

where a is in degrees and the corresponding directivity is given by ⎛h⎞ D = 2.94 ⎜ ⎟ + 1.15 ⎝M⎠

(5.44)

The variation of optimum values of a and arm length per unit wavelength is shown in Fig. 5.14, and corresponding directivity variations are plotted in Fig. 5.15. In these figures, the asterisks show some of the actual data compared and solid line that of either a second or third order polynomial fit to the actual data [14, 15].

FIG. 5.14

Optimum angles a versus V-dipole arm length.

10 MM calculations Polynomial fit Weeks curve

8 7

Directivity

6 5 4 3 2 1 0 0.00

FIG. 5.15

0.50

1.00

1.50

2.00

2.50

3.00

Directivity versus V-dipole arm length.

206


Therefore, for V-dipoles with 0.5 h/l 3.0, either Fig. 5.14 or Eq. (5.40) may be considered to obtain the optimum value of a. Similarly for directivity, Fig. 5.15 or Eq. (5.44) may be used for calculations. The radiation characteristics of asymmetrically fed wire antenna (off centre), with overall length (l) less than a half wavelength (i.e., l < l/2) is almost independent from feed point location along the wire. However, if overall length (l) of wire antenna is greater than l/2 (i.e. l > l/2), the feed point and current distribution undergoes a phase reversal maintaining almost sinusoidal current distribution, and this situation finally causes influences on input impedance as well as radiation pattern of the antenna. The input impedances of centre fed and off-centre fed dipoles are approximately related as (see [16]) Zcf = Zocf cos2(kDl)

(5.45)

where Zcf = input impedance of symmetric centre driven dipole Zocf = input impedance off-centre driven dipole Dl = displacement from the centre through the arm k = wave number

SLEEVE WIRE ANTENNA Sleeve antenna is another form of wire antenna with advantages like mechanically simple, broadband, purely vertically polarized and omnidirectional radiation in horizontal plane. The sleeve antennas are widely used in mobile communication and broadcast systems [17]. The basic geometrical configuration of a sleeve antenna is shown in Fig. 5.16. The sleeve antennas are formed by adding sleeve(s) to a monopole or a dipole antenna. Accordingly, there are two types of sleeve antennas: sleeve monopole antenna and sleeve dipole antenna. The addition of sleeves increases the operating bandwidth of the antenna. The reason behind it is that if short tubes (or sleeves) are added around a conventional monopole antenna,

Sleeve

Feed

FIG. 5.16

Sleeve antenna geometry.


207

the resulting antenna shows satisfactory impedance matching with its feeding device, over a wider frequency range than that of a single monopole antenna [18].

Sleeve Monopole It is the addition of sleeves to a monopole antenna, and is fed by a co-axial line (see Fig. 5.17). In 1947, Bock et al. designed an elementary sleeve antenna, useful for high frequency applications. 2a

Sleeve dia 2b

H

L

Ground plane Coaxial line

FIG. 5.17

More elementary sleeve monopole.

Later in 1966, as an attempt to improve bandwidth further, Poggio and Mayes proposed a sleeve monopole, in which the feed has been elevated from the ground plane into the sleeve itself as shown in Fig. 5.18. The radiating element of diameter 2a, protruding out of the enclosing cylindrical sleeve of diameter 2b is an extension of the centre conductor of the co-axial feed line, whose outer conductor is at a distance l from the ground plane. The cylindrical sleeve having length L is also shorted to the ground plane. The total height (H) of proposed new sleeve antenna is set to resonate at approximately one-quarter wave length at the lowest frequency. In principle the length of sleeve may be any portion of the total length of the monopole from greater than zero to the entire radiating portion of the antenna. However, in practice the sleeve portion may be only from 33% to 50% of the height of monopole, because the current at the virtual feed point changes only slightly as the overall monopole height varies from l/4 to l/2 [19]. Poggio and Mayes results show that first sleeve monopole resonance occurs at a frequency where monopole length H is approximately l/4. In addition, a VSWR of less than 8 to 1 is achieved over 4:1 frequency band by suitably adjusting the length l that is displacing the feed point within the sleeve. It was also observed that significant improvement in VSWR is possible only at the cost of the bandwidth.

208


FIG. 5.18

Sleeve monopole antenna.

The other optimum pattern design of a sleeve monopole is: (H–L/L) = 2.25 and b/a = 3.0. The reader interested in knowing about pattern optimization of sleeve monopole may see [19]. In order to overcome the problem of high VSWR of sleeve monopole (i.e. VSWR 8:1), K.G. Thomas and his team proposed a new top loaded dual sleeve antenna having features of broadband operation. It has been found the antenna features excellent radiation characteristics within a broad impedance bandwidth of 4.2:1, covering 0.5 to 2.1 GHz for VSWR less than 2:1. This new antenna is found to be a suitable option for broadband mobile and vehicular communication [20]. The proposed antenna is fed by a co-axial transmission line of characteristic impedance 50 W. It is like gap-coupled two layer-stacked sleeve monopole antennas (Sleeve A and Sleeve B) of length L1 and length L2 respectively (Fig. 5.19). The exterior of sleeve A acts as the radiating element while interior B acts as the outer conductor of the feed. A discontinuity (gap) is created in the outer conductor of the feed just above the ground

FIG. 5.19

Proposed dual sleeve antenna.


209

plane, which is significantly responsible for the broadband impedance matching. Therefore, wider impedance bandwidth can be obtained by optimizing the gap. Generally, a teflon spacer of dielectric constant 1.8 is used to introduce gap between sleeves A and B. The inner conductor of feed line is extended beyond the upper end of sleeve A and then it is surrounded by sleeve B. The upper end of sleeve B is attached to a hollow cylindrical section of diameter D and height h. The sleeve B is shorted at a point at a distance d down from the upper end of the sleeve. The bottom faces of cylindrical top section (of diameter D) and height h acts as the top loading. The diameter of the circular ground plane and the cylindrical top section is maintained as the same; therefore the section comprising Sleeve A and Sleeve B acts as a capacitor plate antenna in the lower frequency band and produces input resistance identical to near quarter wave resonance. The position of the metallic short is optimized by using a short circuit plunger, presenting a very high impedance at the lower end of upper sleeve section B and thereby decouples it in the upper frequency band operation. The photograph of proposed antenna is shown in Fig. 5.20.

FIG. 5.20

Photograph of experimental dual sleeve antenna.

Design Specifications and Experimental Results The performance characteristics of the dual sleeve antenna have been measured and found in good agreement with simulated ones [20]. The obtained results reveal that the first resonant of antenna occurs at a frequency where monopole length (H) is approximately l/4. Total length of antenna is 0.23l and the diameter of circular ground plane feed is 0.083l, where l is wavelength in frequency range 0.5 to 2.1 GHz. A VSWR less than 2:1 is noticed over the entire frequency band. The cylindrical top loading significantly reduces the lower edge of the frequency band from 0.75 to 0.5 GHz with a resonance near 0.6 GHz. The line section (L1 + L2) acts as the conventional sleeve monopole and also operates in frequency band 0.5–2.1 GHz. It is also observed that return-loss performance of antenna with larger ground plans is degraded over the frequency band. The radiation pattern represents perfect omnidirectional patterns in the H plane and typically dipole pattern in E plane. The deviation from typical monopole pattern in E plane is attributed to the small ground plane size of the antenna.

210


Sleeve Dipole Antenna Sleeve dipole antenna is closely resembles an asymmetric dipole and hence can be analyzed in a similar manner [see Fig. 5.21(a)]. It is essentially the same as that of a base-driven monopole antenna above ground plane. By introducing the outer sleeve the excitation gap voltage (maintained by the feeding line) is moved upward from the conducting plate z = 0 to z = zf. Therefore, image theory yields an equivalent structure in which two generators carry equal voltage at z = ±zf.

FIG. 5.21

Sleeve dipole configuration and approximate equivalents.

As shown in Fig. 5.21(b), because of the linearity of Maxwell’s equations the total current in the antenna system will be equal to the sum of the currents maintained independently by each generator in each of the two asymmetric excited radiating elements [21]. Therefore the total current at the input to sleeve IA (zf) is then approximately sum of currents at the points z = ±zf, from the two configurations in Fig. 5.21(c) (upper and lower halves). Since both the structures are identical at their feed, the input current is IA = Ias(zf) + Ias(–zf)

(5.46)

Then the input admittance YA to the sleeve will be ⎛ I as (z f ) + I as ( − z f ) ⎞ YA = ⎜ ⎟ VA ⎝ ⎠

or

⎛ I as (z f ) ⎞ YA = Yas ⎜ 1 + ⎟ ⎜ I as ( − z f ) ⎟⎠ ⎝

(5.47)


where

Yas =

1 Z as

=

211

2 Z1 + Z 2

which is useful for determining the impedance of asymmetric dipoles where IA = input current at the feed of the sleeve dipole [Fig. 5.21(a)] Ias(±h) = currents of asymmetric structures at z = ±h [Fig. 5.21(c)] Zas = input impedance of centre-fed sleeve dipole Z1 and Z2 = impedances due to symmetrical antennas of half lengths L1 and L2 Analysis shows that the frequency response of sleeve dipole antenna is much superior to either that of a half wavelength or full wavelength dipoles.

Open-sleeve Dipole Antenna The sleeve dipole antenna can also be approximated with an open-sleeve dipole by replacing tubular sleeve with two conductors close to either side of the driven element. According to Barkley [22], open sleeve antenna is a variation in the physical arrangement of the conventional sleeve antenna. It consists of a dipole with two closely spaced parasitic elements of length approximately one half of a centre-fed dipole. In fact, so many parameters arise which can be varied to allow a wide range of choice between VSWR performance and operating bandwidth. However, Barkley made a parametric study on impedance characteristics by varying the length and spacing of open-sleeve monopole antenna. The gain and radiation characteristics of open-sleeve antenna are found similar to a conventional cylindrical sleeve antenna. Later, H.E. King and J.L. Wong [23] conducted study on a balun-fed open sleeve dipole in front of a metallic reflector for operation in 225 MHz to 400 MHz frequency range, especially for a satellite system application. They made parametric study on the VSWR response as a function of dipole and sleeve diameters, sleeve length and sleeve to dipole spacing. In addition, radiation patterns, gain and effects of mutual impedance were also determined in order to derive suitable parameters for the proposed antenna system operating over a 1.8:1 frequency band. Basic configuration of open-sleeve dipole antenna is shown in Fig. 5.22, in which, proposed antenna structure has been incorporated with a balloon, and dipole is fed by a co-axial line. The design parameters of the open-sleeve dipole antenna for lowest VSWR for frequencies 225 MHz and 400 MHz are given in Table 5.3. Both the dipole and the sleeve were constructed in circular cylindrical elements with a copper-clad 0.141 in. diameter, semi-rigid co-axial cable as the feed line. The balanced line of the balloon is also a part of the semi-rigid cable (but without the centre conductor), and the short-circuit of this line is coincident with the reflector surface. The dipoles are screwed into the feed terminals while the sleeves were supported by Styrofoam. The angle a at the feed point was made 90° for all dipoles, except for 1 18 dipole. However, for the 1 18 dipole, a was inadvertently chosen as 45°.

212


Feed point details (a) Front view

D D

(b) Top view

Reflector surface

Co-axial feed (c) Side view

FIG. 5.22 TABLE 5.3

The open-sleeve dipole antennas with a flat reflector.

Design parameters of open-sleeve antenna at 225 MHz and 400 MHz

Parameters

Value at 225 MHz

Value at 400 MHz

D (diameter) H (length) total L (length) S Sd

0.026l = 3.45 cm 0.385l = 51.20 cm 0.216l = 28.73 cm 0.0381l = 5.067 cm 0.163l = 21.68 cm 90° (for all dipoles) 45° (dipole 1.125 in)

0.047l = 3.53 cm 0.684l = 51.34 cm 0.385l = 28.88 cm 0.0677l = 5.071cm 0.29l = 21.75 cm 90° (for all dipoles) 45° (dipole of 1.125 in)

a

Performance Characteristics VSWR and coupling characteristics Though VSWR tests were conducted taking both dipole and sleeve of the same diameter at two values of dipole to reflector spacings 6 in. and 8.6 in. But it was noticed that the latter


213

one showed significantly better VSWR performance than the first one. Therefore, 8.6 in. spacing was considered a reasonably good choice for the antenna and it was considered for the entire measurements. The reason behind choosing 8.6 in. spacing is that it is equal to 0.29l at f = 400 MHz and it is the position where radiation pattern beam bifurcation starts. The VSWR characteristics of antenna as a function of frequency for different parameters are: (a) dipole and sleeve diameters, (b) sleeve spacing, and (c) sleeve length are shown in Figs. 5.23, 5.24 and 5.25 respectively. Figures 5.24 and 5.25 show the VSWR responses for 3/4 in. diameter open-sleeve dipole with sleeve spacing and length respectively [23].

D = 1–1/8 (Conventional dipole without sleeves) VSWR

0 = 3/8 1/2 3/4 1–3/8 1–1/8

2

1 200

FIG. 5.23

250

300 Frequency (MHz)

350

400

VSWR response of open-sleeve dipole for various dipole and sleeve diameters. 6 Sleeve spacing S = 1.05 S = 1.80 S = 1.29

5

VSWR

4

Sleeve length L = 11.88 in.

3

Dipole length H = 21.7 in.

2

All dimensions are in inches.

1 200

250

300

350

400

Frequency (MHz)

FIG. 5.24

VSWR for 3/4 in. dia open-sleeve dipole with sleeve spacing as parameter.

214

Antenna and Wave Propagation 9 L = 14.5 in.

8

L = 13.5 in.

S = 1.29 in. H = 21.7 in.

7

VSWR

6 5 L = 12.75 in. 4 3

L = 11.88 in. L = 11.00 in.

2 1 200

250

300

350

400

Frequency (MHz)

FIG. 5.25

VSWR for 3/4 in. dia open-sleeve dipole with sleeve length as parameter.

Group of plots shown above indicate that VSWR characteristics of an axially or a transverse mound dipole in front of a 60 in. diameter cylinder is nearly same as those of a dipole mounted in front of a flat reflector. The VSWR response becomes flatter and the required dipole length shorter as the diameter increases. The wide band characteristic has been confirmed by showing the VSWR response of conventional dipole of diameter 1 18 in. From Fig. 5.24, it is clear that sleeve spacing has relatively little effect on VSWR response for diameters less than 3/8 in. However both Figs. 5.24 and 5.25 illustrate how the VSWR response can be modified at the upper end of the frequency band at the expense of the midband region. The amount of coupling between two adjacent sleeves (terminated in one of three conditions: open circuit, short-circuit and 50 W load) was found less than –20.6 dB in the 225 MHz–400 MHz frequency range. Radiation pattern, directivity and gain The radiation pattern and gain of proposed antenna in front of a reflector were measured on a half-scale model, where dipole to reflector, sleeve to reflector and dipole to reflector spacing were made up half size. The measurements conducted at three test frequencies: 450 MHz, 600 MHz and 800 MHz, indicate that both E plane and H plane patterns are similar at these frequencies plotted. The E plane pattern shows minimum power of » –32.5dB at angle 100° for frequencies 450 MHz and 600 MHz; however, it is reduced to 40 dB for 800 MHz [23]. The directivity of antenna in each case is computed by the integration of the E plane and H plane patterns, assuming the patterns were azimuthally symmetric, i.e. DE =

∫

Q 0

E -pattern and DH =

∫

Q 0

H -pattern

and then final directivity D is taken as an average of both, i.e.

(5.48)


D=

DE + DH

215 (5.49)

2

The antenna gain is then calculated by subtracting ohmic losses from the directivity D, i.e. G = D – PL

(5.50)

where PL is an ohmic loss. The gain of the open-sleeve dipole antenna for different frequencies is tabulated in Table 5.4. TABLE 5.4

Gain of the open-sleeve dipole antenna for different frequencies Frequency (MHz)

Gain (dB) Theoretical Measured

450 600 800

8.62 9.03 7.24

9.1 9.1 7.0

Therefore construction of open-sleeve dipole antenna is a simple and effective approach for achieving a 1:8:1 operating bandwidth and further bandwidth broadening is also possible without cast of radiation pattern alteration.

BEVERAGE ANTENNA OR WAVE ANTENNA The Beverage antenna is a long-wire antenna of length greater than one wavelength, however some designer claimed that its length > 0.5l is sufficient enough. Unlike regular long-wire antenna, the Beverage antenna is mounted close to the earth’s surface (typically < 0.1l) [24]. Beverage antenna was first used as directional antenna in 1992 however its technical description by H.H. Beverage (after whom it is named) was published as “The wave-antenna for 200-metre Reception” in QST Magazine in the same year. In general, the length of Beverage antenna varies from 8 to 10 ft. Basic configuration of a Beverage antenna carries a single travelling wave is shown in Figs. 5.26(a) whereas current and phase distributions along antenna are shown in Figs. 5.26(b) and (c), where IR is relative current and DA is distance along antenna. The current distribution is uniform while there is a linear variation Wave RL

Feed

(a) Basic configuration FIG. 5.26 Contd.

216


IR

DA

(c) Phase variation

(b) Current distribution FIG. 5.26

Beverage antenna.

in phase along the antenna. Beverage antennas are terminated at the far end with a resistance RL equal to the antenna characteristic impedance and shows uni-directional pattern; however, unterminated Beverage antennas offer bi-directional radiation.

Principle of Operation In general, the electric field of a wave travelling along any perfect conducting surface is normal to the surface, however they undergo forward tilt near the surface, in case the surface is an imperfect conductor, such as earth’s surface/ground. If E is electric field intensity of a wave travelling along imperfect conductor (tilted wave) in x-direction, then x-component of electric field intensity (Ex) will enter the surface and dissipated resulting in to heat [25]. However, normal component (Ey) continues to travel along the surface and responsible for propagation [Figs. 5.27(a) and (b)]. Horizontal components provide the means of generating an RF current in the conducting wire. Therefore, Beverage antennas work on vertically polarized waves arriving at low angles of incidence. Beverage antenna consists of a long horizontal wire terminated in its characteristics impedance is shown in Fig. 5.27(c). In this case of Beverage antenna as waves incident/travel toward the receiver, the horizontal component of field (Ey), induced emfs along the antenna, and all these add up in the same phase at the receiver. On the other hand, the waves arriving from the opposite directions are largely absorbed in the termination. Therefore, the antenna exhibits a directional pattern in the horizontal plane with maximum response in the direction of termination.

E Y

E Direction of propagation


Y Ey Ex

Perfect conductor

(a)

FIG. 5.27

X

Perfect conductor

(b)

(a) Electric field on ground; (b) Component of electric field E.

X


217

E Y Ey Tx

Ex

A

B Rx

Terminated end X Ground

(c)

FIG. 5.27

(c) Machanism of radiation.

The Beverage antenna is highly directional, responsive to low-angle signals, has little noise pick-up, and produces excellent signal to noise ratios. Applications of Beverage antenna range from 15 to 500 KHz and 3 to 30 MHz. Beverage receiving antenna requires a lot of space, as it is several wavelengths long, mounted near the ground and oriented towards desired reception. A ballon is required at the juncture of the wire (antenna) and coaxial feed line. High resistance (nearly 600 W) is needed to terminate the end of antenna. Beverage antenna provides good gain and directivity, but its efficiency is poor, that why it is not suitable as a Tx antenna. Theoretical length of Beverage antenna should be close to a factor known as the maximum effective length (MEL), which is defined as MEL =

M ⎛ 100 ⎞ − 1⎟ 4⎜ ⎝ K ⎠

(m)

(5.51)

where l = operating wavelength (m) K = velocity factor and expressed in per cent Beverage antennas work best over lossy ground, which does not make a very good ground connection. The Beverage antenna should be installed at a height of 6 to 10 ft. off the ground and it should be level with the ground over its entire length. In case the ground is not flat enough, a height 6 to 10 ft. can also be considered above the average terrain elevation [26].

RHOMBIC ANTENNA The rhombic antenna is a highest developed long-wire antenna [see Fig. 5.28(a)]. It is a nonresonance antenna and consists of four similar conductors joined together in a rhombus form/ diamond shape. A rhombic antenna can also be made of two obtuse-angle V antennas that are placed side by side, erected in a horizontal plane. The rhombic antenna is terminated with load resistor RL of suitable such that to match the feeding Tx line. For a typical rhombic antenna designs: a = 14.4°, L = 6l and H = 1l load resistor RL are typically on the order

218


of 700 to 800 W. The terminated resistor should be non-inductive and capacitance should be negligible. In particular, terminating resistance of 800 W can be used if antenna is to be used as a receiving antenna. The antenna carries outward travelling waves are absorbed in the matched load. The reason behind its radiation is that the separation between the lines is longer compared to wavelength. Rhombic antenna is also called diamond antenna, because its principle of operation is based on travelling wave radiator. Its acute angles and leg length determine completely the physical form and size of the antenna. In general, rhombic antenna is fed by a balance line and terminated load resistance is adjusted to set up travelling waves in the all four legs of the antenna. Rhombic antenna offers maximum gain along its axis, i.e., line joining feed and termination points and its polarization exists in the horizontal plane. In practice only half of power is radiated rest half is dissipated in the terminating resistor. The radiation pattern of antenna is unidirectional; however, it can be converted into bidirectional by removing the terminated resistance. Antenna of this kind found suitable for long distance shortwave reception of horizontally polarized waves. In case, we choose a = 0.08f, the beams of the rhombic antenna [see Fig. 5.28(a)] numbered 1, 2, 3, and 4 will be aligned, where f is the angle of maximum radiation [27] and 0.371 ⎞ ⎛ G = cos−1 ⎜ 1 − ⎟ LM ⎠ ⎝

FIG. 5.28(a)

Basic configuration of rhombic antenna.

(5.52)


219

There are several designs of rhombic antenna: Alignment design: The maximum of main lobe is considered with the desired elevation angle a. The design parameters for an alignment design of rhombic antenna are Height of antenna above ground H =

Length of each leg L =

M

5.53(a)

4 sin B 0.5 M sin 2

5.53(b)

B

Tilt angle q = 90 – a

5.53(c)

Maximum E-design: The maximum relative field intercity E is obtained at the desired elevation angle a for constant antenna current. The design parameters are given as

H=

L=

M

5.54(a)

4 sin B 0.371 M

5.54(b)

sin 2 B

q = 90 – a

and

5.54(c)

0.371

= 0.74 of 0.5 the value for the maximum E-design. Assuming uniform current, in general, the electric field intensity E in the vertical plane coincides with the rhombic axis is given by [28]

That is, only the length is different in the alignment design method being

E=

[cos R {sin( Hr sin B )} {sin (Z Lr )}2 ]

Z

(5.55)

in which a = elevation angle w.r.t. ground 2q = angle between legs ÿ ÿ yÿ = 1 – sin q cos a ÿ

Hr = Lr =

2Q H

M 2Q L

M

The length of legs (L) varies from 4l to 8l, due to which major lobes of radiation pattern changes from 17° to 24° only [see Fig. 5.28(b)]. However, the typical value of power loss is 35% to 50% in terminating resistor. The directivity varies from 13 dB to 20 dB and power gain is of the order of 17 dB to 18 dB. As a result, a rhombic antenna operates satisfactorily over a wide frequency range.

220


90Y

Horizontal plane

75 60

120

45

135

30

150

15

165

180

345

195

330

210 315

225 300

240 255

FIG. 5.28(b)

270

285

Radiation pattern of rhombic antenna.

Advantages · The rhombic antenna is much easier to construct and maintain than other wire antennas of comparable gain and directivity. · The rhombic antenna is useful over a wide frequency range. Only small changes in gain, directivity, and characteristic impedance occur with a change in operating frequency. · The rhombic antenna also has the advantage of being non-critical as far as operation and adjustment are concerned. · It is a highly directional broadband antenna with greatest radiated/received power along the axis or longer diagonal. · The voltages present on the antenna are much lower than those produced by the same input power on any resonant antenna. This is particularly important when high transmitter powers are used or when high-altitude operation is required. · The vertical angle of radiation is low and hence these are suitable for long distance F-layer ionosphere propagation.

Disadvantages · A fairly large antenna citation is required for its erection as legs are made at least 1l or 2l long at the lowest operating frequency. However, when larger gain and directivity are required, legs of length 8l to 12l are also used. Therefore, they are suitable only when a large area is available.


221

· Rhombic antennas are used for long-distance sky wave propagation at low vertical angles of radiation (less than 20°). Since horizontal and vertical patterns depend on each other, if a rhombic antenna is designed to have a narrow horizontal beam, the beam is also lower in the vertical direction. Therefore, obtaining high vertical-angle radiation is impossible, except with a very broad horizontal pattern and low gain. · A considerable amount of the input power is dissipated in the terminating resistor though it is necessary to make the antenna unidirectional. · The efficiency of rhombic antenna is decreased significantly because of matched termination as it produces large numbers of lobes.

SOLVED EXAMPLES Example 5.1 Obtain inclined angle and directivity of a V-dipole antenna of arm length h = 0.75l. Show that directivity of V-dipole is greater than the directivity of 1.5l long linear dipole (3.4 dB). Solution: a = –149.3(0.75)3 + 603.4(0.75)2 – 809.5(0.75) + 443.6 = 112.87° D = 2.94 ´ 0.75 + 1.15 = 3.355 = 5.26 dB. Hence proved. Example 5.2

Find the radiation resistance of a 0.25 l long dipole antenna. 2

Solution:

2

⎛ dl ⎞ ⎛ 0.25 M ⎞ We know that Pr = 790 ⎜ ⎟ = 790 ⎜ ⎟ = 790 × 0.0625 = 49.4 : ⎝M⎠ ⎝ M ⎠

Example 5.3 Calculate the maximum radiation intensity and directivity of a l/2 dipole antenna carrying current 2.5 Å rms value at f = 200 MHz. Also, find the maximum effective area responsible for radiation. Solution:

2 We know that I rms =

I 02

2 Maximum radiation intensity,

⇒ I 02 = 2 × 2.5 × 2.5 = 12.50

U m = 4.8 I 02 = 4.8 × 12.50 = 60 WSr −1

And directivity is already defined as equal to 1.64 or 2.25 dB. The operating wavelength: l = c/f = 150 cm. The maximum effective area, Aem = 0.13 × M 2 = 0.13 × 150 2 = 3.0 × 103 cm 2 . Example 5.4 Calculate the radiation efficiency of above dipole, if it manufactured using copper wire of radius 2.0 cm.

222


Solution:

The radiation efficiency of a l/2 is given by

1

I= 1+

M

fN

T

a

× 9.6 × 10 −4

Given: a = 20 mm = 0.02 mm, l = 150 cm = 1.5 m.

m = m0 = 4p ´ 10–7 = 12.56 ´ 10–7 and s = 5.7 ´ 107 Therefore the value of factor

Hence

M

fN

a

T

=

I=

150 ⎛ 12.56 × 10 −7 × 2 × 10 8 ⎞ ⎜ ⎟ = 157 ´ 10–3 ⎟ 2 ⎝⎜ 5.7 × 10 7 ⎠

1 1 + 1.57 × 10

−5

× 9.6 × 10 −4

= 0.9998

h = 99.98% Example 5.5 Find the change in impedance of folded dipole antenna, when it is designed using l/2 dipoles of radii 1.5 and 2.0 cm. Also, find the input impedance when separation between element 10 cm is considered. Solution:

In general, Zin of a folded dipole antenna is 300 W. 2

2

2.0 ⎞ a ⎞ ⎛ ⎛ In the first case Z in = 73 ⎜ 1 + 2 ⎟ i = 73 ⎜ 1 + ⎟ i = 397.44 : 1.5 ⎠ a1 ⎠ ⎝ ⎝

Hence, the change in impedance; DZin = 97.44 W 2

2

log (s/a1 ) ⎞ log (10/1.5) ⎞ ⎛ ⎛ 2 Again Z in = 73 ⎜ 1 + ⎟ = 73 ⎜ 1 + ⎟ i = 73 × (1 + 1.1805) log ( / ) log(10/2.0) s a 2 1 ⎠ 1⎠ ⎝ ⎝

Zin = 347.089 W Hence, the change in impedance, DZin = 47.0879 W. Example 5.6 In order to improve the performance of a transmitter, a wire antenna is designed with 10 numbers of dipoles in series. Calculate the following, if it is to be operated at 150 MHz: (i) Angle of maximum radiation (ii) Total length, and (iii) Radiation resistance


Solution:

223

(i) Angle of maximum radiation

cos R m =

(ii) Total length L =

n− 1 n

=

9 10

492(n − 0.05) f MHz

⇒ R m = cos−1 (0.9) = 25.4°

=

492(10 − 0.05) 150

= 32.64 ft

(iii) Radiation resistance RR = [73.12 + 69 log10(n)] W = [73.12 + 69 log10(10)] W = 142.12 W Example 5.7 Obtain the input impedance of a 0.45 l long wire antenna to be operated at 1.5 MHz. The antenna is fixed at a height of 1.85 m from the ground. Assume the resonant resistance is 60 W. Also, find the directivity in the direction of 30° from the axis of antenna, if Z0 = 50 W. Solution:

We know that the input impedance is given by

Z in

⎛ 2Q ⎞⎤ ⎡ ⎢ 60 + j 50 tan ⎜ M × 0.45 M ⎟ ⎥ ⎡ R + jZ 0 tan (C l) ⎤ ⎝ ⎠⎥ ⎢ = Z0 ⎢ L ⎥ = 50 ⎢ ⎛ 2Q ⎞⎥ ⎣ Z 0 + jRL tan (C l) ⎦ ⎢ 50 + j 60 tan ⎜ × 0.45 M ⎟ ⎥ ⎢⎣ ⎝ M ⎠ ⎥⎦

⎡ 60 + j 50 tan (162°) ⎤ = 50 ⎢ ⎥ = 1.74[26.84 – j19.9] ⎣ 50 + j60 tan (162°) ⎦

Hence

Zi = (46.6 – j34.6) W

Directivity D =

120 Rrw sin R m 2

=

120 60 × sin (90° − 30°) 2

= 2.67 = 4.26 dB

Example 5.8 A 0.4 l long wire antenna is made up of a conductor with cross-section area of 36.0 cm2 and characteristic impedance of 55 W. The antenna is supposed to be operated at 25 MHz at height 2.0 m from the ground. Calculate resonant load terminated resistance (RL) and input impedance (Zin). Also, find the change in RL value if the height of antenna increases by 20%. Solution:

The resonant load resistance at height h from the ground is given as

⎛ 4h ⎞ 2 RL = 138 log10 ⎜ ⎟ and A = Q r = 36 ⇒ r = d ⎝ ⎠

36

Q

= 3.38

224


⎛ 4 × 2 × 10 2 RL = 138 log10 ⎜ ⎜ 6.77 ⎝

Hence

⎡ ⎢ 286 + j 50 tan Input impedance, Zin = 50 ⎢ ⎢ ⎢ 50 + j 286 tan ⎢⎣

⎞ ⎟ = 138 log10 (118.17) = 286 : ⎟ ⎠

⎛ 2Q ⎞⎤ × 0.4 M ⎟ ⎥ ⎜ ⎝ M ⎠⎥ = (7.38 − j 67.04) : ⎛ 2Q ⎞⎥ × 0.4 M ⎟ ⎥ ⎜ ⎝ M ⎠ ⎥⎦

In the second case, when height h increases, i.e., 2 ´ 1.2 = 2.4 ⎛ 4 × 2.4 × 10 2 ⎞ ⎛ 960 ⎞ RL = 138 log10 ⎜ ⎟ = 138 log10 ⎜ ⎟ = 297 : ⎜ ⎟ 6.77 ⎝ 6.77 ⎠ ⎝ ⎠

Therefore, the change in value of RL, DRL = 11 W. Example 5.9 Obtain the height for the load resistance of 250 W, if the diameter of the wire is 2.6 cm. Also, find position of a point where antenna directivity is 2.5 dB. Solution:

The resonant load resistance at height h from the ground is given by ⎛ 4h ⎞ RL = 138 log10 ⎜ ⎟ ⎝ d ⎠

250

or

138L

⎛ 4 × h × 10 2 ⎞ = log10 ⎜ ⎟ ⎜ ⎟ 2.6 ⎝ ⎠

1.8L = log10(153.5 h) Þ h = 0.4 m = 40 cm Directivity, D =

120 RL × sin 2R max 1/2

or

120 ⎛ ⎞ R max = sin −1 ⎜ ⎟ ⎝ 250 × 1.78 ⎠

= sin −1 (0.52)1/2 = 31.28°

Example 5.10 Find the incident angle a and directivity of a V-dipole antenna of arm length h = 1.6l. Solution: will be

The arm length h is equal to 1.6l Þ h/l > 1.5. Therefore, the incident angle a

B

⎛h⎞ = 13.39 ⎜ ⎟ ⎝M⎠

2

⎛h⎞ − 78.27 ⎜ ⎟ + 169.77 ⎝M⎠

= 13.39(1.6)2 – 78.27(1.6) + 169.77 = 78.82°


Example 5.11 7.5 dB. Solution:

225

Design a V-dipole antenna to be operated at f = 20 MHz, with directivity

Directivity,

D = 7.5 dB = 10 log (X) Þ D = 5.62 ⎛h⎞ 5.62 = 2.94 ⎜ ⎟ + 1.15 ⎝M⎠

Therefore,

1.52 ´ l = 1.52 ´ 15 = 22.8 m The inclined angle (a) = 13.39(1.52)2 – 78.27(1.52) + 169.77 = 81.74°

a = 81.74°

or

Example 5.12 What will be maximum effective length of a Beverage antenna to be operated at 100 MHz, if the velocity factor is considered to be 50%? Solution:

We know that maximum effective length is given by

MEL =

M ⎛ 100 ⎞ 4⎜ − 1⎟ ⎝ K ⎠

=

M ⎛ 100 ⎞ 4⎜ − 1⎟ ⎝ 50 ⎠

=

3 4

= 0.75 cm

Example 5.13 An electric dipole is formed with opposite charges of (+) and (–), separated at distance 2l. Show that the electric potential at a distance r from the centre of the dipole is V =

ql cos R 2QF r 2

where q is an angle between the radius r and axis. Find the actual values if q = 15 pF, r = 500 m and 2l = 6 cm at an angle of 60°. Solution:

Let us consider the dipole, formed as shown in Fig. 5.29.

FIG. 5.29

Dipole formed by two opposite charges.

226


Then the potential V at distance r will be 1⎞ ⎛1 − ⎜ ⎟ 4QF ⎝ r1 r2 ⎠ q

V =

in which r1 = r – l cos q Hence V =

r2 = r + l cos q.

and

⎛ r + l cos R − r − l cos R ⎞ ql cos R ⎜ ⎟= 4 QF ⎝ 2 Q Fr 2 r 2 − l 2 cos2R ⎠ q

In particular

V = Example 5.14

15 × 10 −14 × 3 × 10 −2 × cos 60° 2 × 3.14 × 8.8574 × 10

−12

× 500

2

=

4.045 25 × 10 4

= 1.61 ´ 10–3 V

An antenna carrying current of 4 A produces a uniform field as follows: ⎧⎪ 2V/m for R = 30° − 80° and G = 0° − 90° E= ⎨ elsewhere ⎪⎩ 0

at a distance of 100 m from its centre. Find solid beam angle, directivity, radiation resistance and loss resistance for 90% efficiency. Solution:

Solid beam angle : A =

∫

90° 0°

80°

∫

sin R dR dG

30°

80°

Q ⎡ Q ⎤ = − [cos 80° − cos 30°] = 1.087 Sr ⎢ − cos R ⎥ 2 ⎣ 2 ⎦ 30° D=

Rr =

I= or

Rl =

4Q :A

1 I2

4 × 3.14

=

1.087

:A r2 Rr

Rr + Rl

Rr 9

=

E2 Z

⇒

7.208r 9

= 1

I

= 11.55 = 10.63 dB

1

22

4

377

1.087 × 100 2 × 2

=1+

Rl Rr

= 0.78 :

=

1 9

=

Rl Rr

=

1 9

= 7.208 :

=

Rl Rr


227

20 I

Am −1 . r Find the efficiency of antenna if loss resistance is 20% of the radiation resistance. Also, find their values. Example 5.15

Solution:

An isotropic antenna is characterized by the field pattern E =

Given E =

20 I r

Am −1 E2

Hence power density Pd =

I 400 I 2

Pd =

or

for free space

r × 120 Q

If Pt is power over sphere, then it will be equal to pr2Pd and as per circuit theory it must be equal to I2R. That is, I2R = 4pr2Pd, where R is radiation resistance and hence R=

4Q r 2 Pd I

4Qr2

=

2

I

×

2

400 I 2 r

× 120 Q

2

=

400

Hence

Rl = 13.33 ´ 0.02 = 0.267 W

Therefore, efficiency I =

Rr Rr + Rl

=

13.33

=

13.33 + 0.02

30

1 1 + 0.02

= 13.33 :

= 98%

Example 5.16 Find directivity, gain, effective aperture area and solid beam angle for a centre-fed dipole of length l/10. Assume that the field distribution across the antenna is E(q) = sin q and Rl = 0.4 W. Solution:

(i) The solid beam angle : A =

(ii) The directivity D =

4Q :A

=

4Q × 3 8Q

∫∫ Q 4

Rr Rr + Rl

=

∫

Q 0

sin 3R =

8Q 3

= 1.5 = 1.76 dB

2 ⎛ I av (iii) The radiation resistance Rr = 790 ⎜ ⎜ I ⎝

(iv) The efficiency I =

sin 2R d : = 2Q

2 2 ⎞ 2 ⎛1⎞ ⎛ 1 ⎞ ⎟ LM = 790 ⎜ ⎟ ⎜ ⎟ = 1.975 : ⎟ ⎝ 2 ⎠ ⎝ 10 ⎠ ⎠

1.975 1.975 + 0.4

(v) Gain G = hD = 0.833 ´ 1.8 = 0.967 dBd

= 83.34%

= 8.38 Sr

228


(vi) Effective aperture Ae = I Am = 0.833

M2 :A

= 0.833

3M 2

= 0.099 M 2

8Q

in case l = 3.2 cm. Hence, Ae = 0.099 ´ 3.22 = 1.014 cm2 Example 5.17 (a) Design a maximum E-type Rhombic antenna for an elevation angle a = 17.5° to be operated at l = 3 cm. (b) Design an alignment type rhombic antenna for an elevation angle a = 17.5° to be operated at l = 3 cm. Solution:

(a)

ÿ

a = 17.5° and l = 3.0 H=

M

4 sin B

0.5 M

L=

4 sin B 2

=

=

3.0

= 2.5 m

4 sin 17.5o

0.5 × 3.0 2

o

4 sin 17.5

=

1.5 0.0904

= 16.5 m

q = 90° – 17.5 = 72.5° Angle of maximum radiation 0.371 × 3.0 ⎤ ⎡ o G = cos−1 ⎢1 − ⎥ = 21.17 16.5 ⎣ ⎦

(b)

ÿ

a = 17.5° and l = 3.0

H=

L=

M

4 sin B

0.371 M 4 sin B 2

=

=

3.0

= 2.5 m

4 sin 17.5o

0.371 × 3 2

o

4 sin 17.5

=

1.113 0.0904

= 12.3 m

q = 90° – 17.5 = 72.5° Angle of maximum radiation 0.371 × 3 ⎤ ⎡ o G = cos−1 ⎢1 − ⎥ = 24.37 12.3 ⎣ ⎦

Example 5.18 Calculate the relative electric field pattern in the axial direction for a rhombic antenna of elevation angle a = 17.5° and to be operated at l = 3 cm. Solution:

We know that E=

cos G [sin (Hr sin B )][sin (Z )]2

Z


Given: a = 17.5°, q = 72.5° and l = 3 cm ÿ

(a) Alignment method

L

Lr = 2Q

M

Hr = 2Q

Z

=

H

M

= 2 × 3.14 ×

16.5 3.0 2.5

= 2 × 3.14 ×

M

(1 − sin 72.5 cos 17.5) 2

= 34.54

= 5.23

=

(1 − 0.954 × 0.954) 2

= 0.545

Hence E=

E=

cos 72.5 [sin (5.23 × sin 17.5)] [sin (0.545 × 3.454)]2 0.545

0.3007 × 0.02744 × 0.1079

= 16.225 Vm −1

0.545

(b) E-Max design

Lr =

E=

=

2Q L

M

=

2 × 3.14 × 12.3 3.0

= 25.75

cos 72.5 [sin (5.23 × sin 17.5)] [sin (0.545 × 25.75]2 0.545

0.3007 × 0.02744 × 0.0589 0.545

= 8.95 Vm −1

OBJECTIVE TYPE QUESTIONS 1. The resonance antenna has zero (a) Input reactance (c) Input impedance

(b) Input resistance (d) None of these

2. One of the following combinations is true for Hertzian dipole: (a) Hr = Hq = 0 and Ef = 0 (b) Hr = Hf = 0 and Eq = 0 (c) Er = Ef = 0 and Hf = 0 (d) None of these 3. The induction field surrounding dipole varies as (b) 1/r2 (a) 1/r3 (c) 1/r (d) None of these

229

230


4. Radiation resistance of Hertzian dipole is (a) Equal to radiation resistance of half wave dipole (b) Lesser than radiation resistance of half wave dipole (c) Greater than radiation resistance of half wave dipole (d) None of these 5. The directivity of an ideal dipole is found to be (a) 1.0 (b) 1.5 (c) 1.64 (d) None of these 6. Folded dipole antenna function at (a) Only at odd harmonics (c) Both of these

(b) Only at even harmonics (d) None of these

7. The apex angle of V-antenna varies between 36° and 72° for leg length (a) l/2 to 3l/2 (b) 5l to 8l (c) 8lÿ to 2l (d) None of these 8. Beverage antenna was first used in the year (a) 1992 (b) 1995 (c) 2002 (d) None of these 9. The length of a Beverage antenna is (a) Greater than l (b) Lesser than l (c) Lies between 0.55lÿ and 5l (d) None of these 10. Rhombic antenna is (a) Bidirectional (c) Both (a) and (b)

(b) Unidirectional (d) None of these

11. The height of a Rhombic antenna is given by H = (a) Alignment method (c) Both (a) and (b)

M

as per the 4 sin B (b) Maximum E-method (d) None of these

12. The directivity of a Rhombic antenna varies (a) 13 dB to 20 dB (b) 20 dB to 25 dB (c) 25 dB to 30 dB (d) None of these 13. The radiation resistance of a folded dipole antenna found to be (a) 350 W (b) 292 W (c) 73 W (d) None of these 14. A half-wave is to be operated at 250 MHz. What will be its length if the velocity factor of antenna element is 0.85? (a) 0.3 m (b) 1.2 m (c) 0.51 m (d) None of these 15. The following is/are wideband antenna(s): (a) Folded dipole antenna (b) Marconi antenna (c) Discone antenna (d) None of these


231

16. A folded dipole antenna is conveniently fed by (a) Co-axial cable (b) Tx line (c) Flat ribbon type Tx line (d) None of these 17. Radiation resistance in a given direction is (a) Power radiated per m2 (b) Field transmitted per m2 (c) Power radiated per unit solid angle (d) None of these 18. For a dipole of length (l), the number of lobes in the radiation pattern will be (a) 1 (b) 2 (c) 4 (d) None of these 19. A balun is virtually a/an (a) Impedance transformer (c) Attenuator

(b) Frequency supporter (d) None of these

20. The inverted antenna can be used for reception of waves of frequency (a) 10 GHz (b) 60 GHz (c) 90 GHz (d) None of these

Answers 1. 6. 11. 16.

(a) (a) (c) (c)

2. 7. 12. 17.

(b) (c) (a) (c)

3. 8. 13. 18.

(c) (a) (b) (d)

4. 9. 14. 19.

(c) (c) (c) (a)

5. 10. 15. 20.

(b) (b) (a) & (c) (b)

EXERCISES 1. What are the advantages and application of rhombic antenna? Design a rhombic antenna of leg length 1.5l and elevation angle 35°. 2. What is the suitable frequency range of applications of long wire antenna? 3. Write the expression for the maximum field for a terminated long wire antenna. 4. Describe the principle of operation of folded and V-dipole antennas. Explain the various parameters of antennas. 5. Compare half-wave dipole, folded dipole antenna and V-dipole antennas in terms of designs and radiation characteristics. 6. What are the Marconi and Hertz antennas? What are their radiation patterns and impedances? List the differences between them.

232


7. Describe the principle of operation of rhombic antenna. Explain the various parameters of antenna. 8. Sketch the phase and current distributions of Beverage antennas. 9. Why is rhombic antenna used? Draw its neat diagram and explain its special features. What happens to the main lobe of rhombic antenna if its frequency is doubled? 10. What are the advantages of dual sleeve antenna over a sleeve monopole antenna? 11. Show that the radiation resistance of a l/2 dipole antenna is 73 W. 12. Describe folded dipole antenna. Show that the input impedance of this antenna is » 300 W. Mention its applications as a receiving antenna. 13. Describe sleeve dipole and open sleeve dipole antennas. Compare the l/2 dipole, sleeve dipole and open sleeve antennas. 14. Describe a method to calculate the directivity and gain of open sleeve dipole antenna. 15. Describe working principle and applications of Beverage antenna. 16. Derive the expression for the radiation efficiency of a l/2 dipole antenna. (Hint: See Eq. 5.22.) 17. Describe the characteristics of folded dipole antenna. Write the expression for its input impedance in different cases. 18. What are the resonant and non-resonant antennas? Sketch their radiation patterns. Write the expression for the field strength of these antennas. 19. Describe the characteristics of various V-dipole antennas. Highlight their applications in various communication systems. 20. Find the direction of maximum radiation of a rhombic antenna of leg length 0.6l. 21. What is rhombic antenna? Explain its design procedure with reference to height. 22. A rhombic antenna above the ground is to be designed for a main beam maximum at an elevation angle of b. Determine the rhombic configuration required for this angle. 23. How does rhombic antenna differ from the Beverage antenna? Write the formulae involved in the design of the rhombic antenna. 24. Find the radiation resistance of an antenna in a medium of er = 2.2, if its electric 10 I V field pattern is E = , where I is current and r is distance. r m 25. Show that average power radiated from a Hertzian dipole of length dl is equal to 40p2(dll)2 I 02 .


233

REFERENCES [1] Thiele, G.A. and W.L. Stutzman, Antenna Theory and Design, John Wiley & Sons, New York, 2001. [2] Kraus, J.D., Antennas, 2nd ed., McGraw-Hill, New York, 1988. [3] Loo, Y.T. and S.W. Lee, Antenna Handbook, Van Nostrand Reinhold, New York, pp. 27–21, 1988. [4] IEEE Standards on Antennas, “Methods of testing”, 48, IRE, 2S2, 1948. [5] Stratton, J.A., Electromagnetic Theory, McGraw-Hill, New York, p. 537, 1941. [6] Schelkunoff, S.A. and H.T. Friis, Antennas: Theory and Practice, John Wiley & Sons, Inc., p. 338, 1952, New York. [7] King, R.W.P., The Theory of Linear Antennas, Harward University Press, Cambridge, M.A., 1956. [8] Amman, M.J. and Z.N. Chen, “A wide band shorted planar monopole with bevel”, IEEE Trans Antennas Propagate., Vol. AP. 51. No. 4, pp. 901–903, April 2003. [9] Thiele, G.A., “On the accuracy of the transmission line model of the folded dipole antenna,” IEEE Trans Antennas Propagate, Vol. AP-28, No. 5, pp. 700–703, May 1980. [10] Austin, B.A. and A.P.C. Fourier, “Numerical modelling and design of loaded broadband wire antennas,” Proc. IEEE 4th Int. Conf. on HF Communication systems and Techniques, 284, pp. 125–128, 1988. [11] Clark, A.P. and A.P.C. Flourier, “An improvement to the transmission line model of the folded dipole antenna,” IEE, Proc-H, Vol. 138, No. 6, Dec. 1991. [12] Prasad, K.P., Antenna and Wave Propagation, 2nd ed., Satya Prakashan, New Delhi, 1996. [13] Richmond, J.H., “Computer program for thin wire structures in a homogenous conducting medium,” NTIS, Springfield, VA., 22131, NASA, Contractor Rep., CR. 2399, July 1973. [14] Thiele, G.A. and E.P. Ekelman, “Design formulas for V-dipoles,” IEEE Trans., Antennas and Propagate, Vol. AP. 28, No. 4, July 1980. [15] Iizuku, K., “The array of two travelling wave V-antenna as a space craft antenna,” IEE, Proc., Vol. 65, No. 7, pp. 64–65, May 1976. [16] Yagi, H., “Beam transmission of ultra short waves,” IEEE Proc., Vol. 72, No. 5, pp. 634–645, May 1984. [17] Li, J.Y. and Y.B. Gan, “The characteristic of sleeve antenna,” Progress in Electromagnetic Research, Symposium 2005, China, Hangzhou, pp. 23–26, August 2005.

234


[18] Bock, E.L., J.A. Nelson and A. Dorne, Sleeve Antennas in Very High Frequency Techniques, McGraw-Hill, New York, pp. 119–137, 1947. [19] Poggio, A.J. and P.E. Mayes, “Pattern bandwidth optimization of the sleeve monopole antenna,” IEEE Trans Antennas Propagate, Vol. AP. 14, No. 5, pp. 643–645, Sep. 1966. [20] Thomas, K.G., et al., “Wide band dual sleeve antenna,” IEEE Trans Antennas Propagate, Vol. No. 3, March 2006. [21] Weeks, W.L., “Antenna Engineering,” McGraw-Hill, NY, Sec. 2.6, pp. 161–180, 1968. [22] Barkley, H.B., “The open-sleeve as a broadband,” U.S. Novel Post Graduate School Monterrey, CA, Teach Rep. 14, AD-82036, June 1955. [23] King, H.E. and W.L. Wing, “An experimental studies of balun fed open-sleeve dipole in front of a metallic receiver,” IEEE Trans Antennas Propagate, Vol. AP. 20, No. 3, pp. 201–204, March 1972. [24] King, R.W.P., “Asymmetric driven antennas and sleeve dipole,” Proc., IRE, Vol. 38, pp. 1154–1164, Oct. 1950. [25] Kraus, J.D., Antennas—For all applications, Tata McGraw-Hill, New Delhi, 2005. [26] Johnson, R.C. (Ed.), Antenna Engineering Handbook, McGraw-Hill, New York, 1993. [27] Wolf, E.A., Antenna Analysis, Chap. 8, John Wiley, New York, 1967. [28] Beck, A.C. and L.R. Lowry, “Horizontal rhombic antennas,” Proc., IRE, 23, pp. 24–46, January 1953.

C H A P T E R

6

Loop Antennas

INTRODUCTION The loop antennas are very simple, inexpensive and versatile antennas. They are just wire antennas, but compulsorily not a straight wire; they may be of any shape—circular, square, rectangular, hexagonal and triangular as well as many more configurations. However, the circular/square loop antenna is very common, simplest in construction and easy in analysis. Basically, loop antenna is a radiating coil of any convenient cross-section of one or more than one turns carrying sinusoidal current. A loop of more than one turn is also called frame. The overall directional performance of the antenna can be significantly enhanced by selecting proper phasing between turns/coils. In general, there are two types of loop antenna—small loop and large loop. It is assumed that the periphery of small loop antenna is less than the one wavelength (i.e., C < l) however, a large loop antenna periphery is greater or equal to one wavelength (i.e. C ³ l). It is found that the radiation of a small loop antenna is equivalent to radiation of an infinitesimal magnetic dipole whose axis is normal to the plane of the loop. The field pattern of a small circular loop of radius a can be determined very easily by equating its area with the area of a square loop of side d (see Fig. 6.1). That is, d2 = pa2. The ordinary loop antenna is designed in such a way that its periphery is smaller than wavelength. Because currents are found to be of same magnitude and phase throughout the loop. The radiation resistance of loop antenna is smaller than the loss resistance; hence its radiation efficiency is poor and this is the reason why loop antenna is mostly used as receiver not as a transmitter. One of popular method of improving the radiation resistance hence efficiency of loop antennas is increasing its perimeter and number of turns. In addition, radiation resistance of loop antenna can also be increased by inserting ferrite core into loop, which is termed ferrite loop antenna.

Historical View During 1915–1930, the first loop antennas were used in receivers to minimize dependence on long wire antennas. Later in 1938, they appeared again and used to fully eliminate the 235

236


a

(i) Square loop

(ii) Circular loop

(a) Basic configurations

(i) Square

(ii) Rectangular

(iii) Circular

(iv) Triangular

(b) Geometries

FIG. 6.1

Basic configuration of loop antenna and the various geometries.

need of long wire antennas. The first high performance loop antenna termed box loop antenna (as it was wound on a 40² square box frame) was designed by Ray Moore in the mid-1940s; this later became popular as the Moore loop antenna. The next major advances in loop antenna designs were brought in the 1960s by Nelson at M.I.T. The main advantages with the new antenna were its free movement in the vertical and horizontal planes. This loop was 35" on a side and wound on a wood frame. These antennas have the alt-azimuth feature (available as a kit). Joe Worchester (1970–1977) developed the ‘Space Magnet’, a small 12² ferrite rod loop antenna using a bipolar junction transistor amplifier. This was probably the first loop antenna commercially available to the hobbyist, at a cost of about $45.00 at that time. This antenna also used a Faraday shield around the ferrite bar. However, Ralph Sanserino (1970– 1985) designed a 2-ft air core box loop using a differential amplifier. The amplifier was also used in the radio ferrite loop antenna. In the 1980s, Mackay Dymek and Palomar designed small ferrite antennas primarily for the broadcast band applications. In subsequent years, the following antennas were designed by antenna engineers; each utilized the Nelson alt-azimuth feature: (i) A 23², high performance, ‘Space Magnet’-like, ferrite rod loop was designed using differential amplifier by Radio West during 1979–1985. (ii) Quantum loop, a small ferrite rod antenna less than 1¢ in size (length), with a high gain (40 dB) amplifier, was designed by Gerry Thomas in 1990. (iii) The high performance, solidly built first air core loop antenna was designed by Kiwa in 1992. He integrated IC amplifier opto-isolated regeneration and varactor diodes with antenna for better performances. (iv) A high performance transformer coupled non-amplified 35² spiral wound antenna was designed by Moore in December 1994. Since then varieties of loop antennas have been continuously designed and developed for various purposes.

Loop Antennas

237

PRINCIPLE OF OPERATION In order to discuss the principle of operation of loop antenna, let us consider a single loop rectangular antenna; it is such that its plane is vertical and free to rotate around the ZZ axis (see Fig. 6.2). From Fig. 6.2 it is clear that out of four arms (AB, DC, AD, and BC) two arms (AB and DC) act as horizontal antenna while the other two arms (i.e., AD and BC) act as vertical antenna.

FIG. 6.2

Equivalence of a loop antenna to a rectangular antenna with rotation axis (where E1 and E2 are the voltage induced in vertical arms AD and BC).

In terms of this construction, radiation from a loop antenna can be described in two ways: (i) If the plane of the loop is right angle to the direction of arrival of vertically polarized waves [see Fig. 6.3(a)] then the same voltage will be induced in both the arms. These voltages will produce currents of equal magnitude and opposite phase in the loop, hence they will be cancelled out. This happens because during normal position of loop antenna, plane with respect to incoming waves, both sides (rather all the four arms) are at equidistance from the radiator, therefore no emf is induced. If at all any emf exists, that would be due to horizontally polarized downcoming waves and it could be neglected as its magnitude is very small. That is in horizontal arms no emf is produced whatever may be orientation, as result no radiation from the loop in this plane. (ii) If the loop is rotated by 90° such that plane of the loop is along the direction of arriving waves [see Fig. 6.3(b)], the voltage induced in each vertical sides will not be cancelled out. This is because of the involvement of distance between two vertical sides (i.e. CD). And waves take some time to travel this distance, introducing a definite phase difference (say, a) between induced fields E1 and E2. Therefore, rms value of emf induced in two vertical sides will be the same in magnitude, but different in phase by a. Therefore resultant induced emf across the vertical ZZ axis would be (E1 – E2) and it will be produced around the loop. The value of induced emf will be maximum, when the

238

FIG. 6.3


(a) Perspective view of loop antenna; (b) Plane view of loop antenna, when the plane of loop is (^r) to direction of incoming waves.

plane of the loop is along the direction of incoming waves from the transmitter. In any other position (except 90°), E1 and E2 will not be in phase; hence the resultant is zero and therefore no radiation (Figs. 6.4 and 6.5). The radiation pattern of a small loop antenna is free from the exact size of the loop and it is similar to the radiation pattern of Hertzian dipole with a minor difference that E and H are interchanged. Therefore a small loop is surrounded by a magnetic field everywhere at right angle to the loop and hence referred to as a magnetic dipole [1]. D

C

AD A

B (a)

FIG. 6.4

BC

(b)

(a) Side view of loop antenna; Plane view of loop antenna, when the plane of loop is (||) to the direction of incoming waves.

FIG. 6.5

Vector difference of E1 and E2.

Loop Antennas

239

In general, the induced emf in any direction q from the loop antenna is given by Eq = Erms cos q

(6.1)

where q is the angle between the plane of the loop and direction of wave arrival and Erms is rms value of electric field E. The value of Eq depends on the height of the vertical side (h), width of the loop (d) and operating wavelength (l). Loop antenna is very suitable for direction finding applications because when loop is rotated 360° around vertical axis ZZ, the maximum radiation appears twice, first at q = 00, 180° and then at q = 90°, 270°.

Radiation Fields In order to treat a small loop as a short magnetic dipole, let a small loop of area A carrying uniform current I be replaced by an equivalent magnetic dipole of length l, which carries fictitious magnetic current Im as shown in Fig. 6.6.

FIG. 6.6

Magnetic equivalence of small loop.

Here qm = pole strength at the end of dipole p = qm ´ l (p is magnetic dipole moment) Im = Imoejwt The magnetic current is related to pole strength by I m = − N or

∫

I mo e jX t dt = − N

∫

dqm dt

dqm dt

dt

which gives qm = −

Im

jXN

(6.2)

240


Equating magnetic moment of loop (I ´ A) with the magnetic moment of dipole, i.e. qml = IA

Im ⎞ ⎛ ⎜− ⎟l =I ×A ⎝ jXN ⎠

or

I m = − j 240 Q 2

IA

Ml

which can be re-written in retarded form as follows:

[I m ] = − j 240 Q 2

[I ]

Ml

A

(6.3)

Equations (6.3) describes the relation between loop area A and current I to its equivalent magnetic dipole of length l and fictitious current Im. where

or

r⎞ ⎛ [I m ] = I mo e jX ⎜ t − ⎟ c⎠ ⎝

(6.4a)

r⎞ ⎛ [I ] = I o e jX ⎜ t − ⎟ c⎠ ⎝

(6.4b)

Therefore the retarded magnetic vector potential, F, of the magnetic current (Im) can be given by

G N F= 4Q or

Fz = az

∫∫∫

N 4Q

∫

[J m ] r

dv

+ l/2

[I m ]

− l/2

r

dz

(z-component of F)

(6.5)

Solving Eq. (6.5) with the help of Eqs. (6.3) and (6.4) gives ⎛

r⎞

N I mo jX ⎜⎝ t − c ⎟⎠ Fz = le 4Q r E=

Therefore

1

N

(∇ × F )

(6.6)

(6.7)

Solving Eq. (6.7) in polar co-ordinate (r, q, f) system, the f-component of E is obtained as EG =

[I m ] l sin R ⎛ jX 1 ⎞ + 2⎟ ⎜ 4Q r ⎠ ⎝ cr

(6.8)

Loop Antennas

241

Therefore, at large distance r >> l (i.e., far-field region). jX [ I m ] l sin R

EG =

=

4 Q cr

Substituting

I m = − j 240Q 2

we get

EG =

Therefore

HR =

j[I m ] l sin R 2r M

IA

Ml

120 Q 2 [I ] sin R

A

r

M2

EG

=

I (=120 Q )

(6.9)

Q [I ] sin R A rM 2

(6.10)

The electric and magnetic fields of loop and short dipole are given in Table 6.1. TABLE 6.1

Comparison of far-fields of small loop and short dipole

Types of field

Loop

Electric field

EG =

Magnetic field

HR =

120 Q 2 [I ] sin R

Dipole

A

M

r

Q [I ] sin R

L

r

M

2

2

ER = HG =

j 60 Q [I ] sin R

L

r

M2

j[I ] sin R

L

r

M2

Since Ef and Hq are the functions of q (the angle measured from the polar axis as shown in Fig. 6.7) and independent from f, the radiation pattern of a small dipole is of doughnut shape.

FIG. 6.7

Radiation pattern of small dipole antenna.

242


INDUCED EMF OF LOOP ANTENNA In previous section, we have seen that emf induces between vertical sides of loop, provided fields in the sides are in phase of a. In order to derive expression for emf in any loop antenna, the value of a must be known, particularly when the plane of loop is at an angle q with respect to the direction of arrival of signals. Let us consider vertical sides AD and BC to be represented by two-point source (similar to two-point source arrays) separated at distance d, with the waves (1, 2) incident at any instant at angle q, as shown in Fig. 6.8.

FIG. 6.8

Loop antenna (as two-point source) and wave arrivals.

Therefore, similar to array analysis, the phase difference between rays 1 and 2 will be

B=

2Q

M

= Path difference =

Q d cos R M

If at any instant the electric field arrived at origin O is Em sin w t, then the fields arriving at R will lead and at P lag by angles +a and – a, respectively. Therefore, the resultant emf across the loop will be the difference of emfs induced in arms AD and BC, i.e.

ef (q) = Em h sin(w t + a)|AD – Em h sin(wt – a) = Em h [sin(w t + a) – sin(w t – a ) = 2Em h(sin a) cos w t = 2Em h a cos w t (since a is very small compared to sin q)

or

e f (R ) ≈

2Q hd cos R

M

Em cos X t

(6.11)

If there are N number of turns in the loop and area A = hd, then instantaneous emf around the loop can be given by

Loop Antennas

e f (R ) =

2Q AN cos R

M

Q⎞ ⎛ Em sin ⎜ X t + ⎟ 2⎠ ⎝

Q⎞ ⎛ e f (R ) = Vm sin ⎜ X t + ⎟ 2⎠ ⎝ in which

or

Vm =

Vrms =

2Q AN cos R

M 2Q A N cos R

M

243 (6.12a)

(6.12b)

Em

(6.12c)

Erms

(6.12d)

⎛ 2Q AN ⎞ with all the parameters having their usual meaning. The term ⎜ ⎟ is known as effective ⎝ M ⎠ height of loop. When it multiplies by field strength in mV/m, gives the induced voltage in mV provided it is taken in maximum response position, i.e. cos q = 1. Equation (6.12a) is the general expression for the instantaneous value of emf at the centre of the loop and it is clear that induced ef (q) is an alternating emf of frequency w/2p. Also, there is p/2 phase difference between ef (q) at the centre of loop and EM field being received at the antenna.

RADIATION PATTERN OF LOOP ANTENNA To describe the radiation pattern of a loop antenna, let us assume a circular loop of radius r carrying current I is represented by a square loop of side d, such that the area of the square loop is the same as the area of the circular loop, i.e., d2 = pa2 as shown in Fig. 6.9. We know that the fields from sides AB and DC must be ignored as they are equal and opposite in nature. Therefore, only two sides, AD and BC, will be responsible for radiation and they may be considered to be two short dipoles. That is, individual dipoles AD and BC act like two isotropic point sources in Y-plane (see Fig. 6.10). Now, the far-field radiation pattern due to isotropic sources AD and BC with reference to centre point O will be (see [2]) Ef = Field component due to AD + Field component due to BC = – E0e+iy/2 + E0e–jy/2 = –2jE0 sin ⎛ C d sin R ⎞ EG = − 2 jE0 sin ⎜ ⎟ 2 ⎝ ⎠

Z

2 (6.13)

j 60Q [I ] L where y = bd sin q is phase difference and ER = , is electric field amplitude of M dipole. Here, L of short dipole is equivalent to d, i.e., L » d and q = 90°, because the pattern is measured from x-axis instead of z-axis.

244


FIG. 6.9

(a) Equivalence of circular loop to square loop; (b) its orientation in (r, q, f) coordinate system.

FIG. 6.10

Therefore

EG =

hence

HR =

AD and BC dipoles as two-point sources.

120 Q 2 [I ] sin R A r M2

Q [I ] sin R A rM 2

provided d << l

(6.14a)

(6.14b)

where A = d2 is the area of square loop. Comparison of Eqs. (6.9) and (6.14) reveals that the radiation fields of circular loop and square loop are identical, provided they have equal area.

Loop Antennas

245

LARGE LOOP ANTENNA As we have seen, the far-field radiation fields of circular loop and square loop antennas are ⎛ M2 ⎞ the same, provided the loops are identical and their area is such that ⎜ A < ⎟ . So, in ⎜ 100 ⎟⎠ ⎝ general, the properties of loop antenna depend on area and the shape has no effect. The radiation pattern of these antennas depend on the angle q not on f, i.e., the radiation characteristics of electrically small loop antennas, which have a perimeter much less than l, are insensitive to loop shape and depend only on the loop area. Also, the radiation from the small loop is maximum in the plane of loop and is zero along the axis normal to the loop. These facts are consequences of the current amplitude and phase being constant around the loop, provided the loop perimeter length L is electrically small. However, for the loops of considerable perimeter (greater than l) the current amplitude and phase vary with positions around the loop and cause performance variation with changing the size. Equivalently, a fixed size large loop antenna shows performance changes with varying frequency, and acts like a resonant antenna. The radiation pattern of a large loop antenna depends on both q and f. From the construction point of view, large loop antennas have either a circular or square perimeter, and both are operated near the first resonant point, if perimeters of circular loops are equivalent to square loops. A detailed explanation on a square loop antenna is given here. A square loop of one wavelength l perimeter is considered, as it can be analyzed using the same techniques that we generally use for other resonant wire antennas. Since the size of loop is 1l, it is reasonable to assume that current distribution around the loop is sinusoidal. Then, the current distribution is continuous around the loop (— curves). Since the loop is fed at the centre of a side parallel to x-axis (see Fig. 6.11), the sinusoidal current can be expressed as ˆ 0 cos(C x ′) I1 = I 2 = − xI

where x′ ≤

I 3 = I 4 = − yˆ I 0 sin (C y ′)

where y′ ≤

M 8

M 8

(6.15) (6.16)

Here (x¢, y¢) represent a point at the midle of the side. The vector potential is A=N

e− j C r 4Q r

∫

loop

Ie j C (rˆ − r ′) dl

(6.17)

ˆ + yj ˆ + zk ˆ ), r¢ where l is length of side of square and rˆ is unit radial vector and equals (xi is position vector of point (x¢, y¢, z¢) and r is distance of observation point. Using Eqs. (6.15) and (6.16), the solution of Eq. (6.17) gives the simplified form of vector potential as

246

Antenna and Wave Propagation Y Sinusoidal current distribution Current magnitude

I2

X I4 I3

I1

FIG. 6.11

A= N

Square loop of one (l) perimeter and (l /4) side length.

e− jCr 2 2 I 0 4Qr

C

⎡ ⎛ Q cos : ⎞ ⎟ ⎢ cos ⎜ 4 ⎝ ⎠ ⎢ xˆ 2 ⎢ sin H ⎢ ⎢ ⎛ Q cos : ⎞ sin ⎜ ⎢ ⎟ ⎢ ˆ ⎝ 4 ⎠ ⎢− y sin 2 : ⎢⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎛ Q cos : ⎞ ⎛ Q cos : ⎞ ⎪⎫ ⎥ ⎪⎧ × ⎨cos : cos ⎜ ⎟ − cos ⎜ ⎟⎬⎥ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎭⎪ ⎥⎦ ⎩⎪

⎧⎪ ⎛ Q cos H ⎞ ⎛ Q cos H ⎨cos H sin ⎜ ⎟ − cos ⎜ 4 ⎠ 4 ⎝ ⎝ ⎩⎪

= Ax xˆ + Ay yˆ (say)

⎞ ⎫⎪ ⎟⎬ ⎠ ⎭⎪

(6.18) cos g = sin q cos f

in which

cos W = sin q sin f Therefore the far-zone electric field components will be G ER = − jX AR = − jX AR = − jX (Ax xˆ Rˆ + Ay y R) = jX (Ax cos R cos G + Ay cos R sin G ) E y = − jX AG = − jX ( − Ax sin G + Ay cos G )

Similarly

(6.19) (6.20)

Substituting the value of Ax and Ay from Eq. (6.18), we get the simplified values of Eq and Ef as follows:

ER =

jI 0I e − j C r 2Qr

⎡ sin G sin C ⎤ cos G cos B B B B C C C − − − cos R ⎢ ( cos sin ) ( sin cos ) ⎥ 1 1 2 1 − C12 ⎣ 1 − B1 ⎦ (6.21)

Loop Antennas

EG =

jI 0I e − j C r

where B =

2Q r

Q 4

247

⎡ cos G sin C ⎤ sin G cos B cos R ⎢ (B1 cos B − sin B ) + ( C1 sin C − cos C ) ⎥ 2 2 1 − C1 ⎣ 1 − B1 ⎦ (6.22)

sin R sin G

and

C =

Q 4

sin R cos G

ÿ a1 = sin q sin f and b1 = sin q cos f A close observation of Fig. 6.11 reveals that there are planes, xy, xz and yz, where farfield radiation takes place effectively and therefore needs to be analyzed. Basically there are three cases. Case A: The far-fields in x-y plane, which is the plane of loop (an E-plane) known as principal plane and is obtained by putting q = 90°; hence

Q⎞ ⎛ ER ⎜R = ⎟ = 0 2⎠ ⎝

Q ⎞ jI Ie− jC r ⎛ EG ⎜ R = ⎟ = 0 2⎠ 2Q r ⎝

(6.23)

⎡ ⎛ Q cos G ⎞ ⎤ ⎟ ⎢ sin ⎜ ⎥ 4 ⎛ Q sin G ⎞ ⎛ Q sin G ⎞ ⎫⎪ ⎠ ⎧⎪ ⎢ ⎝ ⎥ sin G cos ⎜ ⎟ − sin ⎜ ⎟⎬ ⎢ ⎛ Q cos G ⎞ ⎨⎩⎪ ⎥ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎭⎪ ⎢ ⎜ ⎥ ⎟ Q⎢ ⎝ 4 ⎠ ⎥ ⎢ ⎥ 4⎢ ⎛ Q sin G ⎞ ⎥ cos ⎜ ⎟ ⎢ 4 ⎧ ⎫ cos cos Q G Q G ⎛ ⎞ ⎛ ⎞⎪⎥ ⎝ ⎠ ⎪ ⎢+ × ⎨cos G sin ⎜ ⎟ − cos ⎜ ⎟⎬⎥ 4 ⎠ ⎭⎪ ⎥ ⎢ ⎛ Q sin G ⎞ ⎝ 4 ⎠ ⎝ ⎩⎪ ⎜ ⎟ ⎢ ⎥ ⎣⎢ ⎦⎥ ⎝ 4 ⎠

(6.24) For f = 0° and 180° along x-axis, Ef = 0, this is true also since both the sides 3 and 4 alone have patterns that are zero in the broadside direction; this is because the current distributions are in opposite phase around the mid-point. And along y-axis, i.e., q = f = p/2, Eq. (6.24) reduces to EG = −

jI 0Ie − j C r 2 Qr

1 2

R =G =

Q 2

= −

jI 0 Ie − j C r 2Q r

= Ey

(6.25)

248


Case B:

The xy-plane, which is an E-plane Ef(f = 0) = 0

ER (G =0) =

and

jI 0Ie

− jCr

2 Qr

(6.26)

⎛Q ⎞ ⎛Q ⎞⎤ ⎡ ⎢ sin R sin ⎜ 4 sin R ⎟ − cos ⎜ 4 sin R ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎥ ⎢ ⎢ ⎥ cos R ⎢ ⎥ ⎢⎣ ⎥⎦

(6.27)

i.e., Eq = 0, for q = 90°, 270°, i.e., along y-axis. Eq ¹ 0, for q = 0°, 180°, i.e., along x-axis. Hence from Eq. (6.27), we get ER

R =0

= −

jI 0I e− C r 2Q r

= Ex

(6.28)

which is similar to far-field radiation along y-axis in case of xy-plane. Case C:

In yz-plane (H-plane)

Q⎞ ⎛ ER ⎜ G = ⎟ = 0 2⎠ ⎝

(6.29)

Q ⎞ − jI 0I e− j C r ⎛ ⎛Q ⎞ EG ⎜ G = ⎟ = cos ⎜ sin R ⎟ 2⎠ 2Q r ⎝ ⎝4 ⎠

(6.30)

and

Q⎞ ⎛ which indicates that EG ⎜ G = ⎟ ¹ 0 for any value of q. 2⎠ ⎝ Q⎞ − jI 0I e− j C r ⎛ Therefore, EG ⎜ G = ⎟ = , which is the same as the electric field along 2 ⎠R = 0 2Q r ⎝ the x-axis (Eq. 6.28). Comparison of field levels along x and y directions, indicates that the field in the y-direction is 2 times that in the x-direction, i.e., ( 2 E y = E x ) . The radiation patterns of the square loop antenna in these planes are plotted in Fig. 6.12. So, it can be concluded that in case of 1l square loop, radiation is maximum normal to the plane of the loop (i.e., along the z-axis) and it is polarized parallel to the loop side containing the feed. However, in the plane of the loop, there is a null in the direction parallel to the side containing the feed point (i.e., along the x-axis) and there is lobe in the direction normal to the side containing the feed (i.e., along the y-axis). But in the case of a small loop, there is a null on the axis and maximum (uniform) radiation in the plane of the loop. It has also been found that the input impedance of 1l square loop antenna (wire radius 0.001l) is about 100 W and resonance occurs for a 1.09l perimeter. The gain is 3.09 dB, which is lesser than 3.82 dB, the gain of a straight wire 1l dipole antenna.

Loop Antennas

FIG. 6.12

249

Principal radiation pattern of a one-l square loop antenna.

LOOP ANTENNA PARAMETERS The important parameters of a loop antenna are: radiation resistance, loss resistance, total radiated power and directivity [3], and they are expressed as follows: 2

4

⎛ A ⎞ ⎛C ⎞ 1. Radiation resistance of a single turn loop Rr = 31200 ⎜ 2 ⎟ = 197 ⎜ ⎟ (:) ⎝M ⎠ ⎝M⎠ ⎛ NA ⎞ 2. Radiation resistance of N turns Rr = 31200 ⎜ 2 ⎟ ⎝M ⎠

3. Total radiated power P = 30 Q 2 I m2 4. Loss resistance Rl = N

Ll

f N0

d

QT

C

M

= 60 Q 3 I m2

a

M

(:)

5. Radiation efficiency (N turns) I =

1 ⎛ Rl ⎞ + 1⎟ ⎜ R ⎝ r ⎠

=

where the ratio of Rl/Rr for copper conductor is

6. Quality factor (Q) =

2

2Q f 0 L Rr + Rl + Rc

=

f0 'f HP

Rr Rl + Rr

3430 3.5 C 3 fMHz Nd

250


7. Bandwidth (BW) 'f HP =

S

8. Signal-to-noise ratio

=

N

f0 Q

Pr N

(Hz)

=

Pt Aet Aer r M 2ITs 'fHP 2

Directivity: In general, the directivity of a circular loop antenna with uniform current distribution is given by Foster’s expression as follows [4]:

D=

2C M

⎡ 2 ⎛ C sin R ⎞ ⎤ ⎢ J1 ⎜ ⎟⎥ ⎣ ⎝ M ⎠ ⎦ max

∫

2C

M 0

(6.31)

J 2 (y) dy

There are two particular cases:

C

1

3 , the directivity Dmax = , which is the same as the 3 2 directivity of a small electric dipole. This is because the pattern of small loop is equivalent to short electric dipole.

(i) For small loop of

(ii) For a large loop of

M

≤

C

⎛C ⎞ ≥ 5, directivity D = 0.682 ⎜ ⎟ . M ⎝M⎠

Maximum Effective Area and Gain As usual, the maximum effective area of a loop antenna is Aem =

and

M2 0.682 = × C M = 5.42 × 10 −2C M 4Q D 4 × 3.14 Gain =

4Q Aem I

M2

In the above equations A = area of loop C = 2 pa N = number of turns in loop Im = peak or maximum current in loop of area A Ll = circumference/perimeter of loop

(6.32)

(6.33)

Loop Antennas

251

For a small square loop of arm length l, C = 3.5 l and Ll = 4 l d= L= Rc = Ts = r= h=

diameter of loop wire inductance (H) conductor loss Antenna temperature (K) distance from the observation point (m) radiation efficiency factor

The instantaneous expression for electric and magnetic fields at a large distance (r) from a loop of radius a are EG =

60QC [ I ] a

J ′ and HR =

r

QC[ I ] a 2r

J′

(6.34)

where J¢ = J1(ba sin q) represents pattern of antenna and others are constant.

MULTI-TURN LOOP ANTENNA In the previous section, we discussed all the parameters of loop antenna. The radiation efficiency of the loop antenna is calculated in terms of ohmic resistance and radiation resistance. In general, radiation resistance is very less than the ohmic resistance, therefore h is low and heavily depends on the ohmic loss [5], i.e.

I=

Rr

≈

Rr + Rl

Rr Rl

as Rr << Rl

In order to increase the radiation efficiency, a multi-turn loop antenna is chosen. In which the ohmic resistance of a small loop antenna is taken to be same as that of a straight conductor. The length of this conductor is equivalent to length of uncoiled loop, and calculation is done under skin effect [6]. A multi-turn loop of close spacing can be treated as a system of parallel wires, and the distribution of current over conductor cross-section is determined by two effects skin effect and proximity. A multi-turn loop antenna of radius b with spacing between turns 2c is shown in Fig. 6.13(a). The loops have essentially N turns, and each carrying same current I, and the total length of the loop is much lesser than the free-space wavelength. If the current is confined to a thin layer near the wire surface, it is known as skin effect. If the effect causes a nonuniform distribution of current in the layer, this is called proximity effect. If the skin depth (d) in a conductor is very small compared to the wire radius, i.e. d << a, then the resistance per unit length of a system of n equally spaced parallel conductors carrying equal currents in the same direction is given by n

Rn = 2aRs

Q

∑ ∫R

m =1

=0

K m2 (R ) dR :/m

(6.35)

252


b

2c

a

FIG. 6.13(a)

Multi-turn loop antenna.

where Km = Surface current on mth perfect conducting wire carrying total current of 1 A ⎛ XN ⎞ Rs = Surface resistance equals ⎜ 0 ⎟ ⎝ 2T ⎠

1/2

The additional loss due to proximity effect is then ⎛R ⎞ R p = R0 ⎜ n − 1 ⎟ = ( Rn − R0 ) ⎝ R0 ⎠

(6.36)

where R0 is the skin effect resistance alone and equals nRs 2Q a

: /m

(6.37)

For close spacing between the conductors, the loss due to proximity effect can be more than double the ohmic resistance of the system of conductors. Hence, the ohmic resistance of the loop antenna can be given as Rl = Circumference ´ Resistance per unit length = 2pb ´ Rn = 2pb ´ (Rp + R0) where Rn = Rp + R0 nRs ⎛ Rp ⎞ ⎛ Rp ⎞ Rn = ⎜ + 1 ⎟ × R0 = ⎜ + 1⎟ × 2Q a ⎝ R0 ⎠ ⎝ R0 ⎠

Therefore

nRs nbRs ⎛ R p ⎛ Rp ⎞ ⎞ Rl = 2Q b ⎜ + 1⎟ × = + 1⎟ ⎜ 2Q a 2Q a ⎝ R0 ⎝ R0 ⎠ ⎠

(6.38)

Loop Antennas

253

Rr = Radiation resistance [1]

and

= 20p2n2(bb)4

(6.39)

Therefore the radiation efficiency is obtained as

IA =

Under no proximity effect,

Rr

=

Rr + Rl

Rp R0

20Q 2 n2 (bC ) 4

(6.40)

nbRs ⎛ Rp ⎞ 20Q n (bC ) + + 1⎟ ⎜ a ⎝ R0 ⎠ 2 2

4

=0.

After simplification, Eq. (6.40) reduces to 1

IA = 1+

(6.41)

8.48 × 10 −10 (f MHz T r )1/2 n(b′)3 a′

where a¢ = a/l and b¢ = b/l. ÿ sr = ratio of conductivities of wire and copper The variation of radiation efficiency hA with domensionless quantity (K) and the number of elements (N) is shown in Fig. 6.13(b). 100 N=8

Radiation efficiency (%)

N=6 N=4 10

N=8 N=6 N=4 N=3 N=2 N=1

1

0

FIG. 6.13(b)

N=3

0.1

1.0

without proximity effect with proximity effect

N = number of turns

6

10

K

The variation of radiation efficiency of the small loop antennas vs factor K for different values of N.

254


If the matching network is used along with antenna, it introduces losses as large as ohmic loss; therefore overall efficiency of antenna is modified to [7] E = h AE m

(6.42)

where Em is the radiation efficiency of matching network.

IMPEDANCE OF A LOOP ANTENNA In general, the radiation pattern of a loop antenna is calculated by integrating the Poynting vector over a large sphere. This gives total radiated power P, which is then equated to the square of the effective current on the loop times the radiation resistance. This radiation resistance is the value that appears at the loop terminals connected to the transmission line (and not the entire length of perimeter of the loop). Another formula is needed to calculate the reactive part of impedance. This method is complex, as it involves Bessel’s function and many approximations are also need to be considered. External devices such as phase-shifter/multiple feeds are needed to obtain uniform current distribution and proper phase condition over the loop [8]. In this section, approximate closed and simple formulas are used to determine the input impedance of a regular loop configuration, where the ratio of a/l is limited up to 0.8 for the reactive part and up to 0.5 for the resistive part. The proposed method is based on an equivalent transmission model, where a loop of fixed perimeter is equated with a shortened line of finite length of varying characteristic impedance. Let us assume that the length of Tx line is equal to half of C, whereas wire radius a remains the same. The loop is considered to be extended in such a way that it forms a Tx line (Fig. 6.14). The characteristic impedance of the line is taken as the average characteristic impedance of the loop. Hence it is clear that input impedance determined in this way is approximately equal to the reactive part of impedance of the loop.

FIG. 6.14

Circular loop and its equivalence to two-wire short-circuited Tx line.

Loop Antennas

l is

255

We know that the input impedance of a finite lossless (i.e., a = 0) Tx line of length ⎡ Z cos(C l) + jZ 0 sin (C l) ⎤ Z in = Z 0 ⎢ L ⎥ ⎣ Z 0 cos(C l) + jZ L sin (C l) ⎦

(6.43)

If the line is short-circuited, ZL = 0, and hence Zin = jZ0 tan bl = jXL where Z0 is the characteristic impedance of the line, C =

(6.44)

2Q

M

(wave number), and l is length

of Tx line and equals the perimeter of loop 2. If the area confined by the Tx line is taken to be equal to the area of the loop, the characteristic impedance (Z0) of Tx line is equal to the average impedance of the loop wire ⎛s⎞ Z 0 = 276 log ⎜ ⎟ : ⎝a⎠

(6.45)

where s is the spacing between coaxial wires/loop area.

Resistive Part of Impedance Resistive part of input impedance (R) is related to the radiated power, and heavily depends on the perimeter of loop and almost independent from the radius of wire. Actually, resistive part of the input impedance (R) is computed using moment method, and investigations on its variation against (C/l) deduced that it follows a formula of the form ⎛ CC ⎞ R = a tan b ⎜ ⎟ ⎝ 2 ⎠

(6.46)

where a and b are new constants which have different values for different loop configurations (see Table 6.2). The reactive part of input impedance for different loop configuration has also been calculated using above approximation and found to be relatively in good agreement with results computed using moment method for (C/l) » 0.8 [8]. Good accuracy of the results obtained from this method is restricted to (C/l) » 0.5. Typically, for a small circular loop, Eq. (6.46) reduces to

or

⎛ CC ⎞ R = 1.793 tan 3.928 ⎜ ⎟ ⎝ 2 ⎠

(6.47)

R = 20 b2A2

(6.48)

256


TABLE 6.2 S.No. 1 2 3 4 5 6

Constants a and b for different loop geometries for different values of (C/l) Configurations of loop Circular Square (centre fed) Square (corner fed) Triangular (base fed) Triangular (top fed) Hexagonal

(C/l) £ 0.2 A b 1.793 1.126 1.140 0.694 0.688 1.588

3.928 3.950 3.958 3.998 3.995 4.293

0.2 £ (C/l) £ 0.5 a b 1.722 1.073 1.065 0.755 0.667 1.385

3.676 3.271 3.452 2.632 3.280 3.525

IMPEDANCE OF LOOP ANTENNA IN CONDUCTING MEDIUM In the previous section, we have described the input impedance, and the resistive and reactive parts of loop antenna, using equivalent Tx line model. But the properties such as the conductivity of the medium have not been taken into consideration, which may effectively change the impedance of the antenna. Galejs [9] in his paper reported that a thin layer of insulation helps to maintain a uniform current flow around a loop antenna, in case it is immersed in a conducting medium. Later, it was also suggested that the presence of the insulating material could be neglected while computing the impedance of loop antenna; however, Kraichman [10] first analyzed a loop antenna located in a conducting medium. Therefore, the present section describes a method to calculate the impedance of a loop antenna located in a conducting medium using tabulated functions. Let a circular loop antenna having uniform current distribution across its wire is immersed in a medium of conductivity (s) as shown in Fig. 6.15, where am, ai and a0 are the medium, inner and outer radii of loop, and r is radius of conductor wire [11].

FIG. 6.15

Circular loop geometry.

Loop Antennas

257

Then, the impedance of loop is given by (see [8]) Z = jXN am ai

where H = − jXNT =

(1 − j )

∫

Q

e − jH R

cos Z dZ = (Re + jX m )

R

0

T >> 1 and d is skin depth. XF

is propagation constant for

E

(6.49)

R 2 = am2 + ai2 − 2 ai am cos Z

(6.50)

If y = R/d, the real and imaginary parts of impedance from Eq. (6.49) can be separated as

XN ai am E XN ai am = E

Re =

Xm

∫

Q

f (y )

0

y

∫

Q

g( y)

0

y

cos Z dZ

(6.51)

cos Z dZ

(6.52)

where f(y) and g(y) are newly introduced functions and defined as follows:

f (y) = e− y sin y = −

∞

∑ n =1

g(y) = e − y cos y = 1 +

∞

∑ n =1

⎛ nQ ⎞ (2) n/2 ( − y) n sin ⎜ ⎟ ⎝ 4 ⎠ n!

(6.53)

⎛ nQ ⎞ (2) n/2 ( − y) n cos ⎜ ⎟ ⎝ 4 ⎠ n!

(6.54)

If the ratio (ai/am) lies between (1 < ai/am £ 0.95), the value of y may be accurately approximated as y=

R E

=

ai am

4(1 − cos Z )

E

2

=

2 ai am E

sin

Z 2

(6.55)

Therefore, from Eqs. (6.53) and (6.54), the value of Re [Eq. (6.51)] reduces to Re = − XN

where K n −1 =

∫

Q 0

Z⎞ ⎛ ⎜ sin ⎟ 2⎠ ⎝

∞

(2)(3 n − 2)/2

n =1

n!

ai am ∑ n −1

cos Z dZ .

⎛ nQ ⎞ sin ⎜ ⎟ ⎝ 4 ⎠

⎛ ⎜− ⎜ ⎝

n

ai am ⎞ ⎟ K E ⎟⎠ n −1

(6.56)

258


Substituting the value of g(y) into Eq. (6.54), Eq. (6.52) gives X m = XN ai am

−

∫

Q

cos Z

0

R

XN ai am E

∫

dZ ⎛ nQ ⎞ ( − y)n −1 cos ⎜ ⎟ ⎝ 4 ⎠ cos Z dZ = X m1 + X m 2 n!

∞

Q

∑ (2) n / 2

0

n =1

(6.57)

which consist of two terms. The first term is recognized as the external inductance of a direct current loop, i.e., (6.58) Xm1 = wL1 where L1 = N ai am

∫

Q

cos Z

0

R

dZ

Equation (6.58) can be solved in terms of elliptic integrals and L1 is found equal to ⎤ 1 ⎞ ⎡⎛ q2 ⎞ ⎛ L1 = N ai ⎜ 1 + ⎟ ⎢⎜ 1 − ⎟ K − E⎥ p ⎠ ⎣⎢⎝⎜ 2 ⎠⎟ ⎝ ⎦⎥

in which p =

ai am

2 p

, q=

(1 + p)

, K=

∫

dG

Q /2

(1 − q sin G ) 2

0

2

1/2

(6.59)

and E =

∫

Q /2 0

(1 − q 2 sin 2G )1/2 dG ,

and the second term

X m2 =

XN ai am E

∫

Q 0

Q

∑ (2) n/2

n =1

⎛ nQ ⎞ ( − y) n −1 cos ⎜ ⎟ ⎝ 4 ⎠ cos Z dZ n!

(6.60)

Substituting the value of y from Eq. (6.55) and after simplification yields X m 2 = X L2 = XN

or

∞

(2)(3n − 2)/2

n =1

n!

ai am ∑

∞

(2)(3 n −2)/2

n =1

n!

L2 = N ai am ∑

⎛ nQ ⎞ cos ⎜ ⎟ ⎝ 4 ⎠

⎛ ⎜− ⎜ ⎝

⎛ nQ ⎞ ⎛ cos ⎜ ⎟ ⎜− ⎝ 4 ⎠ ⎜⎝

n

ai am ⎞ ⎟ I E ⎟⎠ n −1 n

ai am ⎞ ⎟ I E ⎟⎠ n −1

(6.61)

where In–1 is another constant function [11]. Therefore, the total external impedance of the loop antenna can be expressed as Z = Re + jX (L1 + L2 )

where Re, L1 and L2 are defined in the above equations.

(6.62)

Loop Antennas

259

FERRITE ROD ANTENNA As already mentioned, the main disadvantage with a simple loop antenna is its low efficiency. This is so because loss resistance of a small magnetic loop antenna is comparable to its radiation resistance. Expression to calculate efficiency of a loop antenna indicates that antenna efficiency can be increased by increasing the radiation resistance and in turn the circumference of the loop. But this method makes antenna heavy. Fortunately, ferrite rod has the properties of increasing magnetic flux, the magnetic field, the open-circuited voltage and hence the radiation resistance. This is because of the high permeability m and high resistivity of the ferrite material. The high resistivity reduces the core losses and raises the quality factor of a coil. Therefore, by inserting a ferrite rod into the circumference of the loop, antenna efficiency could be increased. The parameters of n-loop ferrite antenna are modified as follows [12]: The induced maximum emf

V=

2Q

M

EANF Ner

(6.63)

⎛ N NA ⎞ ⎛ NC ⎞ 2 Radiation resistance Rr = 31200 ⎜ er 2 ⎟ = 20Q 2 ⎜ ⎟ N er ⎝ M ⎠ ⎝ M ⎠ 2

Effective length le =

2Q

M

ANF Ner

as V = Ele

4

(6.64)

(6.65)

where mer = effective relative permeability of the ferrite and its value ranges from 100 to 10000. The mer depends upon the choice of material and the size and shape of the rod. Higher length to diameter ratio of the ferrite rod offers a high permeability m, which is desirable for better performance. The effective relative permeability of ferrite loop in terms of relative intrinsic permeability of unbounded ferrite material mfr is given by

Ner =

Ne Nfr = N0 1 + D(Nfr − 1)

(6.66)

where mfr is much larger than 1 (i.e., mfr >> 1) and D is demagnetization factor and found to be different for different geometries. The value of D for an ellipsoid of length 2l and radius a (l >> a) is given by 2

⎤ ⎛ a ⎞ ⎡ ⎛ 2l ⎞ D = ⎜ ⎟ ⎢ ln ⎜ ⎟ − 1⎥ ⎝l⎠ ⎣ ⎝a⎠ ⎦

(6.67)

However, for a sphere, D is equal to 1/3. The basic configuration and radiation mechanism of ferrite rod antenna are shown in Figs. 6.16 and 6.17 respectively. Ferrite rod antenna is a broadside radiator; maximum radiation is in direction perpendicular to the antenna axis provided antenna lies at 90° w.r.t. incoming waves (see Fig. 6.18). Because this particular orientation induces maximum voltage in the antenna coil, the performance

260


FIG. 6.16

Basic configuration of ferrite rod antenna. Magnetic field lines along rod antenna

Coils around rod

FIG. 6.17

Radiation mechanism of ferrite rod antenna.

Ferrite rod antenna

FIG. 6.18

Radiation pattern of ferrite rod antenna.

of the antenna is maximum when coil is positioned at the centre of the rod and degraded in case coil moves toward the end of the rod. Therefore, coil is placed at the quarter point from one end of the rod. Because of losses in ferrite, when we try to use ferrite loop antenna as Tx antenna, the power dissipated heats up the material until it decomposes/melts. This is because ferrite behaviour tends to vanish (me falls to unity), when we apply a large field or try to transmit significant power levels. This is the reason why ferrite rod antenna makes an excellent Rx antenna, not Tx antenna, except the power level to be transmitted is quite low (typically less than a watt or so).

Loop Antennas

261

APPLICATIONS OF LOOP ANTENNA Loop antennas are used in RF and aircraft receivers, direction finding and also as UHF transmitters. Loop antennas are also used in pagers and handheld transceivers, which is probably due to loop’s low efficiency, which is not as important as SNR. Most of the applications of loop antennas are found in HF, VHF and UHF bands, and they are used even in microwave frequency range as probes in field measurement and as directional antennas for radio-propagation. Large loop antennas significantly used in directional arrays are helical arrays, yagi-arrays and quad arrays. As far as ferrite rod antennas are concerned they are mostly used in pocket transistor radios. Ferrite loop antenna is a form of RF antenna and found suitable in portable transistor, broadcast receiver as well as in many hi-fi tuners for LW, MW and SW frequency bands. As the antenna is tuned, it usually forms the RF tuning circuit for the receiver, enabling both functions to be combined within the same components, thereby reducing the number of components and hence the cost of set. Ferrite rod antenna is directive and it best operates only when the magnetic force lines fall in line with the antenna. That is the antenna has a null position where the signal level is at a minimum when the antenna is in the direction of the transmitter. Though we can compare ferrite loop antenna with RF antenna, its efficiency is much less than that of a large RF antenna. The performance of the ferrite also limits the frequency response and it is only up to MHz levels. Mostly it is used as Tx; however it can also be used as Tx antenna where efficiency is not an issue and transmitted power level is low. As they are more compact than RF antennas, this can be an advantage, and as a result they can be used in RFID applications.

SOLVED EXAMPLES Example 6.1 Calculate the voltage induced by a plane wave of electric field E = 0.03 Vm–1 at operating frequency 2.0 MHz for a (i) vertical antenna of height 8 m and (ii) loop antenna of 1.5 m2 and 10 turns, if the plane of the loop is in the plane of propagation of the wave and also, (iii) find total power radiated by the loop. Solution:

Given

E = 0.03 V/m, l = 300/2 m, h = 8 m N = 10, q = 0° and A = 1.5 m2

Therefore (i) Magnitude of induced voltage in vertical antenna is given by | Vrms | = | Erms h sin wt | = Erms h = 0.03 ´ 8 = 0.24 V (ii) Vrms for the loop is Vrms =

=

2 Q Erms AN

M

cos R V

2 × 3.14 × 0.24 × 1.5 × 10 150

cos 0° = 0.151 V

262


2

⎛ A⎞ ⎛ 1.5 ⎞ (iii) Radiation resistance Rr = 31200 ⎜ 2 ⎟ = 31200 ⎜ = 2.08 : 2 ⎟ ⎝M ⎠ ⎝ 150 ⎠ Therefore, the radiated power PT =

1 V2

=

4 Rr

1 4

×

0.1512 2.08

= 2.74 mW

Example 6.2 A circular loop frame has 150 turns such that the periphery of each turn is 1.5 m. Determine maximum field strength so that receiver tuned at 10 MHz inducing a voltage of 10 mV. Consider loop is oriented at 60° from the direction of the transmitter and quality factor of the loop is 85. Solution:

Given: Q = 85, N = 150, 2pr = 1.5 p; therefore r = 0.75 m. A = pr2 = 3.14 ´ 0.752 = 1.766 m2 f = 10 MHz Þ l = 30 m, Vin = 10 mV and q = 60°

We know that Vrms = Erms =

and

Vin

=

Q

10 −2 85

= 1.176 × 10 −4 V

Vrms M

=

2Q AN cos R

1.176 × 10 −4 × 30 2 × 3.14 × 1.766 × 150 × 0.5

Vm −1 = 4.232 μ Vm −1

Em = 2 × Erms = 2 × 4.232 = 5.985 Vm–1

Hence

Hm =

Em 377

=

5.985 377

= 1.587 × 10 −2 μ Am −1

Example 6.3 A square loop of arm length 0.8 m has 20 turns. The winding wire is characterized with resistance 10–5 W and inductance 0.5 mH. It is supposed to be turned by a capacitor at resonance with a wave defined by E = 250 sin (8p ´ 10–5) mVm–1. Find the rms value of voltage developed across the capacitor when the waves incident at 60° w.r.t the plane of loop. Solution:

Area, A = 0.8 ´ 0.8 = 0.64 m2 N = 20 R = 10 W

L = 0.5 mH

E = 250 sin (8p ´ 10–5) Then

Em = 250 ´ 10

therefore

Em =

Em 2

=

–6

Vm

250 × 10 −6 2

t = Em sin wt (say)

–1

= 176 × 10 −6 Vm −1

w = 2p ´ f = 8p ´ 105 rad s–1. Therefore, the voltage across the capacitor will be

Loop Antennas

(Vrms )c = Vrms Q =

2Q E rms AN cos R X L

M

R

263

X 2 E rms AN cos R L

=

c

R

In which c is velocity of wave and 3 ´ 108 ms–1. Therefore,

(Vrms )c =

(8Q × 10 5 )2 × 176 × 10 −6 × 0.64 × 20 × 0.5 × 0.5 × 10 −3 3 × 108 × 10.5

=

3.554 × 10 5 3.15 × 10 9

= 1.128 ´ 10–4 V = 0.1128 mV Example 6.4 Design a loop antenna which induced a maximum emf of 0.55 mV across it at a frequency of 1 MHz when it is placed in a field region, where the rate of flow of energy is 5 mVm–2. Solution:

The operating wavelength l = 300/1 = 300 m

⎛ A , where Rr = 31171 ⎜ 2 We know that Wmax = 4 Rr ⎝M Let us consider a circular loop antenna of area A is used V2

Rr =

Hence,

V2 4 Wmax

=

(0.55)2 × 10 −6 4 × 5 × 10 −3

⎡ 15.125 × 10 −6 × (300) 4 ⎤ A= ⎢ ⎥ 31171 ⎥⎦ ⎣⎢

2

⎞ ⎟ . ⎠ in this case.

= 15.125 μ: 1/2

= 1.982 m 2 1/2

⎛ 1.982 ⎞ A = Q a2 = 1.982 ⇒ a = ⎜ ⎟ = 0.79 m ⎝ Q ⎠ i.e., radius of loop is 0.79 m. Therefore, perimeter of loop = 2pa = 2 ´ 3.14 ´ 0.79 = 49.612 cm.

Example 6.5 A 3-turn circular loop antenna of radius is 12 cm made up of copper of radius 1.2 mm. The antenna carries a current of 10 mA when being operated at f = 2 MHz. Calculate the radiated power, percentage radiation efficiency, effective area and gain of the antenna. Solution:

N = 3, a = 0.12 m, d = 2.4 ´ 10–3 m, Im = 10 ´ 10–3 A, l = 300/2 = 150 m.

The total radiated power P = 60 Q 3 × I m2 × n × =

a

M

60 × (3.14)3 × (10 −2 )2 × 0.12 × 3 150

W = 0.4458 mW

264


The radiation efficiency (I) =

1 ⎛ Rl ⎞ + 1⎟ ⎜ ⎝ Rr ⎠

where for copper

Rl

3430

=

Rr

C

3

=

3.5 fMHz Nd

3430

= 1.99 × 10 6

3

⎛ 2 × 3.14 × 0.12 ⎞ 3 ⎜ ⎟ × 2 × 1 ⎝ ⎠

2 × 3 × 2.4 × 10 −3

Therefore radiation efficiency 1

(I) =

C

=

M

≈

(1.99 × 10 + 1) 6

2 × 3.14 × 0.12

1

= 5.025 × 10 −5 = 5.025 × 10 −3 %

(1.99 × 10 6 )

= 5.02 × 10 −3 <<

150

1 3

.

Hence Dmax = 3/2. Maximum effective aperture Ae =

Gain G =

M2 150 2 = = 1194.26 m 2 4Q D 4 × 3.14 × 1.5 4Q × Ae ×

M

I

2

=

4 × 3.14 × 1194.26 × 5.025 × 10 −5 150

2

= 3.35 × 10 −5

= 10 log 3.35 – 5 = 0.25 dB. Example 6.6 A circular loop antenna with uniform in phase current has diameter D. Find (i) radiation pattern for D = l/10; (ii) Rr and directivity of the antenna for D = l/3. Solution:

(i) The far-field pattern of a circular loop is given by

M ⎛ 2Q ⎞ ⎛Q ⎞ × × sin R ⎟ = J1 ⎜ sin R ⎟ J1 (C a sin R ) = J1 ⎜ 6 ⎝ M ⎠ ⎝3 ⎠ 3

1⎛Q 1 ⎛Q ⎞ ⎞ ⎛Q ⎞ J1 ⎜ sin R ⎟ is a Bessel’s function of first order and equals ⎜ sin R ⎟ − ⎜ sin R ⎟ + ... 2⎝ 3 8⎝3 ⎠ ⎠ ⎝3 ⎠

For

C

M

≤

1 3

⇒

C

M

=

2Q

M

×

M 20

=

Q 10

= 0.314

Loop Antennas

265

1 ⎛Q ⎛Q ⎞ ⎞ J1 ⎜ sin R ⎟ ≈ ⎜ sin R ⎟ = 0.523 sin R 2 ⎝3 ⎝3 ⎠ ⎠

Therefore, far-field pattern can be found as (E1)q=0 = 0 dB (E)q=30° = 0.523 ´ 0.5 = –5.83 dB Similarly (E3)q=60° = –3.44 dB

(E4)q=90° = 0.523 ´ sin 90° = –2.81 dB

and

The resultant pattern is shown in Fig. 6.9.

FIG. 6.19

Radiation pattern of circular loop antenna for Example 6.6.

(ii) We have

C

M

=

2Q

M

×

M 6

= 1.0472 >

1 3

M⎞ ⎛a⎞ ⎛1 Therefore, Rr = 3720 × ⎜ ⎟ = 3720 × ⎜ × ⎟ = 620 : 6⎠ ⎝M⎠ ⎝M ⎛ 2 × 3.14 × M ⎞ D = 0.682 ⎜ ⎟ = 0.7141 = − 1.462 dB M ×6 ⎝ ⎠

Example 6.7 Find skin-effect resistance of a 4-turn circular loop antenna of copper wire to be operated at f = 2 MHz. The radius of loop’s wire is 1.2 mm. Also, find radiation efficiency, if loop radius is 0.45 m and resistance per unit length is 0.207 W.

nRs ⎛ XN ⎞ We know that Rs = ⎜ 0 ⎟ and R0 = 2Q a ⎝ 2T ⎠ 2

Solution:

266


⎛ 2 × 3.14 × 2 × 10 6 × 4 × 3.14 × 10 −7 ⎞ Rs = ⎜ ⎟ = 0.372 m: ⎜ ⎟ 2 × 5.7 × 10 7 ⎝ ⎠

4 × 0.372 × 10 −3

R0 =

Hence

IA

We know that

where a′ =

b′ =

a

M b

M

=

=

150

0.45 150

T =1 T

Rp

0.0095

R0

IA

0.1975

=

1.488 7.536

= 0.1975 :

⎡ 8.48 × 10 −10 (f MHzT r )1/2 ⎛ R p ⎞⎤ = ⎢1 + + 1 ⎥ ⎜ ⎟ n (b′)3 a′ ⎢⎣ ⎝ R0 ⎠ ⎥⎦

1.2 × 10 −3

T′ = =

2 × 3.14 × 1.2 × 10

−3

−1

= 0.08 × 10 −4

= 3.0 × 10 −3 Rp = (Rn – R0) = 0.207 – 0.1975 = 0.0095 W

= 0.048

⎡ 8.48 × 10 −10 (2 × 1)1/2 × 1.048 ⎤ = ⎢1 + ⎥ 4 × 27 × 10 −9 × 0.08 × 10 −4 ⎦⎥ ⎣⎢

−1

= [1 + 1.45 × 10 +3 ] −1

= (1455.43)–1 = 0.06878%. Example 6.8 Find resistance, inductance and impedance of the loop antennas to be operated at f = 1 MHz, if the shape of the loop is (i) Small circular loop of radius 0.15 m (ii) Square loop of sides 0.12 m (iii) Triangular loop of arms 0.28 m (iv) Hexagonal loop of arms 0.25 m Assume that the radius of the wire is constant in all cases and is equal to 2.5 ´ 10–3 m. Solution:

We have

C

M

=

2 × 3.14 × 0.15 300

= 0.00314 , which is less than 0.2.

Therefore, XL = jwL = Z0 tan (bl), in which case ⎛S⎞ Z 0 = 276 log ⎜ ⎟ ⎝r⎠

where l = C/2, S = area of loop, r = radius of wire.

Loop Antennas

⎛ 3.14 × (0.15)2 (i) Z 0 = 276 log ⎜ ⎜ 2.5 × 10 −3 ⎝

267

⎞ ⎟ = 276 log (0.0283 × 10 3 ) = 400.69 : ⎟ ⎠

⎛ 2 × 3.14 × 2 × 3.14 × 0.15 ⎞ X L = 400 tan ⎜ ⎟ = 400 tan 0.0098 = 0.06895 300 ⎝ ⎠ L=

Hence,

0.06895

= 10.97 nH

2 × 3.14 × 10 6

2

2 2 2 ⎛ 2Q ⎞ 2 20 × 4 × (3.14) × {3.14 × (0.15) } R = 20 ⎜ A = ⎟ 300 2 ⎝ M ⎠

=

788.768 × 0.05 90000

= 0.0044 :

(ii) Side of square is 0.120 m long. ⎛ 0.122 ⎞ Z 0 = 276 log ⎜ ⎟ = 209.87 : ⎜ 2.5 × 10 −3 ⎟ ⎝ ⎠

4l ⎞ ⎛ 2Q × Z in = jX L = jZ 0 tan ⎜ ⎟ 2⎠ ⎝ M ⎛ 4Q l ⎞ ⎛ 4 × 3.14 × 0.12 ⎞ = jZ 0 tan ⎜ ⎟ = jZ 0 tan ⎜ ⎟ 300 ⎝ M ⎠ ⎝ ⎠ = 0.0184 L=

0.0184 2 × 3.14 × 10

6

= 2.93 × 10 −9 = 2.93 nH

and ⎛ CC ⎞ R = a tan b ⎜ ⎟ ⎝ 2 ⎠ where a = 1.126 and b = 3.950. ⎛ 2 × 3.14 × 2 × 0.12 ⎞ Therefore, R = 1.126 tan 3.950 ⎜ ⎟ 300 ⎝ ⎠

log R = log 1.126 + 3.95 log tan (0.005024) = log 1.126 + 3.95 log (8.768 ´ 10–5) log R = 0.052 – 16.0254 = –15.973 Hence R = 1.062 ´ 10–16 W » 0.

268


(iii) Triangular loop’s arm length is a = 0.28 m Area =

3 4

3

a2 =

4

(0.28)2 = 0.034 m2

⎛S⎞ ⎛ 0.034 ⎞ = 312.856 : Z 0 = 276 log ⎜ ⎟ = 276 log ⎜ −3 ⎟ ⎝r⎠ ⎝ 2.5 × 10 ⎠ 3a ⎞ ⎛ 2Q ⎛ 3.14 × 3 × 0.28 ⎞ X L = Z 0 tan ⎜ × ⎟ = 312.856 tan ⎜ ⎟ = 0.0494 2 ⎠ 300 ⎝ M ⎝ ⎠ 0.0494

Hence

L=

and

⎛ CC ⎞ R = a tan b ⎜ ⎟ ⎝ 2 ⎠

2 × 3.14 × 10 6

= 7.86 × 10 −9 = 7.86 nH

where a = 0.694 and b = 3.998. Therefore ⎛ 2 × 3.14 × 3 × 0.28 ⎞ 3.998 ⎛ 3.14 × 0.28 ⎞ R = 0.694 tan 3.998 ⎜ ⎟ = 0.694 tan ⎜ ⎟ × 300 2 100 ⎝ ⎠ ⎝ ⎠

log R = log 0.64 + 3.998 log tan (3.14 ´ 0.0028) log R = – 0.1586 – 15.254 = –15.4086 Hence R = 3.9026 ´ 10–16 W » 0. (iv) Length of hexagonal arm is a = 0.25 m.

Area =

6 3 4

a2 =

6 3 4

(0.25)2 = 0.162 m 2

⎛S⎞ Z 0 = 276 log ⎜ ⎟ = 276 log ⎝r⎠

⎛ 0.162 ⎞ = 500.27 : ⎜ −3 ⎟ ⎝ 2.5 × 10 ⎠

⎛ 2 × 3.14 × 6 × 0.25 ⎞ X L = 500.27 tan ⎜ ⎟ = 0.13708 300 × 2 ⎝ ⎠

Hence

L=

0.13708 2 × 3.14 × 10 6

= 21.83 × 10 −9 = 21.83 nH

Loop Antennas

Therefore,

269

⎛ 2 × 3.14 × 3 × 0.25 ⎞ R = 1.588 tan 4.293 ⎜ ⎟ 300 ⎝ ⎠

log R = log 1.588 + 4.293 log (0.000274) Hence, R = 8.088 ´ 10

–16

or

log R = 0.2008 – 15.293 = –15.092

W » 0.

