Transmission Lines & Wave Guides

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VEL TECH MULTI TECH TECH HIGHTECH

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STUDY MATERIAL TRANSMISSION LINES AND WAVEGUIDES DEPARTMENT OF

ECE

JUNE – 2010

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Vel Tech Vel Tech Multi Tech Dr.Rangarajan Dr.Sakunthala Engineering College Vel Tech High Tech Dr. Rangarajan Dr.Sakunthala Engineering College VEL TECH

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SEM - V

INDEX UNITS

PAGE NO.

I.

Filters

06

II.

Transmission Line Parameters

51

III. The Line at Radio Frequency

95

IV.

Guided Waves Between Parallel Planes

138

V.

Waveguides

179

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# 42 & 60, Avadi – Veltech Road, Avadi, Chennai – 62. Phone : 044 26840603 26841601 26840766

email : [email protected] website : www.vel-tech.org www.veltechuniv.edu.in

R S ∗ Student Strength of Vel Tech increased from 413 to 10579, between 1997 and 2010. ∗ Our heartfelt gratitude to AICTE for sanctioning highest number of seats and highest number

of courses for the academic year 2009 – 2010 in Tamil Nadu, India. ∗ Consistent success on academic performance by achieving 97% - 100% in University examination results during the past 4 academic years. ∗ Tie-up with Oracle Corporation for conducting training programmes & qualifying our students for International Certifications. ∗ Permission obtained to start Cisco Networking Academy Programmes in our College campus. ∗ Satyam Ventures R&D Centre located in Vel Tech Engineering College premises. ∗ Signed MOU with FL Smidth for placements, Project and Training. ∗ Signed MOU with British Council for Promotion of High Proficiency in Business English, of the University of Cambridge, UK (BEC). ∗ Signed MOU with NASSCOM. ∗ MOU’s currently in process is with Vijay Electrical and One London University. ∗ Signed MOU with INVICTUS TECHNOLOGY for projects & Placements. ∗ Signed MOU with SUTHERLAND GLOBAL SERVICES for Training & Placements. ∗ Signed MOU with Tmi First for Training & Placements.

VELTECH, VEL TECH MULTI TECH engineering colleges Accredited by TCS VEL TECH, VEL TECH MULTI TECH, VEL TECH HIGH TECH, engineering colleges & VEL SRI RANGA SANKU (ARTS & SCIENCE) Accredited by CTS. Companies Such as TCS, INFOSYS TECHNOLOGIES, IBM, WIPRO TECHNOLOGIES, KEANE SOFTWARE & T INFOTECH, ACCENTURE, HCL TECHNOLOGIES, TCE Consulting Engineers, SIEMENS, BIRLASOFT, MPHASIS(EDS), APOLLO HOSPITALS, CLAYTON, ASHOK LEYLAND, IDEA AE & E, SATYAM VENTURES, UNITED ENGINEERS, ETA-ASCON, CARBORANDUM UNIVERSAL, CIPLA, FUTURE GROUP, DELPHI-TVS DIESEL SYSTEMS, ICICI PRULIFE, ICICI LOMBARD, HWASHIN, HYUNDAI, TATA CHEMICAL LTD, RECKITT BENKIZER, MURUGAPPA GROUP, POLARIS, FOXCONN, LIONBRIDGE, USHA FIRE SAFETY, MALCO, YOUTELECOM, HONEYWELL, MANDOBRAKES, DEXTERITY, HEXAWARE, TEMENOS, RBS, NAVIA MARKETS, EUREKHA FORBES, RELIANCE INFOCOMM, NUMERIC POWER SYSTEMS, ORCHID CHEMICALS, JEEVAN DIESEL, AMALGAMATION CLUTCH VALEO, SAINT GOBAIN, SONA GROUP, NOKIA, NICHOLAS PHARIMAL, SKH METALS, ASIA MOTOR WORKS, PEROT, BRITANNIA, YOKAGAWA FED BY, JEEVAN DIESEL visit our campus annually to recruit our final year Engineering, Diploma, Medical and Management Students.