Example 6.9 Find the relative permeability of a single-turn ferrite spherical loop if the relative permeability of the un-balanced ferrite material (mfr) is 85.76. Also, find the radiation resistance if the circumference of loop is 3.55 l. Solution:

We know that the demagnetization factor for spherical rod is 0.333; therefore

Ner =

Nfr 85.76 = = 2.94 1 + D(N fr − 1) 1 + 0.333(85.76 − 1) 4

⎛C ⎞ Rr = 20 Q ⎜ ⎟ N er2 = 20 (3.14)2 (3.55) 4 (2.94)2 = 27.07 × 10 4 : ⎝M⎠ 2

Hence,

Example 6.10 Find the radiation resistance and the efficiency of a 10-turn ellipsoidal ferrite loop antenna of length 35 cm and radius 3.5 cm. The antenna is to be operated at 50 MHz and loss resistance is only 2.5% of the radiation resistance. Assume that mfr is 85.76. Solution:

The demagnetization factor ⎛a⎞ D=⎜ ⎟ ⎝l⎠

Ner =

2

⎡ ⎛ 2l ⎞ ⎤ ⎛ 3.5 ⎞ ⎢ ln ⎜ ⎟ − 1⎥ = ⎜ ⎟ ⎣ ⎝a⎠ ⎦ ⎝ 35 ⎠

2

⎡ ⎛ 2 × 35 ⎞ ⎤ −3 ⎢ ln ⎜ ⎟ − 1⎥ = 19.95 × 10 3.5 ⎠ ⎣ ⎝ ⎦

Nfr 110 = = 34.64 1 + D(Nfr − 1) 1 + 19.95 × 10 −3 (110 − 1)

A = pa2 = 3.14 ´ (3.5/100)2 = 3.85 ´ 10–3 ⎛ N NA ⎞ Rr = 31200 ⎜ er 2 ⎟ = 31200 ⎝ M ⎠ 2

Rl = 42.81 ×

Therefore

I=

Rr Rr + Rl

=

2

⎛ 34.64 × 10 × 3.85 × 10 −3 ⎞ ⎜ ⎟ = 42.81 : ⎜ ⎟ 362 ⎝ ⎠

2.5 100

42.81 42.81 + 1.070

= 1.070

= 0.9756 = 97.56

270


OBJECTIVE TYPE QUESTIONS 1. The radiation characteristics of a small loop antenna is equivalent to radiation of an infinitesimal magnetic dipole, whose axis is (a) Normal to the plane of loop (b) Parallel to the plane of loop (c) Both (d) None of these 2. A loop antenna is preferred for use as receiving antenna rather than the transmitting one because of its (a) High efficiency (b) High gain (c) Low efficiency (d) None of these 3. Moore loop antenna was designed on a square box frame, later which was famous as (a) Box loop antenna (b) Box square antenna (c) X-ray antenna (d) None of these 4. Quantum loop antenna is integrated with a high gain 40 dB amplifier; it is an example of (a) Small loop antenna (b) Long wire antenna (c) Ferrite rod antenna (d) None of these 5. When a loop antenna is freely rotated around its axis the maximum radiation appears at (a) (0°, 180°) and (90°, 270°) (b) (30°, 180°) and (60°, 270°) (c) (90°) and (180°) (d) Only at 145° 6. Foster’s expression deal with the directivity of a (a) Rectangular loop antenna (b) Square loop antenna (c) Circular loop antenna (d) None of these 7. The radiation efficiency of a ferrite loop antenna is greater than a loop antenna by the multiple factor (b) N er2 (a) F er2 (c) mer (d) None of these 8. The value of relative effective permeability (mer) of ferrite material ranges from (a) 100 to 10,000 (b) 10 to 1,000 (c) 50 to 500 (d) None of these 9. Loop antenna is a type of (a) Directive antenna (c) Both of these

(b) Low efficiency antenna (d) None of these

10. What will be the directivity of a circular loop antenna of radius l? (a) 5.7 (b) 1.5 (c) 0.745 (d) None of these 11. What will be the radiation resistance of a circular loop antenna of radius l? (a) 4000 W (b) 17200 W (c) 73.56 W (d) None of these

Loop Antennas

271

12. The directivity of small loop antenna is 1.5, because its radiation field is maximum at angle q (a) 90° (b) 60° (c) 100° (d) None of these 13. What will be the radiation efficiency of a 1 m dia loop of 10 mm dia copper wire at 103 KHz (a) – 40.5 dB (b) –50 dB (c) 82.75 dB (d) None of these 14. A 10-turn ferrite rod antenna operating at 2 MHz. The length of the rod is 15 cm, radius of the rod is 1.5 cm and diameter of a wire wound on the rod is 1.5 mm, what will be radiation efficiency if mer = 60? (a) 5.7 ´ 10–4 (b) 15.7 ´ 10–2 (c) 2500 (d) None of these 15. The induced maximum emf across ferrite antenna terminal increases by a multiple factor of (a) mer (b) mer + 1 (c) 25 mer (d) None of these

Answers 1. (a) 6. (c) 11. (d)

2. (c) 7. (b) 12. (a)

3. (a) 8. (a) 13. (a)

4. (c) 9. (c) 14. (d)

5. (a) 10. (d) 15. (a)

EXERCISES 1. What are the advantages of loop antenna over monopole antenna? 2. Describe the applications of loop antenna in wireless communication. How does it take the place of monopole antenna in wireless communication? 3. If a small loop of area A carries current I is radiating in free space, show that farzone electric field is given by E =

120 Q 2 [ I ] sin R ⎛ A ⎞ ⎜ 2 ⎟ . Draw its radiation pattern r ⎝M ⎠

also. 4. Describe the array theory of loop antenna. Derive the expression for the induced voltage for n-turn loop antenna operating at frequency fo. 5. What are the characteristic features of circular and square loop antennas? Write the expressions for their far-fields. 6. Describe the large loop antenna. Write the expressions for far-fields in different planes of the antenna.

272


7. Show that for a square loop antenna, the radiation resistance is Rr = 31200(a/l)4, where a is side of loop antenna, whereas for the elliptical loop antenna, it is Rr = 1950 (pab/l2)2, where a and b are the semi-major and semi-minor axes of the ellipse. 8. What are the advantages of multi-turn loop antenna? Show that the total ohmic resistance for n-turn circular loop antenna with loop radius b, wire radius a and turn separation 2c is RL =

⎛ Rp ⎞ Rs ⎜ + 1⎟ . a ⎝ R0 ⎠

nb

9. Using suitable approximations, find the resistance, inductance and impedance of a centre-fed loop antenna to be operated at f = 1.5 MHz, if the shapes of loop are (a) equivalent triangle of side = 0.3 m. (b) Hexagonal loop of arm = 0.28 m. 10. Find the radiation efficiency of a 12-turn ferrite ellipsoidal antenna, operating at f = 150 MHz, under the following specifications: Rt = 1.5% of radiation resistance Radius (a) = 0.5 m Length (l) = 1.2 m mfr = 90. Also, find the directivity of the antenna, if gain of the antenna is 15 dB. 11. Define the loop antenna. Mention the disadvantages of the loop antenna. 12. Define the directivity of a circular loop antenna. Explain both small and large loop antennas with suitable examples. 13. Derive the expression for the far-field components of a small loop antenna. 14. Find the voltage induced by a plane wave of field strength of 10 mVm–1 and frequency of 10 MHz in (i) vertical antenna of height 8 m; (ii) 10-turn loop antenna of area 2m2, if the loop is in the plane of travelling of the wave. [Ans: (i) 80 mV; (ii) 4.18 mV] 15. Calculate the field strength induced in a 10-turn loop antenna operating at 300 MHz. The loop antenna area is 2.5 m2 and it is found that maximum potential difference across antenna is 3.5 mV. Assume antenna response is maximum along travelling path of the wave. [Ans: E = 0.2512 Vm–1] 16. Find the change in induced voltage of 12-turn loop antenna operating at 20 MHz, if orientation of antenna changes 45° off from original position. Assume area of loop is 1.5 m2 and field across it is 0.02 Vm–1. [Ans: V1 = 0.15 V, and V2 = V1 cos 45° = 0.106, hence DV = 0.044] 17. Similarly find change in induced field strength if voltage across the antenna is 0.05 V. [Ans: E1 = 6.64 mVm–1t–1, E2 =

E1 o

cos 45

=

6.64 0.707

= 9.39 mVm −1t −1 ]

Loop Antennas

273

18. A 15-turn loop antenna each with 1.5 m2 area operating at 15 MHz. Calculate the peak value of the magnetic field intensity H of the radio wave, which induces an emf of 20 mV (rms) in the antenna. Also find changes in the value of H, if 5 turns are added to the loop. [Ans: H1 = H2 =

2 Vrms 2Q f N0 AN H1 × N1 N2

Am −1 =

=

0.02828 2662.09

1.062 × 15 20

= 1.062 μAm −1

= 0.7965 ]

19. Find the radiation efficiency of a 4-turn circular loop antenna operating at f = 100 MHz. The radius of loop is l/20 and radius of antenna wire is 10–4 l, and conductivity of wire is 5.7 ´ 10–7 Sm–1. Assume that the antenna is radiating into a medium having dielectric constant er = 5.5 and spacing between turns is 5 times greater than the radius of antenna wire.

XN0 2T

⎛ S2 ⎞ and Rr = 31200 N 2 ⎜ 4 ⎟ ] ⎜M ⎟ Rr + RL a ⎝ ⎠ 20. A 8-turn copper loop antenna operating at 50 MHz is connected to a 50 W Tx line. The radii of loop and its wire are l/25 and l/250 respectively, and spacing between turns is 0.01l. Determine directivity, radiation efficiency and gain of antenna, assuming Rp/R0 = 1.7. [Hint: D = 1.5 (1.761 dB), Rr(8) = 6.282 W, Rs = 18.609 ´ 10–4, RL = 4093 ´ 10–4 W, hr = 93.88% and G = 1.408 = 1.46 dB]

[Hint: I =

Rr

, Rs = RL =

b

REFERENCES [1] Kraus, J.D., Antennas: For All Applications, Tata McGraw-Hill, New Delhi, 2005. [2] Prasad, K.P., Antenna and Wave Propagation, 2nd ed., Satya Prakashan, New Delhi, 1996. [3] Carr, Joseph J., Practical Antennas: Handbook, 4th ed., McGraw-Hill/TAB Electronics, 2001. [4] Foster, D., “Loop antennas with uniform current,” Proc. IRE., Vol. 32, pp. 603–607, Oct. 1944. [5] Thiele, G.A. and W.L. Stutsman, Antenna Theory and Design, John Willey & Sons, New York, 2001. [6] Smith, G.S., “Radiation efficiency of electrical small multi-turn loop antenna,” IEEE Trans. Antenna Propagate, Vol. 20, No. 9, pp. 656–657, Sept. 1972. [7] Wheeler, H.A., “Fundamental limitations of small antennas,” Proc. IRE, Vol. 35, pp. 1479–1484, Dec. 1947.

274


[8] Awadallo, K.H., et al., “A simple method to determine the impedance of a loop antenna,” IEEE Trans. Antenna Propagate, Vol. AP. 32, No. 11, pp. 1248–1251, Nov. 1984. [9] Galejs, J. , “Admittance of insulated loop antennas in a dissipative medium,” IEEE Trans. Antenna Propagate, Vol. 13, pp. 229–235, March 1965. [10] Kraichman, M.B., “Impedance of a circular loop antenna in an infinite conducting medium,” J. Res. NBS, Vol. 66D, pp. 499–503, July–August 1962. [11] Benning, C.J., “Impedance of a loop antenna in a conducting medium,” IEEE Trans. Antenna Propagate, Vol. 13, pp. 242–243, March 1966. [12] Rumsey, V.H. and W.L. Weeks, “Electrically small ferrite loaded loop antenna,” IRE Convention Record, Vol. 4, Part I, pp. 165–170, 1956.

C H A P T E R

7

Metal-Plate Lens Antennas

INTRODUCTION Lens antenna basically a radiator which basic principle of operation is based on the collimated action of a simple optical lens. Like a parabolic reflector, a lens antenna is an antenna which is applied at microwave frequency and its working manner is same as that of a reflector antenna. That is, a lens antenna is also fed by horn/dipole antenna from its focal point and produces collimated rays on another (right) side of lens. The basic operating principle of a lens antenna is illustrated in Fig. 7.1. Plane wave front

Spherical wave front

Direction of plane wave propagation Source

Collimating metallic strip

FIG. 7.1

Basic operating principle of lens antenna.

From Fig. 7.1, it is clear that one can convert spherically radiated microwave energy into a plane wave in a desired direction by using a point source and collimated lens antenna. 275

276


The point source can be regarded as a gun that shoots the microwave energy toward the lens and the collimating lens forces all radiated segments of the spherical wave front into parallel paths. Usually a collimated lens is made of a dielectric material of finite dielectric constant (er). However, collimated lens can also be constructed using materials having refractive index less than unity at RF, so that the focusing properties can be achieved. Although parabolic reflector and lens antennas have same applications yet the main uses of lens antenna are only at higher frequencies, because at lower frequencies the lens antenna becomes heavy. Actual frequency range of lens antenna starts at 1.0 GHz, but its greatest uses are at and above 3 GHz. Lens antenna functions on the principle of equality of path lengths and illustrates the principle of reciprocity theorem.

Metal-Plate Lens Antennas A metal-plate lens antenna is an attractive alternative to dishes or large horns being used for portable microwave operation. There are other similar types of microwave lenses, namely dielectric lens and Fresnel lens; however, metal plate lens is the easiest to build, cheap, easy to construct and integrate, and lightest to carry. The metal plate lens antenna is constructed using series of thin metal plates with narrow air gap between them, such that the curvature of the edges of the plates forms the lens and the space between the plates forms a series of wave-guides. The input and output edges of these wave-guides are shaped simply to change path lengths and hence form the lens surface. The metal lenses were first originally described by KBIVC at the 1992 Eastern VHF/ UHF Conference for 10 GHz. Later, lens antennas were designed by Angel Vilaseca and found suitable for VHF communications. After that, Kock [1] described the design process, working principle and applications of metal-plate lens antenna using HDL_ANT computer program, which provides adequate gain enhancement and doubles the range of Gunnplexer system. The working principle of metal-plate lens is similar to ordinary lens, only refraction in this case occurs at the interface of two plates in which light travels at different speed changing the direction of travelling the beam of light. If the beam constitutes many rays of light, each one may be bent. The rays at the edge of beam are bent more; they end up parallel to the centre rays, which are hardly bent. This will happen only if each ray takes exactly the same time to travel from its source (i.e., focal point of the lens) to its destination. Since light travels more slowly in glass, a lens is made thicker at the middle to slow down the rays with a shorter path and thinner at the edges to allow the rays with longer paths to catch up. The simple arrangement of a lens along with travelling rays is shown in Fig. 7.1. The curvature of the lens to form the beam exactly is an ellipse. However, for small bending angles a circle is almost identical to an ellipse and nearly all optical lenses are grounded with spherical curves. The metal-plate lens differs from an optical lens is that the phase of the EM waves travel faster in a wave-guide than in free-space. Thus the curvature of the metal lens is the opposite of an equivalent optical or dielectric lens. Therefore, concave lens is same as convex lens in this case or vice versa. We can still avoid the use of circular aperture instead of ellipse, as long as we don’t require bending the rays too sharp; that is the reason why the lens is fed with a small horn.


277

Lens Antenna Design First of all, the dimensions of feeding horn are calculated by using Eplane =

56 AM e

deg and H plane =

67 AM h

deg

and the gain of horn antenna is G = 10 log (4.5 AleAlh) dB where Ale and Alh are aperture dimensions in E and H planes per wavelength respectively. The focal length of lens is given by

f =

Lensdiameter ⎛W ⎞ 2 tan ⎜ E ⎟ ⎝ 2 ⎠

where WE is the width of lens in E-plane. However, the spacing between the metal-plates is related to index of refraction as follows:

M ⎤ ⎡ N = ⎢1 − l ⎥ 2 dS ⎦ ⎣ in which Ml =

1/2

M0 is wavelength in the lens Fr

dS = spacing between plates m = refractive index = l1/l0 ÿ

and the focal length f of the two lens surfaces are calculated as follows: 1 ⎞ ⎛ 1 − = (N − 1) ⎜ ⎟ f R2 ⎠ ⎝ R1

1

where f is the focal length, and R1 and R2 are radii of curvature. For any single curved surface, one radius approaches infinity [see Fig. 7.2(a)]. The single curved lens along with symmetrical double curve is shown in Fig. 7.2(b). The radius of curvature, as calculated above, is for the surface and thus the central plate which has full curvature. The rest of the plates must be successively wider and have smaller radii so that the edges of all the plates form a spherical lens surface. However, for very large lenses the size may be reduced by stepping the width of the plates into zones which keep transmission in phase. In particular construction of metal plate lens antenna is simple. Metal-plates of aluminium foil spaced by Styrofoam of thickness 0.75 in. are formed metal-plate lens antenna. Final

278


FIG. 7.2

Basic geometry of E-plane metal-plate lens: (a) single curve; (b) double curve.

antenna is a block of foam-there is no need to shape the foam to the lens size. A metal lens antenna works only in the E-plane. This is parallel to the elements of a dipole or Yagi antenna but perpendicular to the wide dimension of a wave-guide. In order to achieve maximum gain the metal-plate must be normal to the wide dimension. The horn should point through the centre of the lens, but the focus is not as critical as a dish. The lens focuses the beam more tightly but does not change the beam direction, because tilting the lens will not steer the beam. Paul’s results for 10 GHz [2] reveal that the best gain was with the horn slightly closer to the lens than calculated. It might be due to edge effects. This effect may be eliminated if the size of the plates is chosen slightly larger than calculated and resulted the gain a bit higher. All the lenses were designed to be fed with the standard Gunnplexer horn which has small matched phase-centres. The gain measurements on an antenna range are shown in Table 7.1. It was found that the lenses perform with about 50% efficiency, if we consider them as having a round aperture. The corners do not contribute significantly, even though they are made square for convenient fabrication and mounting. The metal-plate lens antennas are also found useful at frequencies of 5.76 GHz and 24 GHz, with foam thicknesses of around 35 mm and 8 mm respectively. TABLE 7.1 Antennas Gunnplexer horn Horn + 150 mm Horn + 300 mm

Gain and efficiency for different lens antennas Focal distance — 8 in. 21 in.

Gain 17.5 dBi 20.9 dBi 27.4 dBi

Efficiency 57% 45% 50%


279

TYPES OF LENS ANTENNA Basically there are two types of lens antenna: E-plane and H-plane lens antennas. In addition, there are two special lens antennas. Luneburg and Rotman lens antennas. E-plane lens antenna: It is also known as fast lens antenna; here the electric path length is decreased by the lens. H-plane lens antenna: It is also known as dielectric or delay lens antenna, here electric path length is increased by the lens. Dielectric lenses are further divided two types: (a) Non-metallic type lens—polystyrene and lucite (b) Metallic or artificial dielectric type lens. All delay lens antennas may be regarded basically as end-fire antennas with the poly-rod and monofilar axial mode helix as the rudimentary form.

E-plane Lens Antenna The function of operation of E-plane lens antenna depends upon the acceleration of waves by the lens. In this type of lens, the metal plates are parallel to E-plane or plane of the electric fields. E-plane metal plate lens antenna makes use of wave-guide theory, where three wavelengths lc, lg and l0 are related to each other (see Fig. 7.3). If v is the velocity of wave propagation in the x-direction between two parallel conducting plates of large size, then it can be given as v=

v0 2 ⎡ ⎛M ⎞ ⎤ ⎢1 − ⎜ 0 ⎟ ⎥ ⎢⎣ ⎝ Mc ⎠ ⎥⎦

1/2

As lc = 2a, for the dominant mode TE10 wave, then

v=

Þ

v0

2 ⎡ ⎛M ⎞ ⎤ = ⎢1 − ⎜ 0 ⎟ ⎥ ⇒ v0 ⎝ 2 a ⎠ ⎦⎥ ⎣⎢

v

2 ⎤ 1/2

⎡ ⎛M ⎞ ⎢1 − ⎜ 0 ⎟ ⎥ ⎢⎣ ⎝ 2 a ⎠ ⎥⎦

m = [1 – (l0/2a)2](1/2)

1/2

=

1

N

(7.1)

where v0 = velocity of wave in free space. a = spacing between plates l0 = free space wavelength m = equivalent index of refraction of a medium constructed by many such parallel plates at constant spacing

280

Antenna and Wave Propagation z

E

O

y

a x

FIG. 7.3

Basic configuration of E-plate lens antenna.

Equation (7.1) indicates that m is always greater than unity and the spacing between plates must not be less than a critical value, i.e.,

a=

M 2

. This implies that a ³ l/2.

Therefore, while designing an E-plane metal plate lens antenna, it is compulsory to construct the antenna from parallel plates only. It is also found that the velocity of wave propagation v between the plates is always greater than the free-space velocity v0. The geometry of E-plane metal plate lens, shown in Fig. 7.2, contains r, which is found from Fermat’s principle as follows: r=

L (1 − N ) 1 − N cos R

(7.2)

That is, when (m < 1), Eq. (7.2) represents an equation of ellipse. If primary antennas were a line source perpendicular to say, Fig. 7.2, all the plates would be identical and the lens surface would be in the form of an elliptical cylinder. It must also be noted that waves entering the lens at point P (L to R) obey Snell’s law of refraction; however, it is not necessarily the case for wave entering at point P¢, where the metal plates constrain the wave to travel between them. The major disadvantage of this antenna is that it is frequencysensitive (i.e., BW is relatively small), and the total BW is given by (see [3])

'f =

2NE (1 − N 2 ) t

(7.3)

where t = thickness of concave portion of lens antenna ÿ d = electrical path difference of OQQ¢ and OPP¢; if d = 0.25l, then

'f =

50 N (1 − N 2 ) t

(7.4)


281

Hence, for m = 0.5 and t = 6l, the bandwidth Df is found to be 5.5%, which is very small. The only possible way to decrease the frequency sensitiveness of E-plane metal plate lens antenna is zoning. A zonal E-plane metal plate antenna is shown in Fig. 7.4.

FIG. 7.4

Zonal E-plate lens antenna.

Here the thickness (z) of a zone step is given by

M0

z=

(7.5)

1 − N

Therefore the focal length (L) for nth zone is equal to Ln = L + (n – 1)z

(7.6)

where n = 1, 2, 3, …. Zoning of an antenna is important in two contexts: first, it saves weight and second, it increases the bandwidth. The bandwidth of a zonal E-plane metal plate lens antenna is given by 'f =

50 N

(7.7)

(1 + K N )

where K is the number of zones, the zone on the axis of the lens being counted as the first zone. Under the above condition, i.e., m = 0.5, and K ~ t/2 = 3, the bandwidth of this zoned lens is found to be 10%, i.e., nearly double than the conventional lens antenna, i.e.,

'f =

50 × 0.5 1 + 3 × 0.5

=

25 1 + 1.5

= 10%

282


H-plane Metal-Plate Lens Antenna (a) Non-metallic dielectric lens antenna Non-metallic dielectric lens antenna is similar to the optic lens, hence they may be described by the ray analysis method of optics and Fermat principle. If this lens has to convert a spherical wave from a source (at focus) into a plane wave front, then all the ray’s paths from O to the plane surface of the lens should have equal electric length (see Fig. 7.5). A planoconcave lens can solve the purpose, provided it is fed with an isotropic antenna.

FIG. 7.5

Non-metallic dielectric lens antenna.

Then from Fig. 7.5, the electric path length OPP¢ must be equal to OO¢QQ¢ or simply OP = OQ. Therefore r

M0

=

L

M0

+

r cos R − L

Ml

⇒ r=

L (N − 1)

N cos R − 1

(7.8)

where

N=

M0 free space wavelength = wavelength in line Ml

Equation (7.8) is an expression of hyperbola with focal length (l) and radius (r) and represents the curvature of the lens in polar co-ordinate and describes the required shape of the lens. Polystyrene (er = 2.5 and m = 1.6) and polyethylene (er = 2.2 and m = 1.6) are important materials for dielectric lens antenna deigns. If the wavelength is comparable to the lens aperture at RF, the wave emanating from such lens antenna produces radiation pattern with uniformity of illumination just like a parabolic reflector. If the lens antenna is circular and of diameter D, the radiation of uniform illustration is like that of paraboliod and gain is approximated by G = 6(D/l)2. A large focal length provides more uniform illustration rather than a short focal length lens. However practically it is difficult to get uniform illumination for a lens antenna.


283

(b) Metallic or artificial dielectric type lens The concept of artificial lens antenna was first proposed by Kock in 1948 in a paper entitled, ‘Metallic delay lens’. The design procedures for artificial lens antenna are similar to ordinary dielectric lens. The major difference between ordinary and artificial lens antennas is that ordinary lens antenna consists of molecular particles of microscopic size, while artificial one consists of discrete metal particles of macroscopic size. The metal particles may be of any configuration: spheres, disks, strips or rods. But mostly spherical metal particles are considered due to its simplicity and ease of analysis [4]. A plano-convex lens constructed with metal spheres and arranged in a 3-dimensional array or lattice structure as shown in Fig. 7.6. This arrangement simulates the crystalline lattice of an ordinary dielectric substrate but on a much large scale. These spheres are analogous to oscillating molecular dipoles of an ordinary dielectric lens antenna.

FIG. 7.6

3-D lattice of dielectric substrate used for lens.

There are two basic design requirements of an artificial lens antenna: · The size of metal particles should be small compared to the operating wavelength. This is to avoid resonance effect. The maximum particle size (parallel to the electric field) less than l/4 is found satisfactory. · The spacing between the particles should be less than l, this is to avoid diffraction effect.

ANTENNA ANALYSIS AND DESIGN In order to design an artificial lens antenna, it is necessary to know the effective index of refraction in addition to the other parameters. The refractive index of a lens can be measured experimentally with a slab of the material or calculated it theoretically. In the present section a method to calculate the refractive index is described by considering a spherical metal particle, because they are light in weight and can be more readily analyzed.

284


In Fig. 7.7, an uncharged spherical particle is placed in an electric field E, which induced two opposite charges: (+q) and (–q). If these induced charges are separated at distance l, the arrangement is an electric dipole of dipole moment (m), which is equal to ql, i.e., m = ql.

FIG. 7.7

Spherically charged particle placed in an electric field.

Therefore, the potential due to this dipole at distance (r), where r >> l, will be (see [5])

V =

ql cos R

(7.9)

4 QF 0 r 2

and the polarization of the artificial dielectric lens will be

P = Nql where N = total number of spheres/cm3 l = unit vector of length (l) joining the charges Since polarization P is related to electric displacement density (D) and electric field intensity (E) as follows: D = F E = F0 E + P

hence effective dielectric constant of lens (e) is F = F0 +

P E

= F0 + N

ql E

(7.10)

where e0 is the free space permittivity. The potential at a distance r from the point charge is defined as the work done per unit charge by an external agent in transferring a test charge from infinity to that point and is expressed as follows: V =−

∫

r ∞

E dl


285

Therefore in the present case, i.e., in a uniform field the potential will be V =−

∫

r ∞

E cos R dr = − Er cos R

(7.11)

where q is the angle between radial vector and the field direction. Then the potential V0 outside the sphere placed in an originally uniform field will be

ql cos R

V0 = − Er cos R +

(7.12)

4 Q F0r 2

Therefore as r ® a, i.e., at the circumference of the sphere, V ® 0, i.e.,

Er cos R =

ql cos R 4 QF 0 r

2

⇒

ql E

= 4 QF 0 a3

(7.13)

Hence, from Eq. (7.10), F = F 0 + 4QF 0 Na

3

⇒

F F0

= 1 + 4 Q Na3

er = 1 + 4pNa3

or

(7.14)

where er = e/e0 and defined as the effective relative permittivity of artificial dielectric lens. Therefore, if the effective relative permeability of the artificial dielectric is unity, the index of refraction is given by

N = Nr Fr = [1 + 4Q Na3 ]1/2

(7.15)

However, the magnetic field lines of a radio wave are deformed around the sphere, because highfrequency fields penetrate to only a very small distance in good conductor. The effective relative permeability of an artificial dielectric of conducting sphere is given by mr = (1 – 2pNa3); hence

N = [(1 + 4Q Na3 )(1 − 2Q Na3 ]1/2

(7.16)

Equation (7.16) reveals that m¢ is smaller than the m¢0. Since in most of the cases the relative permeability is unity, in general m¢ can be taken as F r . The values of m¢ for different dielectric spheres are listed in Table 7.2. TABLE 7.2 Type of particle

Different values of m¢ for different dielectric spheres

Relative permittivity (er)

Relative permeability (mr )

Index of refraction

Sphere

(1 + 4pNa3)

(1 – 2pNa3)

m = [(1 + 4pNa3) ´ (1 – 2pNa3)]1/2

Disk Strip

(1 + 5.33Na3) (1 + 7.85N¢W3)

~1 ~1

(1 + 5.33 Na3)1/2 (1 + 7.85N¢W3)1/2

286


All the values given in Table 7.2 are reliable only for er < 1.5; however for er > 1.5, N becomes sufficiently large and particles start interacting because of their close spacing [6].

DISTRIBUTION OF ILLUSTRATION ON THE APERTURE OF THE LENS In order to find the variation of field intensity in the aperture of spherical lens, let us consider an annular section of strip width (dr) on the annular zone of radius (r) at angle (q), as shown in Fig. 7.8. From the figure it is clear that for an isotropic point source (primary antenna) and given focal length L, the field at the edge of the line (q = q1, say) is lesser than at the centre (q = 0). The variation of field intensity in the aperture plane of the lens can be determined by calculating the power per unit area passing through an annular section of the aperture as a function of the radius (r). Referring to Fig. 7.8, the total power passes through the considered annular section can be given by W = 2pr dr Pr

(7.17)

where Pr is power density/Poynting vector at radius (r).

FIG. 7.8

Annular section of strip width (dr) on annular zone of radius (r) at angle (q).

This power must be equal to the power radiated by the isotropic source over the solid angle = 2p sin q dq. That is W = 2p sin q dq U

(7.18)

where U is radiation intensity of the isotropic source. Equations (7.17) and (7.18) yield ( p) S =

where r = r sin q as q ® 0.

(N cos R − 1)2 ((N − 1)2 (N − cos R ) L2

U

287


PS =

(N − 1)2 (N − 1) (N − 1)L 2

2

U=

U

(7.19)

L2

Therefore, the ratio of (Pr) to (Pr)q=0 is given by

(PS )R (PS )R =0

=

(N cos R − 1)2 (N − 1)2 (N − cos R )

Hence, in the aperture the field intensity is equal to ⎡ (PS )R ⎤ = ⎢ ⎥ E0 ⎢⎣ (PS )R =0 ⎥⎦

ER

1/2

=

(N cos R − 1)

1

(N − 1)

(N − cos R )

(7.20)

where Eq/E0 is referred to as relative field intensity (Er) at a radius (r). For m = 1.5 ES =

(1.5 cos R − 1)

1

(1.5 − 1)

(1.5 − cos R )

(7.21)

Therefore, for q = 20°, Er = 0.7 and for q = 40°, Er = 1.4. Therefore for nearly uniform aperture illumination, an angle (q1) to the edge of the lens (even less than 20°) is essential unless the pattern of the primary antenna is an inverted type, i.e., one with less intensity in the axial direction (q = 0) than in direction of off-axis. However uniform aperture illumination may be replaced by a tapered illumination in order to suppress minor lobes. But the disadvantage of this method of producing a taper is that the lens becomes bulk. However, the bulkiness can be reduced by choosing a lens of smaller (q1) and large focal distance.

LUNEBURG LENS ANTENNA Luneburg lens is a live example of artificial dielectric lens antenna. It is designed to provide scanning performance that is free from the direction of radiated beam as well as its own spherical geometry. Lenses of Luneburg type may be constructed on foamed dielectric containing many small glass spheres whose spacing are varied to achieve the relation m(r). They may also be made of spherical shells of graded dielectric constant fitted one within the other; however, at least ten steps with equal increments of refractive index are desirable. Twodimensional lenses may be made using conducting plates with a variable spacing adjusted to provide the required refractive index variation, assuming that propagation within the lens takes place as in a TE10 wave-guide. If a Luneburg sphere is cut in half and a reflecting sheet placed on the flat side, Luneburg reflector lens antenna results in an incoming wave at an angle of incidence qi brought to a focus at the corresponding angle of reflection qr = qi. Basically, two primary feeds are used for the Luneburg lens antenna: (a) open ended wave-guide and (b) tapered slot antenna. Especially, tapered slot antenna makes the antenna

288


integrated with an entire RF system. A rotationally symmetric parallel plate Luneburg lens requires a specific refraction index in order to focus the wave of the primary feed at the lens aperture [7]. Here, a periodic regular metal post structure, which acts as a metallic artificial isotropic dielectric, is used. Y.J. Port used geometrical optics (GO) and aperture integration method in order to predict the far-field pattern of Luneburg antenna. It was found that the proposed antenna offers wide coverage with multiple primary feeds at very low cost. Also, the complete antenna system is made of metal and hence can be cheaply and massively produced using such a precise casting method. The Luneburg lens antenna is a spherical symmetric delay type lens formed with a dielectric having index of refraction (m), which depends on radius as follows [see Fig. 7.9(a)]: 2 ⎡ ⎛r⎞ ⎤ N = ⎢2 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ R ⎠ ⎥⎦

1/2

(7.22)

where R = radius of sphere r = radial distance from centre of sphere When r ® 0, N = 2 − N max at centre of sphere. When r ® 1, m = 1 = mr at principal axis.

O

FIG. 7.9(a)

Basic geometry of Luneburg lens antenna.

Using Luneburg lens antenna signals can be received simultaneously from many directions since space is available on the sphere to place feed horns or other receiving devices. For steering a single beam, the receiver/transmitter can be switched to different feed horns, or a single movable feed horn can also be used. The variable refractive index can be obtained with an artificial dielectric material or with concentric shells of dielectric having different indices of refraction. Full Luneburg lens provides beam steering in both polar co-ordinates (i.e., q and f). However, for steering in only one co-ordinate (f), a plane (parallel sided) section through the centre of sphere can be used. In general the beam is no longer the same in both co-ordinates due to vignette in the q-direction.

Maxwells Fisheye Lens A lens structure similar to Luneburg construction but different in index presented by Maxwell is known as Maxwell fisheye lens. The lens structure is a sphere and it is assumed to be


289

embedded in an optical material with the same index of refraction as appears at the outer surface of the lens (e.g. water in the case of a fish with the idealized Maxwell’s lens). The refractive index of the Maxwell fisheye sphere is given by (see [8])

N(r ) =

2 ⎛r⎞ 1+ ⎜ ⎟ ⎝R⎠

(7.23)

2

which indicates that m(r) decreases from a value of 2 at the centre to 1 at the outer surface. This lens has a property of creating an image of an object on the sphere at the diametrically opposed point on the same sphere (i.e., a somewhat far-sighted fish). The different index gradient of the Luneburg has an additional advantage, in that it allows the lens to focus on an incident parallel beam onto a point, as illustrated in Fig. 7.9(b). In/Out

In/Out

FIG. 7.9(b)

Spherical Luneburg lens: Point to parallel beam or vice versa.

Applications In view of the major applications of Luneburg lens antenna especially to automotive radar, a multiple beam antenna is very important for the wide coverage of the vehicles. A new Luneburg lens antenna composed of multiple primary feeds and a parallel plate Luneburg lens with metallic artificial dielectric are found supporting wide scanning, hence suitable for the automotive applications in MM and sub-MM wave frequencies [9]. The major applications of Luneburg antenna include: · A Luneburg lens antenna is commonly used in wide angle scanning applications such as radar reflectors for targets and drones. It has also been used in military applications as tracking radar antennas. · Luneburg lenses are useful in a variety of antenna, satellite-based communication systems and scattering applications. Presently, most Luneburg lenses are used as RCS augmenters. Luneburg type lenses also have ability to form a number of

290


independent beams at different frequencies and powers. Low profile Luneburg lenses fed by a horn antenna have been recently used for a variety of airborne applications. In antenna applications, the main advantages are in their ability to form multiple beams that may point in arbitrary directions and their broadband behaviour. Recently, in 2005, NHK of Japan has developed a hemispherical lens antenna for HDTV transmission via communication satellite. This antenna is also found suitable for a small SNG van at ‘breaking news’ sites. Konkur Ltd. has produced a line of Luneburg lens antenna with dia = 8 m, which was found very suitable for satellite TV and commercial applications. In particular, a hemispherical lens antenna upto 8 m diameter have also been made, with the commercial product line extending to 4 m satellite uplink antennas. This is termed Multsat 1 M Lens antenna.

ROTMAN LENS ANTENNA As we know there are several applications of multiple beams, both in radar and communication systems. In these systems it is often required to cover wide area which is achieved using an antenna array. But a multiple beam forming network requires for controlling the amplitude and phase at each element of the antenna array. Microwave lenses are found to be an important alternative class of multiple beams forming network [10–12]. First microwave lens for wide beam scanning was proposed by Ruze in 1950; later in 1963, it was modified by Rotman for better scanning capabilities. Suggested lens for improving scanning capabilities by Rotman is termed Rotman lens antenna. They are very useful as multiple beam forming networks for a linear array antenna. A Rotman lens has three focal points, and the shape of focus for the position of exciting elements between those focal points is given by a circular arc. Therefore, when exciting elements between focal points are excited, phase errors on the aperture of the linear array antenna occur which need to be minimized [13]. D. Archer in his study on ‘Lens fed multiple beam arrays’, suggested that wide-angle scanning capabilities of these lenses are well-established. However, Rotman himself suggested that the feed curve should be circular and he optimized the design parameters accordingly. Shelton [14], proposed that it is not necessary to have circular focal arc, even a Rotman lens can designed with front to back symmetry. The Rotman approach to design a dielectric lens is well-suited for implementation in strip line or microstrip circuitry. In general, the design of a particular strip-transmission line lens is done in region between air and the dielectric then scaled by the inverse square root of the substrate dielectric constant. Gagron derived a procedure for proper refocusing of the dielectric filled Rotman lens with beam port locations, determined according to Snell’s law. sin C =

sin B

Fr

where a = scan angle of antenna array ÿ ÿ b = corresponding angle of focus inside lens er = relative dielectric constant in parallel plate region of lens


291

The proposed approach is an alternative lens configuration named refracting lens, which may offer a wider field of scan at a given focal length for lenses fabricated in microstrip/ strip line. The theoretical performance of the refracting lens has also been compared with that of conventional design at wider scan angle. It is found that the magnitude of the coupling coefficient for the beam port/array port pairs is 1.36 times (2.7 dB) greater for the refracting lens at a frequency of 10 GHz. P.K. Singhal [15] proposed elliptical refocusing of Rotman type lens and analyzed it using the contour integral technique. Comparison of results indicates that the lens is more compact with elliptical focal arc. Recently at GIT, Georgia, a prototype Rotman lens that operates at mm waves was built. It is first of its kind to operate at frequencies as high as 37 GHz. The device got its name because it is able to focus milimeter (mm) wave energy coming from a particular direction by passing the EM wave energy through a pair of parallel plats that are shaped like a lens. Beam forming is carried out on one side of the plates, fed by a switch array. Energy fed into a specific focal part emerges from the antenna elements and produces a beam along a particular direction. Beside the low cost, compact size and ruggedness, Rotman lens antenna also offers very low throughput loss and side lobe emissions. In the prototype Rotman, side lobe power can be suppressed by a factor of one thousand below the energy of the main beam. The power loss through the lens itself is less than 2 dB.

Applications Applications of Rotman lens antenna include: 1. Autonomous aircraft landing systems Poor visibility caused by heavy fog can keep pilots from seeing enough of the runway to allow a safe landing. A synthetic vision system based on millimetre wave radar could produce images through fog, allowing aircraft to land even when runways are obscured. Such a system would require a reliable, compact and inexpensive antenna (such as Rotman) to be affordable. 2. Synthetic vision for ground vehicles Operators of ground vehicles such as tanks also need to see through fog and smoke, but the vibration and harsh operating conditions limit use of conventional antennas. A Rotman lens antenna could be integrated into the tank’s structure, eliminating the need for an external dish and providing necessary reliability and ruggedness. 3. Automobile collision avoidance systems Collision avoidance systems built into automobiles could provide drivers with warning of approaching vehicles. If implemented in plastic, the Rotman antenna could lower the cost of such systems enough to make them practical.

292


4. Commercial communications Rotman lens antennas could be used in short-range building to building wireless communications. Implementation in plastic could help lower the capital costs for such systems. 5. Missile seekers Low cost, reliability and compactness of Rotman lens antenna could find applications in airborne systems such as missile seekers.

Tolerances in Lens Antenna As dielectric lens is an optical device the differences in the path length may be caused by deviations in thickness from the ideal contour as well as the variations in the refractive index. Assigning an allowable variation of (l/32) rms to the refractive index, we can define the thickness tolerance as follows: 't

−

Md

't

M0

=

1 32

't ⎡ M0 1 't 1 ⎤ − 1⎥ = ⇒ [N − 1] = ⎢ Md ⎣ Md M0 32 ⎦ 32 't =

M0 32(N − 1)

If m0 = 1 is the free space refractive index, then 't =

0.03 M0

Therefore, the tolerance on m will be

'N =

0.03 M0 't

and

NU =

'N

N

'N

=

In a particular case, for m = 1.5, Dt = 4l0, it is found that

0.03 M0

N 't 'N

N

= 0.5% .

Tolerances for different types of lens antenna are listed in Table 7.3. All tolerances in Table 7.3 assume l/32 rms for the individual lens variations and l/64 rms for the reflector variations, where m is the refractive index, t is the thickness of the lens, and tl is the lens thickness in terms of free space wavelength.


TABLE 7.3

293

Tolerances for different types of antennas

Type of antenna

Type of tolerance

Amount of tolerance (rms)

Parabolic reflector

Surface contour

0.016 l0

Dielectric lens (unzoned)

Thickness

⎛ 0.03 M0 ⎞ ⎜ ⎟ ⎝ N − 1⎠

Refractive index

⎛ 3 M0 ⎜ ⎝ Nt A

Thickness

3%

Dielectric lens (zoned)

Refractive index E-plane metal plate lens (zoned)

Thickness Plate spacing

E-plane metal plate lens (unzoned)

Thickness Plate spacing

⎞ ⎟% ⎠

3(N − 1)

N

%

3% 3N

N +1

%

3 M0 1 − N

%

3N (1 − N 2 )tM

%

SOLVED EXAMPLES Example 7.1 Design a plano-convex dielectric lens antenna using a material of dielectric constant er = 8.5 and mr » 1 with diameter 10 l. The antenna is to be operated at a frequency of 5 GHz and F is to be 1. Also find the relative power density at the edge of the lens. Solution: We know that

N = Nr F r = 8.5 = 2.9 M=

3 × 108 5 × 10 9

= 0.06 m = 60 mm

Since F = 1, so from geometry L = d = 10l = 600 mm.

294


where d is the diameter of the lens. Hence from Eq. (7.8), we have r=

L (N − 1)

N cos R − 1

=

600 (2.9 − 1)

⇒ r=

2.9 cos R − 1

1140 2.9 cos R − 1

which gives the relation between q and r of the lens antenna. Basic geometry of the lens antenna is shown in Fig. 7.10.

FIG. 7.10

Lens antenna for Example 7.1.

The corresponding relative power density will be PR P0

=

(2.9 cos 25.5° − 1)2 (2.9 − 1) (2.9 − cos 25.5°) 2

=

1.633 3.61 × 1.997

=

4.25 7.209

= –2.29 dB. Any value of (q) giving satisfactory value of d, i.e. d = 2r sin q, will optimize the design angle. That is q (deg)

r (mm)

r sin q (mm)

0 10 20 22 25 26 27 30

600 614 660 674 699 709 720 754

0 106 225 252 300 311 326 377

Example 7.2 Design an artificial dielectric lens antenna, with a dielectric constant of 1.4, to be used at 3.0 GHz, when the artificial dielectric consists of (i) copper sphere or (ii) copper strip. Consider the radius of the sphere to be 5% of the operating wavelength and the width to be 0.45% of the wavelength.


Solution:

295

(i) We know that for a sphere er = 1 + 4pNa3

Hence

N=

Fr − 1

4 Q a3

At 3 GHz, l = 0.1 m = 100 mm; therefore a = 5 mm N=

1.4 − 1

= 2.55 × 10 5 m −3

4 × 3.14 × (5 × 10) −3 )3

Therefore, the dielectric volume per sphere is 1 2.55 × 10 5 m −2

= 4 × 10 −6 m 3

The volume of the each sphere is given by

4 3

(a)3 =

4 3

× 3.14 × (5 × 10 −3 )3 = 5.2 × 10 −7 m 3

Therefore Volume of dielectric per sphere Volume of sphere

=

4 × 10 −6 5.2 × 10 −7

= 7.7

That is, the volume of dielectric per sphere is greater than the volume of the sphere. (There is space between two adjacent spheres.) Side of dielectric volumetric shapes (4.6 ´ (10)–6)1/2 = 15.9 mm. Therefore, the artificial lens will be as shown in Fig. 7.11.

Cube 2.45 mm

2.45 mm

Sphere 10 mm

FIG. 7.11

Artificial lens for Example 7.2.

296


(ii) For copper strips, er = 1 + 7.85 N¢W2

N=

Fr − 1

7.85 W

=

2

1.4 − 1 −2 2

7.85 × (0.45 × 10 )

= 2500 m 2

as W = 4.5 mm = 0.45 ´ 10–2 m. Then the area covered per strip is

1 2500

= 4 × 10 −4 m 2

So, if the covered area is a square, its sides will be 0.02 m. The ratio of

Square Strip

=

20 4.5

= 4.44.

Therefore four strips can be adjusted in a particular area (see Fig. 7.12). 20 mm

Strip width (4.5 mm)

FIG. 7.12

Four strips in Example 7.2.

Example 7.3 Design an unzoned plano-concave E-plane metal plate lens of the unconstrained type with aperture 10 l squares for use with a 3.5 GHz line square 10 l long. If the source is to be 20 l away from the lens, then (i) find the spacing between the plates, (ii) draw the shape of the lens and find its dimensions and (iii) find the bandwidth of the lens, if the maximum tolerable path difference is l/4. Assume F = 2 and m = 0.6. Solution: (i) As F = 3.5 GHz, l = 0.085 m = 85 mm. From Eq. (7.1), if a is the spacing between the plates, then

a=

M

1

2 (1 − N )

2 1/2

=

85

1

2 (1 − 0.36)1/2

= 53.13 mm

(ii) From Eq. (7.2) r=

L (1 − N ) 1 − N cos R

⇒ rM =

LM (1 − N ) 1 − N cos R

=

20 (1 − 0.6) 1 − 0.6 cos R

=

8 1 − 0.6 cos R


297

Since F = 2, D = L/2 = 5l.

q (deg)

rl

(D/2) = rl sin q

0 10 12 15.25 20

20 19.6 19.3 19.0 18.34

0 3.4 4.026 5.0 6.27

From Fig. 7.13

FIG. 7.13

Unzoned plano-concave E-plane metal plate lens for Example 7.3. tl = Ll + rl cos 15.25° = 20 – 19 cos 15.25° = 1.67

(iii) Hence the frequency bandwidth will be

'f =

2Nf (1 − N ) tM 2

=

2 × 0.6 × 0.25 (1 − 0.36) × 1.67

= 0.28 = 28%

where d = l/4 is the maximum tolerable path difference. Example 7.4 A five-zoned E-plane metal plate dielectric lens is made of material of mr = 0.66 and is to be used at f = 40 GHz. Find zone thickness, extension focal length and percentage operating bandwidth. Also find the percentage of plate spacing tolerance. Solution:

K = 5, er = 3.2. Hence m =

Zone thickness (z) =

M0 1 − N

=

–3 0.66 = 0.81, l0 = 7.5 ´ 10 m

0.075 1 − 0.81

= 0.395 m

Lex = (n – 1)z = (5 – 1) ´ 3.95 = 1.578 m

298


'f =

%tolerance =

50 × 0.81 1 + 5 × 0.81 3N

=

1+N

= 8%

3 × 0.81 1 + 0.81

= 1.34%

Example 7.5 (i) Estimate the tolerance of zoned and unzoned dielectric lens of refractive index 1.5 and thickness t = 0.25l to be used at 5 GHz. (ii) Repeat the same for the E-plane metal plate lens antenna if m = 0.75. Solution:

(i) From the table for zoned dielectric lens, the tolerance is 3(N − 1)

N

% ⇒

3(1.5 − 1) 1.5

= 1.0%

For unzoned dielectric lens, the tolerance (thickness) is 0.03M0

N − 1

%=

0.03 × 6 1.5 − 1

= 0.36%

Tolerance (refractive) is 3

N , tM (ii)

=

3 1.5 × 0.25

= 0.5%

Zoned (E-plane) tolerance is 3N 1+N

=

3 × 0.75 1 + 0.75

= 1.3%

Unzoned (E-plane) tolerance (thickness) is

0.03 × M0 1 − 1.5

=

0.03 × 6 0.5

= 3.6%

The tolerance (plate spacing) is

3N (1 − N ) tM 2

%=

3 × 0.75 (1 − 0.752 ) × 25

= 2.11%

Example 7.6 Find the index of refraction of lens material at radial distance r = 10 cm for the following types of lens: (i) Luneburg lens; (ii) Maxwell’s fisheye, assuming spherical shape of radius 1.0 m.


Solution:

299

(i) Luneburg lens: index of refraction 2 ⎡ ⎛r⎞ ⎤ N = ⎢2 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ R ⎠ ⎥⎦

1/ 2

= [2 − (0.1)2 ] 1/2 = 1.411

(ii) Maxwell’s fisheye: index of refraction ⎛

2 ⎞ ⎛ ⎞ = 1.98 ⎟=⎜ 2 ⎟ ⎝ 1 + r ⎠ ⎝ 1 + 0.01 ⎠

N=⎜

2

2

That is, the index of refraction changes as construction of lens changes. Example 7.7 The wavelength of a wave (f = 400 MHz) increases by 50% if it passes through a metal-plate lens antenna. Find the separation of plates. Solution:

We know that

N=

Ml and ll = 1.5 ´ l0 = 1.5 ´ 0.75 = 1.125 m. M0

Therefore

N=

Ml 1.125 = = 1.5 0.75 M0

M ⎤ ⎡ N = ⎢1 − l ⎥ 2d s ⎦ ⎣

1/2

= ds =

Ml 1.125 = = 45 cm 2 2 × (2.25 − 1) 2 × (N − 1)

Example 7.8 Find the focal length of the metal-plate lens antenna of diameter 40 cm. Assume that the E-plane width of lens is 25% lesser than the diameter. Solution: D = 40; WE = 40 ´ 75% = 40 ´ 0.75 = 30 cm. We know

f =

=

DL ⎛W ⎞ 2 tan ⎜ E ⎟ ⎝ 2 ⎠ 0.4 0.4 = = 1.33 m ⎛ 0.15 × 180 ⎞ 2 tan (8.6°) 2 tan ⎜ ⎟ ⎝ 2 × 3.14 ⎠

300


OBJECTIVE TYPE QUESTIONS 1. The first Rotman lens was designed in (a) 1963 (b) 1950 (c) 1960 (d) 2002 2. The focal length of metal plate lens antenna is _________ of to its diameter. (a) Inversely proportional (b) Directly proportional (c) The inverse square root (d) None of these 3. Lens antennas are mostly used in the frequency range of (a) 50–80 MHz (b) Above 80 MHz (c) 100–150 MHz (d) Above 3.0 GHz 4. A metal plate lens antenna function only in (a) E-H plane (b) E-plane (c) H-plane (d) None of these 5. Lens antenna function on the principle of (a) Equality of phase (b) Phase difference (c) Equality of path lengths (d) None of these 6. Operating bandwidth of the zonal lens antenna is found to be (a) 5% (b) 8% (c) 10% (d) 15% 7. Maxwell fisheye lens antenna is similar to (a) Luneburg lens antenna (b) Rotman lens antenna (c) Both of these (d) None of these 8. Multsat 1 M lens antenna is an example of (a) Metal-plate lens antenna (b) Rotman lens antenna (c) Luneburg lens antenna (d) None of these

Answers 1. (a) 6. (c)

2. (b) 7. (a)

3. (d) 8. (c)

4. (b)

5. (c)

EXERCISES 1. Find the spacing between plates of an unzoned plano-concave E-plane metal lens antenna made of material having er = 1.57 and to be operated at f = 3.75 GHz. 2. Estimate the optimum thickness of concave portion of E-plane metal-plate lens antenna operating with 6.2% bandwidth, if refractive index is 0.57. Also, find the percentage change in operating B/W if the thickness increases by 15%.


301

3. Design a plano-convex dielectric lens antenna for 6 GHz with diameter 20l. The lens material is to be polystyrene and F is unity. Draw the cross-section of lens and also find relative power at the edge of lens. 4. Design an artificial dielectric lens antenna for 3.2 GHz, with a material of refractive index 1.2 where the dielectric consists of (i) copper discs; (ii) copper strips. 5. Design an unzoned plano-concave E-plane metal-plate lens antenna of the unconstrained type with an aperture, of 16l square for use with a 3 GHz line source and 20l length. The source is 30l away from the lens, whose F number is 2. If the refractive index is 0.8: (i) What should be the spacing between the plates? (ii) Draw the shape and find the dimensions, and (iii) What is the BW of the antenna if maximum tolerable path difference is 0.25l? 6. Repeat Example 7.6 with the following specifications: (i) m = 1.75, tl = 0.35 and f = 3 GHz; (ii) m = 0.65 and f = 3.5 GHz. 7. An EM wave of frequency 500 MHz passes through a metal plate lens antenna of m = 1.68. Find the wavelength in the antenna if the spacing between plates is half of the free-space wavelength. [Ans: 109.344 cm] 8. Find the focal length of the metal plate lens antenna of E-plane width 20 cm. Assume the diameter of the antenna is 20% greater than the E-plane width. [Ans: 1.2 m] 9. Describe the advantages of Rotman lens antenna. 10. Describe the similarities and differences of lens antenna with reflector antenna. 11. What are applications of Rotman lens antenna in communication systems?

REFERENCES [1] Kock, W.E., “Metal-lens antennas,” Proc., I.E.R., pp. 828–836, Nov. 1946. [2] Wade Paul, “Metal plate lens antenna,” www.w1ghz.org/antbook, Chapter 3, 1998. [3] Kraus, J.D., Antennas: For All Applications, 3rd ed., Tata McGraw-Hill, New Delhi, 2003. [4] Prasad, K.D., Antennas and Wave Propagation, Satya Prakashan, New Delhi, 1996. [5] Bandhopadhyay, T.K., Antennas, Radio Wave Propagation and Noise, Khanna Pub., 4th ed., New Delhi, 2003. [6] Kock, W.E., “Metallic delay lens,” Bell System Tech., J., 27, pp. 58–82, January 1948. [7] Part, Y.J., et al., “Applications of Luneburg lens antenna to automotive radar in subMM and MM wave frequency,” IFHE, University of Karlsruhe, Germany, 2002. [8] Lawrena, G.N. and S.H. Hwang, “Beam propagation in gradient refractive-index media,” Applied Opt., Vol. 31, pp. 5201–5210, Sep. 1992.

302


[9] www.ecs.umas.edu/Allerton [10] Maybell, M.J., “Ray structure method for coupling coefficient analysis of the twodimensional Rotman lenses,” IEEE Antennas Propagate, Symposium Dig., June 1981. [11] Smith, M.S., et al., “A microstrip multiple beam forming lens,” Radio Electron. Engg., Vol. 54, No. 7/8, pp. 318–320, 1984. [12] Hilton, J., et al., “Lens antenna with amplitude shaping or sine condition,” IEEE Proc., Vol. 136, Pt. H, No. 3, June 1989. [13] Katagi, T., et al., “An improved design method of Rotman lens antennas,” IEEE Trans. Antennas Propagate, Vol. AP. 32, No. 5, May 1984. [14] Shelton, J.P., “Focusing characteristic of symmetrically configured bootlace lenses,” IEEE Trans. Antennas Propagate, Vol. AP. 26, pp. 518, July 1978. [15] Singhal, P.K., et al., “Theoretical investigation on elliptical refocusing of Rotman type lens multiple beams forming,” Journal of Microwaves and Optoelectronics, Vol. 3, No. 4, April 2004.

C H A P T E R

8

Parabolic Reflector Antennas

INTRODUCTION According to the applications of antenna particularly in radar systems, antennas are classified into two categories: optical antenna and array antenna. Optical antenna category comprises antennas that function on principle of optics, and examples are lens antennas and reflector antennas. We have already discussed lens antennas in Chapter 7. Reflector antennas, which are found suitable for radar applications are considered in this chapter. Reflector antennas are modified plane sheet antennas where the backward radiation from the antenna is eliminated using a plane sheet reflector. In general, beams of predetermined characteristics may be produced by means of large, suitably shaped and illuminated reflector surfaces. Reflectors are high gain antennas and routinely achieve gains for in excess of 30 dB in the microwave spectrum. That is why reflectors are very much required for long-distance communication and high resolution radar applications. Depending on feed techniques there is a variety of reflector antennas and each one is suited for particular applications. The most common types of reflector antenna are illustrated in Fig. 8.1(a)–(f). Despite these reflectors, there are many more configurations of reflector antennas—circular reflector, hyperbolic reflector, parabolic cylindrical reflector and Cassegrain reflector, etc., which are suitable for various purposes [1]. Figure 8.1(a) shows a basic parabolic reflector antenna, which collimates radiation from a feed (at the focus) into a pencil beam, proving high gain and minimum beam width. They are built with apertures of many wavelengths to provide high directive radiation. The parabolic reflector functions on the principle of radiation; it reflects the waves originating from a source at the focus into parallel beams. That is, the parabolic reflector transfers the curves wave front from the feed antenna into a plane wave front. Feed antenna is known as primary antenna, while reflector is known as secondary antenna. Figure 8.1(c) represents the active corner reflector antenna. It is nothing but a modified form of flat sheet reflector (aÿ = 180°). Here, two flat sheets intersect at an angle a < 180° and produce sharp radiation beams parallel to the axis. 303

304


Y

F

Z

X

FIG. 8.1(a)

Basic parabolic reflector.

FIG. 8.1(c)

Active corner reflector.

FIG. 8.1(b)

FIG. 8.1(d)

Reference co-ordinate system.

Parabolic reflector antenna.

The active reflector is found most practical where apertures are of 1 or 2 wavelengths. A corner reflector without an exciting antenna can be used as a passive reflector or target for radar waves, but in this application the aperture may be of many wavelengths with a 90° constant corner angle [2]. Reflectors of this specification (a = 90°) have the property that an incident wave is reflected back towards its source and itself acts as a retro-reflector [Fig. 8.1(e)]. Figure 8.1(f) shows an elliptical reflector antenna and produces a diverging beam with all reflected waves passing through the second focus of the ellipse. Elliptical reflector has two focuses: the driven element is fixed at the first focus, F1, however, reflected beams are collimated at focus F2. A hyperbolic reflector also has two focuses, but the first one is a virtual focus. As usual, the reflector is fed at second focus and produces forward radiation. Another similar reflector is circular reflector where the incident waves follow reverse path through the feed point. In this case, all the reflected waves pass through the focal point. A multiple reflector system, also known as Cassegrain antenna, offers one more degree of freedom by shaping


305

Driven element

F2 Aperture 90°

F1

Driven element

FIG. 8.1(e)

Retro-reflector antenna.

FIG. 8.1(f) Elliptical reflector antenna.

the primary beam and allowing the feed system to be conveniently located behind the main reflector. Parabolic cylindrical reflector performs beam collimation in one plane but allows the use of a linear array in the other plane, thereby adding flexibility in the shaping or steering of the beam in that plane. In addition to the above reflectors, there are some special geometry reflector antennas that are used as high-gain antennas with different beam widths in the principal planes. Reflectors shown in Fig. 8.2(a) produce a narrow main beam in the horizontal plane and are found very suitable for VSAT (very small aperture terminal) satellite communication applications. This is because a geostationary satellite avoids interference between adjacent Parabola

Parabola

Circle

(a)

FIG. 8.2

(b)

(c)

(a) Non-circular aperture parabolic reflector; (b) parabolic torus; (c) horn reflector.

306


satellites. The essential condition for this reflector is that the feed antenna must have a broader pattern in the horizontal plane for proper dish illumination. The parabolic torus, shown in Fig. 8.2(b), is curved version of the parabolic cylinder and it is a combination of parabolic and circular cross-section in the principal planes. A popular application for this reflector antenna (employs multiple feeds located along the feed arc) is to produce separate beams for receiving different satellite signals with a single earth terminal. Aperture efficiency of this reflector is too low, but there is a cost saving over using several antennas. The horn reflector antenna, shown in Fig. 8.2(c), is formed by joining a horn with an offset parabolic reflector and it is very useful for terrestrial microwave communication links due to its low side and back lobes.

FEEDING SYSTEMS As is known, the reflector antenna is fed indirectly through another antenna (primary antenna) located at the centre of the reflector. Therefore, it is necessary that the primary antenna be fed properly in order to realize maximum performance, particularly high aperture efficiency and high gain. Because most communication systems operate at microwave frequency, feeds for reflector antennas should be typically some form of flared wave-guide. Only at lower frequencies can dipole be used, particularly in the form of a linear array of dipoles, to feed a parabolic cylinder reflector. But, in general, common feed includes wave-guide slots and open wave-guides; however, the flared wave-guide horns are most widely used in reflector feeding. Various feed techniques of reflector antennas are shown in Fig. 8.3.

Basic Requirements of Reflector Feeding Systems As we know in receiving mode most of the reflectors convert incoming waves into spherical phase fronts centred at the focus. In view of this, the feeds must be point source radiators. In addition, if high aperture efficiency is desired reflector feeds must have the following characteristics: · The feed pattern should be rotationally symmetrical over the desired operating frequencies. · The feed should have a point phase centre and the phase centre should be positioned at the focal point of the reflector. · The feed should be small in order to reduce blockage; usually on the order of wavelength in diameter. · The feed must provide proper edge illumination (@ 11 dB), minimum spill over and correct polarization with minimum cross-polarization (< –30 dB). · The feed must be capable of handling the required peak and average power levels without breakdown under all operational environments. · Additional considerations include operating frequency band and types of antennas used, such as whether the antenna is a single-beam, multi-beam or mono-pulse antenna.


Wave-guide horn

Front feed

307

Dipole disc

Cutler (dual) Vertex (rear) feeds

Cassegrain feed

Offset feed

Simply flared pyramidal horn

Simply flared conical horn

Finned horn

Compound flared multimode horn

FIG. 8.3

Gregorian feed

Corrugated conical horn

Segmented aperture horn

Various types of feed for reflector antennas.

A simple form of reflector feed is a dipole antenna, but low efficiency and high crosspolarization are its limitations. This is because they often integrated with some type of metallic backing to reduce direct feed radiation in the direction of the reflector main beam. Dipolefed reflector functions satisfactorily in the UHF range with an aperture efficiency of about 24%. However, for operating frequencies above GHz levels wave-guide and small horn antennas are preferred as they offer minimum losses. Rectangular/pyramidal wave-guides and horn antennas operating in dominant mode TE10 mode are widely used as feed for reflectors.

308


This is because they meet high power requirements. Circular wave-guides and conical horns operating in the dominant TE11 mode are also used as feed and provide more symmetric principal plane patterns. Both the feeds discussed above have unbalanced principal plane patterns, this is due to the markedly different amplitude distributions that are uniform in the E-plane and taper to zero in the H-plane. The open-ended wave-guide provides a good match to a reflector with q = 59° or F/D = 0.44°; however, conical horn antennas provide optimum gain with 52 to 56% efficiency range [3]. In case of special antenna performance requirements— polarization diversity, multiple beams, high beam efficiency, or ultra-low sides—the feed becomes more complex. Reflector antennas are fed by segmented finned, multimode and/or corrugated horns. The most popular multimode feeds are Potter horns and they operate on the principle of Huygens’ source. Both the modes (TE11 and TM11) are generated in these feeds. A bandwidth of 10% is achieved with diameter > 1.3l and half-power is calculated using 1/26l/df, where df is the larger aperture of the corrugated horn. The narrow bandwidth constraint of the dual mode feed can be overcome by hybrid mode (HE11) feed. In this case, a mixture of TE11 and TM11 modes occur in a natural way and propagates with a common phase velocity. The most popular hybrid mode feed is the corrugated (conical) horn. This feed is very useful for most of today’s microwave reflector antennas [4] (see Fig. 8.4).

FIG. 8.4

Corrugated feed horn antenna.

The corrugated horn has the desirable features for feed antennas of a phase centre, i.e., it is independent (stable) with frequency. The phase-centre of the feed is positioned at the focal point of a reflector system for maximum gain. The phase-centre of the corrugated horn is at the horn aperture centre for small D and moves along the axis towards the throat as D increases, approaching a fixed value at the horn apex for D > 0.7l. The corrugated horn also called a scalar horn, because of its field direction dependence. This name is usually reserved for the large flare angle cases. From Fig. 8.4.


'=

⎛B ⎞ tan ⎜ ⎟ 2 ⎝2⎠

df

309 (8.1)

The –12 dB beam width of a corrugated horn feed is calculated by (BW)−12 dB ≅ 0.8(B )

(8.2)

PARABOLIC REFLECTOR The parabolic reflector antenna is a simplest and popular form of reflector antenna. It is inherently a very wideband antenna. At low frequency the bandwidth of a reflector is obtained by choosing the size of the reflector and therefore it should be at least of several wavelengths. However at high frequencies, performances are limited by the smoothness of the reflecting surface. Surface distortions necessarily must be less than a wavelength to avoid phase errors in the aperture. After all, the bandwidth of a reflector antenna is limited by the bandwidth of the feed antenna rather than the reflector. Reflectors still serve as a basis for many radar antennas in today’s applications, because they offer the maximum available gain and minimum beam widths with the smallest and simplest feeds [5]. The basic geometry of a parabolic reflector antenna is shown in Fig. 8.5, where a parabolic conducting reflector surface, of focal length F, has the feed at the focus F. From the geometrical optic, it is clear that spherical wave emerging from F is transformed into a plane wave travelling in the positive (+z) direction after reflection from the inner surface of parabola (see Fig. 8.6).

FIG. 8.5

Cross-sectional view of a reflector.

310

Antenna and Wave Propagation Axis

Beam Vertex

F

Plane

Parabolic reflector

FIG. 8.6

Ray representation of a parabolic reflector.

The equation describing the parabolic reflector surface’s shape in the rectangular form with reference to point (r, z) can be written as 2

⎛S⎞ 2 ⎜ ⎟ + F + Fz = 0 , 2 ⎝ ⎠

or

r2

r

£ a

= 4F(F – z)

(8.3)

The points (r = 0, z = F) and (r = a) z = (F2 – a2/4F) represent the apex and edges of the reflector respectively. For a given displacement r from the axis of the reflector, r is the distance of the point R on the surface of the parabolic reflector from the focal point (F). With reference to a point R(r, qf) in the polar co-ordinates, the parabolic reflector can be expressed as r=

2F

(8.4)

1 + cos R f

This is because any two rays generated from the focus takes two paths; path 1, FVF (2F) and path 2 (FRA) = r + r cos qf. Therefore, the distance travelled by these rays must be the same, i.e., FVF = FRA 2F = r(1 + cos

⇒ r=

2F 1 + cosR f

q f) ⎛R f ⎞ = F sec2 ⎜ ⎟ ⎝ 2 ⎠


311

Hence, the projection of this distance r onto the aperture plane is

S = r sin R f = From DFRA,

2F sin R f 1 + cosR f

AF = FR sin

At apex (qf = 0°),

r = F

and

= 2 F tan

qf r

Rf 2

(8.5)

= 0

At edge of reflector (qf = q)

r=

2F

r

and

1 + cos R

= a

(8.6)

Therefore, from Eq. (8.5), we get

a = 2F tan

⇒

F D F D

=

R 2

or

1

2

= 2F tan

= 0.25 cot

4 tan R /2

= 0.25 cot

D

R 2

or

R 2

R 2

⎡ 1 ⎛ D ⎞⎤ R = 2 tan −1 ⎢ ⎜ ⎟ ⎥ ⎣ 4 ⎝ F ⎠⎦

(8.7)

The referred axis symmetric parabolic reflector is specified completely in terms of two parameters—the diameter D and the ratio of focal to diameter length (F/D). The ratio (F/D) gives the shape (curvature rate) of the antenna and is known as aperture number. When (F/D) approaches infinity the reflector becomes planar, i.e., the normally incident waves are reflected back as plane waves. In contrast, if (F/D) is 0.25, the focal point lies in the plane passing through the rim. The ratio of (F/D) will also have an impact on the cross-polarization level. The cross-polarization decreases as (F/D) increases and in case of flat reflectors it reduces to zero. The main objective in designing the reflector antenna is to match the feed antenna pattern to the reflector and have the feed pattern about 10 dB down in the direction of rim, i.e., F(qf = q) = – 10 dB. Feed antennas for this purpose can be designed for an (F/D) ratio between 0.3 and 1.0. The focal distance (F) of a reflector antenna can be expressed easily in terms of D and height H. Therefore, from Eq. (8.3), for r = D/2 and z = F – H0. Therefore, from Fig. 8.5 the focal length (F) is 2

⎛D⎞ ⎜ ⎟ = 4F ( F − F + H 0 ) ⎝2⎠

Þ

F=

D2 16 H

(8.8)

312


That is, when F/D = 1/4, Eq. (8.5) gives H0 = D/4, i.e., H = F. The variation of angle q with aperture number (F/D) is given in Table 8.1. TABLE 8.1

Values of angle q for different aperture numbers

S. No.

F/D

q

1

0.25

90°

2

0.30

79.6°

3

0.33

73.7°

4

0.40

64.0°

5

0.50

53.4°

6

1.0

28.1°

The following two properties of a reflector antenna make it very useful: (i) All rays leaving the feed point F are collimated after reflection from the inner surface and travel parallel to the reflector axis (i.e., z-axis). (ii) All the path lengths from the focal point (F) to the reflecting surface are the same and equal to 2F. Since the total path length is constant (2F), the phase of waves arriving in the aperture plane from a point source at the focus will also be constant. Hence, the parabolic reflector with a feed that has a point phase centre at the focal point will produce uniform phase across the aperture plane. The aperture amplitude distribution, however, will not be uniform.

FIELD DISTRIBUTION ON AN APERTURE OF PARABOLIC REFLECTOR In order to derive the expression for far-field distribution on the aperture of reflector, let us consider a narrow strip of width dy and dr, located along the radial line of the reflector, in Cartesian and Polar co-ordinates respectively (see Figs. 8.7 and 8.8). Both cases have isotropic sources; a line source and a point source at their respective focuses. So, if Py is the power density at y, then the power P in the strip of width dy will be P = dy, Py = dq . P¢ when P¢ = the power per unit angle per unit length in the z-direction. Therefore, from the above equation Py P′

=

dR dy

=

dR d (R sin R )

=

1 d (R sin R ) dR

=

1 d ⎛ 2F sin R ⎞ ⎜ ⎟ dR ⎝ 1 + cos R ⎠


FIG. 8.7

Cross-section of cylindrical paraboloidal reflector.

FIG. 8.8

313

Cross-section of paraboloidal of revolution.

which, after simplification, yields 1 + cos R ⎡ Py ⎤ ⎢ ′⎥ = 2F ⎣ P ⎦R

Therefore

1 ⎡ Py ⎤ = ⎢ ′⎥ 2F ⎣ P ⎦R =0

Hence, the ratio of power density at q to power density at q = 0° can be expressed as PR P0

=

1 + cos R

(8.9)

2

Since the field intensity of a radiator is proportional to the square root of power density, Eq. (8.9) is reduced to ⎡1 + cos R ⎤ = ⎢ ⎥ E0 2 ⎣ ⎦

ER

1/2

(8.10)

where Eq /E0 is the relative field intensity at a distance y from the axis and y = R sin q. Similarly in the case of paraboloidal of revolution the total power P through the annular section of radius r and width r is given by Pr = 2pr dr = 2p sin q dqÿ U

(8.11)

This is because the total power (P) must be equal to the power radiated by the isotropic source over the solid beam angle 2p sin q dq. Here Pr is the power density at a distance r from the axis in W-m2 and U is radiation intensity in W-m2. Therefore, from Eq. (8.11)

314


PS U

2

=

sin R ⎡ (1 + cos R ) ⎤ = ⎢ ⎥ U 2L ⎛ dS ⎞ ⎣ ⎦ S⎜ ⎟ d R ⎝ ⎠

as r = R sin q

(8.12)

Hence, the ratio of power density at q to the power density at q = 0° can be given as ⎛ PR ⎞ ⎡ (1 + cos R ) ⎤ ⎜ ⎟= ⎢ ⎥ 2 ⎦ ⎝ P0 ⎠ ⎣

2

(8.13)

Since the field-intensity of a radiator is proportional to the square root of power density, Eq. (8.13) reduced to ⎡1 + cos R ⎤ = ⎢ ⎥ E0 2 ⎣ ⎦

ER

(8.14)

where Eq /E0 is the relative field intensity at a distance r from the axis and r = R sin q.

PARABOLIC REFLECTOR ANTENNA PARAMETERS The radiation from a large paraboloid with uniformly illuminated aperture is essentially similar to the radiation from a circular aperture. This is true if both of them have same diameter D in an infinite metal plate and a uniform plane wave is incident on its surface. The radiation field pattern for such a uniformly illuminated aperture can be calculated in terms of the normalized field pattern Ef as follows [1]: ⎡⎛ Q D ⎞ ⎤ J1 ⎢⎜ ⎟ sin R ⎥ 2M ⎣⎝ M ⎠ ⎦ EG = QD sin R

(8.15)

where D = diameter of reflector antenna (m) ÿ ÿ f = angle w.r.t. normal to aperture J1 = first-order Bessel’s function and J1(x) = 0, when x = 3.83 Therefore, the angle q0 (say) to the first nulls of the radiation pattern can be calculated as follows:

QD ⎡ 3.83 M ⎤ sin R 0 = 3.83 ⇒ R 0 = sin −1 ⎢ ⎥ M ⎣ QD ⎦ If q0 is very small, the above equation yields

R0 =

1.22M D

(rad)

or

R0 =

70 M D

(deg)

(8.16)


315

Therefore, for a large circular aperture, the first null beam width will be BWFN =

140 M D

(deg)

(8.17)

The beam width between half-power points is defined as HPBW =

58 M

(deg)

D

(8.18)

The directivity Dr of a large uniformly illuminated aperture is defined as Dr = 4Q

Area

Q

1

(8.19)

M2

Hence for a circular aperture, Dr = 4Q .

4

2

D .

M2

⎛D⎞ = 9.87 ⎜ ⎟ ⎝M⎠

2

(8.20)

Similarly, for rectangular/square apertures First null beam width (BWFN) = L2

115 M L

deg

(8.21)

2

⎛L⎞ = 12.6 ⎜ ⎟ 2 M ⎝M⎠ Power gain (G) over l/2 dipole = 7.7(L/l)2

Directivity (Dr) = 4Q

(8.22) (8.23)

The maximum achievable gain for an aperture antenna is given by Gmax = 4p/l Ap, where Ap is the aperture area. This is possible only under ideal circumstances; however, in practice, these conditions are rarely met, and the modified expression of gain is 2

Gmax = F a

4Q

M2

Ap

(8.24)

where ea is aperture efficiency and its values lie between 0 and 1. The spherical spreading loss at the aperture edge (dB) is given by

⎧ ⎫ ⎪ ⎪ 1 ⎡1 + cosR ⎤ ⎪ ⎪ 20 log ⎨ 1 + = − 20 log ⎢ ⎥ 2⎬ 2 ⎣ ⎦ ⎛F⎞ ⎪ ⎪ 16 ⎜ ⎟ ⎪ ⎪ ⎝D⎠ ⎭ ⎩

(8.25)

The above expression shows that (F/D) influences the amount of spherical spreading loss. And it varies from 0.5 to 6.0 dB for (F/D) ranging from 1.0 down to 0.25.

316


In the case of flat reflector (i.e., aperture number ¥), the angle of the beam scan (q1) equals the feed tilt angle (q 2); however, for curved reflectors the beam scan angle will be less than the feed tilt angle. In general, scanning is quantified with a factor known as beam derivation factor (BDF) which is expressed as BDF =

R1 Beam scan angle = Tilt angle R2

(8.26)

i.e. BDF is maximum at unity for a flat reflector and decreases with decreasing (F/D) for axis-symmetric and offset reflectors. However, for small displacement d (as shown in Fig. 8.9) another approximate expression can be used [6].

BDF =

FIG. 8.9

⎛ 4F ⎞ 1 + 0.36 ⎜ ⎟ ⎝ D ⎠ ⎛ F⎞ 1 + ⎜4 ⎟ ⎝ D⎠

−2

(8.27)

Beam scanning of a reflector antenna by feed displacement.

Lateral feed displacement introduces a planar phase front tilted with respect to the aperture plane responsible for beam scanning in a direction opposite to the displacement.

Polarization Loss Efficiency The definition of polarization loss efficiency is based on the integration of Poynting vector with the antenna aperture. The power flow across the antenna aperture is assumed to be equal to the radiated power. Polarization loss efficiency is defined as the ratio of the co-polarized power flow across the antenna aperture to the total power, i.e.


I=

∫

2Q 0

∫

R 0

∫

2Q 0

∫

R 0

317

| Emp (R G ) |2 S 2 sin 2R dR dG (8.28)

| Emp (R , G ) |2 + | Ecp (R , G ) |2 S 2 sin R dR dG

where r2sin q dq df (= dSa) = surface element on a sphere radius Emp = aperture electric field with main polarization Ecp = aperture electric field with cross-polarization

r

and they are expressed in matrix form as follows: ⎛ Emp ⎞ ⎛ sin G cos G ⎞ ⎜ ⎟=⎜ ⎟ ⎜ Ecp ⎟ ⎜⎝ cos G − sin G ⎟⎠ ⎝ ⎠

⎛ ER ⎞ ⎜ ⎟ ⎜ EG ⎟ ⎝ ⎠

in which Eq and Ef are the components of far-fields of the antenna. The polarization loss efficiency is found to be maximum at q0 = 65° and decreases as q0 gets smaller; this is due to decrease in feed taper. PLE of parboiled reflector antenna is slightly greater than the polarization efficiency of Cassegrain gain antenna [7]. However, the result of using far-fields radiated by the front-fed paraboloid gives completely different results because it takes into account the axial surface currents.

PARABOLIC CYLINDRICAL ANTENNA A parabolic cylindrical reflector fed by a line source can accomplish the flexibility of steering the elevation or the azimuth beam at a modest cost. The line sources (feed) may be of many types, ranging from a parallel plate or a slotted wave-guide to a phased array using standard designs [8]. The geometrical configuration of a parabolic cylinder antenna is shown in Fig. 8.10.

FIG. 8.10

Basic configuration of parabolic cylindrical antenna.

318


In Fig. 8.10 the reflector system has the contour defined by z = y2/4F, where x, y, z and F are shown properly. The feed is on the focal line F–F¢ and a point on the reflector surface is located with respect to the feed centre at x and r = F sec2 y/2. In parabolic cylinder antenna the power density falls off as r rather than r2; this is because the feed energy diverges on a cylinder instead of a sphere in other reflectors. That is why the space taper is defined as follows: ⎡ (4F/D)2 ⎤ Space taper = 20 log ⎢ 2⎥ ⎣⎢1 + (4F/D) ⎦⎥

is halved in decibels. Figure 8.10 indicates that at angle q from broad side the primary beam intercepts the reflector at (F tan q) past the end of the vertex. Since the peak of the primary beam from a steered line source lies on a cone, the corresponding intercepts on the right and left corners of the top of reflector are further out at (F sec2 y/2 tan q) and the corners of a parabolic cylinder are seldom rounded in practice. The main drawback of parabolic cylinder antennas is the large blockage in case of symmetrical shapes. That is the reason they are often built offset. However, if properly designed and fed by an offset multiple-element line source, parabolic cylinder antennas can offer excellent performance.

MULTIPLE-REFLECTOR ANTENNA It is a reflector antenna that minimizes the shortcomings of general paraboloidal reflectors by adding a secondary reflector. The contour of the added reflector determines the pattern of power distribution across the primary reflector and thereby gives control over amplitude in addition to phase in the aperture. This is very helpful to produce very low spillover or to produce a specific low-side lobe distribution. A multiple reflector is also known as Cassegrain antenna. Figure 8.11 shows a Cassegrain reflector antenna where the feed is placed at one Axis

Feed

F

Plane

Parabolic reflector

FIG. 8.11

Cassegrain reflector antenna where feeds are placed at different focuses.


319

focus of the hyperboloid and the parabolic focus at the other. All the parameters shown in Fig. 8.11 have their usual meanings. The parameters of Cassegrain antenna are related as follows: tan

Zr

= 0.25

2 1

+

tan Z v

Dm

=

fm 1

tan Z r

Þ

Ds tan

yv

+ tan 1 −

and

fs Ds fs

=2

tan Z v tan Z r

(8.29)

fm

4

=2

tan Z v + tan Z r

Þ

1 Dm

Ds

yr = 2fs tan yv tan yr 1 e

=

2 Lv

(8.30)

fv

In which the eccentricity (e) of the hyperboloid is given by

e=

⎛Z + Zr ⎞ sin ⎜ v ⎟ 2 ⎝ ⎠

⎛Z − sin ⎜ v 2 ⎝

Zr ⎞ ⎟ ⎠

and the equivalent focal length (fe) and magnification m is given by (see [9]) fe =

Dm ⎛ Zr ⎞ ⎜ tan ⎟ 4 ⎝ 2 ⎠

and

m=

fe fm

=

(e + 1) (e − 1)

Thus the feed has to be designed to produce suitable illumination within subtended angles (±yr) for the longer focal length. The typical aperture efficiency of the Cassegrain antenna lies between 55% and 60%. Gregorian antenna is another antenna similar to the Cassegrain antenna, where an ellipsoidal sub-reflector is used in place of a hyperboloid. Gregorian reflector is a form of dual reflector antenna that offers perfect focus and has a concave sub-reflector located beyond the virtual feed point. The main drawback of this antenna is that it suffers from large blockage if they are symmetrical. However, this may be minimized by choosing the diameter of the sub-reflector as equal to that of the feed. This is true only if ⎡ 2M f m ⎤ D ≅ ⎢ ⎥ ⎣ k ⎦

1/2

in which k is the ratio of the feed-aperture diameter to its effective blocking diameter; generally this is slightly less than 1. By using a polarization twist reflector and a sub-

320


reflector made up of parallel wires, the blocking can be reduced significantly. This is because the sub-reflector is transparent to the secondary beam with its twisted polarization. In the particular case of a dual-reflector, blockage can be eliminated by offsetting both the feed and the sub-reflector [10].

Advantages of Dual Reflector These are: (i) Both types of dual reflector offer the major advantage of having the feed conveniently located near the apex of the main reflector, which in turn eliminates the long transmission line run and associated losses. (ii) Another advantage of dual reflector over single reflector is its low-noise earth terminal applications. The feed radiation not intercepted by the sub-reflector of a dual-reflector (e.g. spillover) is directed toward the low-noise sky region rather than the more noisy ground, as happens in the case of spillover of a single reflector. The increased effective focal length provides several additional advantages. These are: (a) Cross-polarization in far-field pattern improves with larger focal length to diameter ratio. (b) There is less spherical spreading loss at the rim of the main reflector, and finally the main beam scan performance is improved. Dual-shaped reflectors are in wide use for high gain axis symmetric systems such as satellite communication. High efficiency of 85% is achieved with a 1.5 m main reflector with a dual reflector (offset) operated at 31 GHz.

SOLVED EXAMPLES Example 8.1 A parabolic reflector antenna of diameter 1.22 m is used at 28.56 GHz from a geostationary satellite. If F/D = 0.5, and the measured field patterns have the following beam widths: HP = 56° and BW–10 dB = 104°. Calculate (a) the angle from the centre to the edge of the reflector, and (b) the edge illumination (EI). Solution:

(a) We know that, 1 ⎛ D ⎞ ⎞ −1 ⎛ −1 ⎛ 1 ⎞ R 0 = 2 tan −1 ⎜ ⎟ = 2 tan ⎜ ⎟ = 2 tan ⎜ ⎟ = 53.1° ⎝ 4F ⎠ ⎝ 4 × 0.5 ⎠ ⎝ 2.0 ⎠

(b) Since half of beam width (i.e., BW/2) = 52°, the 10 dB down point will fall inside the reflector; hence let us assume the feed pattern has fallen to 11 dB down at the rim, i.e., FT = 11 dB. Therefore, the spherical loss will be ⎡1 + cos R ⎤ ⎡1 + cos 53.1 ⎤ Lph = − 20 log ⎢ ⎥ = − 20 log ⎢ ⎥ = − 1.9 dB 2 2 ⎣ ⎦ ⎣ ⎦

Hence, EI = FT – Lph = 11 – (–1.9) = 12.9 dB.


321

Example 8.2 Find the half power of a Potter horn used as a feed for parabolic reflector antenna, of diameter of 1.5 m. The operating frequency is 300 MHz and the flare distance is 1.48 m. Solution:

The operating wavelength, M =

3 × 108 300 × 10 6

= 1.0 m

The diameter of the horn is 1.5 m > 1.3lÿ » 1.3. Therefore, the half-power =

1.26 M df

=

1.26 × 1.0 1.48

= 0.852 = – 0.6989 dB.

Example 8.3 Find the displacement in phase-centre of a corrugated horn of flare distance 1.5 m for the flared angle, a = 38°. Also find the –12 dB beam width. Solution:

The beam width (BW) −12dB = 0.8 × B = 0.8 ×

Therefore, ' =

⎛ 38o tan ⎜ ⎜ 2 2 ⎝

1.5

38 × Q 180

= –2.755 dB.

⎞ ⎟ = 0.75 tan (19o ) = 0.2582 m. ⎟ ⎠

Example 8.4 A circular parabolic antenna is operated at 3.0 GHz with a power gain of 30 dB. Calculate the diameter as well as the half-power beam width of the antenna.

Solution:


3 × 1010 3 × 10 9

= 10 cm = 0.1 m.

Power = 30 dB = 1000 units Since

⎛D⎞ Gp = 6 ⎜ ⎟ ⎝M⎠

2

⇒

D

M

=

Gp 6

=

1000 6

= 12.9

= D = 12.9l = 12.9 ´ 0.1 = 1.29 m HPBW =

70 × M D

=

70 × M 12.9 × M

or

129 cm.

= 5.426o

Example 8.5 Calculate the beam width between first nulls (i.e., FNBW) of a 2.5 m paraboloid reflector being used at 6 GHz. Also, find out its gain in dB. Solution:


c f

=

300 × 10 6 6 × 10 9

= 5 × 10 −2 m

322


FNBW =

140 M D

=

14 × 5 × 10 −2 2.5

= 2.8o

2

⎛D⎞ ⎛ 2.5 ⎞ Gain, G p = ⎜ ⎟ = 6 ⎜ = 6 (0.05 × 103 )2 = 1.5 × 10 4 = 41.761 dB −2 ⎟ M × 5 10 ⎝ ⎠ ⎝ ⎠

Example 8.6 Find the angular aperture for a parabolic reflector of diameter 12 m, in case the aperture number is 0.55, and also find the position of the focal point with reference to the reflector mouth. Given: D = 15 m, F/D = 0.55.

Solution:

F

We know that

Hence

= 0.25 cot

D tan

R 2

or

R 0.25 = = 0.4545 2 0.55

q

or

0.55 = 0.25 cot

R 2

R = 24.44 2

or

= 48.88°

Therefore, the angular aperture 2q = 2 ´ 48.88 = 97.76° As F/D = 0.55, therefore F = 0.55 ´ 12 = 6.6 m. Example 8.7 Calculate the space taper and beam deviation factor (BDF), if aperture number of a reflector antenna is 0.25. Also calculate the angular aperture and diameter of reflector antenna for a focal length of 8.4 m. We know that

Solution:

F D

Therefore

= 0.25 cot

R 2

R 0.25 = = 1 ⇒ R = 90° 2 0.25

cot

Thus angular aperture, 2q = 180° As

F D

= 0.25 ⇒ D

⎡ F⎤ 1 + 0.36 ⎢ 4 ⎥ ⎣ D⎦ BDF = −2 ⎡ F⎤ 1 + ⎢4 ⎥ ⎣ D⎦

F 0.25

=

8.4 m 0.25

= 33.6 m = 33.6 m

−2

=

1 + 0.36 [4 × 0.25] −2 1 + [4 × 0.25]

−2

=

1 + 0.36 1+1

=

1.36 2

= 0.68 m = –1.675 dB


323

2

⎡ ⎛ F⎞ ⎤ ⎢ ⎜4 ⎟ ⎥ ⎢ ⎝ D⎠ ⎥ = 20 log Space taper = 20 log ⎢ 2⎥ F ⎛ ⎞ ⎢1 + 4 ⎥ ⎜ ⎟ ⎥ ⎢ D ⎝ ⎠ ⎦ ⎣

⎛ 1 ⎞ ⎜ ⎟ = 20 log (0.5) = – 6.020 dB. ⎝1 + 1⎠

Example 8.8 Find the difference between the relative power and field intensity at the angles 30° and 45° on a parabolic surface, assuming that the antenna is fed at the focal point. Solution:

Given

q

= 30°

1 + cos 30 1 + 0.8634 ⎛ P ⎞ 1 + cos R = 0.933 = – 0.302 dB We know that, Pr = ⎜ R ⎟ = = = 2 2 2 ⎝ P0 ⎠

⎡1 + cos 45o ⎤ Er = = ⎢ ⎥ E0 2 ⎢⎣ ⎥⎦ ER

1/2

⎡ 1 + 0.707 ⎤ = ⎢ ⎥ 2 ⎣ ⎦

1/2

= 0.923 = – 0.6876 dB

Therefore difference, D = (– 0.6876 + 0.302) = 0.386 dB

OBJECTIVE TYPE QUESTIONS 1. Reflector antennas are very suitable for the following applications: (a) Mobile (b) Radar and mobile (c) Satellites (d) All of these 2. The gain of reflector antenna is found to vary between (a) 5–6 dB (b) 30–35 dB (c) Above 50 dB (d) None of these 3. Which of these is true for a parabolic reflector antenna? (a) High gain and high beam width (b) High gain and low beam width (c) Low directivity and circularly polarized (d) None of these 4. Reflector antenna is not fed directly, it is fed by another antenna located at focal point, which is called (a) Feed antenna (b) Focal antenna (c) Primary antenna (d) Secondary antenna 5. Apex angle of a active corner reflector antenna is always (a) 90° (b) Less than 90° (c) Less than 100° (d) Less than 180°

324


6. If the incident wave from a reflector antenna reflected back towards its source and has vertex angle 90°, the antenna is known as (a) Retro-reflector (b) CP reflector (c) Double reflector (d) Parallel reflector 7. The Potter horn functions on the basic principle of (a) Fermat’s principle (b) Huygens’s principle (c) Principle of optics (d) None of these 8. The corrugated horn is also known as (a) Scalar horn (b) Vector horn (c) Active horn (d) None of these 9. The (–12 dB) beam width of a corrugated horn of vertex angle of 50° will be (a) 0.579 dB (b) 1.796 dB (c) 3.07 dB (d) 0.6977 dB 10. The ratio (F/D) of a reflector antenna is known as (a) Aperture number (b) Focal number (c) Both of these (d) None of these 11. Which of the following is (a) The cross-polarization (b) The cross-polarization (c) The cross-polarization (d) None of these

true? decreases as (F/D) decreases decreases as (F/D) increases is independent from (F/D)

12. What will be the directivity of a circular antenna of diameter 2l? (a) 100° (b) 20° (c) 39.5° (d) 0° 13. The spherical spreading loss lies between (a) 0.5 and 6.0 dB (b) 6.0–10 dB (c) 1.2–1.9 dB (d) None of these 14. The polarization loss efficiency of a reflector antenna is maximum at an angle of (a) 50° (b) 60° (c) 65° (d) None of these 15. The aperture efficiency of a Cassegrain antenna is (a) 20–30% (b) 30–40% (c) 45–60% (d) 55–60% 16. What will be the half-power of a Potter horn used as a feed for parabolic reflector of diameter = 3 m, at l = 2 and flare distance = 1.55 m? (a) 2.11 dB (b) 3.12 dB (c) 4.0 dB (d) None of these


325

Answers 1. 6. 11. 16.

(d) (a) (b) (a)

2. (b) 7. (b) 12. (c)

3. (b) 8. (a) 13. (a)

4. (c) 9. (d) 14. (c)

5. (d) 10. (a) 15. (d)

EXERCISES 1. Show that relative field intensity due to a parabolic reflector at an angle q from the focal point is expressed as Er = cos q/2. Also find the field intensity at angle, qÿ = 23° if the electric field at q = 0 is 25 mVm–1. 2. Find the FNBW, directivity and power gain of a square corner reflector antenna of size 1.5 m. Assume operating wavelength is l = 12.5 cm. 3. Find the same parameters as given in Exercise 2 for a circular reflector antenna of diameter 10.5 m at operating frequency of 3.5 GHz. 4. Calculate the angular aperture and focal length for parabolic reflector antennas with aperture numbers 0.35 and 0.50. 5. Obtain BDF and spherical spreading loss of a parabolic reflector antenna of aperture number F/D = 0.42. 6. Show that the aperture efficiency is 2/3 for a rectangular aperture with a uniform amplitude distribution in a particular direction. 7. An antenna operates at 150 MHz and has a physical aperture area of 100 m2, a gain of 25 dB, and a directivity of 24.5 dB. Compute (a) maximum effective aperture; (b) aperture efficiency and (c) radiation efficiency. 8. Describe various feed mechanisms of reflector antennas. Show that the aperture number is F/4D, where F is focal length and D is diameter of the antenna.

REFERENCES [1] Kraus, J.D., Antennas for All Applications, Tata McGraw-Hill, New Delhi, 2003. [2] Wood, P.J., “Reflector antenna analysis and design”, IEE/P Peregrines, London and New York, 1980. [3] Rudge, A.W., et al., Handbook of Antenna Design, Vol. I, Van Nostrand Reinhold, New York, Chap. 3, 1988. [4] Johnson, R.C. (Ed.), Antenna Engineering Handbook, 3rd ed., McGraw-Hill, New York, 1993.

326


[5] Slattern, C.J., “The theory of reflector antennas,” Air Force Cambridge Res. Lab, a FCR L-66-761, Physical Review Paper, 290, 1966. [6] Lo, Y.T., “On the beam deviation factor of a parabolic reflector”, IEEE Trans., Ant., and Prop.; Vol. AP. 8, pp. 347–349, May 1990. [7] Mehmet, S. and Y. Erdem, “Polarization losses in reflector antennas”, IEEE, Trans. Antenna and Propagate, Vol. AP. 33, No. 8, pp. 899–904, August 1985. [8] Setion, Leo, Practical Communication Antennas with Wireless Applications, Prentice Hall, PTR, NJ, 07458. [9] Hansen, R.E., Microwave Scanning Antennas, Academic Press, New York, Peninsula, Los Altos, CA, 1985. [10] Hanan, P.W., “Microwave antennas derived from the Cassegrain telescope”, IRE, Trans., Vol. AP-9, pp. 140–153, March 1961.

C H A P T E R

9

Yagi Antenna

INTRODUCTION A Yagi–Uda antenna is an array system of a dipole and closely spaced similar elements. In the array only one element is fed which is termed the driven element, whereas the others are mutually coupled with the driven element. In addition, a reflector is placed on the rear side and one or more directors are placed on the front side of the driven element. That is, in Yagi antenna, dipole acts as the driven element, and the other elements operate as reflector/ director. Yagi antennas are directional along the axis perpendicular to the dipole in the plane of the elements, from the reflector through the driven element and out via the director(s). In general, the length of elements of the antenna decreases by 5% from reflector to the last director. A simple form of Yagi–Uda antenna with reflector length (0.49l), driver length (0.47l) and director length (0.45l) each spaced only (0.04lÿ apart) is shown in Fig. 9.1, where the radius of each wire is 0.001l. Supporting rod

Driven element

Direction of radiation

Director Reflector Feed

FIG. 9.1

Configuration of three elements in YagiUda antenna.

The history of Yagi antenna started with Shintaro Uda’s experiments conducted on the use of parasitic reflector and director elements in 1926 and published entitled On the Wireless 327

328


Beam of Short Electric Waves. He first measured the radiation pattern and gain with a single reflector and a single director and later with a single reflector and as many as 30 directors. He found the highest gain with reflector about l/2 in length and spaced l/4 from the driven element, whereas the best director’s length was about 10 per cent less than l/2 with optimum spacing about l/3. George H. Brown demonstrated the advantages of close spacing, and proposed that reflector to the driven element spacing also needs to be reduced. Hidetsugu Yagi, Professor at Tohoku University presented a paper along with Shintaro Uda at the Imperial Academy on the Projector of Sharpest Beam of Electric Waves in 1926 and later in 1927 both together presented a paper on On the Feasibility of Lower Transmission by Electric Waves. The narrow beam width of short waves produced by the guiding action of the multi director periodic structure (they named Wave Canal) had encouraged them to suggest it for short wave power transmission. Later in 1928, Yagi published another article on Beam Transmission of Ultra Short Waves (IRE), in which he acknowledged the work of Professor Uda with his 9 papers published on antennas at a wavelength of 4.4 m. This paper has been reprinted in 1984 in its original form in the proceedings of IEEE. Since then it has became customary throughout the world to refer to this radiator as a Yagi–Uda antenna. The Yagi–Uda antenna as a directive antenna was also called a “wave projector” or “wave canal” [1–3].

PRINCIPLE OF OPERATION In order to understand the working principle of Yagi–Uda antenna, let the driven element be a resonant half-wave dipole. If a parasitic element is positioned much closer and it is excited by the driven element with the roughly equal amplitude, so the field incident on the parasitic element will be [4] Einc = Edriver

(9.1)

As a result, a current is excited on parasitic element, equal in amplitude and opposite in phase to the incident wave and, finally, radiated as electric field. This excited current is tangential to the wire, because the electric field arriving at the parasitic element from driver is tangential to it. Since the total electric field tangential to a good conductor is zero, the field radiated by parasitic element should be such that the total tangential field on it is zero, i.e., Einc = Eparasitic = 0 Hence, Eq. (9.1) gives

Epara = – Edriver = – Einc

DESIGN PARAMETERS Length of Elements A basic Yagi–Uda antenna consists of three elements, namely reflector, driver and director. Lengths of these elements are calculated as follows:

Yagi Antenna

Length of reflector =

500 f MHz

Length of director =

ft

475

Length driven element =

f MHz

455 f MHz

329

ft

ft

Directivity The dual end fire beam characteristics of Yagi antenna can be changed to more desirable single end fire beam by choosing suitable design parameters. The E-plane pattern of the antenna is equal to the H-plane pattern multiplied by array factor which is that of a half-wave dipole. The maximum directivity of this antenna is found to be 9 dBi or 7dBd for reflector spacing between 0.15l and 0.25l. Director-to-director spacing is typically 0.25l and 0.35l, with larger spacing being common for long arrays and closer spacing for shorter arrays. Typically, the reflector length is l/2, and the director length always 10 to 20% shorter than its resonant length. The exact length being rather sensitive to the number of directors ND and the inter-director spacing SD. A multi-element Yagi–Uda antenna is shown in Fig. 9.2. Sd Lr

Ld

L

Sr

FIG. 9.2

Z

h

Multi-element YagiUda array antenna.

Gain Gain in Yagi antenna is related to its boom length (Ld), but for a parasitic array there is a smaller increase in gain per element as directors are added to the arrays (Sd must be fixed), since the Yagi antenna is not uniformly excited (Fig. 9.2). The addition of directors up to 5/6 only provides significant increase in gain. Adding one director in 3 elements (i.e., 4 elements Yagi) provides 1 dB gain enhancement, particularly adding one director to increase from 9 to 10 elements in array increases 0.2 dB gain. However, addition of reflectors results in fractional increase in gain and hence usually not preferred. Gain (dBd) versus reflector to the driven element spacing is plotted in Fig. 9.3. The main effect of reflector is on feed point impedance and on the back lobe of the antenna. The gain in Yagi–Uda antenna made with isolated dipole is around 2.5 dB, whereas l/2 elements

330


FIG. 9.3

Measures gain in dBd of a dipole and reflector element for different spacing SR.

are replaced by a flat plate, the gain is found to be 3 dB. Thus, a single wire like reflector element is almost as effective as flat plate in enhancing the gain of a dipole. Figure 9.4 shows the gain versus the number of elements N in the array, including one reflector and one driven element, with an inter-element spacing for all the elements of SR = SD = 0.15l. The gain is mostly controlled by the director element, by changing their length and spacing in between, both are interrelated. The more sensitive parameter is the director length, which becomes more critical as the boom length increases. 13 12 11 10 Gain (dB)

9 8 7 6 5 4 3 2 1 1

2

3

4

5

6

7

8

9

10

11

Number of elements, N

FIG. 9.4

Gain of a typical YagiUda antenna versus total number of elements.

Yagi Antenna

331

A Yagi–Uda antenna has relatively low input impedance and narrow bandwidth. However both can be improved at the expenses of others (such as gain and magnitudes of lobes). One way to increase the input impedance without affecting other parameter is to use an impedance step-up element as feed (two-element folded dipole with a step-up ratio of about 4). Frontto-back ratio of about 30 (»15 dB) can be achieved at wider than optimum element spacing, and for optimum designs, the minor lobes are about 30% or less of maximum lobes [5]. Input impedance and radiated field In order to describe the input impedance of a Yagi antenna, let two elements Reflector and driven element of a Yagi–Uda antenna be radiated in free space. Let l1 be the length of the driven element and l2 be the length of reflector and d is spacing between reflector and driven element, as shown in Fig. 9.5. So, if V1 and V2 are the voltages applied to the driven element and reflector respectively, then they can be expressed in terms of corresponding induced currents and input impedances as follows: V1 = l1Z11 + l2Z12

V2 = l1Z21 + l2Z22

and

where l1 = current in the driven element l2 = current in the reflector Z11 and Z22 = self-impedances of driven element and reflector Z12 and Z21 = mutual impedances between driven element and reflector V2, Z2

V1, Z1

l2

l1

l2

l1

d Reflector

FIG. 9.5

Director

Voltage and current representation across 2-YagiUda antenna.

As we know, in case of Yagi antenna only driven element is fed, reflector is coupled from voltage induced in driven element, hence V2 = 0. Therefore, I1Z11 = –I2Z12

332


or

I 2 = − I1

Z 21 Z 22

As Therefore

= − I1

Z 22

Z21 = Z12 V1 = I1 Z11 + − I1

Þ

Z12

I1 =

Z12 Z 22

Z12

V1 2 ⎞ ⎛ Z12 ⎜ Z11 − ⎟ ⎜ Z22 ⎟⎠ ⎝

And therefore

I2 =

V1 Z12 2 ( Z11 Z 22 − Z12 )

Hence, the input impedance (the ratio of voltage to current) at the driven element can be given by Z1 =

⎛ Z2 = ⎜ Z11 − 12 ⎜ I1 Z 22 ⎝

V1

⎞ ⎟ ⎟ ⎠

(9.2)

That is, the input impedance of driven element modifies because of presence of parasitic element (i.e. reflector). The field distribution around the antenna can be obtained by letting a constant current in driven element and estimating the current induced in the reflector. Therefore, the field radiated from antenna in two principal planes will be I2 ⎡ ⎤ E (R ) = k ⎢ I1 + C d cos R ⎥⎦ ⎣

(9.3a)

I2 ⎡ ⎤ E (G ) = k ⎢ I1 + C d cos R ⎥⎦ ⎣

(9.3b)

in the vertical plane and

in the horizontal plane. The angles q and fÿ are the point of locations from the centre of antenna in the vertical and horizontal planes, respectively, and k is a constant.

DESIGN OF YAGIUDA ANTENNA Experimental investigations by Viezbicke at the NIST have produced a wealth of information on Yagi–Uda antenna design. The main objective is to determine optimum design for a specific boom length is given by [6]. LB = S R + h

Yagi Antenna

333

Boom length from 0.2 to 0.4l was included in the report. Numerous data are included to facilitate the design of the antenna of different lengths to yield maximum gain. Design criteria are also presented for stacking Yagi–Uda array either above the other or side-by-side. Although the graphs do not provide all possible designs, but they do accommodate most practical concerns. A Yagi–Uda antenna with at least several directors is an end fire travelling wave antenna that supports surface wave of the slow wave type (c/v) > 1. That is, the driverreflector pair launches a wave onto directors that shows the wave down such that the wave phase velocity v is less than that of velocity of light in free space (c). The phase delay per unit distance along the axis of the array in the forward direction is greater than that of the ordinary end fire condition. The standard graphs can only be used to design arrays with overall lengths of 0.4l, 0.8l, 1.2l, 2.2l, 3.2l and 4.2l with corresponding gains of 7.1, 9.2, 12.25, 13.4 and 14.2 dB respectively, and with d/l ratio between 0.001 and 0.04. The heart of the design data is given in Table 9.1 and Figs. 9.6 and 9.7. TABLE 9.1

Optimized uncompensated lengths of parasitic elements for YagiUda antennas of six different lengths

d/l = 0.0085 s12 = 0.2l Length of reflector (l1/l) l3 ⎧ ⎪ l4 ⎪ l5 ⎪ ⎪ l6 ⎪ l7 ⎪ ⎪ Length l8 ⎪ of l9 ⎪ ⎪ directors l10 ⎨ ⎪ (l ) l11 ⎪ l12 ⎪ ⎪ l13 ⎪ l14 ⎪ ⎪ l15 ⎪ l16 ⎪ ⎪⎩ l17

0.4

Length of Yagi–Uda antenna (l) 0.8 1.20 2.2 3.2

4.2

0.482 0.442

0.482 0.428 0.424 0.428

0.482 0.428 0.420 0.420 0.428

0.482 0.432 0.415 0.407 0.398 0.390 0.390 0.390 0.390 0.398 0.407

0.482 0.428 0.420 0.407 0.398 0.394 0.390 0386 0386 0.386 0.386 0.386 0.386 0.386 0.386 0.386

0.475 0.424 0.424 0.420 0.407 0.403 0.398 0.394 0.390 0.390 0.390 0.390 0.390 0.390

Spacing between directors (sij/l)

0.20

0.20

0.25

0.20

0.20

0.308

Directivity relative to l/2 dipole (dB)

7.1

9.2

9.2

12.25

13.4

14.2

Design curve, Fig. 9.6

A

B

B

C

B

D

334


FIG. 9.6

Standard design curve to determine length of elements of YagiUda arrays [1].

FIG. 9.7 Increase in optimum length of parasitic elements as a function of metal boom diameter. The specified information is usually the centre frequency, gain d/l0 and D/l0 ratios, and it is required to find the optimum parasitic elements lengths (all directors and reflectors). The spacing between the directors is uniform but not the same for all designs. However, there is only one reflector at 0.2l away from director and it is same for all the designs. The characteristics of Yagi–Uda antenna are affected by all of the geometrical parameters of the array, number of elements, lengths of elements and their spacing, etc. Table 9.2 represents characteristics of several Yagi–Uda antennas, which have been obtained using the moment method. The data listed well agrees with the experimental data [6, 13].

Yagi Antenna

TABLE 9.2

Characteristic performances of equally spaced YagiUda array antennas Element lengths

Number of elements (N1) 3 4 4 4 4 5 5 5 5 6 6 6 7 7 7

H-Plane

Spacing Reflector, LR Driver, L Directors, Gain Front(wave(wave(waveLD (dB) to-back lengths) lengths) lengths) (waveratio lengths) (dB) 0.25 0.15 0.20 0.25 0.30 0.15 0.20 0.25 0.30 0.20 0.25 0.30 0.20 0.25 0.30

335

0.479 0.486 0.503 0.486 0.475 0.505 0.486 0.477 0.482 0.482 0.484 0.472 0.489 0.477 0.475

0.453 0.459 0.474 0.463 0.453 0.476 0.462 0.451 0.459 0.456 0.459 0.449 0.463 0.454 0.455

0.451 0.453 0.463 0.456 0.446 0.456 0.449 0.442 0.451 0.437 0.446 0.437 0.444 0.434 0.439

9.4 9.7 9.3 10.4 10.7 10.0 11.0 11.0 9.3 11.2 11.9 11.6 11.8 12.0 12.7

5.6 8.2 7.5 6.0 5.2 13.1 9.4 7.4 2.9 9.2 9.4 6.7 12.6 8.7 8.7

ÿ

E-Plane

Input HPH SLL H HPE SLLE impe- (degrees) (dB) (degrees) (dB) dance (ohms) 22.3 + j15.0 36.7 + j9.6 5.6 + j20.7 10.3 + j23.5 25.8 + j23.2 9.6 + j13.0 18.4 + j17.6 53.3 + j6.2 19.3 + j39.4 51.3 – j1.9 23.2 + j21.0 61.2 + j7.7 20.6 + j16.8 572 + j1.9 35.9+j21.7

84 84 64 60 64 76 68 66 42 68 56 56 58 58 50

–11.0 –11.6 –5.2 –5.8 –7.3 –8.9 –8.4 –8.1 –3.3 –9.0 –7.1 –7.4 –7.4 –8.1 –7.3

66 66 54 52 56 62 58 58 40 58 50 52 52 52 46

–34.5 –22.8 –25.4 –15.8 –18.5 –23.2 –18.7 –19.1 –9.5 –20.0 –13.8 –14.8 –14.1 –15.4 –12.6

Conductor diameter = 0.005l

HANSENWOODYARD END FIRE ARRAY In long distance communication, an ordinary antenna does not meet the requirements of high gain as well as directive characteristics; hence, it is necessary to design new types of antennas that could meet the demands of many applications. There are two ways to achieve this goal: One is to enlarge the dimension of radiating element and another is to enlarge the dimensions of antenna system without necessarily increasing the size of individual radiators. That is, to form an assembly of radiating elements in proper geometrical configuration. This new concept of antenna, formed by multielement is referred to as an array. This approach is of great importance, since the performance characteristic of antenna system can be regarded as produced by an individual element. The element of an array may be of any form, but identical elements are always convenient, simpler and more practical. An array of identical elements having identical magnitude and each with a progressive phase are referred to as a uniform array. Basically, there are two types of arrays, namely broad side array and end fire array. In the broad side array, direction of maximum radiation of an antenna directed normal to the axis of array (q = 90°). This is a linear array of several isotropic radiators of same magnitude and phase, therefore, d = 0 and y = d cos q, where d is the phase difference of adjacent sources, and y is total phase difference of the fields from adjacent sources. To make y = 0, q must be equal to (2k + 1)p/2, where k = 0, 1, 2, 3, …. The field is, therefore, maximum when qÿ = p/2 or 3p/2. In order to have the maximum radiation normal to the axis, it is necessary that all the elements should have the same phase excitation as well as the same amplitude excitation.

336


The separation between elements may not be constant value. To ensure that there is no maximum radiation in other directions (except normal to axis), the separation between the elements should not be equal to multiple of l (i.e. d ¹ nl). Instead of having the maximum radiation along perpendicular to axis of array, it may be desirable to direct it along the axis of array. That is, it is necessary that antenna should radiate towards only one direction (either q = 0° or 180°, Fig. 9.8). An array of this type is referred to as an end fire array. Under the condition y = 0 and q = 0, S = – dr, i.e. for an end fire array, the spacing between radiators is related progressively by the same amount as the spacing between radiators in radians. To direct the maximum radiation towards q = 0°

y = kd cos q + b|q=0° or b = –kd

(9.4a)

and maximum radiation is desire towards q = 180°

y = kd cos q + b|q=0° or b = +kd

FIG. 9.8

(9.4b)

Array of n-point sources.

Thus, the end fire radiation is accomplished when b = m kd for q = 0° and 180° respectively. The end fire array under condition b = – kd, produces a maximum radiation in the direction q = 0°, but not provides the maximum directivity. Hansen and Woodyard [7] proposed that a larger directivity can be achieved by increasing the phase changes between the radiators. They proposed that the required phase shift (d) between closely spaced elements of a very long array should be equal to ⎛

C = − ⎜ kd + ⎝

2.94 ⎞ Q⎞ ⎛ ⎟ = − ⎜ kd + ⎟ N ⎠ N⎠ ⎝

for maximum radiation in q = 0° direction

(9.5a)

Yagi Antenna

⎛

C = ⎜ kd + ⎝

2.94 ⎞ ⎛ Q⎞ ⎟ = ⎜ kd + ⎟ N ⎠ ⎝ N⎠

337 (9.5b)

for maximum radiation in q = 180° direction. Equations (9.5a) and (9.5b) are known as increased directivity condition for the end fire radiation. In order to realize the increase in directivity under this condition, it is necessary to have the following values of y. (a) For maximum radiation in q = 0° direction

| Z | = |kd cos R + C |R = 0° =

Q N

= |kd cos R + C |R =180° = Q

(9.6a) (9.6b)

(b) For maximum radiation in q = 180° direction

| Z | = |kd cos R + C |R =180° =

Q N

= |kd cos R + C |R =0° = Q

(9.7a) (9.7b)

where N is the number of elements. The condition of |y| = p/N in Eq. (9.6a) or |y| = p in Eq. (9.7a) is realized by the use of Eqs. (9.5a) and (9.5b) respectively. For an array of N-element, the condition of |y| = p is satisfied by using Eq. (9.5a) for q = 0° and Eq. (9.5b) for q = 180° and for chosen element spacing of ⎛ N − 1⎞ M d=⎜ ⎟ ⎝ N ⎠ 4

(9.8)

i.e., if the number of elements in array (i.e., N) is very large, Eq. (9.8) can be approximated by d = l/4. Thus for a large uniform array, the Hansen and Woodyard condition can only yield an improved directivity provided the spacing between the elements is approximately l/4. In case of Yagi–Uda array, if the partial boom length h, measure from the driver to the furthest director is very long (h >> l), the Hansen and Woodyard condition requires that the phase difference between the surface wave and free space at the director furthest from the driven element to be approximately p/180o. Thus h lg – h b = p ⎛ 1 1⎞ − 2h ⎜ ⎟ = 1 or ⎜ Mg M ⎟⎠ ⎝

M M =1+ Mg 2h

(9.9a) (9.9b)

where bg a guided phase constant along the forward direction of the axis of the array and lg is corresponding wavelength. Note that bg = – b cos q0 » b(c/v) and c/v > 1, implies that |cos q0| > 1. Equation (9.9b) has shown as the upper dashed lines in Fig. 9.9.

338

FIG. 9.9


Variation of relative phase-velocity (c/v) = l/lg, with h/l, for minimum gain surface wave antennas. Where HWHansen and Woodyard condition, EPEhrenspeck and Poehler values, and LBLower bond.

However, Ehrenspeck and Poehler in their experiment obtained that the optimum terminal phase difference for Yagi antenna is about 60°, rising to about 120° for 4l < h < 8l, and then approaching 180° for h > 20l. This is shown by solid curve in Fig. 9.9. Examining Table 9.1, for those with a director spacing SD = 0.2l, indicates that the directors do tend to be shorter for longer boom lengths.

ANALYSIS OF YAGI ANTENNA Available literatures reveal that there had been many experimental investigations rather than theoretical formulation of the Yagi–Uda array, there are however many expectations. A paper published by Winkinshaw in 1946 dealt with only short (4 directors) Yagi–Uda antennas. A theoretical and experimental paper by Green gives design data for several practical Yagi antenna arrays. This method of the classical one in which the antenna was considered as a special kind of central driven dipole array and actual exciter elements are short circuited at the terminals. Green has analyzed arrays with up to ten elements using this method, and found satisfactory results. Serracchioli and Levis theoretically investigated the phase velocity of long end fire uniform dipole arrays. And found that phase velocity depends on element’s radius and length as well as spacing between elements. Mailloux developed a method for analyzing the behaviour of finite reflectors of less 20 elements Yagi array. His method essentially consists of truncating a semi-infinite array, and then calculating the reflection coefficient at the end of the array. Soon after Mailloux, R.W.P King applied his three-term theory to uniform Yagi–Uda array, where the phase velocity and radiation patterns were calculated in terms of complex current distribution on all the elements of array. The calculation of radiated electric field by the elements is based on rigorous equations and the results are

Yagi Antenna

339

found to be in good agreements with experimental reported results. In addition, the results enable one to show clearly the dependence of the far field patterns on the phase velocity along the array. Let us consider a Yagi–Uda array of multilinear elements, as shown in Fig. 9.10, in which it is found that an efficient representation for the current on pth element is as follows: N

I p (z ) =

∑

I n p cos(2n − 1)

n =1

FIG. 9.10

Qz

(9.10)

L

Coordinate system used for analysis of YagiUda array.

This is series of odd-ordered even modes and represents that the current goes to zero at the ends of element p. This is found to be a suitable approximation for elements whose diameter is small in terms of l. An integral expression for the electric field radiated by array may be derived from the vector potential. This integral expression is

Ez (x, y, z ) =

where and

M N /F 8Q 2 j

D +2

N

∑ ∑I ∫ np

p =1

G(x, y, z/x ′, y′, z ′) =

n =1

L/2 − L/2

exp[ − jkr ] r5

G(x, y, z/x ′, y′, z ′) cos (2n − 1)

Q z′ Lp

[(1 + jkr )(2r 2 − 3a2 ) + (kar )2 ]

r = [(x¢ – x)2 + (y¢ – y)2 + a2 + (z¢ – z)2]1/2

dz ′

(9.11)

(9.12) (9.13)

where (x¢, y¢, z¢) represents a point on source region on qth element and (x, y, z) represents an observation point on pth element as shown in Fig. 9.11.

340


FIG. 9.11

Showing distance from an observation point on pth element to source region on qth element. Insert shows relation between z¢ and q, when observation point and source region are on the same element.

It has been found that z¢ is an efficient variable of integration when the current on wire is observed at an observation point on qth element under condition p ¹ q. However, it is more efficient to use q as the variable of integration, since the integrand varies rapidly, when z = z¢.

System of Linear Equation Let us consider a Yagi–Uda array composed of a single reflector and driven element and D directors, also let each element of different length and each have N modes. Using Eq. 9.11, the first part of system equations is then found to be D +2

N

∑ ∑ Cm , n p I n p = 0 ,

m = 0, 1, 2, 3, ... N ´ D

(9.14)

p =1 n =1

These equations are generated by requiring that tangential component of electric field E be zero at N points on each director. That is, Et is zero at a total of N ´ D points on the

341

Yagi Antenna

directors. The matching point on a single director is shown in Fig. 9.12. The next N equations are similar to the previous N ´ D equations since Et vanishes at N points on the reflector element. Thus D +2

N

∑

∑ Cm, n p I n p = 0 ,

m = (N ´ D) +1, N ´ (D + 1)

(9.15)

p =1 n =1

The last N equations are generated by using the boundary conditions on the driven element as shown in Fig. 9.13. D +2

N

∑ ∑ Cm , n p I n p = 0 ,

m = N ´ (D + 1), N ´ (D + 2) – 1

(9.16)

e = D + 2, m = N ´ (D + 2)

(9.17)

p =1 n =1

N

∑ I ne = 1 ,

and

n =1

ZN–1

ZN ZN–1

ZN–2

Z3

Z3

Z2

Z2 Z1

Z1

x

x

IAMP

FIG. 9.12

Parasitic element with N matching points along its axis.

FIG. 9.13

Driven element with N1 matching points along its axis.

On the driven element the tangential E boundary condition is only enforced at N–1 points even though there are N modes. The Nth equation on the exciter arises from the constraint on the terminal, current value [8].

Far-field Pattern The far-field pattern of a single element Yagi–Uda antenna, shown in Fig. 9.1, is given by ER (R ) =

jXN

4Q r0

exp[ − jkr0 ] sin R

∫

L/2 − L/2

I (z ′) exp ( jkz cos R ) dz ′

(9.18)

342


Since the current is expressed as a Fourier series, Eq. (9.18) reduces to ER (R ) =

− jL N /F

Q r0

N

(2n − 1) I n cos(Q L ′ cos R )

n =1

(2n − 1)2 − (2L ′ cos R )2

exp[ − jkr0 ] sin R ∑ (−1) n

(9.19)

where L¢ = L/l. Since r0 = [r – (r sin q cos f + y sin q sin f + z cos q)], let us define a factor called the pattern factor F(q, f)

F (R , G ) = L p sin R exp[± jk (x p sin R cos G + y p sin R sin G + z p cos R )] N

(2n − 1) I np cos (Q L ′cos R )

n =1

(2n − 1)2 − (2L ′ cos R )2

× ∑ (−1)n

(9.20)

Thus, if the total number of elements is M each with N modes, the total pattern factor FT (q, f) can be written as N

FT (R , G ) = sin R ∑ L p exp[± jk (x p sin R cos G + y p sin R sin G + z p cos R )] × p =1

N

(2n − 1) I npc cos (Q L ′ cos R )

n =1

(2n − 1)2 − (2L ′ cos R )2

∑ (−1) n

with

M = D+2

(9.21)

These expressions are very simple, and applicable if the number of modes on each element is the same; however, if not so they are necessarily more complicated. In practice, it is found desirable to retain more modes on the driven element than required for the parasitic elements. The method is simple and not limited to the number of elements. This method can also be applied to non-linear planar array of any number of elements, arbitrary spacing and length. Good agreement is seen to exist with the experimental pattern. There are various advantages of this method over experimental, particularly saving of time and money. In addition, there is possibility of obtaining antennas optimized with respect to directivity, side lobe levels and bandwidth. As well as far-field pattern of a Yagi–Uda array is concerned, it may be tuned or adjusted for a particular frequency in two ways; one way is to vary the director spacing while maintaining the element length and reflector spacing constant. The second one is to vary the director length while holding all other parameters constant. It is so because they alter the phase–velocity along the array. The usual effects of altering the reflector spacing or length are mostly to change the level of the back lobe and to control the impedance of array. Further, changing the length of the driver will have a negligible effect on the pattern, but it may change the input impedance of the array.

CIRCULAR POLARIZATION FROM THE YAGIUDA ARRAY In general circularly polarized radiation is achieved, if two Yagi–Uda antennas are crossed (elements at right angles on the same boom) with the driven element fed in phase quadrature,

Yagi Antenna

343

or both driven elements are fed in phase but with one array displaced l/4 along the boom with respect to the other. Another alternative is to use the crossed directors with a monofilar axial mode helix antenna fed by a single coaxial transmission line feed. The term ‘crossed’, mean two identical Yagi antennas where the first one is placed in vertical position and the second in horizontal position (V–H arrangement), with the angle between them always 90°. It is also possible to achieve CP by placing these antennas on X arrangement, the first one with angle of 45° and the second with angle 135° using ‘cross boom’. A typical V–H and X arrangements of Yagi–Uda antennas is shown in Fig. 9.14.

Vertical

Cross boom

X Yagis

Horizontal

Mast

FIG. 9.14

Circularly polarized YagiUda antennas.

In X arrangement the metallic cross boom does not affect the antenna’s efficiency in contrast with the V–H arrangement where the cross boom affects the radiation of antenna as it is between horizontal antenna’s elements. This is also because in V–H arrangement half of the horizontal elements are ‘shadowed’ by the cross boom. But this problem (unwanted radiation) is overcome by using the cross boom made up of non-metallic material (like plastic, wood, PVC, etc.). As a result, there would be no significant effect between the cross boom and the horizontal antenna and both the antennas can be mounted in vertical and horizontal positions. In addition, the V–H arrangement offers a small advantage of having a ‘switchable’ antenna system and useful to change polarizations: vertical (for mobiles), horizontal (D xing) and circular polarization (satellites) for purposes. The V–H arrangement also needs a minimal number of coaxial relays, to make the system ‘switchable’ between vertical– horizontal–RHCP polarizations, whereas X arrangement needs comparable more coaxialrelays (and phase-stubs) for this purpose.

YAGIUDA LOOP ANTENNAS As is well known the performance characteristics of a very small loop antenna is similar to that of a short dipole and in the far-field region it has a null along its axis. The radiation

344


along its axis increases as circumference of the loop increases and it reaches maximum at above one wavelength. Thus, loops can be used as basic elements of Yagi–Uda array as shown in Fig. 9.15. And hence by proper selection of loop size and their spacing a unidirectional beam can be obtained along the axis of the loops and the array. The general performance characteristics of a Yagi–Uda loop array is controlled by the geometrical parameters: reflector, feeder and director size and spacing between elements. x

2a Sik

z Reflector

Directors

y

FIG. 9.15

YagiUda array of circular loops.

A coaxial loop array with two elements (driven and reflector) is the simplest configuration of a Yagi–Uda loop array. Its operating principle is similar to the ordinary Yagi–Uda array. The Yagi–Uda loop array of two elements resulted in a 1.8 dB higher gain than a corresponding two-dipole array. In addition, the radiation characteristics of the loop arrays mounted above the ground are less affected by the electric properties of the soil as compared with those of dipole array [9]. However, the two-loop array resulted in the elimination of corona problem at high altitudes. Shen and Raffoul studied the Yagi–Uda circular loop array by travelling the wave approach, assuming a constant current on loops with equal and progressive phase shift along the directors. The resonant properties of circular loops (element of a coaxial loop array) and the effect of the various array parameters on its characteristics are described in the following section [10]. Consider a coaxial Yagi–Uda array consisting of N circular loops of arbitrary dimensions, as shown in Fig. 9.16. Let ai, bi and dij be the radii of wire element, the loop and element spacing respectively, with ai << bi and ai << l, and l is operating wavelength. If only second loop is excited, resulting loop current, input admittance, radiation pattern, and forward and backward gains are given by [11] m0 ⎡ ni ⎤ 1 + jXCgi ⎥ cos nG I i (G ) = ∑ Yin2 V2 cos nG + Vi ⎢ ∑ n n =0 ⎣ n = m0 +1 Zii ⎦

(9.22)

Yagi Antenna

345

z N

bN

Directors

1

Driven element y O

Reflector x

FIG. 9.16

Coaxial YagiUda array of circular loops. Y2 = I2/V2

(9.23)

∞ ⎡ ER ⎤ ⎡cos R ⎤ N ⎡sin G ⎤ I n n exp ( − jkr ) ⎢ ⎢ ⎥ = − ⎥ ∑ kbi exp(jkdi1 cos R ) ∑ j I n (J n −1 (xi ) ± J n +1 (xi )) ⎢ ⎥ 4r n =0 ⎢⎣ EG ⎥⎦ ⎣⎢1 ⎦⎥ i =1 ⎣⎢cos G ⎦⎥

(9.24) Gd , r =

QI0 4[Re (Y2 )] V2

N

2

∑ kb1 exp ( ± jkdii )

i =1

2

I i1

(9.25)

where k and h0 are the propagation constant and characteristic impedance of free space respectively. The admittance matrix Yijn is the inverse of the impedance matrix Zijn , Cgi is the gap capacitance, xi = kbi sin q, and the subscripts d and r refer to the forward and backward gains along the array axis [12]. The expressions for loop currents and input admittance of the considered array involves impedance and admittance matrices whose dimensions depend on the number of array elements. Loops are basically resonant elements and their currents are mostly due to resonant modes. Therefore, it is clear that elements of a loop array resonate at certain modes, depending upon their size and size of exciting loop. The radiation pattern, input impedance and forward and backward gains can be obtained by retaining only dominant mode for resonant loops and, dominant and two adjacent modes for resonant loops, respectively.

346


EFFECTS OF ARRAY PARAMETERS Basically, there are two parts of a Yagi–Uda array: the first one is the combination of reflector and feeder, and second is rows of directors. From available reports it is clear that changing the reflector length and spacing does not make any significant effect on the forward radiation pattern. Therefore, the reflector size and spacing can be used to optimize other parameters without affecting the forward gain. To study the effect of various array parameters on the array characteristics, a twelve-element loop array is considered. The effects of the exciter and reflector size and spacing has shown that the optimum directive gain Gd is obtained for condition kb1 = 1.05, kb2 = 1.1 and d12/l = 1.1. Thus, selecting the reflector size kb1 = 1.05, to maximize the array gain, its spacing is selected to be 0.1l, which gives the maximum gain. The exciter size, kb2 = 1.1, gives resistive input impedance only for an isolated exciter element. The radius of wire loop a is same for all the elements and is related to the radius of the exciter (b2) by 2 ln[2p(b2/a)] = 11 [12].

Effect of the Reflector In the previous section it is mentioned that spacing and size of reflector have negligible effect on the forward gain. However, the reverse gain and the input admittance are very sensitive to reflector size and spacing from driver. Thus, reverse gain and the input admittance of loop array can be controlled by varying the size and location of reflector without affecting its forward radiation.

Effect of the Directors The size and spacing between elements have large effect on the forward gain and results in increase or decrease in forward gain.

Effect of Exciter and Wire Cross-section As far as forward gain is concerned, it is nearly unaffected with driver size and wire crosssection, but reverse gain and input admittance are very sensitive to these parameters. Thus, they can be used to control the reverse gain and input admittance without affecting the forward radiation. In particular, it was found that for practically useful radiation pattern, director circumference must be less than the exciter. The effect of various parameters on the characteristics of array is summarized in Table 9.3. The director of size kb = 0.9 seems to be optimum director size whereas director directive forward gain (Gr) has the largest value for any director’s spacing. In addition, any average rise in Gr by adding a director to the array is the largest and the directive gain increases with increasing the director spacing. For director of size kb = 0.7, the forward gain is almost independent of the director spacing dD for all dD ³ 0.15l. In addition, there are no side lobes in front of director for both E- and H-planes when dD ³ 0.25l. The 3-dB beam

Yagi Antenna

TABLE 9.3

347

Effects of elements size and spacing on the antenna parameters

Parameters characteristics

Reflector spacing

Reflector size

Exciter size

Wire-area Director size cross-section and spacing

Forward gain

Negligible effect Large effect Large effect

Small effect

Small effect

Small effect

Large effect

Large effect Large effect




Backward gain Input admittance

width of the H-plane pattern found to be larger than that of the E-plane pattern. The results are reported in [12], which can be used to design arrays with different wire radius, loop radius and separation between elements. Hence, for maximum array gain, the director of size kb = 0.9 is the optimum choice and array gain generally increases with increasing director spacing but the ratio of increase, decreases with the number of directors. Furthermore, for a given array length, the array gain increases with the array element. Since gain is constant for the director size kb = 0.7 and dD ³ 0.5l, the array length can be used as a parameter to optimize the other parameters.

ADVANTAGES AND APPLICATIONS The Yagi–Uda antenna is a widely used due to its high gain capability, low cost and ease of construction. The Yagi antenna is one of the more popular antennas used in the frequency range of microwaves. It is a narrow bandwidth antenna; however, bandwidth of Yagi antenna can be improved by using feeds other than a dipole, such as a folded dipole. The driving point impedances of the both (dipole and folded dipole) are usually reduced considerably from their self-impedances by mutual coupling effect. Further, increased gain can be achieved by arraying/stacking Yagi antennas. Maximum gain results for a separation of almost one wavelength. Thus, for a given application if somewhat narrow bandwidth can be tolerated, the Yagi–Uda antenna can provide good gain (approx. 9–12 dB) at low cost. Applications would include any system usually less than 1000 MHz in the HF and SHF portions of the spectrum where antenna gain and directionality are factors. Avionic systems and VHF radar were very suitable technical areas for Yagi–Uda antenna applications. Optimum values of design parameters ranges are as follows: Driven element Directors Separation between directors Radii of directors Separation between driven elements and parasitic

0.45–0.49l 0.4–0.45l 0.3–0.4l 0.15–0.25l and 0.15–0.25l

In addition, there are some specific characteristics of Yagi–Uda antennas: · If the length of antenna is 6l or more, the overall gain is independent of the director spacing.

348


· The size and spacing of the directors have a large effect on the forward gain, backward gain and input impedance. · The reflector size and spacing have negligible effect on the forward gain and large effects on the backward gain and input impedance. · More than one reflector slightly improves the directivity of the antenna. · The addition of more directors will increase the gain of the antenna, but it is up to 5 directors only. · The input impedance of the driven element can be increased by using folded dipole. This is an advantage as the Yagi antenna generally has low input impedance and the antenna needs to match with the Tx line impedance.

SOLVED EXAMPLES Example 9.1 Solution:

Design a 4-element, Yagi–Uda antenna to be operated at 100 MHz.

Length of reflector =

Length driven element = Length of director =

500 100

475 100

= 5 ft

= 4.75 ft

455

= 4.55 ft 100 and length of 4th elements = 4.55 ´ 0.95 = 4.3225 ft. Example 9.2 Design a Yagi–Uda five-dipole array for operation at 500 MHz, fed by 300 W 2-wire Tx line. Solution: Length of reflector = 0.475l = 0.475 ´ l (300/500) = 0.285 m = 28.5 cm Length of driver = 0.460l = 0.460 ´ 60 = 27.65 cm Length of director (1) = 0.440l = 0.440 ´ 60 = 26.4 cm Length of director (2) = 0.430l = 0.430 ´ 60 = 25.8 cm Length of director (3) = 0.400l = 0.400 ´ 60 = 24.0 cm Spacing between reflector and driven element = 0.25l = 0.25 ´ 60 = 15 cm Spacing between directors = 0.31l = 0.31 ´ 60 = 18.6 cm Width of element = 0.01l = 0.01 ´ 60 = 0.6 cm Width of metal boom = 0.01l = 0.01 ´ 60 = 0.6 cm Length of array = 27.6 + 3 ´ 18.6 = 83.4 cm Example 9.3 Find the length of a Yagi–Uda antenna with a directivity 9.2 dB at f = 50.1 MHz, if desire diameter of coupling element = 2.54 cm. Solution:

f = 50.1 MHz = l = 598.8 cm, hence d/l = 2.54/598 = 4.24 ´ 10–3

Yagi Antenna

349

Therefore, from Table 9.1, the total number of elements be 5 (1 reflector, 1 driver and 3 directors) and spacing between reflector and driver and between directors = 0.2l. Hence, the total length of array = (0.2l + 3 ´ 0.2l) = 0.08lÿ = 0.8 ´ 598 = 478.4 cm. Example 9.4 of l/4. Solution:

Find the directivity of H–W uniform array of 12 elements with separation

The directivity of H–W uniform array of n-element array is: d⎞ ⎛ D = 1.789 ⎜ 4 n ⎟ = 1.789 × 12 = 21.468 = 13.32 dB ⎝ M⎠

Example 9.5 Design a Yagi–Uda array of linear dipoles to cover TV channels from 54 MHz to 216 MHz, designated at f = 216 MHz. The gain of array should be 12.35 dB (over dipole). Assume elements diameter is 0.95 cm and supporting boom diameter is 1.90 cm. Solution:

Given f = 216 MHz, l = 1.3889, m = 138.89 cm

Gain of said antenna would be 12.35 – 2.15 = 10.2 dB Therefore, the total number of elements in the array are 6 (1 reflector, 1 driver and 4 directors) Spacing between reflector and driver (S12) = 0.2l = 0.278 m Spacing between directors = 0.25l = 0.25 ´ 1.389 = 0.3495 m Total length of array = (0.278 m + 0.3495 m) = 1.676 m For a desired d/l = 0.95/1.3889 = 0.0068 ® d = 0.0068l Therefore, from Fig. 9.6, when we draw vertical line through d = 0.0068l, we find that the uncompensated lengths are found to be: I¢1 = 0.482l, I¢3 = 0.432l, I¢4 = 0.420l, I¢5 = 0.412lÿ and I¢6 = 0.403l For a desired D/l = 1.90/1.3889 = 0.0137 ® D = 0.0137l Therefore, from Fig. 9.7, when we draw horizontal line corresponding to D = 0.0137l, increase in length found to be 0.0095l. Thus, we find the design lengths of array: I1 = (0.482 + 0.0095)l = 0.495l = 0.684 m I3 = (0.432 + 0.0095)l = 0.4415l = 0.614 m I4 = (0.420 + 0.0095)l = 0.4295l = 0.598 m I5 = (0.412 + 0.0095)l = 0.4215l = 0.586 m I6 = (0.482 + 0.0095)l* = 0.4125l = 0.574 m However, the length of driven element would be 0.5l = 0.69449 m.

350


OBJECTIVE TYPE QUESTIONS 1. The Yagi–Uda antenna was invented and described in (a) Japanese (b) China and India (c) USA (d) None of these 2. The simplest Yagi–Uda antenna consists of three elements, namely reflector, driven element and director, which one is a resonant half-wave dipole at operating frequency: (a) Reflector (b) Director (c) Driven element (d) None of these 3. Name the antenna structure in which elements are arranged at same LOS level and collinearly spaced at 0.0l. (a) Yagi and Log-periodic antennas (b) Horn antenna (c) Rhombic antenna (d) None of these 4. The parasitic elements of Yagi–Uda antenna are excited by the current flow in the driven element. The phase and currents flowing due to induced voltage depends on the (a) Spacing between elements (b) Length of elements (c) both (a) and (b) (d) None of these 5. What should be the length (in feet) of reflector and director of a Yagi–Uda antenna operating at 250 MHz (a) 2, 1.82 (b) 2.82, 2.5 (c) 1, 2 (d) None of these 6. In case of Yagi–Uda antenna, there exists a voltage node at (a) End of director (b) 0.25l point on driven element (c) Centre of each parasitic element (d) None of these 7. The three-element Yagi–Uda antenna also known as (a) Radiation antenna (b) Phase antenna (c) Directivity antenna (d) Beam antenna 8. Due to its high gain and beam width, Yagi–Uda antenna termed as (a) Super directive antenna (b) Super gain antenna (c) Both (a) and (c) (d) Super beam antenna 9. In case of Yagi–Uda antenna, the field radiated by the parasite is such that the total tangential field on it is (a) Greater than zero (b) Lesser than zero (c) Zero (d) Between zero and unity 10. The radiations pattern of a Yagi–Uda antenna array is (a) Broadside (b) End-fire (c) Collinear (d) None of these

Yagi Antenna

351

11. Yagi antenna can produce circular polarization if (a) Two Yagi antennas crossed with the driven elements is fed in phase quadrature (b) Feed the crossed directors with a monofilar axial mode helical antenna (c) Both (a) and (b) (d) None of these 12. The gain of a Yagi–Uda antenna can be increased by (a) Adding more directors (b) Adding more reflectors (c) Both (a) and (b) (d) None of these

Answers 1. (a) 6. (c) 11. (c)

2. (c) 7. (d) 12. (a)

3. (a) 8. (c)

4. (c) 9. (c)

5. (a) 10. (b)

EXERCISES 1. What is a Yagi–Uda antenna? How does it differ from a rhombic antenna? 2. Describe the principle of operation of Yagi–Uda antenna. Explain its properties with reference to directivity and bandwidth. 3. Sketch a 3-element Yagi–Uda antenna. Describe the importance of length of reflector and its separation from driven element with regard of performances of Yagi–Uda antenna. 4. Describe the effects of following parameters on the performance of Yagi–Uda antenna: (a) Length of elements, reflector, driven element and director. (b) Spacing of elements, reflector, driven element and director. (c) Total number of elements. 5. What are the applications of Yagi–Uda antenna? Sketch its radiation patterns in two principal planes. 6. What is a Yagi–Uda antenna? Explain its general characteristics. 7. Draw the structure of 3-element Yagi–Uda antenna and give the dimensions and spacing between the elements in terms of wavelength. 8. Design a 3-element, Yagi–Uda antenna to be operated at 80 MHz. 9. Show that the input impedance at the driven element two elements Yagi–Uda array can be given by Z1 = Z11 −

2 Z12

Z 22

10. Write the expressions for the radiated fields of Yagi–Uda antenna.

352


11. Describe the gain, directivity and beamwidth of Yagi–Uda antenna. 12. Describe the importance of Hansen–Woodyard condition. 13. How to obtain CP radiation from Yagi–Uda antenna? 14. What do you mean by Yagi–Uda loop antenna? How does it differ from Yagi–Uda dipole antenna? 15. Design a Yagi–Uda five dipoles array for operation at 450 MHz, fed by 300 W 2-wire Tx line. 16. Find the directivity of H–W uniform array of 8 elements with separation of l/4. 17. Find the number of elements in 12.5 dB, H–W uniform array if elements separation is l/4. 18. Find the length of a Yagi–Uda antenna with a directivity 9.2 dB at f = 50.1 MHz, if desire diameter of array element = 1.90 cm. 19. Design a Yagi–Uda array of linear dipoles to cover TV channels from 54 MHz to 216 MHz, designated at f = 216 MHz. The gain of array should be 14.4 dB (over dipole). Assume elements diameter is 0.90 cm and supporting boom dia is 2 cm. 20. Design a Yagi–Uda array of linear dipoles to cover TV channels from 88 MHz to 216 MHz, designated at f = 216 MHz. The gain of array should be 12.35 dB (over dipole). Assume elements diameter is 0.95 cm and supporting boom diameter is 1.90 cm.

REFERENCES [1] Yagi, H., “Beam transmission of ultra short waves,” Proc., IRE, Vol. 26, pp. 715–741, June 1928. [2] Thiele, G.A., “Analysis of Yagi–Uda types antennas,” IEEE, Trans. and Propagate, Vol. 1 (AP.17), pp. 24–31, Jan. 1969. [3] Ehrenspeck, H.W. and H. Poehler, “A new method for obtaining maximum gain for Yagi antenna,” IRE, Trans. and Propagate, Vol. AP.7, pp. 379–386, Oct. 1959. [4] Uda, S. and Y. Mushiak, Yagi–Uda Antenna, Saski Printing and Publishing Company, Sendai, Japan, 1954. [5] Green, H.E., “Design of data for short and medium length Yagi–Uda array,” Institution of Engineers (Australia), Electric. Engg., Trans., pp. 1–8, March, 1966. [6] Viezbicke, P., “Yagi antenna design,” NBS Technical Note 688, US Government Printing Office, Washington DC, Dec. 1976. [7] Hansen, W.W. and J.R. Woodyard, “A new principle in directional antenna design,” Proc. of the Institute of Radio Engineers, Vol. 26, No. 3, pp. 333–345, March 1938. [8] Thiele, G.A., “Calculation of the current distribution on a thin linear antenna,” IEEE, Trans., Antennas and Propagate, (Comm), Vol. AP. 14, pp. 648–649, Sept. 1966.

Yagi Antenna

353

[9] Adachi and Y. Mushiak, “Studies large circular loop antenna,” Sci. Rep, Research Institute of Tehoku University (RITU), Vol. B9.2, pp. 79–103, 1959. [10] Shen, A. and L. Shafai, “Properties of coaxial loop array,” IEEE Trans., Antenna and Propagate, Vol. AP. 28, pp. 547–550, July 1978. [11] Ito, S., N. Inagaki and T. Sekiguchi, “An investigation of the array of circular loop antennas,” IEEE Trans. Antenna and Propagate, Vol. AP. 19, pp. 469–76, July 1971. [12] Shoamanesh, A. and L. Shafai, “Design data for coaxial Yagi array of circular loops,” IEEE Trans. Antennas and Propagate, AP. 27(5), pp. 711–713. Sept. 1979. [13] Stutzman, W.L. and G.A. Thiele, Antenna Theory and Design, Chap. 5, John Wiley Inc., New York, 1981.

C H A P T E R

10

Log-periodic Antenna

INTRODUCTION A log-periodic antenna is an antenna having a structural geometry such that its impedance and radiation pattern are logarithmic function of operating frequency. To increase the bandwidth of Yagi–Uda antenna array, the log-periodic techniques are generally used; this new developed antenna is known as log-periodic antenna. The log-periodic dipole antenna has certainly been one of the simplest and most useful log-periodic antennas developed in last decades. Various log-periodic antennas are available, which provide unidirectional, bidirectional and omni-directional pattern and either linear or circular polarizations. Unidirectional antennas have moderate directive gain and possibility of high gain has also been proposed through the applications of array techniques or by using the antennas as primary sources for large reflectors. In practice, variations over the frequency band of operation are minor, hence logperiodic antennas are usually considered to be frequency independent antennas. A series of frequency independent antennas were developed at the University of Illinois in the late 1950s and 1960s. Several antenna geometries were examined and those that produced broadband behaviour led to the determination of the properties necessary for wide bandwidth. Simplest log-periodic antennas consist of only parallel wire segments is known as logperiodic dipole array antenna and is shown in Fig. 10.1. The log-periodic dipole antenna (LPDA) is a series of parallel wire dipoles of successively increasing lengths outward from the feed point at the apex. Although LPDA has similar directivity range (7–12 dB) as Yagi–Uda, they are achievable and maintained over much wider bandwidth. However, there are major differences between them. A successful method of constructing LPDA is shown in Fig. 10.2. Using a coaxial cable feed, phase reversal of 180° is achieved between adjacent elements [1, 2]. This feed arrangement provides a builtin broadband balun resulting in a balanced overall system. The elements and feed line both are made of piping. The coaxial cable is brought to the feed through the hollow part of one of the feed line pipes. The outside conductor of the coax is connected to that conductor, while its inner conductor is extended and connected to the other pipe of the feeder line. 354


FIG. 10.1

355

Log-periodic dipole array antenna.

Feed point

Array elements

FIG. 10.2

Method of construction of LPDA.

MATHEMATICAL ANALYSIS AND DESIGN PARAMETERS A particular LPDA may be completely described in terms of its parameters. As shown in Fig. 10.1, a wedge of enclosed angle a bounds the dipole lengths. The scale factor t for the LPDA is defined as U =

Rn +1 Rn

< 1

(10.1)

356


a

Right triangles of enclosed angle

show that Ln /2

tan B =

Hence,

L1 R1

Rn

Ln

= ... =

=

Rn

Ln +1 /2

=

(10.2)

Rn +1

Ln +1

= ... =

Rn +1

LN RN

(10.3)

Equations (10.1) and (10.3) yield Rn +1

U =

Rn

Ln +1

=

(10.4)

Ln

Thus, the ratio of successive elements position is equal to the ratio of successive dipole lengths. The spacing factor (s) for the LPDA is defined as

T=

dn

(10.5)

2 Ln

Combining Eqs. (10.4) and (10.5), we find that all the dimensions are scaled by U =

Rn +1

=

Rn

Ln +1 Ln

d n +1

=

dn

where spacing between elements is given in Fig. 10.1, in which dn = Rn – Rn+1. Since Rn+1 = tRn, So dn = (1 – t)Rn

Ln

tan(B )

From Eq. (10.2)

Rn =

Hence

d n = (1 − U )

2

(10.6)

Ln

2 tan (B )

(10.7)

Substituting the value of dn into Eq. (10.5) gives

T = or

dn 2 Ln

T = tan

=

1 − U 4 tan (B )

− U⎞ ⎜ ⎟ ⎝ 4T ⎠

−1 ⎛ 1

(10.8)

Like in other log-periodic antennas, there is also an active region for the LPDA, where the few dipoles near to l/2-dipoles support much more current to the other radiating elements. It is clearly viewed that the operation of LPDA is similar to that of a Yagi–Uda antenna.


357

The longer dipole behind the most active dipole (with largest current) behaves as a reflector and the adjacent shorter dipole in the front acts as a director. The radiation is then off of the apex. The operational frequency band limit of LPDA is necessarily determined by the frequencies at which longest and shortest dipoles are half-wave resonant, i.e.,

L1 ≈

ML 2

and LN ≈

MU

(10.9)

2

where lL and lU are the wavelengths corresponding to the lower and upper frequencies limits. Since the active region is not confined completely to one dipole, extra dipole can be added to each end of array to ensure adequate performance over the frequency band. The number of additional dipoles required is function of the antenna parameters (t, s), but it is sufficient for non-critical applications. In an ideal case where there is only one dipole, active region reduces to one dipole only, the operating bandwidth B = fmax – fmin, would be given by the difference of the resonant frequency of the first and the last dipoles. If N is total number of dipoles, the ratio of fmax/fmin would be equal to the ratio f max f min

=

lN l1

1

=

U

(10.10)

N −1

Since in an active region there are Na = N1 + N2 dipoles, where N1 is the number of dipoles shorter than l/2 dipole and N2 is the number of dipoles larger than resonant dipoles, Eq. (10.10) reduces to f max f min

lN

= IB

(10.11)

l1

where hB < 1 represents the efficiency of the antenna. If hB is known as the total number of dipoles in the antenna can be calculated in terms of fmax/fmin and hB as follows: N =1+

log(lmax /lmin ) log(1/U )

= 1 + Na +

log(f max /I B f min )

=

log(1/U )

log(f max /f min )

(10.12)

log(1/U )

If lN – k1lmax and l1 – k2lmin, where generally k1 > 0.5 and k2 < 0.5 [3]. In this condition, Eqs. (10.11) and (10.12) yield

IB = and

k2 k1

, N1 =

log(2 k1 ) log(1/ U )

and

N a = N1 + N 2 =

N2 =

log(1/I B ) log(1/U )

log(1/2 k2 ) log(1/U )

(10.13)

358


Efficiency hB, i.e., the factor k1 and k2 should be evaluated for each pair of values t and s . The criterion of evaluation of hB, considering an active region limited to the dipoles, whose current level at the dipole input is 10 dB lower than the value at the input of the dipole where such current is maximum, by assuming k1 = 0.5, i.e., N1 = 0. It has been found that the choice of N1 = 0 does not change very much, the overall number of active dipoles Na = N1 + N2, except changing the length L of the antenna and number of longer dipoles, which have a strong influence on the cost of the antenna. The length of antenna L can be given by

L = Rmax − Rmin =

lN − l1 2

cot B = Mmax

1 − U N −1 1 − U

T U

N1

(10.14)

and it depends in a relevant way on N1. The optimization of the antenna deals with the finding the values of L and N pairs, which results in the wanted design objects (gain, bandwidth and VSWR) and choosing the pairs of design parameters which result in the best cost of compromise. The optimum design of a LPDA, however, is a compromise among different requirements on various antenna parameters which have to be considered in addition to gain, bandwidth and cost and size. The input VSWR should also be controlled and kept maintained below an acceptable value. For high power applications, the primary interest is the knowledge of the distribution of the voltage gradient along the antenna length and its dependence on the antenna parameters.

Feed Techniques Basically there are two methods of feeding the LPD array antenna structure, as shown in Fig. 10.3. In both the methods, dipoles are attached to the feed lines such that their respective terminal impedances appear in parallel across the line at the logarithmic varying interval Ds. In the first method the currents in the element have the same relationship as the terminal phases, if the elements are closely spaced, the phase progression of the currents in the array is to the right. Since this type of excitation is a condition for end-fire array in that direction,

(a)

(b)

FIG. 10.3

Two methods of excitation array.


359

it produces an end-fire beam in the direction of the longer elements and interference effects to the pattern results. On the other hand, in the second method the elements are connected in such a way that a phase constant of p radians are added to the terminal phase of each element. This is accomplished mechanically by twisting the feeder 180° between each element. It is clear that the short closely spaced elements almost are in phase opposition and hence radiate very little energy and their interference effects are negligible. In the remaining part of the array, where elements are long and spacing between them is more, the elements will radiate in the direction of phase progression so that the radiation pattern is end-fire in the direction of shorter elements. Under this mode of excitation, all the most energy radiating elements will be those near resonant length and their combined radiation pattern is forward the array apex as desired. When operating frequency is increased the radiating elements appear closer to the apex by an amount proportional to change in l so that the normalized array length remains approximately constant and antenna shows frequency independent behaviour over any desired bandwidth.

Effect of Truncation on Efficiency and Radiation Pattern To use as a practical broad band radiator, array could be truncated so that at all frequencies above some lower limit (determined by maximum physical dimension of the array) the properties of truncated antennas are identical to those of the infinite structure and they perform in same way too. This imposes a requirement for rapid attenuation of the incident waves along the feeder so that only negligible energy is reflected from the truncated end. If it is so it follows that the array characteristics will be periodic in periods of bandwidth (1/t, 1), and then antenna may broadband to any desired degree by simply adding elements according to the general scheme. If an antenna array is equivalent to a linear passive fourterminal network with input terminal at the apex and output at the truncated end. Then it may be similar to network shown in Fig. 10.4.

PR Pt

Pi Antenna array Pr

Input

Output

FIG. 10.4

Input

Output

Array as a two-port network.

360


It is required that at the truncated antenna the transmitted power Pt, through the network be only a negligible part of the net power in (Pi – Pr), i.e., the transmission coefficient S12 of the network, must be small. It is compulsory that the antenna geometry must efficiently convert the power incident (Pi) on the feeder to radiated power PR. The radiating efficiency of the antenna according to Fig. 10.4 can be defined as

I=

PR Pi − Pr

In terms of the scattering parameters this can re-written as

IR =

1 − S11

2

− S12

1 − S11

2

2

(10.15)

The efficiency defined in this way does not take into account the conductor losses in the elements, therefore it is true radiating efficiency only in the lossless cases. The experimental set-up for the measurement of radiation efficiency was first developed by G.A. Deschamps. The radiation efficiency as a function of array size was measured using an apparatus for several arrays. This was accomplished by starting with a few short elements closer to the apex in place and then measuring the scattering coefficients of the network as each additional element was snapped in place. Response of radiation efficiency versus array size is shown in Fig. 10.5. From which it is clear that by the time an element of length (l/2) is added approximately 80% of the power is being radiated [1]. Experiments conducted for the input impedance reveal that there is variation in value of input impedance vs a for a given value of t. The array elements were arranged in such a way that the terminals of the shortest dipole

FIG. 10.5

Radiation efficiency vs antenna size.


361

appeared at the junction between the coaxial cable and balance line. Truncation of the array at both ends provides upper and lower frequencies limit accordingly. The input impedance in this frequency limits behaves in a manner characteristic of the design parameters a and t. Several combinations of antenna were investigated and results are tabulated in Table 10.1. TABLE 10.1

at ÿ ÿ ÿ ÿ ÿ ÿ a (deg) 10 20 30 40

Behaviour of input impedance (R0) and VSWR vs a for different values of t

ÿ ÿ ÿ ÿ tÿ = 0.95

ÿ ÿ tÿ = 0.89

ÿ ÿ ÿ tÿ = 0.8

R0

VSWR

R0

VSWR

R0

VSWR

75 62 58 —

1.32 1.4 2.02 —

82 70 64 58

1.22 1.20 1.48 1.82

98 88 80 70

1.82 1.68 1.57 1.6

Particularly for a 75 W coaxial cable feed, array having tÿ = 0.89 and a » 17°, then VSWR results on the line is about 1.2:1. Radiation patterns were also measured for the same value of parameters and it was found that the characteristics of radiation pattern also vary with a for different values of t. The cross-polarization for the antennas were found considerably better than 20 dB below maximum pattern. The estimated directive gain as function of a and t is also shown in Fig. 10.6.

FIG. 10.6

Directive gain vs tÿ and a.

362


PERFORMANCE CHARACTERISTICS AND DESIGN OF LPDA Power Gain The power gain G0 of LPDA and its relation with various design parameters has been described very well in [4]. It has been found that for each values of power gain G0, there exist an optimum pair of values of s and t. For the same value of t there are different values of G0. Difference in the gain values seems more important for the smaller values of t. The dependence of the power gain on the parameters of antenna are found as: t (0.82 to 0.96 for corresponding N = 12 and 16), sÿ (0.1 to 0.22) and Z0 in the range from 50 to 500 W [4]. All the calculations have been made assuming two dipoles behind the l/2 dipole, i.e., N1 = 2, except for the case of Z0 = 500 W. In this particular case for the smaller value of t (< 0.84), N1 = 4 has to be considered in order to avoid consistent variation of gain with frequency particularly. The variation of gain with frequency for LPDA parameters t = 0.78 and s = 0.16 are shown in Fig. 10.7, where fN is resonating frequency of longest dipole for Z0 = 500 W [2]. G(dB/iso) 12 11 10 9 8 7 6 5 4 3 2

fN

FIG. 10.7

fN–1

fN–2

f

Variation of gain with frequency of operation.

As far as N1 is concerned, the degradation of the gain is being observed, when the frequency of operation is moved between the frequencies of resonance of dipoles number N and N–3. Minimum frequency has been taken as the frequency of operation for which the gain is 0.5 dB lower than maximum gain. With regard to N2, it has been observed that when t < 0.92, the gain decreases slowly by N2 only when N2 is smaller than critical value, otherwise decreases abruptly when N2 is larger. When t £ 0.92, the extension of active


363

region is taken up as the N2 value increases which results in a 0.25 dB loss of gain with respect to maximum gain. In addition, it is also found that · the uses of thicker dipoles are very needed for high power applications and improve the G0 although this effect is more pronounced for lower values of t. · higher gain values are obtained for rather low Z0 values (particularly for higher values of t) which are lower than the Z0 values and practically achievable with parallel transmission lines, especially in case of high power applications. The value of power gain G0 also depends on overall length of the antenna as well as overall number N of the dipoles, which are the parameters affecting the cost as well as weight of the antenna. The behaviour of power gain G0 versus overall length L, as function number N of dipoles for fmax/fmin = 4, Z0 = 100 W and t ranges (0.82–0.96) are shown in Fig. 10.8, where length L is normalized with lmax [2].

FIG. 10.8

The behaviour of power gain G0 versus overall length L.

Input Impedance and VSWR As the construction of LPDA is based on log-periodic law, the input impedance of an ideal LPD antenna (N = ¥ dipoles) should vary in a periodic manner with frequency. As it is well known that a suitable matching is required between antenna and feed line to maintain the minimum VSWR. The input feed line of characteristic impedance Z0 should match to the

364


average value (Zap) of Zip input impedance of LPDA, where the magnitude of the residual VSWR is found minimum. The average impedance (Zav) is expressed as Zav = Rav + jXav, the value of Rav could be intercepted as an equivalent characteristics impedance of dipole feed line whose value Z0 is modified into Rm. This is because of the capacitive load arising due to smaller dipoles fixed before the active region. The approximate expression for Rm is given as [5] Rm =

Z0 ⎛ Z U ⎜1 + 0 ⎜ 4Z aT ⎝

1/2

⎞ ⎟ ⎟ ⎠

(10.16)

where Za is the equivalent characteristics impedance of dipoles and

⎡ ⎛h⎞ ⎤ Z a = 120 ⎢ ln ⎜ ⎟ − y ⎥ ⎣ ⎝a⎠ ⎦

(10.17)

where h and a are the dipole’s half length and its radius respectively. And the factor y is equal to y = [1.9 ´ 10–3 (h/a) + 6.8589t – 7.0 ´ 10–4 Z0 – 4.1693]

(10.18)

The value of y decreases with decreasing (h/a) and t, and increasing Z0. The periodicity of Zi with the frequency is rather good for tÿ = 0.82 and becomes poorer when increases t. The minimum residual VSWR is decreased by decreasing Z0 and increasing t. An optimum value exist for (h/a), which corresponds to low value of (h/a) for tÿ = 0.9 and increases when tÿ = 1. The variation of the VSWR, for tÿ = 1 may be, however, partially modified with respect to the ideal case of N = ¥, because of limited numbers of dipoles used in the calculation as well as in the active region.

Design of LPDA The design of any antenna configuration is ultimate aim of the antenna designers. The antenna must be designed under certain specifications to meet the demand applications. The most appropriate design procedure for a log-periodic antenna was first given by R.L. Carrel [5] in year 1961. The basic configuration of a log-periodic array is specified in terms of its design parameters: t, aÿ and sÿ [design Eqs. (10.4)–(10.8)]. From these equations, it is clear that once any two of them are specified, the third one can be found easily. There are some additional equations that can be used to design the log-periodic array antenna. The first one is bandwidth of active region Bar which relates a and t as follows [6]: Bar = 1.1 + 7.7(1 – t)2 cot a

(10.19)

However, in practice, designed bandwidth (Bs) is slightly greater than the required bandwidth (B), these two bandwidths are related


Bs = BBar = B[1.1 + 7.7(1 – t)2 cot a]

365 (10.20)

where Bs = deesigned bandwidth B = desired bandwidth Bar = active region bandwidth The total length (L) of the antenna configuration from the shortest (lmin) to the longest (lmax) elements is given by L=

Mmax ⎛ 4

1 ⎞ ⎜1 − ⎟ cot B Bs ⎠ ⎝

Mmax = 2lmax =

in which

u

(10.21)

f min

The centre to centre spacing (s) between the two conductors, each of identical diameters (a) is determined by ⎛ Z ⎞ s = a cosh ⎜ 0 ⎟ ⎝ 120 ⎠

(10.22)

where Z0 is the characteristic impedance of feed/transmission line. The complete geometry of LPDA is defined by the number of dipole elements (N), the lengths of shortest and longest elements (LN and L1), spacing between elements and operating frequency.

E-plane Pattern The expressions for the E-plane pattern of a log-periodic antenna were first given by Carrel [5] in which the factor sin2 q caused the calculated E-plane beam width to be too narrow resulting an excessively high estimate of directivity. Later the modified expression for the E-plane pattern was given as [7] PE (R , G ) =

1 sin R

N

[cos(C hn cos R − cos C hn )]

n =1

sin C hn

∑ iAn

× exp(j C x n sin R cos G )

(10.23)

Comparison of theoretical values of gain for various LPDA is given in Table 10.2 [7]. However, variation in gain of LPDA with a and t are shown in Fig. 10.9. The curves are a modification of those originally presented by Carrel [5] that has been found to have a gain that erroneously high. From available data it is concluded that Carrel’s original curves were more in error (for gain) for lower values of t than for higher values. Thus, the 11 dB and 6 dB contours in Fig. 10.9 are 1 dB and 2 dB lower respectively than those in Carrel’s [5].

366


TABLE 10.2

Comparison of values of gain for different log-periodic antennas

Case

N

t

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab

12 12 16 12 8 8 8 8 8 8 8 8 8 8 8 8 12 12 12 12 12 12 12 12 12 12 12 12

0.8 0.88 0.94 0.93 0.82 0.82 0.82 0.82 0.86 0.86 0.86 0.86 0.9 0.9 0.9 0.9 0.82 0.82 0.82 0.82 0.86 0.86 0.86 0.86 0.9 0.9 0.9 0.9

s

0.14 0.16 0.18 0.146 0.1 0.125 0.15 0.175 0.1 0.125 0.15 0.175 0.1 0.125 0.15 0.175 0.1 0.125 0.15 0.175 0.1 0.125 0.15 0.175 0.1 0.125 0.15 0.175

G1 (dB) 6.1 8.1 10.4 9.1 5.9 6.4 6.6 6.4 6.6 7.1 7.5 7.5 7.2 7.7 8.2 8.3 6.2 6.5 6.5 6.4 6.9 7.3 7.5 7.4 7.8 8.3 8.6 8.8

G2 (dB)

G3 (dB)

G4 (dB)

G5 (dB)

6.3 8.3 10.5 9.0 6.2 6.6 6.7 6.2 7.0 7.4 7.5 7.4 7.8 8.0 8.4 8.8 6.2 6.5 6.7 6.3 7.0 7.4 7.7 7.7 8.0 8.5 8.7 8.8

6.8 8.4 10.4 8.9 7.0 7.4 7.4 6.8 7.4 7.7 7.8 7.7 7.8 8.0 8.5 8.9 7.0 7.3 7.5 6.9 7.4 7.8 8.1 8.0 8.1 8.6 8.8 8.9

— — — — 7.9 8.4 8.5 7.9 8.3 8.7 8.9 8.7 9.0 9.2 9.6 9.5 — — — — — — — — — — — —

— — — — 8.0 8.3 8.3 7.7 8.3 8.6 8.7 8.6 8.7 8.9 9.3 9.7 — — — — — — — — — — — —

The value of Gmax is greater than the values of gain contour at the optimum s line in top portion. The Gmax versus t curve probably represents an upper bound on the LPDA gain that could be achieved in practice for feeder impedances of 100 W or greater. Since LPDA is a very popular broadband antenna of simple configuration, low cost and light weight, it is better to give design details and illustrates them by examples (see solved Example 10.1). The bottom portion of Fig. 10.9 shows a gain that is derived from the data in [2], where N is number of dipoles and may be 12 to 47.


FIG. 10.9

Variation in gain of LPDA with

s

367

and t.

TYPES OF LOG-PERIODIC ANTENNAS The various types of log-periodic antennas used in communications are 1. 2. 3. 4.

Log-periodic Log-periodic Log-periodic Log-periodic

toothed planar antenna toothed wedge antenna toothed trapezoid antenna Yagi–Uda array antenna

Log-periodic Toothed Planar Antenna Before discussing the log-periodic toothed planar antenna, let us put some attention on bow tie antenna (Fig. 10.10). Bow tie antenna is a planar version of the finite bi-conical antenna. It has a bidirectional pattern with main broad beams perpendicular to the plane of antenna. They are linearly polarized antennas and used as a receiving antenna for UHF TV channels frequency, behind the wire grid ground plane, to reduce the back lobe. Since currents are abruptly terminated at the ends of the fins, the antenna has limited bandwidth. The log-periodic toothed planar antenna is one of the first log-periodic antennas, that is similar to the bow-tie antenna except for the teeth. In log-periodic toothed antenna, the teeth act to disturb the current that would flow if the antennas were of bow tie type construction.

368


~

FIG. 10.10

The bow tie antenna.

In log-periodic toothed planar antenna, currents flow out along the teeth of the antenna. From Fig. 10.11, the ratio of edge distances are constant and given by scale factor U =

Rn +1 Rn

<1

(10.24)

<1

(10.25)

and the slot width is expressed

T=

an Rn

These relations are true for any value of n, the number of teeth. The parameter t gives the period of operation of the structure. Thus, one would expect periodic pattern and impedance

FIG. 10.11

Self-complementary LPTP antenna.


369

behaviour with the same period. For example, if frequencies fn+1 and fn from adjacent periods lead to identical performance then fn f n +1

=U

or

fn < fn+1

log fn+1 = log fn – log(t)

or

(10.26)

Thus, the performances of antenna are periodic in a logarithmic fashion and that is why it is called log-periodic antenna. If the teeth sizes of these antennas are adjusted properly, the structure can be made self-complementary. Again from Fig. 10.11, it is also clear that:

g

+

b

= 180°

b

+ 2d =

b

=

and

a

(10.27)

If the structure is self-complementary:

a

=

g

and

d

(10.28)

Hence, from Eqs. (10.27) and (10.28), we get aÿ = 135° and b = 45° for a self-complementary log-periodic toothed planar antenna. If the widths of the teeth and gaps are equalized, then an

T = Using Eq. (10.25) and solving for

s

Rn

=

Rn +1 an

gives

T= U

(10.29)

This relationship and self-complementary features of the antenna are very popular in practice. Thus, LPTA should have performance (impedance and pattern) that varies periodically with frequency. The self-complementary types of antenna do lead to performances that do not vary greatly for frequencies between periods, that is, for fn < f < fn+1. The properties of the log-periodic toothed planar antenna depend on t. It has been found that 3 dB beam-width increases with increasing t, from 30° at t = 0.2 to 75° at t = 0.9. The radiation pattern shows two lobes with maxima in each normal direction to the plane of the antenna. The radiation is linearly polarized parallel to the teeth edges. Most of the current appears on teeth that are about a l/4 long (active region). The frequency band of operation is set by the frequencies where the longest and smallest teeth are a quarter wavelength long [8].

Log-periodic Toothed Wedge Antenna The geometrical configuration of log-periodic toothed wedge antenna is shown in Fig. 10.12. It is unidirectional and linearly polarized log-periodic antenna. The radiation patterns are nearly frequency independent for 30° < y < 60°. The bandwidth of the wedge antenna version is similar to the sheet antenna version, but input impedance reduced with decreasing y.

370


FIG. 10.12

Log-periodic toothed wedge antenna.

For a planar case (i.e., yÿ = 180°) the self-complementary antenna of input impedance (188.5 W) shows only 165 W impedance, whereas the wedge form with y = 30° has impedance of 70 W. As y is decreased, the impedance variation over a period of the structure increases. For example a 3:1 variation occurs for y = 60° relative to the geometrical mean. The polarization is directed along y-axis for an on-axis radiation. There is x-directed small crosspolarized component also arises from the radial current mode. Typically, this cross-polarized level is 18 dB down from the co-polarized level on the axis, which indicates a strong excitation of the transverse current mode associated with frequency independent behaviour.

Log-periodic Toothed Trapezoid Antenna From the construction point of view, log-periodic toothed trapezoid antenna is similar to logperiodic toothed wedge antenna, only the toothed edges are replaced by straight edges. This modification of the geometry turns out to be of little consequence in the performance of the antenna. The performance of this antenna is similar to its curved edge version (selfcomplementary, Fig. 10.11). This antenna can be formed by bending the planar version into a wedge, creating an antenna similar to that of Fig. 10.12. The radiation patterns of these two antennas, wedge and trapezoid antennas are similar, but latter one has better impedance performance with only about a 1.6:1 variation over a period for y = 30°. Figure 10.13 shows the geometrical configuration of log-periodic toothed trapezoid antenna. In a similar fashion, if solid metal sheets of trapezoid antenna are replaced by thin wires, in which thin wires are shaped to follow the edges of the sheet antenna, this new version of antenna is called log-periodic wire antenna. The log-periodic trapezoid wire antenna has similar performances to that of sheet version antenna. It is linearly polarized and unidirectional antenna. Measurements have shown that for a wedge angle y = 45°, E and H planes, 3 dB beam-width are 66° with the gain of 10.2 dB and FBR 12.4 dB. Another version


FIG. 10.13

371

Log-periodic tooth trapezoid antenna.

of log-periodic antenna is zig-zag wire antenna, which exists in both planar and wedge shapes. Figure 10.14 shows the geometrical configuration of log-periodic trapezoid wire and zig-zag antennas.

(a) Trapezoid

FIG. 10.14

(b) Zig-zag

Log-periodic wire antenna.

Log-periodic YagiUda Array Antenna The log-periodic Yagi–Uda array consists of individual cells. Each cell resembles the basic Yagi–Uda antenna of different sizes; number of elements of different sizes. The size and length of the element differ by a geometric constant t [9] is defined by U =

Ln +1 Ln

=

d n +1 dn

(10.30)

372


Like Yagi–Uda antenna each cell has at least three elements: driven element, director and reflector elements. Basic configuration and design parameters of a log-periodic Yagi–Uda array are shown in Fig. 10.15.

FIG. 10.15

Log-periodic YagiUda array antenna and design parameters.

The driven element in each cell is fed by a common transmission line. The method of excitation of driven element is not in a continuous and transposed manner as in the conventional dipole periodic dipole array. But in-between elements, there are parasitic as in the logperiodic array with parasitic elements. The advantages of log-periodic Yagi–Uda array is that it provides wider bandwidth comparison to LPDA. The feed line is two parallel wire lines, where driven elements on one side of the antenna connected to one line and the elements on the other side are connected to other line. Design parameters In order to design a log-periodic Yagi–Uda array let us consider nth cell of overall length of cn. The cell contains a driven element, director and reflector element of lengths Ln, Ld and Lr respectively. The spacing between director and driven elements is Snd, and between driven element and reflector elements is Snr. Then a particular log-periodic Yagi–Uda antenna can be completely described by the following parameters; (a) The ratio of the cell cn–1 located near the feed point of the antenna than cell cn, as geometric constant (t) U =

cn −1

(10.31)

cn

(b) The ratio of the driven element spacing between the (n–1)th and nth cell to driven element length of the nth cell


San Lna

=

1 − U 2 tan (B /2)

373 (10.32)

(c) The ratio of the spacing between the diameter and the driven element to the spacing between the driven and the reflector element H =

n Sda

Sarn

(10.33)

(d) The ratio of the driven element to the reflector element length in the same cell is;

C =

Lna Lnr

(10.34)

The geometric ratio (t) of the antenna is established by the ratio of the director to reflector lengths within a given cell. However, there are some limits with these parameters; the lower limit is determined by pattern break-up, which occurs when (t) is less than 0.8. The upper limit is set by the practical length of the antenna. A good front-to-back ratio may be obtained when the cell length c n is approximately equal to 0.2 t long at its resonant frequency. The pattern gain and input impedance of log-periodic Yagi antenna are low. An efficient LPYA may be realized by increasing the cell length cn toward a half wavelength. Another parameter that describes the antenna characteristics is g. The value of g should be equal or slightly less than unity. The pattern break-up and large excursion in gain will appear if the value of g is significantly different from unity. This is a specific design parameter for three-element log-periodic Yagi–Uda array. The values of geometric constant t must be greater than 0.8, when more number of director elements are needed to accommodate in an individual cell. Since these individual cells will have a narrower frequency band-width, more cells are required to obtain continuous pattern coverage over a particular frequency range. The directivity of antenna may be improved to certain extent by adjusting the relative spacing of the director and newly added parasitic elements in the cells. Principle of operation The phase centre or apparent source of radiation of the log-periodic Yagi–Uda antenna appears to be confined near the driven element of each active cell. At a particular frequency, the active radiation region may be confined, mainly to a single cell. As the frequency is increased or decreased, an adjacent cell becomes increasingly more active until it is the main source of radiation. Between these periods, the active region is composed of two cells. Two sources of radiation exist, and because of this, the radiation pattern may be expected to be perturbed by the array factor of these two sources.

374


LOG-PERIODIC YAGIUDA ARRAY DESIGN AND PERFORMANCE CHARACTERISTICS A three-element log-periodic Yagi–Uda dipole array was designed and investigated with the following specifications (Fig. 10.15). Geometric constant (t) = 0.9 Sn/Ln = 0.853, b = 1.0 Þ Ln = Lnr g = 0.8 Lnd = 0.45l, Lnr = 0.5l Snd = 0.2l, Snr = 0.25l The first cell of this antenna has the following dimensions: L1d = 3.12 inches, L1 = 3.46 inches, L1r = 3.46 inches, S1d = 1.39 inches and S1r = 1.73 inches. (vii) The driven element and parasitic element of the antenna were made from the rods of diameter 0.125 inches [1]. The driven element of antenna is excited by an infinite balun made from a 50 W co-axial cable through two parallel wire transmission feeders. Each line of feeder is of 0.25 inch outside the diameter and spaced by 0.35 inch. (i) (ii) (iii) (iv) (v) (vi)

The performance characteristics of the designed array antenna were measured for the frequency range of 1100 to 1250 Mc/s. And it was found that The 3-dB beam width in E-plane = 42° The 3-dB beam width in H-plane = 50° Front-to-back ratio = 10 dB Behaviours of measured absolute gain response vs frequency is shown in Fig. 10.16. An average value of 10.1 dB with ±1 variation over frequency range 1050 to 1400 Mc/s has been observed. Impedance characteristics indicate that the antenna tends to be inductive which could be regarded to the mutual coupling effect between the driven and parasitic elements. The variation of VSWR response vs frequency is shown in Fig. 10.17. It is obvious that the minimum value of VSWR reflects proper matching between device and feed and, hence, good response of the antenna. From Fig. 10.17, it is also seen that the value of VSWR is minimum (1.5–2.2) in two frequency ranges (1080–1120 Mc/s) and (1300–1450 Mc/s). Hence, it can be concluded that the considered antenna provides a good response at these frequency ranges and performance of the antenna deteriorates for other frequencies.

APPLICATIONS OF LOG-PERIODIC ANTENNA Applications of LP antennas have been limited to frequency domain applications mostly as a transmitter. LPAs have been used in a couple of applications because of their high directivity and wide-band behaviour. LPAs also are popular in communication applications because their high directivity allows the signals to be concentrated into specific area without wasting


FIG. 10.16

375

Absolute gain vs frequency.

FIG. 10.17

VSWR vs frequency.

energy on the outskirts. Log-periodic antennas are used as a low cost alternative to conventional Yagi–Uda antennas operated at VHF/UHF. They are also used for electromagnetic interference (EMI) testing and electromagnetic capability (EMC); like ANSI C 63.4, FCC-15, FCC-18, EN 55022 emission testing and IEC 61000-4-3 immunity testing. In EMI applications they are used to pick up unwanted RF signals and measure the strength of interfering fields. When combined with monitoring receivers LPAs make it possible to intercept unauthorized radio transmitters or to determine the angle of incidence (direction finding) and the polarization plane of EM waves.

376


SOLVED EXAMPLES Example 10.1 Design a LPDA that operates over the entire VHF TV and FM broadcast bands, which span the 54 to 216 MHz frequency range for 4:1 band. Assume the design gain chosen to be 6.5 dB. Solution:

The values of design parameters

t

= 0.822

t

and

and

s

s

for optimum design are [Fig. 10.8]:

= 0.149

⎛1 − U ⎞ −1 ⎛ 1 − 0.822 ⎞ o We know that B = tan −1 ⎜ ⎟ , hence B = tan ⎜ ⎟ = 66.6 ⎝ 4T ⎠ ⎝ 4 × 0.149 ⎠ The length of the longest dipole is determined first. At the lowest frequency of operation (i.e. 54 MHz) the dipole length should be near a half-wavelength, so

L1 = 0.5lL = 0.5 ´ 5.55 = 2.78 m The shortest dipole length should be in the order of LL = 0.5 l1 = 0.694 m at 216 MHz. The LPDA element lengths are computed until a length in the order of 0.694 m is reached. To be specific, design lengths are found from L1 using Ln+1 = tLn. For example, L2 = tL1 = 0.822 ´ 2.78 = 2.29 m

L3 = tL2 = 0.822 ´ 2.29 = 1.88 m L4 = tL3 = 0.822 ´ 1.88 = 1.54 m

Similarly, L5, L6, L7, L8, L9 are 1.27 m, 1.04 m, 0.85 m, 0.705 m and 0.579 m respectively, i.e., the array is terminated with nine elements since L9 = 0.579 m is less than 0.694 m length for the highest operating frequency. Elements could be added to either end to impedance performance at the band edges. The element spacing for this antennas array is found from dn = 2s Ln = 2 ´ 0.149 Ln, using different values of Ln (1, 2, 3, ..., 7). We can find the corresponding values of dn and they are equal to 0.828 m, 0.682 m, 0.560 m, 0.0.459 m, 0.378 m, 0.310 m, 0.256 m and 0.210 m respectively. These dipole lengths and spacing completely specify the LPDA as shown in Fig. 10.1. The outline of the antennas fits into an angular sector of angle a = 33.3°. Example 10.2 Design a log-periodic dipole antenna to cover all the VHF TV channels from 55 MHz to 220 MHz. The required directivity is 9 dB and input impedance is 50 W. The elements should be made of aluminium tubing with 2.0 cm outside diameters for the largest element and the feeder line and 0.48 cm for the smallest element. These diameters yield identical (l/d) ratios for smallest and largest elements. Solution: From Fig. 10.8, for the given value of D = 10.0 dB, corresponding values of t = 0.925 and s = 0.17. We know that B = tan

− U⎞ ⎜ ⎟ ⎝ 4T ⎠

−1 ⎛ 1


B

Hence,

377

⎛ 1 − 0.925 ⎞ −1 = tan −1 ⎜ ⎟ = tan (0.1103) = 6.294 ≈ 6° 4 × 0.17 ⎝ ⎠

Bar = 1.1 + 7.7(1 – 0.9252) cot (6°) = 1.512 Bs = BBar =

Mmax =

L=

f max

× Bar = (220 × 1.512/55) = 6.04

f min v f min

= 300/55 = 5.454 cm

Mmax ⎛ 4

1 ⎞ 5.454 ⎛ 1 ⎞ o o ⎜1 − ⎟ × cot (B ) = ⎜1 − ⎟ × cot (6 ) = 10.825 cm 4 6.04 B ⎝ ⎠ s ⎠ ⎝

The total number of elements N = 1 +

log( f max /f min ) log(1/U )

=1+

log(4) log(1/0.925)

= 18.70

That is, total number of elements is 18 or 19.

T =

T U

=

0.17 0.925 hmax amax

We know where

= 0.176

and

= 2.73 ×

100 2

lmax =

Mmax 2

= 2.73 cm

= 1365

⎡ ⎤ ⎛h ⎞ Z a = 120 ⎢ log ⎜ max ⎟ − y ⎥ ⎢⎣ ⎥⎦ ⎝ amax ⎠ y = [1.9 ´ 10–3(h/a) + 6.8589t – 7 ´ 10–4 ´ Z0 – 4.1693] = [1.9 ´ 10–3(1365) + 6.8589 ´ 0.925 – 7 ´ 10–4 ´ 500 – 4.1693] = 4.733

Hence,

Za = 120[log(1365) – 4.733] = 298.44 W ⎛ Z ⎞ S = a cosh ⎜ 0 ⎟ = 2 × cosh (50/120) = 2.18 cm ⎝ 120 ⎠

Hence the design parameters are

a

= 6°, L = 10.825 cm, N = 18/19 elements, Za = 298.44 W and S = 2.18 cm.

Example 10.3 A log-periodic dipole array is to be designed to cover frequency range from 84 to 200 MHz and has a gain of 7.5 dB. Find the required element length and spacing for optimum design and also find out the total length of antenna.

378


For the given gain of 7.5 dB, corresponding value of

Solution:

t

= 0.86 and

sÿ =

0.1510.

We know that B = tan

B

Hence,

lmax =

Mmax

=

2

v 2f min

=

⎛ 1 − 0.862 ⎞ −1 = tan −1 ⎜ ⎟ = tan × 4 0.159 ⎝ ⎠

300 2 × 84

= 1.785 m and lmin =

U =

We also know that Hence

− U⎞ ⎜ ⎟ ⎝ 4T ⎠

−1 ⎛ 1

Ln +1 Ln

⎛ 0.138 ⎞ ⎜ ⎟ = 12.24° ⎝ 0.630 ⎠

Mmin x 2

=

v 2 f max

=

300 2 × 200

= 0.75 m

⇒ Ln +1 = U Ln

L2 = tL1 = 0.862 ´ 1.785 m = 1.539 m L3 = tL2 = 0.862 ´ 1.539 m = 1.326 m L4 = tL3 = 1.143 m, L5 = t L4 = 0.985, L6 = 0.849 m and L7 = 0.732 m

i.e., the array is terminated in seven elements, since L7 = 0.732 m is less than 0.75 m, length for the operating frequency. We also know the spacing between elements is given by, dn = 2sLn, as s is 0.159, dn = 0.138 Ln. For each value of L, the corresponding values of d are: d1 = 0.568 m, d2 = 0.489 m, d3 = 0.422 m, d4 = 0.363 m, d5 = 0.313 m, and d6 = 0.269 m. 6

Hence the total length of the antenna is ∑ d n = 2.424 m . And outline of the antenna

fit into an angle (a = 24°).

1

Example 10.4 Design a log-periodic Yagi–Uda array taking following specifications: t = 0.8, and San /Lna = 0.85, b = 0.88 and g = 0.9, if the spacing between active element and reflector is 10 cm for a reflector length of 25.67 cm. Solution:

We know that apex angle of the antenna is related by San Lna

=

1 − U 2 tan (B /2)

⇒ tan (B /2) =

Hence,

2 ×

ÿÿa

= 13°

If the following are parameters of nth cell Lna = length of active element Lnd = length of director Lnr = length of reflector

1 − U San Lna

=

1 − 0.8 2 × 0.85

= 0.118


379

Snda = spacing between director and active element Snar = spacing between active element and reflector Xnd = distance of diameter from the apex of the antenna Xna = distance of active element from the apex of the antenna Xnd = distance of reflector from the apex of the antenna n n + Sarn = Sarn (1+ Sda /Sarn ) = 10.0(1 + 0.9) = 19 cm Then the length of nth cell = Sda

Hence, the lengths of (n – 1)th and (n + 1)th cells are Cn–1 = t Cn = 0.8 ´ 19 = 15.2 cm As and

b

and

Cn+1 = Cn/t = 19/0.8 = 23.75 cm

= 0.88 = Lna/Lnr Þ Lna = 0.88 ´ 25.67 = 22.59 cm Lnd =

t

´ Lna = 0.88 ´ 22.59 = 19.88 cm

Similarly, the lengths of (n – 1)th and (n + 1)th cells and spacing between elements of them can be determined.

OBJECTIVE TYPE QUESTIONS 1. The bow-tie antenna is also known as: (a) Log-periodic antenna (b) Yagi–Uda antenna (c) Bifin (d) Helical antenna 2. Which is not true for a bow-tie antenna: (a) It is a linearly polarized biconical antenna (b) The bi-directional pattern normal to the plane of the antenna (c) The limited bandwidth (d) Preferably used as transmitter 3. One of the first designed log-periodic antennas is (a) Log-periodic toothed planar antenna (b) Log-periodic toothed wedge antenna (c) Log-periodic toothed dipole array antenna (d) None of the above 4. A log-periodic antenna is so named because (a) Its impedance and radiation pattern are logarithmic function of operating frequency. (b) Its impedance is logarithmic function of operating frequency. (c) Radiation patterns are logarithmic function of operating frequency squared. (d) Both (b) and (c). 5. Because of its special geometry a log-periodic antenna acts as (a) An almost frequency independent antenna. (b) A partially frequency dependent antenna. (c) A wide beam-width antenna. (d) None of the above.

380


6. The frequency band limits of LPDA is calculated by the frequencies at which longest and shortest dipoles are (a) Full wave resonant (b) Half wave resonant (c) Quarter wave resonant (d) None of the above 7. What is the length of third element of a three-element 500 MHz log-periodic antenna for t = 0.822? (a) 0.202 m (b) 0.978 m (c) 8.95 mm (d) none of the above 8. What is the apex angle a for a log-periodic antenna of gain of 7.5 dB and corresponding values of t = 0.865 and s = 0.158? (a) 1.98° (b) 10.78° (c) 23.6° (d) 12.05° 9. One of the very useful multi-band HF receiving antennas is: (a) Dipole antenna (b) Log-periodic antenna (c) Rhombic antenna (d) Helix antenna 10. LPAs have been used in many applications because of their (a) High directivity (b) Higher B/W (c) High directivity and wide band-width (d) None of the above 11. Which is not true for log-periodic toothed wedge antenna? (a) Unidirectional and linearly polarized (b) Cross-polarization level is 18 dB down the co-polarization (c) Input impedance is 70 W (d) Frequency independent for any value of y. 12. Performance characteristics of log-periodic wedge and trapezoid antennas differ in (a) Radiation pattern (b) Polarization (c) Input impedance (d) None of the above 13. Which has wider band-width? (a) Yagi–Uda array antenna (c) Log-periodic antenna

(b) Log-periodic Yagi–Uda array antenna (d) None of the above

14. The directivity of log-periodic Yagi–Uda array antenna can be improved further: (a) By adjusting the relative spacing of the directors (b) By adding parasitic elements in the cells (c) Both (a) and (b) (d) None of the above 15. Which is not correct for log-periodic Yagi–Uda array? (a) The 3-dB beam width in E-plane 42° (b) The 3-dB beam width in H-plane 50° (c) Front-to-back ratio is 10 dB (d) None of the above


381

16. The HPBW of LP toothed planar antenna relates as (a) Increases with decreasing t (b) Increases with increasing t (c) Independent with value of t (d) None of the above 17. The radiation pattern of LP toothed planar antenna is linearly polarized (a) Parallel to teeth edge (b) Normal to teeth edge (c) At angle 80° from the teeth plane (d) None of the above 18. The number of lobes contained in the radiation pattern of LP toothed planar antenna is (a) One (b) two (c) three (d) None of the above

Answers 1. 6. 11. 16.

(c) (b) (d) (b)

2. 7. 12. 17.

(d) (a) (c) (a)

3. 8. 13. 18.

(a) (d) (b) (b)

4. (a) 9. (b) 14. (c)

5. (a) 10. (c) 15. (d)

EXERCISES 1. If in an LPDA design apex angle (2a) in Fig. 10.1 is considered as a, then show that B = 2 tan

− U⎞ ⎜ ⎟ ⎝ 4T ⎠

−1 ⎛ 1

2. Design an optimum LPDA to operate from 470 MHz to 980 MHz with 9 dB gain. Add one extra element to each over the length of antenna. 3. Design an LPDA each with directivities of 9 dB and characteristics impedance 86–25W, taking into account the following additional specifications: (a) Frequency coverage (54–216 MHz), use dimensional tubing with outside diameters of 1.905 cm and 0.476 cm for the largest and smallest elements respectively. [Ans: Apex angle a = 5.71°, No. of elements = 25/26, lmax = 2.778 m, dmax = 1.905 cm, S = 2.42 cm. Total length of array is 11.538 m.] (b) Frequency coverage (54–88 MHz), using diameters of 1.905 cm and 1.117 cm for the largest and smallest elements respectively. [Ans: Apex angle a = 5.71°, No. of elements = 13, lmax = 2.778 m, dmax = 1.905 cm, Total length of array is 8.1187 m.]

382


4. Construct a log-periodic Yagi–Uda array antenna, if the centre/nth cell of the antenna specifications are as follows: n

n

a = 14°, Sa /La = 0.876 , b = 0.57 and g = 0.9. The length of reflector is 25 cm and

it is spaced 12 cm away from the driven element. 5. Describe feed mechanism of log-periodic antenna. 6. Find the angle of apex of a log-periodic Yagi–Uda antenna, if geometric ratio t is 0.973 and spacing between two successive cells is 0.854 time the active length. 7. Find the spacing between two successive cells, (n+1)th and nth. If length of reflector element is 20 cm, b = 0.976, geometric ratio t is 0.973 and apex angle of the antenna is 0.26 radians.

REFERENCES [1] Isbell, D.E., “Log-periodic dipole arrays,” IRE Trans. on Antenna and Propagate, Vol. AP. 8, pp. 260–70, May 1960. [2] Vito, G.De. and G.B. Stracco, “Comments on the design of log-periodic dipole antenna,” IEEE, Trans. on Antenna and Propagate, Vol. AP. 21, pp. 303–308, May 1960. [3] Smith, C.E., “Log-periodic antenna design handbook,” Smith Electronics, Inc., Cleveland, Ohio, Res., Rep. 1966. [4] Vito, G.De. and G.B. Stracca, “Further comments on the fesign of LPDA,” IRE Trans. on Antenna and Propagate, Vol. AP. 22, pp. 714–718, Sept. 1974. [5] Carrel, R.L., “The fesign of log-periodic antennas,” IRE, Int., Com., Rec., Vol. I, pp. 61–75, 1961. [6] Balanis, C.A., “Antenna Theory: Analysis and Design,” 2nd ed., John Wiley & Sons, Inc., 2003. [7] Butson, P.C. and G.T. Thompson, “A note on the calculation of the gain of LPDA,” IEEE, Antenna and Propagate (Com), Vol. AP. 24, Jan. 1976. [8] Rumsey, V., “Frequency Independent Antennas,” Academic Press, New York, 1966. [9] Barbana, N., “Log-periodic Yagi–Uda array,” IEEE, Antenna and Propagate (Com), Vol. 14, pp. 235–238, March 1966.

C H A P T E R

11

Horn and Cone Antennas

INTRODUCTION Horn antennas are flared open waveguides, which produce an ideal wave front with a larger aperture and provide greater directivity. Horn antennas are very popular antennas in the microwave frequency region and operate satisfactorily at the frequencies up to 1 GHz. The horns have the property of effecting a smooth transition from a waveguide to free space. A horn antenna has the advantages of providing high gain, low VSWR and wide bandwidth in a small, rugged package with no adjustments needed, and has a wide enough beam to be easily pointed under adverse conditions. In addition, they are of low weight and rather easy to construct them. Horn antenna may be considered as a transformer from the impedance of a Tx line to the impedance of free space, 377 W. An optimum horn is a shortest one that approaches maximum gain. Conical horn is an optimum horn which also provides the desired gain with minimum amount of material. Horn antennas have been found useful since 1940 [1]. Horn antennas are not new, the first horn antenna was constructed and operated at 60 GHz, by Indian physicist J.C. Bose in 1897. It was a pyramidal horn, later that was named ‘collecting funned’. One of the most important applications of horn antennas is being used as a feed for a reflector antenna. The feed for a parabolic dish is often specified as horn antenna having a pattern which is 10 dB down at the edge of the dish. The horn antenna can be used as a transmitting/receiving antenna or as a gain standard with gains from 9.5 to 22 dBi. Its wide bandwidth and predictability is ideal for many broadband applications. A horn antenna is the ideal choice for rover station.

Waveguide and Horn Antennas In general, a waveguide is one type of Tx line, capable of guiding EM waves at high frequency with minimum amount of power loss [2]. So, if one dimension (side) of waveguide is greater than a half wavelength, the waves propagate through it with extremely low loss. 383

384


At the same time, if the end of waveguide is simply left open, the wave will radiate out from the open end. It also offers radiation into open space, if it is fed at one end and the other end is kept open like an opened Tx line. In this process a small portion of the incident wave is radiated and remaining portion is reflected back by open circuit. This is due to discontinuity at the edges, which matches the waveguide to free space very poorly. These discontinuities could be reduced by flaring of waveguide and the entire energy incident in forward direction radiated subject to proper impedance matching. Generally, waveguide impedance differs from the space impedance, hence flaring of waveguide is done which decides matching of impedance, and also provides more directive radiation, wider directivity and narrow bandwidth. There are different ways of flaring the waveguide and, hence, transforming into horn antennas. The waveguide supports a finite number of propagating modes, whereas free space supports very large/infinite number of propagating modes. Practical waveguides have the larger dimension greater than a half wavelength, to allow wave propagation, but smaller than a wavelength, to suppress higher-order modes which can interfere with low-loss transmission. Thus, the aperture of an open-ended waveguide lesser than a wavelength does not provide much gain. That is, for high gain a larger aperture is desirable, not a large waveguide. So, if the size of waveguide is expanded/tapered into a larger aperture, the more gain is achieved while preventing undesired modes from reaching the waveguide. To achieve maximum gain for a given aperture size a and hence, maximum efficiency, the taper must be long enough so that the phase of the wave is nearly constant across the aperture.

Various Configurations of Horn Antennas Broadly, horn antennas are classified as rectangular and circular type horn antennas. Under the assumption that rectangular waveguide is energized with a TE10 dominant electric field (along y direction). The vertical and horizontal planes refer to E- and H-plane, respectively. The E plane contains electric fields, whereas H-plane contains magnetic fields and when corresponding planes are flared (that is E or H), resulted horn antenna is called E-plane or H-plane sectoral horn antennas. On the other hand, when both the planes (E and H) are flared, it is referred to as a pyramidal horn antenna. Most commonly used horn antennas are illustrated in Fig. 11.1. In Fig. 11.1, arrows of the fields indicate their directions, and the length represents the magnitude of the field intensity. The field variation across the aperture of the small flared rectangular horns is similar to the sinusoidal distribution of the TE10 mode across the original waveguide. In Fig. 11.1, horn antennas, shown in LHS, are rectangular horns, whereas all the horn antennas shown in RHS are conical horn antennas. They are formed when the circular waveguides carrying TE11 mode, flared in similar manner as rectangular waveguide for rectangular horns. The conical horn shown in Fig. 11.1(f) is similar to pyramidal horn [Fig. 11.1(d)]. The horn antenna shown in Fig. 11.1(g) is excited in the TEM mode by a vertical radiator, whereas horn antenna [Fig. 11.1(h)] is excited in the TE10 mode by a small horizontal loop antenna. Both the horn antennas are non-directional in the horizontal plane and they are termed bi-conical horn antennas.

Horn and Cone Antennas Rectangular horns

385

Circular horns

Horn

Waveguide

Aperture

Throat (a) Exponentially tapered pyramidal

(e) Exponentially tapered

y x z (b) Sectoral H-plane

(f) Conical Axis

y x z

(c) Sectoral E-plane

(g) TEM bi-conical Axis

y

E

x z

(d) Pyramidal

FIG. 11.1

(h) TE01 bi-conical

Various types of horn antennas.

In addition, there are special purpose horn antennas, including those with a dielectric or metallic plate lens in the aperture to correct for the phase error and those with metallic ridges inside the horn to increase the bandwidth. These special purpose horn antennas are illustrated in Fig. 11.2. For a simple horn, the 10 dB beam width is roughly 1.8 times the 3 dB beam width. Whereas the other characteristics of the main beam may be approximated by using Keller’s universal horn patterns, defined by ⎛R ⎞ PdB (R ) = 10 log ⎜ ⎟ ⎝ R0 ⎠

2

(11.1)

386


E

(a) Box horn

P

(a) Compound horn

E

ar

ola ab

E

E

(c) Asymmetric horn

(d) Hog horn

E

(e) Rounded waveguide

p

E

(f) Waveguide with dielectric

E

(g) Pointed waveguide

FIG. 11.2

(h) Waveguide with disc

Special purpose horn antennas.

where PdB(q) = the relative power at angle q from the centre of the beam ÿ ÿ ÿ ÿ q0 = the angle where the power is 10 dB down from maximum value ÿ

The horn shown in Fig. 11.2(a) is a compound horn antenna and offers broadband characteristics. They are used to provide a good broadband impedance matching to free space by properly designing the E-plane flare length and angle. The box horn illustrated in Fig. 11.2(b) shows a narrower pattern in the H-plane than a sectoral horn of the same aperture. The other horn antennas such as asymmetric horn and hog, horn antennas shown in Fig. 11.2(c)–(h), have plane patterns either broader or narrower than the open-ended waveguide. The hog horn Fig. 11.2(d) combines a 90° direction change and aperture enlargement and phase correction in the H-plane. This horn is used to illuminate ‘cheese’ or pill-box antennas. The asymmetrical horn, shown in Fig. 11.2(c), is used as an offset feed for parabolic cylindrical antennas [3].


387

HORN ANTENNA PARAMETERS The cross-sectional view of a rectangular horn antenna is shown in Fig. 11.3.

FIG. 11.3

Cross-sectional view of a rectangular horn antenna.

Therefore, from geometry [2] cos

R R0 R = = 0 2 R0 + E lH

sin

R A A = = 2 2(R0 + E ) 2 lH

tan

R A/2 A = = 2 2 R0 R0

and

lH = R0 + d

where d is path difference (in m) between the rays travelling along the side and along the axis of the horn. Usually, the value of d0 lies between 0.1l and 0.4l. However, in the E-plane of the horn d £ 0.25l and in H-plane d is about 0.4l, and since E approaches zero at the edges. A = aperture of horn and ⎛ A ⎞ −1 ⎛ R0 ⎞ qH = flare angle = 2 tan −1 ⎜ ⎟ = 2 cos ⎜ ⎟ ⎝ lH ⎠ ⎝ 2 R0 ⎠

Since, d << R0, ( R0 + E )2 = R02 +

A2 4

388


A2

⇒ E 2 + 2R0E = R02 +

4

⇒ R0 ≈

A2 8

(11.2)

Equations (11.1) and (11.2) are the design equations of horn antenna. Again, from cos

RH R0 R = = 0 2 R0 + E lH ⇒ R0 + E =

R0

cos (R H /2 )

R0

⇒ E =

cos (R H /2 )

− R0

(11.3)

For a constant length R0 the directivity of the horn increases (i.e. beam width decreases) as the aperture A and flare angle qH increases. The largest flare angle for which path length d does not exceed a certain value, offers maximum directivity. The corresponding horn dimensions are termed optimum horn dimensions and they are expressed as E0 =

R0 cos (R H /2)

− R0

(11.4)

and R0 =

E 0 cos(R H /2)

1 − cos(R H /2)

(11.5)

If the aperture and flare angle become so large that d approaches to 180°, the field at the edges of the aperture is in opposite phase to the field on the axis. For very large horn angles the ratio (R0/lH) closes to unity that is the effect of the additional path length d on the distribution of the field magnitude is negligible. When d = 180°, phase reversal at the edges of the aperture decreases the directivity. Horn antenna with axial horn length, R0 = 10l and flare angle 25° provide the maximum directivity. Directivity The directivity of a horn antenna is expressed as D=

4Q Ae

M

=

4QF h Ap

M

(11.6)

where Ae and Ap are effective and physical aperture areas, eh is aperture efficiency and l is operating wavelength. Generally, aperture of the horn antenna is taken » 1.0l, and corresponding reported efficiency is 60%, therefore, Eq. (11.6) reduces to D=

7.5 Ap

M

⎛ 7.5 Ap ⎞ = 10 log ⎜ ⎟ dBi ⎝ M ⎠

(11.7)


389

Particularly, for a pyramidal horn, directivity D = 10 log (7.5aElaHl) dBi where aEl = E-plane aperture in l aHl = H-plane aperture in l FNBW The first null beam width for different types of horn apertures are given as 115

Rectangular aperture = Circular aperture =

(11.8a)

R0M

140

(11.8b)

DM

Optimum E-plane rectangular horn = Optimum H-plane rectangular horn =

115 aE M 172 aH M

(11.8c) (11.8d)

HPBW The half-power beam width for different types of horn apertures are given as Rectangular aperture =

Circular aperture =

51

(11.9a)

R0M

58

(11.9b)

DM

Optimum E-plane rectangular horn =

56 aE M

Optimum H-plane rectangular horn =

67 aH M

(11.9c) (11.9d)

where all the parameters have their meaning. The dimensions of a rectangular pyramidal horn for optimum gain are given as [4] A = 3M0 ah

and B = 3M0 ae

(11.10)

where ah (correspond to A) and ae (correspond to B) known as slant-lengths of the horns. The effective area is close to 50% of its aperture area, and its gain is given as

Gain (dB) = 8.1 + 10 log

AB

M02

(11.11)

390


For example, A horn of A = 9 inch and B = 4 inch, operating at 80 GHz will have a gain of 22.2 dB.

H-PLANE SECTORAL HORN ANTENNA As already mentioned, if the horn serves to flare only in the H-plane dimension, it is called H-plane sectoral horn antenna. Hence, it is fed from a rectangular waveguide of interior dimensions a and b (a > b). The geometrical configuration and cross-sectional view of the considered horn antenna are shown in Fig. 11.4. y

x b

a b

FIG. 11.4(a)

FIG. 11.4(b)

A

z

Geometry of H-plane sectoral horn antenna.

Cross-section of H-plane sectoral horn antenna.

In which lE, R, R1 and RH as well as qH are geometrical parameters and they are related as

⎡⎛ l ⎞ 1⎤ RH = (A − a) ⎢⎜ E ⎟ − ⎥ 4⎦ ⎣⎝ A ⎠

(11.12)

391


and

⎡ ⎛ x ⎞⎤ R = [ R12 + x 2 ]1/2 = R1 ⎢1 + ⎜ ⎟ ⎥ ⎝ R1 ⎠ ⎦ ⎣

2 ⎡ 1⎛ x ⎞ ⎤ = R2 ⎢1 + ⎜ ⎟ ⎥ 2 ⎝ R1 ⎠ ⎥ ⎣⎢ ⎦

1/2

Since x << R1, that holds only if A/2 << R1, then 1 ⎛ x2 ⎞ ⎜ ⎟ 2 ⎝⎜ R1 ⎠⎟

R − R1 ≈

(11.13)

Therefore, aperture phase-variation in the x direction can be given by ⎛ C ⎞ 2 ⎜ ⎟x

e − j C (R − R1 ) = e − j ⎝ 2R1 ⎠

(11.14)

and hence the aperture electric field distribution inside the aperture leads to ⎛ C ⎞ 2

Eay

⎛ Q x ⎞ − j ⎝⎜ 2 R1 ⎠⎟ x = E0 cos ⎜ ⎟e ⎝ A ⎠

and zero elsewhere. The aperture distribution phase error as a function of position (x) is therefore given by

E1 =

C 2 R1

x2

(11.15)

and (i) Since the maximum value of ‘x’ is (A/2), the maximum phase error will be

E 1max =

C

2

Q ⎛ A⎞ ⎜ ⎟ = 2R1 ⎝ 2 ⎠ 4M

⎛ A2 ⎜ ⎜ R1 ⎝

⎞ ⎟ ⎟ ⎠

(11.16)

(ii) The complete far-zone radiated electric field in (q, f) direction is

⎡⎛ C b ⎞ ⎤ sin ⎢⎜ ⎟ sin R sin G ⎥ ⎛Q R ⎞ e ⎛ 1 + cos R ⎞ ˆ ⎣⎝ 2 ⎠ ⎦ (11.17) ˆ E = j C E0 b ⎜ 1 ⎟ ⎜ ⎟ (R sin G + G sin R ) b C 4 2 r C C ⎡ ⎤ ⎝ ⎠ ⎝ ⎠ ⎢ 2 sin R sin G ⎥ ⎣ ⎦ − jC r

where r is far-field distance and I(q, f) is a constant of function (q, f) [5]. The complete radiation is rather cumbersome; let us examine the principal plane patterns. In the E-plane f = 90°, and the normalized radiated electric field is therefore given by ⎡ ⎢ sin ⎛ 1 + cos R ⎞ ⎢ FE (R ) = ⎜ ⎟⎢ 2 ⎝ ⎠⎢ ⎢⎣

⎪⎧⎛ C b ⎞ ⎪⎫ ⎤ ⎨⎜ ⎟ sin R ⎬ ⎥ ⎪⎩⎝ 2 ⎠ ⎪⎭ ⎥ ⎥ Cb sin R ⎥ 2 ⎥⎦

(11.18)

392


The second factor is the pattern of a uniform line source of length b along the y-axis, as one would expect from the aperture distribution. Similarly, in the H-plane, f = 0°, and the normalized H-plane pattern is given by

FH (0) =

1 + cos R 2

fH (R )

(11.19)

in which f H (R ) ∝

∫

A/2 − A/2

⎛ Q x′ ⎞ − jC cos ⎜ ⎟e ⎝ A ⎠

′ R12 + x ′2 e j C sin R x

dx ′

(11.20)

where x¢ is value of x, along (+A) direction. (iii) The H-plane beam width for an optimum H-plane sectoral horn, for A >> l is given by

HPBW ≅ 1.36

M A

= 78°

M

(11.21)

A

where l is operating wavelength. In the H-plane sectoral horn antenna, as the phase error increases, the beam width and number of side lobes increases.

E-PLANE SECTORAL HORN ANTENNA E-plane sectoral horn antenna is formed by flaring a rectangular waveguide in the E-plane. Figure 11.5 illustrates the geometry and cross-sectional view of the E-plane sectoral horn antenna. y x B

a b

z

a

FIG. 11.5(a)

Geometry of E-plane sectoral horn antenna.


FIG. 11.5(b)

393

Cross-sectional view of E-plane sectoral horn antenna.

The geometrical parameters are related as follows: ⎡ ⎛ l ⎞2 1⎤ RE = (B − b) ⎢⎜ E ⎟ − ⎥ 4⎥ ⎢⎣⎝ B ⎠ ⎦

and

1/2

(11.22a)

2 ⎡ ⎛B⎞ ⎤ lE = ⎢ R22 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦

(11.22b)

⎛ B ⎞ R E = 2 tan −1 ⎜ ⎟ ⎝ 2 R2 ⎠

(11.22c)

With the similar procedure like H-plane sectoral horn antenna, the parameters of the E-plane horn antenna can be expressed as ⎛ C ⎞ 2 the aperture phase error E 2 = ⎜ ⎟y ⎝ 2 R2 ⎠

and maximum phase error E 2max =

Q 4M

⎛ B2 ⎜ ⎜ R2 ⎝

(11.23) ⎞ ⎟ ⎟ ⎠

(11.24)

The aperture electric field distribution for E-plane horn ⎛ C ⎞ 2

Eay

⎛ Q x ⎞ − j ⎜⎝ 2 R2 ⎟⎠ y = E0 cos ⎜ ⎟e ⎝ B ⎠

(11.25)

394


The normalized H-plane pattern follows with f = 0° is found from

⎛ 1 + cos R ⎞ FH (R ) = ⎜ ⎟ 2 ⎝ ⎠

⎡⎛ C a ⎞ ⎤ cos ⎢⎜ ⎟ sin R ⎥ ⎣⎝ 2 ⎠ ⎦ ⎡⎛ C a ⎞ ⎤ 1 − ⎢⎜ ⎟ sin R ⎥ 2 ⎠ ⎣⎝ ⎦

2

(11.26)

The second factor in this expression is the pattern of a uniform phase cosine amplitude tapered line source of length a. Similarly, the E-plane pattern follows with f = 90° can be expressed as ⎡1 + cos R ⎤ FE (R ) = ⎢ ⎥ f E (R ) 2 ⎣ ⎦

where

f E (R ) ∝

∫

B/2 − B/2

e

′ R22 + y′2 e j C sin R y

− jC

(11.27)

dy′

(11.28)

where y¢ is value of y along (+B) direction. The half-power beam width relationship for the optimum horns is expressed as

HPE = 0.94

M B

= 54°

M B

(11.29)

The optimum value of A and B are given by A=

3M R1 and B =

2M R2

PYRAMIDAL HORN ANTENNA The pyramidal horn (E–H plane horn) antenna is a most popular form of rectangular waveguide (Fig. 11.6). As it is flared in both E and H planes, it leads to narrow beam widths in both principal planes, forming a pencil beam. The principal plane patterns are the same as those obtained from the previous sectoral horn calculations. The E- and H-plane patterns of the pyramidal horn equal the E-plane and H-plane patterns of E- and H-planes sectoral horn antennas respectively. The gain of pyramidal horn antenna is defined by G=

4Q

M2

where ep = Aperture efficiency. » 5% for optimum pyramidal horn

F p AB

(11.30)

Horn and Cone Antennas y

395

x

x lH B

a

z

b

R1

a

A

FIG. 11.6(a)

A

z

RH

Overall geometry of pyramidal horn antenna.

FIG. 11.6(b)

Cross-sectional view of H-plane pyramidal horn antenna. x

lE

b

R2

B

z

RE

FIG. 11.6(c)

Cross-sectional view of E-plane pyramidal horn antenna.

Therefore, the gain of an optimum gain pyramidal horn is G = 0.51

4Q

M2

F p AB

(11.31)

Many applications for horns require a specified gain to be realized at a known operating frequency. Generally, the optimum gain design approach is used, because it renders the shortest axial length for the specified gain. The directivity of pyramidal horn can be estimated by using

DdB =

26000 HPE° HPH°

(11.32)

The aperture electric field is obtained by combining the results for H-plane and E-plane sectoral horns, i.e.

⎛Q x ⎞ Eay = E0 cos ⎜ ⎟ exp ⎝ A ⎠

⎧⎪ ⎡ C ⎛ x 2 y2 ⎞ ⎤ ⎫⎪ + ⎟⎥ ⎬ ⎨− j ⎢ ⎜⎜ R2 ⎟⎠ ⎥⎦ ⎪ ⎪⎩ ⎢⎣ 2 ⎝ R1 ⎭

(11.33)

396


To specify the required gain G at the operating wavelength l and specify the connecting waveguide dimensions a and b, the design parameters of the antenna can be found as: A = 0.45 M G B=

R1 =

1 4Q

×

(11.a)

GM 2 0.51 A

A2

(11.c)

3M

RH = R1

( A − a) A

a⎞ ⎛ = R1 ⎜ 1 − ⎟ A⎠ ⎝

b⎞ ⎛ RE = R2 ⎜ 1 − ⎟ B⎠ ⎝

2 ⎡ ⎛B⎞ ⎤ lE = ⎢ R22 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦

R2 =

B2 2M

(11.d)

(11.e)

2 ⎡ ⎛ A⎞ ⎤ lH = ⎢ R12 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦

and

(11.b)

1/2

(11.f)

1/2

(11.g)

(11.h)

Using above design expressions, the pyramidal horn antenna can be designed for any operating frequency. All the dimensions must be taken in centimetres. The variation of aperture efficiency and directivity for the standard gain horn antennas in frequency range 8.2 to 12.4 GHz is shown in Fig. 11.7, in which solid line shows directivity and dotted line shows efficiency variation. It is found that directivity increases with frequency while efficiency decreases with frequency due to increasing phase errors. Horn antennas operate satisfactorily over a bandwidth of about 8 to 14 GHz, but the performance is optimum only at the design frequency. Directivity of antenna varies from 21.3 to 23 dB, whereas corresponding efficiency lies between 30 and 55%.

Polarization of Pyramidal Horn In this section the polarization properties of pyramidal horn fed by a rectangular waveguide (in TE10 mode) with a y-directed electric field has been considered. In contrast to radiation intensity, the polarization ratio of the pyramidal horn is simply defined as

Horn and Cone Antennas 23.5

0.60

23.0

0.55 D 0.50

22.5

0.45 22.0

Ep 0.40

21.5

21.0

Aperture efficiency (Ep)

Directivity (D), dB

397

0.35 0.30 8

9

10

11

12

13

Frequency (f ), GHz

FIG. 11.7

Directivity and efficiency of the standard gain rectangular horn.

Ph = −

ER EG

= − tan G

(11.34)

where Eq and Ef are the far-fields of pyramidal horn [6] and they are related to fÿ as follows: Eq = K sin f

(11.35a)

Ef = K cos f

(11.35b)

where K is a constant and function of r and q. From the expressions of E-plane and H-plane sectoral horn antennas (Eqs. 11.17 and 11.19) and (Eqs. 11.26 and 11.27), it is clear that they are quite different from the fields of the pyramidal horn and yet have the same polarization ratio. Note that the fields are everywhere linearly polarized, but not in the same direction. Since the horn antennas are intended to produce a y-polarized linear field in the main beam, it is appropriate to use a dipole antenna as a test receiving antenna. Since the fields of y-directed dipole is given as [7] Er = 0 ER =

and

EG =

(11.36a)

− jXN I 4Q r

− jXN I 4Q r

l cosR sin G e− jkr

(11.36b)

l cos G e− jkr

(11.36c)

398


Therefore, the polarization ratio Pd =

ER EG

= cos R tan G

(11.37)

If the same degree of impedance matching for an antenna is maintained as we vary its polarization properties, then the ratio of actual power received to the power received under the most favourable circumstances of matched polarization is defined as polarization matched factor. It is also called the polarization efficiency (r) and its value lies between 0 £ r £ 1. In general, the polarization match factor shows how well a receiving antenna of effective length h is matched in with the polarization of incoming waves. In this particular case, the polarization matched factor between considered dipole and any one of the horn antenna can be expressed as

S=

(1 + cos R tan 2G )2 (1 + tan 2G ) (1 + cos2R tan 2G )

(11.38)

In principal E-plane, f = p/2 and in the principal H-plane f = 0, therefore from Eq. (11.38), r = 1. By differentiating Eq. (11.36) with respect to tan2q, it is found that the largest rate of change of r with q occurs near the direction q = 0, same as the waveguide opening into a plane, where tan2fÿ = 1, i.e., f = p/4, Hence, under this condition, Eq. (11.38) reduces to

S=

(1 + cos R )2 2(1 + cos2R )

as f = p/4

(11.39)

Typically, pyramidal horn antennas have E- and H-plane beam-widths much smaller than the minimum polarization beam width. That is why in many situations we neglect polarization beam-width in the main beam. It is a different matter for horns other than pyramidal, however, the E-plane sectoral horn generally has a large H-plane radiation intensity beam width and the H-plane sectoral horn has a large E-plane radiation intensity beam width. Therefore, it is compulsorily required to consider polarization effects in the main beam of the E- and H-plane sectoral horns. Circular horn fed by a circular waveguide For an optimum-gain, circular horn’s diameter should be: D = (3lc l0)1/2

(11.40)

If so, the gain of circular horn antenna shown in Fig. 11.8(a), is given by ⎛QD⎞ Gain (dB) = 20 log ⎜ ⎟ − 2.82 ⎝ M0 ⎠

(11.41)

Here, lc is the aperture-length of the circular horn antenna as illustrated in Fig. 11.8(b).


399

lc

D

D

lc

(a) Basic configuration

FIG. 11.8

(b) Cross-sectional view

Circular wave-guide fed circular horn antenna.

BI-CONICAL ANTENNAS Basically, there are two types of bi-conical antenna: (i) Finite bi-conical antenna (ii) Infinite bi-conical antenna The various types of wire antennas such as dipole, folded dipole, V antenna and other similar antennas have already been discussed. The dipole antenna is a large energy storage and narrow bandwidth, high Q antenna. The narrow band width is due to rapid changes of input impedance with frequency. However, V antennas have totally different characteristics; they store small energy and offer wide bandwidth. Geometrical configurations of V, bi-conical V and curved bi-conical V antennas are shown in Fig. 11.9.

FIG. 11.9

Various types of bi-conical antennas.

Bi-conical antenna is very similar to V antenna. The main characteristics of bi-conical antennas may be listed as follows: (i) They are like Tx lines. (ii) Most of energy radiated, very little energy reflected.

400 (iii) (iv) (v) (vi) (vii)


Non-resonant and low-Q antenna. Unidirectional antenna. Input impedance remains constant over wide frequency range. Well impedance matched antenna. The VSWR found less than 2 over this bandwidth.

Like helical antenna, the constant impedance curved bi-conical V antenna is a travelling wave antenna. Typically, the VSWR £ 1.5 over a 2 to 1 bandwidth and this behaviour remains same, even in an array, because of the small mutual coupling of the elements. An infinite bi-conical antenna is similar to a uniform Tx line and acts a guiding structure for a traveling outgoing spherical waves. Figure 11.10 illustrates an infinite biconical antenna and an analogous infinite uniform Tx line. The characteristic impedance Zc of both the structures are constant and also equal to input impedance Zi (i.e., Zc = Zi), because they are of infinite length. In particular, the input impedance of infinite biconical antenna is purely resistive (i.e., Zi = Ri) and defined as Z i = Ri = 120 ln[cot (R h /2)]

(11.42)

where qh = cone angle.

FIG. 11.10

(a) Infinite bi-conical antenna and (b) analogous infinite uniform Tx line.

Radiation Pattern of Bi-conical Antennas In order to describe the radiation pattern of a bi-conical antenna, let us refer to Fig. 11.11. As we see, the structure is infinite, it can be treated as a Tx line. When a time varying voltage is applied across, the gap currents flow radially out from the gap along the surface of the conductors, as a result encirculating magnetic field Hq is generated. If TEM mode propagation (all the fields transverse to direction of propagation), takes place the electric


FIG. 11.11

401

Field and current components of an infinite bi-conical antenna.

field would be perpendicular to the magnetic field and be q-directed. Therefore, this field can be expressed G H = HG Gˆ G E = ER Rˆ G J =0

and in the region between cone From the first Maxwell’s equation, i.e.,

where

∇ × H = F D + J

(11.43)

∂Ee jX t D = F E = F = jXF E ∂t

(11.44)

If (Er, Eq, Ef) and (Hr, Hq, Hf) are the components of E and H along r, q and fÿ directions, then from Eq. (11.43) r-component is reduced to

1

∂

r sin R ∂R

(sin R HG ) = jXF Er = 0 ⇒

(sin q Hf) = constant

HG ∝

or

1 sin R

∂ ∂R HG =

(sin R HG ) = 0

(11.45)

K sin R (11.46)

402


Since the structure acts as a guide for spherical wave propagation, we can re-write Eq. (11.46) as e− j C r

HG = HR

1

(11.47)

4 Q r sin R

Again from Eq. (11.43), the q-components can be expressed as

1 ∂

(rHG ) = jXF ER

r ∂r

(11.48)

Substituting the value of Hf from Eq. (11.47), we get ER =

1 ⎛ e− j C r 1 rH ⎜ R ⎜ ∂ 4 sin jXF r r ⎝ Qr R −1

HR ⎛ C ⎞ =⎜ ⎟ × sin R ⎝ XF ⎠

⎞ −1 HR ⎟= ⎟ sin jXF r R ⎠

⎛ 1 e− j C r ⎜ ⎜ ⎝ ∂r 4 Q r

⎛ 1 e− j C r ⎞ ⎜ ⎟ = I HG ⎜ ∂r 4Q r ⎟ ⎝ ⎠

⎞ ⎟ ⎟ ⎠

(11.49)

where h = b/we. Equation (11.49) indicates that Eq = hHf, which is a basic condition for TEM mode propagation, that is, our consideration is confirmed. Since, field components vary as 1/sin q, Eq. (11.46), the radiation pattern can be obtained by using F (R ) =

ER ER h

=

sin R h sin R

, Rh < R < Q − Rh

which is normalized to unity at its maxima occurs on the surfaces of conductor (i.e. q ® qh). Input impedance As we know the input impedance (zin) is the ratio of the terminal voltage and current. Therefore, in order to find input impedance we must find the voltage and current first, so referring to Fig. 11.12(a), the voltage can be found by integrating the electric field along a constant radius r and it will be: V (r ) =

Q −R h

∫R

h

ER r dR

(11.50)

When the potentials on the top and bottom of cone are positive and negative respectively, the electric field lines extend from the top to the bottom cone as indicated by arrows in Fig. 11.12(b).


(a)

FIG. 11.12

403

(b)

(a) Finite bi-conical antenna; (b) Energy flow in finite bi-conical antenna.

Again from Eqs. (11.49) and (11.50)

I HR − j C r Vr = e 4Q

Q −R h

∫R

h

Q −R h

dR

I HR − j C r ⎡ R ⎤ = e ⎢ ln tan ⎥ sin R 4Q 2 ⎦R ⎣ h

or Vr =

I HR − j C r ⎛ R h ⎞ ln ⎜ cot e ⎟ 4Q 2 ⎠ ⎝

(11.51)

G G Since the boundary condition on Hf at the conducer surface is J x = HG , the total current on one cone is found by integrating the current density Jx around the cone, i.e. I (r ) =

∫

2Q 0

J x dSx =

∫

2Q 0

HG r sin dG = 2Q rHG sin R

⎛ e− jC r ⇒ I (r ) = (2Q r sin R ) ⎜ H 0 ⎜ 4Q r ⎝

1 ⎞ e− j C r H0 ⎟= sin R ⎟⎠ 2

(11.52)

Hence, from Eqs. (11.51) and (11.52), input impedance (Zin) is Z in =

V (r ) I (r )

=

I ⎛ Rh ⎞ Rh ⎞ ⎛ ln ⎜ cot ⎟ = 120 ln ⎜ cot ⎟ Q ⎝ 2 ⎠ 2 ⎠ ⎝

(11.53)

404


Since, Eq. (11.53) is not function of r, therefore it may be considered the values of Zr for any value of r between 0 and r. In other words, the input impedance is constant along the radial direction of bi-conical antenna for a fixed value of qh. In addition, there are special values of Zin. For angle qh less than 20° ⎛ 2 ⎞ Z in = Z 0 = 120 ln ⎜ ⎟ : ⎝ Rh ⎠

(11.54)

where qh is in radians. · if qh = 1°, Zin = (568 + j0) W · if qh = 50°, Zin = (91 + j0) W The input impedance is real, because there is only a pure travelling wave, in addition, the structure is infinite. And there are no discontinuities, which would generate the relative components in the impedance by causing reflections, hence setting up standing waves. The variation of input resistance (Ri) with cone angle indicates that Ri decreases as cone angle qh increases. At angle qh = 90° the input resistance Ri is around 100 W. In another case, when bi-conical antenna is modified and the lower portion is replaced by a large ground plane, the input resistance reduces to half the value. It is known as single cone antenna and its input resistance (Ri) at q = 90° is found to be about 50 W. However, bi-conical antenna is formed when maintained antenna is enclosed in hypothetical sphere. Again from a finite bi-conical antenna of radius r, shown in Fig. 11.12, where energy flowing near the cone is reflected but with energy escaping perpendicular to the axis in the equatorial region. That is when the radiating wave reaches at the circumference of shell, part of energy reflected back and stored in, and rest of energy radiated perpendicular to the axis than close to the cones. The input impedance (Zin) in this case is given by

Z in = Z c

⎛ 2Q r ⎞ Z c + Z m tan ⎜ ⎟ ⎝ M ⎠ ⎛ 2Q r ⎞ Z m + Z c tan ⎜ ⎟ ⎝ M ⎠

(11.55)

where r = radius/cone-length (m) l = operating wavelength Zc = 120 ln cot (qh/4) and Zm = Rm + jXm where the values of Rm and Xm are given in [8]. The measurement result shows that the lowest VSWR over a given bandwidth can be obtained with the largest cone angle. Typically, for cone angle, < 20° to 60° <, the VSWR values lie between 2 and 5. Both uniform Tx line and bi-conical antenna have similar current distributions, decreasing from left to right. The virtual resistances of the bi-conical antenna radiate energy whereas


405

the real resistance of the uniform Tx line absorbs energy and results in heat loss. The coaxial feed is found very suitable for bi-conical antenna, and makes these antennas convenient for mounting on masts. Maximum radiation is appeared in the horizontal plane inside an imaginary sphere of radius h (say) just enclosing the antenna; TEM waves exist together with higher-order modes created at the ends of the cones. These higher-modes are major causes of the antenna reactance, because the ends of the cones cause reflections that set up standing waves which lead to a complex input impedance. The increases of Xe (Zin) can be held minimum over a progressively wider bandwidth by increasing qh, however at the same time, Re(Zin) becomes more sensitive to frequency variations. A modified flat-version of the bi-conical antenna is bow-tie antenna. The bow-tie antenna of 60° provides a VSWR < 2 over a 2 to 1 bandwidth for the length L equal to 0.8l at the centre frequency. Total radiated power The total radiated power by a bi-conical antenna is given by

⎛ R ⎞ ⎪⎫ ⎪⎧ PT = 2 QI |H 0 |2 ln ⎨cot ⎜ ⎟ ⎬ ⎪⎩ ⎝ 4 ⎠ ⎪⎭ where H0 is magnetic field intensity at the centre of antenna.

BROADBAND SLOTTED CONE ANTENNA The present section describes a new type of broadband antenna, which was first proposed by Wanselow et al. [9]. The antenna is named as slotted cone antenna and has the following properties: 1. Light weight, compact profile, and UHF radiator. 2. Nearly half smaller than the conical and discone antennas. 3. Radiates primarily vertical polarization and like a vertical monopole yields an omnidirectional pattern in the horizontal plane. The slotted cone antenna is nothing but an altered form of the cone antenna. It consists of an inverted cone with a small longitudinal or axial slot cut in one side (Fig. 11.13). Basically, it is electric hybrid radiating structure and appears like a shunt fed, top loaded, wide plate (low Z0) Tx line antenna. In its lower mode of operation, the vertical conecylindrical feed is utilized to excite the structure. An approximate antenna circuit for the cone antenna in its lower mode of operation is shown in Fig. 11.14, where d1 and d2 represent distances and h1 and h2 are heights respectively. Theoretical model In order to describe the slotted cone antenna, let us consider approximate circuit diagrams shown in Fig. 11.14. Since the sketch depicts the antennas with one cone unwrapped, it is

406


d2

Edge view of radiating end plate

h2

d1 h1 Feed

FIG. 11.13

Slotted cone antenna model.

FIG. 11.14

Approximate antenna circuit.

better to consider this radiating structure as a Tx line loaded antenna. If so, at its lowest resonance frequency, the antenna may be considered as a combination of a large number of quarter wave elements with increasing lengths from (h1 + d1) to (h2 + d2). As a result, solid surface of increasing height and length (slopping wide plate) is formed, which leads to consider only the approximate behaviour of impedance near its first resonance. Since circuit concepts have been considered, energy and power relations need to be considered for the complete theoretical analysis. Cut-off frequency and bandwidth The low cut-off frequency (fc) of the slotted cone antenna can be obtained in terms of quarter wave resonance frequencies of the structure. If f1 and f2 are the resonance frequencies of the Tx line of lengths (h1 + d1) to (h2 + d2) respectively, then the low cut-off frequency (fc) will be equal to the geometric mean of ratio of f1 and f2, i.e.,

fc =

f1

(11.56)

f2

where f1 and f2 are defined as f1 =

f2 =

c 4(h1 + d1 )

F re

c 4(h2 + d 2 )

F re

(11.57a)

(11.57b)

where ere is effective relative dielectric of the antenna material and c is velocity of light in free space. It has been determined that at cone angle of » 60°, the antenna performance is


407

satisfactory. At low cut-off frequency the antenna height (h2) is found to be about (1/8l) at cone angle of 60°. At much lower and higher cone angles, antenna performances deteriorate and become similar to those of a resonant monopole antenna. The impedance characteristic of the slotted cone antenna shows that it acts like a high-pass filter. A 50 W impedance matching is achieved when the cylinder feed is transformed through a 45° cone to the input connector [10].

VSWR, HPBW and Polarization A typical VSWR characteristic of the SCA over a 7 to 1 bandwidth is shown in Fig. 11.15. It reveals that the VSWR values lie between 1.2 and 2, exception to first and second octaves where the mismatch rises to about 2.5 to 1, which leads to impedance matching in the antenna structure.

FIG. 11.15

Variation of VSWR with frequency slotted cone antenna.

The variation of the omni-directional HPBW as a function of frequency is shown in Fig. 11.16 which shows that the beam-width decreases as frequency increases. Up to 4 octave frequency, however a further increase is noticed between four and five octave frequencies. The slotted cone antenna is a vertically polarized antenna, though the amount of polarization is small. A considerable amount of cross-polarization (horizontal polarization) also found at frequencies near the upper edge of the second octave and above. For further details of slotted cone antenna, the readers may refer [9] and [10].

408


Half-power elevation beam width (degrees)

70

60 50 40 30

20 10 0

fc

2fc

3fc

4fc

5fc

Frequency

FIG. 11.16

Variation of 3-dB elevation beam width as a function frequency.

ADVANTAGES AND APPLICATIONS OF HORN ANTENNA The advantages of horn antenna are its simplicity in construction, easy method of feeding, larger gain as well as versatility. Horn antennas are fed by co-axial cable. The path-difference d lies between 0.20l and 0.45l. The optimum axial length is 10l, whereas flared angle varies from 15° to 40°. The directivity of pyramidal/conical antenna is more as they have more than one flared angles. It is very suitable for high frequency application with minimum loss of power. The most conveniently horn antennas are used as feed for reflector/dishes and lens antennas, which are very useful for large radio-astronomy, satellite tracking and mobile communication. It is also used as reference antenna for calibration and gain measurement of test antennas. The bi-conical antenna represents one of the canonical problems in the antenna theory, and its model is well suited for examining usual characteristics of dipole type antennas. A bi-conical antenna acts as guide for spherical waves like uniform Tx line acts as guide for the plane waves. This is because characteristic impedance of biconical antenna is nearly same as that of Tx line. VSWR less than 2 can be easily achieved over frequency band provided cone slant length selected between (l/4) and l. Balance feeder of impedance about 150 W is required to feed the biconical antenna of cone angle up to 60°. The half of cone angle is selected between 30° and 60°. The weight of biconical antenna can be maintained lower by constructing it in cage form. The radiation resistance of biconical antennas is found to be approximately in between 100 and 570 W for cone angle 2° to 50°. The bandwidth of biconical antenna increases as cone angle increases, whereas VSWR has inverse characteristics. The VSWR for larger cone angles over 2 to 1 bandwidth are found,


S.No.

Cone angle (deg)

VSWR

1 2 3

20 40 60

< 5 < 3 < 2

409

Bi-conical antennas are very useful for point-to-point communications. They are used in transmitter and receiver especially at airport to maintain the communication with aeroplanes coming from any directions.

SOLVED EXAMPLES Example 11.1 A pyramidal horn antenna of mouth length 10l cm is fed by a rectangular waveguide in TE10 mode. Determine the design parameters of the antenna at operating frequency 2.5 GHz.

M=

Solution:

3 × 1010 2.5 × 10

9

=

30 2.5

= 12 cm

As we know that the typical values of d in E- and H-planes are 0.20l and 0.35l. Therefore R0 =

A2 8E

=

(10 M )2

100 M

8 × 0.2 M

1.6

⎛ A ⎞ −1 R E = tan −1 ⎜ ⎟ = tan R 2 ⎝ 1⎠

= 62.5 × 12 = 7.50 m

⎛ 10 M ⎞ –1 ⎜ ⎟ = tan (0.08) = 4.6° × M 2 62.5 ⎝ ⎠

62.5 ⎛R ⎞ ⎛ ⎞ R H = cos−1 ⎜ 1 ⎟ = cos−1 ⎜ ⎟ = 6.3° ⎝ 62.5 + 0.375 ⎠ ⎝ lH ⎠

lH = 2R1 tan qE = 2 ´ 62.5 l ´ tan (6.3) = 13.8 ´ 12 = 1.66 m Therefore, R1 = 7.5 m

and

lH = 1.66 m

qE = 4.6° and qH = 6.3°

Ans.

Example 11.2 A rectangular pyramidal horn has slant (aperture) lengths, aEl = 1.2 and aHl = 1.1 per wavelength. Calculate directivity, dimension of horn and gain, if operating frequency is 3 GHz. Solution:

M=

c f

=

3 × 1010 3 × 10 9

= 10 cm

410


D = 10 log (7.5 aEl ´ aHl) dB

Directivity

= 10 log (1.2 ´ 1.1 ´ 7.5) = 9.95 dB A = 3 × 10 × 1.2 = 6 cm

B=

2 × 10 × 1.1 = 4.6 cm

AB = 6.0 ´ 4.6 = 2.8.14 cm2

⎛ AB ⎞ G(dB) = 8.1 + 10 log ⎜⎜ 2 ⎟⎟ ⎝ M0 ⎠ ⎛ 28.14 ⎞ = 8.1 + 10 log ⎜ ⎟ = 2.6 dB ⎝ 100 ⎠

We also know that G = h D

I=

G D

2.6 dB

=

9.95 dB

= –7.35 dB

h = 18%, i.e., poor radiation.

or

Example 11.3 Calculate the maximum phase error and HPBW of H-plane and E-plane sectoral horn antennas of length 10 l, operating at lÿ = 2 cm. Solution:

We know that for H-plane and E-plane sectoral horn antennas

E1 m = ⎜

2 ⎛ Q ⎞ A ⎟ ⎝ 4 M ⎠ R1

and

HPBW = 78°

2 ⎛ Q ⎞B E 2m = ⎜ ⎟ ⎝ 4 M ⎠ R2

and

HPBW = 54°

M A

M B

Given R0 = 10l, Hence, A = 3 × 2 × 10 × 2 = 11.95 cm B = 3 × 2 × 10 × 2 = 8.944 cm

\

E 1m = E 2m =

3.14 4 × 2 3.14 4 × 2

× ×

(11.95)2 10 × 2 (8.944)2 10 × 2

=

=

448.39 160.00 251.18 160.00

= 2.80 cm

= 1.57 cm


2

(HPBW) H = 78° ×

11.95 2

(HPBW) E = 54° ×

8.944

411

= 13.05° = 12.075°

Ans.

Example 11.4 Obtain input impedance Zb and pattern factor of infinite bi-conical antenna, of cone angle 30°. Assume rotation angles 10, 50 and 80°. Also find change in value of (Zb), in case cone angle is reduced to 15°. Solution:

Rh ⎞ ⎛ We know that Z b = 120 ln ⎜ cot ⎟ 2 ⎠ ⎝ = 120 ln (cot 15°) = 292.35 W

In case qh reduces to 15°. ⎛ 2 ⎞ ⎛ 2 × 180° ⎞ Z b′ = 120 ln ⎜ ⎟ = 120 ln ⎜ ⎟ 15 ⎝ ⎠ ⎝ Rh ⎠

= 120 ln (24) = 381.36 DZ = Z¢b – Zb = 89.02 W The pattern factor is given F (R ) =

sin R

sin 30°

F (10°) =

sin 10°

F (50°) = and

sin R h

F (80°) =

sin 30° sin 50° sin 30° sin 80°

= = =

0.5 0.174 0.5 0.760 0.5 0.988

= 4.584 dB = − 1.852 dB = − 2.943 dB Ans.

Example 11.5 Determine polarization ratio of (i) pyramidal horn, and (ii) dipole antenna along (q, f) = (30°, 60°) direction. Also find polarization matched factor in the same direction. Solution:

Given q = 30°

and

f = 60°

ph = – tan f = – tan 60° = –1.732 ⎛ 1 ⎞ (ph )dB = 10 log ⎜ ⎟ = 10 log (0.577) = − 2.385 dB ⎝ 1.732 ⎠

412


(pd)dB = 10 log (cos q tan f)

and

= 10 log (cos 30° ´ tan f) = 10 log (0.866 ´ 1.732) = 1.760 dB Polarization material factor P =

=

(1 + cos R tan 2 G )2 (1 + tan 2 G )(1 + cos R tan G )2

(1 + cos 30° tan 2 60°)2 (1 + tan 2 60°) (1 + cos2R tan 2G ) 2

⎡1 + 0.866 × (1.732)2 ⎤ ⎣ ⎦ = ⎡(1 + 1.7322 )(1 + 0.8662 × 1.7322 ) ⎤ ⎣ ⎦

=

12.9 3.999 × 3.25

=

12.9 12.998

= 0.9924

= – 0.033 dB Example 11.6 Find the optimum directivity and gain of a circular wave horn antenna of diameter 1.25 m and aperture length 1.8 m, at operating wavelength 20 cm. Solution:

We know that D = (3lcl0)1/2 = (3 ´ 1.8 ´ 0.20)1/2 rad = 1.03 rad = 60° ⎛QD⎞ gain = 20 log ⎜ ⎟ − 2.82 ⎝ M0 ⎠

and

⎛ 3.14 × 1.03 ⎞ = 20 log ⎜ ⎟ − 2.82 = 21.35 dB 0.20 ⎝ ⎠

Example 11.7 Design a rectangular pyramidal horn antenna to be operated at l0 = 20 cm with required gain of 22 dB, if dimensions of the connecting waveguide are 25 cm and 20 cm. Also find the gain of this antenna. Solution:

Given l0 = 20 cm, gain = 22 dB = 160 numbers

Then flared aperture dimension A = 0.45 l G = 0.45 ´ 20 160 = 113.84 cm


B=

GM 2

1

160 × 20 × 20

=

4Q 0.51aH

413

4 × 3.14 × 0.51 × 113.84

64000

=

729.21

= 87.77 cm

Then the length of antenna is R1 =

A2 3M

=

(113.84)2 3 × 20

= 216.0 cm

a⎞ 25 ⎞ ⎛ ⎛ RH = R1 ⎜ 1 − ⎟ = 216 ⎜ 1 − ⎟ = 168.57 cm A⎠ 113.84 ⎠ ⎝ ⎝ R2 =

B2 2M

=

(87.77)2 2 × 20

= 192.58 cm

b⎞ 20 ⎞ ⎛ ⎛ RE = R2 ⎜ 1 − ⎟ = 192.58 ⎜ 1 − ⎟ = 148.69 cm B⎠ 87.77 ⎠ ⎝ ⎝ 1

Flared length

2 ⎡ ⎛ A⎞ ⎤2 2 ⎢ lH = R1 + ⎜ ⎟ ⎥ = [2162 + (56.82)2 ] 1/2 = 223.35 cm ⎢⎣ ⎝ 2 ⎠ ⎥⎦ 1 2 1 ⎡ ⎛B⎞ ⎤2 2 2 2 2 lE = ⎢ R2 + ⎜ ⎟ ⎥ = [(192.58) + (43.66) ] = 197.47 cm ⎢⎣ ⎝ 2 ⎠ ⎥⎦

and gain = 8.1 + 10 log

AB

M

2 0

⎡113.84 × 87.44 ⎤ = 8.1 + 10 log ⎢ ⎥ 20 × 20 ⎣ ⎦

= 8.1 + 10 log (24.885) = 22.05 dB 4Q ⎛ ⎞ Gain can also be calculated GdB = 10 log ⎜ 0.51 2 AB ⎟ = 22.025 dB M ⎝ ⎠

Example 11.8

Comment on the “radiation mechanism of horn antenna”.

Solution: In a waveguide, in general, only small portion of incident waves are transformed into radiation and remaining part reflected back by other end (open circuit). This is because of poor matching between open discontinuity and free space. As a result, because of diffraction

414


at the edge there is non-directional radiation of low intensity (i.e., poor radiation). However, if these abrupt discontinuities are replaced by a gradual transmission all the incident waves in forward direction could be radiated. This is achieved by terminating/flaring open end of waveguide with proper impedance matching. Finally, the diffractions at the open end are reduced resulting in increased directivity and lower beam width radiation. Example 11.9 Calculate the terminal impedance of a conical antenna of 2° total angle, which operates against a very large ground plane. Solution:

We know that

⎧⎪ ⎛ R ⎞ ⎫⎪ Z = 120 ln ⎨cot ⎜ ⎟ ⎬ ⎝ 4 ⎠ ⎭⎪ ⎩⎪ However, when q is very small (i.e. q < 2°) ⎛R ⎞ 4 cot ⎜ ⎟ = ⎝4⎠ R

Z = 120 ln (4/q)

Hence,

= 120 ln [4/2] = 83.18 Example 11.10 A 250 W twin-lead uniform Tx line is connected to a bi-conical antenna. Find (a) the cone angle that is enough for impedance matching. (b) What is VSWR, if the length of antenna arm is such that it offers input resistance of 25 W? Solution:

We know that ⎧R ⎫ ⎧R ⎫ Z = 120 ln cot ⎨ ⎬ ⇒ 250 = 120 ln cot ⎨ ⎬ ⎩4⎭ ⎩4⎭ ⎧R ⎫ ⎧R ⎫ cot ⎨ ⎬ = 8.031 ⇒ ⎨ ⎬ = 7.097 ⎩4⎭ ⎩4⎭

q = 28.4°

or Reflection coefficient (* ) =

R − Z in R + Z in

=

VSWR =

25 − 250 25 + 250

1 + 0.846 1 − 0.846

= 0.818

= 11.98


415

Example 11.11 Find total radiated power by a bi-conical antenna with cone angle 45°, if the magnetic field intensity at the centre of antenna is 2 mA/m. Solution:

⎛ R ⎞ ⎪⎫ ⎪⎧ PT = 2QI | H0 |2 ln ⎨cot ⎜ ⎟ ⎬ ⎝ 4 ⎠ ⎪⎭ ⎩⎪

⎧⎪ = 2Q × 377 × | H0 |2 ln ⎨cot ⎩⎪

⎛ R ⎞ ⎫⎪ ⎜ ⎟⎬ ⎝ 4 ⎠ ⎭⎪

= 2 ´ 3.14 ´ 4 ´ 10–4 ´ 377 ln (5.027) = 4.057 ´ 10–5 377 = 1.529 ´ 10–2 watts

OBJECTIVE TYPE QUESTIONS 1. Directivity of horn antenna is high because (a) It produces a uniform phase-front with a larger aperture (b) It produces a uniform phase-front with a narrower aperture (c) It is made by heavy material (d) None of the above 2. Relation between axial-length of horn antenna and its aperture is (a) Axial-length (L) is inversely proportional to square of aperture area. (b) Axial-length (L) is proportional to square of aperture area. (c) Aperture area is proportional to the axial-length. (d) None of the above 3. Choose the most appropriate one: “The typical value of path-difference for the optimum performance of the horn antenna is (a) Plane horn, d = 0.25 (b) Conical horn, d = 0.32 (c) H-plane horn d = 0.40 (d) All of the above 4. What will be the largest flare angle for which, d = 0.25l and axial-length is 10l (a) 34° (b) 25.4° (c) 23° (d) 45° 5. Very large flare angle (q) results in the (a) Decreased directivity (b) Decreased gain (c) Increased directivity (d) Increased gain 6. The optimum path difference for E-plane horn antenna is (a) d £ 0.25l (b) d ³ 0.25l (c) d = 0.25l (d) None of these

416


7. The path difference for H-plane horn is even d ³ 0.25l, because (a) Fields become zero at the horn edge (b) Fields go infinite at the horn edge (c) Fields become constant at the horn edge (d) None of the above 8. What is approximate maximum power gain of an optimum horn antenna with a square aperture on a side. (a) 39 dBi (b) 29 dBi (c) 56 dBi (d) None of these 9. The required aperture area for an optimum reflector horn antenna operating at 2 GHz with 12 dBi gain. (b) 34.56 cm2 (a) 34.56 cm2 2 (c) 40.24 cm (d) 47.5 cm2 10. Tick the correct answer: (a) Flaring in the direction of electric field vector (E-plane) (b) Flaring in the direction of magnetic field vector (H-plane) (c) Flaring in the direction of both electric field vectors (E–H plane) (d) All are correct. 11. Horn antenna function is based on (a) The principle of equality of path (b) Huygens principle (c) Both (a) and (b) (d) The Maxwell principle 12. The region existing close to the mouth of horn antenna is (a) Far-field region (b) Near-field region (c) Quasi-static region (d) Electrostatic region 13. Very large flare-angle (q) results in (a) Increased beam width (b) Decreased directivity (c) Decreased gain (d) Both (a) and (b) 14. In practice flare angle varies between 15° and 40°, and corresponding value of beam width (a) 66° (b) 166° (c) 100° (d) 120° 15. The directivity of pyramidal/conical horn antennas is ________ the E-plane/H-plane horn antennas (a) More than (b) Less than (c) Equal to (d) None of these 16. Bandwidth of horn antenna is greater than that of the parabolic reflector antenna, because there are no involvements of resonant element. That is why its directivity is _________ to/than that of parabolic reflector. (a) More than (b) Less than (c) Equal to (d) None of these


417

17. For the standard horn antenna with good performance, the level of cross-polarization should be below co-polarization: (a) 30 dB (b) 50 dB (c) 10 dB (d) None of these 18. The bandwidth of bi-conical antenna can be increased by (a) Increasing its cone angle (b) Decreasing its cone angle (c) Increasing its arms lengths (d) Decreasing its cone angle 19. A single cone antenna with 90° angle offers an input resistance about (a) 50 W (b) 150 W (c) 100 W (d) 60 W 20. Select the correct statement: (a) V-antenna has bi-directional pattern and narrow bandwidth (b) V-antenna has uni-directional pattern and narrow bandwidth (c) V-antenna has uni-directional pattern and wide bandwidth (d) V-antenna has bi-directional pattern and wide bandwidth 21. Which is not a property of bi-conical V antenna (a) Constant impedance characteristics (b) Uni-directional pattern (c) Wide bandwidth (d) Large beam-width 22. Which is not a property of curved bi-conical V antenna (a) Constant impedance characteristics (b) Uni-directional pattern (c) Wide bandwidth (d) None of the above 23. What will be the directivity of a conical horn antenna of radius 2.3 cm at l = 1.2 m. (a) –20.53 dB (b) –34 dB (c) 50 dB (d) 67 dB 24. What is the required aperture area for an optimum rectangular horn operating at 3 GHz with 12 dBi directivity (b) 1.34 m2 (a) 0.0211 m2 2 (c) 0.050 m (d) 1.67 m2 25. What will be the required diameter of a conical horn antenna operating at 3 GHz with 14 dBi directivity (a) 22.18 cm (b) 30.85 cm (c) 56.00 cm (d) 6.34 cm

Answers 1. (a) 6. (a)

2. (b) 7. (a)

3. (d) 8. (b)

4. (b) 9. (d)

5. (a) 10. (d)

418


11. (c) 16. (b) 21. (d)

12. (b) 17. (a) 22. (d)

13. (d) 18. (a) 23. (a)

14. (a) 19. (a) 24. (a)

15. (a) 20. (a) 25. (a)

EXERCISES 1. Design a pyramidal horn antenna of axial length 12 cm to be operated at f = 3 GHz. Assume that the antenna is fed by a rectangular waveguide in its dominant mode. 2. Calculate the directivity and gain of a rectangular horn antenna of aperture lengths 1.5 and 1.8 per wavelength, if the antenna is to be operated at 2.5 GHz. 3. Obtain the input impedance (Zb) and pattern factor of an infinite bi-conical antenna of cone angle 20°, for rotation angles 10°, 30° and 50°. Also find change in value of (Zb), if the cone angle is increased twice. 4. Determine the polarization ratio and polarization matched factor of (i) pyramidal horn, and (ii) dipole antenna along (q, f) = (20°, 50°). 5. Design a rectangular pyramidal horn antenna with a gain 20 dB to be used in radar system at 20 GHz. Also find the gain of the antenna, if the dimensions of connecting apertures are 25 cm and 30 cm. 6. Calculate the length (L), aperture aH and half-angles in both the planes of a pyramidal EM horn for which the aperture aE = 9l. The horn is fed with rectangular TE10 mode waveguide. Assume dE = 0.12 l and dH = 0.45 l, also find (i) HPBW in both the planes, (ii) directivity (iii) aperture efficiency. 7. Show that the characteristic impedance (Z0) of a single cone and ground plane is half of that of a bi-conical antenna. 8. Find the physical aperture of a pyramidal horn antenna with ‘L’ = 10l, dE = 0.25l, dH = 0.45l. Assuming 98% efficiency, also find its directivity. 9. Obtain the terminal impedance of a conical antenna of total 1.8° operating against a very large ground plane. 10. What will be the required diameter of a conical horn antenna operating at 3.5 GHz with 15 dBi directivity? 11. What will be the directivity of an optimum rectangular horn operating at 3 GHz with sides 20l and 8l? 12. What is the required aperture area for a conical antenna at operating at 4 GHz with 14 dBi gain? 13. A 200 W twin-lead uniform Tx line is connected to a bi-conical antenna. (a) Obtain the cone angle that is enough for impedance matching. (b) What is VSWR, if the length of antenna arm is such that it offers input resistance of 30 W?


419

14. Total radiated power by a bi-conical antenna is 200 mW. Find the cone angle of the antenna, if the electric field intensity at the centre of antenna is 0.2 mV/m. 15. What will be the difference in directivity of conical and pyramidal horn antennas at f = 4 GHz, if the effective areas of the antennas are same? Also find the radius of conical horn if its power gain is 12 dB.

REFERENCES [1] Love, A.W., Electromagnetic Horn Antenna, IEEE Press, New York, 1996. [2] Prasad, K.D., Antenna and Wave-propagation, Satya Prakashan, New Delhi, 1996. [3] Rajeshwari, Ch., Antenna Theory and Practice, 2nd ed., New Age International Publication, New Delhi, 2004. [4] Johnson, R.C., Antenna Engineering Handbook, 3rd ed., McGraw-Hill, New York, 1993. [5] Stutsman, W.L. and G.A. Thiele, Antenna Theory and Design, 2nd ed., John Wiley & Sons, Inc., New York, 1981. [6] Balanis, C.A., Antenna Theory, Harper & Row, New York, 1982. [7] Mott, Harold, Polarization in Antennas and Radars, A Wiley-Inter Science Publication, New York, 1986. [8] Kraus, J.D., Antennas: For All Applications, Tata Mc-Graw Hill Publishing Company, New Delhi, 2003. [9] Wanselow, R.D. and D.W. Milligan, “Broadband slotted cone antenna,” IEEE, Trans. Antenna Propagate, Vol. AP-14, No. 2, pp. 179–182, March 1966. [10] Storer, J.E., “The impedance of an antenna over a large circular screen,” J. Appl., Phys., Vol. 22, pp. 1058–1066, Aug. 1951.

C H A P T E R

12

Helical Antennas

INTRODUCTION A helical antenna is a specialized antenna that transmits and receives electromagnetic fields with circular polarization. It is named so because copper wires are helically wound on a core made of an insulative material, thereby enabling the size of the antenna to be minimized. The performance of such a antenna depends on the shape of the terminals on which it is mounted and the matching circuit integrated between the antenna and feed. Helical antennas are designed for use with an unbalanced feed line such as coaxial cable. The centre conductor of the cable is connected to the helical element whereas its shield is connected to a reflector. Helical antennas are commonly connected together in so-called bays of two, four, or occasionally more elements with a common reflector. The entire assembly can be rotated in the horizontal (azimuth) and vertical (elevation) planes, so the antenna system can be aimed toward a particular satellite. In case the satellite is not in a geostationary orbit, the azimuth and elevation rotators can also be operated by a computerized robot which is programmed to follow the course of the satellite across the sky. In general a helical antenna appears as one or more springs/coils mounted against a flat reflecting screen. The reflector is a circular or square metal mesh or sheet whose cross dimension (diameter or edge) measures at least 3l/4. If the helix or reflector is too small (i.e. the frequency is too low) the efficiency of antenna is severely degraded. The input impedance of helical antenna was found to be essentially resistive and constant over a wide frequency range. The helical antenna is a circularly polarized and high directive antenna. The helical antenna or simply axial helix is only a small fraction of half wavelength in radius and acts as a waveguide. The helix behaves like a terminated (matched) transmission line at low frequency (for circumference equal to l/2) and flat wave propagation (VSWR– ¥, i.e. no reflected) takes place along the helix. However, when frequency increases, distribution of waves along the antenna changes dramatically [1–3]. The concept of helical antenna was first described by J.D. Kraus in 1947 in his article “The helical beam antenna”. Helical antennas are basically used to generate electromagnetic energy typically in the form 420

Helical Antennas

421

of radio waves, which are being used for, telephony and data communications at large scale. They are mainly used in mobile stations. The performances of the helical antenna greatly affect the performance of the mobile station. Helical antennas are also used in portable communication devices, such as cellular phones, because of their wideband characteristics and the advantage of requiring less space. A single turn helix is a simplest configuration of helical antenna. If the number of turns in the helix increases (say N), it may be considered as the end-fire array of N-sources. If one turn of the helix is uncoiled (Fig. 12.1), the various geometrical parameters of helix are related as follows: S = spacing between turns = C tan a a = pitch angle = tan–1 (S/C) L = length of one turn = C 2 + S 2 D = diameter of helix (centre to centre distance) C = circumference of helix = pD A = axial length = NL N = total number of turns h = height = axial length = NS and d = diameter of helix conductor

FIG. 12.1

Single turn helix and its parameters.

From the above relations, it is clear that when D = 0 and a = 90°, helix reduces to a linear wire antenna, on the other hand, when a = 0 and S = 0, the helix becomes a loop antenna. However various models of helical antennas are given in Table 12.1 [4].

Reflector Model of Helix The axial helix can be considered as a spring of N turns with a reflector. The size of the reflector R is equal to C and it can be a circle or a square. The wounding of helix decides circular polarization (CP) characteristics, which can be either RHCP or LHCP, however to have maximum transfer of energy, both ends of the link must have the same polarization.

422


TABLE 12.1

Various models of helical antennas and their applications

S.No. Antenna model

Specifications

Applications

1

Model 8357–800

Set of 4–24 inch long helices, frequency 1525– 1660.5 MHz, handle up to 70 W RF power.

In portable satellite communications

2

Model AR < 2 dB, handle up to 70 watts CW, 8613 Series frequency range of 1525–1660.5 MHz

3

Model 9610–800

8-turn helix, RHCP AR < 2 dB. 15-feet length VHF applications with a 28-inch pitch diameter. 100–160 MHz, 6.0–12.8 dBic

4

Model 9610–810

8-turn helix, LHCP AR < 2 dB. 15-feet length VHF applications with a 28-inch pitch diameter. 100–160 MHz, 6.0–12.8 dBic

5

Model 8617–810

Length 10 feet, handle up to 1 kW of RF power. Frequency 236–330 MHz, gain of 12 to 17 dBic.

In uplinks to UHF satellite applications

6

Model 8617–800

Frequency range 310–400 MHz and overall length of 7 feet.

UHF applications

7

Model 9339–800

Frequency range of 2.31–2.36 GHz, 6.5 dBic gain with a 72° beam width.

Radars

8

Model 9392–800

Frequency ranges 1.2–1.35 GHz. LHCP with a beam width of 60°, gain of 11 dBic and handle 100 W CW.

Aerodynamic radomes

9

Model 9171–810

Six-element helix array is fed from a microstrip In mobile satellite power divider, 3.65–6.425 GHz, gain of 15 dBic systems aboard with BW of 12° azimuth and 70° elevation. moving vehicles

In fixed-site L-band satellite applications

If circumference C of a turn is approximately one wavelength (l), and the distance (S) between the turn is approx. 0.25 C. The gain G of the antenna relative to an isotropic radiator (dBi) can be estimated by G = 11.8 + 10 log [Cl2 NS]

(12.1)

The characteristic impedance (Zh) of the resulting Tx line will be ⎛C ⎞ Z h = 140 ⎜ ⎟ (:) ⎝M⎠

(12.2)

Helical Antennas

423

PARAMETERS OF HELIX ANTENNA The most important parameter of a helix is its operating bandwidth, which is directly related with the behaviour of the other parameters: Gain, beam width, impedance and polarization. Gain is inversely proportional to half-power beam width, whereas other parameters are function of number of turns (N), pitch angle (a) and operating frequency (f). Helix antenna parameters are also depend on the ground size and shape, the helix conductor radius, feeding technique as well as supporting structures. The helix may be fed axially, or peripherally, and also at any convenient location on the ground-plane launching structure with the inner conductor of the co-axial line connected to the helix and the outer conductor bonded the ground plane. The terminal impedance of the antenna under different feeding situations is given by ⎛C ⎞ R = 140 ⎜ ⎟ , with axial feed ⎝M⎠

(12.3)

However with peripheral feed ⎛C ⎞ R = 150 ⎜ ⎟ ⎝M⎠

−1/2

(12.4)

These relations are true only, if

0.8 ≤

C

M

≤ 1.2 , 12° £ a £ 14° and number of turns (N) ³ 4

However, by inserting a suitable matching section, the terminal impedance (R) can be made any desired value between 50 W and 150 W.

TYPES OF HELIX ANTENNA The helix antenna generally operates in two modes: the axial mode and normal mode. The axial mode helix provides maximum radiation along the axis of the antenna and it occurs only when the helix circumference is on the order of a wave-length. In the normal mode, radiation is most intense in direction normal to the helix axis. The normal mode of operation occurs only when the helix diameter is small compared to a wavelength. Based on modes of operation, there are two types of antennas.

Axial Mode Helical Antenna It is one of the popular antennas to be used at UHF frequencies and hence widely used for satellite communications. It generally receives attention in the context of very broad band antennas. Axial mode helix radiates as an end-fire antenna with a major lobe along the axis

424


of the helix, and the radiation is close to circularly polarized near the axis. Larger number of turns narrow the main beam of the helix antenna. Typical helix is shown in Fig. 12.2. Specifically, a helix can be described as a monofilar (one-wire) axial-mode antenna, which is very non-critical and one of the easiest of all antennas built so for. The basic helix theory tends to restrict axial-mode operation of the helix to the pitch angles between 12° and 14° and corresponding circumference from 0.8l to 1.2l. However, the number of turns and the conductor size do not make serious effects on the helix performances. Axial mode antenna is most suitable when a moderate gain of up to about 15 dB and circular polarization radiation is required. Helices with less number of turns also perform satisfactorily over the frequency range corresponds to 3/4l £ C £ 4/3l, which gives a bandwidth ratio of BW =

f2 f1

=

C/M2 C/M1

=

4/3 3/4

= 1.78

(a)

(b)

FIG. 12.2 (a) Basic geometry of helical antenna; (b) Geometry and dimension of a left-handed wound helix.

Helical Antennas

425

This is approximately equal to conventional bandwidth ratio (2:1) for a wideband antenna [2]. However, for long helices the upper operating frequency is lower than (4l/3) reducing the bandwidth below 1.78. In the preceding section, an axial mode single winding (monofilar) helix antenna is described in detail emphasizing the design considerations. The parameters of monofilar axial mode-helix are expressed as HPBW =

52 CM NSM

(12.5)

115

FNBW =

CM NSM

(12.6)

If it is assumed that the patterns of both field components are of the same shape and are figures of revolution around the helix axis, the directivity of the antenna is defined as D = 12 NC2M SM

(12.7)

Above relations are true only for 0.8 < Cl < 1.15, 12° < aÿ < 14° and N > 3. The gain of the antenna is defined as G=

26000 HP

2

= 6.2 C M2 NSM

(12.8)

i.e., gain is proportional to f 3. The gain increases with number of turns N that is doubling N will increase gain by 3 dB, however experimental results show that this will be most accurate only for N = 10. It is also clear that gain is function of the circumference, however, it must be noted that helix is constructed to maintain C ~ l and gain is found for C ~ 1.1l [3]. Axial ratio (circular polarization) is given by

AR =

2N + 1 2N

(12.9)

That is quality of circular polarization increases with increasing the number of turns. Minimum value of axial ratio is found to £ 1.2 for 0.8 < Cl < 1.2 for corresponding N > 3. In principle, operation of helix is based on the assumptions that the helix carries pure travelling waves that travel outward from the feed. The electric field associated with these waves rotate in a circle, producing circular polarized radiation off the end of helix. That is why, the input impedance of the helix antenna is purely resistive. Radiation field As per the geometry of helix antenna, radiation pattern can be modelled using array theory, with each turn being an element of the array, and the applying principle of pattern multification (Fig. 12.3). If N is total number of turns, and S is spacing between the turns, hence for long helix (i.e., NSl > 1), the array pattern is much sharper than the single turn pattern and hence

426


FIG. 12.3

Array of point sources representing 1 turn of the helix.

collectively determines the shape of total far-field pattern of the antenna. Since the circumference is approximately equal to 1l, the current of opposite points on a turn is about 180° out of the phase. This cancels the current direction reversal introduced by the half-turn. Therefore, the radiation from opposite points on the helix is nearly in phase, leading to reinforcement along the axis in the far-field region. This radiation mechanism is closely equivalent to that of the one wave-length loop antenna. If current magnitudes on the turns are considered as uniform with progressive phase a, the total radiation pattern can be given by ET (R ) = K cos R

sin (NZ /2)

N sin (Z /2)

(12.10)

where K = normalized constant for a measure pattern cos q = individual element pattern ÿ ÿ ÿ y = bd cos q + a ÿ

If initially the helix is modelled as an ordinary end-fire array, a main beam maxima occurs in the q = 0 direction for y = 0, which gives

a = – bd

(12.11)

which is the phase delay due to axial-propagation corresponding to the distance d along the axis for one turn. However, the current wave follows the helix introducing another (–2p) phase shift. This is because the periphery of a single turn is about a wavelength. Thus, for the ordinary end-fire array of m elements the phase delay becomes

a = bd – 2p m,

(m = 0, 1, 2, 3, …)

(12.12)

which is condition for Hansen–Woodyard end-fire array; this is accounted for with an additional (–p/N) phase delay over the ordinary end-fire array. Thus the element to element phase shift is reduced to ⎛

B = − ⎜ C d + 2Q + ⎝

Q⎞ ⎟ N⎠

(12.13)

This phase-shift leads to a value for the phase-velocity of the travelling waves so the phaseshift of waves in one transit around a turn of length L can be expressed as

Helical Antennas

a = – bn L

427 (12.14)

where bn is a new phase constant associated with wave propagation along the helices. Therefore, from Eqs. (12.13) and (12.14), we get

Q⎞ ⎛ ⎜ C d + 2Q + ⎟ = C n L N⎠ ⎝ or

⎛

C n = ⎜ C d + 2Q + ⎝

Q ⎞1

⎟ N⎠L

(12.15)

Velocity factor This is a new constant defined as the ratio of phase velocity to the free space velocity of wave propagation and it is expressed as p=

v c

=

⇒ Cn =

X /c X /v

=

C0 Cn

C0

(12.16)

p

where v is the phase velocity of travelling waves along the helical conductor and c is freespace velocity. Hence the progressive phase-shift (a) in terms of p can also be expressed as

B=

L C0 p

=

2Q p

LM

(12.17)

Hence

Z

L ⎞ ⎛ = 2Q ⎜ dM cos R − M ⎟ p ⎠ ⎝

(12.18a)

and also from Eqs. (12.15) and (12.16), we get

p=

LM ⎛ 2N + 1 ⎞ dM + ⎜ ⎟ ⎝ 2N ⎠

(12.18b)

So, it is clear that the main beam maximum occurs for q = 0, that is, y = –2p – p/N.

Therefore,

Q⎞ ⎛ sin ⎜ − N Q − ⎟ K ( − 1) N +1 2⎠ ⎝ = ET (R = 0) = K Q ⎞ ⎛ ⎛ Q ⎞ N sin ⎜ − Q − N sin ⎜ ⎟ ⎟ 2N ⎠ ⎝ ⎝ 2N ⎠

(12.19)

428


Normalizing equation (12.19) in such that the maximum is unity, yields K = (–1)N+1 N sin(p/2N) and the final pattern factor reduces to ⎛ Q ⎞ FT (R ) = ( − 1) N +1 sin ⎜ ⎟ cos R ⎝ 2N ⎠

⎛ NZ ⎞ sin ⎜ ⎟ ⎝ 2 ⎠

(12.20)

⎛Z ⎞ sin ⎜ ⎟ ⎝2⎠

y = bd (cos q – 1) – 2p – p/N

in which

Equation (12.20) is useful to calculate both the components of E-fields, i.e., Eq and Ef.

Normal Mode Helical Antenna The word normal means the radiation perpendicular/right angle to helix axis. Therefore, N-helix is also described as broadside antenna, whereas A-helix antenna is an end-fire antenna. Like A-helix, N-helix also radiates circular polarization radiation. For normal mode of operation the circumference of helix must be smaller than the wavelength, whereas for A-helix mode, it must be approximately a wave length (i.e., D << l and L << l as well). Since helix is electrically small, the current is assumed to be constant in magnitude and phase over its length, which results its efficiency low. The far-field pattern of N-helix antenna is independent from number of turns involved and, therefore, it may be obtained by describing one turn. Since pitch angle (a) is one of most important design parameters of helix, it may be considered as special case of loop and dipole antennas as shown in Fig. 12.4. Because, when a = 0, helix becomes loop, whereas when aÿ = 90°, helix turned into linear antenna. Therefore, the far-field pattern of the N-helix antenna can be estimated with the help of far-field patterns of loop and dipole antennas. The far-field of small loop has only f component and given by EG = IC 2 AI

EG =

30 Q 2

M2

e− j C r 4Q r

sin RGˆ = 120 Q × 4Q 2Q D 2 [ I ]

(Q D 2 ) [I ]

e− jC r r

e− jC r 4M 2 4Q r

sin RGˆ

sin R Gˆ

(12.21)

Similarly, far-field of small dipole has only q-component and it is given by ER = jXN[ I ] S =j

2Q c

M

e− jC r 4Q r

N[ I ] S

sin R Rˆ

e− j C r 4Q r

sin R Rˆ

in which S is spacing between turns and it is the length of the small dipole.

(12.22)

429

Helical Antennas

Loop

S

==>

D

Dipole

Dipole

FIG. 12.4

+

Loop

Equivalence of a single turn helix with dipole and loop antennas.

Therefore, the total far-field radiation of one turn helix will be the vector sum of fields due to small dipole of length S (spacing between elements) and small loop antenna of diameter D. That is, ET = Ef + Eq, where Ef and Eq are expressed in Eqs. (12.21) and (12.22). They are in phase quadrature. This is due to presence of j operator in (12.22). The ratio Eqÿ and Ef gives axial ratio of wave radiation from the helix, i.e. AR =

ER EG

=

4XN S

N X NF F

⎛ 2Q ⎞ 2 ⎜ ⎟Q D M ⎝ ⎠

=

2S M

QD

2

=

2S M

CM 2

(12.23)

Equation (12.23) corresponds to axial-ratio of polarization ellipse greater than or less than unity and the value of axial-ratio lies between 0 and ¥. In particular, there are three limiting cases, when AR = 0 (with S = 0), this corresponds to a small loop with horizontal polarization and AR = ¥ (with d = 0) corresponds to a short dipole with vertical polarization. The special case occurs, when AR = 1, that is Eq and Ef are equal and corresponding polarization is circular and this case occurs only, when C = Q D = 2.5 M

Since

L = C2 + S2

S2 + C 2 – L 2 = 0 Sc2 + 2Sc M − L2 = 0

(for CP)

This gives Sc =

−2M ±

⎡ = M ⎢ −1 ± ⎢ ⎣

Also as,

S = C sin B ⇒

(2M )2 + (2 L )2 2 2⎤ ⎛L⎞ 1+ ⎜ ⎟ ⎥ ⎝M⎠ ⎥ ⎦

Bc

⎛S ⎞ = sin −1 ⎜ c ⎟ ⎝L⎠

(12.24)

430


Therefore,

Bc

⎡ −1 ± 1 + (L )2 M = sin −1 ⎢ LM ⎢ ⎣

⎤ ⎥ ⎥ ⎦

(12.25)

Usually, the N-helix oriented vertically and operated in such a way that axial ratio is greater than unity, leading vertically polarized radiation. Like axial-ratio, the pitch angle (a) also describes polarization characteristics of N-helix. For example, when a = 90°, we have a dipole antenna and the corresponding polarization is linear and vertical. Radiation resistance The radiation resistance of N-helix is nearly same as for a small dipole and equal to 2

⎛I ⎞ ⎛2⎞ × 790 × ⎜ av ⎟ hM2 = 395 ⎜ ⎟ hM2 2 ⎝ I ⎠ ⎝Q ⎠ (For sinusoidal current distribution.)

RrH =

1

(12.26)

The advantage of the helix over straight stub or Marconi antenna is that its inductance can resonate the antenna. The electrical performance of axial mode-helix is influenced by its design parameter’s values (L, S, a). The performances of A-helix is satisfactory over a wide range of pitch angles but it is optimum for 12° < a < 14°, however, helices are usually constructed with pitch angle a = 13°. As far as the number of turns is concerned, as more turns are added, gain of antenna increases and the polarization axial ratio decreases, i.e. cross-polarization increases. However, tapering the helix by gradually reducing the diameter near the end of the helix improves impedance, radiation pattern and also the degree of polarization. The N-helix antenna also referred to as resonant stub helix antenna, and pattern is almost omni-directional. The advantage of stub-helix over a straight-wire monopole of same height is that helix acts as an inductor, tending to cancel the capacitance inherent in electrically short-antennas. The N-helix antenna is very popular in small transceivers such as handheld personal radios.

HELICAL ANTENNA WITH DIFFERENT FINITE GROUND SIZE In a particular case, when a helical antenna is designed to operate at 1.7 GHz under specifications; antenna length (L) = 684 mm, antenna radius (a) = 28 mm, pitch angle (a ) = 13.5° and wire radius (r) = 0.3 mm and total number of turn is 16. Then unfolded length of the wire comes about to 3 m [5]. This is the case where antenna ground plane is infinite. Analysis shows that the gain of antenna rages (12.2 ± 1.5 dBi) in frequency range (1.21–2.12 GHz). The radiation characteristics show that side lobes are low only in the narrow band near 1.225 GHz, whereas at higher frequencies side lobes pronounced as high as 6 dB below the main lobe. In an attempt to improve the performance of helical antenna Djordjevic et al. [6] found that radiation pattern and gain of helical antenna strongly depend on the shape and size of

Helical Antennas

431

the ground conductor (infinite). So, they concluded that performance of helical antennas can be improved by shaping the ground conductor. And they proposed three types of ground conductor: square, cylindrical cup and truncated cone and they are shown in Fig. 12.5. 2a 2r p

b

(a)

(b)

D2

h

h D (c)

FIG. 12.5

D1 (d)

Helical antennas: (a) Infinite ground plane, (b) square conductor, (c) cylindrical cup, and (d) truncated cone.

The optimum dimensions of these conductors are as follows: Square patch size = 1.5l Cup dimension D = 1l and h = 0.25l Truncated cone D1 = 0.75l and D2 = 2.5l and h = 0.5l Among these cases, truncated cone acts as a reflector and offers highest gain. The benefit in gain is found to be 3 dB higher than infinite ground plane [7]. The value of h has been chosen up to 2l and in a few cases cone height was selected approximately same as the height of helical antenna and gain up to 8 dB higher than an infinite ground plane was noticed. However, for tall cones the length of helix (L) has practically no influence on the gain. This fact indicates that the cone is the main source of radiation and acts like a horn antenna.

432


HEMISPHERICAL HELIX In general, hemispherical helix is a substitute of spherical helical antenna which is difficult to maintain in vertical position over a flat ground plane. It is constructed by winding the helix on the surface of a polystyrene hemisphere and rest very steadily on the ground plane. Basic geometry of a hemispherical helix is shown in Fig. 12.6, where the curved portion of the thin wire is given by ⎛ G ⎞ R = cos−1 ⎜ − 1 ⎟ Q N < G < 2Q N for RHCP ⎝QN ⎠

G ⎞ ⎛ R = cos−1 ⎜ 1 − ⎟ Q N < G < 2Q N for LHCP QN ⎠ ⎝ in which a = radius of hemisphere N = twice the number of turns of the helix h = height from the ground plane b = radius of outer conductor r = radius of wire and equal to radius of inner conductor

FIG. 12.6

Hemispherical helical antenna.

The proposed antenna is fed with a co-axial cable at the periphery of the hemisphere. In particular 3-turn hemisphere (axial mode) was designed and analyzed using the moment method [8], with specifications a = 1.95 cm, h = 0.39 cm, rw = 0.05 cm and b = 4.5 rw. The obtained result reveals that performances of hemisphere antenna are very similar to that of spherical helical antenna, hence it is also termed a standing wave antenna.

Helical Antennas

433

Parameters The variation of impedance with C/l, indicates that resistance decrease from 550 W to 180 W, for C/l = 1.05 to 1.55. It is also seen that relatively large reactance (–200 to 180 W) makes it significantly different from the reactance of the cylindrical helical antenna. Whereas variation in gain with C/l shows that the gain is maximum (@ 7.6 dB) at C/l = 1.33, accompanied by two more peaks at C/l = 1.1 and C/l = 1.45, having gains of 5.8 dB and 6.2 dB, respectively. While 3-dB gain bandwidth is observed for C/l ranging from 1.08 to 1.5 (Figs. 12.7 and 12.8).

FIG. 12.7

Measured and computed input impedance of 3-turn hemispherical antenna.

FIG. 12.8

Measured gain of 3-turn hemispherical antenna.

434


Radiation pattern measurement shows that it is very similar to those of spherical helical antenna. Particularly, in range of – 40° to 40°, the magnitude differences of field component Eq and Ef, found to be < 3 dB, while phase-difference is 90° (with a difference < 15°). The measurement of axial ratio with the elevation angle q for C/l = 1.2, shows that angular coverage with circular polarization (AR < 3 dB) is about 90° (– 40° < q < 50°). It is also found that an increase in number of turns N of helix is reduced the axial ratio [9]. Measured 3-dB axial ratio bandwidth for 3-turn hemispherical helix is listed in Table 12.2. TABLE 12.2 S.No. 1 2 3

Axial-ratio and BW of 3 turn hemispherical helix Parameters

3-dB axial ratio bandwidth (in C/l) Lowest axial-ratio (in C/l) % bandwidth

Values 1.16–1.25 1.2 7.5

It is clear that hemispherical antenna offers extremely good C.P. over wider beam-width, same as spherical helical antenna. The purity of circular polarization radiation in general higher in axial direction (about 5 dB better within the range of –15° < q < 15°). The values of power gain varies from 9.1 to 9.8 dB, which is very different from the behaviour of power gain of a cylindrical helical antenna, which increases significantly with number of turns as well as frequency. The major disadvantage with axial-mode hemispherical antenna is that the gain cannot be increased significantly by increasing the number of turns. In addition, increased number of turns of helix deteriorates circular polarization and also reduced C.P. beam-width. The hemispherical antennas found very useful in array construction suitable for satellite reception in satellite communication.

APPLICATIONS OF HELICAL ANTENNA Helical antennas find applications in the following: 1. In space communications for telephone, television data transmission, and also being employed on satellites and at ground stations. 2. In global position satellites. 3. Because of its CP, high gain and simplicity, it is also used as feed in parabolic reflector arrays. 4. In mobile communications. In addition, helical antennas are also found suitable in medical applications, particularly for heating cancer tumours in microwave hyperthermia (i.e. for treatment of tumours having different sizes and location in various places of human body). In order to improve performance of microwave antenna employed in microwave hyperthermia, Thomy et al. [10] have proposed a helical antenna system, which consists of double helix known as helical applicator (Fig. 12.9).

Helical Antennas Helix 2

Helix 1

25 mm Dielectric sheating

Metallic helix

FIG. 12.9

435

25 mm

Junction point of outer conductor and helix

Metallic helix with inner conductor

Geometry of dual helical applicator.

The first helical antenna is the inner conductor rolled up around the teflon sheath whereas second one is soldered at the outer conductor and rolled up around the cable. Thermotherapy system consists of a microwave generator (f = 915 MHz and maximum power 100 W), whereas microwave radiometer functions around 3.0 GHz and used to measure temperature. V. Thomy analyzed antenna using FDTD and calculated delivered energy in terms of SAR and scattering parameter S11. The variation of S11 with frequency shows that the reflection coefficient below –10 dB can be achieved provided matching is good. This reveals that at least 90% of the input power is delivered to the surrounding (i.e., in tumours) and hence killed them by increasing the temperature up to high level.

Helical Antenna for G.P.S. Applications As we know that highly circular polarization and omni-directional radiation characteristics of helical antenna makes it a dominant candidate for satellite applications: G.P.S. positioning. However, increasing the number of satellites in space reduces the positioning time and increases the reliability of G.P.S. In order to overcome this difficulty a fractional turn qualifilar helical antenna was introduced which is compact size antenna and very suitable in mobile handsets for G.P.S. applications. Another helical antenna, bifilar antenna was also introduced that covers a wide range of frequency, including two satellite positioning systems and can also support G.P.S. (1.575 and 1.2274 MHz) and GLONASS systems. Bifilar helical antenna is composed to helices equally spaced circumferentially on a cylinder and fed with equal amplitude signals with relative phase of 0° and 180°, whereas qualifilar helices antenna composed of two bifilar helical antennas equally spaced circumferentially on a cylinder and fed with equal amplitude signals with relative phase of 0°, 90°, 180° and 270°. The optimum angle of BHA is found to be approximately 40o and input impedance is approximately 250 W [11]. The G.P.S. application also needs to cancel the effects of ionospheric perturbation which can be done by using dual band receiver. The requirements of G.P.S. and GLONASS antennas include, frequency range 1.2 to 1.65 GHz, RHCP (with AR @ 3dB) and gain is 4 dBi minimum at zenith. The variation of S11 parameter with frequency shows that both the frequencies, 1.2 GHz and 1.65 GHz, have same values (@ – 8.0 dB) of reflection coefficients. Whereas minimum value observed at 1.1 GHz (@ –30 dB) and exactly (–10 dB) seen at 1.4 GHz [12]. The performance characteristics of BHA are shown in Table 12.3.

436


TABLE 12.3

Performance characteristics of BHA at two frequencies

S.No.

Parameters

1.2 GHz

1.6 GHz

1

Axial ratio

2 3 4 5

Gain HPBW FBR C/l

Axial ratio has 160° of BW with 5 dB maximum 5 dB 140° 15 dB 0.5

Axial ratio has 160° of BW with 5 dB maximum 7.5 dB 100° 10 dB 0.5

SOLVED EXAMPLES Example 12.1 A 10-turn axial mode helix antenna is constructed under specifications: a = 13°, f = 8 GHz and er = 3.45. Find the gain, HP and AR of the antenna. Solution:

Given N = 10, M =

3 × 1010 8 × 10 9

= 3.75 cm

S = C tan a = 3.75 tan 13° = 0.796 cm Hence,

HP =

65° C

N

M

S

=

M

65° 3.45 3.75

0.796 10 × 3.75

= 48°

The gain of the antenna will be 2

2

0.796 ⎞ ⎛C ⎞ ⎛ S ⎞ ⎛ 3.45 ⎞ ⎛ G = 6.2 ⎜ ⎟ ⎜ N ⎟ = 6.2 ⎜ ⎟ ⎜ 10 × ⎟ 3.75 ⎠ ⎝M⎠ ⎝ M⎠ ⎝ 3.75 ⎠ ⎝

= 11.1 = 10.5 dB Axial ratio AR =

2N + 1 2N

=

21 20

= 1.05 = 0.212 dB

Example 12.2 Consider a 4-turn stub helix operating at f = 883 MHz. Find axial-ratio, if h = 0.168l and D = 0.5 cm. Solution:

M=

3 × 1010 883 × 10 6

= 34 cm

Helical Antennas

3.14 × 0.5

C = lD =

h

The turn spacing S =

N

0.168M

=

34

437

= 0.046 l

= 0.042l

4

The helix length L = [(NC)2 + h2]1/2 = [(4 ´ 0.046l)2 + (0.042l)2]1/2 = 0.25l 2(S/M )

AR =

(C/M )

2

2 × 0.042

=

(0.046)2

= 38 = 15.8 dB

2

⎛h⎞ Rr = 640 ⎜ ⎟ = 640 × 0.168 = 18 : ⎝M⎠

Example 12.3 A 12-turn axial helix antenna of circumference l has turn spacing of l/4. Determine HPBW, BWFN and directivity and also examine polarization characteristic of the antenna. Solution:

Given N = 12, Cl = 1, Sl = 1/4

HPBW =

52 C

M

N

S

M

115

FNBW = C

M

N

=

S

=

52 1 1 12 4

115 3

=

52 3

= 30°

= 66.39°

M

Directivity = D = 12CM2 × N × SM = 12(1)2 × 12 ×

1 4

= 36 =15.563 dBi Axial ratio =

2N + 1 2N

=

25 24

= 0.178 dB, i.e. high circularly polarized.

Example 12.4 Design an array of 12-axial mode helical antennas each having 10 turns. The spacing between each turn is 0.125l. Also, find the directivity of the array and then total effective area. Solution:

Given N = 6, Sl = 0.125 and Cl = 1

Directivity of an axial mode helical antenna of periphery one wavelength is given by

438


Directivity D = 12Cl2 NSl D = 12 ´ 12 ´ 0.125 = 18 = 12.56 dB Therefore, effective area of each helix is DM 2

Ae =

4Q

=

18M 2 4 × 3.14

= 1.44M 2

Hence, the total area of array formed by 6 numbers of helical antennas is (Ae)T = Ae ´ N = 1.44l2 ´ 6 = 17.2l2

D=

4Q Ae

M2

=

4 × 3.14 × 17.2 × M 2

M2

= 216 = 23.35 dBi

Example 12.5 Design a circular polarized N-helix antenna of each turn-length (L) = 1.8l. And also calculate the radiation resistant at a height of 0.06l from the ground plane. 2

Solution:

⎛2⎞ ⎛ 2 ⎞ Rr = 395 × ⎜ ⎟ hM2 = 395 × ⎜ ⎟ ⎝Q ⎠ ⎝ 3.14 ⎠

Since L = 1.8l, Sc = M ⎡ −1 ± ⎣⎢

2

1 + (LM )2 ⎤ = M ⎡ −1 ± ⎢⎣ ⎦⎥

= l[–1 + 2.05] = 1.06l

Bc and

× (0.06)2 = 0.6 :

1 + (1.8)2 ⎤ ⎦⎥

(Taking +ve sign)

⎛S ⎞ ⎛ 1.06 ⎞ = sin −1 ⎜ c ⎟ = sin −1 ⎜ ⎟ = 36.04° ⎝L⎠ ⎝ 1.8 ⎠ 2 × S × M =

C=

2 × 1.06 × M = 1.45l

Example 12.6 Find the phase velocity (v) of a travelling wave along the helical conductor, if C = l, aÿ = 12°, and N = 12. Solution:

We know that S = C tan a = lÿ tan 12° = 0.213l L = [C2 + S2] = [(0.213l)2 + l2]1/2 = 1.022l

Hence,

p=

where Sl » dl in Eq. (12.19)

LM 1.022 = = 0.815 ⎛ 2N +1 ⎞ ⎛ 25 ⎞ SM + ⎜ ⎟ 0.213 + ⎜ ⎟ ⎝ N ⎠ ⎝ 12 ⎠

Helical Antennas

439

Hence, v = pc = 0.815 ´ 3.0 ´ 108 = 2.445 ´ 108 m/s, which is lesser than the free-space wave velocity, i.e., velocity decrease in the antenna. Such a wave is referred to as slow wave. However, if the helix parameters vary overrather large ranges (5° < a < 20° and 0.6l < C < 4l/3), the phase-velocity is adjusted automatically to maintain increased directivity. Example 12.7 Describe the characteristics of 3-turn helical antenna with reference to circular polarization with a suitable example. Solution: We know that helix of 3 turns will have circular polarization provided the circumference of antenna lies between 3/4l < C < 4/3l. So, in case C = 0.5 metre (19.68 inch) the smallest wavelength lm = 0.75 C = 0.375 metre and corresponding frequency of operation is 800 MHz. The largest wavelength lmax = 1.33 C = 67 metres and corresponding frequency will be 450 MHz, hence fractional BW = 56%. The input impedance is primarily real and can be approximated in ohms as Zin = 140 C/l. The pitch angle of helix varies between 12° and 14°, however its typical value is 13°. The axial ratio for helix antennas decreases as the number of loops N increases, whereas gain increases with frequency. For example, for N = 10 turns helix that has 0.5 metre circumference and pitch angle 13° (i.e., S = 0.13 metre), the gain is 8.3 (9.2 dB). The normalized radiation pattern of 10-turn helical antenna is shown in Fig. 12.10. The wire radius (r) has practically no influence on the antenna performance in wide range 0.00025 < (r/l) < 0.025.

FIG. 12.10

Normalized pattern of helical antenna (dB).

Example 12.8 Calculate the gain of a helical antenna relative to an isotropic radiator if C = 3.50l, N = 12, and antenna pitch angle is 13.5° at operating wavelength 5 cm.

440


Solution:

We know that S = C tan

a

= 3.5 ´ 5.0 ´ tan 13.5° = 4.2 cm

⎡⎛ 3.5M ⎞2 ⎤ Therefore, the gain = 11.8 + 10 log ⎢⎜ ⎟ × 12 × 4.2 ⎥ = 39.7 dB. ⎢⎣⎝ M ⎠ ⎥⎦

OBJECTIVE TYPE QUESTIONS 1. Input impedance of the antenna is constant over a wide range of frequency (a) Yagi antenna (b) Helical antenna (c) Patch antenna (d) None of these 2. Pitch angle (a) of helix is related to spacing between turns (S) as follows: (a) S = C tan a (b) C = S tan a (c) SC = tan a (d) None of these

3. When pitch angle (a) of a helix approaches to 90°, it functions as a (a) Multi-turn helix (b) Loop antenna (c) Linear antenna (d) None of these 4. When pitch angle (a) of a helix approaches to 0°, it functions as a (a) Multi-turn helix (b) Loop antenna (c) Linear antenna (d) None of these 5. Wounding of wire around the helix is characterized as (a) Circular polarization of the antenna (b) Polarization of the antenna (c) Bandwidth of the antenna (d) None of the above 6. Gain of helical antenna is (a) Directly proportional to HP beam width (b) Inversely proportional to HP beam width (c) Independent from HP beam width (d) None of the above

7. Normal mode of operation of helical antenna occurs when the helix diameter is (a) Small compare to wavelength (b) Large compare to wavelength (c) Equal to wavelength (d) None of these 8. Axial mode of operation of helix antenna restricted for the pitch angle (a) Between 12° and 14° (b) Between 14° and 16° (c) Between 16° and 18° (d) None of these

Helical Antennas

441

9. The gain of axial mode helix is related to the operating frequency (f) (a) Directly proportional to frequency (f) (b) Inversely proportional to frequency (f 2) (c) Directly proportional to frequency (f 3) (d) None of the above 10. The input impedance of the helix antenna is found to be (a) Purely inductive (b) Purely resistive (c) Inductive as well as resistive (d) None of these 11. The far-field of N-helix depends on number of turns (N) involved as follows: (a) Directly proportional to N (b) Inversely proportional to N (c) Independent from the N (d) None of these 12. In case of normal mode helix antenna when spacing between turns (S) = 0 and axial ratio (AR) = 0, helix antenna acts as a loop antenna with (a) Vertical polarization (b) Circular polarization (c) Parallel polarization (d) Horizontal polarization 13. Increasing the number of turns of axial-mode hemispherical antenna results in (a) Increased bandwidth (b) Reduced bandwidth (c) Increased CP beam width (d) Reduced CP beam-width 14. The microwave radio-meter consists of helix used to measure (a) Pressure (b) Beam-width (c) Temperature (d) None of these 15. Axial mode of operation of helical antenna occurs when the helix diameter is (a) Comparable to wavelength (b) Large compare to wavelength (c) Equal to wavelength (d) None of these

Answers 1. (b) 6. (b) 11. (c)

2. (a) 7. (a) 12. (d)

3. (c) 8. (a) 13. (d)

4. (b) 9. (c) 14. (c)

5. (a) 10. (b) 15. (a)

EXERCISES 1. Estimate the number of turns of helix antenna, if propagating wave reduces its velocity upto 20% in it. The antenna is designed in such a way that circumference is equal to operating wavelength and pitch angle is 12°. 2. Find the gain, half-power and axial ratio of a 12 turn axial-mode helix antenna, if the antenna is to be operated at l = 5 cm, with L = 12° and C = 3.5 cm. 3. Design a 5 turn stub-helix antenna to be used at f = 9 GHz, if radius of turn is 10.25 cm and Ll = 0.18.

442


4. Design a 3-GHz circularly polarized helix antenna each having turn-length L = 18 cm. Also, calculate the radiation resistance at the height of 0.8l from the ground plane. 5. Design an array of 8 axial mode helical antenna each having 5 turns. Calculate the directivity and total effective area of antenna system, if the spacing between turns is 1.5 cm. 6. What are the advantages of hemispherical antenna? And why it is called standing wave antenna? 7. Describe the design procedure of a hemispherical antenna with a suitable example. 8. Describe the characteristics of hemispherical antenna with reference to its impedance and gain against circumference. 9. Write the expressions to approximate the upper portion of hemispherical antenna in terms of number of turns involved. 10. What are the major disadvantages of hemispherical helix? 11. Describe the characteristics of bifilar helical antenna needed for its applications in global positioning system. 12. Draw the merit between helical antennas with infinite and finite ground planes. 13. What do you mean by finite ground helical antennas? Describe their various types. Conical reflector ground plane offers better performance than the types of finite ground helical antennas why. 14. Compare axial and normal mode helical antennas.

REFERENCES [1] Prasad, K.D., Antenna and Wave Propagation, Satya Prakashan, New Delhi, 1988. [2] Balanis, C.A., Antenna Theory, 2nd ed., Wiley India, New Delhi, 2007. [3] King, H.E. and J.L. Wong, “Characteristics of 1 to 8 wavelength uniform helical antennas,” IEEE, Trans., Antennas and Propagate, Vol. 3, AP. 28, pp. 82–91, 1980. [4] http://www.seaveyantenna.com [5] Djordjevic, A.R., M.B. Bazdar, T.K. Sarkar and R.F. Harrington, Analysis of Wire Antenna and Scatters; Software User’s Manual, Artech House, Boston, 2002. [6] Djordjevic, A.R., A.G. Zajic and M.M. Ilic, “Enhancing the gain of helical antenna by shaping the ground conductor,” IEEE Antennas and Wireless Propagation Letter, Vol. 5, pp. 138–140, 2006. [7] [email protected], Helix for Wi-Fi Applications. [8] Hui, H.T., K.Y. Chan, and E.K.N. Yung and X.Q. Shing, “Co-axial feed axial mode hemispherical helical antenna,” Electronics Letter, Vol. 35(23), pp. 1982–1983, Nov. 1999.

Helical Antennas

443

[9] Hui, H.T., K.Y. Chan, and E.K.N. Yung, “The low profile hemisphere helical antenna with circular polarization radiation over a wide angular range,” IEEE Trans. on Antennas and Propagate, Vol. 51, No. 6, pp. 1415–1418, June 2003. [10] Thomy, V., J. Carlier, L. Dubois, J.C. Camart and J. Pribetich, Helical Antennas for Medical Applications, University of Science and Technology, B.P. 69. LILLE, France, 2006. [11] Nakano, H., Helical and Spiral Antennas: A Numerical Approach, Dept. of Electrical Engg., Hosei, University of Tokyo, Japan. [12] Cheng, W.L., T.H. Chang and J.F. Kiang, “Helical Antenna for G.P.S. Applications,” National University of Taiwan, ROC, IEEE, pp. 3329–32, 2004.

C H A P T E R

13

Microstrip Antenna

INTRODUCTION Microstrip antennas have been one of the most innovative topics in recent antenna technology, and increasingly finding applications in wide range of modern microwave systems. Microstrip antenna is not a new, its idea dates back to the 1950s, but no serious attention was paid to this radiator till the 1970s. However since then several developments started taking place in this area and microstrip antenna became a key component for modern communications; mobile and satellite communications, remote sensing, sensors and also in optical systems. Microstrip antennas are most common form of printed antennas, where only a portion of the metallization is responsible for radiation. A microstrip antenna (MSA) in its simplest form is a layered structure with two parallel conductors separated by a thin dielectric substrate and the conductor. The upper portion is termed patch that is responsible for radiation and lower portion acts as a ground plane. This is the reason microstrip antenna is also known as patch antenna. Figure 13.1 shows the various shapes and sizes of radiating patches.

Rectangular

Elliptical

Square

Circular

Equilateral

Triangular

FIG. 13.1

Pentagonal

Hexagonal

Various shapes and sizes of radiating patches. 444

Annular ring

Microstrip Antenna

445

The patch antenna belongs to the class of resonant antennas and their resonant behaviour is only responsible for the main challenges in microstrip antenna design achieving adequate bandwidth. The resonant characteristics of MSA also mean that at frequencies below UHF they become excessively large. The trade-off in microstrip antennas is to design a patch with loosely bound fields extending into space and tightly bound to the feeding circuitry. This is to be accomplished with high radiation efficiency, impedance and bandwidth with the desired polarization. The patch antennas typically used for frequencies from 100 MHz to 100 GHz [1–2].

Basic Configurations of Microstrip Antenna A rectangular microstrip antenna and its configuration, as shown in Fig. 13.2, consists of very thin (on the order of 0.01 l0) radiating metallic patch placed at small height h (h << l0) on one side of a dielectric substrate, which other plane is grounded. z y

W

x Patch

L h

Substrate material Ground plane

FIG. 13.2

Basic geometry of a rectangular microstrip antenna.

Generally, an MSA consists of rectangular radiating patch element, fed with a co-axial line in which length L is most critical dimension and slightly less than half-wave length in the dielectric substrate. A variety of dielectric substrates having dielectric constant in the range of 2.2 to 12 can be used to design the microstrip antenna. However, since fringing fields are the main sources of radiation, dielectric substrate with low er is usually preferred because it allows the enhanced fringing fields; hence, enhanced bandwidth. The choice of patch thickness also affects the bandwidth over which antennas are required to be operated. The MSA is designed in such a way that its radiation is maximum normal to the patch (broadside direction), which is accomplished by properly choosing the mode of excitation. The radiating patch is normally of copper and gold, and can assume any virtual shapes. Square, elliptical, and circular patches are commonly used due to simplicity of analysis and fabrication and also offered low cross-polarized radiation [2–4].

446


FRINGING FIELDS AND MECHANISM OF RADIATION Since radiating patches are of finite dimensions, excited fields at the edges of the patch undergo fringing. The amount of fringing is a function of patch dimensions, height as well as dielectric constant of substrate material. Since a microstrip antenna can be assumed nonhomogeneous structure; combination of substrate and air, most of the field lines reside in the substrate and remaining exist in the air. As a result, an effective dielectric constant ere is introduced whose values lie between 1 and er, and is a function of frequency. All the electric field lines are concentrated mostly in the substrate provided L/h and er are greater than unity. Radiation from MSA can be explained by taking a rectangular patch antenna shown in Fig. 13.3. L Patch h

E

Ground plane Feed

FIG. 13.3

Side view of microstrip antenna and fringing fields.

Since the thickness of the patch is very small compared to the operating wavelength, the fields varying along the thickness of the substrate are neglected. However, the fields at the edges of the patch can be decomposed into normal and tangential components. The normal components are out of phase; therefore the far-field produced by them cancels in broad side direction. This is because patch length is l/2. On the other hand, the tangential components are in phase and parallel to ground, hence the resulting fields are combined and provides maximum radiation to path surface, i.e., broadside direction. The electric field configuration of the patch, with no variation along the thickness is shown in Fig. 13.4, where fields vary along the length of the patch. y Slot 1

Slot 2 E

Feed

Patch

Ms

FIG. 13.4

L

x

W

S

Top view showing the fringing E fields and equivalent magnetic current Ms.

Microstrip Antenna

447

Advantages of Microstrip Antenna A microstrip antenna is (a) (b) (c) (d) (e)

compact, low profile and light weight cheap and easily developed thin and planar structure compatible with feed and matching network mountable on missiles, rocket and satellites

Disadvantages of Microstrip Antenna A microstrip antenna has (a) (b) (c) (d)

narrow band width and low efficiency low power handling capability limitation on maximum gain poor isolation between feed and antenna

Applications of Microstrip Antenna Major applications of microstrip patch antennas are listed in Table 13.1. TABLE 13.1

Applications of microstrip patch antennas in various fields

S. No.

Platform

Systems

1

Aircraft

2 3

Missiles Satellites

4 5 6

Ships Land vehicles Others

Radars, navigations, communications, altimeter, and landing systems Radars, fuzing and telemetry Communications, direct broadcasting TV, remote sensing, radars and radio-meters Communications, radars and navigations Mobile satellites and mobile radios As biomedical radiator in microwave therapy As a feed element in large and complex antennas

FEEDING TECHNIQUES OF MICROSTRIP ANTENNA Depending upon the shapes and sizes, the feed mechanism of patch antennas is grouped into Probe feed, microstrip line feed, aperture coupled feed and proximity coupled feed.

448


Probe Feed The probe feed is oldest and popular feed technique for MSA. It is very simple to implement by extending the centre conductor of connector up to the patch (Fig. 13.5).

Substrate Patch

Co-axial connector Ground plane

FIG. 13.5

Top and side views of probe fed MSA.

The desirable impedance matching can be achieved by simply adjusting the position of feeding point. This feed is easy to fabricate and has low spurious radiation. Most of patch antennas, such as rectangular, square and circular patches are generally probe feed along patch centre line in the E-plane. In case of circular patch, the centre of the patch is connected to the ground plane and the feed point is located at short distance out from the centre. This is because the impedance of the patch decreases toward the centre, and it is zero at centre and high (»100 W) at edge. The distance from the centre to the feed point is determined by error and trial basis method, and for a patch to be designed on dielectric substrate (er = 2.6), this distance is to found be about 32% of the disc radius. The main drawback of this technique is that it introduces definite amount of inductance that deteriorates the patch from being resonate, in case patch thickness is » 0.1l or more. In addition, it provides narrow bandwidth and difficult to model because holes are drilled in substrate and connector protrudes outside the ground plane, thus not making it planar. It also generates effective cross-polarization.

Microstrip Line Feed In microstrip line feed technique, a narrow conducting strip is connected to the edge of the patch; however, this technique also permits the feed and patch to be printed on the same substrate and provides planar structure (Fig. 13.6). If so, the width of strip should be smaller than the operating wavelength, height of substrate as well as the sides of patch antenna. The purpose behind inset cut in the patch is to match the impedances of the feed line and patch without any additional matching element, which is done by properly controlling the inset position. Hence, this is an easy feeding technique, as it provides ease of fabrication and simplicity in modelling as well as impedance matching. This feed technique is well suited for arrays

Microstrip Antenna

449

Substrate Patch

Microstrip line feed

Ground plane

FIG. 13.6

Microstrip line fed MSA.

where the feed networks are printed with array elements. In case of higher impedance difference between patch and feed, the impedance of the antenna ZA can be matched to a Tx line of characteristic impedance Z0 ~ 50 W, via quarter wavelength Tx line. In general, the characteristic impedance of a microstrip line decreases by increasing the strip width, as loss resistance is inversely proportional to radius of wire, i.e.,

Rohms =

L 2Q a

Rs

(13.1)

where L = length of the wire a = radius of wire and Rs = surface resistance =

XN 2T

That is, wider the strips lower the characteristic impedance. In addition, as the thickness of the dielectric substrate increases, surface waves and spurious feed radiation also increases and hampers the bandwidth of the antenna. The feed radiation also leads to undesired crosspolarized radiation.

Aperture Coupled Feed In aperture coupled feed technique (see Fig. 13.7), the radiating patch and the microstrip feed line are separated by the ground plane. Coupling between the patch and the feed line is achieved through an aperture/a slot in the ground plane. Generally, a high dielectric material is preferred for the bottom substrate to enhance the binding of fields to the feed line and a thick and low dielectric constant material for the top substrate to promote radiation from the patch. Hence, selection of material in this way leads to increased bandwidth. In order to minimize cross-polarization the coupling aperture/slot is usually centered under the patch. The amount of coupling from the feed line to the patch depends upon the

450


3

Patch

Substrate 1

Aperture/slot 1

4 Microstrip feed Substrate 2

Shapes 1, 2, 3, 4 respresent ground plane.

FIG. 13.7

Aperture coupled feed MSA.

shape size as well as position of the aperture. Since the patch and the feed line are separated by ground plane possibility of spurious radiation is very low. The main disadvantage of the technique is that it is difficult to fabricate antenna due to multiple layers which also increases antenna volume. In addition, aperture coupled antenna structure provides narrow bandwidth and cross-polarization, however overall performance is better than the co-axial feed and microstrip feed line techniques.

Proximity Coupled Feed This feed technique is also known as the electromagnetic coupling feed scheme (Fig. 13.8). It is different from other feed techniques in such a way that microstrip feed line does not

Substrate 1

Patch

Aperture/slot

Substrate 2

FIG. 13.8

Proximity coupled feed MSA.

Microstrip Antenna

451

need contact of the patch, hence less sensitive to etching errors. Impedance matching is done by varying the length of the feed line and the width-to-line ratio of the patch. The main advantage of the technique is that it eliminates spurious feed radiation and provides high bandwidth (as high as 13%), this is due to overall increase in the thickness of the antenna. In addition, scheme provides choices between two different dielectric media—one for the patch and another for the feed line—to optimize the individual performances. Like aperture coupled scheme, fabrication difficulty, narrow bandwidth and cross-polarization are major disadvantages of this feed technique. The two additional feed techniques are: slot coupled and co-planar feed techniques (Fig. 13.9). The first one slot is created in the ground plane and this technique leads to the following advantages: simple in fabrication and ease of integration with active devices and good for heat dissipation. Both patch and slot can be etched in a single step. However, stray radiation from slot and limitation in large feeding network layout are major disadvantages of the technique. In the second, one co-planar wave-guide is inserted in ground plane and the technique leads to following advantages: simple in fabrication and ease of integration with active devices and good for heat dissipation and less stray radiation from the feed. However, requirement of more space and less freedom in large feed network design are main disadvantages of the technique. In addition to these feed techniques, there are other two more feed techniques: gap coupled and two-layered coupled feed techniques which are simple in fabrication and easy in integration with active devices. They are also good for heat dissipation. Substrate Patch

Patch

Slot in ground plane (a)

FIG. 13.9

Co-planar wave-guide in ground plane (b)

(a) Slot coupled feed, and (b) Co-planar wave-guide feed.

Comparison of above feed techniques are summarized in Table 13.2.

452


TABLE 13.2

Performances of different feed techniques

S. No. Parameters

Coaxial feed

Microstrip feed

Aperture feed

Proximity feed

1

Impedance matching

Controlling length and position of probe

Controlling width, length and position of strip

Controlling width of feed line and length of slot

Controlling width, position and length of strip line

2

Fabrication

Soldering and Etching is needed drilling needed

Alignment required Alignment required

3

Bandwidth

2–5%

2–5%

2–5%

13%

4

Reliability and profile

Poor because of soldering. Low weight

Better

Good

Good

Low weight

Bulky

Bulky

5

Spurious radiation

More

More

Moderate

Minimum

6

Crosspolarization

Effective

Less

Poor

Poor

BANDWIDTH ENHANCEMENT AND HIGHER ORDER MODES In general, the bandwidth of a rectangular microstrip antenna in terms of wavelength (l), height (h), and dielectric constant (er) of the substrate is given by

⎛ F − 1⎞ ⎛ h ⎞ ⎛W ⎞ BW = 3.77 ⎜⎜ r 2 ⎟⎟ ⎜ ⎟ ⎜ ⎟ ⎝ Fr ⎠ ⎝ M ⎠ ⎝ L ⎠

(13.2)

Equation (13.2) is valid for (h/l) << 1. Usually, the bandwidth of microstrip antenna is defined as the frequency range relative to the centre frequency for a VSWR less than 2:1. The VSWR stands for voltage standing wave ratio and expressed in terms of reflection coefficient (G) of the input feeding network as follows: VSWR =

1+ * 1 − *

(13.3)

A VSWR of 2:1 implies a reflection coefficient (G) of approximately –10 dB. The reflection coefficient (G) is found from input impedance into the path (Zin) and the characteristic impedance of feed line (Z0 ~ 50 W) by using relation *=

Z in − Z 0 Z in + Z 0

(13.4)

Microstrip Antenna

453

As already mentioned, a narrow bandwidth (~ a few per cent) is a major disadvantage of MSA and prevents its uses in many practical applications. A large number of attempt being made toward the development of creative designs and techniques for improving the band width of the MSA. The most direct approach to enhance the bandwidth is to design path antenna using a thick and low dielectric substrate. But this technique inevitably leads to spurious feed radiation and surface wave generation or feed inductance. Proper impedance matching between feed and patch antenna is a second way of enhancing the band width. This could be achieved by using various matching sections between patch and feed. Spurious radiation may be a concern; however it can be eliminated if the matching network is co-planar with the antenna. Another successful way to enhance the bandwidth of MSA is to use parasitic coupling between patches [5]. In a best way, it is done with two stacked patch elements where bottom patch is fed directly and top one is proximity coupled and finally, results in enhanced band width. Though any feed technique can be used to improve the bandwidth of the patch antenna, but band width upto 5% only has been achieved using probe and microstrip line feeds. Further enhancement in bandwidth (~13%) is observed in case antenna is fed by aperture/proximity coupled feeds. Since there are similarities between dielectric loaded cavity and the microstrip antenna, high order modes resonate in addition to the dominating resonance modes. For the rectangular MSA high resonance modes occur when the distance d between the feed point and the opposite edge of the antenna is a multiple of a half wavelength. The corresponding resonance frequencies fn (say) are given by [3,6]. f nr =

c 2nd

Fr

,

n = 1, 2, 3, …

(13.5)

However, for the circular patch antennas the resonant frequencies fn are given by f nc =

′ )c J m′ ( X mn 2Q ae F r

Since the lowest mode is (1, 1) with X11 = 1.841, the corresponding resonance frequency reduces to fr =

1.8421 c 2Q ae F r

(13.6)

Note that the ground wire forces the electric field to be zero at the centre of the disc therefore no modes with m = 0 are excited. The other values of J¢m(X¢mn) can be found from Bessel’s function. Once a resonance is known the desired radiation patterns and other performance characteristics of the antenna can be determined using above expressions [4].

454


RECTANGULAR PATCH ANTENNA The rectangular patch antenna is a simplest form of patch antenna and usually designed to operate near the resonance. This implies that the imaginary (reactance) part of the input impedance (complex) is zero. The fringing fields responsible for radiation act as an additional length to the patch (Figs. 13.2 and 13.3). The length of a half-wave patch is chosen slightly less than the operating wavelength this is to compensate the length introduced by the fringing fields. The amount of length introduced depends on the substrate media width and height of the patch. The resonance length of a half-wave patch is given by [7] L ≈ 0.49 Md = 0.49

M0 half-wave patch Fr

(13.7)

where ld = wavelength in the dielectric substrate ÿ l0 = free space wavelength ÿ ÿ er = dielectric constant of the patch substrate ÿ

Radiated Fields The radiation pattern of a rectangular patch antenna is rather broad with a maximum radiation in the direction normal to the plane of the antenna. Pattern computation of the patch is easily performed by first representing fringing electric field [4] using M s = 2 Ea × nˆ

where Ms = equivalent magnetic current Ea = fringe electric field The factor 2 appears due to the image of the magnetic current in the electric ground plane this is true only if thickness of substrate (h) is chosen to be small. The far-field components (Eq, Ef) of electric field E for a rectangular patch of length L and width W is computed as follows: Eq = E0 cos f F(q, f) (13.8) Ef = – E0 cos qÿ ´ sin f F(q, f)

(13.9)

in which the first term represents the pattern factor for a uniform line source of width W whereas second term F(q, f) is the array factor for a two-element array along the length of patch (L) corresponding to the edge slots and it is given by [8]

⎡ CW ⎤ sin ⎢ sin R sin G ⎥ ⎣ 2 ⎦ × cos ⎡ C L sin R cos G ⎤ F (R , G ) = ⎢ ⎥ CW ⎣ 2 ⎦ sin R sin G 2 where b is the free space phase-constant.

Microstrip Antenna

455

Design Parameters The resonant frequency of RMSA can be obtained using fr =

c 2 F e (L + 2'l)

(13.10)

where L = length of rectangular patch c = velocity of light ÿ er = effective dielectric constant Fr − 1 ⎡

12 h ⎤ ⎢1 + ⎥ W ⎦ 2 2 ⎣ Dl = extension in patch length

=

Fr + 1

+

⎡ ⎢ (F e + 0.3) = 0.412 l ⎢ ⎢ (F e − 0.258) ⎢⎣

−1/2

⎛W ⎞⎤ + 0.264 ⎟ ⎥ ⎜ ⎜ h ⎟⎥ ⎜ W + 0.8 ⎟ ⎥ ⎜ ⎟⎥ ⎝ h ⎠⎦

The width of the patch is not critical, it can be calculated using W=

c 2 fr

⎛ Fr + 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠

−1/2

(13.11)

Radiation Resistance The radiation resistance of a resonant half-wave patch (L » 0.49 ld) is given by 2 ⎧⎪ F 2 ⎛ L ⎞ ⎫⎪ Rr = 90 ⎨ r ⎜ ⎟ ⎬: ⎪⎩ F r − 1 ⎝ W ⎠ ⎪⎭

(13.12)

Typical value of impedance of a rectangular patch antenna varies from 100 W to 400 W. By widening the patch the factor L/W becomes one and impedance for half-wave patch varies with dielectric constant as follows: ⎛ F r2 ⎞ Z S = 90 ⎜ ⎟: ⎜ Fr − 1 ⎟ ⎝ ⎠

(13.13)

The square patch is a modified form of rectangular patch antenna (i.e. L = W), therefore all the parameters can be determined for the square patch simply by replacing L with W or vice versa. Microstrip antennas can also be treated as resonant half-wave parallel plate Tx line

456


whose characteristic impedance depends on the number of parallel field lines confined between patch and ground plane. Each field line has characteristic impedance equal to intrinsic impedance of free-space. Therefore, characteristic impedance of the microstrip antenna can be given by [9] Zc =

where Z 0 = 120 Q

Z0

(13.14)

n Fr

Nr . Fr

Available results reveal that the input resistance of microstrip antenna does not depend strongly on the substrate thickness, however it can be decreased by increasing the width of the patch as long as (W/L) is less than 2. This is because aperture efficiency of a single patch antenna start decreasing as (W/L) goes beyond 2. Since radiation of microstrip antenna covers half of space, i.e. solid beam area of p-steradian in general, its directivity will be

D=

4Q

Q

= 4 or log 4 = 6.02 dB

Radiation Conductance In general, radiation conductance is defined as

G=

in which

(13.15)

|V0 |2

V0 = hE0

where and

2 Pr

Pr =

|V0 |2 240 Q

∫

Q 0

SR2 sin 3R dR

⎛ k0 ⎞⎤ ⎡ ⎢ sin ⎜ 2 W cos R ⎟ ⎥ ⎝ ⎠⎥ SR = ⎢ ⎢ ⎥ cos R ⎢ ⎥ ⎣⎢ ⎦⎥

However, in particular cases of rectangular microstrip antennas solution of above equations yields the radiation conductance to 1 ⎛W ⎞ G= ⎜ ⎟ 90 ⎝ M ⎠

when W << l.

2

(13.16)

Microstrip Antenna

G=

⎛W ⎞ ⎜ ⎟ 120 ⎝ M ⎠ 1

457 (13.17)

when W >> l.

Directivity Since rectangular patch antenna is a two-slot radiating device its directivity can be expressed as G = 8.2 dB

(13.18)

when W << l.

G = 8(W/l) (13.19) when W >> l. Directivity of patch antenna approximately free from substrate height as long as height is small and resonant frequency is fixed.

CIRCULAR PATCH ANTENNA After rectangular patch antenna, circular patch antenna is a second popular patch antenna used for various purposes in modern technology. It is used not only as a single element but also in array form and it is analyzed by treating the patch, ground plane and the material between the two as a circular cavity. The geometry of the circular patch antenna is shown in Fig. 13.10, in which all the parameters having usual meaning [2, 3].

FIG. 13.10

Cavity model and equivalent magnetic current density for circular patch antenna.

Rectangular patch has two degrees of freedom (length and width of the patch), however a circular patch has only one degree of freedom (radius of patch). Therefore, by changing/ altering the radius, patch does not change the order of the mode; on the other hand, it

458


changes the absolute value of resonant frequency. Various design parameters of circular patch are given below: Effective radius (ae) Under the effect of fringing fields, it is determined by 1

⎡ 2h ae = a ⎢1 + aF r Q ⎢⎣

⎧⎪ ⎛ Q a ⎞ ⎫⎪ ⎤ 2 ⎨ ln ⎜ ⎟ + 1.7726 ⎬ ⎥ ⎪⎩ ⎝ 2h ⎠ ⎪⎭ ⎥⎦

(13.20)

where a = actual radius of the patch h = substrate thickness. Resonance frequency fr =

1.841 c 2Q ae

Fr

(13.21)

where c is velocity of light in free space. Radiated fields The radiation fields of circular disc can be compared with that of the circular loop and can be determined by [3,10] ER = − j

and

EG = − j

K 0 aeV0 2r K 0 aeV0 2r

′ ) e − jK0 r (cos R J 02

(13.22)

e− jK0 r (cos R sin G J 02 )

(13.23)

where J02 = J0(K0ae sin q) + J2(K0ae sin q) J¢02 = J0(K0ae sin q) – J2(K0ae sin q) V0 = voltage across the patch r = distance of observation point K0 = wave number in free space Directivity The directivity of a circular patch antenna in terms of effective radius is defined as Dc =

(K 0 ae )2 120 Gd

where Gd = conductance due to radiated power and

(13.24)

Microstrip Antenna

=

(K 0 ae )2 480

∫

Q /2

2 ′2 + cos2 R J 02 [J 02 ] sin R dR

0

459 (13.25)

For a very small value of the radius, the directivity approaches to 3 (4.8 dB). Quality factor and loss tangent Quality factor is one of the important parameters of patch antenna that represents all kinds of losses such as radiation, conduction and dielectric losses in the antenna. Therefore, total quality factor of the antenna can be expressed as sum of quality factors corresponding to radiation, conduction and dielectric losses. In general total quality factor is expressed as [10] 1 QT

1

=

Qr

+

1

+

Qc

1 Qd

(13.26)

where Qr is the quality factor due to radiation losses and 1 Qr

=

2 Q fr WT Pr

Qc is the quality factor due to conductor losses and 1 Qc

=

2 Q fr WT

=

Pr

h Ed

That is, Qc will be more dominating for a patch antenna of very thin substrate. Qd is the quality factor of the dielectric losses and 1 Qd

=

2 Q fr WT

=

Pd

1 tan E

Therefore, the total quality factor is 1 QT

=

Pr 2 Q fr WT

+

Ed

h

+ tan E

(13.27)

+ tan E

(13.28)

Hence, total effective loss tangent will be E eff =

1 QT

or E eff =

Pr 2 Q fr WT

+

Ed

h

460


Bandwidth In general, the fractional bandwidth of a circular patch antenna is given by 'f fr

=

1 Q

⇒ 'f =

fr

(13.29)

Q

This formula is less useful as it does not take into account the impedance matching at the feed terminal of the antenna. Carver and Mink [3] have proposed a modified expression for bandwidth that takes into account the impedance matching and fractional bandwidth is given by

⎛ s − 1⎞ 'f = fr ⎜ ⎟ ⎝ Q s ⎠

(13.30)

where s = VSWR and Q = total quality factor of the antenna. In general, the bandwidth of the circular patch is proportional to volume of the patch, therefore, for a rectangular patch antenna, it can be expressed as follows: BW » Area ´ Height = Length ´ Width ´ Height =

1 Fr

×

1 Fr

×

Fr =

Fr

The bandwidth of a rectangular patch is inversely proportional to the square root of the dielectric constant of the substrate, i.e., bandwidth increases as dielectric constant decreases. Available results reveal that bandwidth also increases as the substrate thickness increases. Radiation efficiency The radiation efficiency for a circular patch antenna is defined as

Ie =

QT

(13.31)

Qrad

where QT and Qrad are defined in Eqs. (13.26 and 13.27). The circular polarization bandwidth of a microstrip antenna is given by [11]

BW(%) = 12

Axial ratio QT

(13.32)

where axial ratio is in dB and QT is quality factor and its value must be less than 10.

Microstrip Antenna

461

CIRCULARLY POLARIZED MICROSTRIP ANTENNA In general, a single point feed microstrip antenna produces linear polarization. However, it is possible to produce circular polarization by using two feed point to excite two orthogonal modes on the patch with a 90° phase difference between their exciting modes. An elliptical patch feed with a coaxial feed on its radial line and at circumference also produces circular polarization. The radiated fields of an elliptical patch antenna is obtained in terms of tabulated functions ErEq and Ef by taking into account the fringing fields at the edges of the antenna. A coaxial fed elliptical patch antenna is shown in Fig. 13.11, where all the parameters have their usual meaning. The antenna is fed on the edge at f0 = 45° from the semi-major axis at a point (u0, v0) by a coaxial line through the dielectric substrate. The elliptical coordinate u and v are used in substitution of x–y co-ordinate whereas z-axis remains constant/ unaffected.

FIG. 13.11

Elliptical patch antenna with co-ordinate system.

The relations between (uv) and (x, y) are as follows [5]: x = c cosh u cos v y = c sinh u sin v and therefore ellipse is defined as ⎛a +b⎞ u = u0 = ln ⎜ ⎟ ⎝ c ⎠

(13.33)

⎛ tan G 0 ⎞ v = v0 = tan −1 ⎜ ⎟ ⎝ tanh u ⎠

(13.34)

and

462


defines hyparbola, in which 1/2

⎛ b2 ⎞ c = ⎜1 − 2 ⎟ ⎜ a ⎠⎟ ⎝

Design equation An elliptical patch antenna having quality factor Q is designed by using a

=1+1

b

0887

(13.35)

Q

in which a and b are the semi-major and minor axes of the ellipse and a=

pc fr F r

where c is velocity of light in free space and er is dielectric constant. And p is empirical constant whose value lies between 0.27 and 0.29 and fr is resonance frequency. Q is quality factor and for axial-ratio £ 3 dB, it is defined as [2] Q=

where, Mc =

M0 Fe

0.35Mc

(13.36)

36.7h + 016 Mc

and ee is effective dielectric constant.

The left and right hand components of circular polarized field are related to radiated fields in q and f directions as follows [11]:

ER =

ER + jEG

and EL =

2

ER − jEG

(13.37)

2

Therefore, it will be logical to mention that in circular polarization radiation maximum and minimum fields occur when the LHCP and RHCP components of radiated electric fields: EL and ER are alternatively added and subtracted, i.e. Emax = |EL| + |ER|

and

Emin = |EL| – |ER|

Hence, axial ratio for circularly polarized waves can be expressed as follows: AR =

AR =

Emax Emin Emax Emin

=

=

|E L | + |E R | |E L | − |E R | |E R | + |E L | |E R | − |E L |

for LHCP

for RHCP

Microstrip Antenna

463

A square patch antenna fed with 90° hybrids points also produce circular polarization. Using sets of printed dipole and slot proximity coupled to a microstrip feed line with the proper phasing also produce circular polarization. In case of aperture coupled cross-slots/ slits on the radiating patch can also be used to produce CP.

OTHER PARAMETERS Double Tuning A double tuned patch antenna is an antenna which is designed at a single frequency and functions satisfactorily at two another frequencies closed to the design frequency. Double tuning can be obtained from parasitic coupled patch antenna with considerable amount of bandwidth enhancement. Double tuning can also be obtained with a single aperture coupled microstrip antenna by lengthening the coupling slot so that it approaches resonance. A bandwidth more than 20% has been achieved for single element and arrays using this technique. It is also possible to use parasitic elements co-planar with driven elements to produce double tuning as shown in Fig. 13.12. However, the technique may not be suitable for array applications because the array requires spacing greater than half wavelength and the phase centre of the array element is not fixed over its operating bandwidth. Parasitic patch

Active patch

Ground plane Coaxial feed

FIG. 13.12

Microstrip stacked antenna.

Coupling Coupling between two identical patches depends on relative alignment, as they are shown in Figs. 13.13 and 13.14. E-plane arrangement offers the smallest coupling isolation for very small spacing (< 0.10 l) whereas H-plane arrangement exhibits the smallest coupling for large spacing (> 0.10 l) [12]. The mutual conductance between two patches for E-plane arrangement is higher for wider elements; however, it is higher for narrower element in the H-plane arrangement.

464


E

d

E

E

d

E

E

FIG. 13.13

E-plane arrangement.

FIG. 13.14

H-plane arrangement.

Selection of Substrate Material The dielectric constant of substrate material plays a vital role for overall performance of patch antenna. Almost all the properties of an antenna depend on the types of substrate used in designing, particularly the radiation efficiency of the antenna strongly depends on the dielectric constant of the substrate. Selection of material depends on its loss-tangent, temperature gradient and thickness. Ideally, thick and low dielectric constant substrate having low insertion loss is preferred for broadband applications and higher efficiency. However, in contrast the high dielectric constant of FR4 and other common materials require that antenna to be mounted on stand-off using an air space to reduce dielectric constant.

Laminate Composite for MSA Recently, Arlon, proposed a family of laminate composite foam clad R/F which consist of a low permittivity microporous polymeric core bonded to an impermeable copper-clad polymer film overlay that provides a low composite dielectric. Foam clad R/F material has er from 1.15 to 1.30, and low loss tangent from 0.002 to 0.004, so it can be conveniently processed (print and etch) for manufacturing of patch antenna particularly. Foam clad R/F of multiple thicknesses from 0.043 to 0.250 inches are also available which can be used to design microstrip antennas. The specific properties of this product make it suitable for applications exhibiting broadband properties with high speed data transfer. Applications of this material also include products for the RFID readers specially tag market, Wi-Fi 802.11b/g, Wi-Max antenna, base-station antennas and automobile radars. Foam clad also generates extremely low passive inter modulation (PIM), that is why it is used conveniently at locations where multiple antennas are located and simultaneously transmit and receive signals.

Photonic Band Gap Antennas The photonic structure is basically a periodic metallic pattern printed on dielectric substrate for microwave and millimetre wave applications. The photonic structure provides a stop band of EM waves propagating through it. The frequency range of stop band depends upon

Microstrip Antenna

465

the pattern geometry and its dimensions. If the antenna operating frequency falls within this stop band, it is attenuated while propagating through the substrate. Thus, the generation and propagation of surface waves are stopped. Unlike conventional microstrip line the microstrip on PGB substrate shows a stop band over 5.5 GHz to 10.5 GHz in their transmission characteristics. The PGB attributes a new feature of restricting the propagation/generation of surface waves in a microstrip which are applied to improve the performance characteristics of the antenna. Use of PGB structure also leads to wide bandwidth characteristics of the antenna. Suppression or reduction of surface waves reduces side lobes level, improving antenna efficiency and thus antenna gain.

Mobile Antennas Antennas for wireless communication and handsets demand some specific features such as compactness, wideband, multiple band operation, high gain, directivity reception, and uniform radiation pattern, and reduced radiation hazards. The radiation hazard is an important issue from the health point of view. Radiation hazard is measured in terms of a parameter known as specific absorption rate (SAR), which is defined as the rate of energy absorption for unit mass of surface measured in W/kg. The SAR depends on the density and conductivity of the biological substance as well as the magnitude of radiated fields. Mathematically, it is expressed as follows: SAR =

T E2 md

(13.38)

E = r.m.s. value of electric field within the material = conductivity of the material and md = mass-density of the material ÿs

Planar inverted F antenna (PIFA) is highly suitable for mobile communication due to its light weight, small size and ability to receive/radiate both vertically and horizontally polarized waves. Compact printed antennas are other antennas, which are suitable for mobile communication. Numerous geometries have been investigated so as to develop compact microstrip antennas with (i) wide bandwidth, (ii) dual frequency dual polarization, (iii) enhanced gain and (iv) circular polarization, etc. These characteristic performances are achieved by introducing shorting pins, strips, slits or spur lines cut on the patch element. Circularly polarized antennas in recent years also find potential applications in mobile communication and handsets.

Integrated Antennas If a printed antenna is placed close to microwave source or transmitter, fabricated on the same board, then the antenna element becomes a part of oscillator/receiver and acts as an equivalent LC circuit as well as radiating device. This is basic concept of the integrated antenna. Hybrid type antenna using inverted microstrip patch integrated with gunn diode/

466


varactor diode /or FET is the simplest example of integrated antenna. An improved broadband and polarization features of this antenna lead its application as a quasi-optical power combiner. Genetic algorithm (GA) finds many applications in optimizing antennas and it has been applied to continuously optimize the various parameters of the antennas and SNR plus interference ratio of the adaptive antenna array. The GA has also been successfully applied to improve the impedance bandwidth of microstrip antennas. The frequency selective surface (FSS) is low cost printed electromagnetic material used to control surface current on it. This is usually achieved by a two dimensional array of metallic patterns forming printed inductors and capacitors. Various applications of FSS in electromagnetic are found to improve the gain and the pass-band operation of a printed antenna. Hopefully, microstrip antenna will play most important role in order to replace wire communication into wireless communication network in near future. Microstrip antennas are suitable not only for communication systems but also for biological systems. In recent years, rapid decrease in the size of personal communication devices has lead to the need for compact MSA. Several techniques such as using high dielectric material, incorporating a shorting pin and cutting slits/slot in radiating patch have been used to miniaturize the MSAs. There are two important computational tools: Finite difference time domain (FDTD) Maxsoft based and IE3D Zeland Software based on moment method which are mostly used for numerical analysis of MSA and printed antennas. Recent trends and developments in microstrip antennas may be classified as follows [8,12]: · · · · · · · · ·

Wide band and multiband microstrip antennas Compact reduced sized antenna Applications of genetic algorithm (GA) Circular polarization with wide impedance bandwidth and multifunction antennas Microstrip and printed antennas on photonic band gap structures. Applications of frequency selective surface Active integrated antennas Antennas for wireless communication and handsets Multi-band fractal microstrip antennas.

SOLVED EXAMPLES Example 13.1 Find the value of radiation conductance for a microstrip antenna designed on a substrate of dielectric constant 2.2 and operating at 3.0 GHz. Solution: The operating wavelength l = 10 cm Width of the patch W =

30

2

2 × 3

2.2 + 1

= 3.952 cm

Microstrip Antenna

467

which is << wavelength. Therefore, radiation conductance 2

1 ⎛ 3.952 ⎞ −3 G= ⎜ ⎟ = 1.74 × 10 Siemens 90 ⎝ 10 ⎠

Example 13.2 Design a rectangular patch antenna on a dielectric substrate with er = 2.65 to be operated at f = 3.0 GHz. Determine antenna parameters, bandwidth, and VSWR, if antenna is fed by a Tx line of 50 W, assume thickness of substrate is 0.02% of wavelength l. Solution:

M=

Width W =

c f

=

3 × 1010 3 × 10

= 10 cm and, hence, thickness h =

9

c ⎛ Fr + 1 ⎞ ⎜ ⎟ 2 fr ⎝ 2 ⎠

−1/2

=

3 × 1010 2 × 3 × 10

9

⎛ Fr + 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠

−1/2

=

10 100 5

1.825

12 h ⎞ ⎛ F + 1 ⎞ Fr − 1 ⎛ Effective dielectric constant F e = ⎜ r ⎟+ ⎜1 + ⎟ w ⎠ 2 ⎝ 2 ⎠ ⎝ 12 × 0.2 ⎞ ⎛ = 1.825 + 1.325 ⎜ 1 + ⎟ 3.70 ⎠ ⎝

= 1.825 +

1.325 1.284

× 2 = 0.2 cm

=

5 1.35

−1/ 2

−1/2

= 2.85

⎡ ⎛W ⎞⎤ + 0.264 ⎟ ⎥ ⎢ ⎜ (F e + 0.3) h ⎠⎥ × ⎝ Extension in length 'l = 0.412 h ⎢ ⎢ (F e − 0.258) ⎛W ⎞ ⎥ + 0.8 ⎟ ⎥ ⎢ ⎜ ⎢⎣ ⎝h ⎠ ⎥⎦ ⎡ ⎛ 3.7 ⎞⎤ + 0.264 ⎟ ⎥ ⎢ ⎜ (2.85 + 0.30) 0.2 ⎠⎥ × ⎝ = 0.412 ´ 0.2 ⎢ ⎢ (2.85 − 0.258) ⎛ 3.7 ⎞ ⎥ − 0.8 ⎟ ⎥ ⎢ ⎜ ⎝ 0.2 ⎠ ⎦⎥ ⎣⎢ 18.764 ⎤ ⎡ 2.85 + 0.30) × = 0.0824 ´ 2 ⎢ ⎥ 17.7 ⎦ ⎣ (2.85 − 0.258)

=

4.87 45.88

= 0.106 cm

= 3.70 cm

468


Hence,

L=

c 2 F e fr

− 2'l =

3 × 10 3 × 10 × 2 ×

2.85

= –2 ´ 0.106 = 1.27 cm

⎛ F r2 ⎞ ⎛ L ⎞2 (2.65)2 ⎛ 1.27 ⎞ 1019.39 Resistance Rr = 90 ⎜ = 90 = 45.1287 : × ⎟ ⎜ ⎟= ⎜ F r − 1 ⎟ ⎜⎝ W ⎟⎠ 1.65 3.7 22.5885 ⎝ ⎠ ⎝ ⎠ ⎛ Fr − 1 ⎞ ⎛ W ⎞ ⎛ h ⎞ ⎛ 265 − 1 ⎞⎛ 3.7 ⎞⎛ 1 ⎞ Bandwidth B/W = 3.77 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 3.77 ⎜ ⎟⎜ ⎟⎜ ⎟ = 0.05 = 5% ⎝ 2.65 ⎠⎝ 1.27 ⎠⎝ 50 ⎠ ⎝ F r ⎠ ⎝ L ⎠ ⎝ M0 ⎠

Reflection coefficient (G) =

Z in − Z 0 Z in + Z 0

=

45 − 50 45 + 50

= −

5 95

= − 0.053

|G| = 0.053 Hence,

VSWR =

1+ * 1 − *

=

1 + 0.053 1 − 0.053

=

1.053 0.947

= 1.112

Example 13.3 Design a circular patch antenna of radius 2 cm on a dielectric substrate er = 3.2, to be operated at f = 3.0 GHz. Determine: (i) the conductance of the antenna for directivity 6 NV. (ii) the quality factor for an operating bandwidth 5%. Solution:

Since f = 3.0 GHz, l0 = 10 cm

Effective radius ae =

c × 1.8421 2Q × f r F r

=

5 × 1.8421 3.14 ×

3.2

=

9.2 5.62

= 1.64 cm

Circumference extension DC = 2p(a – ae) = 2 ´ 3.14 ´ 0.36 = 2.26 cm.

⎛ 2Q ⎞ × 1.64 ⎟ ⎜ (K 0 ae ) ⎝ M0 ⎠ Gd = = 120 × Dc 120 × 6

(i) Conductance

= (ii) B/W 'f = ' f =

fr QT

or

(0.628 × 1.64) 720 QT =

fr 'f

=

= 1.47 × 10 −3 mh/cm 3.0 0.05

= 60 NV

Microstrip Antenna

469

Example 13.4 A rectangular patch is operating at 5 GHz for circular polarization. Find two resonant frequencies associated with two lengths of the antenna as well as relative ratio of the length. Assume operating bandwidth is to be 3% for VSWR 1.5. Also find the CP nature of antenna radiation. Solution:

We know that ⎛ s − 1⎞ ⎛ 1.5 − 1 ⎞ 'f = fr ⎜ ⇒ Qt = ⎜ ⎟ ⎟ = 13.6 ⎜ ⎟ ⎝ 0.03 1.5 ⎠ ⎝ Qt s ⎠

Therefore, two resonant frequencies

f1 =

fr 1⎤ ⎡ ⎢1 + Q ⎥ ⎣ ⎦

1/2

=

5 1 ⎤ ⎡ ⎢1 + 13.6 ⎥ ⎣ ⎦

1/2

= 4.826 GHz

and 1⎤ ⎡ f2 = fr ⎢1 + ⎥ Q⎦ ⎣

1/2

1 ⎤ ⎡ = 5 ⎢1 + ⎥ 13.6 ⎦ ⎣

1/2

= 5.18 GHz

and relative ratio of the two lengths will be L

=1+

W

1

=1+

Q

1 13.6

= 1.074

That is patch is nearly square

AR =

BW × Q 12

=

3 × 13.6 12

= 3.4 dB

i.e., circular polarization is poor this is because Q is greater than 10. Example 13.5 Design a rectangular patch antenna to be used in satellite communication at a frequency of 1.6 GHz. Assume the dielectric constant of substrate is 10.2 and thickness is 0.127 cm. Also find its resonant input resistance and directivity. Solution:

We know that 1/2

W =

Fe =

c ⎛ 2 ⎞ ⎜ ⎟ 2f r ⎝ F r + 1 ⎠ Fr + 1

2

+

=

Fr − 1 ⎛

2

1/2

2 ⎛ ⎞ ⎜ ⎟ 2 × 1.6 ⎝ 10.2 + 1 ⎠ 30

12 h ⎞ ⎜1 + ⎟ W ⎠ ⎝

−1/2

= 3.962 cm

470


=

10.2 + 1 2

10.2 − 1 ⎛ 12 × 0.127 ⎞ + ⎜1 + ⎟ 2 3.962 ⎠ ⎝

−1/2

= 9.51

⎛W ⎞⎤ ⎡ + 0.264 ⎟ ⎥ ⎜ ⎢ h (F + 0.3) ⎝ ⎠⎥ 'l = 0.412 h ⎢ e ⎢ (F e − 0.3) ⎛ W ⎞ ⎥ ⎢ + 0.8 ⎜ ⎟ ⎥ ⎣⎢ ⎝h ⎠ ⎦⎥ ⎛ 3.962 ⎞⎤ ⎡ + 0.264 ⎟ ⎥ ⎜ ⎢ (9.51 + 0.3) ⎝ 0.127 ⎠⎥ = 0.412 × 0.127 ⎢ = 0.054 cm ⎢ (9.51 − 0.3) ⎛ 3.962 ⎞ ⎥ ⎢ + 0.8 ⎟ ⎥ ⎜ ⎣⎢ ⎝ 0.127 ⎠ ⎦⎥

L = Le − 2'l =

c

− 2'l =

2f F e

15 1.6 F e

− 2 × 0.054 = 2.94 cm

The radiation conductance, as w << l0 2

G=

2

1 ⎛W ⎞ 1 ⎛ 3.92 ⎞ −4 ⎜ ⎟ = ⎜ ⎟ = 4.96 × 10 S 90 ⎝ M0 ⎠ 90 ⎝ 18.75 ⎠

The resonant input resistance Rin = Z in =

1 2G

=

1 2 × 4.96 × 10 −4

= 100.8 :

Directivity D = 8.2 dB as w << l0. Example 13.6 A patch antenna is operating at 3 GHz such that its fractional BW is found to be 3.5% for VSWR 1.6:1. Find its radiation efficiency if it radiates one fourth of power fed to it. Also find the total loss tangent assuming input power 75 MW. Solution: ⎛ 1.6 − 1 ⎞ ⎛ s − 1⎞ 'f = fr ⎜ ⇒ Qt = ⎜ = 13.55 ⎜ Q s ⎟⎟ ⎜ 0.035 1.6 ⎟⎟ ⎝ t ⎠ ⎝ ⎠ Qr =

2 Q fWT Pr

=

2 × 3.14 × 3 × 4 1

= 75.36

Microstrip Antenna

471

Therefore Qt

Ie =

Qr

=

13.55 75.36

= 0.1798 = 17.98

Loss-tangent 1

Ee =

Qr

=

1 13.55

= 0.0133

Example 13.7 Find the conductivity of patch antenna operating at 3 GHz, if quality factor due to its conduction is found to be 30. Find power radiated if total power stored in the antenna is 80 mW. Also find total effective loss-tangent of the patch antenna if loss-tangent due to dielectric loss in the patch is 0.002. Solution:

We know that Ed =

T=

h Qt

=

0.127

1

QN f E

= 0.0042 cm

30

2 d

=

1 3.14 × 3.14 × 4 × 10

−7

× (0.0042)2 × 3 × 10 5

= 4.73 ´ 104 S/m and Pr =

2 Q fWT

=

Qr

2 × 3.14 × 3 × 80 30

= 50.25 mW

The total effective loss-tangent E t = tan E +

Ed

h

+

1 Qc

= 0.002 +

0.0042 0.127

+

1 30

= 0.0684

Example 13.8 Design an elliptical patch antenna on bakelite substrate to be operated at 3.5 GHz, if quality factor corresponding to circular polarization is 25.67. Solution:

We know that the semi-major axis of an elliptical patch is given: a=

pc fr F r

=

0.275 × 30 3.5 4.8r

= 1.076 cm

Therefore from a b

=1+1

0887 Q

=1+1

0887 25

67 = 1.024

472


⇒ b=

1.076 1.0424

= 1.032 cm

Therefore, ⎛ b2 ⎞ c = ⎜1 − 2 ⎟ ⎜ a ⎟⎠ ⎝

1/2

1/2

⎛ 1.032 2 ⎞ = ⎜1 − ⎟ ⎜ 1.9762 ⎟⎠ ⎝

= 0.28 cm

Feed point location ⎛a + b⎞ ⎛ 1.076 + 1.032 ⎞ 28 ⎟ = 2.018 cm u = u0 = ln ⎜ ⎟ = ln ⎜ 0 ⎝ c ⎠ ⎝ ⎠

on the radial line.

OBJECTIVE TYPE QUESTIONS 1. Which technique is not used to enhance the bandwidth of the microstrip antenna (a) Thick and high dielectric substrate (b) Cutting holes/slots (which increases inductance) (c) Adding reactive component (to reduce VSWR) (d) Increasing size of the patch. 2. Narrow bandwidth of patch antenna is useful particularly in (a) Wireless communication (b) Government security (c) Medical applications (d) None of these 3. Increasing height of substrate results in (a) Increase of efficiency and bandwidth (b) Only high bandwidth (c) Only high efficiency (d) Good radiation 4. Surface waves which degraded polarization of patch antenna generated because of (a) Increasing height of substrate (b) Using shorting pin (c) Stacking the antennas (d) None of these 5. Shorting pin loading technique is used to (a) Increase applications of antenna (b) Reduce the size of antenna (c) Improve the polarization (d) None of the above 6. Decreasing the quality factor of the MSA is a popular method of increasing (a) Bandwidth (b) Impedance (c) Radiation (d) None of these

Microstrip Antenna

473

7. MSA is a broadband antenna, however end-fire radiation can also be found by (a) Selecting suitable substrate (b) Selecting suitable mode of excitation (c) Selecting patch (d) None of the above 8. Microstrip antennas provide better performance if dielectric constant of substrate range between (a) 1.0 and 10.0 (b) 2.2 and 8 (c) 2.2 and 12.5 (d) > 13.5 9. The amount of radiation of MSA depends on the (a) Height of the substrate (b) Ratio of length (L) to substrate height (h) (c) Dielectric constant (d) Both (b) and (c) 10. The effective dielectric constant of substrate material lies between 1.0 and er and it is function of frequency as follows: (a) At low frequency it is constant (b) At height frequency it is constant (c) At all frequencies it is constant (d) None of the above 11. Which of the following feed technique offer larger bandwidth (a) Coaxial feed (b) Aperture coupled feed (c) Proximity coupled feed (d) Microstrip line feed 12. The fundamental mode of operation of a rectangular MSA is (a) TM10 and field vary along the length only (b) TE10 and fields vary along the length and width (c) TE11 and field vary along the width only (d) None of the above 13. The dominant mode of propagation of EM waves of RMSA is (a) Quasi-TEM (b) TEM (c) TE and TM both (d) None of these 14. What will be the conductance of RMSA of width 15th times lesser than the wavelength? (a) 1000 S (b) 10–6 S (c) 4.9 ´ 10–5 S (d) None of these 15. For an MSA designed on very thin substrate which is true statement (a) Feed reactance is comparable to the resonant resistance (b) Feed reactance is greater than the resonant resistance (c) Feed reactance is very small compare to the resonant resistance (d) None of the above

474


16. The scan blindness, in general, attributed to the mutual coupling between the array elements (patches). This scan blindness occurs for both the planes (E-plane and H-plane) at grazing angle of (a) 45° and 60° (b) 60° and 80° (c) 90° (d) None of the above 17. What will be the length of RMSA at f =1.6 GHz, if er is 2.2. (a) 6.25 cm (b) 6.87 cm (c) 7.5 cm (d) None of the above 18. What will be the length of CMSA at f = 1.6 GHz, if er is 2.2. (a) 8.5 cm (b) 3.89 cm (c) 7.5 cm (d) 3.589 cm

Answers 1. 6. 11. 16.

(d) (b) (c) (c)

2. 7. 12. 17.

(b) (b) (a) (a)

3. 8. 13. 18.

(a) (c) (a) (d)

4. (a) 9. (d) 14. (c)

5. (b) 10. (a) 15. (c)

EXERCISES 1. What is input impedance of patch antenna at its centre? 2. What are the degrees of RMSA and CMSA? 3. What is dominant mode of propagation of waves in circular wave-guide? 4. What is directivity of CMSA of small radius? 5. Describe the advantages of CP patch antennas. How can you achieve CP from an elliptical patch? 6. Write a note on the following: (a) CP patch antennas (b) Mobile antennas (c) Future scope of patch antennas. 7. Explain the concept of coupling between two identical patch antennas. How does it affect the performance of antenna system? 8. What do you mean by polarization bandwidth? Write the expressions for polarization bandwidth and radiation efficiency. 9. Describe the radiation mechanism of a patch antenna. Mention its advantages and applications. 10. What are the major disadvantages of a patch antenna? Describe methods to overcome them.

Microstrip Antenna

475

11. Describe feed techniques of patch antennas. How do they differ from each other? 12. Describe the design procedure of rectangular patch antenna with a suitable example. 13. Describe the design procedure of circular patch antenna with a suitable example. 14. Describe Tx line model of rectangular patch antenna. 15. Describe cavity model of circular patch antenna. 16. Fractional bandwidth of 8 GHz CP antenna is 5% for reflection coefficient 0.4. Estimating two resonant frequencies associated with the two lengths of the antenna determine the configuration and polarization characteristics of the antenna. 17. A microstrip antenna is fed by a microstrip line such that er = 8 and ratio of width to height is 5.4. Assuming (t/h) = 0, obtain ee and dimension of antenna if operating frequency is 3.5 GHz. 18. Design a rectangular patch antenna on substrate having er = 4.34 and h = 0.18 cm. The antenna is fed by a Tx line of impedance 50 W and to be operated at f = 2.5 GHz. 19. Design a circular patch antenna of radius 1.5 cm on dielectric substrate with e = 23.5 ´ 10–12 F/m. Calculate the minimum VSWR, if the quality factor of the antenna is 50. Assume your own design frequency. 20. Find bandwidth and radiation resistance of a RMSA operating at f = 2.0 GHz if L/W » 0.4l. Assume the substrate thickness is 1.8 mm. 21. Describe the design procedure of a CP elliptical patch antenna with a suitable example. Also explain its polarization characteristics with reference to axial-ratio. 22. Design a 3-GHz elliptical patch antenna on substrate with dielectric constant 2.65 such that effective dielectric constant is 0.5% greater than dielectric constant. And operating frequency is 3.5 GHz if quality factor corresponding circular polarization is 25.67.

REFERENCES [1] Deschamps, G.A., “Microstrip microwave antennas”, 3rd USAF Symp. on Antennas, 1953. [2] Howell, J.Q., “Microstrip antennas,” IEEE, Trans. on Antennas and Propagate, pp. 90–93. Jan 1975. [3] Carver, K.R. and J.W. Mink, “Microstrip antenna technology,” IEEE, Trans, Antennas Propagate, AP-29(1), pp. 02–24, Jan. 1981. [4] Shen, L.C., “Analysis of a circular disc printed circuit antenna,” Proc., Inst Elect. Engs., Vol. 126, pp. 1220–1222, 1979. [5] Yadava, R.L., “Studies on some microstrip radiating structures for circular polarization,” Ph.D. Thesis, IT-BHU, Varanasi, India, 2001.

476


[6] Balanis, C.A., Antenna Theory: Analysis and Design, 2nd ed., John Wiley & Sons, India, 1997. [7] Bahl, I.J. and P. Bhatia, Microstrip Antennas, Artech House, 1980. [8] James, J.R. and P.S. Hall, Handbook of Microstrip Antennas, Vols. 1 and 2, Peter Peregrinus, London, 1989. [9] Deshpande, M.D. and M.C. Pandey, “Input impedance of microstrip antenna,” IEEE, Trans, Antennas Propagate, AP. 30(4), pp. 645–650, July 1982. [10] Balanis, C.A., Advanced Engineering Electromagnetic, John Wiley & Sons, New York, 1989. [11] Milligan, T.A., Modern Antenna Design, McGraw-Hills, Inc., USA, 1996. [12] Richards, W.F., Microstrip Antennas, Van Nostrand Reinhold C., New York, 1988.

C H A P T E R

14

Surface Wave Propagation

INTRODUCTION The concept of electromagnetic (EM) waves came from the remarkable discovery of Faraday and Maxwell that time varying electric fields gave rise to wave travelling in free space with a velocity of 3 ´ 108 m/s. The main advantages of using EM wave are that it needs natural medium to travel from transmitter to receiver, which requires no maintenance. Therefore, the free space is considered to be the best medium for EM wave propagation. The free space is portion of atmosphere far away from the earth, which does not interfere with normal radiation and wave propagation. This is because no fields as well as solid or ionized particles exist in space. It is an important fact that only small fraction of the power radiated by an antenna reaches to the receiving antenna, the remaining is dissipated in the medium. And ratio of power radiated to power received is known as transmission loss, which is usually of the order of 150–200 dB. The EM waves ranging from few KHz to MHz in the EM spectrum are referred as radio waves. When the radio waves radiate from the transmitter it spreads in all directions decreasing in amplitude with increasing distance, which is termed propagation of waves and it follows the inverse square law. This law states that the intensity of radiating waves is inversely proportional to the square of distance from the radiator (E µ r–2) [1], where the electric field is measured in terms of V/m. The following expression relates the power density in any electric system to the electric field intensity; P=

E2 R

where P = power in watts (W) R = resistance (W) E = electric field (V/m) 477

(14.1)

478


However, in the case of radio wave propagation, the resistance (R) is nothing but the free space impedance, that is, 377 W. Therefore, for example, if the field intensity for any medium is 20 V/m, then the power density of the signal will be P = 53 mW/m2.

20 2 V/m 377 :

=

HISTORICAL VIEW Around 19th centeury, mathematician A.N. Sommerfield was the first who verified the observation of a radio phenomenon. In his system he claimed that 90% to 95% of the electric energy is manifested at the transmitters output as current waves with the reminder existing as dissipating EM radiation. Later in 1907, Johann Zenneck, while working to explain Marconi’s trans-oceanic results showed that a unique type of surface wave could travel along the interface between the ground and the air. This is so called Zenneck Wave. James R. Wait commented on Sommerfield’s analysis of surface wave and stated that The field amplitude varies inversely as the square root of the horizontal distance from the source. Sommerfield also distinguished the electrodynamics surface wave and its horizon counterparts the space wave [2]. While citing the analysis of Sommerfield surface wave model, Nikola Tesla asserted that the exact composition of the surface wave transmission depends on the design of the transmitter. It has also been noticed that the phenomenon of surface wave is closely related to that of creeping waves and travelling waves in electromagnetic scattering theory. As the study of radio-propagation progressed, certain modifications were included in it and new results were achieved. In 1937, experimental tests showed that the simple antenna driven at 150 MHz produces 100 times lower field strength than predicted and it has been proposed that Zenneck waves can indeed be generated. The lower the frequency, the lower the propagation losses. The variation of field strength with frequency indicates that a Zenneck wave propagates best at frequency up to 35 kHz and would lose its advantage as frequency increases further. The complex longitudinal propagation phase constant along the earth’s surface for the Zenneck surface wave is given by [3] ⎤ 2Q ⎡ F gr − j 60 M0T Cz = ⎢ ⎥ M0 ⎣⎢ (F gr + 1) − j 60 M0 T ⎥⎦

1/ 2

(14.2)

where l = free space wavelength egr = ground dielectric constant s = conductivity of the earth ÿ

It was also noticed that the phenomenon of surface wave is closely related to that of creeping waves and travelling waves in electromagnetic scattering theory.


479

CHARACTERISTICS OF EM WAVES The EM waves are transverse electromagnetic (TEM) waves, i.e., they contain both electric and magnetic fields in such a way that they are perpendicular to each other and also perpendicular to the direction of propagation. An EM wave maintains its properties even after radiation from the antenna, hence causing polarization. The vector representation of an EM wave is shown in Fig 14.1.

FIG. 14.1

Electromagnetic wave propagation.

The initial polarization is determined by the orientation of the antenna, that is, if the antenna is vertical it produces vertical polarization and similarly horizontal polarization. The frequency of an EM wave is nothing but the number of EM waves that pass through a given point in one second. When the wavelength (distance between two sequent peaks) of EM wave is 0.5 micron, the wave can be detected by our eyes and called light. The characteristics of EM waves are same as light. The ratio of electric to magnetic fields is termed the intrinsic impedance. The value of which is 377 W for free space [4–5]. The power density (the rate of flow of EM energy per unit area) is expressed in terms of Poynting vector and it is equal to the cross product of E and H. Total energy of EM waves is equally distributed between electric and magnetic field vectors and they are 1/2 eE2 and 1/2 mH2, respectively. Therefore, the total energy of EM waves is given by ET =1/2 eE2 + 1/2 mH2. The frequency range of EM waves between 300 MHz at low end and 300 GHz at the high end of electromagnetic spectrum (Fig. 14.2) is considered as the microwave band. The special features such as high frequency of microwaves provide wide bandwidth capability and make possible the following systems. Communication system · Microwave relay · Troposcatter communication

480

5 10

6 10

10

7

Wavelength about 3 football fields long

10

8

9 10

10 10

Wavelength about 3 m or 10 feet long

FIG. 14.2

11 10

12 10

Wavelength about 3 cm or 1 inch long

13 10

14 10

15 10

16 10

Wavelength 400–700 nm

X-rays Gamma rays

Ultraviolet

Visible light

Infrared

Millimet re waves, telemetry

Microwaves radar

Television FM radio

Shortwave radio

AM Radio


17 10

18 10

Hz

Wavelength about 30 x diameter of hydrogen atom

Electromagnetic wave spectrum.

· Satellite communication · Mobile radio communication · Telemetry Radars · Search and velocity measuring · Air traffic control and fire control · Navigations and tracking Electronics welfare jammers Microwave heating:

Microwave ovens and microwaves therapy for cancer treatment.

APPLICATIONS OF EM WAVES Depending upon the frequency, EM waves produce different effects in various materials and devices. Therefore, the different parts of EM spectrum have importance for different purposes. Frequencies below 30 MHz are used for broadcasting and world radio communication, however EM waves between 30 MHz to 300 MHz are found very suitable in many other applications. Waves of frequency range 3 ´ 103 GHz and 3 ´ 105 GHz (infrared) falling on our skin produce sensation of heat. Waves of wavelength 720 mm to 40 mm produce a sensation of colours ranging from violet to red. The waves from 400 nm to 300 nm wavelength range are referred to as ultraviolet radiation. However, waves of wavelength below 10–5 nm are the cosmic rays arriving on the earth’s surface from the entire universe. The waves of frequency 3 ´ 1014 Hz is used for fibre optical communication systems. However, the entire applications of microwave bands can be described as follows:


481

The 450 MHz is very useful for long distance radar search and frequency range 470–870 MHz for UHF TV, whereas cellular phones operate at 900 MHz. Just above 1 GHz, air traffic control transponder operates and allows aircraft to repeat identification code to air traffic control radar. The telemetry system transmits data at frequency less than 2 GHz and provides satisfactory performances. The troposcatter communication systems operate above 2 GHz, where microwave signals are scattered off of the troposphere and achieve long distance communication [6]. The frequency 2.45 GHz is multipurpose frequency and used for microwave ovens, RFID, medical therapy and also in communication systems. The frequency just higher than 3 GHz is very suitable for airport search radar and just below 4 GHz is used for point-topoint microwave rely, which carries thousands of telephone channels and television programmes across the world. The 6 GHz and 4 GHz are well-known frequencies for satellite up- and downlink systems. The studio transmitter link transmits radio and television programmes from the down link studio to the transmitter site at frequency slightly greater than the 7 GHz. Airborne fire control radar operates at 10 GHz and just above that there is another microwave rely band used for telephone transmission. There is another satellite communication band with 20 GHz (downlink) and 30 GHz (uplink) frequencies. A public radar is found comfortably operating at 10.25 GHz and 24 GHz and missile seeker radar performs well at a frequency of 94 GHz.

FUNDAMENTAL EQUATION OF WAVE PROPAGATION As mentioned in the introduction to this chapter, only a small fraction of radiated power is received at the receiver from an isotropic radiator in free space. The received signals, however, must be 10–20 dB above the receiver noise to complete the link between Tx and Rx. The amount of received power depends mainly on transmitted power, gains of Tx and Rx antennas and separation between them. In addition, operating frequency and path attenuation also affect the amount of received power. Therefore, in order to describe the characteristics of wave propagation, it is necessary to derive equation relating to these parameters. The expression containing these parameters is known as Friis free-space wave equation. Let us consider an isotropic radiator transmitting power in free-space, so the medium surrounding is homogenous and non-absorbing of dielectric constant unity. The power density at distance R from the centre of the radiator will be Pd =

where Pd Pt R 4 pR 2

= = = =

Pt 4Q R 2

(14.3)

power density (W/m2) transmitted power (W) distance between transmitter and receiver (km) spherical surface area (m2)

If a directional antenna is used at receiver, the receiving power density is increased by the multiple of gain of the transmitting antenna, i.e.,

482


Pd =

Pt Gt

(14.4)

4Q R 2

where Gt = maximum directive gain of the Tx antenna and 2

⎛D⎞ = 6 ⎜ ⎟ in the case of microwave dish antenna. ⎝M⎠ in which D is larger aperture of antenna and l is operating wavelength. If a is path loss (attenuation) of the medium the power density is modified to Pd =

Pt Gt

4Q R 2

B

(14.5)

Since the transmitted power is spread over a spherical area of many miles, the receiving antenna picks up only a small fraction of the radiated power. The amount of power at the Rx antenna will be area of the receiving antenna (A) times the power density at the Tx antenna Pd =

Pt Gt A 4Q R 2

Since the gain of receiving antenna is Gr =

4Q A

M2

B

(14.6)

, then A =

Gr M 2 4Q

.

Hence the power received at the receiver will be Pr = Pt Gt Gr B

M2

(14.7)

(4Q R 2 )

In Eq. (14.7), all the parameters are known or can be determined easily except path attenuation a. The path attenuation depends upon atmospheric conditions that vary with time and local weather, i.e. E, it can be estimated only. However the path attenuation a can be approximated by any value for corresponding frequency from Table 14.1. TABLE 14.1

The value of a for different frequencies

S. No.

Frequency (GHz)

a (in dB/km)

1 2 3 4 5 6 7

10 20 30 50 100 150 200

0.01 0.02 0.05 0.10 1.00 5.00 25–30


483

Equation (14.7) can also be written as Pr (dBw) = Pt (dBw) + Gt (dB) + Gr (dB) + a (dB) – Lp (dB)

(14.8)

which is known as fundamental equation of free space propagation or Friis free space wave equation. In this equation, the term Lp = 4pR2/l2 is known as free space path loss. This may be defined as the ratio of antenna area one wavelength square to area over which the transmitted power has been spread.

Lp =

Again

4Q R 2

M2

=

4Q R 2 f 2

(14.9)

c2

where c is velocity and f is frequency of the propagating waves. Therefore, the free space path loss directly depends on the frequency of the propagating waves and Tx to Rx separation. If c and R are expressed in and f in MHz, the path loss can be given by Lp = 32.45 + 20 (log10 R + log10 f) (dB)

(14.10)

whereas if R is in miles and f in GHz, the path loss can be expressed by Lp = 97 + 20 (log10 R + log10 f) (dB)

(14.11)

However, especially at microwave frequencies loss of approximately 20 to 30 dB can be observed due to cross-polarization of Tx and Rx antennas.

ELECTRIC FIELD INTENSITY AT FINITE DISTANCE FROM Tx ANTENNA In a particular case, if a horizontal Hertzian dipole antenna is used as a Tx above the horizon, then energy will travel like a wave in free space. Therefore, the amplitude of electric field vector in the radiation field can be given as [1] ER = 60 Q

where r = Imax = dl = l=

I max dl rM

sin R =

60 ⎡ Q I max dl ⎤ ⎢ ⎥ r ⎣ M ⎦

as q = 90°

(14.12)

far-field distance maximum current in the antenna length of the dipole operating wavelength

Also, the power radiated by the dipole is given by ⎡ Q I dl ⎤ Pt = 80 ⎢ rms ⎥ ⎣ M ⎦

Taking I rms =

I max 2

, Eqs. (14.12) and (14.13) yield

2

(14.13)

484


2

⎡ Q I dl ⎤ Pt = 80 ⎢ max ⎥ = 80 ⎣ 2M ⎦

That is

ER =

60 2 r

⎡ ER r ⎤ ⎢ ⎥ ⎣ 60 2 ⎦

2

Pt /80 V/m

For example, at any instant Pt = 1 kW and distance from the Tx r is 1 km (i.e. receiver), then

ER =

60 2 1000

1000/80 V/m = 300 mV/m

However, in general

⎛ mV ⎞ ER ⎜ ⎟ = 173 ⎝ m ⎠

Pt (kW) r (km)

(14.14)

MODES OF WAVE PROPAGATION Basically, there are four major modes that the waves transmitted from a Tx may follow to reach the destination and they are surface wave, space wave, troposphere and ionosphere propagations. Propagating waves follow a particular path which depends on transmitting frequency, distance between Tx and Rx earth’s constants as well as other atmospheric conditions. The first two propagation modes are grouped into ground wave propagation, but they behave differently enough to warrant separate consideration. The surface waves travel in direct contact with earth’s surface and, therefore, suffer a severe frequency dependent attenuation caused by absorption into ground. On the other hand, no part of space wave normally travels in contact with the surface of earth, as it is radiated from an antenna many wavelengths above the ground. The VHF, UHF and microwave signals usually follow space wave propagation modes. The tropospheric waves are lumped with the direct waves up to some extent and troposphere region extends between the earth surface and the stratosphere. Most forms of ground waves take tropospheric paths. The waves of frequencies between 2 MHz to 30 MHz follow ionosphere modes. The ionosphere acts like a reflecting surface to these waves, completing the propagation at any arbitrary point on the earth surface. The peculiar feature of the ionosphere is that molecules of air gases (O2 and N) can be ionized stripping away the electrons under the influence of solar radiation and certain other sources of energy. The properties of the ionosphere are important to microwave technology because of the noise contribution [7].


485

Surface Wave Propagation These types of propagation take place along the curvature of the earth surface (Fig. 14.3), like EM waves are guided by wave-guide or any other guiding structures. The surface waves follow the contours of the earth because of the process of diffraction.

FIG. 14.3

Surface wave propagation.

When the surface wave meets an object and the dimensions of the object do exceeds its l, the wave tends to curve or bend around the object. The smaller the object, the more pronounced the diffraction action would be. The surface wave propagation is also known as Norton’s wave propagation, and is of practical importance up to 2 MHz frequency. Basically, it is the main part of ground wave propagation which is grouped into surface, space, and troposphere wave propagation. The surface wave suffers additional ground attenuation to the same factor as the free space. These ground losses are caused by the ohmic resistive losses in the conductive earth. Surface wave attenuation is frequency dependent and it increases rapidly as frequency increases. In addition, the surface wave propagation is affected by the following factors: heights as well as distance of/between Tx and Rx antennas and the terrain and weather conditions along the transmission paths. Usually, surface waves are produced by the vertical antennas (Tx and Rx) that are closer to the earth surface, hence waves are vertically polarized. The electric field components of the waves are vertical with respect to ground, while the horizontal components are shortcircuited by the ground. While propagating along the surface of earth, waves induce charges in the earth, which travel with the wave and, hence, constitute a current (Fig. 14.4). While carrying this current, earth behaves like a leaky capacitor and, therefore, it can be represented as s resistance in shunt arm with a capacitor. This reveals that earth behaves as a conductor and it may be described in terms of conductivity and dielectric constant. As earth attenuation increases with frequency, ground attenuation is limited for low and medium frequency (upto 2 MHz) only. The attenuation of a surface wave also depends on

486

Antenna and Wave Propagation Signals spreading out from the transmitter

Transmitter antenna

Wave fronts angled downwards allowing them to follow the earth’s surface

FIG. 14.4

Vertical electric fields of surface wave propagation.

the electrical properties of terrain over which the wave travels. Better the conductivity, the lesser the attenuation. All the broadcast signals received during daytime are due to surface wave propagation.

Surface Wave Tilt Tilting is another means of attenuating surface waves, in addition to ground attenuation and diffraction. Tilting of surface waves gradually increases as the wave propagates along the curvature of earth [4]. This increase in the tilt of wave causes more short circuits of the electric field components and hence the field strength goes on reducing, as a result surface wave dies at the certain distance. However, this distance (range of propagation) can be increased by increasing the transmitted power at particular frequency. Because of the ground loss effects, the surface wave is attenuated at much faster rate than the inverse square law. The phenomenon of wave tilting in successive wave front is shown in Fig. 14.5. W1

Tx

T1

W3

W2

T2

W4 T3

T4 T5

Rx W5

Earth curvature

where T1, T2, T3, T4 and T5 are the tilting angles in increasing order. W1, W2, W3, W4 and W5 are successive wave fronts.

FIG. 14.5

Tilting wave fronts in surface wave propagation.

487


For a vertically polarized wave at the surface of the earth, the amount of tilt depends upon earth constants; conductivity (s) and dielectric constant (er). The slight tilt forward of the electric G field G strength is responsible for a small vertically down component of the Poynting vector E × H , significant to furnish the power loss in the earth over which the wave is travelling. The angle of tilt (T) with respect to the normal is given by [7] T = [F r2 + (60 MT )2 ]

(14.15)

where all the parameters have usual meaning. The tilt angles as a function of frequency are tabulated in Table 14.2. TABLE 14.2

Tilt angles as a different function of frequencies

S. No.

Frequency (KHz)

Earth surface (deg)

Sea water (deg)

1 2 3 4

20 200 20,00 20,000

4.3 13.4 32.3 35

2.08 0.13 0.5 1.38

PLANE EARTH REFLECTION In a particular case, if an EM wave is incident at an angle f unto a finite conductivity smooth earth surface, the magnitude and phase of the reflected wave can be determined considering as the reflection at the surface of a perfect dielectric. In another case, when the earth is rough, the reflected wave tends to be scattered and reduction in amplitude comparably higher than the reflection from smooth-earth surface. The famous Rayleigh criterion gives a measure of earth roughness as follows [8]

R=

4QT sin G

M

(14.16)

where R = earth surface roughness ÿ ÿ s = standard deviation of the surface irregularity relative to the mean surface ÿ ÿ fÿ = angle of incidence measured from the grazing angle ÿ ÿ lÿ = operating wavelength Typically, for R < 0.1, there is a well defined specular reflection and the corresponding reflecting surface may be considered as being smooth. However, for R > 10, the surface is definitely rough and the reflected wave has a small magnitude. A surface which is considered rough for high incidence angle (i.e. large f), may approach a smooth surface as the angle of incidence f approaches grazing. And when the incident wave is near grazing over a smooth earth the reflection coefficients approaches –1 for both, horizontal and vertical polarizations. It is known that the earth is not a good conductor as silver and copper. Therefore, in order to describe reflection from the earth surface, its finite conductivity and

488


dielectric constant must be taken into account. Analysis of Maxwell’s equation for a medium, which has a finite conductivity (s) and dielectric constant (er), reveals that there is another term known as equivalent complex dielectric constant (e ¢) and defined as

T ⎞ ⎛ F ′ = ⎜F + ⎟ jX ⎠ ⎝

(14.17)

Therefore, the characteristics of wave propagation in different mediums (in terms of propagation constant g) can be determined by replacing the dielectric constant (e) with complex dielectric constant (e) (Eq. 14.13). For an isotropic radiator above the plane earth, the electric field intensity E can be given as [9] E = E0{1 + Rhe jD + (1 – Rh) Fe jD + ...}

(14.18)

in which Rh = reflection coefficient E0 = field strength for free space propagation

Reflection Coefficient for Vertical and Horizontal Polarizations For a vertically polarized wave incident on the surface of a perfectly smooth earth, the reflection coefficient (Rh) is given by Rv =

Er Ei

=

F rt cos R −

F rt − sin R

F rt cos R +

F rt − sin R

2

2

(14.19)

where Ei = incident field intensity Er = reflected field intensity ert = ratio of e2 and e1 and the dielectric constants of mediums 1 and 2 respectively q = angle of incidence from the normal. However, in case of plane earth reflection, the direction of the incident wave is measured from the earth surface ÿÿ

That is, y = 90° – q, so that cos q = sin y and sin q = cos y. Also if the wave is incident from air onto the surface of earth, e1 = e0 and e2 = e¢ =

T ⎞ ⎛ T⎞ ⎛ . Therefore, Eq. (14.15) may be written as ⎜F + ⎟ = ⎜F − X ⎟⎠ jX ⎠ ⎝ ⎝ Rv =

where F r =

Fr F0

Er Ei

=

(F r − jx ) sin Z − {(F r − jx ) − cos2Z }1/2 (F r − jx ) sin Z + {(F r − jx ) − cos2Z }1/2

= Relative dielectric constant and x =

T 18 × 10 9T . = fMHz XF 0

(14.20)


489

Similarly, for horizontally polarized fields, the reflection coefficient can be obtained from Rh =

Er Ei

=

cosR −

F rt − sin 2R

cosR +

F rt − sin 2R

(14.21)

Therefore, under previous assumptions, Eq. (14.20) could be re-written as Rv =

Er Ei

=

sin Z − {(F r − jx ) − cos2Z }1/2 sin Z + {(F r − jx ) − cos2Z }1/2

(14.22)

From Eqs. (14.21) and (14.22), it is clear that the reflection coefficient is a complex quantity and reflected wave will differ both in magnitude and phase from the incident wave. There are some additional results of reflection factors Rv and Rh. 1. For the horizontally polarized waves, the phase of reflected wave differs from that of the incident wave by nearly p degrees for all angles of incidence. 2. When the angle of incidence is near grazing (y = 0°), the reflected wave is equal in magnitude but p° out of phase with the incident wave for all frequencies and conductivities. 3. Both the magnitude and phase of reflected waves change with the angle of incidence and this change is greater for the higher frequencies and lower conductivities. 4. For vertically polarized wave incidence at grazing angle the electric vector of the reflected wave is equal to that of the incident wave and also 180° phase reversal for all finite conductivities. However, the magnitude and phase of the reflected wave decreases rapidly as the incidence angle increases from grazing. The angle, at which the magnitude approaches minimum and phase, goes through – 90° is known as the Pseuo-Brewster Angle. 5. At the angles of incidence greater than Pseuo-Brewster Angle the magnitude of the reflected wave increases and the phase approaches zero. 6. When y = 90°, i.e., wave is incident perpendicular to the reflecting surface, there is no difference between vertical and horizontal polarizations, and the reflection coefficients Rv and Rh have equal values with 180° phase difference.

REFRACTION AND REFLECTION OF WAVES In general, if a wave strikes from one medium to another medium having different characteristics, three possibilities can be assumed: (i) Some of the energy can be transmitted into the second medium where it may continue at a different velocity. (ii) Some of the energy can be reflected back into the first medium. (iii) Some of the energy can be absorbed by the medium. In some particular cases, all three phenomena—reflection, transmission, and absorption— may also occur to some degree.

490


Refraction When a wave passes from one medium into another medium that has a different velocity of propagation, change in the direction of the wave occurs. The process of changing of direction as the wave enters in the second medium is called refraction. The phenomenon of refraction waves is shown in Fig. 14.6, where a radio wave with constant speed is travelling through the earth’s atmosphere and return to the earth.

Maximum electron density

FIG. 14.6

Refraction of waves.

As the waves enter the dense layer of electrically charged ions, the part of wave that enters the new medium first travels faster than the part of waves that have not yet entered the new medium. This abrupt increase in the velocity of upper part of wave causes the wave to bend back toward the earth. This bending of waves is always toward the medium that has the lower velocity of propagation. Radio waves passing through the atmosphere are affected by certain factors such as temperature, pressure, humidity, and density. These factors can cause the radio waves to be refracted more and more.

Reflection Reflected waves are simply waves that are neither transmitted nor absorbed from the object of the medium on which they strike during propagation from transmitter to receiver. If the reflecting object is smooth and polished, the angle between the incident ray and the normal will be the same as the angle between the reflected ray and the normal. This conforms to the law of reflection which states “The angle of incidence is equal to the angle of reflection”. The amount of incident wave energy reflected from a given surface depends on the nature of the surface, wavelength as well as the angle of incidence. As the angle of incidence


491

increases, the amount of wave energy reflected increases. The reflected energy is the greatest when the wave is nearly parallel to the reflecting surface while it is least, when the incident wave is perpendicular to the surface. In particular, when an EM wave propagates into a conductive medium, the majority of it is reflected. The small amount which does penetrate the surface is quickly absorbed by and being converted into heat by the resistance of the medium. The attenuation occurs at an exponential rate that is for every millimetre deeper into the medium the wave’s power is reduced by an additional power of some constant attenuation factor. Attenuation can also occur in media which are not conductive. Pure water, for example, is not a good conductor (although even slightly impure water can conduct well). Nonetheless, raindrops, which are pure, do absorb and attenuate EM waves. This is due to losses which occur when the electron clouds around the water molecules are distorted by the E field in the EM wave. This results in a complex dielectric constant. This is the basis for cooking with a microwave oven. The water in the food absorbs a large amount of the incident EM radiation, turning it into heat. When radio waves are reflected from flat surfaces, phase shifts in the alternations of waves are observed. Reflection of two waves from the earth’s surface is shown in Fig. 14.7.

A ed ct es le av f Re io w d ra

B

Earth’s surface

FIG. 14.7

Phenomenon of reflection of wave.

In which after reflection, the waves are found to be approximately 180° out of phase from their original relationship. In general, amount of phase-shift that occurs in reflection is not constant. However, it varies with polarization of incident waves and angle of incidence. Radio waves that maintain their phase relationship after reflection normally produces a stronger signal at the receiving end. While the received signals which are out of phase produce a weak or fading signal. The changes in phase between incident and reflected signals are one of the major reasons for fading in propagation.

492


Phase Difference between the Direct and the Ground Reflected Waves In order to find the phase difference between direct and indirect waves, let us refer to Fig. 14.8, where ht and hr are the heights of Tx and Rx antennas spaced away at d distance from each other and R is the reflection coefficient.

FIG. 14.8

Geometry for direct and reflected waves.

The phase difference between the said waves can be found from two rays approximation. Considering only a line of sight and a ground reflection, it can be expressed as (from Fig. 14.6) D = R2 – R1 where R1 can be found from DPQR and R1 = {d 2 + (ht2 − hr2 )} . In order to calculate R2, let us consider reflected rays from the earth surface is coming from virtual point P1, Then PQ1 = P1Q1 = ht. Therefore, from DP1QR, R2 = {d 2 + (ht2 − hr2 )} . Hence phase-difference

'=

2Q

M

= ⎡ {d 2 + (ht2 + hr2 )} − ⎣⎢

{d 2 + (ht2 − hr2 )} ⎤ ⎦⎥

(14.23)

Equation (14.23) is true for limited propagation range only. For large propagation ranges, i.e. for large d we can modify this equation based on

1+x ≈ 1+

x 2

and the result offers most accurate phase difference ' ≈

4Q ⎛ ht hr ⎞ ⎜ ⎟. M ⎝ d ⎠


493

This phase difference (D) and an approximated reflection coefficient give an often-used two-ray model for plane-earth propagation.

FIELD STRENGTH AT FINITE DISTANCE DUE TO GROUND WAVE The simplest formula to determine field strength due to surface wave was first proposed by Maxwell. According to Maxwell’s field strength at distance (d) from the transmitting antenna at operating wavelength l is given by

E=

I ht hr I s Md

(14.24)

where h = intrinsic impedance of the medium and it is 120p for free space ht, hr = effective heights of Tx and Rx antennas Is = antenna currents The examination of above equation reveals that there is no term involved indicates the reduction of field strength due to ground attenuation and atmospheric absorption. However, in practice it is observed that field strength at receiver decreases because of different types of losses present in the propagating medium. Hence, actual field strength at the receiving antenna is less than the given by Maxwell’s Eq. (14.24), as a result this equation fails to determine the field strength for the fairly large distance propagation. In order to overcome the deficiency of Maxwell’s method, Sommerfield proposed a new model by including the factors representing the decay of field strength. He grouped surface and space propagations into a basic ground wave propagation and described that the surface wave predominated at large near distance to the earth, whereas the space wave at larger distance above the earth. In order to have a clear separation between surface and space waves, an electric dipole normal to the earth surface was used as transmitting antenna. In principle the electric field intensity at larger distance from the transmitter (vertical dipole in present case) can be obtained by neglecting higher orders of 1/R1 and 1/R2. As a result, the expressions for the components of E, above a finite conducting plane reduces to [8] ⎡ Ez = j 30 C I dl ⎢cos2Z ⎢⎣

⎛ e− j C R1 e − j C R2 + Rv ⎜ ⎜ R1 R2 ⎝

⎞ e − j C R2 ⎟ + (1 − Rv )(1 − u2 ) F ⎟ R2 ⎠

⎡ ⎛ e− j C R1 e− j C R2 ⎢sin Z cos Z ⎜ + Rv ⎜ R1 R2 ⎢ ⎝ E S = − j 30 C I dl ⎢ ⎢ sin 2 Z ⎞ e− j C R2 ⎛ ⎢× F ⎜1 + ⎟ 2 ⎟⎠ R2 ⎜⎝ ⎢⎣

⎤ ⎥ ⎥⎦

(14.25)

⎤ ⎞ ⎟ − cos Z (1 − Rv ) u 1 − u2 cos2Z ⎥ ⎟ ⎥ ⎠ ⎥ ⎥ ⎥ ⎥⎦ (14.26)

494


where Ez and Er Rv R1 and R2 I dl u2

= components of E along z and r directions respectively. = plane wave reflection coefficient for the vertically polarized waves. = distance from the Tx and its image to observation point P respectively = antenna current = length of dipole and = 1/er – jx other parameters have their usual meaning

From Eqs. (14.25 and 14.26), it is clear that total electric field (Ez, Er) consists of two parts. The first part contains inverse-distance term and free from attenuation factor F and the second term contains the additional attenuation factor F. The first term is refereed to as a space wave whereas second term referred to as surface wave. Combining Eqs. (14.25 and 14.26) and separating into for these two types of waves gives

(ET )space = [ Ez2 + E S2 ] 1/2 space = j 30C I dl ⎡⎣cos4 Z + sin 2Z cos2Z ⎤⎦

⎛ e− j C R1 e − j C R2 + Rv ⎜ ⎜ R1 R2 ⎝

⎛ e− j C R1 e − j C R2 = j 30C I dl cos Z ⎜ + Rv ⎜ R1 R2 ⎝

⎞ ⎟ ⎟ ⎠

1/2

or

⎞ ⎟ ⎟ ⎠

(14.27)

Similarly, 1/2

(ET )surface = ⎡⎣ Ez2 + E S2 ⎤⎦ surface = j 30 C I dl (1

− Rv ) F

⎛ e − j C R2 ⎡ sin 2Z 2 2 Z − ⎢1 2u + (u cos ) ⎜ 1 + ⎜ R2 2 ⎢⎣ ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(14.28)

(by descending u2)

Field Strength due to Vertically Polarized Wave When the vertical dipole is at the surface of the earth, the total electric field (given in Eq. 14.28) due to surface wave part gets reduced to (ET )surface = j 30 C I dl (1 − Rv ) F

e− j C R ⎡ ⎢ pˆ (1 − u2 ) + nˆ cos Z R ⎢⎣

⎛ sin 2 Z ⎜1 + ⎜ 2 ⎝

⎞ ⎟ u 1 − u2 cos2 Z ⎟ ⎠

⎤ ⎥ ⎥⎦

(from Eqs. 14.27 and 14.28) in which, pˆ and nˆ are the unit vectors parallel and perpendicular to the dipole antenna. R is the distance between Tx and point of observation and R >> l.


495

The attenuation factor F is defined as

F = ⎡⎣1 − j QX e−X {erfc (j X )}⎤⎦

⎡ ⎧ 2 F = ⎢1 − j QX e−X ⎨ ⎩ X ⎣

or

∫

2 ⎫⎤ e− v dv ⎬⎥ jX ⎭⎦

∞

(14.29)

In Eq. (14.29), w is defined as

X=

⎤ − j C Ru2 (1 − u2 cos2 Z ) ⎡ sin Z ⎢1 + ⎥ 2 2 2 ⎢ ⎥ − u (1 u cos ) Z ⎣ ⎦

(14.30)

It is seen that the attenuation factor F is complicated function of distance, frequency, dielectric constant and conductivity of earth along which waves propagate. The value of F is near to unity for the distance R within a few l, however, the value of F approaches unity as distance approaches zero. For the waves incident at grazing angle (yÿ = 0), the absolute value of F is referred to as ground wave attenuation and represented by A. That is A = |F| or A = [1 − j QX e −X {erfc ( j X )}]

Z =0

A = [1 − j Q p1 e−X {erfc ( j p1 )}] where p1 = X Z =0 , and it is complex quantity. Hence, p1 = pejb (say)

(14.31)

where p and b are new constants known as numerical distance and phase constant, respectively. These two constants are determined in terms of frequency, distance and earth’s constants. Evaluating w for y = 0, using Eq. 14.30 and comparing the result with Eq. 14.31 yields

pv =

and

with

QR cos b Mx

⎛ F + 1⎞ b = 2b2 − b1 = tan −1 ⎜ r ⎟ ⎝ x ⎠ ⎛F ⎞ b2 = tan −1 ⎜ r ⎟ ⎝ x ⎠ ⎛ F − 1⎞ b1 = tan −1 ⎜ r ⎟ ⎝ x ⎠

(14.32a)

(14.32b)

(14.32c)

(14.32d)

496


Field Strength due to Horizontally Polarized Waves When a horizontally polarized wave is incident at the surface of the earth of finite conductivity, a complex form of electric field expression is obtained [5]. The field expression indicates that waves are horizontally and vertically polarized in the f = 90° and f = 0° directions respectively. For any other values of f between 0° and 90°, the electric fields (waves) are elliptically polarized. The attenuation of the propagating wave in this case can also be determined in terms of numerical distance and phase constant. This is done using the same ground attenuation A as it was used for vertically polarized wave. However, the relation between numerical distance p and phase constant b is slightly different from the previous case. They are related as follows:

QR x M cos b1

(14.33a)

in which

b = 180° – b1

(14.33b)

and

⎛ F − 1⎞ b1 = tan −1 ⎜ r ⎟ ⎝ x ⎠

(14.33c)

ph =

In the above equation l is operating wavelength and s conductivity of the earth (mhos/cm). The value of p for any finite distance R is different in vertical and horizontal polarizations. The numerical distance p will be greater for horizontal polarization than the vertical polarization, hence greater attenuated waves. This is due to the involvement of x in the expression of pv and ph. That is why, at low and medium frequencies ( ≅ 2 MHz) only vertically polarized surface wave polarization is possible, through vertical dipole antennas. However, at higher frequencies, attenuation of surface waves is very high for both the polarizations; as a result, surface wave propagation is limited to very short ranges. In these frequency ranges propagation is possible only by space waves using the elevated antennas.

Relation between A, p and b Actually, the value of numerical distance p (i.e. pv and ph) depends upon the frequency of wave, conductivity and dielectric constant of the earth and also distance of the observation point from the radiator. It varies with these parameters as follows [10]: p µ R ∝

f

∝

1

Te

However, the phase-constant b is a measure of the power factor angle of the earth (the actual


497

power factor angle is b2). In general, the numerical constant p for all values of b is given by following empirical formula [7]: A=

2 + 0.3p 2 + p + 0.6p2

− sin(b)

p 2

e

−

5 8

p

(14.34)

However, in particular cases · For b < 5°, this equation gets reduced to

A=

and

p=

2 + 0.3 p 2 + p + 0.6 p2 2 0.582 d km f MHz

T (mS/m)

(14.35a)

(14.35b)

· For b < 5° and p < 4.5, (i.e., for short numerical distance), the attenuation factor A, approximated to A = e − (0.43 + 0.01p

2

)

That is A varies almost exponentially with p for this case. · b > 5° (i.e. for large numerical distance) A=

1 2p − 3.7

which shows that A is inversely proportional to p, hence the field intensity of the surface wave will also vary inversely with the square of the distance from the Tx. The variation of attenuation factor A with numerical distance p for different values of b is shown in Fig. 14.9. The observation of Fig. 14.9 indicates that (a) For p £ 0.1, there is no significant variation in A and it almost mentioned unity. (b) For p (> 0.1 and < 1.0), attenuation factor A reduces slowly with increasing the value of p. (c) For p > 1 and < 10, attenuation factor A decreases rapidly with increasing p. (d) For p > 10, attenuation factor A almost inversely proportional to p, as a result, the filed strength of surface wave decreases with the square of the distance from the Tx. When earth’s dielectric constant er, s and frequency f of wave are such that x >> er, then the power factor angle approaches zero, and earth behaves as resistive impedance to the wave-propagation. On the other hand, earth behaves as capacitive impedance, when b = 90°. It has also been noticed that the earth that behaves as a conductor at low frequencies giving rise to Joulean-heat loss, acts as a dielectric at very high frequencies [7]. The corresponding maximum distance covered by the surface wave of frequency f is obtained by

498

Antenna and Wave Propagation 1.0

b = 0° b = 30° b = 60° 0.1 b = 90° b = 180°

A

0.01

0.001 0.01

0.1

1.0

10

100

1000

p (Numerical distance)

FIG. 14.9

Variation of ground attenuation factor A with p and b.

dmax =

100 3

fMHz

km

(14.36)

Typically, value of the maximum distance is between 90 km and 125 km corresponding to frequency 1.5 MHz to 0.5 MHz.

MULTI-HOP TRANSMISSION The surface wave propagation is limited to the frequency up to 2 MHz, as it follows the earth’s curvature. And propagating waves lost their energy over certain distance. Therefore, long distance communication cannot be made through this mode of propagation. Alternate mode of propagation which is suitable over long distance is multi-hop propagation (Fig. 14.10), in which waves of frequency range around 2 MHz–30 MHz are reflected from ionosphere layer and return back to the earth completing the communication. However, since in the day time the lower frequencies of this band are highly attenuated, the frequency range 10 MHz to 30 MHz is only suitable for long distance communication/broadcasting. Similarly, during the night hours, higher frequencies around 30 MHz are not at all reflected back to the earth, so that somewhat lower (< 30 MHz) frequencies are utilized for the propagations (Fig. 14.11).


499

North F2 Layer

Day Double hop

Single hop

Earth T

R

South

FIG. 14.10

Single and double hop transmission.

FIG. 14.11

Day and night views.

The signals at receiver are vector sum of all the waves reflected from ionosphere and follow different paths. The longest single hop propagation is obtained when the radiating waves are tangential at the earth surface. In general long distance high frequency communication completed via two to four hop-transmission paths, and each contributes appreciable energy to the receiver. The maximum distance covered by the ionosphere wave in a single hop transmission is 2000 km and 4000 km for E and F layers respectively.

Effect of Ground The surface wave is very dependent upon the nature of the ground, at which it propagates. Ground attenuation, terrain roughness and the dielectric constant all affect the wave propagation. In addition to this, the ground penetration varies becoming greater at lower frequencies, which shows that it is not just the earth conductivity that is of importance. At high frequencies it may not affect much, but at lower frequencies penetration means that ground strata down to 100 metres may have an effect. Despite, all these variables it has been noticed that terrain with good conductivity offers best performance. Thus, soil type and the moisture content are also of importance. The mashy land is good and salty seawater is best for surface wave propagation. However, sandy terrain and city centres are by far the worst. This means that sea paths are optimum, subject to variations due to the roughness of the sea, which causes path losses being slightly dependent upon the weather [9].

Effects of Polarization As we have seen, polarization of the antenna has major effect on the surface waves. Vertical polarization offers considerably less attenuation than horizontally polarized signals. In some cases, this difference can amount to several tens of decibles. This is the reason why medium

500


waves broadcast stations use vertical antennas even if they have to be made physically short by adding inductive loading. That is why ship making use of the MF marine bands often use inverted L antennas as they are capable of radiating significant amount of vertically polarized waves.

SOLVED EXAMPLES Example 14.1 Find the maximum power received at a distance of 20 km over the free space. The communication system consists of transmitting antenna with a gain of 30 dB and receiving antenna with a gain of 25 dB w.r.t. isotropic antennas, assume the Tx antenna is fed by an input power of 200 W at the operating frequency 75 MHz. Solution: Given Pt = 200 W = 46.02 dB, dkm = 20 fMHz = 75 and gains of antennas = 30 dB and 25 dB We know that Pr(dB) = Pt(dB) + Gr(dB) + Gt(dB) + a – Ls(dB) in which Ls(dB) = 32.45 + 20 log dkm + 20 log fMHz = 32.45 + 20 log 20 + 20 log 75 = 32.45 + 26.02 + 37.50 = 96 dB Therefore,

Pr(dB) = 30 + 25 + 0 – 96 – 46.02

(assuming a is 0)

= 5.02 dB = 1.78 W. Example 14.2 Find the free space loss for the communication at a distance of 150 km from a transmitter operating at 50 GHz. Also calculate the power received at the receiver if Pt = 200 W and gains of Tx and Rx antennas are 30 dB and 25 dB. Solution:

We know that Ls(dB) = 32.45 + 20 log dkm + 20 log f MHz = 32.45 + 20 log 150 + 20 log 50 ´ 103 = 32.45 + 43.52 + 55.92 = 132 dB

and

Pr(dB) = Pt(dB) + Gr(dB) + Gt(dB) + a – Ls(dB) Pr(dB) = 200 + 20.75 + 20.75 + 150 ´ 0.02 – 132 = 112.5 dBW

Example 14.3 Determine electric field intensity at a distance of 20 km from a 150 kW transmitter employing a short vertical antenna.


Solution:

501

We know that Pt ⎛ mV ⎞ ER ⎜ = 173 ⎟ = 173 d km ⎝ m ⎠ = 105.94

150 20

mV m

Example 14.4 Repeat the above question, if the field strength value is 300 mV/m at a distance of 1 km from the transmitter for a radial power of 1 kW. Solution:

In this case

150 mV ⎛ mV ⎞ ER ⎜ = 183.71 ⎟ = 300 20 m ⎝ m ⎠ Example 14.5 Find the standard deviation of surface irregularities of earth, if surface roughness is 185 dB and an EM wave of frequency = 25 GHz is incident at angle 37° from the normal. Solution:

We know that surface roughness is given by R=

4QT sin G

M

3 × 1010

in which, R = 185 dB = 1.267, M =

T =

Hence

or T =

25 × 10 9

1.267 × 1.2 4 × 3.14 × sin 53

o

RM 4Q sin G

= 1.2 cm and f = 90° – q =

1.5204 10.03

= 0.152 cm

Example 14.6 A microwave signal of frequency 1.6 MHz is propagating through a surface of conductivity s = 2.17 S/m and e = 1.56. Determine the loss-tangent and equivalent complex dielectric constant in this case. Solution:

We know that the equivalent complex dielectric constant is

e¢ = e(1 + s /jwe) where s/we is known as loss-tangent. =

2.17 2 × 3.14 × 1.6 × 10 × 1.5 × 8.854 × 10 6

= 1.63 ´ 104 = 42.13 dB

−12

=

2.17 133.45

× 10 6

502


F ′ = 1.5 × 8.854 × 10

Therefore,

−12

⎛ 1.63 × 10 4 ⎞ 1 + ⎜ ⎟ ⎜ ⎟ j ⎝ ⎠

= (13.28 ´ 10–12 – j2.16 ´ 10–7) Example 14.7 Find the reflection coefficient for vertically and horizontally polarized fields, when EM waves are incident at angle 30° from surface through a medium having e = e0 onto surface of e = 1.5 e0. Solution:

q = 60°.

Given q is angle from the normal. In present case given angle is 30°, therefore

Rv =

=

Er Ei

=

1.5 cos 60 o − [1.5 − sin 2 60 o ]1/2 1.5 cos 60 o + [1.5 − sin 2 60 o ]1/2

0.75 − [1.5 − 0.75]1/2 0.75 + [1.5 − 0.75]1/2

= − 0.072 = 11.42 dB

Similarly

Rh =

Er Ei

=

cos 60° − [1.5 − sin 2 60°]1/2 1/2

cos 60° + ⎡⎣1.5 − sin 2 60°⎤⎦

= − 0.26 = 5.77 dB

Example 14.8 Find the electric filed strength at operating wavelength l = 3.5 cm, when Tx and Rx antennas of heights 10 m and 15 m are 10.2 km apart. Antenna current is to be 10 mA. Solution:

Given l = 3.5,

d = 10.2 km

ht = 10 m and hr = 15 m We know that

E=

I ht hr I s 120 × 3.14 × 10 × 15 × 10 −2 = Md 3.5 × 10 −2 × 10.2 × 103 =

565.5 357

= 1.584

V m

≅ 20 dB

Example 14.9 In a mobile communication heights of Tx and Rx are in ratio of 2 : 3. The transmission of wave establish with E = 5.5 V/m at f = 9 GHz. Find actual height of antennas, if they are 2.5 km apart from each other and carries a current of 5 mA. Solution:

Let ht be 2 k and hr be 3 k.


Then

E=

⇒ 5.5 =

I ht hr I s 377 × 6k 2 × 5 × 10 −2 = Md 3.33 × 10 −2 × 2.5 × 103 11310

× 10 − 4 k 2

8.325

⎡ 5.5 × 8.325 ⎤ k= ⎢ 10 4 ⎥ ⎣ 11310 ⎦

Therefore

503

ht = 12.72 m

and

1/2

= 6.36

hr = 19.08 m

Example 14.10 Obtain the electric field intensity at 10 km from a 1.5 MHz Tx antenna having E = 3 V/m over a ground path with s = 3.0 mS/m and er = 7.0. Solution:

We know that numerical distance p=

A=

Hence,

Eg =

2 0.582 dKm f MHz

=

T (mS/m) 2 + 0.3p

2 + p + 0.6p

E0 A d

=

2

=

0.582 × 10 × 1.52 3

= 4.365

2 + 0.3 × 4.365 2 + 4.365 + 0.6 × 4.3652

3.0 × 0.186 10

= 55.8

= 0.186 = − 7.304 dB

mV m

Example 14.11 Find the distance at which electric field intensity is 5 square times the field intensity at field intensity at unit distance from a Tx antenna. Assume propagation is taking place over a ground path with s = 4.0 mS/m and f = 8.5 MHz and attenuation factor is about 0.2. Also find the value of E0 and Eq. Solution:

Given s = 4, er = 7.5 and f = 8.5 MHz p=

Hence

A=

2 0.582d km f MHz

T (mS/m)

=

0.582 × d × 8.52 4

2 + 0.3 × 10.51d 2 + 10.51d + 0.6 × (10.51d )2

= 0.2

2 + 10.5d = 0.4 + 2.1d + 221d2

d=

+ 8.4 ±

8.42 − 4 × 22 × − 1.6 44

= 10.51d

504


d=

+ 8.4 ±

70.56 + 140.8 44

= 0.521 km

Let k be field intensity at unit distance from the Tx antenna.

E0 A

ER =

Hence,

d

⇒ 5k 2 =

k × 0.2 0.521

k = 0.076 Therefore,

E0 = k = 0.076 = 76 mV/m

and

Eg = 5k2 = 5(0.076)2 = 29.4 mV/m

Example 14.12 Calculate the free-space path loss for an antennas system operating at 6 GHz, that is, 25 miles from the Tx antenna. Solution:

We know the free-space path loss is given by Lp = 97 + 20(log10 R + log10 f) (dB) = 97 + 20(log10 25 + log 6) (dB) = 140 dB

Example 14.13 A radio Tx operating at frequency 1.7 MHz provides electric field intensity of 0.6 mV/m at a distance of 20 km. The ground is characterized with s = 5 ´ 10–5 mS/cm and er = 10. Determine the transmitted power, if Tx antenna has an efficiency of 50% and produces radiation field that is proportional to the cosine of the angle of elevation. Solution:

Given Eg = 0.5 mV/m, f = 1.7 MHz, s = 5 ´ 10–5 mS/cm, er = 10 and d = 20 km.

We know that

tan b =

Fr + 1

x

=

(F r + 1) × 1.7 × 106 1.8 × 1012 × 5 × 10 −5

= 0.207

b = tan–1(0.207) = 11.7° The numerical distance, p =

=

Hence,

A= =

Q d cos b 3.14 × 20 × 103 × 1.7 × 106 × cos(11.7) = xM 1.8 × 1012 × 5 × 10 −5 × 176.47 10676 1588.23

× cos(11.7°) = 0.067 × 0.979 = 6.55

2 + 0.3p 2 + 0.3p + 0.6p 3.965 34.29

2

=

2 + 0.3 × 6.55 2 + 0.3 × 6.55 + (0.6 × 6.55)2

= 0.11 = − 9.369 dB


Eg =

E0 A d

300 P

=

505

A

d

P × 0.11

⇒ 0.6 = 300

20 2

⎡ 12 ⎤ ⇒ P = ⎢ ⎥ = 0.1322 kW = 132.23 W ⎣ 33 ⎦

Since the antenna efficiency is 50%, the transmitter must deliver power 2P = 264.46 W Example 14.14 Find the effective area of a receiver located 30 km from a transmitter radiating 600 W having a gain of 30 dB. Assume that the receivers absorb power of 2.5 mW and other losses are negligible. Solution: Gain 15.5 dB = 35.48 We know that

Pd =

WT 4Q d2

GT

Therefore, received power

Pr = Pd Ae =

Ae =

Ae =

4 Q d 2WT WT GT

28260 21288

=

WT 4Q d2

GT Ae

4 × 3.14 × [(30 × 10)3 ]2 × 2.5 × 10 −6 600 × 35.48

= 1.327 m 2

Example 14.15 Determine the propagation parameters p, b, A and E0 when a horizontally polarized wave of f = 7.5 MHz, incident on a surface defined with s = 5 ´ 10–5 mS/cm, P = 20 kW, D = 15 km and er = 8. Also find the field intensity at distance of 20 km, if the transmitting antenna have an efficiency of 85%. Solution:

where

Numerical distance

p=

Q Rx M cos b1

⎛ F − 1⎞ b1 = tan −1 ⎜ r ⎟ ⎝ x ⎠

506


in which

x=

T

1.8 × 1012 × 7.5 × 10 6

=

1.8 × 1012 × 5 × 10 −2 7.5 × 10 6

= 12

⎛ F − 1⎞ −1 ⎛ 8.5 − 1 ⎞ b1 = tan −1 ⎜ r ⎟ = tan ⎜ ⎟ = 32° x ⎝ x ⎠ ⎝ ⎠

b = 180° – b1 = 180° – 32° = 148° cos b1 = cos 32° = 0.84 p=

A=

3.14 × 15 × 10 3 × 12 40 × 0.84

2 + 0.3p 2 + p + 0.6 p2

=

= 16.82 × 10 3

2 + 0.3 × 16.82 × 103 2 + 16.82 × 10 3 + 0.6 × (16.82 × 10 3 )2

= 4.1 × 10 −8 = − 1.872 dB

E0 =

Eg =

300 20 15 × 0.85

= 97.014 mV/m

E0 A 300 23.52 × 4.1 × 10 −8 = 2.98 × 10 −6 mV/m = d′ 20

= 2.98 nV/m

i.e., propagation is very poor. Example 14.16 A police radar transmitter employs a vertical grounded half-wave antenna with a directivity of 1.45 as compared to a short-dipole. Obtain the field intensity, if

s = 1.5 ´ 10–4 mS/m, Pt = 50 kW, f = 2.0 MHz and er = 12 Solution:

Given D = 1.45, l = 150 m

x=

1.8 × 1012 × 7.5 × 10

6

T

=

1.8 × 1012 × 1.5 × 10 − 4

⎛ F + 1⎞ −1 b = tan −1 ⎜ r ⎟ = tan x ⎝ ⎠ p=

2 × 10

6

⎛ 12 + 1 ⎞ ⎜ ⎟ = 5.5° ⎝ 135 ⎠

Qd 3.14 × 83 cos b = cos 5.5° 135 × 150 xM

= 0.013 × cos 5.5 = 0.0134

= 1.35 × 10 2

mhos m


Maximum distance

Attenuation factor A =

d=

2 + 0.3p 2 + p + 0.6p

2

100 3 2

=

=

100 1.2

507

= 83 km

2 + 0.3 × 0.0134 2 + 0.0134 + 0.6 × 0.01342

= 0.9952

Therefore

Eg =

E0 =

E0 DA 300 50 = × 1.45 × 0.9952 = 36.88 mV/m d′ 83

300 50 83

= 25.56 mV/m

OBJECTIVE TYPE QUESTIONS 1. The inverse square law state that (a) Field varies as square of velocity of propagation. (b) Power varies as square of velocity of propagation. (c) Power density inversely proportional to the square of the distance from the Tx antenna. (d) Electric intensity is inversely proportional to the square of the distance from Tx antenna. 2. What will be free space loss for 1 GHz wave propagation at a distance of 10 km from the transmitter? (a) 112.44 dB (b) 12 dB (c) 90.45 dB (d) 65.56 dB 3. What will be power received in a radio-wave communication in which gains of antennas are 30 dB and 25 dB? Assume transmission loss in the medium is 114 dB. (a) 112.44 dB (b) 12 dB (c) – 49.05 dB (d) 65.56 dB 4. What will be electric field intensity Eq at a distance of 1 km from a transmitter which transmitted power is 1 kW? (a) 173 mV/m (b) 12 dB (c) 1.73 mV/m (d) 773 mV/m 5. Reflected waves from any surface are approximately 180° out of phase from their original relationship. The amounts of phase-shifts are constant, but it depends on (a) Polarization of waves (b) Angle of incidence (c) Both (a) and (b) (d) None of these

508


6. Radio waves passing through the atmosphere are affected by (a) Humidity (b) Temperature (c) Pressure (d) All of these 7. The possible effect of diffraction is observed that it (a) Reduces the radio range (b) Increases the radio range (c) Extends the radio range beyond the visible horizon (d) None of the above 8. Which is not a part of atmosphere (a) Troposphere (c) Ionosphere

(b) Stratosphere (d) Ground sphere

9. The amount of reflection waves from any surface depends upon (a) Wavelength of waves (b) Angle of incidence (c) Roughness of reflecting surface (d) All of these 10. Radio waves can be made to encircle the Earth, by diffraction (a) Using high power transmission (b) Low frequency transmission (c) High-frequency and high power (d) High-frequency and low power 11. What will be electric field intensity at 40 km from a Tx antenna, transmitting power of 150 kW? (a) 52.89 mV/m (b) 152.89 mV/m (c) 521.89 mV/m (d) All of these 12. Electric field intensity at operating wavelength 3.5 cm is found to be approximately 20 dB at distance 10.2 km from a Tx of height 10 m. What will be the height of Rx if antenna carrying current 10 mA? (a) 52 m (b) 15 m (c) 21 m (d) 100 m 13. The free space is considered to be most appropriate medium for EM wave propagation because (a) It needs no maintenance (b) It does not interfere with normal radiation (c) Both (a) and (b) (d) None of these 14. In general, the frequency of radio wave ranges (a) kHz to MHz (b) MHz to GHz (c) Both (a) and (b) (d) None of these 15. If the field intensity for a medium is 15 V/m, then the power density of the signal will be: (b) 100 mW/m2 (a) 23 mW/m2 2 (c) 53.0 mW/m (d) None of these 16. The propagation losses of wave propagation depend on the (a) Increase with frequency (b) Decrease with frequency (c) Constant with frequency (d) None of these


509

17. The main cause of polarization of EM waves is (a) Wave travels with velocity of light (b) Properties of EM wave changes after radiation from the antenna (c) An EM wave maintains its properties even after radiation from the antenna (d) None of the above 18. A wave propagating follows particular path that depends on (a) Transmitted frequency (b) Separation Tx and Rx (c) Atmospheric conditions (d) All of these 19. Which of the following propagation also known as Norton’s wave propagation (a) Ionospheric wave propagation (b) Tropospheric wave propagation (c) Surface wave propagation (d) Space wave propagation 20. The angle at which the magnitude of reflected waves from the earth surface approach minimum with phase of 90° is termed as (a) Critical angle (b) Snell angle (c) Pseudo-Brewster angle (d) None of these

Answers 1. 6. 11. 16.

(c) (d) (a) (a)

2. 7. 12. 17.

(a) (c) (b) (c)

3. 8. 13. 18.

(c) (d) (c) (d)

4. 9. 14. 19.

(a) (d) (a) (c)

5. 10. 15. 20.

(c) (d) (d) (c)

EXERCISES 1. Describe the fundamental principles and characteristics of EM waves. 2. Describe briefly the applications of EM waves. 3. Derive the fundamental equation of wave propagation. 4. Show that the electric field intensity at the finite distance (d) from a transmitter can be expressed by ER =

15 d

2Pt V 5

m

, where all the parameters have usual meanings.

5. What do you mean by polarized wave propagation? Derive the expression for reflection coefficient for vertical polarized waves. 6. Deduce the expression for phase-difference between direct and reflected waves for the ground propagation. 7. Write down the expression to calculate the field intensity at a finite distance from a transmitter at the ground.

510


8. Find the maximum power received over a distance of 50 km from a transmitting antenna having gain of 30 dB at 50 MHz. Let the gain of receiving antenna be 15 dB w.r.t. isotropic antennas and input power is 200 W. 9. Describe the surface wave propagation at long and medium wave frequencies. “Surface wave propagation is one of the most reliable methods of the radio-wave communication”. Justify the statement. 10. What is radio-wave propagation? Discuss the effect of ground on this propagation in medium and short wave bands. 11. An EM wave of frequency 40 GHz is incident at angle q from the normal. Find the change in value of q for standard deviation of surface irregularities of earth 0.158 cm and 0.165 cm, assume surface roughness is around 1.55. 12. What is the reflection coefficient for vertically and horizontally polarized fields, when EM waves are incident at angle 40° through a medium having e = e0 onto surface e = 3.5 e0. 13. In a communication link transmission of wave establish with E = 6.5 V/m at f = 9.8 GHz. Find actual height of Rx antenna, if antenna is separated at 5 km from each other and current appears in the system is 5 mA. Let the Tx antenna height be 3 times more than that of Rx antenna. 14. Describe the effects of the earth’s parameters on the ground wavepropagation. 15. Discuss the importance of surface wave propagation for communication purposes. 16. Describe the effect of frequency, earth constant and earth curvature on surface wave propagation. 17. A vertically polarized wave of frequency = 7.5 MHz, incident on a surface defined with s = 6 ´ 10–5 mS/cm, P = 20 kW, D = 25 km and 5 er = 8. Also find the field intensity at distance of 20 km, if the transmitting antenna has an efficiency of 95%. 18. What do you mean by tilting of wave and tilt angle? How does tilt angle affect the field strength received at a finite distance from the transmitter? 19. Define the ground wave attenuation. Show that attenuation factor A is function of frequency and earth constant. 20. Define attenuation factor A, numerical distance p and phase-constant b and (i) Write down the relation between them for different polarizations. (ii) Describe the importance of these parameters on surface wave propagation. 21. In order to establish communication using two half-wave dipole antennas at 100 MHz, 1 kW power is transmitted. Find the far-field distance between antennas if maximum power received at destination is found to be only 5%. 22. With a neat diagram describe the phenomenon of multi-hop transmission. 23. Calculate the power density reaching the moon’s surface from a 1.5 MW pulse generator located at the earth surface. The antenna gain is 6.5 dB and distance


511

between earth and moon is 4.1 ´ 105 km. Assume the characteristics impedance of medium to be 377 W. 24. Find the open circuit voltage induced in a half-wave dipole when power of 15 W at 200 MHz is radiated from another half-wave dipole separated at 60 km away. Assume that the antennas are positioned for optimum communication.

REFERENCES [1] Sarwate, V.V., Electromagnetic Fields and Waves, New Age International Publications New Delhi, 2007. [2] www.radio-electronics.com [3] www.tfcbooks.com [4] www.mike-wilis.com/PF.htm. [5] Green, D.C., Radio System Technology, Longman Scientific & Technical, Essex, U.K. 1990. [6] Scott, A.W., Understanding Microwaves, John Wiley & Sons, Inc., 1993. [7] Prasad, K.D. and M. Prasad, Antenna and Wave Propagation, Satya Prakashan, New Delhi, 1998. [8] Jordan, E.C. and K.G. Balmin, Electromagnetic Waves and Radiating Systems, Prentice Hall of India, New Delhi, 1988. [9] JPL’s Wireless Communication Reference, Websites. [10] Chatterjee, R., Antenna Theory and Practice, 2nd ed., New Age Publications, Delhi, 2004.

C H A P T E R

15

Tropospheric and Space Wave Propagations

INTRODUCTION In the previous chapter, we discussed all the details about surface wave propagation and saw that this propagation is limited only up to 2 MHz frequency. The wave propagation at frequency > 2 MHz or so follows space path hence termed space wave propagation. At higher frequencies, especially at VHF, UHF and microwave frequencies, the space wave is like a surface wave, but it is radiated many wavelengths above the surface. The space wave is made up of two components: direct and ground reflected waves. There is always a phaseshift between these two component waves because the two signal paths have different lengths. There are some general rules which are applicable in these situations: · A phase-shift of an odd number of half-wavelengths causes the components to add, as a result increases the signal strength (constructive interference). · A phase-shift of an even number of half-wavelengths causes the components to subtract, thus reducing signal strength (destructive interference). · Phase shift other than half-wavelengths add or subtract according to relative polarity and amplitudes. The direct and ground reflected waves show interference which results in a multi-path phenomenon. The form of multi-path phenomenon that is perhaps most familiar in ghosting seen in television reception. In mobile communication, multi-path phenomenon is responsible for existence of dead zones and picket fencing. A dead zone exists when destructive interference between direct and reflected (multiple reflected) waves drastically reduce signal strengths whereas picket fencing occurs as mobile unit moves through successive dead zones and signal enhancement (or normal) zones and it sounds like a series of short noise bursts. A dead zone most often noticed at VHF and above frequencies when the vehicle is stopped, which can be overcome by moving the antenna half-wavelength [1]. At VHF, UHF and microwave frequencies space wave propagation is limited to socalled LOS (line of sight) distance. Like surface wave, space wave reception is also affected 512


513

by l, ht, hr, d and both terrain and weather conditions along the transmission path. The portion of the energy received because of diffraction around the earth’s surface and refraction in the upper atmosphere also included in the space waves. In a particular case, when both the antennas are located normal to the earth surface (at 90°), the direct and ground-reflected waves cancel each other and transmission is entirely due to surface waves. However, in other case (at angle other than 90°) direct and reflected waves combine and produce the resultant signal. If we consider the effect of the earth surface, the expression for the received signal becomes more complicated than in case of free space propagation. The main effect is that signals reflected off the earth surface may partially cancel the line of sight waves. Space-communication up to 100 km is preferred in commercial communication [2]. The loss of signal over direct path is characterized with a parametric term n that is defined as n=

Sr Sf

(15.1)

where n = signal loss coefficient Sr = signal level at the receiver in the presence of a ground reflection component Sf = free space signal strength over direct path, in case of reflection However, in terms of height of antennas and distance between them, the signal loss coefficient is defined as n = 2 sin

2Q h1 h2

M D1

(15.2)

where D1 = direct path length hr and ht = Rx and Tx antenna’s heights l = operating wavelength The reflected signal contains both amplitude and phase changes, and typical phase change is 180°. However, the amplitude changes depend on the frequency as well as characteristics of reflected signal. The characteristics of reflected signal is defined by reflection coefficient given by

rc = pejf

(15.3)

where rc = reflection coefficient p = amount of amplitude change ÿ f = phase-change ÿÿ

Typically, for smooth high-reflectivity surface and a horizontally polarized microwave signal at shallow angle of incidence, the rc is close to –1. Because of direct incidence of signal at the earth surface, the phase-change of the reflected signal at the receiver is at least p radians.

514


TROPOSPHERE The troposphere consists of atmospheric regions adjacent to the earth’s surface and extending up to about 12 km. It is the region where most of clouds are formed. The structure of atmosphere can be specified as follows: 8 to 10 km at polar latitude, 10 to 12 km at moderate latitude and up to 16 to 18 km at the equator. The temperature of this region decreases with height at the rate of about 6.5°C per km and falls a minimum value about –52°C at its upper boundary. However, the percentage of the gas components remains constant with increase in height, yet the vapour components sharply decrease with height. The stratosphere lies just above the troposphere. After a certain critical height (known as troposphere), temperature remains uniform through the narrow belt and begins to increase afterwards. The troposphere accounts for 80% of the entire air mass. It is an inhomogeneous dielectric medium, in which pressure decreases with height, and refractive index as well as velocity of propagation also varies with height (Fig. 15.1) [3].

Troposphere bending

Visual horizon Radio horizon Curvature of earth

FIG. 15.1

Tropospheric bending.

TROPOSPHERE WAVE PROPAGATION In troposphere, slight bending of radio waves occurs and causes signals to return to earth beyond the geometric horizon. Troposphere bending is evident over a wide range of frequencies, although it is most useful in the VHF/UHF regions. Radio signals can be trapped in the troposphere, travelling a longer distance than normal before coming back to the earth surface (Fig. 15.2). Instead of gradual changes in the atmospheric conditions, sometimes distinct regions are formed and regions that have significantly different densities try to bend radio waves passing between regions. However, in a non-homogeneous atmosphere whose index of refraction decreases with height, rays of sufficiently small initial elevation angle are refracted downward with a curvature proportional to the rate of decrease of the index of refraction with height. If the radius of curvature is less than the radius of the earth, such rays reach a maximum height and are confined, or trapped between this height and the earth’s surface. This process


515

Warm air region Duct width

Cool air mass Tx

Rx Ground

Curvature of earth

FIG. 15.2

Formation of ducting.

is referred to as trapping and the region of the atmosphere within which it occurs is called a duct, this is because of the analogy with wave-guide propagation. Duct may be defined as an atmospheric structure that traps rays within a few minutes of arc of the astronomical horizon, so that they cannot escape from the atmosphere, but are periodically bent back down, so as to follow the curvature of the earth. The term duct was not introduced until the 1940s, when radar operators in World War II began observing returns from objects far beyond the normal horizon. The most spectacular atmospheric-refraction phenomena are produced by ducts. The duct acts like a lens, focusing these rays into a crude image at this distance. These focusing effects produce the vertical exaggeration of images in superior mirages. Bending of waves is produced by a steep thermal inversion. The inversion can do this if its lapse rate is more negative than a critical value (near –0.11°/m, for typical conditions). That is, the air temperature must increase with height at a rate greater than 0.11° per metre to produce a duct. In fact, such inversions occur quite often in the lower atmosphere and sometimes are called super-refracting, because the curvature of a horizontal ray in it exceeds the curvature of the earth. Ducts are usually formed over water, but they can also form over land. The wider the ducts the longer the travelling distances. Troposphere ducting supports contacts of 1528.55 km or more over land and up to 4023.36 km over oceans, especially during summer, winter and fall months. Troposphere ducting is the most common type of enhanced propagation at UHF. Out of different mechanism of troposphere wave propagation such as diffraction, abnormal reflection and refraction, and troposphere scattering, the normal refraction is the main mechanism for most of troposphere propagation phenomenon. The dielectric constant (hence the refractive index) of the atmosphere which varies above the earth and set mostly by the moisture contains is a primary factor in the troposphere refraction. When the wave passes between mediums of different densities, its path bends by an amount proportional to the difference in densities. Especially, at UHF and microwaves two cases of tropospheric propagation are observed.

516


(a) Since, air density normally decreases with altitude, the upper portion of radio beam travels slightly faster than the lower portion, resulting in small amount of reflection. Such propagation provides slightly longer distance than the actual distance to the radio horizon. This phenomenon is known as simple refraction. A special case of refraction, super refraction occurs in areas where warm land air goes out over cooler sea (Fig. 15.3). The VHF/UHF/microwave communications up to 200 miles has been observed in these areas. Ducting is the special case of super refraction. In Fig. 15.4, D1 is distance in radio-horizon range, whereas D2 is distance over which communication is possible. High temp.

Low temp.

Refracted path

Tx

Rx Direct path

Hot land mass Cool sea

FIG. 15.3

Super refraction phenomenon.

(b) Another situation in troposphere propagation occurs when cooler air from the earth flows out over warmer seas, termed sub-reflection (Fig. 15.5). Hereby, EM waves bend away from the earth’s surface, reducing the radio-horizon by the amount from 30% to 40%. Antenna placement is critical for duct propagation. The Tx and Rx antennas must be · Inside the duct physically (as in airborne cases). · Able to propagate at an angle such that the signal gets trapped inside the duct. All the troposphere propagation depends upon air-mass temperatures, dielectric constant (hence refractive index) and humidity shows diurnal (over the course of the day) variation caused by the local rising and setting of sun. Distant signals may vary 20 dB in the strength over a 24-hour period. TV, FM broadcast and other VHF signals propagate through troposphere, especially along seacoast paths at some times while being weak [4].


Temperature inversion zone

Troposphere wave

Space wave h Earth’s surface

D1 D2

FIG. 15.4

Duct phenomenon.

Low temp.

High temp.

Rx

Tx Direct path

Refracted path

Cool land mass Hot sea

FIG. 15.5

Sub-refraction phenomenon.

517

518


RELATION BETWEEN THE RADIUS OF CURVATURE AND CHANGE OF DIELECTRIC CONSTANT As already mentioned, refraction of radio waves occurs due to change in dielectric constant (refractive index) of the atmosphere’s layers. The dielectric constant of air is slightly greater than unity and presence of water vapour increases the dielectric constant still further. For this reason, the dielectric constant of atmosphere decreases to unity at greater heights where the air density approaches zero [5]. In order to derive the relation between radius of curvature of the earth and the change of dielectric constant (e), let v be velocity of radio wavepropagation at a height H from the earth. Then from Fig. 15.6

S dR = v dt or

FIG. 15.6

dR dt

=

v

S

(15.4)

Radio-propagation at height H.

Also, the velocity v can be expressed as follows: v = (N0 F 0 F r ) −1/2 = c F r

(15.5)

in which m0, e0 and er are usual constants of a medium. Since at height slightly greater than H (i.e., H + dH, say), H + dH » dr, therefore The corresponding velocity, v + dv ≈ (S + dH )

dR dt

, that is


dv

Hence

=

dH

S =

v dv/dH

=

c Fr 1 dF r − cF −3/2 2 dH

dR

=

dt

= −

S= −

Again from Eq. (15.6)

dR

=

dt dv

dv

dv

2 d F r /dH

dN

N

dH

2

(15.7)

d F r /dH

= −

dN

(considering er = 1)

2

c

where v =

c

= − v

dH

(15.6)

S

Fr

dH

= −

dH

,

v

c 1 dN

N N dH

=

c

N

= −

v dN

N dH

as m = 1

dH

519

(15.8)

Therefore from Eqs. (15.6) and (15.8)

S=

or

v dt dR

S= −

=

v dR /dt

=

v dv/dH

= −

v v d N /dH

1

(15.9)

d N /dH

From Eqs. (15.7) and (15.9), it is clear that the radius of curvature (r) is function of the rate of change of the dielectric constant er and permeability m with H and therefore it varies hour to hour, day to day and also season to season. However, it is convenient to consider the ray paths as straight line rather than curved as actually they are, and also to compensate for the curvature by using a larger value for the effective radius of the earth (Fig. 15.7). Hence, from Fig. 15.7, dH = BO – AO = (kr0 + H) (sec qe – 1) Since, in ÑAOB,

AO = kr0 + H cos R e =

AO BO

⇒ BO =

AO cos R e

(15.10) (15.11a)

= AO sec R e

(15.11b)

Since, distance D, and dH are very small than the effective earth radius (kr0), hence qe is also very small, therefore

520


FIG. 15.7

Curved paths become straight lines when effective radius of earth is considered.

sec R e =

1

=

cos R e

1

R2 1 − e 2

=1+

R e2 2

(15.12)

Hence from Eq. (15.10), the value of dH reduces to ⎛R 2 dH = kr0 ⎜ e ⎜ 2 ⎝

⎞ ⎟ ⎟ ⎠

taking H = 0

Again as qe is small

R e = sin R e =

D (kr0 + H )

≅

D kr0

Therefore dH =

D2 2 kr0

On the other hand, when actual path of ray is considered, then from Fig. 15.6, dH = 2dH D

or

2

=

D2 2 r0 1 r0

D2

−

−

2S 1

S

(15.13)


1 kr0

=

k=

1

1

−

r0

S

1 1 −

521

r0

(15.14)

S

Substituting the value of r from Eq. (15.9), we get k=

1 dN 1 + r0 dH

(15.15)

where r0 = radius of the earth r = radius of curvature k = factor expresses the degree of curvature along any given path and defined as effective earth’s radius factor However, in particular case of two hypothetical spheres, both centred at the exact centre of the earth. The value of k can also be approximated as k=

r1 r0

=

Radius of sphere other than earth surface Radius of sphere at the earth surface

(15.16)

Again, from Eq. (15.14), for a radius of curvature equal to four times the radius of the earth, the effective radius of the earth is 4/3 times the actual radius (i.e. k = 4a/3). However, from Eq. (15.16) the value of k = 1, indicates a straight line and k > 1 and k < 1 indicates refraction and sub-reflection of the waves respectively. Index of refraction (n) Basically, refractive index of atmosphere, measures the differential properties between adjacent zones of air and it is gradually decreases with height above the earth surface. Under standard temperature and pressure, the value of n is equal to F and at the surface near sea level it is » 1.0003. However, in homogeneous atmosphere it decreases by 4 ´ 10–8 per mile of altitude. In order to avoid complexity, refractive index is scaled up by a parameter known as refractivity which is given by N = (n – 1) ´ 106. The rate of change of N with height h is termed gradient of refractive index and directly concerned with the bending waves. The value of n and N vary with altitude, and in particular N tends to vary from about 280 to 320. There are two methods to calculate N e ⎛ 77.6 ⎞ ⎛ ⎞ N=⎜ Hn ⎟ ⎟ ⎜ P + 4810 T T ⎝ ⎠ ⎝ ⎠

(15.17a)

522


and ⎛ 77.6 ⎞ ⎛ 5 e ⎞ N=⎜ ⎟ ⎜ 3.73 × 10 ⎟ T2 ⎠ ⎝ T ⎠ ⎝

in which P = T = e = Hn =

(15.17b)

atmospheric pressure in millibars temperature (K) saturation vapour pressure of atmospheric water in millibars relative humidity expressed as a decimal fraction

TROPOSPHERE SCATTERING AND TROPOSCATTERING PROPAGATION Troposcattering propagation is also known as forward scattering propagation and is of practical importance to VHF, UHF and microwave frequencies. As the name suggests it is another type of extended range tropospheric propagation. The main mechanism involved in troposphere scattering propagation is scattering and reflection from common volume of the troposphere occupied by transmitting and receiving beams. In normal situations, air is not uniform, there are eddies, thermals and turbulences, etc. where the air has slightly different pressure and humidity and hence a different refractive index. The nature of irregularities is not clearly known, but they are considered as a mixture of blobs, small lays, etc. which continuously vary in number, shapes, position, velocity and physical characteristics. When a radio wave propagating through the troposphere meets turbulence, causes an abrupt change in velocity. As a result, a small amount of the energy is scattered in a forward direction and returns to the earth at distances beyond horizon. This phenomenon is repeated many times in the path, as the waves meet other turbulences. The horizon is the limit of communication distance, whereas radio-horizon is actually about 15% more than the horizon. The total received signal is an accumulation of the energy received from each of the turbulences. This scattering mode of propagation enables VHF and UHF signals to be transmitted far beyond the normal LOS [6]. The turbulence that causes the scattering can be visualized as a relay station located above the horizon. It receives the transmitted energy and then reradiates it in a forward direction to the point beyond the line of sight distance. The high gain receiving antenna of diameter less than 50l is used to capture the signals. The magnitude of received signal depends upon the number of turbulences causing scattering in the forward direction and also the gain of the receiving antenna. The scattered common volume formed by the intersection of the antenna beams needs to be in the troposphere, so there is a limit to the propagating range. The angle at which the receiving antenna must be aimed to capture the scattered energy is called the scatter angle. The common scattered volume and scatter angle are shown in Fig. 15.8. The signal take-off angle determines the height of the scatter volume and the value of scatter angle and it is directly related with them [7].

523

Tropospheric and Space Wave Propagations Lost scatter

No scattering

Scatter volume Scatter angle Forward scatter

Longest path Shortest path

Back scatter

T

FIG. 15.8

R

Tropospheric scatter propagation.

Transmission Loss The Tx loss in this mode of propagation increases with distance between Tx and Rx antennas faster than the free space loss. It is also increases with frequency and the loss is approximately proportional to the third power of the frequency up to about 3 GHz and that the increase is less rapid at higher frequencies. Troposcattering propagation is generally more reliable than ionospheric propagation. Typical path length and frequency range are 200 to 500 km and 300 MHz to 5 GHz respectively. Typical value of Tx loss for a 250 km long path at frequency of 150 MHz with 20 dBi antenna is » 140 dB (included antenna gain). The empirical formula for calculating the Tx loss is given by [1] L = M + 30 log f + 10 log d + 30 log q + LN + Lc – Gt – Gr where

M= q= LN = Lc = Gt and Gr = f= dkm =

factor, typically lies between 19 and 40 dB and depends on climate scatter angle (radians) height of the common scattered volume the aperture-medium coupling loss antennas gain frequency (GHz) distance in km

Above parameters are defined as follows: Scatter angle, q = qe + qt + qr in which

(15.18)

Re =

1000 d Re

d = distance between Tx and Rx and Re is effective earth radius

qt and qr = horizon angles of the transmitter and receiver LN = 20 log (5 + g H) + 4.34 g h

524


with H=

and

ÿÿ

10 −3R d 4

, h=

10 −6R 2 Re 8

g = constant = 0.27 km–1 and

Lc = 0.07 e0.055(Gt+Gr)

M-CURVES AND DUCT PROPAGATION We have discussed about usual reflection phenomenon in troposphere, causing troposhperic propagation. In addition, due to abrupt changes in refractive index (because of weather conditions), there are possibilities of abnormal reflections too. These abnormal reflections offer propagation range beyond normal ground wave range and are best handled by using modified index curves. The term modified is used to indicate that the actual refractive index has been modified to account for the curvature of the earth. M-curves are curves that show the variation of modified index of refraction with height. When the modification is done, straight rays above a curved path come out as curved rays above a flat earth. M-curves are useful in predicting at least roughly, the transmission path that is usually expected. There are following regards for M-curves: · When the modified index of refraction increases linearly with height, standard propagation occurs and M-curves are straight line with a positive slope. · If the slope of M-curves decreases near the earth surface, the rays curved upward over the flat earth and substandard propagation results. · If the slope of M-curves increases near the surface of the earth, the upward curvature of ray is less and super standard condition is achieved with greater coverage. · If M-curves become vertical (no change of modified index with height), the rays over the flat earth are straight and very large coverage is possible. In a special case, when the modified index decreases with height over a portion of the range of height, the rays are curved downward, this case is referred as duct propagation. In duct propagation, waves are guided along the duct, like being guided in wave-guide. The upper and lower frequencies limits of duct propagation are 10 GHz and 50 MHz respectively. The upper limit sometimes observed even at higher frequencies in air borne radar and other military equipment applications. Basically, there are two types of ducts. Surface duct In this case, the lower side of duct is at the surface of earth. Over the land areas surface ducts are produced by radiation cooling of the earth. Elevated duct If the inverted portion of the M-curves is elevated above the earth surface, the lower side of the duct is also elevated, the duct is called elevated duct and occur due to subsidence of large air masses. As well as position of antennas is concerned they should be elevated within the duct, otherwise (below or above it), the signal will be very small, resulting in poor communication.


525

Like ordinary wave-guide propagation, there is a critical frequency, below which duct propagation is not possible. Most commonly, a duct profile is calculated with the modified refractive index (M) as follows: M=

77.6 ⎛ 4807 × e ⎞ ⎜P + ⎟ + 0.15 h T ⎝ T ⎠

(15.19)

in which h = The altitude (in m) and e is the vapour pressure and given by ⎛ 7.5Td ⎞ e = 6.1078 × 10 ⎜ ⎟ ⎝ 237+ Td ⎠ where Td is the dew point temperature.

However, a general approach to approximate modified refractive index, in term of refractive index n(h) is given by h⎤ ⎡ M (h) = ⎢ n(h) − 1 + ⎥ × 10 6 r⎦ ⎣

(15.20)

in which r is the radius of the earth. A series of duct profile is shown in Fig. 15.9, where the diffraction in various types of ducts is indicated. The rate of change of M with h, i.e., dM/dt referred as a duct gradient. The typical values of duct gradients lie between 0.1 (weak gradient) and 2 (strong gradient). A large negative value of vertical gradient is obtained when (i) Temperature increases with the height. (ii) Water vapour decreases as height increases, i.e., dry air above a humid air layer, etc.

a

c

b

e

d

D

D

D

D

FIG. 15.9 (a) M profile for various types of ducts showing a standard atmosphere, (b) evaporative duct (c, d) surface ducts, and (e) an elevated duct. The depth of the duct D is shown for each type of duct.

526


The size (i.e., depth and width) and strength (i.e. gradient and depth) determine the minimum frequency and duct angle of the duct propagation, the longest wavelength that can be propagated estimated by

Mmin = 0.66 × A × D 'M where A = = D = DM =

(15.21)

3.77 ´ 10–3 for a surface duct and 5.66 ´ 10–3 for an elevated duct depth of the duct (m) maximum difference in the M within the duct.

The critical angle at which signal can be a duct is given by

Gc = 7.39 × 10

−2

⎡ dM × ⎢ ⎣ dh

⎤ × D⎥ ⎦

1/2

(deg)

(15.22)

Duct Propagation Loss When both the Tx and Rx antennas are within the duct region, the signal is propagated with very low losses [8]. However, for frequencies greater than a critical value fc = 1572/D18 GHz, the first mode attenuation coefficient is found £ 0.03 dB/km. Both the attenuation coefficient a and critical frequency fc increases slightly with increasing the duct layer thickness. The expression for the basic Tx loss within the duct is given by Ld ³ (92.45 + 20 log fGHz + 10 log dkm + 0.03 dkm + A) dB

(15.23)

where A is aperture to medium coupling loss. This loss is considerably less than the free-space loss. Because in free space, radio waves normally spread out in the orthogonal directions of the propagation, the power density decreases with 1/d2 whereas in the duct density varies with 1/d only. Practically, it is found that the energy radiated at elevation angles and propagated along a uniform duct. The corresponding coupling loss is given as Ac = − 10 log

2R c :

dB

2|qc| < W

(15.24)

2|qc| ³ W

= 0.0 dB

where W is half-power point antenna beam width. However, when one or both terminals are within the elevation range of the duct, but beyond the duct’s horizontal extent, the coupling loss is given by Ac = − 10 log

= 0.0 dB

D DL

dB

D/dL £ W

(15.25)

D/dL ³ W

where D is duct thickness and dL is the distance from the end of the duct to the terminal beyond the duct [9].


527

Diffraction Diffraction is bending of radio waves path when they encounter opaque object. The amount of diffraction and effects that it causes is frequency dependent. High frequency (» 3 GHz) waves are rarely diffracted in the normal world that surrounds us. At frequency greater than 3 GHz, the wavelengths are so small as compared to obstacle size, hence large attenuation of the signal occurs and this phenomenon is also based on diffraction of radio waves. Terrain or man-made objects intervene in the path between microwave stations cause diffraction and effective amount of signal attenuation (Fig. 15.10). There is minimum clearance required to prevent severe attenuation (upto 20 to 30 dB) from the diffraction. The required clearance can be calculated from the Huygens–Frenzel wave theory.

Hill

Antenna

FIG. 15.10

Terrain masking of VHF and higher frequency signals.

Let us consider Fig. 15.11, in which A and B are representing source and destination respectively. B1–B2 is a partial spherical surface, on which all the rays have same phaseknown as iso-phase plane. The ga are rays incoming to plane (B1–B2), whereas gb outgoing from the plane. At receiver, strength of signal is algebraic sum of all rays gb. The ray impact points on plane B1–B2 from radii Rn called Frenzel zones. The value of radii (Rn) varies with frequency and distances D1 and D2 as follows:

⎡ N ⎛ D1 D2 ⎞ ⎤ Rn = M ⎢ ⎜ ⎟⎥ ⎣ FGHz ⎝ D1 + D2 ⎠ ⎦

1/2

(15.26)

where N = inter (1, 2, 3, …) M = constant of proportionality equal to 17.3, if Rn in metre and D1 and D2 are in km and frequency is in GHz.

528


(a) Object

d2 d1

L1

h

L2

d

(b) FIG. 15.11

(a) Fresnel zone geometry; (b) Diffraction phenomenon.

If frequency is in GHz and distances are in miles, then

⎡ N ⎛ d1d2 Rn = 72.1 ⎢ ⎜ ⎣ FGHz ⎝ d1 + d2

⎞⎤ ⎟⎥ ⎠⎦

1/2

(15.27a)

However, if the radius of first critical Frenzel zone (R1) is known, the radius of nth Fresnel zone can be calculated using Rn = R1 n [10].


529

Diffraction loss In order to quantify diffraction loss, first a dimensionless parameter v is defined as follows: ⎡ 'd ⎤ v = 2⎢ ⎥ ⎣ M ⎦

1/ 2

where Dd = d1 + d2 – d. In a special case, such diffraction from the single obstacle can be modelled as a knife-edge and corresponding loss can be calculated from

L (v) = 6.9 + 20 log [ v2 + 1 + v]

(15.27b)

SPACE WAVE PROPAGATION The space waves, also called direct waves, travel directly from the transmitting antenna to the receiving antenna as shown in Fig. 15.12. In order to have a good space wave propagation, both the antennas must be seen by each other. That is there must be line of site path between them. The maximum line of sight distances between two antennas depend on the heights of each antenna. The direct waves are useful primarily only in the VHF, UHF and microwave frequencies. TV and FM radio broadcasts are received as direct waves. Satellite communications are best example of direct communication. This is because a satellite has a tremendous line of sight from its vantage point in space and many ground stations can communicate through a single satellite. Because, a typical transmission path is filled with buildings, hills and other obstacles, direct waves are reflected by these obstacles, as a result signals at the receiver follow different directions. That is, waves are not in phase, they may reinforce or cancel each other. This situation is known as multi-path propagation. Tx antenna

Direct wave

Rx antenna

Reflected wave

FIG. 15.12

Direct and reflected waves representation.

530


The multi-path effects for the mobile radio lead the path loss [7] L∝

dn

(15.28)

hr ht

where d = path length hr and ht = heights of antennas (Tx and Rx) n = path loss exponent which indicates the rate at which the path loss increases with distance The value of n in different environment is given in Table 15.1. TABLE 15.1

Value of n in different environment

Environment

Path loss Exponent, n

Free space Urban area cellular radio Shadowed urban cellular radio Building line of sight Obstructed in building Obstructed in factories

2.7 3 1.6 4 2

2 to to to to to

3.5 5 1.8 6 3

Space wave patterns The total electric field due to space wave for a vertical dipole is given by

Etotal space = EG (space) = [Ez2 (space) + E S2 (space)]

which is equal to

⎛ e− j C R1 R e− j C R2 j 30 C l dl cos Z ⎜ + v ⎜ R1 R2 ⎝

which contains two terms: First term

e − j C R1 R1

position of dipole whereas the second term

⎞ ⎟ ⎟ ⎠

(15.29)

represents direct waves, originated at the

Rv e− j C R2 R2

represents reflected waves because of

factor Rv. When the dipole is situated away from the earth, the incident wave is essentially a plane wave and space wave field is same as total ground wave field. Whereas, when the dipole is located close to the earth, the incident wave will not be plane wave and total reflected field will be the sum of space as well as surface wave fields. The vertical radiation pattern of a vertical dipole at the surface of the Earth apparent that the main effects of the finite conductivity on the patterns occurs at the low angles, where the strength of space wave


531

field is much reduced from its value over a perfect conducting surface. The reason behind this is the phase of the reflection factor Rv which varies rapidly for angles of incidence near the Pseudo-Brewster angle, and the phase of Rv is always –90° at this angle. Above Brewster angle the phase of Rv is nearly zero, whereas below this angle (near grazing incidence) the phase of Rv approaches –180°. This rapid change in phase of reflection coefficient near the critical Pseudo-brewster angle is responsible for many of the propagation characteristics peculiar to vertical polarization. On the other hand, the surface wave field of a horizontal dipole in the plane normal to dipole axis is calculated by replacing the Rv by Rh in vertical dipole field expression, i.e., ⎛ e− j C R1 e − j C R2 h Espace = j 30 C l dl ⎜ + Rh ⎜ R1 R2 ⎝

⎞ ⎟ ⎟ ⎠

(15.30)

The factor cos y is absent, because the horizontal dipole by itself is a uniform radiator in the plane normal to its axis. The effects of finite conductivity in this case is much lesser than in the vertical dipole case, because the reflection factor Rh never deviates much from the value –1 (in case of perfect conductor) [5].

SPACE WAVE PROPAGATION PARAMETERS Basically, there are three parameters for space wave propagation: Effective earth radius, line of sight distance and field intensity. The detailed description of effective earth radius has already been given under. Relation between the radius of curvature and change of dielectric constant, therefore rest two of them are considered here.

Line of Sight (LOS distance) In line of sight communication both the antennas are in visual contact with each other. The height of antennas determines the LOS distance which also known as optical horizon. In general, space communication is possible only up to or slightly beyond the line of sight distance. In order to obtain LOS distance, let us refer to Fig. 15.13, in which it is clear that the LOS distance d = d1 + d2, where d1 and d2 are referred to as radio-horizon distances, and defined as the distance travelled after several reflection from Tx to the designation at earth surface. From ÑTO¢O and ÑRO¢O d1 = [(ht + a)2 − a 2 ]1/2 =

2aht

d2 = [(hr + a)2 − a 2 ]1/2 =

2ahr

(Neglecting ht2 and hr2 as they are << 2a). Therefore, d = 2 a ( ht +

hr )

(15.31a)

532


FIG. 15.13

LOS distance and radio horizon distance.

As radius of the earth (a) is 6370 km d = 3.57 ( ht +

hr ) km

(15.31b)

In a particular case, if radius of curvature R of wave path is 4 times the actual earth’s radius, then we known the effective earth radius (a¢) will be a¢ = 4a/3 as a¢ = ka, where, k is effective radius factor. In this condition, the LOS distance modified to d = 4.12 ( ht +

hr ) km

(15.31c)

That is, in this case, LOS distance will be greater than the previous case. The choice of k depends on various geographical conditions and system specifications, but normally not less than 0.5 to 0.7. The values of k for different propagation conditions are listed in Table 15.2. The above equation is again modified, if parameters involved in the equation is given in feet. d = 1.4142 ( ht +

hr ) miles

(15.31d)

From Eq. (15.31d), it clear that the LOS distance can be increased by increasing the heights ht and hr. The units of radio-horizon measurements are:

dnmi = 1.23( hft ) nmi and dkm = 1.30( hft ) km

(15.32)


TABLE 15.2

Perfect

533

Values of k for different propagation conditions

Ideal

Propagation conditions Average Difficult Surface layers ground fog

Bad

Weather

Standard atmosphere

No surface layer or fog

Standard light fog

Typical

Temperature zone, no fog, no ducting good atmospheric mix day and night

Dry, mountains, no fog

Flat tempera- Coastal water ture, some Tropical fog

Coastal water

k-factor

1.33

1.0–1.33

0.66–1.0

0.5–0.4

0.66–0.5

Fog moisture over water

Field Strength As already mentioned, the space wave propagation is completed by the two waves: direct and ground reflected waves. Direct waves suffer negligible attenuation compared to the ground-reflected waves. In order to obtain field strength at the receiver, let us consider Fig. 15.14, where all the parameters have usual meaning. As the distance travelled by the direct and ground reflected wave is approximately same and if earth is a perfect conductor, the field strength at receiver Rx will be E0. However, if the reflection at the ground is specified by the reflection coefficient R, then the magnitude of reflected field will be R ´ E0. Now from ÑTRP, d12 = (ht − ht )2 + d 2 ⎡ = d ⎢1 + ⎢⎣

or

2 ⎛ ht − hr ⎞ ⎤ ⎜ ⎟ ⎥ d ⎝ ⎠ ⎥⎦

⎛ h − hr ⎞ d1 = d + ⎜ t ⎟ ⎝ 2d ⎠

1/2

2 ⎡ ⎤ 1 ⎛ h − hr ⎞ = d ⎢1 + ⎜ t ⎟ + ...⎥ 2⎝ d ⎢⎣ ⎥⎦ ⎠

2

and similarly from right triangle DTAB, (ht + ht )2 + d 2 = d22

or

⎛ h + hr ⎞ d2 = d + ⎜ t ⎟ ⎝ 2d ⎠

2

534


T

d1

ht

E0

P

O

R

hr

d2

Tx

Rx d d

ht

hr

A

B

d1 = direct wave path-length d2 = ground reflected path length E0 = field-strength at R due to direct rays

FIG. 15.14

Direct and ground reflected waves.

Therefore, the path difference between direct and indirect rays will be ' d = d2 − d1 = d +

'd =

(ht + hr )2 2d

− d −

(ht − hr )2 2d

=

2ht hr 2d

+

2ht hr

2ht hr 2ht hr 2d

d

(15.33a)

d

Hence, the phase-difference between direct and indirect rays will be Phase-difference (a) =

4 Q ht hr

Md

radians

(15.33b)

This phase difference is caused due to the path difference. But, in addition to this, there is another phase difference of 180° due to reflection from the ground, say b (i.e. bÿ = 180°). Hence, the resultant phase difference (q) will be

q=a+b


535

Therefore, the resultant field strength at receiving point R can be given by ER = E0 (1 + Ge–jq) ER = E0 (1 + * cos R )2 − (j sin R )2 = E0

1 + * 2 − 2 * cos R

where G = reflection coefficient, if the earth is assumed to be perfect, i.e., G = 1 and

b = 180° or p radians. Then E R = E0

R R (Q + B ) ⎛ ⎞ − 1 ⎟ = 2 E0 cos 1 + 12 − 2.1 ⎜ 2 cos2 = 2 E0 cos 2 2 2 ⎝ ⎠

ER = 2 E0 sin

B 2

= 2E0 sin

4 Q ht hr 2d M

= 2E0 sin

2Q ht hr dM

Since d >> h (height of Tx or Rx antennas) and sin q = q, as q is very small Under this consideration,

E R = 2E0 sin

4Q ht hr 2d M

≈

4Q E0 ht hr dM

(15.34)

If Ef is field intensity of the direct ray, i.e., free space field intensity at unit distance, then E0 = Ef /d, but Ef = 7 P V/m, in which P = Effective radiated power in watts. Hence, E0 = 7 P /d V/m. Substituting this value into ER gives ER =

ER =

4Q × 7 × ht × hr d M 2

88 ht hr

Md2

P=

87.97 P ht hr d2M

(15.35)

P V/m

which is very useful to calculate the field intensity at the receiver spaced away at distance d from the Tx if all other parameters are known.

SOLVED EXAMPLES Example 15.1 Find the percentage change in wave propagation velocity, if the waves pass from air into a medium of dielectric constant er = 3.5. Solution:

We know that v=

c Fr

536


For air er = 1, hence v =

For medium, er = 3.5 v =

3.0 × 108

= 3.0 × 108 m/s

1 3.0 × 108

= 1.6 × 108

3.51

i.e., velocity decreases. Change in velocity Dv = 1.4 ´ 108 m/s and hence, % change = 46.67%. Example 15.2 The electric field intensity of a plane wave propagation is defined by E = G 10 y cos(109t + 30z) V/m. Determine dielectric constant of medium, if mr = m0. Given

Solution:

G E = 10 y cos (109t + 30z) V/m w = 109 rad/sec

i.e.

v=

So the phase velocity

X C

=

b = 30 rad/m

and

10 9 30

=

108 3

m/sec

2

Since, v is also equal to

9 × 10 −16

F =

2

⎛1⎞ ⎛ 3 ⎞ 1/NF or NF = ⎜ ⎟ = ⎜ 8 ⎟ = 9.0 × 10 −16 ⎝v⎠ ⎝ 10 ⎠

4 × 3.14 × 10 −7

⇒ Fr =

F F0

=

9 × 10 −16 × 36 × 3.14 4 × 3.14 × 10 −7 × 10 −9

= 81

Example 15.3 Calculate the value of the factor by which the horizon range of a Tx line is modified. Assume the gradient of refractive index of air near the ground is 0.065 ´ 10–6 m–1 and radius of the earth is equal to 6370 km. Solution:

We know the modified factor =

⇒ k=

1 dN 1+r dH

100 586

=

1 1 − 0.065 × 10

−6

× 6.37 × 10

6

=

1 1 − 0.065 × 6.37

= 1.724

Therefore, r¢ modified horizon range is kr = 1.724 ´ 6370 = 10982.75 km

=

1 1 − 0.414


537

Example 15.4 In order to establish the communication with 35 W transmitter at 100 MHz, the height of Tx and Rx antennas are selected 50 m and 35 m respectively. Find the LOS up to which communication can be maintained and also the field strength at the receiver end. Solution:

Given, P = 35 W, f = 100 MHz or l = 3 m ht = 50 m

We Know that d = 4.12 ⎡⎣ ht +

and

hr = 35 m

hr ⎤⎦ km

d = 4.12 ⎡⎣ 50 + 35 ⎤⎦ = 4.12 [7.07 + 5.9] = 53.50 km and field intensity ER =

=

88 Pht . hr

Md

2

88 35 × 50 × 35

V/m =

9086 × 10 2 8586.75

3 × (53.5)2 × 10 6

V/m

× 10 −6 = 1.058 × 10 − 4 = –3.754 dB

Example 15.5 A TV transmitter is designed to establish communication at a distance of 50 km from it. The height of Tx antenna is 100 m and transmits a power of 45 W at 90 MHz. Find the height of receiver and field intensity at the Rx antenna. Solution:

Given, P = 45 W, f = 90 MHz or l = 10/3 m ht = 170 m

and

d = 4.12[ 100 + hr ] km ⇒

d = 50 m

50 4.12

= [ 100 + hr ] km

[12.13 – 10.00]2 = hr Þ hr = 4.54 m Now

ER =

=

88 P ht hr

Md

2

2.6 × 10 5 8325 × 10 6

V/m =

88 45 × 100 × 4.54 3.33 × (50)2 × 10 6

V/m

= 3.12 × 10 −5 V/m

Example 15.6 Find the maximum range of communication between the antennas of heights 120 ft and 60 ft. What will be the radio-horizon in this case? Solution:

We know that

LOS distance d = 1.4142 ⎡⎣ hr + hr ⎤⎦ miles

= 1.4142 ⎡⎣ 120 + 60 ⎤⎦ = 1.4142 [10.95 + 7.74] = 26.43 miles

538


Radio-horizon drh = 1.4142 hft = 1.4142 120 = 15.5 miles In other units d rh = 1.23 hft = 1.23 120 = 13.48 nmi = 1.30 hft = 1.30 120 = 14.24 km

Example 15.7 A TV Tx antenna mounted at a height of 40 m and transmitted 1.6 kw power isotropically at 6 m wavelength. Determine (i) the maximum line of sight range, (ii) the field strength at the Rx antenna mounted at a height of 5 m at a distance of 12 km, and (iii) the distance at which the field strength reduces to 20 mW/m. Solution:

We know that

(i) d LOS = 1.3[ 120 + 16] = 1.3[10.9544 + 16] = 19.44 km (ii) E R =

(iii) d 2 =

88 P ht hr

Md

2

88 P ht hr

M ER

V/m =

=

88 1600 × 5 × 40 6 × (12)2 × 10 6

88 1600 × 5 × 40 6 × 20 × 10 −3

= 8.14 × 10 −4 V/m

= 76.594 × 31.622 = 2422 km

i.e., d = 49.22 km Example 15.8 Calculate the aperture-medium coupling loss to establish the communication between the two antennas (Tx and Rx) of gain 20 dB. Solution:

We know that the aperture coupling loss Lc = 0.07 e0.055 (Gt +Gr )

Given

Gt = 20 dB = 100 Gr Lc = 0.07 e0.055 (100t +100) = 0.07 × 5987 = 4191.18

= 36.22 dB Example 15.9 Obtain the height of the common scatter volume for a radio communication between two antennas spaced away at 20 km. Assume they are normal to earth surface. Solution:

We know that the height of the common scatter volume (LN) is given by LN = 20 log (5 + gH) + 4.34gh

in which g is constant and » 0.27 km–1 H=

10 −3R d 4

and h =

10 −6R 2 Re 8


539

q = q e + qt + q r

where

ÿÿ

and

Re =

1000 × d km Re

Rr = Rt =

Q 2

1000 × 20

=

(4/3) × 6730

=

2692

= 2.22 rad

= 1.57

Hence,

ÿ q = 2.22 + 1.57 + 1.57 = 5.37 rad

and

H=

h=

6000

10 −3 × 5.37 × 20 4

= 26.85 × 10 −3

10 −6 × (5.37)2 × 897.3 8

= 3.2344 × 10 −2

Therefore, LN = 20 log (5 + 0.27 ´ 26.85 ´ 10–3) + 4.34 ´ 0.27 ´ 3.23 ´ 10–2 = 13.9920 + 0.0378 = 14.03 km Example 15.10 Find the Tx loss, if propagation is to establish at operating frequency 1.2 GHz. Assuming M = 30 dB. Solution:

We know that

L = M + 30 log (fGHz) + 10 log (dkm) + 30 log (qrad) + LN + LC – Gt – Gr = 30 + 30 log (1.2) + 10 log (20) + 30 log (5.37) + 14.03 + 36.22 – 40 = (30 + 2.375 + 13.0 + 21.89 + 11.47 + 36.22 – 40) = 75 dB Example 15.11 Find the critical values of frequency and angle at which duct propagation can take place up to 500 m duct thickness. Also find the minimum frequency, which can be propagated in this case. Assume duct gradient = 1.15 and maximum difference within duct is 0.15. Also find the coupling loss, if half-power beam width of antennas is 15°. Solution:

We know that fc =

or

1572 1.8

D

=

1572 (500)1.8

GHz

log fc = log 1572 – 1.8 log 500 = 3.196 – 4.858 = –1.662 fc = 10–1.662 = 0.02178 Ghz = 21.78 MHz

and

fc = 7.39 ´ 10–2 ´ [|1.15| ´ 500]1/2(deg) = 1.78° lmax = 0.66 ´ 3.77 ´ 10–3 = 0.66 ´ 3.77 ´ 10–3

ΔM ´ 500 0.25 ´ 500 = 0.622 m

540


f =

c

M

=

3 × 10 8

= 4.823 × 108 MHz

0.622

or

f = 4.823 ´ 102 MHz

The coupling loss

⎡ 2R ⎤ ⎡ 2 × 1.78 ⎤ Ac = − 10 log ⎢ c ⎥ = 10 log ⎢ ⎥ 15 ⎣ : ⎦ ⎣ ⎦ = 6.25 dB Example 15.12 Calculate the radius of the first Fresnel zone for a 2.5 GHz signal at a point, that is, 12 km from the source and 18 km from the destination. Repeat the calculation if distances are in miles. Solution:

Given

f = 2.5 GHz, D1 and D2 = 12 and 18 km

⎡ N ⎛ D1 D2 ⎞ ⎤ Rn = M ⎢ ⎜ ⎟⎥ ⎣ FGHz ⎝ D1 + D2 ⎠ ⎦

1/2

⎡ 1 ⎛ 12 × 18 ⎞ ⎤ = 17.3 ⎢ ⎜ ⎟⎥ ⎣ 2.5 ⎝ 12 + 18 ⎠ ⎦

1/2

⎡ 1 ⎛ 12 × 18 ⎞ ⎤ = 72.1 ⎢ ⎜ ⎟⎥ ⎣ 2.5 ⎝ 12 + 18 ⎠ ⎦

1/2

= 17.3 0.4 × 7.2 = 29.4 km

= 72.1 0.4 × 7.2 = 122.35 miles

Example 15.13 Suppose we want to establish a 915 MHz link between two points which are a straight-line distance of 25 km apart. However, 5 km from the one end of link (L1), there is a ridge of height 100 m. Find the diffraction loss (dB). Solution:

From Fig 15.11(b). d1 = [(2000)2 + (100)2]2 = 20000.25 d2 = [50002 + 1002] = 5000.99

l = 0.33 m, v = 3.89 Dd = 1.248 m ⎡1.249 ⎤ v = 2⎢ ⎥ ⎣ 0.33 ⎦

1/2

= 3.89

L(v) = 6.9 + 20 log [(v2 + 1)1/2 + v] L(v) = 6.9 + 20 log [(7.906 + 1)1/2 + 3.89] = 23.64 dB


541

Example 15.14 Determine the multi-path loss for a mobile link at distance of 25 km in free space and urban areas. Assume the height of Tx and Rx antennas to be 50 and 40 m. Solution:

We know that multi-path loss is given

For free space

L=

dn ht × hr

(25 × 10 3 )2

=

50 × 40

= 312.5 × 103

= 5.5 dB For urban area L=

dn ht × hr

=

(25 × 103 )2.8 50 × 40

LdB = 2.8[log(25) + 3 log 10] – (log 2.0 + 3 log 10) = 2.8[1.4 + 3] – (0.3010 + 3) = 9.02 dB Example 15.15 (a) Calculate the free space field strength in the plane of a half wave dipole transmitting power of 26 W at a distance of 15 km, (b) path loss for a propagation at 60 MHz, (c) also find the change in path loss, if dipole antennas are replaced by isotropic antennas and distance between them is increased up to 20 km. Solution:

(a) We know that E=

7.02 P d

=

7.02 26 15 × 10

3

= 2.386 × 10 −3 V/m

E = 2.386 mV/m (b)

M=

300 60

= 0.5 m

⎛ 15 × 10 3 ⎞ ⎜ 18 + 20 log ⎟ dB ⎜ 0.5 ⎟⎠ ⎝ = (18 + 20 [log 3 + 3 log 10)] dB = (18 + 20 [0.4771 + 3]) dB

d⎞ ⎛ LdB = ⎜ 18 + 20 log ⎟ dB = M⎠ ⎝

= 87.54 dB (c)

d⎞ ⎛ LdB = ⎜ 22 + 20 log ⎟ dB = M⎠ ⎝

⎛ 20 × 10 3 ⎞ ⎜ 22 + 20 log ⎟ dB ⎜ ⎟ 5 ⎝ ⎠

= (22 + 20{log 4 + 3}) dB = (22 + 20{0.6020 + 3}) dB = 94.04 dB Therefore change in loss = 94.04 – 87.54 = 6.50 dB

542


Example 15.16 Find the path difference, phase difference and field intensity at unit distance from the Tx, if the propagation is characterized by following parameters:

l = 0.5 or f = 600 MHz P = 205 W, d = 2.5 km and ht and hr = 50 and 45 m Solution:

We know that the path difference 2ht × hr

'l =

d

=

2 × 50 × 45 2.5 × 10 3

= 1.8 m

The total phase difference between rays is ⎛

2Q

⎝

M

R = ⎜Q +

⎞ × 1.8 ⎟ = 8.2Q = 25.75 radians ⎠

Field intensity E=

7.02 P d

=

7.02 20 2.5 × 10 3

= 12.52 mV/m

Also find the transmitted power provided field intensity at the receiver antenna is 1 mV/m. E=

88 P × ht × hr

Md

2

or 10 −3 =

88 × 50 × 45 0.5 × (2.5 × 10 3 )2

P

2

⎡ 0.5 × 6.25 × 10 3 ⎤ −3 2 P= ⎢ ⎥ = [15.78 × 10 ] 88 50 45 × × ⎣⎢ ⎦⎥

Hence, P = 2.49 ´ 10–4 W

OBJECTIVE TYPE QUESTIONS 1. The space wave is made up of direct and ground reflected waves with a finite phase. When this phase difference is an odd multiple of wavelength, the resultant signals causes (a) Constructive interference (b) Distractive interference (c) Mixed of both (d) None of these 2. The space wave is made up of direct and ground reflected waves with a finite phase. When this phase difference is an even multiple of wavelength, the resultant signals causes (a) Constructive interference (b) Distractive interference (c) Both (a) and (b) (d) None of these


543

3. The reason behind picket fencing is (a) Interference of waves (b) Phase-difference of waves (c) Multi-path of waves (d) All of these 4. A dead zone exists when destructive interference between direct and reflected (multiple reflected) waves drastically reduces signal strengths. A dead zone can be removed if (a) Antenna is moved a wavelength (b) Antenna is moved a half wavelength (c) Antenna is rotated 90° (d) Length of antenna doubled 5. In a two-way communication if both the antennas (Tx and Rx) are at 90° to the earth surface then the responsible mode for the propagation is (a) Direct waves (b) Reflected waves (c) Ionosphere waves (d) Surface waves 6. The signal loss coefficient of space wave propagation is (a) Directly proportional to wavelength (b) Directly proportional to path length (c) Directly proportional to height of antennas (d) None of these 7. The troposphere layer of atmosphere extended up to ______ from earth’s surface. (a) 80 km (b) 50 km (c) 100 km (d) None of these 8. The temperature of troposphere layer of atmosphere related with height above earth surface (a) Decreases with height (b) Increases with height (c) Remains constant with height (d) None of these 9. The area between the earth and the warm air of troposphere is termed (a) Duct zone (b) Fencing zone (c) Dead zone (d) Critical zone 10. The phenomenon responsible for tropospheric wave propagation is (a) Reflection (b) Refraction (c) Diffraction (d) Normal reflection 11. TV broadcasting takes place along the (a) Surface of earth (b) Duct zone (c) Troposphere zone (d) 10 km from the earth 12. The effective radius of the earth is 4/3 time the actual radius only if (a) A radius of curvature is equal to four times the radius of the earth (d) A radius of curvature is equal to two times the radius of the earth (c) A radius of curvature is equal to the radius of the earth (d) None of these

544


13. The radio-horizon is actually about _______ percentage more than the horizon (a) 1% (b) 4% (c) 15% (d) 20% 14. Which is incorrect for M curves (a) When the modified index of refraction increases linearly with height, M curves are straight line (b) When the slope of the M curves decreases near the earth surface, the rays curved upward over the flat earth (c) When the slope of the M curve increases near the surface of the earth super standard condition is achieved with greater coverage (d) None of these 15. The highest frequency for duct propagation is limited between frequency ranges (a) 5 GHz (b) 10 GHz (c) 40 MHz (d) 150 MHz 16. The satellite communication in an example of (a) Same path communication (b) Same wave communication (c) Reflected wave communication (d) Direct wave communication 17. The path loss exponent indicates the rate at which the path loss increases with distance in mobile communication. The value of path loss exponent for free space is (a) 1.0 (b) 2.0 (c) 3.0 (d) 4.0 18. In space wave communication both the antennas should be in LOS. The height of antennas determines the LOS distance which other name is (a) Direct distance (b) Space horizon (c) Optical horizon (d) None of these 19. The maximum radio range for tropospheric propagation between antennas of heights 100 ft and 50 ft respectively (a) 24.14 miles (b) 25.09 miles (c) 100 miles (d) 96 miles 20. The free space multi-path loss for a mobile links at distance of 25 km from the Tx is 5.5 dB. If the height of Tx 50, the height of Rx to be (a) 200 m (b) 100 m (c) 80 m (d) 40 m 21. Estimate the aperture-medium coupling loss for a communication between the two antennas (Tx and Rx) of gain 20 dB and 15 dB respectively. (a) 87.87 dB (b) 90 dB (c) 109 dB (d) None of these


545

22. What is the radius of the first Fresnel Zone for a 3 GHz signal at a point that is 12 km from the source and 18 km from the destination? (a) 34.89 km (b) 0.98 km (c) 100.5 km (d) 26.80 km 23. Determine the multi-path loss for a mobile links at a distance of 20 km in urban area if the heights of Tx and Rx antennas are 50 and 40 m. (a) 45.45 dB (b) 8.75 dB (c) 35.78 dB (d) 100 dB

Answers 1. 6. 11. 16. 21.

(a) (c) (c) (d) (a)

2. 7. 12. 17. 22.

(b) (d) (a) (b) (d)

3. 8. 13. 18. 23.

(c) (a) (a) (c) (b)

4. 9. 14. 19.

(b) (a) (d) (a)

5. 10. 15. 20.

(d) (d) (b) (d)

EXERCISES 1. Describe the troposphere and troposphere wave propagation. 2. Find the dielectric constant of a medium, if change in velocity of wave propagation is found to 1.56 ´ 108 m/s, when the waves pass from air to it. 3. A plane Gwave is propagating in a dielectric medium with the electric field intensity E = 20 y cos (1.5 ´ 109t + bz) V/m, find the velocity of waves if the dielectric constant of medium is 80. 4. Derive a relation between radius of curvature (r) and rate of change of height with relative permittivity and permeability. 5. Describe the importance of tropospheric and scatter propagations in microwave communication. 6. In order to communicate between two points which are at a direct distance of 25 km apart, a ridge of height 150 m is placed at 5 km from the one end of point (L1). Find the diffraction loss (dB). 7. Find the path-difference, phase-difference and field intensity over unit distance from a Tx at f = 500 MHz . Assume transmitted power is P = 25W, antennas are at a distance of 50 km and heights of antennas are ht and hr = 50 and 40 m. 8. “Microwave communication is only due to tropospheric propagation.” Justify the statement. 9. Write short notes on the following: (i) Index of refraction; (ii) M curves, and (iii) LOS

546


10. Define the terms surface and elevated ducts and duct gradient. Also describe duct propagation. 11. Explain how troposphere ducts are formed? Describe the duct propagation loss. 12. Describe the phenomenon of diffraction in radio wave propagation. What is the method of calculating diffraction loss? 13. Find the maximum range of communication between the antennas of heights 150 ft and 60 ft. What will be the radio horizon in a case of operating frequency 300 MHz? 14. A television Tx antenna mounted at a height of 120 ft and transmitted 1.8 kW power isotropically at 200 MHz. Determine (i) the maximum line of sight range, (ii) the field strength at the Rx antenna mounted at a height of 15 ft at a distance of 10 km, and (iii) the distance at which the field strength reduces to 80% to initial value. 15. Find the Tx loss for a communication at operating frequency 1.5 GHz, assuming M = 35 dB. 16. Obtain the height of the common scatter volume for a radio communication between two antennas spaced over a distance of 25 km. Assume they are normal to earth surface. 17. Describe the space wave propagation with reference to a dipole antenna as a transmitter. Write the expression for field pattern of vertical and horizontal dipoles. 18. Define LOS distance. Show that it is given by d = 1.4142[ ht + ht and hr are the heights of Tx and Rx antennas.

hr ] miles, where

19. The gradient of refractive index of air at the earth surface is 1.065 ´ 10–5m–1. What will be new horizon if the radius of earth is 6400 km. 20. In a communication at 240 MHz, the heights of Tx and Rx antennas are selected 50 m and 30 m respectively. Find the LOS distance and field strength at the receiver, if transmitted power is 50 W. 21. A TV transmitter is designed to establish direct communication at 100 MHz over 10 km from its location. Find the height of transmitter and field intensity at the receiver antenna, if the height of Rx antenna is 12 m and transmitted power of 45 W. 22. The space wave propagation is completed by the direct and reflected waves. Show 4Q hr ht , where that the phase difference between these two waves is given by B = Md all the parameters have usual meaning. 23. Show that for space wave propagation the field intensity at the receiver is given by ER =

88 Pht hr

Md2

V/m


547

REFERENCES [1] www.rfip.eu/radiopropagation. [2] Hafmann, B., et al., Global Positing System—Theory and Practice, 2nd ed., Shar Library, USA, 1993. [3] Riehl, H., Introduction to the Atmosphere, 3rd ed., Mc-Graw Hill, NY, 1978. [4] Webster, A.R. and T. Ueno, “Tropospheric microwave propagation—An X band diagnostic systems,” IEEE, Trans., Antennas Propagate, Vol. AP-28, pp. 693–699, 1980. [5] Jordan, E. and K. Balmin, Electromagnetic Waves and Radiating Systems, PrenticeHall, New York, 1968. [6] Mc-Donald, V.H., “The cellular concept”, Bell System Technical Journal, 58(i), 15–41, 1979. [7] Prasad, K.D., Antenna and Wave Propagation, Satya Prakashan, New Delhi, 1988. [8] Andrew, L.M., “VHF and microwave propagation characteristics of ducts,” VK3KAQ, Vol. 3–5, Jan. 2007. [9] Dougherty, H.T. and B.A. Hart, “Recent progress in duct propagation predictions,” IEEE, Trans., Antennas and Propagate, Vol. AP. 27, No. 4, pp. 542–548, July 1979. [10] Misme, P., “Affaiblissement de transmission en propagation guide par conduit atmospherique”, Ann Telecommun., Vol. 29(3–4), pp. 1–10, 1974.

C H A P T E R

16

Ionospheric Propagation

INTRODUCTION Ionosphere is defined as the upper portion of the atmosphere, because it tends to be easily ionized by solar and cosmic radiation phenomenon. Several sources of energy cause ionization of the upper atmosphere. Cosmic radiation from outer space causes some degree of ionization, but the majority of ionization is caused by solar energy. Basically, there are two principal forms of solar energy: Electromagnetic (EM) radiation and charged solar particles. Since EM radiation travels with the speed of light (i.e. 3 ´ 108 m/s), solar events that release radiation causes charges to the ionosphere about 8 minutes later. On the other hand, as charged particles have finite mass, travelling at a considerably slower speed, they require two or three days to reach the earth. Since various sources of both EM radiation and charged particles exist on the sun, solar flares can release huge amounts of radiation and particles. These events are unpredictable and sporadic. Solar radiation also varies over an approximately 27-day period, which is the duration the earth takes to complete one rotation of sun. That is, same source of radiation will face the earth once every 27 days, and hence events tend to be somewhat repetitive. The state of the ionosphere is found varying from hour to hour, day to day and season to season in much the same way as does the weather [1–2]. The properties of ionosphere depend upon free electron density, which in turn depends upon altitude, latitude, season and primarily solar conditions. The earth receives a lot of energy from the Sun in the form of radiation (=1370 watts per square metre). During the day hours the bulk of the ionization takes place at altitudes (90 km and 1000 km), where the electron density is found around 104 electrons/cc. The signals from the rockets and satellites penetrate into ionosphere, while the signals from the earth-based transmitters reflect from the ionosphere. In particular, at night signals from earth-based transmitters in the broadcast frequency range are reflected back, and in the day time these reflected signals are very weak, and hence neglected. These reflected signals become stronger as frequency of waves increases, and in particular frequency range (10 MHz to 30 MHz), a point is reached where the waves cease to be reflected back, but instead, 548


549

penetrate the ionosphere to be lost in outer space. Thus, there is a range of frequencies roughly, 3 MHz to 30 MHz, where although the radio waves are greatly attenuated, long distance transmission still occurs because of reflections of signals from the ionosphere. The strength of these signals mainly depends upon frequency of waves as well as the conditions of the ionosphere. A number of ionospheric stations have been set up in various parts of the world; they gather and record information about the ionosphere in those regions. The same information is used for future predictions, analyzing conditions and estimating optimum frequency for communication between any two points on the earth’s surface at any given time.

Historical Views The discovery of the radio wave propagation came from the sincere efforts of many people from different fields. In 1864, James Clerk Maxwell proposed that a charging magnetic field produces an electric field and vice versa. Thus, electromagnetic waves are capable to selfpropagate. For most of RF propagation, EM waves are visualized by the rays (Poynting theory) in the direction of wave propagation. Later in 1901, Marconi, successfully demonstrated Maxwell’s discovery by sending radio waves across the Atlantic Ocean. However, in 1902, Oliver Heaviside and Arthur–Kennelly proposed that there exists a conducting reflective layer that is only bouncing radio waves back to earth. This reflective layer is known as ionosphere and plays an important role in radio-wave propagation.

STRUCTURE OF IONOSPHERE As per ionization density variation, ionosphere composed of several layers in which ionization density reaches a maximum or remains roughly constant. These layers consist of electrically charged gas atoms known as ions. These ions are formed by a process called ionization. Ionization occurs when high energy UW waves from the sun enter the ionosphere region, strike a gas atom, and literally knock the free electrons of the atom. The atom converted into positive ions remains in space along with free electrons. If this process continues, UW waves intensity decreases because of the absorption of energy by the free electrons, and an ionized layer is formed. The rate at which ionization occurs depends on the density of atoms in the atmosphere as well as frequency and intensity of UW waves. In turn, these parameters vary with the activity of the sun; therefore accordingly several layers are formed at different altitudes. These layers are called D, E and F layers. The F layer is further divided into F1 and F2 layers. Lower frequency UW waves penetrate the atmosphere the least; hence they produce ionized layer at the higher altitudes, whereas high frequency UW waves penetrate deeper and produce ionized layers at the lower altitudes. The density of the ionized layer very much depends on the elevation angle of the sun, which changes frequently. This is the reason the height and thickness of the ionized layer vary with the time of day and even the season of the year. When the radiation is removed, free electrons and positive ions recombine, consequently positive ions return to their original neutral state. Between the hours of early

550


morning and late afternoon, the rate of ionization exceeds the rate of recombination and hence density of layer starts decreasing. During this period the density of ionized layers reaches its greatest value and exerts maximum influence on wave propagation. The density continues to decrease from evening to night, and reaches lowest value just before sunrise. As the sun rises density again begins increasing and reaches its maximum at noon. On the other hand, the variation of electron density with altitude [Fig. 16.1(a)] indicates that as the height deceases, the atmospheric pressure and ionization density increases until a height is reached where the ionization density is highest. Below this particular height, though pressure continues to increase, ionization density starts decreasing because the radiation has been absorbed in the process of ionization. Ionospheric propagation 5000

Altitude in km

1000 F2

Day

500 Night 300

F1 E

100

D 50 10

FIG. 16.1(a)

2

10

3

4

5

10 10 10 Electron density/cc

6

10

7

10

Typical variation of electron density with altitude.

Different layers of ionosphere The typical heights of ionospheric layers from the earth’s surface are as follows: D E F1 F2

45–55 miles 65–75 miles 190–120 miles 200 miles (50–95 miles)

50–90 km 90–140 km 140–250 km 250–400 km


551

The presence/absence of these layers and their height from the earth vary with the position of the sun [3]. Radiation in ionosphere is greatest at noon; however, it is lowest in the night. Typically, the D and E layers of ionosphere disappear (or reduce) and F1 and F2 combine in the night. Further details about D, E, F1 and F2 layers are given below. D-layer This is the lowest layer of the ionosphere. The process of ionization in this layer is slow, because all forms of solar energy that are responsible for ionization are severely attenuated by the higher layers. The reason behind that D-layer is much denser than the E and F layers, and also density of air molecules allow ions to recombine and form neutral atoms very quickly. The D-layer has the ability to refract signals of low frequencies, however higher frequency signals pass normally and hence attenuated. The amount of ionization in the layer is proportional to the elevation angle of the sun, so ionization density reaches maximum at midday. In general the D-layer appears in sunlight and disappears at night. In addition, the D-layer exists mostly during the warmer months of the year because of the greater height of the sun above the horizon and the longer hours of day light. The D-layer reflects VLF waves for long-distance communication as well as LF and HF waves for short-range communications. On the other hand, it exhibits a large amount of absorption of medium and short-wave signals hence not suitable for medium and short-wave propagations. E-layer This is one of the important regions of ionosphere for radio communication. General characteristics of E-layer are similar to D-layer. The E-layer also known as Kennelly–Heaviside layer, because the existence of this layer was first studied by these two scientists. In general, during the year, the E-layer is absorptive and does not reflect signals, however in the summer months, it does support propagation. The E-layer exists as an effective radio wavereflector only during the day. The maximum electron density of the layer varies between 4.5 ´ 105–5 ´ 104 from day to night. This layer has the ability to refract signals of frequency as high as 20 MHz and establish communication link up to 1500 miles. The critical frequency of the layer ranges from 3 to 5 MHz at noon and it reduces to 1 MHz at night, however virtual height remains constant about 110–120 km. The critical frequencies for the E-layer can be calculated by using fcE = K

4

cos Z

(16.1)

where K is a constant that depends upon the intensity of the radiation from the sun and is zenith angle of the Sun [4].

y

Sporadic E-layer The ionospheric layer associated with E-layer is sporadic layer. Since the presence of sporadic E-region is very much irregular; it is termed sporadic E-layer and denoted by Es. It is believed that the reason behind its formation is the bombardment of solar particles. It usually

552


occurs in the form of clouds, ranging height from about 1 km to several hundred km across. The occurrence of Es-layer may be observed in day and night hours as well as in any season of the year. The characteristics of Es-layer are no way affected by sun radiation. Es is a very thin layer (around 90 to 130 km) of high ionization density and found fairly stable with height from the ground. In polar regions the Es-layer occurs mainly at night and does not vary with seasons, while in equatorial zones it occurs predominately in day hours. There is no certain reason behind its formation, but studies reveal that it may be due to meteoric ionization, or may be due to vertical transport of ion clouds. This layer is often observed in the lower region of VHF and sometimes in higher frequency regions. The Es-layer is very helpful for long-distance scatter propagation of VHF waves and sometimes produces M-type of reflections between lower layers of E and F. The electron density of Es-layer is 10 times higher than that of E-layer whereas critical frequency approximately doubles the critical frequency of the normal ionospheric layer. F-Layer The F-layer was first proposal by Appleton in 1925, so also called Appleton layer and its height ranges from 140 to 400 km in atmosphere from the earth. The average height of this layer is around 270 km and virtual height is about 300 km. This is only layer which always remains ionized irrespective of hours of the day as well as season of the year. As F layer is closest to the sun, ionization density is maximum at midday and recombination occurs slowly after sunset. However, a fairly constant ionized layer is always present in this layer even in the night. Characteristics of F layer depend upon hours of the day and the elevation of the sun. As soon as the sun rises in the morning, the temperature of atmosphere begins increasing and F-region split up into two layers: F1 layer and F2 layer. The F1 layer is approximately extended from 140 km to 250 km with average height 220 km above the earth. The F2 layer ranges from 250 km to 400 km in day hours having highest electron density, however at night height of F2 layer falls to 300 km where it combines with F1 layer and merges into a single layer. F1 layer is basically formed by ionization of oxygen atoms and density is higher in summer than winter. Most of HF waves penetrate F1 layer, but reflected from the F2 layer. The critical frequency for F1 layer varies from 5 MHz to 7 MHz, whereas electron density ranges between (2–4.5) ´ 105 cc/cm3. F2 layer is formed by the ionization of UV, X rays and corpuscular radiations. The ionization density of F2-layer is affected largely by earth’s magnetic field, ionosphere tides, winds and storms. The ionization density of this layer heavily varies with solar activity and changes from sun spot minimum to sun spot maximum. The height and degree of ionization of F2 layer varies over the course of the day, with the season of the year as well as 27 days sunspot cycles. Propagation in the F-layer is capable of skip distances up to 2500 miles on a single hop, hence suitable for long-distance short-wave communications. The structure of atmosphere is shown in Fig. 16.1(b); however basic differences between above layers are listed in Table 16.1.


553

Atmosphere G-region Outer

F2 Layer F1 Layer E Layer

(c)

0 11

Stratosphere

25 0– 40 0

km

re he sp o Ion

Tropopause

50 –9 0

km

Ion os ph ere (b)

22 0k m

D Layer

km

15

km

Earth

(a)

FIG. 16.1(b)

(a) Troposphere up to 15 km (b) Ionosphere from 50–400 km (c) Outer atmosphere above 400 km

Typical heights of different layers of ionosphere.

TABLE 16.1 D

Comparison between different layers Es

F*

Occurrence Only light hours Day light hours of the day and day hours

Both in day and night hours

Both in day and night hours

Nmax (cc/cm3)

1014 to 106

105 to 4.5 ´ 105 (day) 5 ´ 103 to 104 (night)

10 times greater than E-layer

(2 to 4.5) ´ 105 (F1) (3 to 20) ´ 106 (F2)

fc

100 KHz

3 to 5 MHz

Higher than E-layer

5 to 7 MHz (F1) 5 to 12 MHz (F2)

Formation

Photo-ionization Ionization of all gases by soft X-ray radiations

Importance Not suitable for HF comm.

E

Long distance radio-propagate.

Meteoric By ionization of O2 ionization (F1). Ionization of UV Turbulent motion and X-rays (F2) of air molecules Not suitable for long distance communication

Not suitable for HF wave propagation (F1). Reflecting medium for HF radio-waves (F2)

F1—Follows Champman’s law of variations. F2—Does not follow Champman’s law of variations. And it shows a number of irregularities as located in the highest parts of atmosphere.

554


G and C regions A region further away from the 400 km from the earth is G region whereas the C region ranges from 50 km to 70 km inside D-layer. Since the density of electrons in the D-layer are not sufficient to effect appreciable bending of radio-waves and hence waves highly attenuated while passing through the layer. In C-layer the diurnal variation of absorption is maximum at noon and seasonal variation is maximum in summer.

Propagation Effect as a Function of Frequency High frequency (HF) (3030 MHz) This includes citizen band (CB) radio at 27 MHz, which is an example of frequency reuse planning, and often propagated via sky wave and can be heard hundreds of miles away. Several segments of the HF band are used for amateur radio, military ground purposes and over the horizon communications. VHF and UHF (30 MHz3 GHz) There is very little ionosphere propagation; most of the waves travel by LOS and groundbounce propagation. Applications of this frequency band include: (i) (ii) (iii) (iv) (v) (vi)

Mobile communication. Broadcast FM radio. Aircraft radio, cellular/PCS telephones. Family radio service (FRS), pagers, Police and fire departments of govt. Global positioning system (GPS).

These bands also support satellite communication because these bands are the region where satellite communication begins and signals can penetrate the ionosphere with minimal loss. Super High Frequency (SHF) (3 GHz30 GHz) Use strictly LOS propagation. Applications of this band include: (i) Satellite communication. (ii) Direct broad cast satellite television. (iii) Points-to-point links. The extra high frequency (30 GHz300 GHz) It is also called millimetre wave band and has wider band width. The wave propagation in this frequency range strictly follows LOS path. Precipitation and gaseous absorption may only be significant issues for this band.


555

MEASURES OF IONOSPHERE PROPAGATION The Critical Frequency It is the highest frequency that can be reflected back by an ionosphere layer, when a signal strikes as a vertical (90° with respect to surface) incident wave. Usually, it is denoted by fc and mathematically given as: (16.2)

fc = 9 N max

where fc is in MHz and Nmax = Maximum electron density in cc/m3. The critical frequencies are different for different layers at particular angle of incidence. The lowest value of fc is 3 MHz during the night hours and highest values varies from 10 to 15 MHz during the day. In general, there will be no communication at wave frequency f > fc, but the waves may also reflect back to earth and establish the link only when the angle of incidence is sufficiently glancing, so that mmax = sin imax is satisfied, otherwise the waves will penetrate the layer and disconnect the communication link. Therefore, the condition for radio-frequency greater than fc to be reflected will be sin i > N max >

or

sin i >

1 −

1 −

f c2 f2

,

81 N m

f2

where fc = 81 Nm

(16.3)

In general, as the angle of incidence goes on decreasing and approaches zero (i.e. vertical incidence), the electron density keeps on increasing and reaches maximum. That is, the highest frequency that can be reflected back by the ionosphere is one for which refractive index m becomes zero [1, 6]. The three frequencies transmitted at the same angle and entering in an ionosphere layer is shown in Fig. 16.2. It is clear that 10 MHz wave is refracted quite sharply than the 20 MHz and both returned to earth. The second one travels longer distance than the first one. The 100 MHz wave is obviously greater than the critical frequency for that ionized layer and therefore penetrates into space. The critical frequency is measured from an ionogram, which is a CRT oscilloscope and displays the variation in ionosphere height with function of frequency.

The Angle of Incidence This is the angle at which a wave strikes the particular ionosphere in such a way that its sum with angle at which wave leaves the transmitter is 90°. Figure 16.3 shows three radio waves of same frequency entering the ionosphere layer at different angles.

556

Antenna and Wave Propagation 100 MHz

Ionosphere

20 MHz 10 MHz

FIG. 16.2

Different frequencies entering at same angle in ionosphere layer.

Ionosphere

C A

B

Critical angle

FIG. 16.3

Earth

Different frequencies entering at different angle in ionosphere layer.

The wave A strikes the layer nearly vertical, so it is bent slightly but possesses through the layer and finally lost. When the wave is bent to an angle that is lesser than vertical (wave B), it strikes the layer and is reflected back to earth. The angle made by the wave B is termed critical angle for that particular wave (frequency). That is, any wave entering at angle greater than this angle will penetrate the ionosphere and lost in space. In third case, the wave C strikes the ionosphere at the smallest angle at which it can be reflected and still return to earth. The wave striking at any angle smaller than this angle will be retracted, but will not return to earth. However, the value of critical angle may vary, if the frequency of transmitted wave is changed. In general, as the frequency of the radio waves is increased, the critical angle reduced for refraction to occur. That is, critical angle ic µ f–1.


557

Maximum Useable Frequency We have seen that the critical frequency is the frequency of communication between two points that is mainly characterized by the ionization density. In other words, critical frequency is a property of an ionosphere. However, there are other frequencies known as maximum frequency, lowest frequency, and optimum frequency that must also be taken into account for a successful communication. These frequencies depend upon the height of ionosphere layer, earth radius as well as angle of incidence other than the ionosphere density. The highest frequency, which can be reflected back to earth only at a specific angle of incidence rather than the vertical, is known as maximum useable frequency (MUF). The value of MUF differs for different layers of ionosphere and normal value lies between 8 MHz and 35 MHz, however in case of unusual solar activity MUF may be as high as 50 MHz. In general, the MUF is approximately 3 times greater than the critical frequency. The values of both the MUF and fc vary geographically and increase at latitudes close to the equator. For a successful ionosphere propagation, the following conditions must be satisfied: N = Nmax and f = fmax, i.e. maximum useable frequency. Therefore the angle of incidence may be given as [3, 7]

fc2

1 −

sin i =

2 fMUF

fc2

fc2

2 f MUF

2 f MUF

or

sin 2 i = 1 −

or

fMUF = sec i. * fc

= cos2 i

(16.4)

which is known as ‘Secant law’. From Eq. (16.4), it is clear that MUF for a particular layer is greater than the critical frequency by a factor sec i, where i is angle of incidence. The maximum angle of incidence of a wave for a layer at the 400 km height from the earth may be calculated as i = imax and sin imax = Hence

R R+h

=

6370 6370 + 400

=

6370 6770

= 0.941

imax = sin–1(0.941) = 70.22°

However the angle of incidence corresponding to this limiting distance is found to be around 74° for F-layer. Hence, the MUF for this particular case will be fMUF = sec 74°f

or

fMUF = 3.6 fc

(16.5)

Wave propagations higher than the MUF are normally retracted so slowly that they return to earth beyond the desired location or pass through the ionosphere and finally lost. Variations in ionosphere only may alter the predetermined MUF. This is true only for the radio waves being retracted by the highly variable F2 layer. The value of MUF is highest around noon

558


when ultraviolet waves from the sun are the most intense. As recombination of free electrons and particles started in the evening hours, MUF falls sharply and reaches lowest at night.

Calculation of MUF There are two cases where f MUF are determined: short-distance and long-distance communications. The short-distance communication is limited to 500 km and in this case earth may be considered as flat surface, however, in long distance (more than 500 km) communication the earth need to be considered as a curved surface. MUF from flat surface In order to calculate MUF in flat surface, let us consider a thin ionized layer with sharp ionization density gradient, which gives mirror-like reflection of radio waves (Fig. 16.4). From DPQS cos i =

QS

h

=

PQ

(16.6)

D2 h2 + 4

Q

i

r

h

P

D/2

FIG. 16.4

S

R

Reflection of waves in ionosphere.

The maximum useable frequency for which the wave is to be reflected from the ionosphere and return to the earth are f = fm, r = 90° and N = Nmax. Therefore,

N = sin i = 1 −

fc2 2 fMUF

or

cos2 i =

fc2 2 fMUF

=

4h2 4h2 + D2

[from Eq. (16.5)]


2 fMUF = fc2

559

4h2 + D2

MUF = fc

4 h2

⎛D⎞ 1+ ⎜ ⎟ ⎝ 2h ⎠

2

(16.7)

where D = propagation path distance h = height of ionospheric layer fc = critical frequency for particular layer. MUF for curved earth surface In this case, let us consider the reflecting layer as concentric with earth surface and height of layer remains the same (i.e., h). The transmitted wave leaves the transmitter tangentially to the earth and strikes the ionospheric layer at angle i. If 2q is the angle subtended by the transmission distance D at the centre of the earth O (Fig. 16.5) then, it can be written 2q = D/R or D = 2Rq, where D is termed skip-distance. Then from DQTP, PT = R sin q and OT = R cos q QT = OQ – OT = h + R – R cos

q

PQ = PT 2 + QT 2 = ( R sin R )2 + (h + R − R cos R )2 2

Therefore,

(h + R − R cos R )2 fc2 ⎛ QT ⎞ cos2 i = ⎜ = = ⎟ 2 ( R sin R )2 + (h + R − R cos R )2 f MUF ⎝ PQ ⎠

FIG. 16.5

Incidence of a wave for a layer at the height h.

(16.8)

560


From Fig. 16.5, it is clear that the curvature of the earth limits both the MUF and propagation distance D and the limit is obtained when waves leave the transmitter at grazing angle (i.e. < OPQ = 90°). When D is maximum, q is maximum, therefore

cos R =

OP QO

=

R R+h

1

=

h⎞ ⎛ ⎜1 + ⎟ R⎠ ⎝

The actual value of q is very small as D is very small compared to the radius of the earth. Hence, h⎞ ⎛ cos R = ⎜ 1 + ⎟ R⎠ ⎝

−1

h h ⎛ ⎞ = ⎜1 − + ... ⎟ ∴ << 1 R R ⎝ ⎠

R2 h 2h as q is small 1 − =1 − ⇒ R2 = 2 R R 2h

D2 = 4 R 2R 2 = 4 R 2 ×

Therefore

Dmax =

That is

R

(16.9)

= 8 hR

8 hR

(16.10)

and h = D2/8R, which gives cos q (1 – D2/8R2) and sin q = q = D/2R. Therefore from Eq. (16.8):

fc2 ( f MUF )2max

=

⎡ ⎛ D2 ⎢h + R − R ⎜1 − ⎜ 8 R2 ⎢⎣ ⎝

⎞⎤ ⎟⎟ ⎥ ⎠ ⎥⎦

2

⎡ 2 2 ⎧ ⎛ ⎢ R 2 D + ⎪⎨h + R − R ⎜ 1 − D ⎜ ⎢ 4R 2 8 R2 ⎪⎩ ⎝ ⎣⎢

2⎤

=

⎞ ⎪⎫ ⎟⎟ ⎬ ⎥ ⎥ ⎠ ⎭⎪ ⎦⎥

⎛ 2 D2 ⎞ ⎜⎜ h + ⎟ 8R ⎠⎟ ⎝

2

⎛ D2 D2 ⎞ + ⎜4 + ⎟ ⎜ 4 8R ⎟⎠ ⎝

2

1

which gives

(MUF) max

2 ⎞2 ⎤ 2 ⎡ D2 ⎛ D ⎢ + ⎜h + ⎟ ⎥ ⎜ ⎟ ⎥ ⎢ 4 8 R ⎝ ⎠ ⎦ ⎣ = fc 1/2 ⎡ D2 ⎤ ⎢h + ⎥ 8 R ⎥⎦ ⎢⎣

(16.11)


561

Lowest Useable Frequency The lowest useable frequency (LUF) is sufficient to establish the communication between the two points. Basically, a wave propagation whose frequency is below LUF, refracted back to earth, but at shorter distance than the desired distance, hence no communication. For a given angle of incidence and set of ionospheric conditions, the LUF for a successful communication depends not only on rate of refraction, atmospheric absorption, and signal to noise ratio [8], but also on absorption in the D layer during day hours. Whereas at night LUF is primarily limited by increased noise at lower wave frequencies (Fig. 16.6). Ionosphere

LUF

Frequency below LUF

Transmitter

FIG. 16.6

Receiver

Lowest frequency enough to establish communication.

In addition, the LUF of any wave propagation depends on following factors: (i) Effective radiated power, (ii) the required field strength, which basically depends upon radio noise at the receiver site. The value of day hour LUF is normally much higher than the night time LUF and further increases during SID’s. The value of LUF is estimated from the measurement of noise level at the receiver location and also from the value of sky-wave absorption in particular transmission path.

Optimum Working Frequency As we know, MUF is maximum and LUF is minimum frequencies, which can be used to establish the communication between any two points on the earth. However, neither the MUF nor the LUF is a practical operating frequency. Radio waves at the LUF complete the desired propagation, with low SNR that causes high level probability of multi-path propagation, i.e., more losses. On the other hand, radio signal operating at or near the MUF can result in frequent signal fading and may dropout whenever ionospheric variations alter the nature of the propagation path. Therefore, the most practical operating frequency is one that we can

562


rely on, with the least numbers of difficulties and should be high enough to avoid the problems of multipath loss, absorption and noises at the lower frequencies. But at the same time it should not be so high as to result in the adverse effects of rapid changes in the ionospheric conditions. Due to the above fact it is therefore beneficial to use the frequency somewhat (» 15%) lower than the predictable MUF, such frequency is called optimum working frequency. Like MUF, optical working frequency also varies considerably with time of day, season as well as months of the year. However, in practical, it is not possible to change the frequency of communication according to situations. Hence for continuous communication, it is necessary to use more than one frequency; each one for the day and night applications and even sometimes a third frequency may also be selected for transition period. In general, the lower frequency in the night and higher frequency in the day time is preferred; this is because the waves of lower frequency are bent round more quickly than that of higher frequencies. In general, optimum working frequency may be any value from 0.50 to 0.85 times the MUF. The typical optical working frequency for day hours is 6.45 GHz whereas for night it is 5.0 GHz.

Skip Distance The skip distance is the distance from the transmitter to a point where the waves first return to earth after reflection from the ionosphere. The length of skip distance is the function of frequency of the wave, angle of incidence, and also the degree of ionization present in the path of propagation. Skip distance can also be defined as maximum distance travelled by a wave at maximum useable frequency after reflection from the ionosphere. Higher the operating frequency greater the skip distance and for a frequency less than fc of a layer, skip distance is zero [9].

Relation between Angle of Incidence and Skip Distance For a wave propagation at frequency greater than the critical frequency (f > fc), effects of the ionosphere appear in terms of the angle of incidence at the ionospheric layer. The range of communication (i.e., distance between Tx and Rx) decreases as the angle of incidence decrease. And this phenomenon continues until an angle of incidence is reached at which the distance becomes minimum. A minimum angle is found, beyond which the wave penetrates the layer and does not return to earth. The skip distance and dead zone are shown in Fig. 16.7. In this figure, wave (1) corresponds to large angle of incidence i and small electron density N such that

N = sin i = 1 − and

m

81 N f2

(16.12)

is slightly greater unity. The wave returns to earth after slight penetration into the layer. In case of wave (2), the angle of incidence is further decreased, hence sin i and m decreases still more and N also becomes comparatively large. As a result, the wave (2) penetrates more compare to wave (1)


563

Angle with which reflection does not occur, i.e., waves escape

4 3

Lower edge of ionosphere

i=0 i4

i3

i2

i1 1

2

Dead zone or silence zone Skip distance or skip zone

B

C

Transmitter

FIG. 16.7

Skip distance representation.

before it reaches the earth. In the third case, when angle of incidence is small enough so that m = sin i, cannot be satisfied even by maximum electron density of the layer, hence other waves (i.e., 3, 4, …) penetrate in the layer and never return to earth. For wave propagation on flat earth at frequency f = fMUF, the skip distance is calculated by

Dskip

⎛f ⎞ = 2h ⎜ MUF ⎟ ⎝ fc ⎠

2

− 1

(16.13)

where h = height of ionospheric layer fc = critical frequency of the layer

Virtual Height The virtual height is defined as the apparent height of an ionospheric layer estimated from the time interval between the transmitted signal and the ionospheric echo at the receiver at vertical incidence, provided the velocity of propagation is 3 ´ 108 m/s over the entire path. In other words, the virtual height is a height to which a wave sent vertically upward and travelling with the speed of light would reach taking the same two ways travel time as does the actual wave reflected from the layer. The concept of virtual height in ionospheric propagation can be best understood by considering Fig. 16.8, where a wave is refracted from the ionospheric layer as happens in case of refraction from the prism. Due to this refraction, the transmitted wave takes curve path, and bends down gradually and finally return to destination. Hence,


Projected height

True or actual height (h)

564

i

Actual path

F

FIG. 16.8

Lower edge of ionosphere tc

Actual path

Virtual height (h)

R

O

Virtual and actual heights of an ionized layer.

the actual path of the wave via ionosphere is curve while virtual path is triangular through point D, i.e., TD and DR. The height of point D from the earth surface (say point O) is referred as virtual height of the ionized layer. The virtual height is always greater than the actual height. The virtual height has great advantage of being easily calculated and also very useful in transmission path range calculations. There are two cases, in which virtual height is calculated: (i) flat earth and (ii) curved earth. (i) For flat earth propagation, the ionospheric conditions are assumed to be symmetrical for both incident and reflected waves. Referring to Fig. 16.9, the transmission path range (D) can be calculated as follows:

tan B f = or

D=

HO

=

TO

h D/2

2h

(16.14)

tan B f

(ii) Referring to Fig. 16.10, the transmission path range for the curved earth can be calculated from DTOH, where sin i R

As The angle OTH, (p –

=

sin(Q − B ) ( R + h)

sin(p –

=

sin B

(16.15)

R+h

a) = sin a

a) = p – (i + q) Þi = a – q

(16.16)


FIG. 16.9

Virtual height calculation for flat earth.

FIG. 16.10

565

Virtual height calculation for curved earth.

Therefore, from Eqs. (16.15) and (16.16), we get

sin(B − R or Again, from Fig. (16.10)

Therefore, where

R)

=

sin B R+h

⎛ R sin B ⎞ R = B − sin −1 ⎜ ⎟ ⎝ R+h ⎠

a + b = 90 Þ a

(16.17)

= 90°

⎛ R cos C ⎞ ⎤ ⎟⎥ ⎝ R + h ⎠⎦

⎡

R = ⎢(90 − C ) − sin −1 ⎜ ⎣

cos

b = sin (90° – b)

Now, since

Arc = Radius ´ Angle, i.e. D = R ´ 2q

Hence,

⎡ ⎛ R cos C ⎞ ⎤ D = 2R ⎢(90° − C ) − sin −1 ⎜ ⎟⎥ ⎝ R + h ⎠⎦ ⎣

(16.18)

An instrument known as Ionosode is normally used to measure the virtual height. Ionosode is a short-wave transmitter tunable through the whole short-wave range. It transmits various

566


frequency pulses (pulse duration = 150 ms) whose echos are analyzed by the means of radar. In order to measure virtual height, a pulse-modulated radio wave is vertically transmitted and reflected signal from the ionosphere is received back close to the transmission point. The time T duration required for the round trip is noted and then virtual height is determined by using

h=

cT 2

where h = virtual height c = velocity of light (m/s) T = time period (s) Ionosode automatically plots the variation of virtual height with operating frequency, which is known as ionogram.

REFRACTIVE INDEX OF THE IONOSPHERE An ionosphere, in its normal state, contains positive ions and free electrons, in such a way that the mass of free electrons are negligible than the mass of ions. Therefore, the propagation of the radio waves is influenced by these electrons only and as a result electrons of ionosphere are set in motion. Due to this motion electrons start vibrating along the path parallel to the electric field of radio waves and represent a current proportional to amount of vibration. The actual current flowing through any volume of space consists of both the current components: capacitive current and inductive current. The capacitive current leads the voltages by 90°, while inductive current lags the voltages by 90°. In the present case, the effect of earth’s magnetic field on the vibration of ionospheric electrons lags behind the electric field of the propagating wave, thus the resulting current is inductive current, i.e. the component of conduction current obtained by motion of free electrons. Thus, the presence of free electrons in space decrease the actual current and hence the dielectric constant er. As a result, the path of propagating radio waves are bent toward earth, i.e. from higher electron density to lower electron density, and finally waves returned to earth completing the communication link [9]. If E = E0 sin wt V/m is the electric field intensity of radio waves across a cubic metre of space, then the force exerted on each electron will be F = – eE = – eE0 sin

wt

(16.19)

where e = charge of electron in coulombs E0 = maximum amplitude of electric field w = angular frequency. ÿ

If this force causes a velocity v to each electron, then the electrons will start moving with acceleration dv/dt in the direction opposite to the field (F). Therefore, according to Newton’s law of motion

567


− Ee = m ×

Ee

dv = −

dv dt

dt

(16.20)

⎡ eE ⎤ v = ⎢ 0 ⎥ cos X t ⎣ mX ⎦

(16.21)

m

After simplification, this yields

Again if N is the electron density per cubic metre, then the instantaneous electron current can be given by ie = − Ne v ⎛ Ne2 ie = − ⎜ ⎜ mX ⎝

A m2 ⎞ ⎟ E0 cos X t ⎟ ⎠

(16.22)

Appearance of term cos wt in Eq. (16.22), indicates that ie lags behind the electric field E by 90°. In addition to the ionized area, there is always found unionized air in space, i.e., apart from induction current ie there will also be distance point current id which satisfies Maxwell’s equation Ñ ´ H = Jc + Jd = sE + ¶D/¶t. Therefore, the distance point current (also known as displacement current) in the present case can be expressed as

id =

dD dt

=

d dt

(F 0 E ) = F 0 E0X cos X t

(16.23)

Therefore, the total current iT which flows through a cubic metre of an ionosphere is ⎡ Ne2 ⎤ iT = ie + id = (E0X ) cos X t ⎢F 0 − ⎥ = F eX E0 cos X t mX 2 ⎥⎦ ⎢⎣

(16.24)

where ee = e0 – Ne2/mw2 and is termed the effective dielectric constant of the ionosphere volume. Therefore, the relative dielectric constant er = ee /e0 can be written as Fr =

⎡ Ne2 ⎤ = ⎢1 − ⎥ F0 mF 0X 2 ⎦⎥ ⎣⎢ Fe

So, if we substitute m = 9.107 ´ 10–31 kg, and e = 1.602 ´ 10–18 C w = 2pf, f is operating frequency, and e0 = 8.854 ´ 10–12 F/m.

(16.25)

568


Equation (16.25) reduces to ⎡

F r = ⎢1 −

⎣

81 N ⎤ ⎥ f2 ⎦

(16.26)

Therefore, the refractive index for such medium can be defined by n=

c vp

=

Velocity of light in air Phase-velocity in the medium

But, for a loss-less medium, where mr is unchanged by the presence of electrons, i.e., mr = 1, the phase velocity is equal to c/ F r . Therefore, for such medium the refractive index n can be expressed as n = F r . 1

That is

81 N ⎤ 2 ⎡ n = ⎢1 − ⎥ f2 ⎦ ⎣

(16.27)

which indicates that the refractive index of ionosphere is less than one for air and this reduction further increases with the number of free electrons (N). While deriving Eq. (16.27), it was considered that electrons do not undergo any inelastic collision during their movement and hence there is no dissipation of energy. In general, effective conductivity of ionosphere has also not been considered because of reflection (refraction) in F layers (where the collision frequency is very small) hence conductivity is also correspondingly low; as a result, there is no conductive loss of energy too.

EFFECT OF THE EARTH MAGNETIC FIELD As we have seen, the characteristics of radio waves remain unaffected by earth magnetic field, if it is propagated through un-ionized medium. While electrons and ions in the ionized ionosphere are influenced by the fields of travelling EM waves as well as the earth’s magnetic fields, which cause the charged particles to set in motion along the circular or spiral paths. Then a force is exerted on the vibrating electrons and twisted their paths. Finally, this phenomenon reacts on the incident radio waves and split them into the ordinary and extraordinary waves. This phenomenon is known as magneto ionic splitting. The properties of the ordinary and extra-ordinary waves differ only in F-layer and they are as follows: (i) They travel through ionosphere along slightly different paths, as they bend with different amount by the ionosphere. (ii) The velocity of these waves is different. (iii) The rate of absorption of energy is also different. (iv) Both the waves have elliptical polarization and they rotate in opposite direction. (v) The critical frequency of extra ordinary wave (fce) is always greater than the critical frequency of ordinary waves (fc). And amount of different is approximately half the gyrofrequency (fg).


569

Gyrofrequency A frequency of time period equal to the period of revolution of an electron in its circular orbit under the influence of the earth’s magnetic field of flux density B is termed as gyrofrequency. Since for an electron the earth’s magnetic flux density (B) is 5.0 ´ 10–3 Wb/m. Therefore the gyrofrequency (fg) will be equal to fg = Be/2pm. Substituting the value of m, e and B, we obtain

fg =

5 × 10 −3 × 1.6 × 10 −19 2 × 3.14 × 9.1 × 10 −31

MHz = 1.47 MHz

The impact of the gyrofrequency on the incident waves and then on the polarization can be described as follows [9, 10]: (i) If f > fg: In this case the electrons follow an elliptical path and this path gets narrower as frequency increases further. That is a high frequency plane polarized wave leads elliptically polarized wave as it reflected from the ionosphere [Fig. 16.11(a)]. (ii) If f < fg: In this case the electrons vibrate in small loops making several loops just like a stretched spiral and hence the polarization of propagating wave does not affect much [Fig. 16.11(b)]. (iii) If f = fg: This is the case where resonance phenomenon and oscillating electron receive more and more energy from the incident waves. Due to which, the velocity of these vibrating electrons increases and they describe larger and larger orbits. In this situation, there is a high possibility of inelastic collision and hence large amount of energy is dissipated from the incident radio waves. This loss of energy is high at frequency equal to or near gyrofrequency (fg), this is the reason why propagation between frequency range 1–2 MHz is avoided during day hours in D-region. On the other hand, at the frequencies beyond 2 MHz the attenuation (loss of energy) is minimum and corresponding propagation is preferred for communication. In addition the earth’s magnetic field also affects polarization characteristics of the incident radio waves. At the average magnetic field intensity » 40 A/m, the ionosphere behaves like an isotropic media (i.e., different er values in different direction) [Fig. 16.11(c)].

(a) f > fg

FIG. 16.11

(b) f < fg

(c) f = fg

Impact of gyrofrequency on the incident waves.

570


Gyromagnetic Field (GMF) This is the magnetic field which originates from the rotation of the molten iron core of our planet. The GMF is weaker near the polar regions and strongest near the equatorial regions and at night hours (i.e. opposite to the sun). GMF extends millions of kilometres into space. The shape of GMF is like a water drop with the head pointing toward the sun. This magnetic field produces the magnetic flux lines between poles, which make us comfortable to navigate by use of a compass. The GMF plays a major role in the dynamics of the earth’s atmosphere and without protection of GMF our planet surface may go under a constant bombardment of these charged particles. Because of the importance of the GMF in trapping, charged particles are found necessary for sky wave propagation and even a short term variability of the GMF influences propagation.

REGULAR AND IRREGULAR VARIATIONS IN THE IONOSPHERE Being the upper portion of atmosphere, the ionosphere absorbs large amount of radiant energy from the sun. The radiation from the sun carries ultraviolet, a, b and cosmic rays and produces free electrons and positive as well as negative ions in the ionized region. That is why the characteristics of the ionosphere depend mainly on the angle of the sun rays throughout the day. The ionospheric variations mainly affect the virtual height and critical frequency because they are directly related to electron density. Basically, there two types of ionospheric variations: (i) normal (regular) and (ii) abnormal (irregular) variations. However both the variations have significant effects on the radiowave propagation. The former appears daily diurnal, seasonal and yearly, i.e. in cycles and can be predicted in advance. Seasonal variation in the ionosphere occurs due to earth revolution around the sun. Here, the relative position of the sun moves from one hemisphere to the other, changing seasons. Seasonal variation of D, E and F1 layers is due to the highest angle of sun’s rays. Hence, the ionization density of these layers is greatest during summer. On the other hand, F2 layer follows different patterns; its ionization is greatest in winter and least in summer. As a result, operating frequency of radio propagation in F2 layer is higher in the winter than in summer.

Eleven-year Sunspot Cycle and 27-day Sunspot Cycle Sunspot is one of the most notable phenomena on the surface of the sun. It is a regular appearance and disappearance of dark (shaped area). The exact nature of sunspots is not known, but it appears like they are caused by violent eruptions on the sun and characterized by unusually strong magnetic fields. These sunspots are responsible for variations in the ionization level of the atmosphere, hence all the characteristics. The occurrences of sunspots are of course unexpected and the lifespan of individual sunspot is variable, however the regular cycle of sunspot has not been observed. This cycle has both a minimum and maximum


571

level of sunspot activity that occurs approximately every 11 years. During the period of maximum sunspot activity the ionization density of all layer increases. Because of this, absorption in the D layer increases and the critical frequencies for the E, F1 and F2 layers become higher. In these situations, higher operating frequencies must be used for long distance communications. During the period of minimum sunspot activity, lower frequencies are suitable at night and higher frequencies are rarely used for long-distance communication. It is also observed that critical frequency closely follows the sunspot cycle with minimum and maximum values at 6 MHz and 12 MHz respectively. In particular, the critical frequency fc of E layer varies from 3.1 MHz to 3.8 MHz. Therefore, the prediction of ionosphere characteristics based on eleven years sunspot cycle can be made in advance. As the sun rotates on its own axis, these sunspots are visible at 27-day intervals, that is, the approximate period sun takes to make one complete rotation. The 27-day sunspot cycle causes variations in the ionization density of the layers on a day-to-day basis. The fluctuations in the F2 layer are greater than from any other layer, that is why precise prediction on a day-to-day basis of the critical frequency of the F2 layer is not possible. Therefore in calculating frequencies for long-distance communication, allowances for the fluctuations of the F2 layer must be considered. The most common irregular variations are sporadic E layer, sudden ionospheric disturbance ionospheric storms, tides and winds, fading and whistlers, etc. These irregular variations in ionospheric conditions have drastic effect on wave propagation and directly affect communication capabilities without any warning. The sporadic E layer has already been described; remaining variations are considered in the next section.

Sudden Ionospheric Disturbances (SID) It is the most startling ionospheric irregularities, and also known as Mongel–Dellinger effect. These types of disturbances may occur without warning and may prevail for any length of time, from a few minutes to several hours. The phenomenon is caused by sudden bright eruptions on the sun that produces a large increase in the ionizing radiations that reach up to the D layer. As a result, ionization density of the layer increases causing complete absorption of HF radio waves. The intensity of SID tends to be greatest in the region where sun’s radiation is normal to the earth. SIDs are caused mainly due to radiation of intense UV rays from the solar flares, which penetrates the upper portion of ionosphere and causes very strong increase in ionization in the D region. SID always occurs in day light hours only in D layers not in E, F1 and F2 layers. This sudden ionosphere disturbance is often accompanied by disturbances in terrestrial magnetism and earth currents. It has also been noted that during SID the VLF propagation actually improves.

Ionospheric Storms Ionospheric storms are nothing but disturbances in the earth’s magnetic fields. The reason behind ionospheric storms is the high absorption of sky waves and abnormal changes in critical frequency of E and F2 layers. Ionospheric storms are associated with both solar

572


eruptions and rotation of the sun. It is believed that ionospheric storm results from particle emission from the sun, particularly a and b rays. Particles radiated from a solar eruption have a slower velocity than UV light waves produces by the solar eruption. The major effects of ionospheric storms are a turbulent ionosphere and very erratic sky wave propagation. The critical frequencies are lower than normal, particularly for the F2 layer. Ionospheric storms affect the higher F2 layer first and reduce its ions density, however lower layer is not seriously affected by the storms unless the disturbances are high. The practical influences of ionospheric storms are that the range of frequencies that can be used for communications is much lower than normal and hence communications are possible only at the lower working frequencies. During ionospheric storms signal strengths drop and fluctuate quite rapidly and long time variation in the ionosphere corresponds to variations in the solar activity. Ionospheric storms appear simultaneously in the day and night hours and extend 3 to 4 days and then only come to normal condition. In general, it re-occurs in the interval of 27 days and it is highest near polar region and least at the equators.

FADING Fading is the distortion/fluctuation that a carrier-modulated telecommunication wave experiences over certain propagation media. It can also be defined as undesirable variations in the intensity or loudness of the received waves. It is most troublesome and serious problem in receiving radio waves. There are several conditions that can produce fading, such as variations in the heights and density of ionization layers. Fading is a common characteristic of the radio signals, but more likely with high frequency signals. In case of severe fading, the field strength of radio signals varies from 10 dB to 20 dB. The various types of fading affecting radio-wave propagation are: (i) (ii) (iii) (iv) (v)

Polarization fading Interference fading Selective fading Absorption fading Skip fading

Polarization fading Polarization fading occurs when radio waves meet refraction from the ionosphere or reflection from the earth’s surface, because they cause random changes in the polarization of wave. Because of changes in polarization of receiving waves, the polarization of wave with respect to antenna is constantly changing giving rise to changes of signal level at the receiver and produces polarization fading. Interference fading It is one of the most serious types of fading in radio wave propagation. Because of continuous fluctuations in the ionosphere, the path length of each wave refracting from ionosphere


573

undergoes small variation and relative phase of the waves reaching at the receiver vary in a random fashion. Therefore, the amplitude of their resultant varies continually, which is termed interference fading. The reasons behind interference fading are as follows: (i) Because of fluctuations of height of ionic layer, a radio wave of single frequency is used. (ii) It may also occur due to interference between upper and lower frequencies sky waves. (iii) It may also occur due to interference between sky waves reaching the receiver via different hops or paths. (iv) It may also occur due to interference between the ground and sky waves, particularly at the lower end of HF bond. Selective fading Selective fading (resulted due to multipath propagation) is frequency dependent since each frequency arrives at the receiving point via a different radio path. When a wide band of frequencies is transmitted simultaneously, each frequency varies in the amount of fading. When selective fading occurs, all frequencies of the transmitted signal do not retain their original phases and relative amplitudes. This fading causes severe distortion in the signal and limits the total signal transmitted. Amplitude modulated signals are more distorted by this fading rather than single side band signals, that is why selective fading can be reduced by using exalted carrier reception (ECR) and also signals side band (SSB) systems. Absorption fading This type of fading results from the absorption of the RF energy in the ionosphere. Absorption fading occurs for a larger period than other types of fading because absorption takes place slowly. Skip fading Skip fading occurs at a distances near the skip distances and results due to variation in height or density of ionized layer, which moves the receiving point in out of the skip zone. In general, fading on ionosphere layer is due to multipath propagation between Tx and Rx. Multipath fading is reduced by techniques called space diversity and frequency diversity. In space diversity two or more receiving antennas are spaced away, and since fading does not occur simultaneously at both antennas, hence enough output is almost always available from one of the antennas to provide a useful signal and continue the communication. On the other hand, in frequency diversity pair of transmitters and receivers are used and both of them tuned to a different frequency, where the same information being transmitted simultaneously over both frequencies. One of the two receivers will almost always provide a useful signal, hence continue the communication.

574


ATTENUATION FACTOR FOR IONOSPHERIC WAVE PROPAGATION In order to find attenuation factor (a) for a wave propagation through ionosphere the medium must be a lossy medium. If it is so, attenuation factor (a ) for such medium can be given as [4, 11] ⎡ NF B =X⎢ ⎢ 2 ⎢⎣

⎛ ⎜ 1+ ⎜⎜ ⎝

⎞⎤ ⎛T ⎞ ⎟⎥ 1 − ⎜ ⎟ ⎟⎟ ⎥ XF ⎝ ⎠

1/2

2

(16.28)

⎠ ⎥⎦

⎡ ⎤ Ne2 where F = F 0 F r = F 0 ⎢1 − 2 2 ⎥ F 0 m(V + X ) ⎥⎦ ⎢⎣

ÿs and

m

=

Ne2V

= Equivalent conductivity =

m0

and c =

m(V 2 + X 2 )

1

NF

Substituting the values of e,

m

and

s

into Eq. (16.28) yields 1

⎡ NFF B =X⎢ 0 0 r ⎢ 2 ⎣⎢

⎛ ⎜ 1+ ⎜⎜ ⎝

⎛T ⎞ ⎜ ⎟ ⎝Xc ⎠

2

⎞⎤2 ⎛ − 1⎟ ⎥ = X ⎜ 1 ⎟⎟ ⎥ c ⎜2 ⎠ ⎦⎥ ⎝

1

F r2

T X 2 c2 2

+

⎞2 − 1⎟ ⎟ ⎠

1

X ⎡1 = ⎢ c ⎢2 ⎣

F r2 +

⎡ ⎢(1 − ⎣

Fr )

V⎤ X ⎥⎦

2

F r ⎤⎥ 2 − 2⎥ ⎦

(16.29)

where u = collision frequency and er = dielectric constant. Equation (16.29) is the general expression of attenuation factor (a) for any propagation at frequency f. However, for frequency not too close to MUF, where s/wc ££ 1. The value of a reduces to

B=

or

B=

T

N

2

e

=

T 2

N0 120QT 60QT = = F0Fr Fr 2 Fr

60Q Ne2V

F r m(V 2 + X 2 )

(16.30)

(16.31)


575

Equation (16.31) shows that for the frequency w >> u the attenuation varies approximately as the inverse square of the frequency, hence use of maximum possible high frequency is desirable. However, in general, attenuation is negligibly small, except in the region near the lower edge of the ionosphere (i.e., D layer) Faraday rotation and group delay Faraday rotation is the random rotation of a wave polarization vector as it passes through the ionosphere. It is most pronounced below 10 GHz and makes a certain amount of polarization loss on satellite links. Most satellite communication systems use circular polarization, because it limits Faraday rotation, hence polarization loss. Group delay occurs when the velocity of propagation differs from the velocity of light c for a wave passing through the ionosphere. This is an important concern for the wide band systems, because the group delays do vary with frequency and proportional to 1/f2. This distortion of wide band signals is called dispersion [11].

ENERGY LOSS DUE TO COLLISION IN IONOSPHERE The radio waves travelling through the ionosphere possess kinetic energy and this kinetic energy is imparted to the free electrons, as a result these free electrons are set in vibration. These vibrating electrons collide with ionospheric gas molecules from time to time though gas pressure is low, and electrons transfer kinetic energy to gas molecules. The energy extracted by the free electrons and lost in the collision depends upon the probability of collision, the amount of energy lost per collision as well as number of electrons. The rate of collision of electrons with neutral particles in the ionosphere is termed as collision frequency. The effect of the collision frequency decreases as frequency of radio wave propagation increases. That is why higher frequency radio waves do not play a major role as large in collisions as low frequencies do. The probability of collision of an electron with the gas molecules depends upon the gas pressure [12]. Therefore, at high gas pressure, the probability of collision is more, hence more is the absorption of energy suffered by a wave passing through the ionosphere. This is the reason that most of the energy loss caused by ionospheric radio wave propagation takes place at the lower layer of the ionosphere where the gas pressure is maximum, i.e. in the lower part of the E and D layers. Since remaining part of the ionosphere has very low atmospheric pressure, the radio waves suffer very small energy loss in this part. The energy lost per collision depends upon the velocity acquired by the electron in its vibration. This velocity is proportional to the electric field intensity and inversely proportional to the frequency of radio waves. Thus, the absorption increases as frequency decrease. Basically there are two types of ionospheric absorption; non-deviative and deviative. The term non-deviative comes from the fact that the absorption is heaviest near lower layer of the ionosphere where the effects of refraction on the travelling wave are very small. The signals passing through this absorbing medium will travel practically in a near straight line a non-deviating path. Therefore, it is called non-deviative absorption, because the signals are

576


not being bent in any direction while being attenuated. The non-deviating absorption is calculated in terms of the absorption coefficient, which is given by [13]

K = 1.15 × 10 −3

NO f2

(16.32)

where K is the absorption coefficient, N is the electron density, v is the collision frequency of electrons with neutral particles and f is the frequency of the radio wave propagation. On the other hand, deviative absorption occurs when the signals spend for a longer period of time within the absorbing medium and meanwhile get refracted. The absorption coefficient in this case depends heavily upon the refractive index of the signal as well as the frequency of collisions (between electrons and neutral particles). The deviative absorption is highest wherever the signal wave deviates the strongest from its expected straight-line path, which occurs typically at/near the signal wave reflection point. Refractions in the lower ionosphere layer where the electron collision frequency is high, experience higher absorption and can be rather high if the frequency of wave propagation decreases. Deviating absorption is usually not as dominant as non-deviative absorption but plays nevertheless an important role in propagation. The earth’s magnetic fields also affect the energy lost through absorption, and this energy loss tends to be high for radio propagation at frequency closer to the gyrofrequency (» 1.417 MHz). Due to this finite energy absorption (loss), the ionosphere may be considered to have a finite but small conductivity. Thus, for a wave propagation, ionosphere acts like a dielectric medium having small conductivity and relative dielectric constant less than unity.

SOLVED EXAMPLES Example 16.1 Find the critical frequency at normal incidence for a layer with maximum electron density 1.45 ´ 108 electrons/cm–3. Solution:

We know that, critical frequency for a layer is given fc = 9[Nmax]1/2 Hz

where Nmax = 1.45 ´ 108 cm–3 = 1.45 ´ 108 ´ 106 m–3 = 1.45 ´ 1014 m–3 fc = 9[1.45 ´ 1014 m–3]1/2 Hz = 108.374 MHz Example 16.2 Estimate the maximum electron density of an ionosphere layer for a critical frequency 5.5 GHz. Solution:

We know that for a critical frequency the density of a layer is expressed as [N max ] =

f2 81

=

(5.5 × 10 9 )2 81

= 3.735 × 1017 electrons/m −3


577

Example 16.3 A high frequency communication between two points spaced away 220 km has to be established via ionosphere at critical frequency 90 MHz. Calculate the height of layer from the earth surface so that the 120 MHz frequency can be transmitted in communication. Solution:

We know that maximum useable frequency is given by ⎡ = fc ⎢1 + ⎢⎣

f MUF

⎛ D⎞ ⎜ ⎟ ⎝ 2h ⎠

2 ⎤ 1/2

⎥ ⎥⎦

2 ⎡ ⎛ 2200 ⎞ ⎤ 120 = 90 ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 h ⎠ ⎥⎦

h2 =

(1100)2 [(4/3)2 − 1]

1/2

Þ h = 125.4 km

Example 16.4 A mobile link has to be established between two points spaced away 1500 km via ionosphere layer of density 4.5 ´ 106 cm–3 at a height 150 km. Calculate the maximum frequency which can be communicated. Solution:

We know that critical frequency for a layer is given fc = 9[Nmax]1/2 MHz = 9[4.5 ´ 106 ´ 106]1/2 = 19.09 ´ 106 = 19.09 MHz

Hence, f MUF

2 ⎡ ⎛ 1500 ⎞ ⎤ = 10.09 ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 × 150 ⎠ ⎥⎦

1/2

= 10.09 26 = 97.34 MHz

Example 16.5 Find the percentage change in critical frequency of an ionosphere layer, if weather conditions increase its density up to 1.5 times. Estimate the values assuming initial density 5.75 ´ 107 cm–3. Solution:

We know that fc1 fc 2

1/2

⎛N ⎞ =⎜ 1⎟ ⎝ N2 ⎠

1/2

⎛ 1 ⎞ =⎜ ⎟ ⎝ 1.5 ⎠

fc 2 = 1.22 fc1 ⇒ 'f = 22.5%

fc1 = 9 5.75 × 10 7 × 10 − 6 = 68.25 MHz

Hence,

fc2 = 1.22 ´ 68.25 MHz = 83.5 MHz Nmax2 = 1.5 ´ 5.75 ´ 107 = 8.625 ´ 107 electrons/cm–3

578


Example 16.6 Determine the frequency of wave propagation through an ionosphere layer with refractive index of 0.6 and electron density 3.25 ´ 105 electrons/cc. Also, determine the corresponding change in frequency, if the refractive index increases by 35%. Solution:

We know that 81 N ⎤ ⎡ N = ⎢1 − ⎥ f2 ⎦ ⎣ 0.62 = 1 −

1/2

81 × 3.25 × 10 5

⇒ 1 − 0.36 =

f2

2.63 × 10 5 f2

⎡ 2.63 × 10 7 ⎤ 3 f = ⎢ ⎥ = 6.4 × 10 KHz 0.64 ⎢⎣ ⎥⎦

m2 = 1.35 ´ m1 = 1.35 ´ 0.6 = 0.81 Therefore,

and

⎡ 81 N ⎤ f′= ⎢ 2⎥ ⎣ 1 - N2 ⎦

1/2

⎡ 81 × 3.25 × 10 7 ⎤ = ⎢ ⎥ 1 − 0.812 ⎣⎢ ⎦⎥

1/2

= 8.744 × 10 3 KHz

Df = 2.344 ´ 103 Khz % change = 36.7

Example 16.7 Determine the change in refractive index of ionosphere propagation at f = 900 Hz if maximum electron density changes from 2.5 ´ 103 to 3.5 ´ 103 electrons/m–3. Solution:

⎡

N = ⎢1 − ⎣

81 N1 ⎤ ⎥ f2 ⎦

⎡ 81 × 2.5 × 10 3 ⎤ N1 = ⎢1 − ⎥ 81 × 10 4 ⎢⎣ ⎥⎦

1/2

⎡ 81 × 3.5 × 10 3 ⎤ N2 = ⎢1 − ⎥ 81 × 10 4 ⎣⎢ ⎦⎥

1/2

Dm = m1 – m2 = 0.064

1/2

= [1 − 0.25]1/2 = 0.87

= [1 − 0.35]1/2 = 0.81


579

Example 16.8 A 10-MHz satellite communication is taking place through ionosphere layer with refractive index 0.975 and height 500 Km. Calculate the ground range of this propagation assuming 10 GHz as MUF and 2.24 GHz as critical frequency. Also find the change in this distance, if MUF changes to 12 GHz.

Solution:

81 N1 ⎤ ⎡ We know that N = ⎢1 − ⎥ f2 ⎦ ⎣

N max =

0.05 × 1014 81

1/2

81 N max ⎤ ⎡ , Hence, 0.95 = ⎢1 − ⎥ (10 7 )2 ⎦ ⎣

= 6.2 × 1010

electrons m3

The corresponding critical frequency 2.24 MHz. When earth is flat The ground range = Skip distance 2 ⎡ ⎤ ⎛f ⎞ Dskip = 2 h ⎢ ⎜ MUF ⎟ − 1 ⎥ ⎢ ⎝ fc ⎠ ⎥ ⎥⎦ ⎣⎢

Dskip

⎛ 10 7 = 2 × 500 ⎜ ⎜ 2.24 × 10 6 ⎝

⎞ ⎟ ⎟ ⎠

2

− 1 = 2 × 500 (4.46)2 − 1

Dskip1 = 4.35 ´ 103 km Dskip2

2 ⎡ ⎤ ⎛ 12 ⎞ ⎢ = 1000 − 1 ⎥ = 5.26 × 103 km ⎢ ⎜⎝ 2.24 ⎟⎠ ⎥ ⎣ ⎦

DDskip = 9.13 ´ 102 km When earth is curvature Dskip

2 ⎡ ⎤ ⎛ D 2 ⎞ ⎢ ⎛ f MUF ⎞ = 2⎜ h + − 1⎥ ⎟ ⎜ ⎟ ⎜ ⎥ 8 R ⎟⎠ ⎢ ⎝ fc ⎠ ⎝ ⎢⎣ ⎦⎥

⎛ (4.35 × 10 3 )2 = 2 ⎜ 500 + ⎜ 8 × 6370 ⎝

⎞ ⎟ ⎟ ⎠

2 ⎡ ⎤ ⎢ ⎜⎛ 10 ⎟⎞ − 1 ⎥ ⎢ ⎝ 2.24 ⎠ ⎥ ⎣ ⎦

= (500 + 371) ´ 4.35 = 7577.7 km

580


The maximum skip distance = (Dskip )max = 2 2 hR = 2 2 × 500 × 6370 = 5048 km

Example 16.9 Obtain the transmission path range for ionosphere propagation at a height of 250 km from the earth surface. The Tx antenna is elevated at an angle of 25° and earth radius is 6370 km on flat surface. Also find the change in height of layer from earth, if propagation is considered along the curve path, over the same range and elevation angle. Solution: We know that (i) For a flat earth surface Transmission path range D =

2h tan C

=

2 × 250 tan 25°

=

500 0.466

= 1072.25 km

(ii) For the curved surface the transmitting path range

⎡ ⎛ R cos C ′ ⎞ ⎤ D = 2 R ⎢(90° − C ) sin −1 ⎜ ⎟⎥ ⎝ R + h′ ⎠ ⎦ ⎣ where h¢ = New height of layer. b = New beam elevation angle. Hence,

⎡ Q ⎛ 6370 × cos 25° ⎞ ⎤ − sin −1 ⎜ 1072.25 km = 2 × 6370 ⎢(90° − 25°) × ⎟⎥ 180° ⎝ 6370 + h′ ⎠ ⎦ ⎣

⎡ ⎛ 5773.18 ⎞ ⎤ 1072.25 km = 12740 ⎢1.134 − sin −1 ⎜ ⎟⎥ ⎝ 6370 + h′ ⎠ ⎦ ⎣ ⎛ 5773.18 ⎞ sin −1 ⎜ ⎟ = 1.0498 ⎝ 6370 + h′ ⎠ ⎛ 5773.18 ⎞ ⎜ ⎟ = sin (1.0498) (rad) ⎝ 6370 + h′ ⎠ ⎛ 5773.18 ⎞ ⎛ 1.0498 × 180° ⎞ ⎜ ⎟ = sin ⎜ ⎟ 3.14 ⎝ 6370 + h′ ⎠ ⎝ ⎠

h¢ = 6651.13 – 6370 = 281.13 Hence

Dh = 31.13 km


581

Example 16.10 For a mobile communication over a height of 120 km via ionosphere layer with Nmax = 2.22 ´ 105 electrons/m3, the maximum useable frequency estimated to be is 6.5 KHz. Find the optimum working frequency (OWF), critical frequency and elevation angle of beam and path range. Solution: (i) The critical frequency fc = 9

N max = 9 2.22 × 10 5

= 9 22.2 × 10 4 = 42.42 × 102 Hz = 42.42 KHz 2 ⎡ ⎤ ⎛ 6.5 ⎞ ⎥ = 278.86 km (ii) The skip distance Dskip = 2 × 120 ⎢ ⎜ 1 − ⎢ ⎝ 4.242 ⎟⎠ ⎥ ⎣ ⎦

(iii) Optimum working frequency = it is 85% of MUF = 0.85 ´ 6.5 KHz = 5.525 KHz (iv) The elevation angle tan C =

b=

2h D

=

2 × 120 278.86

= 0.8606

tan–1(0.8606) = 41°

Example 16.11 In a communication, the pulses of transmitted frequency are received back after a period of 5.5 ms. Find the virtual height of the layer. Solution: Virtual height h = vT/2 in which v is velocity of wave = 3 ´ 108 m/s T = 5.5 ´ 10–3 s h=

So,

3 × 108 × 5.5 × 10 −3 2

= 8.25 × 10 5 m = 8.25 × 10 2 km

Example 16.12 An ionosphere propagation takes place at a critical frequency of 90 MHz. Find the MUF, if transmitter transmits signals at 36° from the earth surface and maximum electron density of the layer. Solution: The angle of transmission is 36° from the surface, therefore, angle of incidence to reflecting layer will be i = 90° – 36° = 54° MUF = sec i, fc = sec 54°.90 = 153.12 MHz Therefore,

N max =

fc2 81

=

90 × 90 × 10 6 81

= 108 electrons/volume

582


Example 16.13 Calculate the maximum range obtainable in a single hop transmission via ionosphere layer at height of 450 km, if the earth radius is 6370 km. Solution:

The maximum range in a single hop transmission is 6370 ⎞ ⎛ R ⎞ −1 ⎛ d = 2R cos−1 ⎜ ⎟ = 2 × 6370 cos ⎜ ⎟ ⎝R +h⎠ ⎝ 6370 + 450 ⎠

= 12740 cos–1 (0.934) = 12740 ´ 20.93 = 12740 × 20.93° ×

3.14 180

= 4651.53 km

and Example 16.14 refractive index

d = 8 hR = 8 × 450 × 6370 = 4788.7 km

If the refractive index m of air is a constant, show that the modified mm will still vary with height and that dmm/dh = 0.1578 unit/metre.

Solution: Since dielectric constant of air and water are 1.0 and 80, respectively, which show that the dielectric constant moisture can vary over a wide range. Accordingly, the refractive index of air will be higher than one and will vary with height. Therefore, to account the presence of moisture, refractive index of air is modified to mm = (m – 1 + h/Re) ´ 106. However, at great height dielectric constant (er) does not vary with height, which gives a constant value of m but dmm/dh increases at the constant rate of 0.15748 unit per metre. In typical atmospheric conditions, near the earth surface, er and hence m decreases with height and this rate is found to be only 0.11811 units per metre. Using above viewpoints the radius of earth is modified to kRe = (0.15748/0.11811) Re or Rm = 4/3 Re. Example 16.15 Describe optimum working frequency (OWF), how does it relate to critical frequency? If the critical frequency for a layer is 43.2 MHz, find critical angle at 15 MHz. Solution:

fMUF = fc = 43.2 MHz OWF = 85 ´ 43.2 = 36.72 MHz

⎡ fc2 ⎢ 1 − 2 Critical angle at 15 MHz, i = sin ⎢ f ⎣ −1

⎤ ⎥ = sin −1 ⎥ ⎦

⎡ 43.22 ⎤ ⎢ 1 − ⎥ = 36.88° ⎢ 152 ⎥ ⎣ ⎦


583

OBJECTIVE TYPE QUESTIONS 1. Ionization in the atmosphere is caused by (a) EM wave radiation (b) Cosmic radiation (c) Solar radiation (d) None of these 2. Ionosphere extends from earth surface (a) Above 50 km (b) Above 100 km (c) Above 120 km (d) None of these 3. The earth receives energy from the Sun per square (a) 1375 W (b) 1370 W (c) 370 W (d) None of these 4. The height of D layer from earth’s surface is (a) 50–90 km (b) 90–140 km (c) 140–200 km (d) None of these 5. Kennelly–Heaviside layer is (a) D layer (b) E layer (c) F1 layer (d) None of these 6. Appleton layer is (a) D layer (b) Es layer (c) F layer (d) None of these 7. Champman’s law of ionospheric variations relates (a) D layer (b) F1 layer (c) F2 layer (d) None of these 8. The highest frequency that can be reflected by the ionosphere is one which has m = 0 (b) MUF (a) fc (c) OWF (d) None of these 9. When the frequency of ionospheric wave propagation increases, the critical angle (a) reduces (b) increases (c) constant (d) None of these 10. Generally, MUF is greater than the critical frequency (a) 2 times (b) 3 times (c) 5 times (d) None of these 11. Optimum working frequency may be calculated by (a) 1.5 MUF (b) 0.85 MUF (c) 2.5 MUF (d) None of these 12. The variation of virtual height with operating frequency is known as (a) IONGRAM (b) IONOSODE (c) Both (a) and (b) (d) None of these 13. The attenuation of propagating signal (in ionosphere) is high at (b) f > fc (a) f = fc (c) f < fc (d) None of these

584


14. In case of severe fading, the field strength of signal varies between (a) 05 dB and 20 dB (b) 10 dB and 20 dB (c) 30 dB and 40 dB (d) None of these 15. For the angular frequency w >> n collision frequency, the attenuation varies (a) f –2 (b) f 2 (c) (d) None of these f

Answers 1. (c) 6. (c) 11. (b)

2. (a) 7. (b) 12. (a)

3. (b) 8. (a) 13. (a)

4. (a) 9. (a) 14. (b)

5. (b) 10. (b) 15. (a)

EXERCISES 1. Define the maximum and minimum useable frequencies for long-distance radio communication. Derive an expression for the MUF in case of a thin ionosphere layer over a plane earth. 2. Describe the ionosphere reflection of radio waves. Derive an expression for critical frequency of a reflecting layer in terms of its ionization density. 3. What are the effects of earth’s magnetic field on radio-wave propagation? Comment on gyrofrequency and show that it should be equal to 1.47 MHz. 4. Derive an expression for the refractive index of the ionosphere in terms of electron density and frequency. 5. Explain the term ‘ordinary wave’ and ‘extraordinary wave’ in the relation of ionospheric propagation. 6. What is fading? Describe the factors that lead to fading in ionospheric propagation. 7. Describe D, E, F, and G layers of the ionosphere. 8. Write notes on: (i) Virtual height (ii) Skip distance (iii) Optimum working frequency. 9. What do you mean by super refraction? Highlight the significances of ionosphere in radio communication. 10. Show that the skip distance D for a critical frequency is 1

⎡⎛ f ⎞ 2 ⎤2 D = 2h ⎢⎜ ⎟ − 1⎥ ⎢⎝ f c ⎠ ⎥ ⎣ ⎦


585

where h = effective height of reflecting layer fc = critical frequency f = wave-propagation frequency 11. An ionosphere layer has a maximum electron density of 5 ´ 1010 electrons per cubic metre. The vertical height of the layer is 100 km. Assuming a flat earth, find the maximum useable frequency for communicating with a receiver situated 200 km away. 2 ⎡ ⎛ D⎞ ⎤ ⎢ Hint: MUF = fc 1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 h ⎠ ⎥⎦

1/2

and

fc = 9 N max

12. Estimate the change in MUF, if virtual height of propagating layer increases by 20%. Also, find the value for maximum electron density 5 ´ 1010 electrons/m3. 13. Find the corresponding change in maximum angle of incidence for a wave propagation at height h = 350 km from the ground. Also find skip distance (D) in km. Hint:

q

= 90° – imax, imax = sin–1(R/R+h) and D = 2R sin

q

14. Describe the properties of ionosphere. What is its height from the earth? Describe the importance of various layers in radio-wave propagation. 15. What are the causes of fading? How can fading be minimized? Describe selective and polarization fadings. 16. Define critical angle of ionosphere wave propagation. What factor does affect it? Show that the maximum angle of incidence of a wave for a layer at the 400 km height from the earth is 70.22°. 17. Define optimum working frequency. Describe its relation with MUF. 18. Define skip-distance, radio-horizon and virtual height. Find the transmission–path distance for ionospheric wave propagation through a layer of height 250 km. The angle of incidence of wave is 20°. 19. Deduce the expression

⎡ ⎛ R cos C D = 2R ⎢(90° − C ) − sin −1 ⎜ ⎝ R+h ⎣ where

⎞⎤ ⎟⎥ ⎠⎦

b = elevation angle R = earth-radius D = transmission path length and h = virtual height

20. (a) How ionosode and ionogram are related with virtual height of ionospheric layer? (b) What do mean by variations in ionosphere? Describe regular variations of ionosphere.

586


21. Show that the conductivity of an ionospheric layer can be expressed as Ne2O

, where n is collision frequency. m(O 2 + X 2 ) 22. List out parameters that are affected by the ionospheric variations. Describe irregular variations of ionosphere.

T =

23. What is SID? Describe Faraday rotation in the ionosphere. 24. How propagation is attenuated in ionosphere? Derive an expression for attenuation factor for ionospheric wave propagation. 25. Write short notes on the following: (a) Ionospheric storms (b) 27 sunspot cycle (c) Even year sunspot cycle.

REFERENCES [1] http:// www.physics.irfu.se [2] http:// spacew.com/www/foe [3] www.tp.umu.se. [4] Jordan, E.C. and K.G. Balmain, in Electromagnetic Waves and Radiating Systems, 2nd ed., Chapter 17, Prentice-Hall of India, New Delhi, 1988. [5] Wait, J.R., EM Waves in Stratified Media, Pergamon Press Inc., New York, 1962. [6] Budden, K.G., Radio Waves in the Ionosphere, Cambridge University Press, New York, 1961. [7] Bowhill, S.A., et al., VLF Ionosphere Radio Wave Propagation, Radio Science, 01, 1356–57, 1966. [8] Mittal, G.K., Radio Wave Engineering—Principle of Communication Systems, 20th ed., Khanna Pub., New Delhi, 2006. [9] Prasad, K.D. and M. Prasad, Introduction to Antennas and Wave Propagation, 2nd ed., Satya Publications, New Delhi, 2007. [10] Sarwate, V.V., EM Fields and Waves, Chap. 16, New Age International Publications, New Delhi, 2006. [11] Scott, A.W., Understanding of Microwaves, John-Wiley & Sons, Inc., New York 1993. [12] http:// en.wikipendia.org/wiki/ionosphere. [13] De Canck, M.H., “Ionosphere Properties and Behaviors—Part 1,” AantenneX Issue No. 110, pp. 01–06, June 2006.

C H A P T E R

17

Antenna Measurements

INTRODUCTION In every communication system, the actual transmission of the waves over the free space is carried out by some sort of antenna systems. And many of them have been described in the previous chapters and it is found that they have their stick specifications. Because of these specifications, different types of antennas are needed in different fields of applications: satellite communications, remote sensing, and radars, mobile and also in personal communications. Though various theoretical models are available to describe the performances of antenna, experimental measurements are still required to validate these data before putting antennas for any particular use. Antenna measurement techniques refer to the art of testing antennas to ensure that they meet particular specifications. The important parameters of the antenna are radiation pattern, impedance, bandwidth, beam width, gain, efficiency, and polarization, etc. Out of these parameters the radiation pattern is of common interest and useful to find the other parameters of antenna. Radiation pattern of antenna is affected by factors such as mounting height and position angle of the antenna, conductivity as well as humidity of medium. Antenna pattern is the response of antenna to a plane wave incident from a particular direction; the relative power density of the wave transmitted by the antenna in a given direction. In the far-field measurement, separating Antenna under Test (AUT) and source antenna at a distance 2D2/l reduce the phase variation (across the AUT) too small enough to achieve a reasonable antenna pattern [1]. In general, the antenna testing is done in the farfield region of the antenna, but as the large antennas require long distance for far-field regions, the near-field technique is also preferred to measure antenna pattern on a surface close to the antenna (3l to 10l). For the compact range method in particular a reflector is used to create a field near the AUT that looks approximately like plane wave. Basically, there are two methods of antenna measurements: indoor and outdoor. Both the methods have their own limitations; the outdoor measurements are not protected from the environmental conditions whereas indoor measurements suffer space restrictions. In general, for the accurate measurements uniform plane waves should incident on the antenna and this 587

588


is possible only if measurements are carried out in far-field region. The region in which antenna measurements are performed effectively is termed antenna ranges (separation between antennas) and it is basically of two types: reflection ranges and free-space ranges. Reflection ranges This is outdoor type test range, where ground is a reflecting surface. The reflection ranges create constructive interference in the region surrounding AUT which is referred as quit zone. In order to have effective communication height of Tx antenna is adjusted while that of Rx antenna is maintained constant. These testing ranges are found suitable for the antenna systems operating in frequency range from UHF to 16 GHz. Free-space ranges This is indoor type test range and is designed mainly to minimize environmental effects. This is most popular test range where the antennas are mounted over tall towers (Fig. 17.1). The main problem of this method is reflection from the ground, which is reduced by (i) Selecting the directivity and SLL of the Tx antenna (ii) Making LOS between antennas obstacle-free (iii) Redirecting or absorbing reflected.

Tx antenna

Rx antenna

Polarization positioner

Polarization positioner

Azimuth positioner

Antenna range

Rotation metre

FIG. 17.1

Free-space antenna measurements range.

Free-space ranges are further classified into elevated ranges, slant ranges, anechoic chambers and compact test ranges. Special indoor test ranges are near-field ranges, and they have several limitations compared to outdoor ranges.


589

Elevated range As we know there are many factors that affect the performance characteristics of antenna, separation between them, their height level and medium surmounting, etc. The separation between antennas for the measurement of a certain antenna depends mainly on the physical size and operating frequency of the antenna. Sometimes the best measurement side is environment where the antenna is used. Basically, elevated range is far-field range, where antennas are placed high enough on tower, buildings or hills, to minimize environmental effect. The range is measured by far-field criterion, i.e., 2D2/l only, however for the measurement of low-side lobe antennas, range greater than 2D2/l may be needed. The range width should be more enough to keep the main beam of the source antenna within the range. Elevated ranges are generally designed to operate mostly over smooth terrain, where antennas are mounted on tower/roots of adjacent buildings. The height of the AUT (HR) is determined by two criteria related to the source antenna (Fig. 17.2). First one, the source antenna should be chosen in such way that the amplitude taper over the AUT is typically no greater than 0.25 dB, this is in order to ensure uniform illumination. Second one is to minimize the range reflections and its first null points toward the base of the test tower, ensuring that the range surface intercepts only side lobe energy. These two criteria determine the height of the AUT to be HR > 4D. R

Direct wave DT

D

HT

FIG. 17.2

HR

Elevated range to minimize reflected waves.

In this range diffraction fences are sometimes used to further minimize the reflections from the ground. But two precautions must be taken: first the diffraction fence should not intercept the main beam of the source antenna and second the top edge of the fence should not straight knife edge, but rather serrated to reduce the edge diffraction. The antenna height and source antenna pattern should be so selected that the main beam of the source antenna does not illuminate the ground between the antennas. In order to keep ground reflection low, source antenna of low side lobes are desirable. The measurement distance between antennas should not be too large. Otherwise, it may cause difficulty to maintain enough the antenna

590


height. The lower limit for the source antenna diameter DT follows from the requirement that the first null of the pattern should fall above the base of the test tower should be DT ≥

1.5M R HR

(17.1)

where l = operating wavelength R = separation between antennas HR = height of AUT above ground (m) However, a narrow beam of the source antenna leads to a low level of reflection but at the cost of a large-amplitude taper in the test zone. For the maximum accepted amplitude taper of 0.25 dB, the upper limit of source antenna diameter is [2]

DT ≥

MR 3D

(17.2)

when D is the diameter of the AUT in metres. Both conditions mentioned above for the source antenna diameter give that the AUT should be mounted at least five times as high as its diameter D. In fact, it is very difficult to eliminate completely the illumination of the ground, especially in the measurement of elevation cuts when the main beam of the AUT is pointing at the ground. However, it is possible to proceed without pointing the AUT below the horizon by tuning AUT upside down after half of the pattern has been measured. To measure the low back lobes it is advantageous to place the AUT higher than the source antenna. Due to reciprocity, the direction of signal propagation does not matter—the AUT can be as well transmitting antenna as the receiving antenna. The main advantage of having the AUT as the receiving antenna is that the data processing and antenna manipulation can occur at one site. The main disadvantage is that the large and wide test range practically may be inconvenient and expensive. In such situations shorter range must be preferred. Slant range In a measurement if the source antenna is fixed near the ground while the AUT is on a tower, the range is celled a slant range. In this way the reflected signals are suppressed. The source antenna points toward the centre of the AUT such that its first null points toward the base. In order to reduce reflections from the ground the tower of the AUT must be constructed with non-conducting materials. The slant range is the hypotenuse of the triangle represented by the altitude of an aeroplane, the distance between the radar antenna and the airplane’s ground track. The importance of slant range is restricted over frequency range from 100 MHz to 200 GHz. The type of antennas that are preferred in this range are simple dipoles to horn antennas of many wavelengths on a side. Example of this range is the distance to airborne radar targets, i.e., an aeroplane flying at high attribute with respect to that of the radar antenna. In general, slant ranges are more compact than elevated ranges.


591

Relation between slant range and ground range The distance between an object and the antenna is equal to the speed of propagation of the waves through the atmosphere multiplied by the time it takes to reach the antenna. This, of course, is the relationship between the speed of EM radiation, time taken and distance travelled. A direct relationship between the slant range and the ground range also exists (Fig. 17.3). Since we know the angles at which the microwaves are propagated, we can use trigonometry to calculate the ground range.

Hn = flying height ÿ ÿ b = depression angle

FIG. 17.3

Slant range and ground range.

In order to view the radar image in the more recognizable ground range configuration, a geometric correction between the two distances is made (Fig. 17.4). ⎡ ( H 2 − Rk2 + 2Re H n ) ⎤ Arc cos ⎢ n ⎥ 2 Rk Re ⎢⎣ ⎥⎦

(17.3)

The slant range to ground range distortion is much more pronounced in airborne SAR systems than in satellite SAR systems. This is a result of the difference in depression angles and the range of the depression angles between airborne and space borne SARs. The slant range to ground range correction may not be necessary in order to create an effective stereo pair from satellite SAR imaginary.

Anechoic Chamber It is a best alternative of outdoor range to provide environment controlled, secure and minimized interference testing. An anechoic chamber consists of a definite volume enclosed

592


Ro, Rk and Rn = Hn = h= d= fk =

FIG. 17.4

Slant range (m) Satellite attitude (km) Illumination angle Earth centre angle Incident angle

Geometric correction.

by microwave absorber walls made by radiation absorbing material (RAM, Fig. 17.5). These walls reduce reflections from the boundary walls and increase the polyurethane foam in the shape of pyramids. RAM basically contains a carbon-impregnated foam that is produced in files, which are often shaped as an array of pyramids or wedges. Pyramidal shapes properly after minimization of front face reflections and work well up to angles of 50° from normal incidence [3]. Indoor reflections can be minimized by lining wall, ceiling and floor with radiation absorbing material (RAM).

Tx SGA

Rx R

FIG. 17.5

AUT

Anechoic chamber and typical RAM tile.


593

These materials are also used as antenna components for reducing side lobes and back lobes of radiation. An ideal absorber also provides an impedance matching for incoming waves at all the frequencies as well as angle of incidence. Pyramids work best at normal incidence, and scatter at a random rough surface if they are large compared to the wavelength. Anechoic chambers are mostly used at microwave frequencies. But advanced materials have been developed which are comfortable to be used in anechoic chamber at frequencies as low as 100 MHz. At higher frequencies reflection coefficient is larger and absorber is thick, whereas at lower frequencies the absorbers have to be very thick. However, at microwave frequencies reflection coefficient found to be around –50 dB at normal incidence provided length of absorber is in the order of few wavelengths. The absorbing materials are expensive for lower frequency range, because the typical size of pyramids is nearly 5¢–6¢ for 100 MHz. In an anechoic chamber, the test area is isolated from interference signals. This can be improved further by shielding. Shielding is also suitable for EMC measurement. An anechoic chamber simulates a reflection loss-free space and allows all weather antennas testing in a controlled environment. Additional advantages of anechoic chamber are: (a) Far-field pattern of a small antenna can also be measured accurately, hence the other parameters, gain, beam width and directivity. (b) By opening the end, anechoic chamber may be combined with outdoor range to measure pattern of large antennas. (c) Large antenna can also be tested on compact range and near-field region which are usually installed in an anechoic chamber. Another type of absorber being used in anechoic chamber is wedge. They are also made of carbon-loaded polyurethane foam. They work well at large incident angles with the wedge direction along the plane of incidence but not as well as pyramids at normal incidence. Basically, there are two types of anechoic chamber: rectangular and tapered (pyramidal), as shown in Figs. 17.6(a) and (b) respectively. However, the design of both is based on the geometrical optics technique with special attention towards minimize specular reflections. Rectangular AC stimulates free-space conditions and minimizes the volume of the quit zone. The pattern and location of source and operating frequency are also taken into account in its design however the receiving antenna (AUT) is isotropic. For the high quality measurement the angle of incident (from the normal) should be in the range of 0° to 70° and corresponding overall width or height of the chamber to be

W ≥

R 2.75

where R is the separation between Tx and Rx antennas. In a tapered AC, the direct and reflected radiations from the walls near the source having finite phase difference appear and contribute to the test antenna. At the lower frequencies in order to keep phase difference low, the source antenna is placed near the apex so that the reflections from the side walls occur near the source antenna. However, at higher frequency it causes difficulty to place the source sufficiently close to the apex and maintains the acceptable phase difference. In such cases, reflections from

594

Antenna and Wave Propagation Amplitude of wavefront

(a)

FIG. 17.6

(b)

Specular reflections from walls of (a) rectangular and (b) tapered anechoic chambers.

chamber walls are suppressed by using high gain antennas at source whose radiation towards walls is minimum. The perpendicular distance ht from the source antenna to the chamber wall should be such that

ht hr =

MR 4

where R is the wavelength and hr is the perpendicular distance from the fixed test antenna to the chamber wall. Compact test ranges In a compact antenna test range planar wave fronts are generated in a very short-range (10–20 metres) from the Tx antenna hence its name. Compact antenna test range (CATR) simulates an infinite range length by generating flat phase front by virtue of one or two reflectors, lens, horns, antenna array or hologram. CATRs are based on one or more reflectors and basically function like a collimating device. The linear dimension of the reflector is usually should be at least three times that of the test antenna so that the illumination at the test antenna sufficiently approximates a planar wave. A CATR is usually installed indoors in an anechoic chamber but very large CATRs are installed outdoor ranges. A CATR with an offset fed parabolic reflector, which transforms the spherical wave radiated by the feed to plane wave in front of the reflector is shown in Fig. 17.7.


595

Range reflector

Test antenna

+ Feed

FIG. 17.7

Schematic representation of compact test range using a reflector and AUT.

Since the beam is collimated the required power is less than on a far-field range. The quiet zone diameter is limited by direction and found about a third of the reflector diameter. Corrugated horn having symmetrical pattern, low side lobes and low-cross polarization are preferred to use as feed in order to minimize amplitude ripple. This is because waves of ordinary feed diffracted from the reflector edge into the test zone causes large ripples. An array feed system can also be used to control beam shape. And beam shaping can also accomplished by feeding a subreflector with a horn antenna. The major drawbacks of compact range are aperture blockage, direction radiation from the source to test antenna, diffraction from the edges of reflector and feed, depolarization coupling between the antennas and wall reflections. The use of offset eliminates aperture blockage and minimize diffractions. The direction radiation from the feed to test zone is suppressed by placing absorbers between feed and test zone. At high frequencies the surface accuracy of reflector is an important factor and should be l/100 or more. Room reflector, edge-diffraction, feed leakage and depolarization cause quality of the test field down, which can be improved by controlling the beam shape as well as diffraction [4]. Besides beam shaping quality of test field can also be improved by reducing diffraction. There are three methods to achieve the same: serrating the edge, rolling the edge, and resistive tapering. The first two methods are very common. The serrated edge is designed to keep the edge diffracted fields outside the quiet zone. The rolled edge illuminates the chamber walls more than the serrated edge but produces better quiet zone field provided the transition from the parabola to the rolled edge is very smooth. The use of a relatively long-focal-length reflector has the additional advantage of minimizing the depolarization effect associated with the curved reflector. In order to have accurate results a great care should be taken in fabrication of reflectors. In this set-up two parabolic-cylinder reflectors having their one-dimensional curvature positioned normal to one another are used. The first reflector produces vertical collimation while the second one horizontal collimation. A spherical wave front created by the SGA hits the first reflector whose curvature is directed inside the paper (Fig. 17.8). The reflected wave front, which is cylindrical, now hits the second reflector with curvature as

596


First reflector

Second reflector

Plane waves

AUT

SGA

FIG. 17.8

Compact antenna test range using two parabolic-cylinder reflectors.

shown in Fig. 17.8. The doubly reflected wave front at the position of the AUT has become planar. Of course, the wave front is not perfectly planar everywhere upon leaving the second reflector. The area where the wave front is planar enough for antenna measurement purposes is called the quite zone. This quiet zone is about 50–60% of the dimensions of the reflector. To reduce the disturbance in the quiet zone caused by diffracted fields from the reflector edges, these edges are either serrated or rolled similar to single reflected CARTs [5]. Merit Dual reflector CATR provides large quiet zone for a given reflector size and cancels the cross-polarization compared to the basic single reflector.

RADIATION PATTERN MEASUREMENT The radiation pattern or power pattern of an antenna is a graphical representation of radiated fields or radiated power is a function of elevation angle q and azimuth angle f for a constant radial distance and frequency. Radiation pattern is three-dimensional patterns, but normally we are interested in two-dimensional cuts taken from three-dimensional pattern and most often these cuts take the form of the radiated power (i.e., field amplitude) as a function of q for constant f. AUT is placed at the origin of cylindrical co-ordinate system such that this antenna acts as Tx antenna and that on a sphere with radius R a probe is moved that receives the transmitted signal by the AUT. At most of the time similar antenna is used as a receiver connected with the probe. If it so, due to reciprocity characteristics of the antennas the measurements are performed with the test antenna placed in the receiving mode. The source antenna is fed by a stable source and the received signal is measured using receiver at different positions from q = 0 to 180°’ (Fig. 17.9). Initially both the antennas (Tx and Rx)

Antenna Measurements 340° 330°

350° 0° 10°

20°

320°

597

30° 40°

310°

50°

300°

60° 70°

290°

80°

280° 270°

90°

260°

100° 110°

250° 240°

120°

230° 220° 210°

130° 140° 200°

190° 180° 170°

160°

150°

Relative power one way (dB)

(a) 0 2 4 6 8 10 2 4 6 8 20 2 4 6 8 30 2 4 6 8 40 Angle (b)

FIG. 17.9

Radiation pattern of an antenna: (a) polar plot, and (b) rectangular plot.

are aligned in the line of their maximum radiation direction, this is done by adjusting angle and heights of the antenna. At the maximum possible, effects of surroundings suppressed/minimized toward increased directivity. In addition the low side lobes of source antenna, clearance of LOS and absorption of energy reaching the range surface also optimized in the measurement.

the are the are

598


In view of better accuracy in the measurement, the precautions need to be taken care for the followings: · Effects of coupling between antennas: inductive or capacitive causes error in measurement. The former one exists at lower microwave frequencies and can be neglected if range of measurement R ³ 10l. Mutual coupling due to scattering and re-radiation of energy by test and source antennas also cause errors in the measurement, hence need to be eliminated. · Effect of curvature of the incident phase front produces phase variation over the aperture of the test antenna and this restrict the range R. For a phase deviation at the edge £ p/8 radians, separation between antennas should be R £ 2D2/l, where D is maximum aperture of the antenna. · Effect of amplitude taper over the test aperture will give deviation of the measured pattern from the actual. This occurs if the illuminating field is not constant over the region of the aperture. Tolerable limit of amplitude taper is 0.25 dB for which decreases in gain is 0.1 dB. · Interference from spurious radiating sources should be avoided.

Concept of Near- and Far-fields In general fields around an antenna are divided into two principal regions: one near the antenna called near-field (Frenzal zone) and the other one at a larger distance £ 2D2/l called the far-field (Fraunhofer zone). In addition, there are several other regions of the radiated fields in the vicinity of the antenna such as reactive-near regions, radiative near-field and far-field regions. At the far-field limit 2D2/l the phase difference between the aperture edge and centre is 22.5° (Fig. 17.10).

FIG. 17.10

Configuration of various fields in the radiation pattern.


599

Far-field Measurements The common method to measure the radiation pattern of any antenna is to build a set-up in such a way that all the possible errors are reduced to an accepted level. As our most of communications take place at large distance, i.e. both Tx and Rx antennas are in far-field region of each other, therefore, it is useful to take measurement in the far-field region only. There are certain advantages of far-field measurement: (a) Coupling and multiple reflections between antennas (Tx and Rx) are not effective. (b) The measured field pattern is valid for any distance in the far-field region; field strength varies according to 1/r only up to a certain distance, after that it becomes constant. (c) If power pattern is required only power (amplitude) measurement is needed, power pattern can determine with field pattern using power–voltage relation. (d) The measured results are not very sensitive to the changes in the location of the phase centre of the antenna and hence the rotation of the AUT does not cause significant measurement errors. The only disadvantage of far-field measurement is the requirement of large distance (test range) between antennas if used antennas are large. Too large test range can result in atmospheric attenuation.

Near-field Measurements This is another type of radiation measurement technique in which AUT is kept in fixed position and radiated field is probed in a plane on a cylinder/sphere close to and front of or surrounding the AUT. Based on coordinate systems, there may be three types of near-field pattern measurements: spherical, planar and cylindrical near-field patterns. Planar near-field antenna measurement is best suitable for highly directive antennas where most of the radiated energy will be directed into the forward direction. Planar near-field technique is very commonly used in antenna measurements. If the scan surface is in the x-y plane and measurement is taken at equally spaced points in both x and y directions, the planarity of scan surface (z-axis) should be l/100 or more. The scan area should be larger than the area of AUT. This area determines the maximum angle qM to which fields can be expected accurately. The spacing between sampling points should be slightly less than l/2, however for directive antennas spacing in order to l or even more can be considered to get main beam of the field pattern accurately. From the measured data the field pattern, gain and polarization of the AUT can also be computed. The spherical near-field antenna measurement is most suitable for omni-directional radiations whereas cylindrical near-field antenna measurement is a good alternative for measuring low directive antennas [1]. Near-field pattern measurements have many advantages compared to the far-field measurements. Since its range is very compact, antennas having large far-field distances can be measured in small space. If the measurement is done in a perfect way, the far-field pattern of very low side lobes (» –56 dB) antennas can also be determined accurately. Near-field

600


measurement can also be used in the diagnostic purposes. Near-field measurement is realized by scanning field close to the AUT on a known surface with known probe antenna of small size and broad beam. The probe is placed at a distance of few wavelength and both amplitude as well as phase are measured at two orthogonal polarizations at the sampling points. From these measured data far-field can also be computed from theoretical field distribution. There are many factors that cause disadvantages to the near-field measurement technique; in accurate probe positioning and connections, reflections, receiver non-linearity, limited area, etc. In addition, since measurements take several hours the stability of equipments needs to be given equal priority. All the precautions need to be taken care to minimize error caused by above factor in order to measure pattern accurately. Generally, small horn antennas are used as a probe antenna. Above all, the most serious errors are caused by multi-reflections between AUT and source antenna. There is a parameter called reflectivity, which measures the amount of reflection appears and it is defined as the ratio of all reflected fields to the directed fields and calculated from the amplitude ripple factor (S) as follows (see [2]): R f = 20 log

⎡10 S/ 20 − 1 ⎤ = 20 log ⎢ S/ 20 ⎥ (dB) Ed + 1 ⎥⎦ ⎣⎢ 10 Er

(17.4)

where Er < Ed. The measured reflectivity depends on various other parameters such as frequency polarization, patterns, and the aspect angle of the probe antenna. It is obvious that measured fields are coupled to antenna, not total fields radiated from the source. Hence, a directive antenna cannot gather all the reflections. So if the peak of main lobe is not directed toward source the coupled directed fields are reduced by a factor 10a/20, where a is the pattern level of the probe toward the source. Hence, the total reflectivity can be found by adding a (dB) to actual reflectivity given by Eq. (17.4), i.e., R fe = a (dB) + 20 log

⎡ 10 S/20 − 1 ⎤ = 20 log ⎢ S/20 ⎥ (dB) Ed + 1 ⎥⎦ ⎣⎢ 10 Er

(17.5)

MEASUREMENT OF REFLECTIVITY There are two methods to measure reflectivity: the VSWR method and antenna pattern comparison (APC) method. In VSWR method a probe antenna having a fixed orientation is moved in the test region making vertical and horizontal scans. The probe of wide beam gathers reflections from a large solid angle. The interference pattern of the direct and reflected waves versus location resembles a standing wave pattern. However, in APC method the patterns are measured at several locations within test region. These patterns are superimposed on one another as a result their main lobe peaks coincide and gives maximum radiation. The maximum deviation of the field at a given angle corresponds to the ripple (amplitude) measured with VSWR method. Because the phase has to be measured accurately, most nearfield ranges are limited at frequencies below 60 GHz.


601

BEAM WIDTH AND DIRECTIVITY MEASUREMENTS The beam width of the antenna is the angle subtended by the 3 dB or 10 dB points from top of the main beam on the both sides of radiation maximum. Basically, there are two types of planes in radiation pattern of an antenna: E-plane and H-plane; each plane contains corresponding field vectors. Accordingly, there are two types of half-power beam widths: q3dB and f3dB. Therefore, once radiation pattern is measured and plotted, beam widths can be calculated, just by choosing a point on the top of radiation pattern then coming 3 dB or 10 dB down and marking points on the both sides of radiation maximum and measuring their angular width. In another way, beam width can be measured without plotting the radiation pattern, but the primary antenna needs to be rotated instead of secondary one. In this case, first of all beam maximum is chosen and corresponding field strength as well as angle is recorded. Now the primary antenna is rotated in either direction until the field strength reaches half the power level of beam maximum or 0.707 time the voltage level and corresponding angle is recorded. The difference between these angular positions is half of beam width for a symmetrical beam; its double is beam width of secondary antenna. By repeating the same procedure and rotating secondary antenna the beam width of primary antenna can also be estimated. As mentioned, the 3-dB beam width in E and H planes is determined using corresponding radiation patterns. And if 3-dB beam widths in these planes are known, directivity can be obtained as follows:

D=

41253

R ER H

≅

72815

R E2 + R H2

(17.6)

where qE and qH are 3-dB beam width in degrees. This method is more accurate for the antenna having negligible side lobes. The other methods of directivity measurement have already been described in Chapter 3.

MEASUREMENT OF RADIATION EFFICIENCY The popular methods of calculating antenna efficiency more accurately and quickly are wheeler method and q-method. These methods relate antenna efficiency to the input impedance rather than far-field integration. First, antenna is represented as series network and real part of the input impedance is equalized to the total antenna resistance which consists of the radiation resistance and the radiation loss. These methods are applicable at the frequencies where the design of an antenna range or anechoic chamber becomes increasingly difficult and expensive to use. Both the methods have been applied successfully to optimizing the efficiency of (a) Multi-turn loop antenna at high frequency for shipboard applications (b) MTL antenna for seismic sensor system and (c) A small conformal array antenna.

602


Wheeler Method This method is the modified method of pattern integration method, where RL + RR is determined from the measurement of the antenna input impedance. Wheeler suggested that enclosing the testing antenna with a conducting sphere of length l/6 in radius eliminates RR from the input impedance of antenna without significantly changing RL. This implies that the conducting sphere cause no changes in the current distribution on the antenna. Hence, the real part of Zin with the sphere in place will have only loss resistance RL. Thus, by conducting two measurements: one without the sphere and another with sphere in place the efficiency of antenna can be determined using Eq. (17.7) [6]

I=

PR PR + PL

=

RR

(17.7)

RR + RL

and then using standard equipment network analyzer/anechoic chamber one can quickly measure RR and RL and therefore efficiency of the antenna. In order to verify the accuracy of this method, an MTL antenna was tested over a frequency range of 160–240 MHz. It was observed that at the highest of band Wheeler efficiency is about 60%, while the other measurement methods predict the efficiency maximum up to 52%. Accuracy of the method decreases generally for the lower frequencies, where loop efficiency is low. The major advantage of this method is that it is quick and eases. Wheeler has also been found to accurately predict the relative efficiencies of two antennas and to also yield a reasonable approximation to the absolute efficiency. Comparison of wheeler efficiency and gain measurement efficiency are shown in Fig. 17.11. 60 Wheeler frequency 50

Efficiency from gain measurements

Efficiency (%)

40

30

20

10

0 1.1f

FIG. 17.11

1.3f

1.5f Frequency

1.7f

1.9f

Comparison of MTL efficiency from Wheeler method and gain measurements.


603

Q-Method In this method, efficiency of the antenna is measured in terms of powers. This method is based on the comparison of measured Qm to ideal Qi of the antenna. The Q-quality factor of a realizable antenna is defined as Qm =

X

× Peak stored energy

Average power radiated + Average power dissipated

If there is another antenna identical to realizable antenna of conductor with perfect conductivity (s = ¥) and zero dielectric loss, i.e., Prad = 0. Then Q-quality factor of this ideal antenna is modified to Qi =

X

× Peak stored energy

Average power radiated

If the current distribution on both the antennas is considered to be same, then the stored energies at both the antennas will also be the same. And the efficiency of realizable antenna is the ratio

Ir = =

Qm Qi Average power radiated

(17.8)

Average power rediated + Average power dissipated

In order to use Eq. (17.8) to calculate efficiencies we must know the quality factor of the ideal loss-less antenna (i.e., Qi), which can be found from [7] and [8] calculated by Chu and Harrington. They calculated the Q-quality factor of a loss-less antenna, which is considered to radiate a number of spherical wave-guide modes. The Qm is determined by measuring the input impedance of the practical antennas and bandwidth can also be determined by same network analyzer which was used in the Wheeler method. In general, if Q-quality factor of any antenna is high it can be expressed in terms of bandwidth and operating frequency of the antenna as follows: q=

f0 'f

where Df is defined as the bandwidth between the frequencies for which the resistance is equal to reactance. In special case if antenna is matched to a Tx line, the half-power frequency that determine Df, occurs for a power reflection coefficient of 0.5, i.e., VSWR = 5.83. To verify the accuracy of this method as well as merit in comparison to Wheeler method a set of measurements has been conducted for five MTL antennas each with different number of turns. All the MTL antennas considered had a total length about 19.5 inches and were tuned to have an impedance of 50 W at frequency about 173 MHz. However, detailed comparison of Wheeler and Q-efficiencies are tabulated in Table 17.1.

604


TABLE 17.1

Comparison of Wheeler and Q efficiencies for MTL antennas

S.No.

No. of loops (N)

(Ka)

Ew (%)

Eq (%)

1 2 3 4 5

6 4 3 2 1

0.056 0.076 0.118 0.156 0.286

6 10 26 46 82

4 12 34 47 122

It is clear that for Ka £ 0.156(K = 2p/l), the Wheeler and Q-efficiencies are approximately the same, whereas for larger Ka, Q-method fails comparatively and Q-efficiency 122% is seen for Ka = 0.286. This is because of radiation of higher order modes by the antenna. In addition, if absolute efficiency is required, Q-method is reasonably good to measure efficiency for Ka £ 0.2, however if only relative efficiency is desired, this method is accurate for Ka somewhat greater than 0.2. Overall, both the methods have been found to accurately predict relative changes in efficiency and to a lesser extent absolute efficiency. However, Wheeler method is limited on the low frequency end, whereas no lower frequency limit seen for Q-method.

POLARIZATION PATTERN MEASUREMENT As we know, in general, the polarization of an antenna is characterized by three parameters, namely axial ratio (AR), tilt angle and the sense of rotation. Axial ratio specially describes types of polarizations: linear, elliptical and circular polarizations. And it is ratio of major to minor axes amplitudes in case of elliptical polarization. The tilt angle (t) describes the spatial orientation of polarization and measured clockwise from the reference direction. If tm is tilt angle of an incident wave that is polarization matched to the receiving antenna, and then the tilt angle tt of a wave transmitted by the same antenna is given by tt = 180 – tm provided a single coordinate system and only one orientation of view are used to characterize the polarization. If there is polarization different between incident waves and receiving antennas, polarization mismatched loss occurs that is measured by polarization loss factor (PLF). In general, the methods that are used to measure the polarization of a radiator/antenna are classified as follows: · The methods that yield complete polarization information, but require no prior information/polarization standard. They are referred to as absolute method. · The methods that yield polarization information, but require a polarization standard or comparison. They are designated as comparison method. · The methods that yield partial polarization information, but do not yield a unique point on the Poincare’s sphere. However, the well known methods of polarization measurement are three-antenna method, phase-amplitude, and polarization pattern methods.


605

Polarization Pattern Method It is clear from the pattern, shown in Figs. 17.12(a) and (b), that there is no information about tilt angle and polarization state. In order to get desired polarization, the polarization of a linearly polarized source antenna is rotated rapidly and at the same time the direction of the AUT is changed slowly and a pattern as shown in Fig. 17.12(c) is obtained. It is referred to as the axial ratio pattern and the method is known as rotating source method; it is preferably used for the testing of nearly circularly polarized antenna (AR £ 1 dB). That is, this method deals with axial ratio, which can be found from the maxima and minima positions corresponding to alignment of the source with the major and minor axes of polarization ellipse respectively. The AUT is used as the transmitted antenna. And field components of any wave linear, elliptical or circular as well as phase-difference between them can be measured using two fixed linearly polarized antennas mounted at 90°. However, the same can also be achieved by using two circularly polarized antennas of the opposite sense of rotation [9]. The phase-amplitude method uses a dual-polarized horn antenna to record simultaneously the amplitude polarization pattern and the relative phase difference between the two orthogonal polarizations. Double-conversion phase-licked receivers are mostly used to perform the amplitude and phase comparison measurements. The polarization of any antenna can be measured conveniently by using it in the transmitting mode and probing the polarization by a dipole in plane that is rotated in the plane of polarization and the received voltage pattern is recorded and analyzed as follows (Fig. 17.13).

Direction of propagation Right hand


Clockwise

(i) Sense of polarization

FIG. 17.12

Contd.

606


FIG. 17.12(a)

Sense of rotation for antenna and polarization ellipse. LHC

Poles represent circular polarizations

Upper hemisphere left-hand sense

Latitude represents axial ratio

45° Linear Equator H represents linear polarizations

Lower hemisphere right-hand sense

FIG. 17.12(b)

Longitude represents tilt angle RHC

Polarization representation on Poincares sphere.


FIG. 17.12(c)

607

Pattern of rotating source method. Test probe rotation

Test antenna

FIG. 17.13

Polarization measurements.

Linear polarization For a linear polarization, the output voltage will be proportional to sin form a figure of eight.

y

and pattern will

Circular polarization For an elliptical polarization, the nulls of the figure of eight are filled and a dumb-bell polarization curve is obtained which is tilted and a polarization ellipse can be drawn as

608


shown by dashed curve in Fig. 17.14. The sense of rotation of the circular and elliptical polarizations can be determined by comparing the responses of two circularly polarized antennas, one left and the other right side rotations. The polarization of the test antenna will be the same as that of one of these two directions for which the response is larger. The axial ratio and tilt angle can be obtained from Fig. 17.14, where y is rotation angle of probe relative to a reference direction.

FIG. 17.14

Polarization pattern of wave.

General Method of Polarization Measurement There are several methods for measuring the polarization characteristic of an antenna. All of the techniques described here work better for some antenna than for others, it is difficult to say, which method is best. It is common to use the AUT to transmit and some standard antenna whose orientation varies as a receiver. It is, of course, equally correct to measure the response of the AUT while transmitting towards its waves of known polarization. Linear component methods There are two parameters: polarization ratio (p) and circular polarization ratio (q), which defines the polarization of the wave. And they are expressed as p=

Ey Ex

=

Ey Ex

e jG

(17.9)

It is clear that the measuring amplitudes of field components and phase between them using two linear polarized receiving antennas, the polarization of an antenna being used in transmitting


609

can be determined effectively. Since only power of the receiver is measured not field strength components, so it is essential that both the antennas must have equal gains and impedances. As a receiver can be switched from one antenna to the other, no difficulties will arise from unequal receiver gains in the two channels. Phase difference f can be measured by slotted line method or by using a calibrated phase shifter. At lower frequencies, dipole antennas and at higher frequencies standard gain horns are used satisfactorily. However, since gain of standard horn antennas is measured (at desire frequencies) prior to their use, this method is not much attractive at higher frequencies. Circular components method For any circular polarized waves, the parameter q is defined as q=

EL ER

=

EL ER

e− jR

(17.10)

Hence, similar to previous procedure, by using two antennas of equal gains and impedance, one LHP and another RHP, the antenna’s polarization can also be measured. Kraus and Rubin have suggested helix antenna as a standard gain antenna for polarization measurements, but it has several limitations, particularly wideband coverage and polarization purity. An alternative to the measurement of phase-difference (q), is the use of a linearly polarized receiving antenna to measure the tilt angle of the polarization ellipse. This overcomes the problem of antenna placement when measuring the phase difference of the wave components. The tilt angle taken together with the rotation sense, obtainable from |q| and the axial ratio, which defined as Axial ratio ( AR) =

1+ q 1 − q

(17.11)

defines the polarization completely. Polarization pattern method This method is based on the determination of polarization matched factor. The modified form of this factor for the two antennas of same rotation sense and the antennas of opposite sense is written as S =

AR12 cos2 (U 1 + U 2 ) + sin 2 (U 1 + U 2 ) ( AR12 + 1)

(17.12)

provided antenna (2) is linearly polarized, i.e. (AR2 ® ¥). Antenna 2 is rotated around a line drawn between the two antennas. For a dipole antenna the rotation axis is perpendicular to the dipole. If t2 = –t1, which corresponds to coincidence between the major axes of the ellipse for the two antennas, then

610


S=

AR12 AR12 + 1

(17.13)

which is maximum. If t2 = –t1 + p/2, which corresponds to the major axis of the linearly polarized antenna coinciding with minor axis of the antenna being tested, then

S=

1 AR12

+1

(17.14)

which is minimum. Since the open source voltage is proportional to the square root of r, hence the ratio of maximum to minimum open-circuit voltage in magnitude will be Vmax Vmin

= AR1

(17.15)

Thus, we have axial ratio of the antenna, undergoing test and, of course, its tilt angle from the known rotation angle of the linear antenna when maximum power is received. A plot of the square root of r from Eq. (17.12) is called the polarization pattern of the antenna, whose polarization is being measured. Figure 17.15 shows the polarization pattern of the two antennas (a) one with axial ratio 2 and (b) another with axial ration ¥ (i.e. linearly polarized antenna).

FIG. 17.15 Polarization patterns for measuring axial ratio and tilt angle of the polarization ellipse. The polarization pattern method lends well to the rapid testing of an antenna’s polarization properties as a function of angle from beam maxima. The main deficiency of this method is its failure to give the rotational sense of the AUT [11].


611

Polarization measurement from amplitude measurement We know that if the modified polarization ratio of transmitting antenna and the conjugate of the modified polarization ratio of a receiving antenna (or vice versa) are plotted on a Poincare’s sphere by means of the Stoke’s parameter, the polarization match factor between the antennas is given by ⎛'⎞ ⎝2⎠

S = cos2 ⎜ ⎟

(17.16)

where D is the angle between the rays from the sphere centre to the two plotted points on the sphere. If we have a transmitting antenna with unknown polarization and receiving antenna with known polarization, then a circle drawn on the Poincare’s sphere with radius (r) and centre at the receiver conjugate polarization point, will pass through the sphere point defining the transmitter polarization. Similarly, for a second antenna, a circle with its conjugate point as centre will also pass through the transmitter polarization point, as a result two circles will at two points on the Poincare sphere. A third receiving antenna can be used to remove the Poincare sphere. Thus, the three circles generated by three Rx antennas paired with the Tx antenna will intersect at one point on the Poincare sphere, therefore, uniquely defining the polarization of transmitting antenna [12]. The polarization match factor r is measured with the measurement of power at the receiving load, which is determined by polarization, antenna gains, transmitted power, etc. That is, additional measurement is required to fully measure the polarization properties of an antenna. A convenient method for this is to use pairs of receiving antennas that have the same gains but are orthogonally polarized (LHP or RHP). Power ratios are then used to determine the polarization of the antenna tested. The method of power ratio can be described by considering the three pairs of antennas: linear vertical and horizontal, linear at 45° and 135° and right and left circular polarizations [see Appendix F].

GAIN MEASUREMENT In order to describe the performance of an antenna the gain is a most important parameter to be measured. Basically, there are two methods of gain measurements: absolute gain method and gain transfer (gain comparison) method. And in this regard there are two types of antenna ranges: free space range and reflection range. Free space range is designed to suppress the contributions from the surrounding environment and it includes slant ranges, elevated ranges, compact ranges, near-field and the anechoic chambers. Reflection ranges designed properly can create a constructive interference in the region of practical antennas, which is referred to as quiet zone. These ranges are outdoor types, where the ground is the reflecting surface and they are usually employed in the UHF region for the pattern measurements of moderately broadband antennas. They are also used for the systems operating in the UHF to 16 GHz frequency region. Usually, free-space ranges are used to measure the gain of high

612


frequency antenna (above 1 GHz). At lower frequencies (i.e., longer wavelength), it is difficult to simulate free-space condition, hence ground reflection ranges are utilized in these frequency ranges, particularly from 1 MHz to 1 GHz. Below 100 MHz directive antennas are physically large and the ground effect becomes increasingly pronounced, that is why the gain of antenna at these frequencies is measured in situ. Antenna gains are not generally measured at frequencies below 1 MHz. Instead, measurements are conducted on the field strength of the ground wave radiated by the antenna.

Absolute Gain Method In absolute gain method, calibration of antennas is done so that the same antennas can be used as a standard antenna in gain comparison method. In this case no prior data of antenna’s gain is required. The experimental set-up for gain measurement is shown in Fig. 17.16.

Tx

Source

Isolator

Directional coupler

Tuner

Attenuator

Rx

Tuner Power meter

Power meter

FIG. 17.16

Block diagram of antenna gain measurements.

For better accuracy of the measurements, the following points must be taken care of: 1. 2. 3. 4. 5. 6. 7.

All antennas meet the far-field criteria; the distance between Tx and Rx, R ³ 2D2/l The antennas are aligned far bore-sight radiation in LOS The measuring system is frequency stable Impedance mismatched in the system components is minimum Polarization mismatch is minimum Reflection from various background and supporting structure is minimum Same criteria need to satisfy for all the method of gain measurement.

The two/three antenna methods, ground reflected range method and radar method are the main techniques that are employed in this measurement method.


613

Two-antenna method In this method, the signal is transmitted from a transmitting antenna of gain Gt and the signal is received by the test antenna of gain Gr placed at far-field distance R. The received power is expressed by

Pr =

Gt Gr Pt (4 Q R)2

M2

⎛P ⎞ ⎛ 4Q R ⎞ Gr (dB) + Gt (dB) = 10 log ⎜ r ⎟ + 20 log ⎜ ⎟ ⎝ M ⎠ ⎝ Pt ⎠

(17.17)

where Pr is the received power and Pt is the transmitted power. When both the antennas are selected identical, then Gr = Gt so that ⎛P ⎞ ⎛ 4Q R ⎞ Gr (dB) = Gt (dB) = 5 log ⎜ r ⎟ + 10 log ⎜ ⎟ ⎝ M ⎠ ⎝ Pt ⎠

(17.18)

Therefore by measuring l, R and Pr/Pt, the gain of any antenna can be obtained. The main disadvantage of this method is that it is difficult to find two antennas of identical gains, so possible error may introduce in the measurement. Three-antenna method In order to overcome the disadvantage of two antenna method the three-antenna method is most suitable method to find gain of the antennas. In this method, total three antennas are used but only two antennas at a time, i.e., 1 and 2, 2 and 3, and 3 and 1, respectively. The following equations are developed for the received and transmitted powers: ⎛P ⎞ ⎛ 4Q R ⎞ Gr (dB) + Gt (dB) = 10 log ⎜ r 2 ⎟ + 20 log ⎜ ⎟ ⎝ M ⎠ ⎝ Pt1 ⎠

(17.19a)

if antennas 1 and 2 are in use ⎛P ⎞ ⎛ 4Q R ⎞ Gr (dB) + Gt (dB) = 10 log ⎜ r 3 ⎟ + 20 log ⎜ ⎟ ⎝ M ⎠ ⎝ Pt 2 ⎠ if antennas 2 and 3 are in use

(17.19b)

⎛P ⎞ ⎛ 4Q R ⎞ Gr (dB) + Gt (dB) = 10 log ⎜ r1 ⎟ + 20 log ⎜ (17.19c) ⎟ ⎝ M ⎠ ⎝ Pt 3 ⎠ if antennas 3 and 1 are in use. Therefore by choosing appropriate value of l, R and Pr/Pt, the gains of three antennas can be measured using the above equations. Since only power ratios are measured there is information about polarization states of antennas in this method.

614


Ground-reflection range method The measurement of the gain of narrow-beam antennas may only be accomplished with reasonable accuracy using above methods. However, for frequency below range (1–2 GHz), gain standards are not generally available except dipoles, whose low directivity often causes excessive measurement error. Hence, in order to reduce the measurement error, it is necessary to consider the calibration of a gain standard with increased directivity at these frequencies. The method of ground-reflection range that can be used to measure the gain of moderately broad beam antennas operated below frequencies 1 GHz is described in this section. The basic test set-up used in this measurement is shown in Fig. 17.17. Horizontal polarization is employed to avoid the wave tilt and rapid variation of reflection coefficient associated with vertically polarized waves reflected from the range surface. This is necessary because the reflection coefficient of the earth depends on the angle of incidence as well as varies rapidly for vertically polarized waves. However, uses of elliptically and circularly polarized antennas are limited, as earth exhibits different reflective characteristics for horizontal and vertical fields. Receiving antenna Transmitting antenna

Rd

hr ht

Rr

Range surface

Transmitting antenna image Ro

FIG. 17.17

Range configuration reflection range gain method.

The field at the receiving end is taken as vector sum of the electric fields due to direct path contribution and reflected path distribution. Hence, the amplitudes of electric field at the receiving antenna due to direct path and reflected path can be given as [13]. 1 2 ⎡ ⎛ M ⎞ ⎤2 Ed = K ⎢ P0 (K1 , Gt )(K 2 Gr ) ⎜ ⎟ ⎥ ⎢ ⎝ 4Q Rd ⎠ ⎥⎦ ⎣

(17.20)


615

1

and

2 ⎡ ⎛ M ⎞ 2⎤2 Er = K ⎢ P0 Gt Gr ⎜ ⎟ r ⎥ ⎢⎣ ⎥⎦ ⎝ 4Q Rr ⎠

(17.21)

where, from the geometry Rd = [ Ro2 + ( hr − ht )2 ]1/2 and Rr = [ Ro2 + (hr + ht )2 ]1/2 , since reflection coefficient approaches unity for small grazing angles and parallel polarized waves. The r2 is known as effective gain factor and it is function of the electrical and geometrical properties of the range surface, radiation pattern of antennas, frequency and polarization of the Tx waves and also the geometry of the test range. The total electric field will be vector sum of Ed and Er at the point over the active region of the receiving aperture provided the antenna heights are adjusted in such a way that these two fields contributions arrive in path at the receiving antenna, i.e. ET = Ed + E r

(17.22)

and, therefore, the corresponding power is 2

rRd ⎤ ⎛ M ⎞ ⎡ 1/2 Pr = P0 Gt Gr ⎜ ⎟ ⎢(K1 K 2 ) + ⎥ Rr ⎦ ⎝ 4Q Rd ⎠ ⎣

2

(17.23)

rRd ⎤ ⎛P ⎞ ⎡ ⎛ 4Q Rd ⎞ 1/2 (GT ) dB = (Gr ) dB + (Gt ) dB = 10 log ⎜ r ⎟ + 20 log ⎜ ⎟ − 20 log ⎢( K1 K 2 ) + ⎥ Rr ⎦ ⎝ M ⎠ ⎣ ⎝ P0 ⎠ (17.24)

From Eq. (17.24), it is clear that the sum of gains (GT) dB is obtained by measuring all the unknown parameters included in RHS of equation except factor r2. The factor r2 can be measured by repeating the measurements with the adjusted antenna height such that the power at the receiving antenna is minimum [14]. In order to satisfy the in phase criteria of Eq. (7.22), it is necessary that ⎡ (2n − 1)Ro ⎤ hr = ⎢ ⎥M 4 ht ⎣ ⎦

(17.25)

The Tx antenna should be placed at such a position that Eq. (17.25) as well as mutual coupling criterion ht ³ 4l should satisfy. The power received at the receiving antenna in this condition will be Pr and calculated by Eq. (17.23), i.e., at a height of Tx antenna such that the receiving antenna is at the first minimum the pattern. If the Tx antenna is replaced at lowest position which satisfies mutual coupling criterion ht ³ 4l as well as the condition, then ⎛ mR ⎞ hr = ⎜ o ⎟ M ⎝ 2 ht ⎠

(17.26)

616


This corresponds to the location of a minimum in the interference pattern of the receiving antenna, i.e. at a height of the Tx antenna such that the receiving antenna is at a field minimum. Then the power received at the antenna is given by 2

Rd′ ⎤ ⎛ M ⎞ ⎡ 1/2 Pr′ = P0 Gt Gr ⎜ ⎟ ⎢(K1′K 2′ ) − r ⎥ Rr′ ⎦ ⎝ 4Q Rd′ ⎠ ⎣

2

(17.27)

The primed parameters in this equation are to distinguish received power from those of first case. where P0 = Power into the terminals of the transmitting antenna Gt = Maximum gain of the transmitting antenna GtK1 = Gain of the transmitting antenna in the direction of the receiving antenna K1, K2 = Antenna directivities (relative to their maximum value) along Rd K¢1, K¢2 = Antenna directivities (relative to their maximum value) along R¢d Gr = Maximum gain of the receiving antenna GrK2 = Gain of the receiving antenna in the direction of the transmitting antenna K = Constant of proportionality Ro = Direct path separation between antennas Ed = Amplitude of the electric field at the receiving antenna due to the direct path wave Er = Amplitude of the ground reflection electric field at the receiving antenna Rr = The effective path length between the Rx antenna and the Tx antenna image r2 = An effective gain factor which accounts for the transfer of energy by means of reflection from the range surface ET = Total electric field at the receiving antenna ht and hr = The height of Tx and Rx antennas l = The wavelength of operation n = A positive integer corresponding to the interference lobe which is peaked on the receiving antenna m = A positive integer corresponding to the location of a minimum in the interference pattern at the receiving antenna. The factor r is supposed to be the same in both the cases. It is preferred to test at horizontal polarization than at vertical, since the factor r2 changes less rapidly with grazing at the horizontal polarization. The value of r2 can be found in terms of above parameters by dividing Pr to P¢r and hence the gains of antennas. The method is directly applied to antennas which coupled only the electric field. Modification must be made for loop and slot antennas, and further modification for the antennas which may be a combination of electric and magnetic field types [15]. Radar techniques The radar technique is a common method to measure the gain of a single antenna. The gain can be determined from the transmitted and received powers using the radar equation, provided the AUT illuminates a target with known radar cross-section. A sufficiently large flat reflector


617

is placed at far enough distance from the AUT, similar to the absolute gain method with identical antennas and the AUT sees its image behind the reflection. Reflecting spheres are widely used as radar calibration targets because they have a know radar cross-section which is independent from the angle. For a perfectly reflecting sphere, s is equal to its physical cross-section pa2. When its radius a > l with a sphere as a target, the gain can be obtained from the radar equation [9] ⎛ 8Q R 2 (GA )dB = 10 log ⎜ ⎜ aM ⎝

⎞ ⎛ Pr ⎞ ⎟ + 5 log 10 ⎜ ⎟ ⎟ ⎝ Pt ⎠ ⎠

(17.28)

where Pr and Pt = received and transmitted powers (W) R = separation between the radar and target (sphere) (m) a = radius of sphere (m) l = wavelength (m) The requirement of very large reflector may account as a drawback of this technique. In a special case, if AUT is short-circuited it will re-radiate the received power. The radar cross-section corresponds to this re-radiation is expressed in terms of GAS as follows [2]:

T r = Ae GAS = 4QT r

GAS =

M

2

M2 2 GAS (m) 4Q

=

2

M

QT r

(17.29)

(17.30)

where GAS = gain of AUT short-circuited Ae = effective area of the AUT (m2)

Gain Transfer (Gain Comparison) Method This method uses two sets of measurements with the test and standard gain antennas. Using the test antenna of gain Gr in receiving mode, the received power Pr is recorded in a matched recorder. The test antenna is then replaced by a standard gain antenna of gain Gs and the received power Ps is again recorded without changing the transmitted power and geometrical configuration. The resonant l/2 dipole (gain » 2.15 dB) and pyramidal horn antenna (gain » ranging from 12–25 dB) are mostly used as standard gain antennas. The calibration uncertainty of the standard antenna gain is ± 0.25 dB. Therefore Pr Ps

or

=

Gr Gs

⎛P ⎞ Gr (dB) = Gs (dB) + log ⎜ r ⎟ ⎝ Ps ⎠

(17.31)

618


Thus by measuring the received power with test and standard gain antennas and knowing gain Gs of the standard gain antenna, the gain of the test antenna can be found.

Gain Measurement of CP Antenna There are two common approaches to measure the gain of CP antennas. In the first one the gains of CP/EP antennas are measured by measuring the partial gains for two orthogonal linear polarizations. First the polarization of the linearly polarized source and standard antenna are set horizontal and measure the gain say (GH), Similarly, (GV) is measured for vertically polarized source and reference antenna. Finally, the total gain of the antenna is expressed as the sum of two partial gains (GT)AUT = GH + GV (GAUT)dB = 10 log(GH + GV)

(17.32)

where GH and GV are the partial gains with respect to horizontal–linear and vertical–linear polarizations. This is a very convenient method, if the antenna possess good polarization purity in the two perpendicular planes. In the other method, we need to design a standard gain antenna which possess CP/EP then perform experimentation. This method is popular in mass productions for power-gain measurements of CP or EP antennas. In order to minimize system disturbance during measurement the two receiving antennas should be mounted back to back on either side of the axis of an azimuth positioned and both of them need to connect through a common switch. If there is no much difference between test antenna and standard antenna, this method is less effected by proximity and multi-path interference [16].

IMPEDANCE MEASUREMENT We know that antenna can be treated as one-port network, based on the circuit theory. Therefore, its impedance can be measured with any impedance measurement techniques like slotted lime method and swept measurement with directional coupler and network analyzers. The most common parameter that is needed to determine impedance of an antenna is VSWR and powers. The equivalent circuit model indicates that there are two types of impedances: self and mutual impedances, associated with any antenna. The equivalent impedance is attached across the two terminals, which are used to connect the antenna to a generator, receiver or a transition’s line. Usually, this impedance is called the driving point impedance. When the antenna is radiating in an unbounded medium free from the other element/obstacles, the driving impedance is same because there is no coupling between AUT and surrounding objects, however, there is coupling between AUT and obstacle. The driving impedance is a function of its self-impedance and the mutual impedance between it and the other sources or obstacles. In practice, the driving point impedance is usually referred to as the input impedance. The presence of ground/earth must be taken into account in determining the input impedance of an antenna. As usual, both the impedances—self and mutual impedances—


Antenna

Source Tx antenna

FIG. 17.18

619

Space waves

Block diagram of impedance measurement.

each have a real and an imaginary part. The real part is designated as the resistance and the imaging part is called the reactance of antenna. Refer to the antenna as a transition device. Usually, under an ideal condition, energy generated by the source should be totally transferred to the radiation resistance Rr, which is used to represent radiation by the antenna. In this condition, maximum power is delivered to the antenna; this case is referred to as conjugate matching. However, in practice, it is not possible; there are conduction dielectric losses due to the lossy nature of Tx line and antenna, as well as the reflections losses at the interface between the antenna and line. As a result, a serious mismatching occurs, causing reflection of energy toward source. The degree of mismatch is a function of antenna input impedance and the characteristic impedance of the line and it determines the amount of power reflected at the input antenna terminal into the line. It is observed that reflected waves from antenna terminal generate new waves known as standing wave, this due to superimposition of incident and reflected waves. The amount of power reflected can be expressed in terms of parameters, known as reflection coefficient and VSWR, which relates to antenna’s input impedance and the characteristic impedance of the line as follows: *

where Za Z0 G S

2

=

Pr Pi

=

Z a − Z0 Za + Z0

2 2

=

S− 1

2

(17.33)

S +1

= input impedances of the antenna = characteristic impedance of Tx line = |G|ejf = Voltage reflection coefficient at the antenna input terminals = VSWR = Voltages standing wave ratio at the antenna input terminal

In fact, Eq. (17.33) is the direct relation between VSWR and input impedance of the antenna. But, it is basically used to compute the VSWR for known Zin of an antenna not vice versa. The measurement of VSWR alone does not provide enough information to compute complex input impedance of the antenna, rather than computing reflection coefficient using Eq. (17.33). So, when reflection coefficient is computed completely with its magnitude and phase, antenna impedance can be determined by using

Z in = Z c

1 − * 1+*

= Zc

1+ *e

jG p

1 − *e

jG p

(17.34)

The phase of reflection coefficient (fp) can be determined by locating voltage maxima or minima along the transmission line. In practice voltage minima is preferred, as it can be

620


measured more accurately than maxima. In addition, the first minima are usually chosen unless the distance from it to the input terminals is too small to measure accurately. The phase of G is calculated by using [17]

G p = [2C x n ± (2n − 1)Q ]

(17.35)

⎡ 4Q ⎤ = ⎢ xn + (2n − 1)Q ⎥ n = 1, 2, 3, … ⎥⎦ ⎣⎢ Mg

where n = number of voltage minima from the antenna input terminal (n = 1, means the first voltage minimum). xn = distance from the input terminals to the nth voltage minimum. lg = guide wavelength (it is twice the distance between two voltage minima/maxima There are several other methods: impedance bridges, slotted lines, and network analyzers methods that can also be utilized to determine the antenna impedance [18, 19].

Mutual Impedance between Dipole Antennas There is another method to measure the mutual impedance between dipole elements above a conducting plane. To understand this measurement procedure, let us consider the arrangement of two dipoles as shown in Fig. 17.19. dipole (1)

dipole (2)

I1

I2 Z12

Z11

Z22

V1

FIG. 17.19

V2

Two coupled dipole antennas.

where I1, V1 and I2, V2 are currents and voltages in antennas 1 and 2 respectively. If Z11 and Z22 are the self-impedances of the antennas and Z12 is the mutual impedance between antennas, then from the network theory, the relationship between the voltages (V1 and V2) terminals currents (I1 and I2) and impedances (Z11, Z22 and Z12) for the coupled antennas can be expressed as: V1 = Z11I1 + Z12I2

(17.36)

V2 = Z22I1 + Z12I2

(17.37)


621

In order to measure the mutual impedance Z12 let the impedances Z11 and Z22 are equal to the isolated impedance of antennas 1 and 2. This approximation is very good except in the case of extreme coupling when antennas are very close to each other. Now consider the input impedance to antenna (1) when antenna (2) is short-circuited. Hence, from Eq. (17.36), we get, V1 I1

⎛I ⎞ = Z S = Z11 + Z12 ⎜ 2 ⎟ ⎝ I1 ⎠

(17.38)

Also, since antenna is short-circuited, V2 = 0, i.e. Z12I1 + Z22I2 = 0 Z12 ⎛ I2 ⎞ ⎜ ⎟= − Z 22 ⎝ I1 ⎠

(17.39)

Hence, Eqs. (17.38) and (17.39) yield Z S = Z11 −

2 Z12

Z 22

Z12 = ± [ Z 22 ( Z11 − Z S )] 1/2

(17.40)

CURRENT MEASUREMENT The measurement of current distribution along an antenna is important because if both the magnitude and phase of the current are known at all the points along an antenna, the farfield pattern of the antenna can be obtained easily. There are a number of methods that are used to measure the current distribution along an antenna. The simplest one requires that a small sampling probe (small loop) to be placed near the test antenna. On the sampling, loop current is induced, which is found to be proportional to the current of the test antenna. There are two ways of connecting loop antenna and measuring device. At very long operating wave-length l the loop and indicator can be connected to a single unit. The output of the loop probe can be detected and transferred as low-frequency signal with high-impedance lines. However, at smaller wavelength l the meter can be connected to a crystal rectifier, because measuring device may be too large to place near the test antenna without disturbing the field. In order to avoid the field distribution near the radiator, the rectifier is attached to the meter using long leads. To minimize the interaction between device and antenna, and induced currents on the leads, the wires are wound on a dielectric support rod in the form of helical choke. The diameter of each turn of helix and spacing between them is needed to be about l/50. The dielectric rod can also be used as a support for the sampling loop. To prevent the dc current on the crystal rectifier, a bypass capacitor is placed along the circumference of the loop.

622


Measurement of Current Distribution for Antenna on a Finite Conducting Earth Usually, the effect of the finite conductivity of the earth on the current distribution of an antenna is found by solving the partial differential equation ∂ 2 Az ∂z

2

⎛ jC 2 − C 2 Az = ⎜ ± ⎜ X ⎝

⎞ ⎟ ZI z ⎟ ⎠

which is derived by equating the tangential components the electric field intensity at the surface of antenna, in which Az = vector potential at the surface of the conductor ÿ b = propagation constant Z = surface impedance Iz = current distribution The vector potential Az depends upon the unknown current distribution and it is an integral function of current distribution. Hence, to determine the current distribution on any antenna, we must express the vector potential in terms of its current distribution and the solve the equation. Available literatures reveal that there is a numerical formula which can be employed to calculate the vector potential, therefore, finding the current distribution of antenna on a finite conductive surface is almost not possible. Alternative way to solve this problem is to do it experimentally. The experimental set-up for current distribution on a finite conducting ground is shown in Fig. 17.20. Most commonly, a dipole is considered as AUT. The probes, which measure the current is installed inside the dipole. The dipole is excited by a signal generator. The loop of the probe is extended just outside the slotted dipole. The output of the probe is measured in electrometer connected by a non-metallic transmission line. The O/P of the probe is recorded from the reading of the electrometer as the probe is slid along the length of the dipole. A field intensity meter is placed at a distance of 30 feet from the probe antenna towards signal generator, to measure the radiated power level of the signal generator. The output of this meter is kept constant in order to maintain power output of the signal generation remains constant during the measurement.

Design Specification For experimentation two dipoles alike in all respects except for physical length were constructs. First one is designed at f = 60 MHz Second one is designed at f = 30 MHz The dipole elements are made of brass tubing cut to length, with a longitudinal slot which permitted the probe loop of an RF detector to extend above the external surface of the tubing. A non-metallic mount is fabricated to hold the test antenna either horizontally or


623

FIG. 17.20 Experimental set-up for measuring current distribution on antennas above ground. vertically polarized at varying distance above the ground. For the 60 MHz dipole this range varied from one inch to quarter wavelength from lower antenna trip to ground, however, for 30 MHz dipole maximum height is limited to 36 inch only during the measurement. Using the above described test set-up, the measurement of current distribution along the length of the dipole for various heights (1, 6, 12, 24 and 36 inches) was carried out. The result shows that for 60 MHz dipole the current distribution is nearly sinusoidal and effect of conductivity of earth is negligible. As the frequency gets lower and lower, the current distribution departs more and more from its sinusoidal distribution [20].

PHASE MEASUREMENT The phase measurement is another important parameter of an antenna without which complete description of the radiation field is difficult. The phase of any antenna is periodic, and it is defined in multiples of 2p. In addition, it is a relative quantity and a reference is a must during measurement for comparison. When an antenna radiates, there is an equivalent point in the antenna geometry, which represents the radiation centre. In the far-field region, the phase pattern of the antenna remains constant with angle when measured with respect to this point. Therefore, the phase centre of an AUT is determined by positioning the rotational axis of the AUT such that the phase pattern within the main beam remains constant. In general, the far-field component of radiation of any antenna is expressed by

624


ˆ (R G ) e jZ (R , G ) En = nE

e− jkr

(17.41)

r

where r = distance of observation point k = propagation constant nˆ = unit vector, For linear polarization nˆ is real and it may be represented by aq and af in the direction of (q , f ) Here, E(q, f) represents the amplitude of the field and y is the phase function describes the phase pattern of the field in the direction of unit vector, i.e. nˆ . It is desirable to assign a reference point to the antenna such that for a given frequency, y(q, f) is independent from the q and f, i.e., y(q, f) is constant. The reference point which makes y(q, f) independent from q and f is called phase centre of the antenna and corresponding radiated fields are spherical waves with ideal equi-phase surface. That is, a phase centre is reference point from which radiation is said to emanate and radiated fields are measured on the surface of a sphere whose centre coincides with the phase centre having the same phase. Basically, there are two techniques that are used to measure the phase pattern. First one measures the phase pattern at short distances from the antenna and the second one measures at long distance from the antenna. Figure 17.21 is designed in such a way that a reference signal is coupled from the transmission line and it is used to compare the phase of the received signal in an appropriate network.

Antenna under test

Source

Movable probe

Reference Phase measurement circuit

Flexible cable

Test

FIG. 17.21

Near-field arrangement for measuring phase patterns.

But for large distances, this method does not provide direct comparison of received signal and reference signal. For the large distance phase pattern, the arrangement shown in Fig. 17.22 can be used in which the signal from the source antenna is received simultaneously by a fixed antenna and the AUT. The phase pattern is recorded as the AUT is rotated while the fixed antenna serves as a reference.


625

Fixed antenna Reference Phase measurement circuit

Distance source

Rotary joint Test Antenna under test

Rotary mount

FIG. 17.22

Far-field arrangement for measuring phase patterns.

The third method to measure the phase pattern of an antenna is differential phase method where no phase reference is needed. In this method, AUT is used as transmitter and two-channel receiver are used. The transmitting AUT is rotated and phase difference is measured using two-channel receiver. Therefore, phase pattern can be calculated from the measured phase difference data. In a slight modification of this method, the phase difference is determined from three power readings, which are measured with a single power meter. The main advantage of this method is that no rotary joints or flexing cables are needed in this measurement system. Rotary joints Distance source

Reference Phase measurement circuit Test Rotary mount

FIG. 17.23

Antenna under test

Differential method of phase measurement.

MEASUREMENTS OF NOISE FIGURE AND NOISE TEMPERATURE The antenna noise temperature is a measure that describes the noise power received by the antenna at a particular frequency. It can be obtained by integrating the product of the antenna directivity and the brightness temperature distribution of the environment over the entire

626


space. The brightness temperature of the environment is dependent on many noise sources: cosmic, atmospheric, man-made and ground. In general, the noise power received at the antenna terminals is equal to (KTaB) in which K is Boltzmann coefficient, Ta is the antenna noise temperature and B is the bandwidth of the system receiver. A simple set-up, where an excess-noise generator is inserted between a matched input termination at temp (T0) and receiver input is used to measure noise-temp and noise-figure. The excess-noise generator may any one from the following [21]: (i) The temperature limited diode with excess noise ratio (ENR) typically 5 dB to 6 dB. where ⎡ T − T0 ⎤ ENR = 10 log ⎢ n ⎥ ⎣ T0 ⎦

in which Tn = Output noise temperature of excess noise source and T0 = Reference temperature ~ 290 K. (ii) The organ gas discharge tube integrated with a wave guide with ENR » 15.5 dB. (iii) Solid state noise source based on the avalanche effect with ENR typically between 15–35 dB. In order to measure the noise-temp and noise-figure accurately two sets of measurements are taken (Fig. 17.24): (a) Measurement without excess noise source. (b) Measurement with excess noise source. Matched termination (R, T0)

Matched termination (R, T0)

FIG. 17.24

Receiver (Tr, Gr)

Excess noise source (ENR, dB)

Power meter (N1)

Receiver (Tr, Gr)

Power meter (N2)

Measurement of noise-figure of a receiver using excess noise source.

In the first case, the receiver is connected to the matched terminal at T0 and the output noise power level (say Na) is measured as follows: Na = KT0BG + KT0(Fr – 1) BG = KBG T0Fr However, in the second case an appropriate excess noise source is inserted between matched termination and input receiver, and output noise power level (say Nb) is measured as Nb = KTnBG + KT0(Fr – 1)BG = KBG[Tn + (Fr – 1)T0]

627


Therefore from the above equations Nb

=

Na

[Tn + (Fr − 1) T0 ] T0 Fr

= Y (say)

(17.42)

After simplification which gives Fr =

Tn − T0

1 ⎛ T − T0 ⎞ =⎜ n ⎟ (Y − 1) T0 ⎝ T0 ⎠ (Y − 1)

⎛ T − T0 ⎞ ⎛ 1 ⎞ 10 log Fr = 10 log ⎜ n ⎟ + 10 log ⎜ ⎟ ⎝Y − 1⎠ ⎝ T0 ⎠

or

Fr(dB) = ENR(dB) – 10 log (Y – 1)

(17.43)

That is, the noise-figure of the receiver can be measured in terms of ENR and Y-parameters. However, in general noise-figure is defined as the “ratio of SNR at input terminal to the SNR at the output terminal”, when referred to a matched input source at a temperature T0. That is Fr =

Si /Ni S0 /N 0

=

Si N 0 S0 N i

=

Tr + T0 T0

T0(Fr – 1) = Tr

(17.44)

This is based on the requirement that the receiver noise is referred to an input termination at T0 = 290 K, where Ni = KBT0

noise input

No = KBGT0 + KBGTr

noise output

in which, K = Boltzmann constant. G = Gain of the device and Tr is the effective noise temperature, which implies that if the receiver is noiseless with gain G then the output noise power due to an input noise source at this temperature will equals the noise power from the output of actual receiver (Fig. 17.25).

Ni = KB T0

T0

Si

G

S0

No = GKB T0 + GKB Tr

FIG. 17.25

Noise-figure of the receiver in terms of effective noise temperature.

628


In a particular case, if the input termination is an antenna, the effective noise temperature need not be equal to 290 K. Therefore, the total effective input noise power to an equivalent noise-free receiver will be ⎛ T + Ta ⎞ N it = KTa B + KTr B = KBT0 ⎜ r ⎟ ⎝ T0 ⎠ Nit = KBT0Fs

or

Nit = Fs B 10 log B − 204 (dB)

(17.45)

where Nit the noise power of system is integrated with antenna and receiver as a whole and Fs is system noise-figure and sum of noise-figures due to antenna Fa and receiver Fr Tr + Ta

Fs =

T0

= Fr + Fa

in which Fr =

Tr T0

and Fa =

Ta T0

In order to find the value of effective noise temperature Tr, let us substitute the value of Tr into Eq. (17.42), which gives [Tn + Tr ] T0 Fr

T ⎞ ⎛ = Y ⇒ YT0 ⎜ 1 + r ⎟ = [Tn + Tr ] T0 ⎠ ⎝

Tr(Y – 1) = Tn – YT0 ⎛ T − YT0 ⎞ Tr = ⎜ n ⎟ ⎝ (Y − 1) ⎠

(17.46)

Once Tr is known, the antenna noise temperature can be obtained using a circuit shown in Fig. (17.26), where it assumed that Gr of the receiver is known, then [21]. N1 = KTABG + KT0(Fr – 1)BG N2 = KTnBG + KT0(Fr – 1)BG Therefore, Y =

N2 N1

=

KTn BG + KT0 (Fr − 1) BG KTA BG + KT0 (Fr − 1) BG

=

Tn + Tr TA + Tr

which gives the antenna temperature as

TA =

Tn + (Y − 1)Tr Y

(17.47)


Antenna

Power meter

Precision RF attenuator LNA

629

N1/N2

Radiometer

T0 Tn Cold load LNA = Low Noise Amplifier

FIG. 17.26

Measurement of noise temperature of an antenna.

SOLVED EXAMPLES Example 17.1 An antenna is pointed towards the side wall of an anechoic chamber. The pattern level varies between –23.5 dB and –24.6 dB as antenna is moved. Find the reflectivity of the antenna. Solution:

We know that the amplitude ripple factor (S) = difference of pattern levels.

S = –24.6 + 23.5 = –1.1 dB and the pattern level (a) = –

(24.6 + 23.5) 2

= − 24.06 dB .

⎛ 10 S1 − 1 ⎞ Therefore, reflectivity is defined as R f = a + 20 log10 ⎜ S1 ⎟ ⎜ ⎟ ⎝ 10 + 1 ⎠

where S1 = S/20 =

1.1 20

= 0.055S and 10 S1 = 10 0.055 = 1.135

Hence,

Rf = –24.06 + (–23.98) = – 48.03 dB

EXAMPLE 17.2 Reflectivity of an antenna that is being used in testing range is 45 dB. Measurement is based on VSWR method. Find the pattern level. Also find percentage of reflection, if the amplitude ripple factor is 1.4 dB and power main lobe peak of the probe is pointed to the source. Solution:

or

We know that

a(dB) = R f

⎛ 10 S/20 − 1 ⎞ R f = a(dB) + 20 log ⎜ S/20 ⎟ ⎜ 10 + 1 ⎟⎠ ⎝ ⎛ 10 S/20 − 1 ⎞ ⎛ 10 0.07 − 1 ⎞ − 20 log ⎜ S/20 ⎟ = − 45 − 20 log ⎜ 0.07 ⎟ ⎜ ⎜ + 1 ⎠⎟ + 1 ⎠⎟ ⎝ 10 ⎝ 10

= – 45 + 23 = –22 dB.

630


We also know that when the main lobe peak of the probe is pointed to the source, a = 0, hence reflectivity of antenna is modified to ⎛E ⎞ R f = 20 log10 ⎜ r ⎟ = 20 log ⎝ ED ⎠

⎛ 10 S/20 − 1 ⎞ ⎜ S/20 ⎟ ⎜ 10 + 1 ⎟⎠ ⎝

⎛ E ⎞ ⎛ 10 0.07 − 1 ⎞ 0.174 ⇒ ⎜ r ⎟ = ⎜ 0.07 = 0.080 ⎟= + 1 ⎟⎠ 2.174 ⎝ E D ⎠ ⎜⎝ 10

Therefore find percentage of reflection, Er = 8%. If ED = 2.57 Wr then Er = 0.080 ´ 2.57 = 0.2056 W Example 17.3 The source antenna and AUT, both are parabolic reflector antennas, which have a diameter of one metre. Estimate the amplitude taper, if operating frequency is 15 GHz and separation between antennas is 2D2/l. Solution:

M=

3 × 1010 15 × 10 9

= 2 cm = 0.02 m

The half-power beam width of a reflector antenna is about

R3dB = At the measurement distance

2D 2

M

70°M D =

=

2 × 1 0.02

70 × 0.02 1

= 1.4°

= 100 m

This corresponds to a transverse distance of 1.4 × Q ×

100 180

= 2.442 m

Therefore, the pattern level of the source antenna is

[0.5/(2.442/2)]

2.

⎡ 0.5 × 2 ⎤ (− 3 dB) = ⎢ ⎥ ⎣ 2.442 ⎦

2

× (− 3 dB) = − 5.50 dB at the edge of the AUT.

Example 17.4 If level of a reflected wave from the ground is 45 dB below the level of the directed wave. Find the amount of possible error in the measurement of (i) Main lobe peak (ii) –15 dB side lobe Solution: Since 1 dB = 20 log x Þ x = 10(1/20) Hence, the reflected power is – 45 dB Þ 10(45/20) = 0.0056


631

(i) The main lobe peak, i.e., power 100% = 1 Error range = 1 ± 0.0056 = + 0.046 dB to – 0.049 dB (ii) – 15 dB side lobes provides the amount reflected power = 10(–15/20) = 0.178

⇒ and

0.178 − 0.0056

= 0.9749 = − 0.143 dB

0.178 0.178 + 0.0056

= 1.031 = 0.134 dB.

0.178

Example 17.5 The maximum allowed phase curvature to measure a very low-side lobe antenna is 5°. Determine the distance between the source and AUT, if maximum aperture of antenna is 10 m and operating frequency is 6.0 GHz. d

D/2

R

R

FIG. 17.27

Solution:

M=

c f

=

Geometrical arrangement for Example 17.5.

0.5 × 108 6 × 10 9

= 0.05 m

If d is distance causing the phase error, then (R + d)2 = R + (D/2)2 which gives R » D2/8d.

2Q

Phase error ('G ) = Therefore, R =

1 8

×

D2 d

M =

1 8

d ⇒ ×

5°Q 180°

=

D 2 × 72

M

2Q

M

d ⇒

⇒ R=

d

M

9D 2

M

=

=

1 72

, i.e. d =

9 × 100 0.05

M 72

= 1.8 × 10 4 m

Example 17.6 Determine the range length, antenna height and source diameter (elevated range) for the measurement of a 1.5 m reflector antenna operating at f = 15 GHz.

Solution:

M=

3 × 108 15 × 10 9

= 0.02 m

632


Therefore 2D 2

R≥

M

=

2 × (1.5)2 0.02

=

2 × 2.25 0.04

= 225 m

As we know that height of the antenna should be five times than the diameter of the antenna, i.e., HR » 5 ´ 1.5 = 7.5 m = HT DT ≥

Hence,

1.5 M R HR

=

1.5 × 0.02 × 225 7.5

= 0.9 m

Example 17.7 An antenna of gain 35 dB is to be tested at 10 GHz. The source antenna of gain 20 dBi is placed 200 m away from it. The receiver sensitivity is –100 dBm. Determine the minimum transmitted power that is needed for a dynamic range of 50 dB. Solution:

M=

c f

=

3 × 108 10 × 10 9

= 0.03 m and R = 200 m

2

2

⎛ M ⎞ ⎛ 0.03 ⎞ −10 ⎜ ⎟ =⎜ ⎟ = 1.42 × 10 = − 98 dB ⎝ 4Q R ⎠ ⎝ 4Q × 200 ⎠

Therefore, Hence

⎛ PR ⎞ ⎜ ⎟ = 35 dBi + 20 dBi − 98 dB = − 43 dB ⎝ PT ⎠dB

Since, minimum needed power for reception is –100 dB and dynamic range is 50 dB. Then Pt = –100 dBm + 43 dB + 50 dB = –7 dBm Þ

Pt = 0.2 mW

Example 17.8 Three antennas A1, A2 and A3 are being tested for their gain at 10 GHz. The transmitted power is ± 4 dBm and received powers are –32 dBm, –36 dBm and –30 dBm for the antenna pairs A1A2, A1A3 and A2A3 respectively. What are the gains of the antennas, if they are separated at a distance of 9 m. Solution:

M=

3 × 108 10 × 10 9 2

= 0.03 m 2

0.03 ⎛ M ⎞ ⎛ ⎞ −5 Therefore, ⎜ ⎟ =⎜ ⎟ = 7.83 × 10 = − 41.064 dB Q × × 4 R 4 3.14 9 ⎝ ⎠ ⎝ ⎠


633

Hence, from Frills Equation ⎛ P ⎞ ⎛ 4Q R ⎞ (GT GR ) = ⎜ T ⎟ ⎜ ⎟ ⎝ PR ⎠ ⎝ M ⎠

2

2

⎛ 4Q R ⎞ ⇒ (GT GR ) dB = ( PT )dB − ( PR )dB + ⎜ ⎟ dB ⎝ M ⎠

Therefore for different pair antennas GA1 GA2 = GA12 = − 32 dBm − 4 dB + 41 dB = +5 dB

(i)

GA1 GA3 = GA13 = − 36 dBm − 3 dBm + 41 dB = +2 dB

(ii)

GA2 GA3 = GA23 = − 30 dB − 3 dB + 41 dB = +8 dB

(iii)

From (ii) and (iii) we get

GA 2

=

GA13 GA 23

⇒

⎛G ⎞ GA1 = ⎜ A13 ⎟ GA 2 ⎝ GA 23 ⎠

⎛G ⎞ 2 GA1GA 2 = ⎜ A13 ⎟ GA2 ⎝ GA23 ⎠

Þ

Þ

Similarly GA1 =

GA1

GA 2 = GA13 GA 23

× GA 2 =

GA 3 =

1 2

2 dB 8 dB GA13 GA1

1/2

× GA23 ⎞ ⎛G ⇒ GA2 = ⎜ A12 ⎟ GA13 ⎝ ⎠

[+5 dB + 8 dB − 2dB] = 5.5 dBi × 5.5 dB = 2 dB + 5.5 dB − 8 dB = − 0.5 dBi

= [2 dB − 0.5 dB] = 1.5 dBi

OBJECTIVE TYPE QUESTIONS 1. The testing antenna is preferred to use in receiving mode because (a) It receives plane wave of uniform plane and phase (b) It receives plane wave of properly polarized (c) It receives maximum power (d) None of the above 2. If the testing antenna is at boundary of far-field zone, the maximum phase error of the incident field from an ideal plane wave is (a) 12.5° (b) 20.5° (c) 24.5° (d) None of these

634


3. The zone full of constructive interference in the region of AUT is referred to as (a) Far zone (b) Near zone (c) Quasi zone (d) Quiet zone 4. Which of the following does not belong to the free-space range (a) Elevated range (b) Slant range (c) Compact range (d) Anechoic chamber (e) None of these 5. In a measurement AUT is mounted at a fixed height whereas Tx antenna is placed close to the ground. It is termed (a) Near-field range (b) Far-field range (c) Elevated range (d) Slant range 6. The far-field region is preferred for antenna measurement, however the near-field region is also used in particular measurement when antennas are close to earth surface (a) 3l to 10l (b) l to 10l (c) 3l to 8l (d) 3l to 20l 7. The lowest frequency up to which an anechoic chamber can be used is (a) 30 MHz (b) 100 MHz (c) 900 MHz (d) 1.2 GHz 8. The main technique involved in the design of anechoic chambers is based on (a) Optics (b) EM field (c) Polarity (d) None of these 9. In tapered anechoic chamber at higher frequency reflections from the walls are suppressed by using (a) Optical lens (b) EM polarizer (c) High gain oscillator (d) High gain antenna 10. Frequency of operation of CATR depends on (a) Size of reflector (b) Surface accuracy (c) Both (a) and (b) (d) Far-field 11. The range of frequency of operation CATRs range between (a) 1 GHz to 100 GHz (b) 1 GHz to 10 GHz (c) 2 MHz to 800 MHz (d) None of these 12. Free-space range is useful to measure the gain of antenna at the frequency (a) 10 MHz (b) 10 MHz to 100 MHz (c) Above 1 GHz (d) None of these 13. At the operating frequency below 0.1 GHz, the directive antennas are physically large, the gain of such antennas is measured using (a) situ (b) Anechoic chamber (c) Both (a) and (b) (d) None of these


635

14. The half-wave dipole and pyramidal horn are universally used as standard gain antenna in the measurement, because they (a) Both possess linear polarization (b) Very directive antennas (c) Isolated from environmental conditions (d) All of the above 15. The extrapolation method is an absolute method and uses three antennas in the measurement. This method includes possible errors cause due to (a) Proximity effect (b) Multipath effect (c) Non-identical antennas (d) All of these 16. If number of antennas used in extrapolation method is circularly polarized, then this method gives the gains and polarizations of (a) Three antennas (b) Any two antennas (c) Any one antenna (d) None of these 17. Extrapolation method fails, if two or more antennas are (a) Linearly polarized (b) Elliptically polarized (c) Circularly polarized (d) None of these 18. If only one antenna is used in this method that is circularly polarized, then this method yields the gain and polarization of (a) Linearly polarized (b) Elliptically polarized (c) Circularly polarized (d) None of these 19. The method measures the gain of moderately broad beam linear antenna (below 1.0 GHz) that coupled only the electric field (a) Gain-transfer method (b) Extrapolation method (c) Ground-reflection method (d) None of these 20. The reflection coefficient of the earth is function of angle of incidence of waves, varies very rapidly for (a) Horizontally polarized waves (b) Vertically polarized waves (c) Both (a) and (b) (d) None of these 21. In the antenna measurement, primary polarization is also known as (a) Desired polarization (b) Undesired polarization (c) Cross-polarization (d) None of these 22. In a measurement at 300 GHz, a reflector antenna of diameter 0.6 m is used, what antenna range will be chosen, if atmospheric attenuation is 10 dB/km. (a) Reflection range (b) Near-field range (c) Anechoic chamber (d) Compact range 23. In the measurement the cross-polarization level limits to below (a) –10 dB (b) –20 dB (c) –30 dB (d) – 40 dB

636


24. A normal mode helical antenna of length 25 mm is used in a handset at 300 GHz. The distance at which near- and far-field ranges meets (a) 160 mm (b) 320 mm (c) 15 cm (d) 2 m 25. It describes the noise power received by the antenna at a particular frequency and product of the antenna directivity and the brightness temperature (a) Antenna noise temperature (b) Noise figure (c) Boltzmann coefficient (d) None of these

Answers 1. 6. 11. 16. 21.

(a) (a) (a) (a) (a)

2. 7. 12. 17. 22.

(d) (b) (c) (c) (d)

3. 8. 13. 18. 23.

(d) (a) (a) (c) (d)

4. 9. 14. 19. 24.

(e) (d) (d) (c) (a)

5. 10. 15. 20. 25.

(d) (c) (d) (b) (a)

EXERCISES 1. What is free-space range? 2. Explain the compact range. 3. What do you understand by gain measurement? 4. Describe polarization and impedance measurements. 5. Define the antenna efficiency. Explain the efficiency measurement. 6. Explain the procedure for measurement of radiation pattern with the help of neat sketch. List the functions of each element involved. 7. Find the directivity of antenna if the 3 dB beam width of the antenna in E and H plans is 20 and 15° respectively. Also find the gain, if antenna is 80% efficient. 8. Estimate the elevated range for a measurement, if an antenna of aperture 2.5 m is used at 20 GHz. 9. In a measurement Tx and Rx antennas are at a distance of 250 m, where minimum required signal level is –100 dBm. Find the minimum radiated power that is needed for a dynamic range 55 dB. Assume the Tx and Rx antenna’s gains to be 20 dBi and 45 dBi respectively. 10. How is the gain of antenna measured by direct comparison method? 11. The level of a wave reflected from the ground is x below the direct wave. What will be the value of x for maximum possible error of ± 0.08 dB in the measurement of main lobe peak?


637

12. In a communication at 5 GHz, antennas are separated at 1.5 km. What will be the maximum allowed curvature in the pattern measurement, if a parabolic reflector antenna of 10 m aperture need to be tested. 13. Explain and deduce the expressions for noise figure and noise temperature of an antenna. 14. Explain the process of measurement of the efficiency of an antenna. 15. Describe the slotted line technique for impedance and VSWR measurements. 16. Describe the method for gain measurement. 17. Explain the method of phase measurement of an antenna. 18. Explain the process of measurement of the polarization of an antenna. 19. The distance between Tx and Rx must be ³ 2D2/l. Justify the statement. 20. The gain of three parabolic antennas: A, B and C is measured using three antenna methods where source antenna and AUT are kept at a distance of 20 m. And transmitted power is ± 5 dBm. The received power are – 40 dBm, – 45 dBm and – 48 dBm for pairs AB, AC and BC, respectively.

REFERENCES [1] Johnson, R.C., et al., “Determination of far-field antenna patterns from near-field measurements,” Proc. IEEE, 61(12), pp. 1668–1694, Dec. 1973. [2] Kraus, J.D., et al., Antennas: For All Applications, 3rd ed., Tata McGraw-Hill, New Delhi, 2002. [3] Emerson, W.H., “Electromagnetic wave absorbers and anechoic chambers through the years,” IEEE Trans Antennas Propagate, AP-21(4) pp. 484–490, July 1973. [4] Balanis, C.A., Antenna Theory: Analysis and Design, John Wiley & Sons, New Delhi, 2007. [5] Hirvonen, T., et al., “A compact antenna test range based on a hologram,” IEEE Trans Antennas Propagate, AP-45(8), pp. 1270–1276, August 1997. [6] Newman, E.H., et al., “Two methods for measurement of antenna efficiency,” IEEE Trans Antennas Propagate, AP-23(4), pp. 457–461, July 1975. [7] Chu, L.J., “Physical limitations of omni-directional antennas,” JAP, Vol. 19, pp. 1163–1175, Dec. 1948. [8] Harringotan, R.G., Time-Harmonic Electromagnetic Fields, McGraw-Hill, New York, pp. 264–316, 1961. [9] Kummer, W.H. and E.S. Gillespie, “Antenna measurements” Proc. IEEE, 66(4), pp. 483–507, April 1978. [10] “IEEE Standard test procedures for antennas,” IEEE Std. 149–1979, published by IEEE, Inc., Distributed by Wiley, 1979.

638


[11] Mott, H., Polarization in Antennas and Radars, John Wiley & Sons, NY, p. 110, 1986. [12] Hatkia, L., “Elliptically polarized waves,” Proc. IRE, Vol. 38, p. 1455, Dec., 1950. [13] Leland, H.H. and A.H. Raymon, “Antenna gain calibration on a ground reflection rang,” IEEE, Antennas and Propagate, AP-21(4), pp. 532–538, July 1973. [14] Carrel, R.L., “Analysis and Design of the Log Periodic–Dipole Antenna,” Univ. Illinois, Urbana AD, 264, 558, p. 131, 1961. [15] Bowman, R.R., “Field strength above 16 gHz,” Proc. IEEE, Vol. 15, June, 1967. [16] Newell, A.C., C.B. Ramon and P.F. Wicker, “Accurate measurement of antenna gain and polarization at reduced distances by an extrapolation technique,” IEEE Trans, Antenna and Propagate, AP.21(4), pp. 418–432, July 1973. [17] Sucher, M. and J. Fox, Handbook of Microwave Measurements, Polytechnic Press of the Polytechnic Institute of Brooklyn, Vol. 1, NY., 1963. [18] ANSI/IEEE Std 148-1959, Reaff 1971. [19] Montgomery, C.G., “Techniques of microwave measurements,” MIT, Radiation Laboratory Series, Vol. II, Chap. 8, McGraw-Hill, NY, 1941. [20] Rashid, A.F., et al., “Quasi-near-zone field of a monopole antenna and the current distribution of an antenna on a finite conductive earth,” IEEE Trans. Antennas and Propagate, AP-18(1), Jan. 1970, [21] Prasad, K.D., Antennas and Wave Propagation, Satya Publications, New Delhi, 1996. [22] Jorden, E.C. and K.G. Balmain, “Antenna fundamentals” in Electromagnetic Waves and Radiating Systems, Prentice-Hall of India, New Delhi, 1990.

Appendix A

SCALAR AND VECTOR POTENTIALS We know that the time-harmonic EM field equations with the current density (J) and source charge density (r) are expressed in the following forms: ∇ × E = − jXN H

(A1.1)

∇ × H = jXF E + J

(A1.2)

S F

(A1.3)

ÑB = 0

(A1.4)

ÑJ = – jwr

(A1.5)

∇E =

The fields radiated by an antenna are created by an impressed current distribution (J). The procedure for finding radiated fields is based on Maxwell’s equations. In a simple approach, this current distribution can be obtained by Maxwell’s equations for the field E and H. In the set of above 5 equations, first two are curl, liner and first order differential equations. They are coupled because the functions E and H are appraised in both the equations are unknown. Hence, the solutions of these equations will result in a significant output, so they must be solved simultaneously. In order to simplify the solutions for the functions E and H, with a given J, we introduce two new functions, namely the scalar and vector potential functions f and A. From Eq. (A1.4), it is clear that the divergence of H is zero, (ÑH = 0), that is the vector field H has only circulation, hence it is often called a solenoid field. As it possesses only a circulation, it can be represented by the curl of some vector function as follows: H=

1

N

∇ × A

639

(A1.6)

640

Appendix A

where A is defined as magnetic vector potential and satisfy the vector identify ÑÑ ´ A = 0. Again from (A1.1) and (A1.6), we find ∇ × E = − jXN

Þ

1

N

∇ × A

Ñ ´ (E + jwA) = 0

(A1.7)

The expression in parentheses is an electric field. It is conservation field, because its curl is zero and behaves as a static field. The scalar electric potential f is defined as (E + jw A) = –Ñf

(A1.8)

This is because Ñ ´ Ñf = 0, for any value f. Solving Eq. (A1.8) for total electric fields gives; E = – (jw A + Ñf)

(A1.9)

Hence, the fields E and H can be determined, if the potential functions are known. Substituting the H from (A1.6) into (A1.2) gives: ∇ × H=

1

N

∇ × ∇ × A = jXF E + J

(A1.10)

We also know that curl of curl, i.e., (1 Ñ ´ Ñÿ ´ A) of a function is defined as Ñ ´ Ñÿ ´ A = Ñ(ÑA) – Ñ2A Þ From (A1.9)

Ñ(ÑA) – Ñ2A = jwme (–jwA – Ñf) + mJ

(A1.11)

Ñ2A + w2meA – Ñ(jwmeÿ + ÑA) = – mJ

The Lorentz condition established a relation between vector potential (A) and scalar potential as follows: ÑA = – jwmef (A1.12) Then Eq. (A1.11) reduces to Ñ 2 A + w2 m A = – m J

(A1.13)

The importance of Eq. (A1.12) leads to conclusion that Eq. (A1.13) involves only A not f. This is known as vector wave equation. This is a differential equation and can be solved for A once the impressed current J is specified. The vector wave Eq. (A1.13) can also be solved by forming three scalar equations, which begin by decomposing A into rectangular components as follows:

G G G ∇2 A = x∇2 Ax + y∇ 2 Ay + z ∇2 Az

(A1.14)

Substituting Eq. (A1.14) into Eq. (A1.13) and equating rectangular components gives Ñ2A x + b 2A x = – m Jx

(A1.15a)

Appendix A

641

Ñ 2 Ay + b 2 Ay = – m J y

(A1.15b)

Ñ2A z + b 2A z = – m Jz

(A1.15c)

where C = NF is known a phase constant for a plane wave. All the three equations in (A1.15) are identical in form, so after solving any one, the other two can be obtained easily. First, let us have the solution for a point source, later an arbitrary source can be formed as a collection of point sources. The differential equation for a point source is ∇2Z ±

C 2Z

= − E (x ) E (y)E (z )

(A1.16)

where y represents a point source at the origin and dÿ is a Dirac delta function. Though the point source is of infinitesimal extent, its associated current has a direction. If the point source current is directed along z-axis, then

y = Az

(A1.17)

Since the point source is zero, other than at the origin Eq. (A1.16) becomes Ñ2y + b2y = 0,

away from the origin

(A1.18)

which is complex scalar wave equation and known as Helm–Holts equation (y has only radial dependence). The common solutions to (A1.18) are e–jbr and e+jbr, which correspond to the waves propagating radically in outward and inward directions, i.e., y = Ce–jbr, where C is constant of proportionality and for a point source it is equal to 1/4pr. That is

Z

=

e− jC r

(A1.19)

4Q r

This is solution to Eq. (A1.16) and is the magnitude and phase variation with the distance r away from a point source located at the origin. If the observation point P is at distance R from the source point

Z

Then,

=

e− j C R

(A1.20)

4Q R

If the source point is a collection of similar point sources weighted by the distribution Jz, the response of Az is the sum of the point sources responses of Eq. (A1.20). This is expressed by the integral over the source volume v Az =

∫∫∫

N Jz

v′

e− j C R 4Q R

dv′

(A1.21)

A similar expression can also be hold for the x and y components. Ax =

∫∫∫ v′

NJx

e− j C R 4Q R

dv′

(A1.22)

642

Appendix A

Ay =

∫∫∫

NJy

v′

e− j C R 4Q R

dv′

(A1.23)

Hence, the general solution will be sum of all the components, which is A=

∫∫∫

NJ

v′

e− jC R 4Q R

dv′

(A1.24)

which is nothing but the solution to the vector wave-equation Ñ 2 A + w2 mA = – mJ So, we can summarize the procedure for finding the fields generated by a current distribution J. First A can be found from Eq. (A1.24), then H field and E field are found from H=

and respectively.

1

N

∇ × A

E = − jX A − j

∇(∇A)

XNF

(A1.25)

Appendix B

LIST OF ANTENNAS EXPRESSIONS The Maxwell equations for the given medium are G G G G G ∂E ∂H ∇ × H = TE + F , ∇ × E= − N ∂t ∂t G G ∇H = 0 ∇E = 0 The electric and magnetic wave equations G G G ∂E ∂2 E 2 and ∇ E = NT + NF 2 ∂t ∂t

G G G ∂H ∂2 H ∇ H = NT + NF ∂t ∂t 2 2

The attenuation factor

⎡ NF B =X ⎢ ⎢ 2 ⎣

⎛T ⎞ 1+ ⎜ ⎟ ⎝ XF ⎠

The phase constant

⎡ NF C =X ⎢ ⎢ 2 ⎣

⎛ T ⎞ 1+ ⎜ ⎟ ⎝ XF ⎠

Beam solid angle : A =

∫∫

Pn (R , G ) d :

2

2

⎤ − 1⎥ ⎥ ⎦ ⎤ + 1⎥ ⎥ ⎦

1/2

1/2

(Sr)

0 0 GHP (deg2 ) Beam solid angle (approx.) : A = R HPGHP (Sr ) = R HP

⎛D⎞ Directivity of circular aperture = 9.8 ⎜ ⎟ ⎝M⎠

2

643

644

Appendix B

Tai and Pereira formula for antenna directivity

DT =

32 ln 2

R E2 + R H2

Effective noise temperature (TE) is related to noise figure (F) as follows F=

TE + T0 T0

Gain of circular aperture (over l/2) = 6(D/l)2 Directivity of short dipole D = 1.5 = 1.76 dBi Radar equation of gain measurement ⎛ 8Q R 2 (GA )dB = 10 log ⎜ ⎜ a.M ⎝

Excess noise ratio

⎞ ⎛ Pr ⎞ ⎟ + 5 log 10 ⎜ ⎟ ⎟ ⎝ Pt ⎠ ⎠

⎡ T − T0 ⎤ ENR(dB) = 10 log ⎢ n ⎥ ⎣ T0 ⎦

Directivity of half-wave dipole D = 1.64 = 2.15 dBi Signal loss coefficient is defined as n = 2 sin

2Q h1 h2

M D1

Impedance of dipole (l/2), Z = (73 + j42.5) Knife-Edge formula for diffraction loss

L (v) = 6.9 + 20 log ⎡ v2 + 1 + v ⎤ ⎥⎦ ⎣⎢ The Line of Site distance

d = 1.4142

(

)

ht + hr miles

Radiation resistance of loop (single turn) R = 197(C/l)4 The focal length of lens antenna

f =

Lensdiameter ⎛W ⎞ 2 tan ⎜ E ⎟ ⎝ 2 ⎠

Appendix B

Distance from the observation point for lens antenna r =

L (1 − N ) 1 − N cos R

The bandwidth of a zonal E-plane metal plate lens antenna 'f =

50 N (1 + K N )

Aperture field intensity for lens antenna ER ⎡ (PS )R ⎤ ⎢ ⎥ E0 ⎣⎢ (PS )R = 0 ⎦⎥

1/2

=

(N cos R − 1)

1

(N − 1)

(N − cos R )

2 ⎛ dA ⎞ Radiation resistance of short-dipole Rrad = 80Q ⎜ ⎟ ⎝M ⎠

2

Maximum effective area of l/2 dipole antenna Ae = 0.13l2 a2 ⎞ ⎛ Input impedance of folded dipole Z in = 73 ⎜ 1 + ⎟ a1 ⎠ ⎝

The length of harmonic antenna L =

492(n − 0.05) f MHz

2

feet

Index of refraction (m) of Luneburg antenna 2 ⎡ ⎛r⎞ ⎤ N = ⎢2 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ R ⎠ ⎥⎦

1/2

Index of refraction (m) of Maxwell’s Fisheye Lens

N (r ) =

2 ⎛r⎞ 1+ ⎜ ⎟ ⎝R⎠

2

Radiation resistance of harmonic antenna Rr = 73.13 + 69 log10(n)

645

646

Appendix B

The angle of maximum radiation

n − 1

cos R m =

n

The maximum directivity of a wire antenna Dm =

120 Rrw sin 2R max

Directivity of V antenna D = 2.94(h/l) + 1.15 Maximum effective length (MEL) of Beverage antenna

M

MEL =

⎛ 100 ⎞ − 1⎟ 4⎜ ⎝ K ⎠

(in metre)

The design parameters of rhombic antenna

M

H=

and L =

4 sin B

0.5M sin 2 B

The scale factor t for the LPDA Rn +1

U =

Rn

=

Ln +1

=

Ln

d n +1

<1

dn

The spacing factor (s) for the LPDA

T =

dn 2 Ln

1 − U

=

4 tan (B )

Designed bandwidth (Bs) of LPDA Bs = B[1.1 + 7.7(1 – t)2 cot a] Periodic frequencies log fn+1 = log fn – log(t) Electric field intensity and induced EMF of loop antenna

EG =

Vm = Total radiated power P = 30Q 2 I m2

C

M

120 Q 2 [ I ] sin R

A

r

M2

2Q AN cos R

M

= 60 Q 3 I m2

a

M

Em

Appendix B

Maximum effective area and gain of a loop antenna Aem = 5.42 × 10 −2C M and G =

4Q AemI

M2

Radiation efficiency of multi-turn loop antenna 1

IA = 1+

8.48 × 10 −10 (f MHzT r )1/2 n(b′)3 a′

Induced maximum emf of ferrite rod antenna V =

2Q

M

EANFNer

⎛ N NA ⎞ ⎛ NC ⎞ 2 Radiation resistance of ferrite rod antenna Rr = 31200 ⎜ er 2 ⎟ = 20 Q 2 ⎜ ⎟ N er ⎝ M ⎠ ⎝ M ⎠ 2

Effective length of ferrite rod antenna le =

2Q

M

4

ANF Ner

2

⎤ ⎛ a ⎞ ⎡ ⎛ 2l ⎞ Demagnetization factor D = ⎜ ⎟ ⎢ ln ⎜ ⎟ − 1⎥ ⎝l⎠ ⎣ ⎝a⎠ ⎦

⎛ F − 1⎞ ⎛ h ⎞ Bandwidth of a rectangular microstrip antenna BW = 3.77 ⎜⎜ r 2 ⎟⎟ ⎜ ⎟ ⎝ Fr ⎠ ⎝ M ⎠ Resonance frequency of circular patch antenna fr =

Resonant frequency of RMS antenna fr =

1.8421 c 2Q ae F r c 2 F e (L + 2'l)

⎛R ⎞ Keller’s Universal formula for horn patterns PdB (R ) = 10 log ⎜ ⎟ ⎝ R0 ⎠ ⎛ A ⎞ −1 ⎛ R0 ⎞ Flare angle (qH) = 2 tan −1 ⎜ ⎟ = 2 cos ⎜ ⎟ ⎝ lH ⎠ ⎝ 2 R0 ⎠

Gain of horn antenna Gain (dB) = 8.1 + 10 log

AB

M02

2

647

648

Appendix B

H Plane beam width for an optimum H plane sectoral horn

HPBW ≅ 1.36 The directivity of pyramidal horn DdB =

M A

= 78°

M A

26,000 HPEo HPHo

Polarization matched factor of the horn antenna S =

(1 + cosR tan 2G )2 (1 + tan 2G ) (1 + cos2R tan 2G )

⎛QD ⎞ Gain circular horn antenna Gain (dB) = 20 log ⎜ ⎟ − 2.82 ⎝ M0 ⎠

Input impedance of infinite biconical antenna Z i = Ri = 120 ln (cot (R h /2))

Total radiated power of a bi-conical antenna

⎛ R ⎞ ⎪⎫ ⎪⎧ PT = 2QI |H0 |2 ln ⎨cot ⎜ ⎟ ⎬ ⎪⎩ ⎝ 4 ⎠ ⎪⎭ Total far-fields ⎛ C d cos R ⎞ (i) Equal amplitudes and phase ET = 2E0 cos ⎜ ⎟ 2 ⎝ ⎠ ⎛ C d cos R ⎞ (ii) Equal amplitudes and opposite phase ET = 2jE0 sin ⎜ ⎟ 2 ⎝ ⎠ N-elements array ⎧ sin n G /2 ⎫ jG ET = E0 ⎨ ⎬e ⎩ sin G /2 ⎭

Broad-side array

⎡ ⎧ (2N + 1)M ⎫⎤ (R max )min = cos−1 ⎢ ⎨± ⎬⎥ 2nd ⎭⎦ ⎣⎩ End-fire array ⎡ (2N + 1)Q ⎤ (R max ) minor = cos−1 ⎢ ± + 1⎥ C nd ⎣ ⎦

and

and

⎛ NM ⎞ (R min )min = cos−1 ⎜ ± ⎟ ⎝ nd ⎠

(R min ) minor = ±

2N M nd

Appendix B

649

BWFN of Chebyshev polynomial patterns

⎡ M Q ⎧1 ⎫⎤ BWFN = 2 sin −1 ⎢ cos−1 ⎨ cos ⎬⎥ 2(m − 1) ⎭⎥⎦ ⎢⎣ Q d ⎩ x0 Angle at which null maxima occurs ⎡ M

⎛ kQ ⎞ ⎪⎫ ⎤ ⎪⎧ 1 cos−1 ⎨ cos ⎜ ⎟⎬⎥ ⎢⎣ Q d ⎝ m − 1 ⎠ ⎭⎪ ⎦⎥ ⎩⎪ x0

R nm = sin −1 ⎢

Directivity of a broadside rectangular array D=

12.56

M2

× Area of aperture

HPBW, directivity and resultant radiation pattern of n-sources binomial array 1.06

HPBW =

n − 1

=

1.06 2L

=

0.75 LM

M

D0 = 1.77 n = 1.77 1 + 2LM

⎛Q ⎞ En = cosn −1 ⎜ cos R ⎟ ⎝2 ⎠

Beam width of major lobe (broad side and end-fire array)

±

114.6 LM

deg and ± 114.6

2 LM

Design parameters of helical antenna ⎛S⎞ S = C tan B , B = tan −1 ⎜ ⎟ and L = C 2 + S 2 ⎝C ⎠

Axial mode helix HPBW =

52 CM

NSM

, directivity D = 12 NC2M SM and gain G =

(Normal mode) Axial ratio, AR =

ER EG

=

2S M

QD

2

=

2S M CM

2

26,000 HP

2

= 6.2 CM2 NSM

⎛2⎞ and radiation resistance = 395 ⎜ ⎟ hM2 ⎝Q ⎠

650

Appendix B

Corrugated horn antenna, half power =

1.26 M

and (BW)–12dB ≅ 0.8(a)

df

Aperture number

F D

= 0.25 cot

R 2

Field-distribution across aperture ⎡1 + cos R ⎤ = ⎢ ⎥ E0 2 ⎣ ⎦

ER

FNBW of circular aperture

BWFN =

140 M D

(deg)

Rectangular/Square apertures (BWFN) =

115 M L

deg , directivity (Dr) = 4Q

L2

M2

⎛L⎞ = 12.6 ⎜ ⎟ ⎝M⎠

⎛L⎞ gain (over l/2) = 7.7 ⎜ ⎟ ⎝M⎠

and

2

2

Equivalent focal length (fe) and magnification m of multi-reflector antenna fe =

Dm ⎛ Z r ⎞ fe (e + 1) = ⎜ tan ⎟ and m = 4 ⎝ 2 ⎠ fm (e − 1)

Radiated field from Yagi antenna in two principal planes I2 ⎡ ⎤ E (R ) = k ⎢ I1 + in the vertical plane C d cos R ⎥⎦ ⎣ I2 ⎡ ⎤ E (G ) = k ⎢ I1 + in the horizontal plane C d cos G ⎥⎦ ⎣

Lengths of these elements of Yagi–Uda antenna Lreflector =

500 f MHz

ft, Ldriven element =

475 f MHz

ft and Ldirector =

455 f MHz

Friis free-space wave equation Pr (dBw) = Pt (dBW) + Gt (dB) + Gr (dB) + a (dB) – Lp (dB)

ft

Appendix B

Electric field intensity at finite distance

⎛ mV ⎞ ER ⎜ ⎟ = 173 ⎝ m ⎠

Pt (KW) r (km)

Angle of tilt (T) with respect to the normal T = [F r2 + (60 MT )2 ]

Maxwell’s field strength at distance (d)

E=

I ht hr I s Md

Maximum distance covered by the surface wave dmax =

100 3

fMHz

km

Longest wavelength in duct

Mmin = 0.66 × A × D 'M The critical angle in duct

Gc = 7.39 × 10

−2

1/2

⎡ dM × ⎢ ⎣ dh

⎤ × D⎥ ⎦

(deg)

Signal to noise ratio

SNR =

Pt Aet Aer (M R)2 kTS 'f

Reflectivity of antenna R f = 20 log

⎡ 10 S/20 − 1 ⎤ = 20 log ⎢ S/20 ⎥ (dB) Ed + 1 ⎥⎦ ⎣⎢ 10 Er

The electric field intensity at distance d from the receiver ER =

88 ht hr

Md2

P V/m

Critical frequency for ionospheric propagation fc = 9 N max

651

652

Appendix B

Maximum useable frequency MUF = sec i × fc or MUF = fc

⎛D⎞ 1+ ⎜ ⎟ ⎝ 2h ⎠

Refractive index

N = sin i = 1 −

81 N f2

Skip distance Dskip

⎛f ⎞ = 2h ⎜ MUF ⎟ ⎝ fc ⎠

2

− 1

Virtual height

⎡ ⎛ R cos C D = 2R ⎢(90o − C ) − sin −1 ⎜ ⎝ R+h ⎣ Gyrofrequency

fg =

Be 2Q m

Attenuation factor for ionospheric propagation

B=

60Q Ne2O

F r m(O 2 + X 2 )

⎞⎤ ⎟⎥ ⎠⎦

2

Appendix C

Skin Effect Properties of Common Metals Material (Temp) Silver (300 k) Aluminium (300 k) Brass (300 k) Copper (300 k) Copper (77 k)

Conductivity s(s/m) 6.17 3.72 1.57 5.80 18

´ ´ ´ ´ ´

where f –1/2 = 1/ f and f 1/2 =

Dept. of penetration d(m)

Surface resistivity Rs(W)

0.0642 f –1/2 0.0826 f –1/2 0.127 f –1/2 0.066 f –1/2 0.037 f –1/2

2.52 ´ 10–7 f 1/2 3.26 ´ 10–7 f 1/2 5.01 ´ 10–7 f 1/2 2.61 ´ 10–7 f 1/2 1.5 ´ 10–7 f 1/2

107 107 107 107 107 f

Relative Permittivity (er) and Relative Permeability (mr) of Materials Material Air (atmospheric pressure) Aluminium and copper Bakelite Barium Glass (plate) Mica Nylon

Relative permittivity (er) 1.0006 1 5 1200 6 6 8

Material PVC (expanded) Quartz Rubber, neoprene Rubber (hard) Rutile (titanium dioxide) Snow, fresh Soil, clay

Relative permittivity (er) ~ 1.1 5 5.2 3.1 100 1.5 14 (Contd.)

653

654

Appendix C

Material Oil, mineral Paper (impregnated) Paper Paraffin Petroleum oil Plexiglas Polyfoam Polystyrene Polyvinyl Chloride (PVC) Porcelain Cobalt Nickel Soft iron Silicon iron Oxygen (STP) Air Aluminium Tungsten

Relative permeability (mr) 2.2 3 8 2.2 2.15 3.4 ~1.05 2.75 2.7 6 250 600 5000 7000 0.999998 1.00000 1.000021 1.00008

Material Soil, sandy Stone, slate Styrofoam Urban ground Vaseline Teflon Water, distilled Water, fresh Water, sea Wood, fir plywood Platinum Manganese Bismuth Mercury Lead Copper Water Hydrogen (STP)

Relative permeability (mr) 10 7 1.03 4 2.2 2.1 80 80 80 2 1.0003 1.001 0.999833 0.999968 0.9999831 0.9999906 0.9999912 1.0

Dielectric Strength of Materials Material Air Barium titnate Bakelite Glass (plate) Mica Neoprene rubber Nylon and pyrex glass Oil, mineral Paper (impregnated)

Dielectric strength (V/m) 3 7.5 24 30 200 12 14 15 50

´ ´ ´ ´ ´ ´ ´ ´ ´

106 106 106 106 106 106 106 106 106

Material Paper Paraffin Petroleum oil Polystyrene Quartz Rubber, neoprene Silicone oil Strontium titanate Teflon

Dielectric strength (V/m) 16 20 12 24 8 25 15 8 60

´ ´ ´ ´ ´ ´ ´ ´ ´

106 106 106 106 106 106 106 106 106

Appendix C

655

Loss Tangent (tan d) of Some Materials A. The Dielectric Loss Tangents for Some Materials Commonly used in Coaxial Cables tan d (at 3 GHz)

Material Air Teflon Polyethylene, DE-3401 Polyolefin, irradiated Polystyrene Polyvinyl formal (Formvar) Nylon Quartz, fused Pyrex glass Water, distilled

15 3.1 3 3.3 1.1 1.2 6 5.4 1.6

´ ´ ´ ´ ´ ´ ´ ´ ´

0.0 10–4 10–4 10–4 10–4 10–2 10–2 10–5 10–3 10–1

B. The Dielectric Loss Tangents for Some Common Materials Material Air Polyolefin, irradiated PTFE (Teflon) RT/Duroid 5880, PTFE microglass Teflon, glass microfibre Teflon, woven quartz Teflon, woven glass Polystyrene, cross-linked Polystyrene, glass microfibre Quartz, fused G10 Epoxy glass Pyrex glass Alumina, 99.5% RT/Duroid 6010.5, PTFE ceramic

tan d (at 100 MHz) 3 2 5 5 6 1.5 2 4 2 8 3 1 2

´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´

0.0 10–4 10–4 10–4 10–4 10–4 10–3 10–4 10–4 10–4 10–3 10–3 10–4 10–3

tan d (at 10 GHz) 3 ´ 1.5 ´ 9 ´ 9 ´ 6 ´ 2 ´ 7 ´ 2 ´ 6 ´ No 7 ´ 1 ´ 2.3 ´

0.0 10–4 10–4 10–4 10–4 10–4 10–3 10–4 10–3 10–5 data 10–3 10–4 10–3

656

Appendix C

Conductivity of Common Materials Material Aluminium Brass Bakelite Carbon steel Clay Constantan Copper Diamond Earth (dry) Ferrite (typical) Germanium Gold Glass Granite Graphite Iron Lead Limestone Marble

s (mhos per metre) 3.82 ´ 107 1.5 ´ 107 10–10 0.6 ´ 107 5 ´ 10–3 0.23 ´ 107 5.88 ´ 107 2 ´ 1013 10–5 100 0.22 ´ 107 4.10 ´ 107 10–12 10–6 7 ´ 104 1.03 ´ 107 107 10–2 10–8

Material Mercury Mica Nichrome Nickel Paper Paraffin Quartz Rubber (hard) Silicon Silver Soil (sandy) Solder Stainless steel Tungsten Water (distilled) Water (fresh) Water (sea) Wax Zinc

s (mhos per meter) 106 10–15 0.1 ´ 107 1.45 ´ 107 10–11 10–15 10–17 10–15 1,200 6.17 ´ 107 1.0 ´ 10-5 0.7 ´ 107 0.11 ´ 107 1.82 ´ 107 10–4 10–3 5 10–17 1.67 ´ 107

Appendix D

Antennas for Different Applications Freq. range Band Services

Antennas

3–10 KHz

Inverted L&T antennas, trideco antenna, velley span antenna and beverage antennas, frame and loop antennas

VLF

Telegraphy for ships, mine and marine communications

30–300 kHz LF

AM broadcast, long distance point to point communications, RFID (125 kHz) and navigation

300– 3000 kHz

MF

AM broadcast navigation, harbour telephone. Automobiles, aircraft and spacecraft, etc. Microwave measurements

3–30 MHz

HF

Short-wave broadcasting, beamed Half-wave antenna, folded dipole, communication, RFID (13.6 MHz) Yagi–Uda, horizontal rhombic, Over The Horizon (OTH) radar loop antenna, V and sleeve antennas, log-periodic dipole antenna

Inverted L&T antennas, folded dipoles, rhombic antennas and umbrella antenna Half-wave and horn antennas as reference antennas

30–300 MHz VHF

FM, TV, radar, short distance Loop, horn, reflectors and Yagi communication, aircraft and ships, antennas. Turnstile and sleeve navigation, radio relay, telephony antennas, circular and ring antennas and telemetry

300– 3000 MHz

TV, short distance communication, radar, LAN, cellular mobile comm., GPS, aircraft navigation, aircraft landing, etc. satellite base station, missiles

(824– 960 MHz)

UHF

657

Biconical and discone antennas, coaxial loop, slotted cylinders and high gain slot as well as helical antennas Panel antenna and Omni-antenna

658

Appendix D

Freq. range Band Services 3–30 GHz

SHF

(2.4– 5.85 GHz)

30–300 GHz EHF

Radar, radio and TV relay links, satellite communication. Wi-Fi/WiMAx, cellular technology DBS systems, GPS, airborne, global weather and earth resource systems, microwave hyperthermia and microwave radiometry. Temperature measurement Mill metric wave radar and amateur.

Antennas Horn antennas—sectoral, pyramidal, conical, biconical and hog horn antennas. Direction pannel, omni and Yagi antennas. Microstrip antennas, fractal antennas, Planar–Inverted F antennas, microstrip slot applicator with a circular aperture. Reflector antennas—dish-parabolic, grid-paraboloidal, cylindrical and torus-type antennas. Lens antennas—dielectric lens, metal plate lens, stepped and unstepped lens antenna, and bowtie antennas.

Appendix E

FREQUENCY SPECTRUM FOR VARIOUS COMMUNICATION SYSTEMS Extended view of frequency applications can be listed as follows: S.No.

Frequency band

Applications

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

30 Hz–6 KHz 30 Hz–6 KHz 450 MHz 470–870 MHz 900 MHz 1 GHz 1.4 GHz 1.530 1.575 GHz 1.8 GHz 2.2 GHz 2.45 GHz 3.2 GHz 3.8 GHz 4 GHz 6 GHz 7 GHz 2 GHz–10 GHz 10.25 GHz 18 GHz

Human voice (speak) Musical instruments Military search radar UHF broadcast TV Cellular telephone bands Fundamental frequency of harmonic multipliers Galaxies emit microwave GPS downlink GPS uplink Space telemetry Troposcatter communications systems Microwave ovens, RFID Airport search radar Point-to-point microwave relay Satellite downlink Satellite uplink Studio Transmitter Link (STL) Microwave relay TWT, Klystron tube, Mixers Police radar band II–satellite downlink 659

660

Appendix E

S.No.

Frequency band

Applications

24 GHz 34 GHz 94 GHz 100 GHz 300 MHz–30 MHz 18 GHz–23 GHz

II–Police radar II–satellite uplink Missile seeker radar Gyrotron tubes Most of microwave systems operate in this range Popular band for microwaves LAN.

21 22 23 24 25 26

Television Very high frequency (VHF) channels Channel number Frequency (MHz)

2

3

4

5

54606672

6

768288

7

8

9

10

11

12

13

174180186192198204210216

Ultra high frequency (UHF) channels* Channel number Frequency (MHz)

14

15

16

17

18

19

470476482488494500506512

20

…

878

…

82

83

884890

For both VHF and UHF channels, each channel has a 6 MHz bandwidth. For each channel, the carrier frequency for the video part is equal to the lower frequency of the bandwidth plus 1.25 MHz while the carrier frequency for the audio part is equal to the upper frequency of the bandwidth minus 0.25 MHz. Examples:

f0 (video) = 54 + 1.25 = 55.25 MHz f0 (audio) = 60 – 0.25 = 59.75 MHz Channel 14 (UHF): f0 (video) = 470 + 1.25 = 471.25 MHz f0 (audio) = 476 – 0.25 = 475.75 MHz Channel 2 (VHF):

* In top ten urban areas in the United States, land mobile is allowed in the first seven UHF TV channels (470–512 MHz).

Radio Amplitude modulation (AM) radio Number of channels : 107 (each with 10 kHz separation) Frequency range : 535–1605 kHz Frequency modulation (FM) radio Number of channels : 100 (each with 200 kHz separation) Frequency range : 88–108 MHz

Appendix E

661

Amateur Bands Band 160 80 40 20 15 10 06

Frequency (MHz)

Band

Frequency (MHz)

1.8–2.0 3.5–4.0 7.0–7.3 14.0–14.35 21.0–21.45 28.0–29.7 50.0–54.0

2 m — — — — — —

144–148 220–225 420–450 1215–1300 2300–2450 3300–3500 5650–5925

m m m m m m m

Frequency Bands for Satellite Communications Frequency band

Uplink (MHz)

Downlink (MHz)

UHF Military C Band Commercial X Band Military Ku Band Commercial Ka Band Commercial Ka Band Military

292–312 5925–6425 7900–8400 14000–14500 27500–30,00 43500–45500

250–270 3700–4200 7250–7750 11700–1220 17700–21200 20200–21200

Cellular Telephone Land mobile systems Uplink : MS to BS (mobile station to base station) Downlink : BS to MS (base station to mobile station) Uplink (MHz)

Downlink (MHz)

United States of America (IS-54):

869–894

824–849

Europe-Asia (GSM): Global Systems for Mobile Communications:

890–915

935–960

Japan (NTT): Nippon Telegraph and Telephone Corporation:

870–885

925–940

Cordless Telephone United States of America: Digital European Cordless Telecommunications (DECT):

46–49 MHz 1.880–1.990 GHz

662

Appendix E

Radar IEEE Band Designations HF (High Frequency): VHF (Ultra High Frequency): UHF (Very High Frequency): L band: S band: C band: X band: Ku band: K band: Ka band: Millimetre wave band:

3–30 MHz 30–300 MHz 300–1000 MHz 1–2 GHz 2–4 GHz 4–8 GHz 8–12 GHz 12–18 GHz 18–27 GHz 27–40 GHz 40–300 GHz

Appendix F

POINCARES SPHERE AND STOKES PARAMETERS Herri Poincare Jules Herri Poincare was one of the greatest mathematical physicists at the end of 19th century who made a series of profound innovations in geometry, the theory of differential equations, electromagnetism, topology, and the philosophy of mathematics.

Poincares Sphere The Poincare sphere, conceived by Herri Poincare in the year 1892, provides a convenient way of representing polarized waves. The upper and lower poles of Poincare sphere represent left and right circularly polarized waves respectively. Points on the equator indicate linear polarization however other points on the sphere represent elliptical polarization. An arbitrary point H chosen on the equator designates horizontal polarization, and the diametrically opposite point V designates vertical polarization. Every point on the surface of the Poincare sphere represents a unique state of polarization and useful to find the general pattern. The change of state of polarization during any optical interaction can be described by a curve on the surface of the sphere connecting the state of polarization. The Poincare sphere provides a very useful way of characterizing polarization state. In addition, it is a simple way of understanding the de-polarization process and of seeing the comparative advantages of using different polarizations for meteorological measurements. Any mixture of different kinds of the polarized wave may be reduced to an equivalent amount of one kind of polarized wave characterized by the orientation and length of axes of an ellipse. It may, of course, happen that the ellipse degenerates to a straight line or circle [1].

663

664

Appendix F

An alternative way of characterizing the polarized wave is by means of the Stokes’ parameter named after scientists G.G. Stokes 2

S0 = E x

2

+ Ey

S1 = E x

2

− Ey

(F1.1a) 2

(F1.1b)

S2 = 2 E x

E y cos G

(F1.1c)

S3 = 2 E x

E y sin G

(F1.1d)

The Stokes’ parameters (S0, S1, S2, and S3) constitute a matrix representation of polarized wave. This kind of representation makes available certain powerful methods for calculating how the state of a beam of polarized wave changes on reflection or on passing through a medium.

Conjugate and Cross-polarized Points If a wave with polarization ratio p is represented by S1, S2, and S3 on the Poincare sphere, then the points S¢1, S¢2, and S¢3 represent conjugate of p, i.e., p*. So p=j

Ey Ex

Ey

e jG then p* = (− j )

Ex

e− jG = ( − 1) j

Ey Ex

e − jG

or p* = j

Ey Ex

as e–jp = cos p – j sin p = –1 + 0 = (–1)

e − j ( −Q +G )

Corresponding Stokes’ parameters are defined as S0′ = E x

2

+ Ey

S1′ = E x

2

− Ey

2

2

= S0 = S1

(F1.2)

S2′ = 2 E x E y cos(Q − G ) = − S2 S3′ = 2 E x E y sin (Q − G ) = S3 From p′ = −

1 p

=j

Ex Ey

e− jG , we find the Stokes’ parameters of the transformed points to be S0′ = E x

2

+ Ey

2

= S0

Appendix F

S1′ = E x

2

2

− Ey

= − S1

665 (F1.3)

S2′ = 2 E x E y cos(−G ) = S2 S3′ = 2 E x E y sin (−G ) = − S3

Stokes Parameters and Poincare Sphere To characterize the amplitude and polarization of a wave, Stokes’ introduced four quantities termed as Stokes’ parameters. They are defined in Eq. (F1.1), in which Ex and Ey are the field components of a wave representing polarization ellipse and f is the phase difference. The parameter S0 gives the amplitude, directly while |Ex| and |Ey| can be found from S0 and S1, then f may be determined from either S2 or S3. From Eq. (F1.3) it is seen that S12 = E x

Hence,

4

+ Ey

− 2 Ex

S22 = 4 E x

2

Ey

2

cos2G

S32 = 4 E x

2

Ey

2

sin 2G

S12 + S22 + S32 = E x = Ex

Þ

4

4

4

+ Ey + Ey

4

4

− 2 Ex + 2 Ex

2

2

Ey

2

2

Ey

2

Ey

2

+ 4 Ex

2

2

E y {cos2G + sin 2G}

= S02

S02 = S12 + S22 + S32

(F1.4)

The description of polarization ellipse reveals that tan B =

Ey

0 ≤ B ≤

Ex

Q

(F1.5)

2

tan 2t = tan 2a cos f

(F1.6)

and

sin 2E = where a = the auxiliary angle ÿ ÿ tÿ = ellipse tilt angle ÿ ÿ dÿ = another auxiliary angle

2 Ex E y Ex

2

+ Ey

2

sin G

(F1.7)

666

Appendix F

Again, from Eqs. (F1.4) and (F1.5), we get

⎛ 2 tan B ⎞ tan 2U = ⎜ ⎟ cos G = 2 ⎝ 1 − tan B ⎠

Þ

tan 2U =

S2

2 1 −

Ey Ex Ey

2

Ex

2

cos G =

2 Ex Ey Ex

2

[from Eq. (F1.1)]

S1

− Ey

2

cos G

(F1.8)

From Eqs. (F1.7) and (F1.1), we get sin 2E =

S3 S0

⇒ S3 = S0 sin 2E

(F1.9)

From Eqs. (F1.7) and (F1.8) and Eq. (F1.4), we get S1 = S0 cos 2d sin 2t

(F1.10)

Again substituting Eq. (F1.10) into Eq. (F1.7) yields S2 = S0 cos 2d sin 2t

(F1.11)

Therefore, expressions for S0, S1, S2 and S3 reduce to S0 = |Ex|2 + |Ey|2 S1 = S0 cos 2d cos 2t S2 = S0 cos 2d cos 2t S3 = S0 cos 2d It is clear that these equations suggest a geometrical interpretation of the Stokes’ parameters. S1, S2 and S3 are the Cartesian coordinates of a point on a sphere of radius S0. The angle 2d and 2t are the latitude and azimuth angles measurement to the point. This interpretation was first introduced by Poincare, hence the sphere is called Poincare sphere [2]. A 3-D Poincare sphere with Stokes’ parameters and latitude and azimuth angles is shown in Fig. F1.1. Since we can describe the amplitude of a wave by S0 and its polarization by S1, S2 and S3, hence any wave can be described by a point on the sphere. To every state of polarization there corresponds one point on the sphere and vice versa. Special cases (1)

For left circular polarization

E x = Ey Hence

and G = +

S0 = 2|Ex| , S1 = S2 = 0 2

and

Q 2 S2 = S0

Appendix F

FIG. F1.1 (2)

The Poincare sphere.

For right circular polarization

Ex = Ey Hence (3)

667

and G = ±

S0 = 2|Ex| , S1 = S2 = 0 2

and

Q 2 S3 = – S0

(a) Left elliptical 0 < f < p and, therefore, from (F1.1d), S3 > 0. All the points for this polarization are plotted on the upper hemisphere of Poincare sphere. (b) Right elliptical

p < f < 2p

S3 < 0.

and

i.e., all the points are in the lower hemisphere. (c) Linear polarization If Ex and Ey ¹ 0, then f = 0, pÿ and S3 = 0 and all linear polarizations are at the equator. For vertical linear polarization, the polarization point is at the (–) x-axis with the intersection of sphere and for linear horizontal it is at the (+) x-axis intersection. The (+) and (–) y-axis intersections correspond to linear polarization with tilt angles (p + 4) and (3p + 4) respectively. Polarization ratio In general, the polarization ratio P (which carries all the necessary information about polarization), is defined as

P=

Ey Ex

=

b a

e jG

(F1.12)

668

Appendix F

where a and b are real numbers and related such as a2 + b 2 = 1 Whereas for circular wave component polarization ratio is termed circular polarization ratio and defined as q=

which is also defined as q =

a − jbe jG a + jbe jG

EL ER

=

L R

e jG

.

Therefore, relation between P and q will be q=

1 − jP 1 + jP

or P = − j

1 − q 1+q

From these relations, it is clear that P is multiplied by j, therefore we can re-write p = jP. Hence, we get q =

1 − P 1+P

or p =

1 − q 1+q

. This new term q is called modified polarization

ratio and relates with electric field components as

q=j

Ey Ex

=

b a

e jG

where Ex, Ey, a, b and f are defined in Fig. F1.2.

FIG. F1.2

Tilted polarization ellipse.

669

Appendix F

Axial ratio The axial ratio of a polarization ellipse in terms of polarization factor is defined as AR =

1+ q 1 − q

whereas axial ratio of a polarization ellipse in terms circular polarization factor is defined as AR =

1+p + 1 − p 1+p − 1 − p

The tilt angle (t) is defined as

U = tan −1

⎡ ⎢1 + ⎢ ⎢ ⎢1 + ⎢⎣

⎛L⎞ ⎤ ⎜ ⎟ sin R /2 ⎥ R ⎝ ⎠ ⎥ ⎥ ⎛L⎞ ⎜ ⎟ cos R /2 ⎥ ⎥⎦ ⎝R⎠

(F1.13)

Solutions to Eq. (F1.13) are U =

1 2

R, Q +

1 2

R

We wish to keep t is the range of 0–p. So, the first form can be used for q positive and the second for q negative. The p and q can also be expressed in terms of Stokes’ parameters and the angles 2d and 2t on the Poincare sphere. Dividing (F1.1d) and (F1.1c) yields: tan G =

S3 S2

⎛S ⎞ or G = tan −1 ⎜ 3 ⎟ ⎝ S2 ⎠

(F1.14)

From (F1.1a) and (F1.1b) gives: Ex =

Ey =

1 2 1 2

(S0 + S1 )

(F1.15a)

(S0 − S1 )

(F1.15b)

Once again S0 + S1 = 2|Ex|2 p = ( j)

Ey Ex

e jG =

2 Ex Ey 2 Ex

2

(j cos G − sin G )

670

Appendix F

p=

or

jS2 − S3

(F1.16a)

S0 + S1

Similarly, q=

S1 − jS2

(F1.16b)

S0 − S3

Substituting the values of S1, S2 and S3 into Eqs. (F1.16a) and (F.1.16b) yields

p=

− sin 2 E + j cos 2E sin 2U

(F1.17a)

1 + cos 2E cos 2U

and

q=

cos 2 E cos 2U − j cos 2E sin 2U

(F1.17b)

1 − sin 2E

Finally, we can find that the Poincare sphere angles in terms of the Stokes’ parameters as follows: ⎛S ⎞ 2E = sin −1 ⎜ 3 ⎟ ⎝ S0 ⎠

(F1.18a)

⎛S ⎞ 2U = tan −1 ⎜ 2 ⎟ ⎝ S1 ⎠

(F1.18b)

and

Some special values of polarization factor P is listed in Table F1.1 given below: TABLE F1.1

The values of P, p and q for different polarizations

Polarization

p

P

q

S1

S2

S3

Right circular Left circular Linear vertical (tÿ = 1/2p) Linear horizontal (tÿ = 0) Linear (tÿ = 1/4p) Linear (tÿ = 3/4p)

1 –1 j¥ 0 J –j

–j J ¥ 0 1 –1

0 ¥ –1 1 –j J

0 0 –S0 S0 0 0

0 0 0 0 S0 –S0

–S0 S0 0 0 0 0

Polarization Efficiency A quantity known as polarization efficiency may be defined for any two polarization states. It is the power transferred between the two polarization states (from a wave to an antenna)

Appendix F

671

relative to what would be transferred if the polarization states were matched. It may be computed from the known positions of the polarization states on the Poincare sphere with the relation ⎛

'⎞

2

I = ⎜ cos ⎟ 2⎠ ⎝ where D is the spherical angle between the two polarization states. The polarization efficiency may also be calculated from equations in which the polarization states are expressed in terms of magnitude and phase of orthogonal components or in terms of axial ratio, tilt angle and sense of polarization [3]. Polarization efficiency, also called polarization match factor, is obviously 0 £ h £ 1. The polarization match factor shows how well a receiving antenna of effective length l is matched with polarization of an incoming wave. In general, if we maintain the same degree of impedance matching for an antenna as we vary its polarization properties, then the ratio of actual power received to that received under the most favourable circumstances of matched polarization is known as polarization match factor. In a wave propagation if two antennas (1, 2), of polarization factors p1 and p2 are used, then the polarization match factor is defined as

I=

(1 + p1 p2 ) (1 + p1* p2* ) (1 + p1* p1 ) (1 + p2 p2* )

(a) If in a transmit-receive system configuration, both the antennas are polarization matched, then h = 1. (b) If two antennas that are so polarized that the received signal is free from crosspolarization, then h = 0. The Stokes’ parameters for a monochromatic wave are given in Eq. (F1.1). However, Stokes’ parameters for quasi-monochromatic wave for time-independent amplitude and phase of wave component are S0 = |ax|2 + |ay|2 S1 = |ax|2 – |ay|2 S2 = 2|ax| |ay| cos f S3 = 2|ax| |ay| sin f where

f = f y – fx

If we compare these parameters to the elements of the coherency matrix [4] defined by

⎡ J xx J xx ⎤ [J ] = ⎢ ⎥ ⎢⎣ J yx J yx ⎥⎦

672

Appendix F

We find that S0 = 2Z0(Jxx + Jyy) S1 = 2Z0(Jxx – Jyy) S2 = 2Z0(Jxy + Jyx) S3 = 2jZ0(Jxy + Jyx)

REFERENCES [1] Kraus, J.D., Antennas, McGraw-Hill, NY, 1950. [2] Mott, H., Polarization in Antennas and Radar, John Wiley, NY, 1986. [3] Hatkio, L., “Elliptically polarized waves”, Proc. IRE, Vol. 38, pp. 1455, Dec. 1950. [4] Born, M. and E. Wolf, Principle of Optics, Pergamon Press, NY, 1965.

Appendix G

THE dB, dBm AND dBi TECHNOLOGY In general, the strength of electrical signals is measured in terms of voltages and currents, hence power by multiplying them, i.e., power = voltage ´ current. However, at microwave frequencies voltages and currents are replaced by electric and magnetic fields. At these frequencies first, powers are measured and then electric and magnetic fields are calculated using measured powers. Power can defined as amplitude/strength of EM waves, and its reference level in general is taken to be 1 mW, because most of equipments: telephone, video-display and computer, etc. need 1 mW power to operate. The range of power in various applications varies between megawatts (106 W) to femowatts (10–15 W). In case signal strength is less than the femotwatts, it will be lost in noise. In order to avoid complexities in calculations, dB and dBm notations are used to express power. The dB represents power in range of watts onward whereas dBm represents power from mW downward. These notations compress the wide range of power used in various applications. In general, the dB of any number is obtained by using: dB = 10 log N. Comparison of decimal numbers, scientific notation and decibel numbers are given in Table G1.1. TABLE G1.1

Decimal numbers, scientific notation and decibel numbers

Number

Scientific notation

dB

Number

Scientific notation

dB

1 10 100 10000 0.1

1.0 101 102 104 10–1

0 10 20 40 –10

0.01 0.001 20 50 80

10–2 10–3 101.3 101.7 101.9

–20 –30 13 17 19

673

674

Appendix G

That is, to convert a decimal number into decibel number (dB), just represent the number as 10 to some exponent and multiply the exponent by 10. In addition, to multiply and divide numbers just add and subtract their dB values respectively. For example, 10 multiplied by 10 (=100); 10 dB + 10 dB = 20 dB (= 100), and 100 divided by 10 (= 10); 20 dB – 10 dB = 10 dB (= 10). The dB equivalent of most numbers can be easily estimated just by using dB values of 2 (= 3 dB), 3 (= 5 dB) and 5 (= 7 dB). Power measured in dB is just the power referenced to 1 watt however dBm power referenced to 1 mW. For example, 1 watt is 0 dB, 10 watts is 10 dB and 1000 watts are 30 dB, similarly, 1 mW is 0 dBm, 10 mW is 10 dBm and 200 mW is 20 dBm respectively and vice versa. Table G1.2 shows the conversion of number to dB and vice versa. TABLE G1.2

Conversion of number to dB and dB to number

Number to dB

dB to Number

Number

dB

dB

Number

3 5 5 = 10/2 80 = 2 ´ 2 ´ 2 ´ 10 500 = 100 ´ 5 = 1000/2

5 7 10 – 3 = 7 3 + 3 + 3 + 10 = 19 20 + 7 = 27 30 – 3 = 27 0 – 10 = –10 3 – 20 = –17

5 6 15 33

3 4 30 2000

–17 –15 –24

–0.02 0.03 0.004

0.1 = 1/10 0.02 = 2/100

Table G1.3 illustrates the common reference values in electronics for dealing with decibels. TABLE G1.3 dBw dBm dBf dBm

Common reference for decibels referenced to 1 watt reference to 1 milliwatt reference to 1 femtowatt reference 1 uV or 1 uV/m

The conversion of some common value of powers to dBm is listed in Table G1.4.

Appendix G

TABLE G1.4 Powers 15 pW 32 nW 20 mW 50 mW 5 mW 125 mW 600 mW 250 W 18 W 4 kW 2.5 W 1.5 kW

Common values of power and dBm mW

dBm

15 ´ 10–9 32 ´ 10–6 20 ´ 10–3 0.05 5 125 6 ´ 102 2.5 ´ 105 1.8 ´ 104 4 ´ 106 2.5 ´ 103 1.56 ´ 106

–78.24 –44.95 –16.99 –13 6.99 21 27.78 53.98 42.55 66.02 34 61.76

675

Glossary

Antenna: A part of a transmitting or receiving system which is designed to radiate/receive electromagnetic waves. An antenna has ability to focus and shape the radiated power in space, i.e. it enhances the power in the desired directions and suppresses the power in other directions. The gain of antenna varies with its size, i.e., large antenna has high gain. Antenna pattern: It is a graphical representation in 3D coordinate of the radiation of the antenna as a function of angular position. Antenna radiation performance is usually measured in two orthogonal principal planes (such as E plane and H plane or vertical and horizontal planes). A plane containing the electric and magnetic field vectors and the direction of maximum radiation, for a linearly polarized wave is termed E and H plane patterns respectively. For a particular plane of antenna, distribution of the off-axis power relative to the on-axis power as a function of angle or position is known as acceptance pattern. Antenna directivity: It is defined as the ratio of the maximum radiation intensity (power per unit solid angle) to the average radiation intensity (averaged over a sphere). The directivity of any radiator except isotropic radiator is always greater than unity. Antenna gain: The maximum gain of antenna is defined as product of the directivity and efficiency. As general efficiency is not 100 per cent, the antenna gain is always less than the directivity. When the reference is a lossless isotropic antenna, gain is expressed in dBi, however if the reference is a half-wave dipole, it is expressed in dBd (and 0 dBd = 2.15 dBi). If direction is not specified, the absolute gain of an antenna corresponds to the direction of maximum effective radiated power. Antenna array: It is grouping of antennas in a particular fashion for improving the overall performance of antenna system. In general half-wave dipoles are used together to control the direction in which most of the antennas power is radiated. Basically, there are two types of arrays: broad side and end-fire array. Yagi–Uda and log-periodic antennas are examples. Antenna bandwidth:

The frequency ranges over which antenna performance is satisfactory. 677

678

Glossary

Aperture: The portion of the plane surface of antenna perpendicular to the direction of maximum radiation through which the major radiation passes. Parabolic and horn antennas are aperture antennas. Adaptive (smart) antenna: An antenna system having circuit components associated with its radiating elements such that the antenna properties are controlled by the receiving waves. Adaptive antennas are used as radar antenna with the side lobe elimination characteristics. The side lobe eliminator antenna consists of a conventional radar antenna where output is coupled with that of much lower gain auxiliary antennas. The gain of the auxiliary antenna is slightly greater than the gain of maximum side lobe of radar antenna. Balanced antenna: An antenna is said to be a balanced antenna with respect to ground when both of its arms have the same electrical relationship to ground. Beam width: The angle (degrees) between the half-power points (–3 dB) on beam pattern of an antenna. Beam steering: Changing the direction of the main lobe of antenna pattern. In radio systems, beam steering is accomplished by switching antenna elements or by changing the relative phases of the RF signals driving the elements. Beverage antenna: A horizontal long-wire antenna used for reception and transmission of low-frequency vertically polarized ground waves. Biconical antenna: A biconical antenna consists of two conical conductors (V shapes), having a common axis and vertex, and extendes in opposite directions. It is simplest structure used to achieve broadband characteristics. Typically, the VSWR £ 1.5 over a 2 to 1 bandwidth and this behaviour remains same, even in an array, because of the small mutual coupling of the elements. Bidirectional array: radiation.

An array radiates in opposite directions of the line of maximum

Blob: A portion of the atmosphere where temperature and pressure differences provide suitable conditions for the refraction of radio waves. Bowtie antenna:

Combination of two V wire antenna and suitable for UHF reception [TV].

Broadside array: An array with direction of maximum radiation normal to the plane of array. Coaxial dipole antenna: An antenna comprising an extension to the inner conductor of a coaxial line and a radiating sleeve which is formed by folding back the outer conductor of the coaxial line. Collinear array antenna: A linear array with their axes lying in a straight line. Basically, there are two types of collinear arrays: horizontal and vertical collinear arrays. Usually, collinear array antennas are found in omni-directional wide beam antennas. Co-polarization:

The polarization in which an antenna have maximum radiations.

Cross-polarization:

The polarization normal to the co-polarization.

Glossary

679

Circular polarization: Polarization in which tip of the electric field vector describes a circle. The magnitude of the electric field vector is constant hence axial ratio is 1 or 0 dB. Corner-reflector antenna: A half-wave antenna with a reflector consisting of two flat metal surfaces joining at an angle behind the radiator/at vertex. Critical frequency: The highest frequency at which a signal may be transmitted directly overhead and reflected back to the earth from the ionosphere. It varies with maximum electron density of ionized layers. dBc:

Power relative to the carrier power.

Dipole antenna: A centre-fed wire antenna whose conductors are straight line. Usually, it is a half wavelength antenna. It is also called doublet and its radiation resistance is 73.6 W. Diffraction:

The process of bending EM waves such that they appear behind an obstruction.

Directional antenna: An antenna designed to transmit and receive RF energy in/from a specific direction(s). That is an antenna that radiates most effectively in only one direction. Direct waves: Waves which propagate in a straight line from the transmitting to the receiving antennas with minimum loss. Director: antenna.

An element placed in front of a driven element to increase the directivity of Yagi

Dish antenna:

An antenna similar to parabolic antenna and shaped like a dish.

Driven array:

An array in which all the elements are fed directly.

Duct propagation: The wave-propagation by bouncing between the earth’s surface and the interface between layers of air having different dielectric constants. In this propagation the waves are guided along the duct, like being guided in wave guide. The upper and lower frequencies limits of duct propagation are 10 GHz and 50 MHz respectively. The upper limit sometimes observed even at higher frequencies in air-borne, radar and other military equipment applications. Effective radiated power (ERP): ERP is the effective radiated power with respect to the directivity of an isotropic radiator. Fading: It is the distortion/fluctuation that a carrier-modulated telecommunication wave experiences over certain propagation media. It can also be defined as undesirable variations in the intensity or loudness of the received waves. It is most troublesome and serious problem in receiving radio waves. In case of severe fading, the field strength of radio signals varies from 10 dB to 20 dB. Far-field region: The region surrounding an antenna where the angular field distribution is essentially independent of the distance from a specified point in the antenna region. It varies with antenna size and frequency of operation.

680

Glossary

Folded dipole antenna: Folded dipole antenna is an extremely practical wire antenna and is also called ultra closed spaced array. It consists of two parallel, closely spaced l/2-dipoles joined together at the outer ends forming a narrow wire loop (d << L and d <<ÿ l). Front-to-back (F/B) ratio: Ratio of the maximum radiation in forward direction an antenna to that of backward direction. Gain: Ratio of radiation intensity in a given direction to that which would be obtained if the power accepted by the antenna were radiated isotropically. It is termed absolute gain, however mostly we deal with relative gain which is defined as ratio of power gain in a given direction to that of a reference antenna in its reference direction. Ground-plane antenna: An antenna which uses ground plane as a simulated ground to produce low-angle radiations. Ground wave: Radio waves travelling along the earth’s surface rather than through the upper parts of atmosphere. They are highly attenuated and suitable only for low frequency (2 MHz). Better the conductivity, the lesser the attenuation. All the broadcast signals received during daytime are due to ground wave propagation. Gyrofrequency: A frequency of time period equal to the period of revolution of an electron in its circular orbit under the influence of the earth’s magnetic field of flux density B is termed Gyrofrequency. Its value is 1.47 MHz. Gyro-magnetic field (GMF): This is magnetic field, which originates from the rotation of the molten iron core of our planet. It is weaker near the polar regions and strongest near equatorial regions and at night hours (i.e. opposite to the sun) GMF extends millions of kilometres into space. Height gain: earth.

Ratio of the field strengths at a particular height to that at the surface of the

Helical antenna: An antenna that is in the form of a helix. When the helix circumference is much smaller than one wavelength, the antenna radiates at right angles to the axis of the helix (normal helix), whereas when the helix circumference is one wavelength, maximum radiation is along the helix axis (axial helix). The pitch angles lies between 12° and 14° with circumference from 0.8l to 1.2l. However, the number of turns and the conductor size do not make serious effects on the helix performances. The axial mode antenna is most suitable when a moderate gain of about 15 dB and circular polarization radiation is needed. Hertz antenna: It is a half-wave antenna that is positioned some distance above earth either vertically or horizontally. The antenna requires a large amount of radiation resistance to radiate sufficient amount of power. Since it is very small, the radiation resistance Rrad is very small and no feeding lines of such low impedance are available to match the dipole. Hertzian dipole is very useful to calculate the field of a large wire antenna. This is done by considering a long wire antenna as a combination of a large number of Hertzian dipoles connected in series.

Glossary

681

Hop: During propagation, a single reflection of the wave and back to earth at a point beyond the horizon is termed hop. Alternate mode of propagation which is suitable over long distance is multi-hop propagation, in which waves of frequency range around 2 MHz to 30 MHz are reflected from ionosphere layer and return back to the earth completing the communication. Horizontal pattern:

The radiation pattern of an antenna parallel to earth surface.

Horizontal polarization: Propagation of waves in such a way that the electric lines of force are parallel to the earth’s surface. Horn antenna: A flared waveguide used as a radiator/antenna. A horn antenna offers broadband characteristics and used to provide a good broadband impedance matching to free space by properly designing the E plane flare length and angle. Half-power beam width (HPBW): The angle between the two points at which the radiation intensity is one-half of the maximum value on the major lobe of radiation pattern. It is commonly referred to as the 3-dB beam width and decreases as antenna gain increases. Isolation: It is a measure of power transfer from one antenna to another and defined as the ratio of the power input to one antenna to the power received by the other antenna. Isotropic radiator: A hypothetical, lossless antenna having identical radiation in all directions. Directivity of this radiator is one and it is used as reference antenna in many applications. Inverted L antenna: section of L shape.

It is half-wave dipole fed by a one-quarter wavelength long vertical

Inverted V antenna: A half-wave dipole erected in the form of an upside-down in V shape and fed at the apex. It is essentially horizontally polarized omni-directional antenna, and also known as a dropping doublet. It is one of travelling wave antenna used in HF band for reception of waves up to 60 GHz. Ionization: The process where radiation from the sun make some of the earth’s atmosphere partially conductive. The majority of ionization is caused by solar energy. Ionosphere: Ionosphere is defined as partially conducting region of the earth’s atmosphere between 50 km and 400 km height, because it tends to be easily ionized by solar and cosmic radiation phenomenon. Ionospheric storms: Ionospheric storms are nothing but disturbances in the earth’s magnetic fields. The reason behind ionospheric storms is high absorption of sky waves and abnormal changes in critical frequency of E and F2 layers. Ionospheric storms are associated with both solar eruptions as well as rotation of the sun. Ionosode: Ionosode is a short-wave transmitter tunable through the whole short-wave range. It transmits various frequency pulses (pulse duration = 150 ms) whose echos are analyzed by means of radar. It is normally used to measure the virtual height. Ionogram: Ionosode automatically plots the variation of virtual height with operating frequency, which is known as Ionogram.

682

Glossary

Linear array: A set of identical radiating elements (e.g. dipole, wire or patch) of uniform magnitude and each with progressive phase arranged along a line at a finite spacing. A linear array has a higher gain than a single radiator and its radiation pattern can be synthesized to meet various antenna performance requirements such as upper side lobe suppression and null fill. Line of sight: Transmission path of direct waves from the transmitting antenna to the receiving antenna. Log-periodic antenna: An antenna array of structural geometry such that its impedance and radiation pattern repeat periodically as the logarithm of frequency. Log-periodic dipole arrays are the most common type, and have a very broad frequency range and unidirectional characteristics. It offers linear or circular polarizations with moderate directive gain. Loading: Connecting an electrical device which is capable of accepting power to match the impedance of an antenna to a transmitter such that maximum power is radiated from a transmitter. Long-wire antenna: An antenna of one wavelength or its multiples at the operating frequency. It is basically of two types: resonant and non-resonant. Loop antenna: A loop antenna is designed in such a way that its periphery is smaller than wavelength, this is because currents are found to be of same magnitude and phase throughout the loop. The radiation resistance of loop antenna is smaller than the loss resistance; hence, its radiation efficiency is poor. Lowest usable frequency: The lowest frequencies that are not penetrated into ionosphere and complete the communication between two terminals. M-curves: Curves that show the variation of modified index of refraction with height. When the modification is done, straight rays above a curved path come out as curved rays above a flat earth. M-curves are useful to predict at least roughly, the transmission path that is usually expected. Microstrip antenna: A microstrip antenna in its simplest form is a layered structure with two parallel conductors separated by a thin dielectric substrate and the conductor. The upper portion is termed patch that is responsible for radiation and lower portion/conductor acts as a ground plane. Major/main lobe: Radiation lobes containing the direction of maximum radiation. Directional antenna having only one main beam is always preferred. Null filling: The process of filling the nulls in the antenna radiation pattern to avoid blind spots in a coverage area. Maximum useable frequency: The highest frequency, which can be reflected back to earth only for specific angle of incidence rather than the vertical. The MUF differs for different layers of ionosphere and normal value lies between 8 MHz and 35 MHz. In general, the MUF is approximately three times greater than the critical frequency.

Glossary

MLPV:

683

Mobile low profile vertical antenna.

Monopole antenna: Monopole antenna is one of the most widely used antennas throughout the RF spectrum ranging from VHF to UHF. Simple structure of monopole coupled with unique properties such as a pure vertical polarization and horizontal omni-directional coverage, hence attracts its extensive possible uses in a variety of applications. Near-field region: Region in close proximity to an antenna in which the electric and magnetic fields do not show a plane wave relationship and the field intensities do not decrease proportionally with the distance away from the source but reduces considerably. Nutating: Moving an antenna feed point in a conical pattern so that the polarization of the beam does not change. Omni-directional antenna: An antenna having an essentially non-directional pattern in a given plane and a directional pattern in any orthogonal plane. It is special types of directional antennas. For wireless LAN antennas, the omni-directional plane is the horizontal plane. Examples of such antennas are slotted, cylinder and turnstile antennas. Parabolic antenna: It is parabolic reflector integrated with a radiating or receiving element at or near its focus, where feed is attached. The feed may be dipole or horn antenna. Basically, it converts spherical wave fronts from the radiating element into plane wave fronts. Parabolic reflector antenna is capable of producing extremely high gains, in the range of 20–30 dBi. Parasitic array: An antenna array containing one or more elements which are not directly excited, but radiate considerably. Path loss:

The ratio of received power to transmitted power for a wave propagation.

Pencil beam:

A narrow and circular beam from a highly directive antenna.

Periodic antenna: An antenna having approximately constant input impedance over a narrow band of frequencies. Periscope antenna: An antenna system in which the transmitting antenna is oriented to produce a vertical radiation pattern and a flat/off-axis parabolic reflector mounted above the transmitting antenna, to direct the beam in a horizontal path toward the receiving antenna. Phased array antenna: A group of antennas in which the relative phases of the respected signals feeding the antenna are varied in such a way that the effective radiation pattern of the array is reinforced in a desired direction and suppressed in undesired directions. Planar array: An antenna system in which all of the radiating elements (both active and parasitic) are in a single plane. Polarization fading: Fading of signals due to polarization rotation of a received signal. The strength of received signal decreases when the incoming wave differs from the polarization of the receiving antenna. Power gain:

The directive gain of an antenna multiplied by its efficiency.

684

Glossary

Power handling: It is the ability of an antenna to handle high power without failure. High power in antenna can cause voltage breakdown and excessive heat (due to conductor and dielectric antenna losses) which would result in an antenna failure. Pyramidal horn: The pyramidal horn (E–H plane horn) antenna is most popular form of rectangular horn. As it is flared in both E and H planes, it leads to narrow beam widths in both principal planes, forming a pencil beam. Gain in antenna varies from 21.3 to 23 dB, whereas corresponding efficiency lies between 30 and 55%. Quad antenna: wavelengths.

An antenna which consists of two square loops which are cut to a quarter

Rabbit ears antenna: A style of antenna used for VHF reception [TV]. Rabbit ears may also be used in combination with a UHF loop antenna. Radio horizon: The greatest distance on the earth surface at which a transmitted wave can be received by the direct path from a receiver located on the ground. Reference antenna: A real antenna that can be used as a basis of comparison with test antenna radiation patterns: dipole and horn antennas. Reflective array antenna: An antenna array in which the active elements are fixed at a predetermined distance from a surface designed to reflect the signal in a desired direction: billboard antenna. Rhombic antenna: An antenna made of four wires of equal length and connected together in the shape of a rhombus. A directional antenna that is composed of long-wire radiators that form the sides of a rhombus, the two halves of which are fed equally in opposite phase at one apex. Shortened dipole:

A dipole antenna made to resonate at a lower frequency using of a coil.

Skip distance: Distances between the points where a radio sky wave leaves the antenna and is successfully reflected and/or refracted back to earth from the ionosphere completing the propagation. Skip zone: The space or region within the transmission range where signals from a transmitter are not received, i.e. between the ground wave and the point where the refracted wave returns. Sloping long-wire antenna: A wire antenna of length greater than one wavelength and supported in an inclined orientation with respect to the ground. Slot antenna: Antenna formed by cutting a slot in a conducting surface of radiator or in the wall of a waveguide. Slotted cone antenna: The slotted cone antenna is an alteration in cone antenna and consists of an inverted cone with a small longitudinal or axial slot cut in one side. Basically, it is electric hybrid radiating structure and appears like a shunt fed, top loaded, wide plate (low Z0) Tx line antenna. In its lower mode of operation, the vertical cone-cylindrical feed is utilized to excite the structure.

Glossary

685

Spoiler antenna: An antenna used for spoiling/changing the radiation pattern of a second antenna so as to reduce the nulls from radiation pattern of the second antenna. Stratosphere: Top-hat: Triatic:

The second layer of the earth’s atmosphere, extending from 10 km to 50 km.

An antenna that is centre-fed and capacitively loaded. A special type of monopole antenna array.

Troposphere: The region of the earth’s atmosphere ranging from 10 km to 15 km above the earth surface. Turnstile antenna: Turnstile antenna consists of two horizontal half-wave antennas mounted at right angles to each other in the same horizontal plane. It is omni-directional and used in VHF communications. Uniform linear array: An antenna array consists of a relatively large number of identical elements arranged in a single line or in a plane with uniform spacing and uniform progressive phase shift along the line. V antenna: A bidirectional antenna, shaped like a V and widely used for communications. It has greater directivity than conventional dipoles. Most of V antennas are symmetrical offers uni- as well as bi-directional radiation patterns. Vertical polarization: Propagation of radio waves such that the electric lines of force are vertical to the earth’s surface. Vertical quarter-wave antenna:

A monopole (whip) antenna that is oriented vertically.

Voltage standing wave ratio (VSWR): The ratio of the maximum/minimum values of standing wave pattern along a transmission line to which a load is attached. The values of VSWR ranges from 1 (matched load) to ¥ for a short or an open load. For most wireless LAN antennas the maximum acceptable VSWR is 2.0, however VSWR of 1.5 or less is excellent. This is approximately the same as a return loss of 14.5 dB, this means most of the signal from the transmitter to the antenna is being radiated (96% radiated and 4% reflected), however a VSWR of 2 (return loss of 9.5 dB) corresponds to 90% radiation and 10% reflection of waves. Whip antenna: A vertical monopole and flexible rod antenna, usually of length between 1/10 and 5/8 wavelengths, supported on a base insulator. Yagi antenna: It is a directive antenna array, consisting of a dipole, reflector and multiple director elements having a gain of 6 to 18 dBi. It is very suitable for MHz frequency range especially for TV reception.

Review Questions

GROUP A 1. What is radiation? 2. What are the two basic classifications of antennas? 3. What are the three factors that define the type, size, and shape of an antenna? 4. What are the terms which are used to identify the points of high current and high voltage on an antenna? 5. What is the need of impedance matching? 6. What are the terms used to identify the points of minimum current and minimum voltage on an antenna? 7. The direction of what field is used to designate the polarization of a wave? 8. What types of polarizations are used at medium and low frequencies? 9. What is the advantage of using horizontal polarization at high frequencies? 10. What type of polarization should be used if an antenna is mounted on a moving vehicle at frequencies below 50 GHz? 11. A radiating source that radiates energy stronger in one direction than another is known as ___________. 12. A flashlight is an example of what type of radiator? 13. What terms are often used to describe basic half-wave antennas? 14. If a basic half-wave antenna is mounted vertically, what type of radiation pattern will be produced? 15. What are the E-plane and H-plane patterns? 16. In which plane will the half-wave antenna be operating if it is mounted horizontally? 687

688

Review Questions

17. Since the radiation pattern of a dipole is similar to that of a doublet, what will happen to the pattern if the length of the doublet is increased? 18. What is the simplest method of feeding power to the half-wave antenna? 19. What is the radiation pattern of a quarter-wave antenna? 20. What is the difference in the amount of impedance between a three-wire dipole and a simple centre-fed dipole? 21. Define the effectiveness ratio. 22. Which has a wider frequency range, a simple dipole or a folded dipole? 23. What is the purpose of antenna stubs? 24. What is the primary difference between major and minor lobes of an antenna pattern? 25. What is the maximum number of elements ordinarily used in a collinear array? 26. Why is the number of elements used in a collinear array limited? 27. How can the frequency range of a collinear array be increased? 28. How is directivity of a collinear array affected when the number of elements is increased? 29. What is the primary cause of broadside arrays losing efficiency when not operating at their designed frequency? 30. When more than two elements are used in a broadside array, how are the elements arranged? 31. How would you represent pattern multification? 32. As the spacing between elements in a broadside array increases, what will be the effect on the major lobes? 33. What are some disadvantages of the end-fire array? 34. Where does the major lobe in the end-fire array occur? 35. To maintain the required balance of phase relationship and critical feeding, how must the end-fire array be constructed? 36. What two factors determine the directivity pattern of the parasitic array? 37. What two main advantages of a parasitic array can be obtained by combining a reflector and a director with the driven element? 38. The parasitic array can be rotated to receive or transmit waves in different directions. What is the name given to such an antenna? 39. What are the disadvantages of the parasitic array? 40. What is the advantage of adding parasitic elements to a Yagi array? 41. The Yagi antenna is an example of __________ array? 42. To radiate power efficiently, a long-wire antenna must have what minimum overall length?

Review Questions

689

43. What is the polarity of the currents that feed the V antenna? 44. What are the main disadvantages of the rhombic antenna? 45. What is the primary reason for the development of the turnstile antenna? 46. Which two composite fields (composed of E and H fields) are associated with every antenna? 47. What composite field (composed of E and H fields) is found stored in the antenna? 48. What composite field (composed of E and H fields) is propagated into free space? 49. What is the term used to describe the basic frequency of a radio wave? 50. What is the term used to describe a whole number multiple of the basic frequency of a radio wave? 51. What is the basic principle of electronic scanning? 52. If a transmitting antenna is placed close to the ground, how should the antenna be polarized to give the greatest signal strength? 53. What are the advantages of Marconi antenna over Hertz antenna? 54. Long-wire antennas are used between frequency ranges ______. 55. What is MUSA? Which antenna is used in it? 56. Directivity of small loop antenna is 1.5, provided that the circumference is ______. 57. Which antennas are suitable for circular polarization applications? 58. What is the range of spacing between elements in Yagi–Uda antenna? 59. Write the relation between design parameters of Horn antenna. 60. What is the main limitation of microstrip antenna? 61. Name the design parameters of LPDA. 62. What is the primary antenna? 63. What is the major reason for the fading of radio waves which have been reflected from a surface? 64. Which layer of the atmosphere has relatively little effect on radio waves? 65. What is the determining factor in classifying whether a radio wave is a ground wave or a space wave? 66. What is the best type of surface or terrain to use for radio wave transmission? 67. What three factors must be considered in the transmission of a surface wave to reduce attenuation? 68. What causes ionization to occur in the ionosphere? 69. How are the four distinct layers of the ionosphere designated? 70. What is the height of the individual layers of the ionosphere? 71. What factor determines whether a radio wave is reflected or refracted by the ionosphere?

690

Review Questions

72. There is a maximum frequency at which vertically transmitted radio waves can be refracted back to earth. What is this maximum frequency called? 73. What three main factors determine the amount of refraction in the ionosphere? 74. What is the skip zone of a radio wave? 75. Where does the greatest amount of ionospheric absorption occur? 76. When a wide band of frequencies is transmitted simultaneously, each frequency varies in different amount of fading. What is this variable fading called? 77. What are three ways of controlling the amount of transmitter-generated EMI? 78. What are three ways of controlling radiated EMI during transmission? 79. What are the two general types of variations in the ionosphere? 80. What is the main difference between these two types of variations? 81. What are the four main classes of regular variation which affect the extent of ionization? 82. What are the three more common types of irregular variations in the ionosphere? 83. What do the letters MUF, LUF, and FOT stand for? 84. When is MUF at its highest and why? 85. What happens to the radio wave if the LUF is too low? 86. What are the disadvantages of operating transmitters at or near the LUF? 87. What are the disadvantages of operating a transmitter at or near the MUF? 88. What is FOT? 89. How do raindrops affect radio waves? 90. How does fog affect radio waves at frequencies above 2 GHz? 91. How is the term ‘temperature inversion’ used when referring to radio waves? 92. How does temperature inversion affect radio transmission? 93. In what layer of the atmosphere do virtually all weather phenomena occur? 94. Which radio frequency bands use the tropospheric scattering principle for propagation of radio waves? 95. Where is the tropospheric region that contributes most strongly to tropospheric scatter propagation? 96. What is meant by radiation pattern? 97. Define the beam efficiency. 98. What are the different types of apertures? 99. What is the limit of G-region? 100. Define the effective height of an antenna. 101. What is meant by front-to-back ratio?

Review Questions

691

102. Define self and mutual impedances of the antenna. 103. What is the reciprocity theorem? 104. What is meant by cross field? 105. What is Mongel–Dellinger effect? 106. What is duality of antenna? 107. Describe Faraday rotation in sky wave propagation 108. What is meant by uniform linear array? 109. Describe the following types of array: (a) Broad side array, (b) End-fire array, (c) Collinear array and (d) Parasitic array. 110. Describe the condition of phase for the end-fire array with increased directivity? 111. A short dipole is also called an elemental dipole, why? 112. A short dipole is called an oscillating dipole, why? 113. List out the applications of loop antennas. 114. List the applications of helical antenna. 115. What are the factors that affect the propagation of radio waves? 116. Define gyrofrequency. 117. What are the main factors that affect the LUF? 118. Duct phenomenon is important for ________. 119. Ionogram is defined as ________. 120. Frequency range suitable for Anechoic chamber is _________.

GROUP B 1. Explain with a diagram how electric and magnetic fields are generated by an antenna. What is the configuration of an ideal antenna? 2. What are the different ways of radio wave propagation is space? Explain with an appropriate sketch. 3. What are different layers in ionosphere? Which one of these in ionization layer? How does it change in length with diurnal and seasonal variations? 4. How does the surface wave propagate? What is the range of frequency over which this propagation occurs? How does this propagation effect by change in the mediums conductivity? 5. How does radio wave propagation occur in various layers of the ionosphere? How does the wave get back to reach ground station at a finite distance from the Tx point? Deduce the condition for the above reflection.

692

Review Questions

6. How do the phase and group velocities of the wave change with the change in ionization density of layers? Do they depend on the frequency of the transmission? 7. Obtain an expression for the distance in surface wave propagation mode in terms of the earth radius and antenna height. 8. In surface wave transmission, calculate the differences of the field at Rx antenna through direct and reflected paths. How is it useful in determining the reception of the transmitted signals? 9. Derive the expressions for the permittivity and conductivity of ionized gas? 10. Show that the directive gain of a Hertzian dipole is = 1.5 sin2q. And for a half-wave dipole, it is ⎛Q ⎞ cos2 ⎜ cos R ⎟ ⎝2 ⎠ 1.64 2 sin R

11. What is LOS? Show that the maximum transmission range in this propagation is d max = 2Re h1 + 2Re h2

in which Re = Radius of the earth, h1 and h2 are the heights of Tx and Rx antennas. 12. What is the retarded time? Why this concept has been introduced in generation of electric and magnetic field vectors? 13. Where is folded dipole used mostly? 14. What special advantages are obtained by using log-periodic antenna? Why it has been named so? 15. Explain with a diagram how parabolic antenna is fed? What is special about such antenna? 16. What is a multipoint beam width in dish antenna? Where is it used mostly? 17. Why is helical antenna proposed to use in some cases? 18. Draw a typical helical antenna and specify its beam width in terms of its diameter, length, turns, spacing, and radiating frequency. 19. Explain how a parasitic element beside a dipole acts as a director or reflector. 20. What is the effect of the additions of reflector and directors to a Yagi array on its impedance, directivity, and bandwidth. 21. State the advantages of Yagi antenna that are popularly used for TV reception. Explain importance of stacking in TV reception.

Review Questions

693

ANSWERS TO REVIEW QUESTIONS GROUP A 1. The phenomenon of transmitting EM waves into the free space is known as radiation. 2. Half-wave (Hertz) and quarter-wave (Marconi). 3. Frequency of operation of the transmitter, amount of power to be radiated, and general direction of the receiving set. 4. Current and voltage loops. 5. To transfer maximum power from one circuit to another. 6. Current and voltage nodes. 7. Electric (E) field. 8. Vertical polarization. 9. Less interference is experienced by man-made noise sources. 10. Vertical polarization. 11. Anisotropic radiator. 12. Anisotropic radiator. 13. Dipole, doublet and Hertz. 14. Non-directional. 15. E-plane and H-plane contain electric/magnetic field vector and direction of maximum radiation. 16. Vertical plane. 17. The pattern would flatten. 18. To connect one end through a capacitor to the final output stage of the transmitter. 19. A circular radiation pattern in the horizontal plane or same as a half wave. 20. Nine times the feed-point impedance. 21. The ratio of effective area to maximum effective area. Its value lies between 0 and 1. 22. A folded dipole. 23. To produce the desired phase relationship between the connected elements. 24. Major lobes have the greatest amount of radiation. 25. Four. 26. As more elements are added, an unbalanced condition in the system occurs which impairs efficiency. 27. By increasing the length of the elements of the array.

694

Review Questions

28. Directivity increases. 29. Lower radiation resistance. 30. Parallel and in the same plane. 31. Enorm = (Pattern of individual isotropic source) ´ (Array factor) 32. They sharpen. 33. Extremely low radiation resistance confined to one frequency and affected by atmospheric conditions. 34. Along the major axis. 35. Symmetrically. 36. Length of the parasitic element (tuning) and spacing between the parasitic and driven elements. 37. Increased gain and directivity. 38. Rotary array. 39. Their adjustment is critical and they do not operate over a wide frequency range. 40. Increased gain. 41. Multi-element parasitic. 42. One-half wavelength. 43. Opposite. 44. The rhombic antenna requires a large antenna site. 45. For omni-directional VHF communications. 46. Induction and radiation fields. 47. Induction field. 48. Radiation field. 49. Fundamental frequency. 50. Harmonic frequency or harmonics. 51. Diverting maximum radiation in the desired direction, by controlling the progressive phase difference between the elements of array. 52. Vertically polarized. 53. The Marconi antenna needs half length comparison to Hertz antenna to produce the same radiation. 54. 500 KHz–30 MHz. 55. Multiple unit steerable antenna; Rhombic antenna. 56. C/l < 1/3 57. Crossed-dipoles, discone, biconical turnstile and helical antennas.

Review Questions

695

58. From 0.10l to 0.20l. 59. Flare angle = tan–1(h/2L), L = h2/8d 60. Narrow bandwidth. 61. Apex angle, scale factor and spacing factor. 62. Antenna used to feed parabolic antenna. 63. Shifting in the phase relationships of the wave. 64. Stratosphere. 65. Whether the component of the wave is travelling along the surface or over the surface of the earth. 66. Radio horizon is about 1/3 farther. 67. (a) Electrical properties of the terrain, (b) polarization of the antenna, and (c) frequency. 68. High energy ultraviolet light waves from the sun. 69. D, E, F1, and F2 layers. 70. D layer is 30–55 miles, E layer 55–90 miles, and F layers are 90–240 miles. 71. Thickness of ionized layer. 72. Critical frequency. 73. (a) Density of ionization of the layer, (b) frequency, (c) angle at which it enters the layer. 74. A zone of silence between the ground wave and sky wave where there is no reception. 75. Where ionization density is greatest. 76. Selective fading. 77. (a) Filtering and shielding of the transmitter, (b) limiting bandwidth, and (c) cutting the antenna to the correct frequency. 78. (a) Physical separation of the antenna, (b) limiting bandwidth of the antenna, and (c) use of directional antennas 79. Regular and irregular variations. 80. Regular variations can be predicted but irregular variations are unpredictable. 81. Daily, seasonal, 11-year, and 27 days variation. 82. Sporadic E, sudden disturbances, and ionospheric storms. 83. MUF is the maximum useable frequency. The LUF is the lowest usable frequency. FOT is commonly known as optimum working frequency. 84. MUF is the highest around noon. Ultraviolet light waves from the sun are most intense. 85. When LUF is too low it is absorbed and is too weak for reception. 86. Signal-to-noise ratio is low and the probability of multipath propagation is greater.

696

Review Questions

87. Frequent signal fading and dropouts. 88. FOT is the most practical operating frequency that can be relied on to avoid problems of multipath, absorption, and noise. 89. They can cause attenuation by scattering. 90. It can cause attenuation by absorption. 91. It is a condition where layers of warm air are formed above layers of cool air. 92. Temperature inversion can cause VHF and UHF transmission to be propagated far beyond normal line-of-sight distances. 93. Troposphere. 94. VHF and above. 95. Near the mid-point between the transmitting and receiving antennas, just above the radio horizon. 96. Radiation pattern is the distribution of radiated power as a function of distance in space with respect to r and q. A graphic representation of radiation pattern shows the variation in actual field strength of the EM wave at all points which are at equal distance from the antenna. It is represented in 2-D and 3-D coordinate systems. 97. The total beam area (WA) consists of the main beam area (WM) plus the minor lobe area (Wm). Thus, WA = WM + Wm. However, the ratio of the main beam area to the total beam area is called beam efficiency. Beam efficiency = SM = WM/WA. 98. There are different types of apertures: Effective aperture: The area over which the power is extracted from the incident wave and delivered to the load is termed as effective aperture. Scattering aperture: It is the ratio of the reradiated power to the power density of the incident wave. Loss aperture:

It is the area of the antenna which dissipates power as heat.

Collecting aperture: Physical aperture:

It is the addition of above three apertures. This aperture is a measure of the physical size of the antenna.

99. Region beyond 400 km. 100. The effective height h of an antenna may be defined as the ratio of the induced voltage to the incident field, i.e., H = V/E. It is related to aperture of antenna. 101. It is defined as the ratio of the power radiated in desired direction to the power radiated in the opposite direction and represented by FBR. 102. Self impedance of an antenna is defined as its input impedance with all other antennas are completely removed, i.e., away from it. Whereas if the presence of nearby antenna no. 2 induces a current in the antenna no. 1, it indicates that presence of antenna no. 2 changes the impedance of antenna no. 1. This effect is called mutual coupling and results in mutual impedance.

Review Questions

697

103. If an emf is applied to the terminals of an antenna no. 1, and the current measured at the terminals of the another antenna no. 2, then an equal current both in amplitude and phase will be observed at the terminal of the antenna no. 1, if the same emf is applied to the terminals of antenna no. 2. 104. Normally, the electric field E is perpendicular to the direction of wave propagation. In some situations the electric field E is parallel to the wave propagation that condition is called cross field. 105. Sudden Ionospheric Disturbance (SID) is known as ________. 106. It is defined as an antenna is a circuit device with a resistance and temperature on the one hand and the space device on the other hand with radiation patterns, beam angle, directivity gain and aperture. 107. Due to the earth’s magnetic fields, the ionospheric medium becomes anisotropic and the incident plane wave entering the ionosphere splits into ordinary and extra ordinary waves. Whenever these modes re-emerge from the ionosphere they recombine into a single plane wave again, changing the plane of polarization, this phenomenon is known as Faraday’s rotation. 108. An array is said to be linear when the elements of the array are spaced equally along the straight line. In addition, the elements are fed with currents of equal magnitude and having a uniform progressive phase shift along the line, then it is called uniform linear array. 109. (a) Broad-side array is an arrangement in which the principal direction of radiation is perpendicular to the array axis and also the plane containing the array element. For broad-side array the phase difference between elements is zero (d = 0), i.e., it is in same phase. (b) End-fire array is defined as an arrangement in which the principal direction of radiation coincides with the array axis. Phase difference between element’s current is (d = bd). (c) In collinear array the antenna elements are arranged coaxially by mounting the elements end to end in straight line or stacking them one over the other with radiation pattern circular symmetry, e.g. omni-directional antenna. (d) In this array only one element is fed, others parasitically excited to reduce the problem of feeding. The power given to one element is electromagnetically coupled, e.g., Yagi–Uda antenna. 110. In an array when d = bd, array produces maximum fields in direction f = 0, but does not give the maximum directivity. However, Hansen and Woodyard showed that a large directivity is obtained by increasing the phase change between the sources so that d = (–bd + p/n) which is referred to as the condition for increased directivity. 111. A short dipole that does have a uniform current will be known as the elemental dipole because considerably shorter than the tenth wavelength maximum specified for a short dipole. Elemental dipole is also called as elementary dipole, elementary doublet and Hertzian dipole.

698

Review Questions

112. A short dipole initially neutral and the moment a current starts flowing in one direction, one half of the dipole require an excess of charge and the other a deficit because a current is a flow of electrical charge. Finally, there will be a voltage between the two halves of the dipole. When the current changes its direction this charge unbalance causes oscillations. Since in such dipole electric charge oscillates, it may be called oscillating electric dipole. 113. Various uses of loop antenna are: · receiving antenna in portable radio and pagers. · as probes for field measurements. · as directional antennas for radio wave navigation. · best uses for direction finding of radio wave propagation. 114. The applications of helical antenna are: · used in space communications for telephone and television. · used in weather satellites and data relay satellites. · in addition, in devices wherever circular polarization needed. 115. (a) (b) (c) (d)

Earth curvature. Earth’s magnetic field. Frequency of the signal. Plane earth reflection.

116. Frequency whose period is same as the period of an electron in its orbit under the influence of the earth’s magnetic flux density B is termed as gyrofrequency. 117. The LUF depends on (a) The effective radiated power (b) Absorption character of ionosphere between transmitter and receiver. (c) The required field strength which in turn depends upon the radio noise at the receiving location as well as type of service involved. 118. UHF and microwave frequencies. 119. Graph between virtual height and frequency. 120. Microwave frequency.

Question Bank with Solutions

1. Find the maximum useable frequency of transmission between two stations 500 km apart, given that electron-density of the reflecting layer is 102 electrons/m2 at an effective height of 240 km. Sol.

The critical frequency of the layer is fc = 9 1012 = 9 ´ 106 Hz. 1/2

f MUF

2 ⎡ ⎛ D⎞ ⎤ = fc ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 h ⎠ ⎥⎦

500 ⎤ ⎡ = 9 MHz ⎢1 + ⎥ 480 ⎦ ⎣

1/2

= 13 MHz

2. Determine the electric field strength at a distance d = 10 km over the rocky land with conductivity s = 1 mS/m and er = 7 from a 3 MHz transmitter with electric field intensity E = 15 ´ 102 mV/m. Sol. and

Numerical distance, p =

2 0.582 d km f MHz

T (mS/m)

= 0.582 × 10 ×

32 1

= 52.38

A = 2+ (0.3p/2) + p + 0.6 p2 = 2 + (0.3 ´ 52.38/2) + 52.38 + 0.6 ´ (52.38)2 = 17.74/2798.044 = 6.34 ´ 10–3

Eg =

E0 A d

(1500 ´ 6.34 ´ 10–3)/10 = 0.95 mV/m = 950 mV/m

3. Find the skip distance for waves of frequency 4.6 ´ 106 Hz at time when the maximum ionization in the E-region has a value of 1 ´ 1011 electrons/m2 at a height of 110 km. Sol.

fc = 9 N max = 9 1011 = 9 × 3.16 × 10 5 Hz = 2.84 MHz 699

700


⎡ (4.6 × 10 6 ⎤ ⎥ = 220 (1.274) = 280.32 km Dskip = 2 × 110 ⎢ 1 − ⎢ 2.8 × 10 6 ⎥ ⎣ ⎦ 4. Two aircraft are flying at an altitude of 4 ´ 103 m and 8 ´ 103 m, respectively. What is the maximum distance along the surface of the earth over which they can have effective point-to-point microwave communication, if radius of earth R = 6.3 ´ 106 m. Sol.

Maximum range for surface wave is = 4.12 ( ht + hr ) km = 4.12[ 4 × 103 + 8 × 10 3 ] km = 4.12[63.25 + 89.44] = 629.09 km

5. Estimate approximately the maximum cut-off wavelength for a duct of height 12 m.

lmax = 0.084 (12)3/2 = 3.49 cm

Sol.

6. The power delivered to an isotropic radiator is 1 kW and the antenna efficiency is 90%. Find the electric field intensity at a distance of 100 km. Sol.

Antenna efficiency h = Power radiated (Wr)/Total power transmitted (Pt) 0.9 = Wt /1 Þ Wt = 900 W

Electric field intensity at a distance of 100 km Erms =

30 Pt d

V/m =

30 × 900 100 × 10

3

= 164.32 × 10 −5 V = 9.64 mV/m

7. What is optimum working frequency for a layer of critical frequency 1 MHz. Also find the critical angle at 15 MHz. Sol.

fMUF = 3.6 fc = 3.6 ´ 12 = 43.2 MHz OWF = 0.85 ´ 43.2 = 36.72 MHz 2

Critical angle at 15 MHz is sin i =

⎛f ⎞ 1 − ⎜ c ⎟ = 1 − 0.64 = 0.6 ⎝ f ⎠

i = sin–1(0.6) = 37° 8. Find the maximum useable frequency for wave propagation if the radio waves of critical frequency is 110 MHz incident at 45° to the D layer of ionosphere. Sol.

We know that a = Angle of transmitted wave from horizontal = 90 – i = 90 – 45° = 45°

fMUF =

fc

sin B

=

110 0.707

= 155.56 MHz


701

9. Calculate the quality factor of 0.5 m long half-wave dipole having a bandwidth of 12 MHz Sol.

Operating wavelength l = 50 ´ 2 = 100 cm

The operating frequency can now be computed by fc = c/l = 3 ´ 1010/100 = 300 MHz Therefore, Q of the antenna is Q = fc/B = 300/12 = 25. 10. An antenna has a radiation resistance and loss resistance of 72 and 10 W respectively. If the power gain antenna is 28, calculate its directivity in dB. Sol.

We know that directivity D = Ap/h

h = Rr /(Rr + Rl) = 72/(72+12) = 72/84 = 0.86

and which gives

D = 28/0.86 = 32.57 = 15.13 dB

11. A police radar antenna produces the narrow beam with both azimuth and elevation beam width equal to 0.5° at frequency 10 GHz. Determine the antenna aperture and gain in dB. Sol.

Azimuth = elevation beam width Dq = Df = 0.5° = 0.5p /180 = 0.00872 rad.

Antenna gain G = 4p/DqDf = 4 ´ 3.14/(0.00872)2 = 16518 = 52.18 dB. Also, antenna aperture area GM 2 16518 × (0.03)2 ⎛ A ⎞ G = 4Q × ⎜ 2 ⎟ ⇒ A = = 4Q 4 × 3.14 ⎝M ⎠

A = 11.84 m2 12. A receiving satellite antenna has an actual projected area to the received beam of 12 square metres. The main lobe of its directional pattern has a length efficiency of 0.8 in the elevation and 0.5 in the azimuth directions. Determine the effective aperture of the antenna. Sol. The aperture efficiency h = length efficiency in elevation ´ length efficiency in azimuth. Therefore, h = 0.8 ´ 0.5 = 0.40 Effective aperture: Ae = h , A = 0.40 ´ 12 = 4.8 square metres 13. Determine the beam width between nulls of a parboiled reflector antenna having a 3 dB beam width of 0.5° and an effective aperture of 5 square metres. Sol. The null to null beam width of most of the antenna response is twice the 3 db beam width. Therefore, null to null beam width = 2 ´ 0.5° = 1.0°. 14. Calculate the length of a half-wave dipole to be cut so as to be resonate at 250 MHz after allowing for required compensation due to length extending effect. Sol.

Operating wavelength l = c/f = 3 ´ 108/250 ´ 106 = 1.2 m.

Therefore length to be cut = 60 – (0.05 ´ 60) = 57 cm

702


15. Design a 3-element Yagi antenna suitable for receiving VHF transmission at 60 MHz. Sol.

Operating wavelength l = c/f = 3 ´ 108/(60 ´ 106) 5 m.

Length of dipole = l/2 = 5/2 = 2.5 m Therefore, length of the dipole = 2.5 – (0.05 ´ 2.5) = 2.375 Length of director = 2.375 – (0.04 ´ 2.375) = 2.28 m Length of reflector = 2.375 + (0.05 ´ 2.375) = 2.49 m Director-dipole spacing = 0.2l = 0.2 ´ 5 = 1 m 16. A focal fed parabolic reflector antenna has the following characteristics of its reflector: Mouth diameter = 2 m and focal length = 2 m If the 3 dB beam width of the antenna has been chosen to be 95% of the angle subtended by the feed at the edges of the reflector, determine the 3 dB beam width and the null to null beam width of the feed antenna. Sol. We know that the angle subtended by the focal point feed at the edges of the reflector q is given by

q = 4 tan–1(D/4f) = 4 tan–1 (2/4 ´ 2) = 56° Therefore, 3 dB beam width = 0.95 ´ 56° = 53.2° Null to null beam width = 2 ´ (3 dB beam width) = 2 ´ 53.2 = 100.8°. 17. A parabolic reflector antenna is fed by feed having a 3 dB beam width of 64°. Calculate focal length, if the mouth diameter of the reflector is 3 m. Assume that the beam width is 0.9 time the subtended angle. Sol.

The subtended angle q given by q = 4 tan–1 [D/4F] Þ

64/0.9 = 4 tan–1 [3/4f] = tan–1 [1/1.333f] = 17.8°

Þ

This gives 1/(1.333 f) = tan (17.8°) = 0.3210

Þ

Focal length f = 1/(1.333 ´ 0.3210) = 2.337 m

18. A small antenna of length 1 cm and diameter 1 mm is used to transmit a signal at 1 GHz near human body. Assuming that the dielectric constant of the body is similar to that distilled water (er = 80) and that the conductivity s can be neglected, calculate the maximum electric field at the surface of human body that is approximately 20 cm away from the antenna. Also find the new distance from the antenna d1 where the signal will be attenuated by 3 dB. The maximum current that appears across the antenna is 10 mA. Sol.

The wavelength of the EM wave within the body

M=

c f Fr

= 3 × 108 /10 9 80 = 3.3 cm


The characteristics impedance of the body is Z 0 =

703

N N0 377 = = = 42 : F F 0Fr 80

Since the dimension of the antenna is significantly less than the wavelength, the electric field intensity | E0 |R = 90° = Z c

I dl ⎛ 2 Q ⎞ I dl 42 × 10 −5 × 10 −2 = Z = = 320 μV/m ⎜ ⎟ c 4Q d ⎝ M ⎠ 2 QM d 2 × 3.14 × 0.033 × 0.2

The attenuation of 3 dB implies that the power changes by a factor of two, and electric field change by a factor of 1.41. The new distance is found to be d1 = 1.414 ´ 0.2 = 0.28 m. 19. Find the HPBW of a short dipole. Sol. We know that the normalized radiation intensity for the electric field of a short dipole is given by Rn(q, f) = (sin2)q i.e.

[Rn(q, f)]max = 1,

when q = p/2

[Rn(q, f)]max = 1/2,

when q = p/4 and 3p/4

Therefore, the HPBW (qHP) for the electric field is p/2. The normalized radiation intensity for the magnetic field of a short dipole is Rn(q, f) = 1. The HPBW of the magnetic field intensity is qHP = p/2. 20. Find the far-field distance of a receiving horn antenna of mouth diameter D = 10 cm, operating at l = 3 cm. Assuming antenna gain Gt = 1.5 and Gr = 1.65 respectively, determine the degradation of the received signal. Sol. The far-field distance; R = D2/2l = (0.1)2/(2 ´ 0.03) = 16.7 cm Beyond this distance the receiving antenna will be in the far-field region of the transmitting antenna. For the loss less cases the effective apertures of both antennas to be Aet = 0.032 ´ 1.5/(4 ´ 3.14) = 1.07 cm2 Aer = (0.03)2 ´ 1.6/(4 ´ 3.14) = 1.18 cm2 From the Friis equation Pr/Pt = (1.07 ´ 1.18)/16.7 ´ 32 = (1.252/2510.01) = 4.98 ´ 10–4 Therefore, the degradation of the received signal 10 log10 (4.98 ´ 10–4) = 33.02 dB

704


21. What is current required to radiate power of 20 W from a loop antenna of circumference l/5. Sol. Assuming a uniform current distribution; the radiation resistance of a loop antenna is given by 2

⎛C ⎞ ⎛ 1M ⎞ Rr = 20 Q 2 ⎜ ⎟ = 20 × 3.142 ⎜ ⎟ ⎝M⎠ ⎝ M5 ⎠

2

= 0.316 W Therefore, the required current will be

I=

2 Pt Rr

=

2 × 20 0.316

= 11.25 A

22. Assume an 80 MHz policy transceiver is faithfully radiating 15 W RF signal through a quarter-wave omni-directional whip antenna from a hill top. Find the field strength for companion transceiver with LOS distance of 20 km (moving in a mobile van). Sol.

The power capital at the source is = 15 W = 11.76 dBW

The path loss at 20 km under LOS will be Lbf = 32.45 + 38.062 + 26.02 = 96.54 dB Therefore, the power of 80 MHz signal at 20 km LOS will be 96.533 – 11.7 dB = 84.78 dBW = 3.14 nano watts (nW) Signal voltage = [377 ´ 3.14 ´ 10–9]1/2 V/m = 1.854 ´ 10–4 V/m Therefore, now if the antenna is in mobile and van has a nominal gain of (say) 1.5 dB over isotropic aerial and is loss-less. It can deliver a signal voltage of = (1.854 ´ 10–4 ´ 3.16) V = 0.586 mV/m 23. A TV transmitting antenna mounted at a height of 100 m radiates 10 kW of power at a wavelength of 5 m. Power is radiated equally in all directions in azimuth. Calculate (a) The field strength at a receiving antenna at height of 9 m at a distance of 10 km. (b) The distance at which the field strength reduces to 2 mV/m. (c) Maximum line of sight range. Sol.

Distance to horizon D = 3550 ⎡⎣ ht + hr ⎤⎦ m = 3550 ⎡⎣ 100 + 9 ⎤⎦ = 46150 m

At a distance of 10 km, the field strength will be Ez =

88 P hd 2

hz hr =

88 10 4 5 × (10 4 )2

9 × 100 = 15.84 mV/m


705

If the field reduces to 2 mV/m at distance d metres: 2 ´ 10–3 = 88(104)1/2 ´ 9 ´ 102/(5 ´ d2) d2 = 88 ´ 102 ´ 9 ´ 102/10–2 = 802 ´ 106 d = 28.32 ´ 102 m = 28.32 km 24. A geostationary satellite is located at a distance of 3000 km with an operating frequency 14.25 GHz. The gain of Tx and Rx antennas are 15 and 20 simultaneously. If the Tx power is 200 kW, find the power received by the Rx antenna. Sol.

Given that Gt = 15, Gr = 20, PT = 20 kW and f = 14.25 GHz

where, l = c/f = 3 ´ 1010/14.25 ´ 109 = 2.1 cm Hence, Pr = PtGTGR(l/4pR)2 = 200 ´ 103 ´ 15 ´ 20 ´ 0.021/(4 ´ 3.14 ´ 3 ´ 106)2 = 6 ´ 107 ´ 10–12 ´ 0.021/1419.783 = 8.87 ´ 10–10 W 25. A satellite earth station antenna having a maximum gain of 80 dB at the operating frequency of 14 GHz, and it generates 10 kW at the O/P. If the feeder loss is 4 dB, determine the EIRP. Sol.

Power at the output of amplifier = 10 kW = 40 dBW Antenna gain = 80 dBW Feeder loss = 4 dBW

So, EIRP = (O/P power) + (Antenna gain) – (Feeder loss) = 80 + 40 – 4 = 116 dBW 26. Determine the power received by a satellite located at 40 × 103 km from the surface of earth. The satellite is operating at a frequency of 11 GHz with EIRP of 21 dBW. The gain of the Rx antenna is 50.5 dB. Sol.

Power received is given by Pr = ERIP + Gr – Pathloss

Given that

EIRP = 21 dB, Gr = 50.5 dB and h = 4 ´ 103 km

2 ⎛ 4 × 3.4 × 4 × 10 7 ⎞ ⎛ 4Q h ⎞ Path loss is given by Pl = ⎜ = ⎜ ⎟ ⎟ ⎜ 2.272 × 10 −2 ⎟ ⎝ M ⎠ ⎝ ⎠

2

= 50.24/2.727 = 18 ´ 109 Pl (dB) = 20 [log(18 ´ 109] = 20 [1.265 + 9] = + 205.30 dB Pr = 21.0 + 50.5 – 205.3 = – 133.8 dBW

706


27. If a satellite is operating at a distance of 2 ´ 104 km from the surface of earth and radiates a power of 4 W. If the gain of Tx antenna is 20 dB, calculate the flux density at the receiving point. Sol.

Flux density at the receiving point is given by

Gd =

Pt Gt 4Q d

2

=

4 × 20 dB (100) 4 × 3.14 × (2 × 10 7 )2

= 8.0 ´ 10–14 W/m2 28. An array consists of 10 vertical elements in a horizontal straight plane. Calculate the angular width of the forward beam in the same plane, if the elements are spaced apart and equally energized in phase. Sol. Since the elements are energized by in phase currents, the array will act as a broadside array and shall have a beam width as given by Beam width = 2q1 = 2l/nd rad = 2l/10 ´ 0.5l = 23° 29. A small dipole antenna is carrying a uniform RMS current of 10 A. Its far zone RMS electric field at distance of r metres in a direction making angle (q) with the conductor is given by E = 200/r sin q V/m. Find the total power radiated and radiation resistance. Sol.

Radiation field of a small dipole is ⎛ 60 Q Erad = ⎜ ⎝ Mr

r ⎞ 200 ⎞ ⎛ sin R ⎟ I max dl sin R cos X ⎜ t − ⎟ = c⎠ r ⎠ ⎝ r⎞ 10 ⎛1⎞ ⎛ ⎜ ⎟ I max dl cos X ⎜ t − ⎟ = c ⎠ 3Q ⎝M⎠ ⎝

Þ Now the power density

⎛ E2 Pr = EH = ⎜ ⎜ 120 Q ⎝

Þ

⎞ ⎟ ⎟ ⎠

⎡⎛ 60 Q ⎞ r ⎞⎤ ⎛ ⎢⎜ ⎟ I max dl sin R cos X ⎜ t − ⎟⎥ C ⎠⎦ ⎝ ⎣⎝ M r ⎠

2

120 Q 2

10 2 ⎡ 60 Q 10 ⎤ R = sin = sin 2R ⎢ ⎥ 2 120 Q ⎣ r 3 Q Q 3 r ⎦ 1


707

Total radiated power PT = Pav dA = Pav 2pr2 sin q dq

PT =

Q

∫R

=0

10 3 3Q r

2

sin 2R 2Q r 2 sin R dR =

Q

∫R

=0

2000 3

sin 2R dR

After simplification PT = 8000/3 = 888.8 W 30. A vertical wire of 2 m long carries a current of 6 A at a frequency of 1 MHz. What will be the radiated field at 30 km in direction q = 90° to the axis of the wire in free space? Sol. ⎛ 60 Q E rad = ⎜ ⎝ Mr

r⎞ ⎞ ⎛ ⎟ I dl sin R cos X ⎜ t − ⎟ c⎠ ⎠ ⎝

r⎞ ⎛ 60 Q ⎞ ⎛ 2 ⎞ ⎛ =⎜ ⎟ ⎜ ⎟ 6 cos X ⎜ t − ⎟ c⎠ ⎝ 3000 ⎠ ⎝ 300 ⎠ ⎝

= 25.13 ´ 10–5, in which f = w (t – r/c) Erad = 251.34 mV/m Maximum radiation electric field intensity = 0.2513 mV/m Therefore, maximum radiation magnetic field intensity H = E/120p = 0.66 mA/m 31. A signal of RMS amplitude of 10 V is applied to the input terminals as of a half-wave dipole. Calculate the radiated power, if the dipole has input impedance of Zn = 73 + j42 W. Sol.

As PT = I2RMS Rv I RMS =

VRMS |Z in |

=

10 732 + 422

= 0.12 A

Hence, PT = (0.12)2 ´ 73 = 1.05 W 32. Parabolic reflectors are used for both Rx and Tx antennas of a microwave circuits are replaced by others having a diameter 1.5 times that of the original. Calculate the overall gain in dB and improvement expected from this modification. Sol. We know that Gp = 6(D/l)2 Let Gp1 and D1 be the gain and diameter of the aerial in the original system. ⎛D ⎞ G p1 = 6 ⎜ 1 ⎟ ⎝ M ⎠

2

708


If the gain of the system with new system is Gp2 and its diameters is D2, then the gain per antenna in dB will be Gain (dB) = 10 log10 (Gp2/Gp1)2 = 10 log10 (1.5 D1/D1)2 = 10 log10 (2.25) = 3.522 dB 33. The radiation intensity pattern of a point source is defined by P = Pm cos2q and has a value only in the upper hemisphere. The space pattern is a figure of revolution of this pattern around the polar (q = 0) axis. Find the directivity Sol.

We know that the total power PT =

∫

2Q

∫

0

PT = Pm

Q /2 0

2Q

∫

∫

0

∫

= 2Q Pm

Q /2 0

P sin R dR dG Q /2 0

cos2R sin R dR dG

cos2R sin R dR = 2 Q Pm ×

1 3

=

2 Q Pm 3

If P is the same as radiated by an isotropic source, then

PT = 4Q P0 ⇒ 2 Pm

The ratio of

P0

Q 3

Pm = 4Q P0

= Directivity = 6 = 7.78 dB

34. (a) Find the directivity of a broad side of array of height h = 12l and length l = 20l. Sol.

We know that the directivity (D) of broadside away is given by D = 12.6

A

M

2

= 12.6 ×

12 M × 20 M

M2

= 3024 = 34.8 dB

(b) A broadband array (shown in the figure) is fed by a twin line, which joins the two dipoles such that the pattern in the horizontal (x–y) plane is that of a single dipole of impedance (Z1 = 73 + j43) W and mutual impedance due to other dipole is (Z2 = –13 – j29) W. Find VSWR and gain of the broadside array of l/2-dipole.


Sol.

709

The total impedance is ZT = (73 + j43) + (–13 – j29) = (60 + j14) W

Turning out the reactance with a series capacitor, the 60 W is transformed by a (l/4) of 200 W twin-line to an impedance Z = 200 (200/60) = 667 W which appears at the junction half between the dipoles. The junction impedance with the lines from both dipoles connected in parallel is 667/2 = 333 W, which is not equal to 280 W. That is, no proper impedance matching and loss of power occurs. The reflection coefficient

Sv =

Z L − Z0 Z L + Z0

=

333 − 280 333 + 280

=

53 613

= 0.086

⎛ 1 + Sv ⎞ ⎛ 1 + 0.086 ⎞ VSWR = ⎜ ⎟=⎜ ⎟ = 1.188 ⎝ 1 − Sv ⎠ ⎝ 1 − 0.086 ⎠

The field gain over l/2 dipole is 2 × 73 (73 − 13)

= 1.56 or 3.9 dB

35. Find the radiation resistance and radiation efficiency of a single turn small circular loop antenna, if the radius of the loop is (l/25) in free space. Sol.

The surface area of the antenna is 2

QM ⎛M ⎞ S = Qa = Q ⎜ ⎟ = 625 ⎝ 25 ⎠ 2

2

710


The radiation resistance 2

2 ⎛ 2 Q QM 2 ⎞ 2 × 3.14 ⎛ 2Q S ⎞ I⎜ 2 ⎟ = Pr = × 377 ⎜ 2 ⎟ = 0.788 : ⎜ M 625 ⎟ 3 3 ⎝ M ⎠ ⎝ ⎠

2Q

The loss resistance for a single term is given by Rl =

a

w N0

b

2T

=

1

2 Q × 108 × (4Q × 10 −7 )

25 × (10 −4 )

2 × 5.7 × 10 7

= 1.053 :

Therefore, radiation efficiency is given by

Ir =

Rr

=

Rr − Rl

0.788 0.788 + 1.053

= 0.43 = 43%

36. (a) The maximum distance d between the elements in a linear scanning array to suppress grating lob is d max =

M 1 + |cos R |

where q is the direction of the pattern maximum. Find the maximum distance between the elements without introducing grating lobes at f = 3.5 GHz, when the array is designed to scan to maximum angle 30° and 70°.

M

. Sol. The considered spacing between the elements is given by d = 1 + cos R 0 Hence, when q = 30°

d30° =

M 1 + cos R

=

M 1 + cos 30

= 0.53 M =

0.53 × 3 × 1010 3.5 × 10 9

= 4.5 cm

and at q = 70°.

d 70° =

M 1 + cos R

=

M 1 + 0.342

= 0.745 M =

0.745 × 3 × 1010 3.5 × 10 9

= 6.387 cm

(b) Describe the procedure of flow of current in a receiving antenna. Sol. On the receiving side, the electric and magnetic fields of receiving wave, strike on antenna and induces a voltage in it. The induced voltage causes a current flow in the receiver apparatus, including transmission structure. Both the transmitting and receiving antennas should have some polarization for matched transmission and reception and it is often called the optimum condition.


711

37. If for a certain input power, the reference and test antennas provide 25 MW and 500 MW power respectively in the direction of maximum power. What is power gain at the test antenna? ⎡ Power to reference antenna ⎤ P(dB) = 10 log10 ⎢ ⎥ ⎣ Power to test antenna ⎦

Sol.

= 10 log (500/125) = 6.021 38. (a) If an antenna radiates 12.5 W of power for an i/p current of 400 mA. What will be its radiation resistance? Sol.

We know that Rr =

2P I

2

=2 ×

12.5 0.42

= 156.25 :

(b) A pyramidal horn antenna has an aperture area of 0.12 ´ 0.06 m2 and operating frequency is 10 GHz. Calculate the beam width and power gain. Sol.

The operating wavelength

M=

c f

=

3 × 10 3 1010

= 0.03 m

Therefore

WB =

Gp =

80 80 80 = = = 20° 4 ⎛ a ⎞ ⎛ 0.12 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ M ⎠ ⎝ 0.03 ⎠ 4.5 × a × b

M

2

=

4.5 × 0.12 × 0.06 0.03 × 0.03

= 36 = 15.563 dB

39. Obtain the half-power (HP) beam width of a 0.24-m parabolic dish antenna operating at 5 GHz, and also calculate the power gain as well Sol.

M=

3 × 108 5 × 10 9

WB = 58 M /D = 2

The power gain

= 0.06 m 58 × 0.06 0.24 2

= 14.5

⎡D⎤ ⎡ 0.24 ⎤ Gp = 6 × ⎢ ⎥ = 6 ⎢ ⎥ = 96 ⎣M ⎦ ⎣ 0.06 ⎦

712


40. For Tx and Rx antennas, the heights are 200 ft and 50 ft respectively, what would be the largest transmission distance in miles? Dmiles = 2 × 200 + 2 × 50 = 30 miles

Sol.

41. What do you mean by stacking in Yagi–Uda antenna? What are its advantages? Sol. In order to further improve directivity of Yagi antenna, stacking is required. Because it improves directivity in the respective plane of stacking, and provide a power gain of 3 dB with each section of Yagi array added. The signal from each Yagi section must be phased correctly before combining them together. Basically, there are two types of stacking: vertical stacking and horizontal stacking. In vertical stacking, Yagi array is placed at l/2 apart vertically, because spacing of l/2 is convenient for inter connecting the array with two l/4 stubs for the impedance matching. Vertical stacking offers greater directivity which is very useful in the reduction of noise and interference, pick up from low angle sources: Ignition from vehicles. On the other hand, in horizontal stacking, Yagi array is placed in horizontal plane one beside the other with minimum spacing between centres l/2. This structure provides a sharper beam in the stacking plane with a gain of 1.8 dB over each section. However, optimum spacing between the centres is about (1 to 5/4l) offers a gain of up to 3 dB. The 3l/4 stub is generally used to combine each Yagi section as well as impedance matching with the feed. 42. Determine the length of a monopole whose input resistance is 50 W, at operating wavelength l = 2.5 m. Sol.

We know that formula for rapid calculation give Rin (monopole) = 12.35 G2.5,

where G = kl for monopole, and = kl/2 for dipole Þ

50 = 12.35 G2.5 log 50 = log (12.35) + 2.5 log G

Þ

1.698 = 1.0916 +2.5 log

Hence,

or

log G = 0.243

G = 1.75 = kl Þ 2p l/l = 1.75 Þ l = 1.75 ´

2.5 6.28

= 0.696 m

43. Determine the maximum input resistance of a dipole of length 0.56l. Sol.

Given

l = 0.56l Rin (dipole) = 11.14 G4.17 ⎛ kl ⎞ = 11.14 × ⎜ ⎟ ⎝2⎠

4.17

⎛ 2M × 0.56 M ⎞ = 11.14 × ⎜ ⎟ ⎝ M × 2 ⎠

4.17


713

Log (Rin) = log 11.14 + 4.17 log 1.7584 = 1.046 + 1.022 = 2.0688 Hence, Rin = 117.188 44. What should be minimum and maximum values of input resistance of a dipole to maintain the VSWR = 2.5, in case dipole is centrally fed via 50 Tx line.

Sol.

⎡ Z in ⎤ ⎢ Z − 1⎥ VSWR − 1 Z in − Z 0 2.5 − 1 ⎥ = = = ⎢ 0 Reflection coefficient [* ] = ⎢ Z in ⎥ VSWR + 1 Z in + Z 0 2.5 + 1 ⎢ Z +1⎥ ⎣ 0 ⎦ 1.5 3 ( Z − 1) Z − 1 3 ⇒ = = r = r =+ Zr + 1 3.5 7 ( Zr + 1) 7

10

Zr =

4

⇒ [(Z )r ]max =

10 4

× 50 = 125 :

3 ⎛ Zr − 1 ⎞ ⇒ 7 Zr − 7 = − 3Z r − 3 ⇒ 10 Z r = 4 and if ⎜ ⎟= − 7 ⎝ Zr + 1 ⎠

(Zr)min = (4/10) ´ 50 = 20 W. 45. A half-wave dipole is radiating into free space such that the origin is at the centre of the dipole and the Z-axis is aligned with the dipole. The dipole is fed with an input power of 100 W. Assuming an overall efficiency of 50%, find the power density at r = 500 m, q = 60° and f = 0°. Sol.

We know that the average power density Wav = I

and power radiated Prad =

1

⎡ cos2 (0.5Q cosR ) ⎤ ⎢ ⎥ 80 Q 2 r 2 ⎣⎢ sin 2 R ⎦⎥ I 02

Rrad I 02

2

We know that Rrad = 73.05 Prad = I 02 =

73.05 × I 02 2 100 73.05

⇒

= 1.3688

50 × 100 100

=

73.05 × I 02 2

714


Hence,

⎡ cos2 (Q /2 cos 60°) ⎤ Wav = 120 Q × ⎢ ⎥ 8 × (Q )2 × 500 ⎢⎣ sin 2 60 0 ⎥⎦ 1.3688

=

20.5332 25 × 3.14 × 10 4

(0.667) = 1.74 × 10 −5 W/m 2

46. In a satellite communication radiated power is 10 W. The characteristics of communication are as follows: (i) Operating frequency f = 100 W (ii) Distance between Tx and Rx = 3.7 ´ 108 m (iii) Tx antenna directivity = 60 dB Determine the magnitude of E and voltage at Rx antenna, if a dipole is used as a receiving antenna with 90% efficiency. Sol.

Prad = 10 W, and r = 3.7 ´ 107 m, D0 = 106 = 60 dB

We know that 2 ⎛ r 2 Emax U max = ⎜ ⎜ 2I ⎝

⎞ ⎟ ⎟ ⎠

Hence, D0 =

4Q × U max Pr

=

2 4 Q ⎛ r 2 Emax ⎜ Pr ⎜⎝ 2I

⎞ ⎟ ⎟ ⎠

⎛ 120Q × 10 ⎞ 2 Emax = D0 × 2 × ⎜ = 4.4 × 10 −7 2 ⎟ Q 4 × r ⎝ ⎠ E = 0.66 ´ 10–4 V/m = 0.066 mV/m From the Frill transmission equation 2

⎛ M ⎞ =⎜ ⎟ GU Gr Pt ⎝ 4Q r ⎠

Pr

Hence 0.03 ⎡ ⎤ Pr = ⎢ 7 ⎥ ⎣ 4Q × (3.7 × 10 ) ⎦

= 6.84 ´ 1014 W

2

× 10 6 × 1.643 × 10


Rr =

V2 8 Rin

715

⇒ V = [8 × Pr × Rin ]1/2

= [8 × 6.84 × 10 −14 × 73.05]1/2 V = 6.34 ´ 10–7 V = 0.634 mV 47. Design a circular loop of constant current such that its field intensity vanishes at angles q = 0° and 90°. And also find its radiation resistance and directivity at operating wavelength l = 2.2 m. We know that

Sol.

Ef = J1(ka sin q) Ef = J1(ka sin q = 0°) = J1(0) = 0 Ef = J1(ka sin q = 90°) = 0

Þ

J1(ka) = 0 ka = 3.84

only if

Radius a = (3.84 ´ l)/2p = 0.61152l = 1.34 m a > 0.5l

So, as

Rr = 60 p2(C/l) = 60p2(2 ´ 3.14 ´ 0.6115 = 2.273 ´ 103 W D = 0.682 (C/l) = 0.682 (2 ´ 3.14 ´ 0.6115) = 2.62 48. Draw the equivalent circuit of a non-resonance small loop antenna in (a) transmitting mode and (b) receiving mode Sol. (a) Tx Mode in which Zin = Rin + jXin = (Rr + RL) + j(XA + Xi) RL = Loss resistance of loop conductance

=

a b

XN0 2I 1

2

Zg RL

Xc Rr

Zin XA

Vg

1

2

716


XA = Inductive reactance a of loop antenna = wLA.

⎡ ⎛ 8a ⎞ ⎤ where LA is inductance = N0 a ⎢ ln ⎜ ⎟ − 2 ⎥ ⎣ ⎝ b ⎠ ⎦ for circular loop of radius a and wire radius b and L A = 2N 0

⎤ a ⎡ ⎛a⎞ ⎢ ln ⎜ ⎟ − 0.774 ⎥ Q ⎣ ⎝b⎠ ⎦

for square loop of side a and wire radius b and Xi = reactance a of loop conductor = wLi with Li = m0(s/2a) where s is the area of the loop. (b) Rx mode When a small loop antenna is used in Rx mode an open circuit voltage is generated across, which can be represented in terms of inductive magnetic field density Bi as follows: Vop = jwpa2B1 Zin

VL

ZL

Vop

When load impedance (ZL) is connected to the output terminals of the loop, the voltage VL develops across the load impedance Z is related as follows: VL = Vop

ZL Z in + Z L

49. A small circular loop is placed at height (l) above an infinite flat perfectly electric conducting ground plane. Find the angle where array factor will be zero. Sol. => =>

Array factor (AF) = 2 cos (bd cos q) = 0

bl cos q = ± np/2, where n = 1, 3, 5, …. (2p/l)l cos q1 = ±p/2 (n = 1)


cos q1 = ±1/4 Similarly,

and

717

q1 = cos–1(±1/4) or q1 = 75.5° => (2p/l) l cos q3 = ±3p/2 (n = 3) cos q3 = (±3/4) or q3 = 41.4° q5 = cos–1(±5/4) = does not exist. or

50. Obtain the current flowing in an elliptical loop (a = 2l, b = 1.2l), when it is radiating a power of 100 W. Assume antenna efficiency 80%. Sol.

The radiation resistance (Rr) = 31.171(p2a2b2/l4) = 31.171 ´ (3.14)2 ´ 22 ´ 1.22 = 17.7 ´ 105 W

Pr =

1 2

Rr |I 0 |2

100 ´ 80% = 17.7 ´ 105 ´ |I0|2 |I 0 |2 =

80 × 2 17.7 × 10 5

= 9.0 × 10 −5 = 9.5 mA

51. Find the inductive reactance of a small loop antenna operating at f = 100 MHz, if it is (a) circular with a = 3l and b = 1.2l, (b) square with a = 3l and b = 0.5l. Sol.

The inductive reactance of a loop antenna (XA) = wLA

(a) where LA = m0a [ln (8a/b) – 2] = 4 ´ 3.14 ´ 10–7 ´ 9 ´ [ln (8 ´ 3/1.2) – 2] = 11.25 mH Hence, XA = 2pf ´ LA = 2 ´ 3.14 ´ 100 ´ 106 ´ 11.25 ´ 10–6 = 7065 W (b) LA = (2m0a/p) [ln (a/b) – 0.774] = 2 ´ 4p ´ 10–7 ´ (9/p) [ln(3/0.5) – 0.774] = 73.28 ´ 10–7 = 7.33 mH Hence, (XA) = wLA = 2pf ´ LA = 2 ´ 3.14 ´ 100 ´ 106 ´ 7.33 ´ 10–6 = 4603.24 W And reactance of loop conductor, Li = (1/2) ´ m0 ´ pa = 1/2 ´ 4 ´ 3.14 ´ 10–7 ´ 3.14 ´ 3 ´ 3 = 177.47 ´ 10–7 = 17.75 mH. 52. Design a 4-element Yagi–Uda antenna to be operated at l = 0.6 m. Sol.

As l = 6, f = 3 ´ 108/6 = 50 MHz.

Therefore, length of reflector (le) = (500/fMHz) ft = 500/50 = 10 ft. Hence length of driven element, first and second directors = 9.5 ft, 9.0 ft, 8.5 ft and 8 ft.

718


53. Derive the expression for power flow in uniform plane wave. Sol.

In time varying electromagnetic field E = E 0 e j wt

and

H = H0 e jwt

or

H=

E0

I

e jX t

The average power density of the wave is

Pav = (1/2) Re (E × H*) =

1 2

Re[E0 e jXt × (E0 /I)e jXt aˆ x aˆ y ]

G = (1/2) Re [E j 0 /I*] az = 1/2 E 2j 0 Re[1/I*] So, average power in terms of electric field intensity Pav = 1/2 E02 Re[1/I*] Similarly, average power in terms of magnetic field intensity is Pav = 1/2 H02 Re[1/I*] . 54. If for some input power the reference and test antenna provide 125 mW and 600 mW power respectively in the direction of maximum power, what would the power gain of test antenna? Sol.

Power gain (dB) = 10 log10 (power to ref. antenna/power to test antenna) = (600/125) = 6.812 dB

Power gain of a horizon dipole is = 3/2 = 1.5. 55. An antenna radiates 12 W of power for an input current of 400 mA. What is it radiation resistance? Sol.

Pr = 2 ´ 12/(400 ´ 10–3)2 =150 W

56. Obtain the beam width (HP) of a 0.24 m parabolic disk operating at 5 GHz. Also, calculate the power gain as well. Sol.

19.8 dB

57. A helix is required to operate at 5 GHz and has the following parameters: S = 0.05 m, D = 0.1 m, l = 0.06 m and n = 15. Find HPBW and maximum power gain Sol.

2.8°, 37.10 dB

58. If the power radiated by an antenna of length 1 cm is 1 W and the radiation wavelength is 15 cm. Calculate current generated in the dipole? Hint: I 0 =

P M 40 Q


719

59. A rectangular horn of dimension 1 m ´ 0.5 m is used to transmit waves, what would be its power gain and beam width if the frequency of wave is 300 MHz. Sol.

2.25 NV, 80°

60. For Tx and Rx antennas at the height of 200 ft and 50 ft respectively, what would be the greatest propagation range in miles. Sol.

30 miles

61. Calculate the electric field at a distance of 1 km from a source of power 10 W. Sol.

17.1 mV/m

62. What is the quality factor of an antenna operating at 20 MHz for a band width of 5 MHz? Sol.

Q=

fr BW

=

20 5

=4

63. If the maximum number of electrons per cubic metre in a strip of an ionization layer is 1011. What would be the highest frequency that would be reflected back from the layer? Sol.

2.85 MHz

64. A half-wave dipole is radiating at 300 MHz, what should be its length for velocity factor 0.87? Sol.

Electrical length (l) =

1 2

×

300 300

= 0.50 m

Hence, its physical length = uf ´ le = 0.50 ´ 0.87 = 0.435 m. 65. An antenna is supposed to receive a LOS signal transmitted form antenna located at a distance of 50 miles from the installation of height 800 ft. What should be the necessary height of the Rx antenna? Sol.

50 ft.

66. The MUF for critical frequency 10 MHz and angle of incidence is 60° will be Sol.

20 MHz.

67. If the diameter of parabolic reflector is doubled, its gain will increase by ⎛D⎞ Gp1 = 6 ⎜ ⎟ ⎝M⎠

2

2

⎛ 2D ⎞ and Gp2 = 6 ⎜ ⎟ = 4Gp1 ⎝ M ⎠

DGp = 30 = 10 log 3 = 4.77 dB Gp2 = 6.020 dB

720


68. What is the highest frequency for which antennas can be made used? Sol.

1012 to 1013 Hz (approximately)

69. What is the condition of the electrons in the transmitting antenna when maximum magnetic field is being transmitted? Sol. Since maximum current is flowing, the electrons have maximum speed up and down the antenna. This is because the electric current produces the magnetic field. 70. How can the electric portion of the electro-magnetic wave be detected? Sol. The electro-magnetic wave can be detected by an antenna similar to the transmitting antenna except that a detector of electric current replaces the voltage source. 71. How can the magnetic field portion of an electro-magnetic wave be detected? Sol. Magnetic waves are best detected by placing a loop of wire (with its ends hooked to a current detector) in path of wave. Changing magnetic field causes current in the loop. 72. How do we make television broadcasts for larger coverage and for long distance? Sol. By using (i) tall antennas which are familiar landmarks in many cities, and (ii) using Artificial satellites—called geostationary satellites. Since television signals are of high frequency and are not reflected by ionosphere, so we use satellites to get them reflected and transmission of TV signals can be used for larger coverage as well as for long distances. 73. Scientists put X-ray astronomical telescope on the artificial satellite orbiting above the earth’s atmosphere whereas they build optical and radio telescopes on the surface of the earth. Why? Sol. X-rays have very high frequency, i.e., smaller wavelength. These rays get absorbed by the earth’s atmosphere whereas optical (visible) radiations can pass through the atmosphere. That is why optical and radio telescopes are installed on the earth’s surface. 74. For an electromagnetic wave, what is the relation between amplitudes of electric and magnetic fields in free space. Sol.

c=

E0 B0

where c is the speed of light in free space. 75. If the charging current for a capacitor is 0.5 A, then what will be the displacement current across its plates? Sol.

Displacement current = Charging current = 0.5 A.


721

76. What is a ground wave? How does it differ from a sky wave? Sol. A signal emitted by an antenna from a certain point can be received at another point of the surface of the earth in two ways. The wave which travels directly following the surface of the earth is called ground wave. The wave that can reach the same point after being reflected from the ionosphere is called sky wave. 77. Why short wave communication over long distances is not possible via ground waves? Sol.

Because the waves got attenuated.

78. Are conduction and displacement currents the same? Sol.

No. They are not same, but they are equal.

79. It is necessary to use satellites for long distance TV transmission but why? Sol. It is so because television signals are not properly reflected by the ionosphere. Therefore, for reflection of signals satellites are needed as reflection is effected by satellites. 80. What is the contribution of the Greenhouse effect towards the surface temperature of the earth? Sol. The infrared radiation emitted by the earth’s surface keeps the earth warm. In the absence of this effect, the surface temperature of earth would be lower. 81. In a communication, if the Tx isotropic antenna transmits power of 10 W through horizontally polarized waves. And the ground has er = 10 and s = 10–3 S/m. Find the net electric field and power intensity at Rx antenna. Assume effective aperture of the antenna is 0.5 m2 and reflection coefficient = 0.512. Sol.

We know that Er =

PI0 ⎡ e− j C R1 e − j C R2 ⎤ +* ⎢ ⎥ = 4 Q ⎣⎢ R1 R2 ⎥⎦

10 × 120 Q ⎡ e − j C R1 e − j C R2 ⎤ + 0.512 ⎢ ⎥ 4Q R2 ⎥⎦ ⎣⎢ R1

Assuming R1 » R2 » 2000 km Er =

300 ⎡ e− j C R1 e − j C R2 ⎤ + 0.512 ⎢ ⎥ = 0.012 V/m = 12 mV/m 2000 ⎣⎢ 1 1 ⎥⎦

The received power density at the Rx antenna is Pr =

|Er |2

I0

=

|12 × 10 −2 |2 377

= 3.88 × 10 −7 W/m 2

722


Therefore the power received by the antenna will be Wr = Pr ´ Ae = 3.88 ´ 10–7 ´ 0.5 = 19.40 ´ 10–3 W 82. Two towers of height 120 m and 80 m, located at a distance of 5 km. What will be the distance of reflection point for ground wave, if the signal frequency is 5 GHz. Also estimate the phase difference between the direct and reflected waves at the receiver due to propagation delay. Sol.

Let us refer to the following figure. R1

h2

h1

R2

x

d

x h2

M=

=

d − x h1

3 × 108 5 × 10

9

h2 d h1 + h2

=

= 0.06 m and C =

G = C (R2 − R1 ) = =

⇒ x=

0.08 × 5 0.120 + 0.080

2Q

M

=

2 × 3.14 0.06

100 Q ⎡ (h + h2 )2 + d 2 − ⎢⎣ 1 3

100 Q ⎡ (120 + 80)2 + 50002 − ⎢ ⎣ 3

= 5003.998 − 5000.159 =

= 2 km

=

100 Q 3

(h1 − h2 )2 + d 2 ⎤ ⎥⎦

(120 − 80)2 + 50002 ⎤ ⎦⎥

3.3838 × 100 Q 3

= 40.17 rad

83. Evaluate the directivity of an antenna, if the radiation intensity is defined as

⎧⎪2 sin 2R . sin 2G , U (R , G ) = ⎨ ⎪⎩0, Sol.

Ans. 6

0 ≤ R ≤ Q, 0 ≤ G ≤ Q elsewhere


723

84. For a 10 turns loop antenna of radius 15 cm operating at 100 MHz, calculate Ae at q = 30° and f = 90°. Sol.

0.2686

85. A large circular parabolic dish antenna has the directivity of 30 dB. Find approximate HPBW of the antenna and its effective aperture at 5 GHz. Sol. Since, parabolic is circular, so if HPBW is qHP, then solid beam angle will be »

Q

4

2 . R HP

4Q

D = 30 = 10 3 =

Ae =

2

Q /4 R HP

G(~D)M 2

= 1000

4Q

4

⇒ R =

0.062 4 × 3.14

1000

= 0.1265 rad

= 0.287 m 2

86. A vertical monopole V antenna of length 40 cm is used to broadcast at AM radio station at 160 m wavelength. Estimate base current and total power radiated by an antenna, if maximum electric field intensity measured at a distance of 40 km is 30 mV/m. Sol.

Length of antenna Operating frequency

=

40 160

=

1 4

i.e., radiator is l/4 – monopole antenna, hence electric field intensity will be

⎛Q ⎞ 60 I m cos ⎜ cos R ⎟ 2 ⎝ ⎠ 60 I m = ER = r sin R r 3 × 10 − 3 =

60 I m 40 × 10

3

⇒ Im =

30 × 40 60

= 20 A

For quarter wavelength monopole the base current is same as Im, i.e., base current is 20 A. And since radiation resistance for monopole is 36.5 W. P=

1 2

I m2 Rin =

1 2(20)2.365

= 7.3 kW

87. A large uniform beam scanning array provides maximum half-wave power beam width of 15°, how many elements in the array are needed, if maximum spacing between

724


elements is l/2. Also estimate maximum variation in the HPBW, during the scanning if the beam is scanned over the full range. Assume both the cases. Sol.

Broad-side array

GHP =

M dN

dN =

GHP =

1

=

L (length of array)

180 × M 15 × 3.14 15Q 180

⇒ 15° =

=

⇒ N=

180 × 2 15 × 3.14

2M

15 Q 180

=

M dN

= 7.6 ≈ 8 nos.

⎛ 180 ⎞ ⇒ dN = 2M ⎜ ⎟ dN ⎝ 15 Q ⎠

2

2

⎛ 180 ⎞ N = 2 × 2⎜ ⎟ = 58 nos. ⎝ 15 × 3.14 ⎠

(i)

GHP =

(ii)

GHP =

M dN

=

2M dN

2

M ⎛ 180 ⎞ d⎜ ⎟ ⎝ 15 × Q ⎠

=

2 × 2

M × 58

=

1 ⎛ 15 Q ⎞ ⎜ ⎟ = 0.0342 rad = 1.94° 2 ⎝ 180 ⎠

= 0.26 rad = 15.04°

Therefore, the variation of HPBW or DHPBW while scanning are 0.04° and 13.04° in broadside and end-fire array respectively. 88. A 5-element array has either inter-element spacing of 0.3l and progressive phase shift of 40°. Find the radiation pattern and direction of maximum radiation of array. Sol.

Z

=

2Q d

M

cos G + 40° = 0.6Q cos G +

40Q

2Q ⎞ ⎛ = ⎜ 0.6 Q cos G + ⎟ rad 180° ⎝ 9 ⎠

Radiation pattern is

E=

⎛ NZ ⎞ sin ⎜ ⎟ ⎝ 2 ⎠

⎛ NZ ⎞ N sin ⎜ ⎟ ⎝ 2 ⎠

=

⎛ 5Z ⎞ sin ⎜ ⎟ ⎝ 2 ⎠

⎛Z ⎞ 5 sin ⎜ ⎟ ⎝2⎠

=

5Q ⎞ ⎛ sin ⎜ 1.5 Q cos G + ⎟ 9 ⎠ ⎝

Q⎞ ⎛ 5 sin ⎜ 0.3 cos G + ⎟ 9⎠ ⎝


725

For the direction of maximum radiation, y = 0, giving 2Q ⎞ 2Q 1 ⎛ ⎛ 10 ⎞ × = ⎜− ⎜ 0.6 Q cos G + ⎟ = 0 ⇒ cos Gm = − ⎟ 9 ⎠ 9 0.6 ⎝ 27 ⎠ ⎝ ⎛ 11 ⎞ Gm = cos−1 ⎜ − ⎟ = 111.74° ⎝ 27 ⎠

89. Design a broadside array of 5 elements which has main beam to side lobe ratio of 10 and smallest possible beam width. Assume inter-element spacing is l. Sol. This is the case of Cheveshev array. The number of elements 2N+1 = 5 that N = 2. Therefore, radiation pattern may be given by T2(x): Main beam to side lobe ratio, R = 10, Hence, ⎡1 ⎤ x1 = cos ⎢ cos−1 (10) ⎥ = 2.3452 2 ⎣ ⎦

Since,

d=

M 2

, Cd =

2Q

M

×

M 2

= 2 ⇒ cos C d = − 1

Since, a=

b=

1 + x1 cos C d 1 − cos C d 1 + x1 1 − cos C d

=

=

1 − x1 2

1 + 2.345 2

=

1 − 2.345 2

= − 0.6726

= 1.6726

Therefore, T2(x) = 2x2 – 1 = 2(a + b cos y)2 – 1 = 2a2 + b2 – 1 + 4ab cos2y + b2 cos 2y which is equal to = a0 + 2a1 cos y + 2a2 cos y Comparing both the side gives a = 2a2 + b2 – 1 = 2(– 0.6726)2 + (1.6726)2 – 1 = 2.7024 a1 = 2ab = 2 ´ – 0.6726 ´ 1.6726 = –2.25 a2 =

b2 2

=

(1.6726)2 2

= 1.3988

Therefore, the individual elements of the array are excited with current amplitudes 1.39 : –2.25 : 2.70 : –2.25 : 1.39

726


90. A helix is required to be operated at 5 GHz and it has the following specifications: S = 0.05 m, D = 0.1 m, n = 15 and l = 0.06 m. Determine the Gp. Sol.

We know that

Gp max =

=

15nS (Q D)2

M3 15 × 15 × 05(3.14 × 0.1)2 0.063

= 5130 = 37.10 dB.

Index

27-day Sunspot cycle, 570

Absolute gain method, 612 Absorption fading, 573 ratio (g), 83 Adaptive antenna, 145 array, 142 Amplitude distribution, 135 Anechoic chamber, 591 Angle of incidence, 555 Antenna(s), 1 coupling, 87 gain, 73 measurement, 587 noise temperature, 84 polarization, 90 resolution, 75 Aperture antenna, 58 coupled feed, 449 distribution phase error, 391 -medium coupling loss, 538 number, 311 AR, 4 Array, 114 factor, 124 Artificial dielectric, 283 Astronomical horizon, 515 Attenuation constant, 22 factor, 574

Axial helix, 421 mode, 423 -propagation, 426 ratio, 4, 18

Bandwidth, 7, 66, 460 enhancement, 452 Beam efficience, 78 efficiency, 79 width, 7, 601 of major lobe, 127 Beverage antenna, 215 Bi-conical antennas, 399 V antenna, 400 Binomial array, 147 Bow tie antenna, 367 Box loop antenna, 236 Brewster angle, 33, 34 Brightness temperature, 84, 626 Broadband antenna, 57, 611 Broadside array, 116, 125 Bunching of the force, 2

Capacitive current, 566 Capture area, 81 Circular components method, 609 patch antenna, 457 727

728

Index

polarization, 90, 342, 607 ratio, 608 Circularly polarized, 19, 461 Collision frequency, 575 Compact test ranges, 594 Complex permittivity, 22 Conical horn, 383 Constitutive parameters of the medium, 24 Corrugated horn, 308 Cosmic radiation, 548 Coupling, 463, 598 loss, 526 Critical angle, 33, 526, 556 frequency, 525, 555 Current distribution, 622 measurement, 621 waves, 478 Cylindrical helical antenna, 434 D-layer, 551 Depth of the duct, 526 Design equation, 462 Diamond antenna, 218 Dielectric lens, 282 delay lens, 279 Different layers of ionosphere, 550 path delay, 145 Diffraction, 527 loss, 529 Directive antenna, 8 Directivity, 7, 69, 78, 329, 458, 601 Dispersion, 575 Displacement current, 567 Dolph–Tchebyscheff or Chebyshev array, 132 Double tuning, 463 Driven element, 118 Duct gradient, 525 propagation, 10, 516, 524 loss, 526 E-layer, 551 E-plane lens, 279 pattern, 365 sectoral horn, 392

Earth magnetic field, 568 Effect of earth on vertical patterns, 130 of ground, 499 Effective aperture, 80 dielectric constant, 567 earth radius, 519 height, 82 radius, 458 Effectiveness ratio (a), 83 Effects of polarization, 499 Efficiency, 73 Electric wave equation, 21 Electrically small antennas, 57 Electronic phased array, 129 Electrostatic, 185 Element, 114 Elevated duct, 524 range, 589 Eleven-year sunspot, 570 Elliptical polarization, 18, 90 reflector, 304 Empirical formula, 523 End-fire antenna, 428 array, 116, 128 Energy loss, 575 Enhanced bandwidth, 445 Extension factor, 196 Extraordinary waves, 568 Extreme coupling, 621

F-layer, 552 Fading, 10, 572 Far-field, 598 region, 8, 59, 623 Faraday rotation, 575 Feed techniques, 358 Feeding systems, 306 Feeding techniques, 447 Ferrite loop antenna, 235 Ferrite rod antenna, 259 Fibre optical communication systems, 480 Field pattern, 78

Index Finite conducting earth, 622 difference time domain, 466 ground size, 430 Flat surface, 558 FNBW, 76 Forward scattering propagation, 522 Foster’s expression, 250 Fractal microstrip antennas, 466 Fraunhofer zone, 598 Free-space ranges, 588 Frenzel zone, 528 Friis free-space wave equation, 481 Function of frequency, 554 Fundamental equation, 481

G.P.S. applications, 435 Gain, 7, 78, 329 comparison, 617 measurement, 618, 611 transfer, 617 Geometric horizon, 514 ratio, 373 Grazing angle, 560 Gregorian antenna, 319 Ground attenuation, 485 reflected waves, 492 -reflection range method, 614 wave, 493 attenuation, 495 Group delay, 575 velocity, 19 Gyro-frequency, 569, 576 Gyromagnetic field, 570

H-plane, 282 sectoral horn antenna, 390 Hansen–Woodyard, 335 Hansen–Woodyard end-fire array, 426 Harmonics, 198 Helical antenna, 5, 420 beam antenna, 420 Helix, 421 hemispherical, 432

Hertz antenna, 194 Hertzian dipole, 183 antenna, 483 Historical, 3 Horizontal collimation, 595 Horizontally, 4 polarized waves, 496 Horn antennas, 383 Hot spots, 6 HPBW, 76, 77 HPBW and FNBW, 7 Huygens–Frenzel wave, 527 Hyperbolic reflector, 304

Impedance bandwidth, 66 of loop antenna, 256 matching, 6, 384, 451, 593 measurement, 618 Index of refraction (n), 521 Induced EMF, 242 Induction field, 8, 185 Inductive current, 566 Infinite, 400 Infrared, 15 Input impedance, 61, 197, 402 and VSWR, 363 resistance, 190 Integrated antennas, 465 Interference fading, 572 Intrinsic impedance, 17, 22 Inverted V-antenna, 203 Ionosode, 565 Ionosphere, 549 Ionospheric propagation, 10 storms, 571 Ions, 549 Iso-phase plane, 527 Isotropic, 8 radiator, 58, 422 source, 70

Keller’s, 385 Kennelly–Heaviside, 551 Kraus formula, 71

729

730

Index

Laminate composite, 464 Large loop, 235, 245 Lens antenna, 275 Line-of-sight, 4 Linear array, 115 polarization, 607 Linearly polarized, 19 Lobes, 8 Log-periodic dipole antenna, 354 toothed planar antenna, 367 toothed trapezoid antenna, 370 toothed wedge, 369 wire antenna, 370 Long-distance communication, 551 LOS (line of sight) distance, 512, 531 Loss area, 81 Loss-tangent, 25 Lowest useable frequency, 561 Luneburg lens, 287

M-curves, 10, 524 Magnetic dipole, 238 Magnetic wave equation, 21 Marconi, 3 antenna, 194, 430 Maximum effective length, 217 radiation direction, 597 useable frequency, 557 Maxwell’s, 16 Fisheye, 288 Measurement of reflectivity, 600 Metal-plate lens, 276 MF marine bands, 500 Microstrip antennas, 444 line feed, 448 Microwave(s), 15 band, 479 Mobile antennas, 465 Modified index, 524 Mongel–Dellinger effect, 571 Monopole, 193 Moore loop antenna, 236

Multi hop propagation, 498 path phenomenon, 512 path propagation, 529 Multiple reflections, 599 -reflector, 318 Multipurpose, 481 Mutual coupling, 149, 615 Mutual impedance, 90, 620

Near-field(s), 8 measurements, 599 Newton’s law of motion, 566 Noise figure, 86, 625 temperature, 625 Non-deviative, 575 Normal mode, 423 helical antenna, 428 Norton’s wave propagation, 485 Numerical distance, 495

Oblique incidence, 29 Omni-directional, 8 HPBW, 407 Optical antenna, 303 Optimum working frequency, 561, 562 Ordinary waves, 568 Orthogonal linear polarizations, 618

Parabolic cylindrical, 317 reflector, 309 Parallel polarization, 30 Parasitic arrays, 117 Parasitic elements, 118 Partial directivity, 70 Pascal’s triangle, 147 Patch antenna, 444 Path attenuation a, 482 Pattern bandwidth, 67 factor, 188

Index Perpendicular polarization, 30 Personal communications, 587 Phase constant, 22, 495 measurement, 623 Photonic band gap antennas, 464 Plane earth reflection, 487 Plane wave, 20 Poincare’s sphere, 611 Polarization, 4, 604 beam width, 398 efficiency, 92, 398 fading, 572 loss efficiency, 316 loss factor, 92 matched, 398 measurement, 611 pattern method, 605, 609 ratio, 92, 396, 608 vector, 92 Power pattern, 78 Poynting vector, 13, 17 Principal, 8 plane, 247 Principle of pattern multification, 425 Probe feed, 448 Propagation, 9 constant, 21 delay, 19 Pseudo-Brewster angle, 489, 531 Pulse-modulated radio wave, 566 Pyramidal horn, 394 /conical antenna, 408 Q-Method, 601, 603 Quality factor, 603 and loss tangent, 459 Quit zone, 588 Radar techniques, 616 Radiated fields, 458 Radiation, 2, 182 absorbing material (RAM), 592 antenna efficiency, 74 bandwidth, 68 conductance, 456

efficiency, 191, 253, 601 field, 8, 58, 239 intensity, 78 pattern, 74, 596 resistance, 69, 430, 455 Radio antenna, 56 waves, 15 Radius of curvature, 519 RAM, 592 Reciprocity theorem, 93, 276 Rectangular microstrip antenna, 445 patch antenna, 454 pyramidal horn, 389 Reflection, 490 coefficient, 29 ranges, 588 total, 33 Reflectivity, 33, 600 Reflector antennas, 303 Refraction, 490, 516 Refractivity, 521 Regular and irregular variations, 570 Resonance, 6 frequencies, 453 Resonant antennas, 57 Retarded magnetic vector potential, 240 Rhombic antenna, 217 Rotating source method, 605 Rotman lens, 290

S-polarized, 34 S11 parameter, 435 Sampling loop, 621 Scatter angle, 522 Scattered equivalent area, 81 ratio (b), 83 Secant law, 557 Selective fading, 573 Self and mutual impedances, 618 Self-complementary antenna, 370 Sensitivity factor, 140, 141 Short-range communications, 551 Side lobe level, 75, 77

731

732

Index

Skip distance, 562 fading, 573 Slant range, 590 Sleeve wire, 206 Slotted cone antenna, 405 Slow wave, 439 Small dipole, 181 loop, 235 SMART antenna, 142 Snell’s law, 30, 280 SNR, 86 and directive gain, 140 Solar eruption, 572 radiation, 548 Solid angle, 71, 78 Sommerfield surface, 478 Space propagation, 9 taper, 318 wave patterns, 530 wave propagation, 512, 529 Specific absorption rate (SAR), 465 Spectrum, 14 Sporadic E-layer, 551 Square loop, 245 Stacked/Rectangular, 137 Standard gain antennas, 617 Standing wave, 619 antenna, 432 Stokes’ parameter, 611 Sudden ionospheric disturbances, 571 Super directive receiving array, 140 -refracting, 515 Surface duct, 524 wave propagation, 9, 485 wave tilt, 486 Tapering, 147 helix, 430 Tchebyscheff, 134 Tolerances, 292 Transmission coefficient, 29

loss, 85, 477, 523 total, 32 Trapping, 515 Troposcattering propagation, 522 Troposphere, 514 propagation, 9 Two-antenna method, 613

Ultraviolet, 15 Uniform linear array, 115 plane wave, 20

V antenna, 204 dipole, 202 Velocity factor, 19, 427 of propagation, 19 Vertically polarized wave, 494 VSWR, 6, 61, 619

Wave equation, 20 number, 16 polarization, 17 Waveguide, 13 Wheeler efficiency, 602 method, 602 Whip antenna, 5 Wider bandwidth, 195 Wire antennas, 181

X-ray, 15

Yagi–Uda, 3, 327 array, 342 loop antennas, 343 Yagi-folded antenna, 197

Zenneck wave, 478 Zonal E-plate, 281

Antenna and Wave Propagation by Yadava

Recommend Documents