Preface to the First Edition This edition is a sincere and co-ordinated effort which we hope has made a great difference in the quality of the material. “Giving the best to VEL TECH


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the students, making optimum use of available technical facilities & intellectual strength” has always been the motto of our institutions. In this edition the best staff across the group of colleges has been chosen to develop specific units. Hence the material, as a whole is the merge of the intellectual capacities of our faculties

across the group of Institutions. 45

to 60, two mark questions and 15 to 20, sixteen mark questions for each unit are available in this material. Prepared By :

Ms. S. Jalaja Asst. Professor. Mr. S. Jebasingh. Lecturer.

EC2305 UNIT I

TRANSMISSION LINES AND WAVEGUIDES FILTERS

9

The neper - the decibel - Characteristic impedance of Symmetrical Networks – Current and voltage ratios - Propogation constant, - Properties of Symmetrical Networks – Filter fundamentals – Pass and Stop bands. Behaviour of the Characteristic impedance. Constant K Filters - Low pass, High pass band, pass band elimination filters - m -derived sections – Filter circuit design – Filter performance – Crystal Filters. UNIT II 9

TRANSMISSION LINE PARAMETERS

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A line of cascaded T sections - Transmission lines - General Solution, Physical Significance of the equations, the infinite line, wavelength, velocity, propagation, Distortion line, the telephone cable, Reflection on a line not terminated in Zo, Reflection Coefficient, Open and short circuited lines, Insertion loss. UNIT III

THE LINE AT RADIO FREQUENCY

9

Parameters of open wire line and Coaxial cable at RF – Line constants for dissipation voltages and currents on the dissipation less line - standing waves – nodes – standing wave ratio - input impedance of open and short circuited lines - power and impedance measurement on lines –  / 4 line, Impedance matching – single and double-stub matching circle diagram, smith chart and its applications – Problem solving using Smith chart. UNIT IV

GUIDED WAVES BETWEEN PARALLEL PLANES

9

Application of the restrictions to Maxwell’s equations – transmission of TM waves between Parallel plans – Transmission of TE waves between Parallel planes. Transmission of TEM waves between Parallel planes – Manner of wave travel. Velocities of the waves – characteristic impedance - Attenuators UNIT V

WAVEGUIDES

9

Application of Maxwell’s equations to the rectangular waveguide. TM waves in Rectangular guide. TE waves in Rectangular waveguide – Cylindrical waveguides. The TEM wave in coaxial lines. Excitation of wave guides. Guide termination and resonant cavities. TEXT BOOK: 1. John D.Ryder, "Networks, lines and fields", Prentice Hall of India, 2nd Edition, 2006. REFERENCES: 1. E.C.Jordan, K.G. Balmain: “E.M.Waves & Radiating Systems”, Pearson Education, 2006. 2. Joseph Edminister, Schaum’s Series, Electromegnetics, TMH, 2007. 3. G S N Raju, Electromagnetic Field Theory and Transmission Lines, Pearson Education, 2006.

UNIT – I PART – A 1. Define Filter? A reactive network that will freely pass desired bands of frequencies while almost totally suppressing other band of frequencies are called as filters. 2. What do you mean by ideal filter? An ideal filter would pass all frequencies in a given without reduction in VEL TECH


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magnitude and totally suppressing all other frequencies. 3. What is cutoff frequency? The frequency which separates a pass band and an attenuation band are called as cutoff frequency. 4. Define the unit of attenuation (or) Define Neper? Attenuation is expressed in decibels or nepers. Nepers is defined as the natural logarithm of the ratio of input voltage or current to the output voltage or current provided the network is properly terminated with Z0. 5. Define Decibel? Decibel is defined as the ten times common logarithms of the input power to the output power. 6. Give the relation between two units of attenuation? The relationship between two units of attenuation can 1 db = 0.115 neper. 7. Give the common types of filters? The four common types of filters are, High pass filter 1. Low pass filter 2. Band pas filter 3. Band stop filter 8. What are the characteristics of ideal filter? The characteristics of ideal filter are: 1. Transmit pass band frequencies without any attenuation. 2. Provide infinite attenuation 3. The transition region between the stop band and pass band would be very small. 4. Throughout the pass band characteristic impedance of the filter match circuit to which it is connected which prevents reflection loss. 9. When networks are said to be symmetric network? When two series arms of a T network are equal or when two shunt arms of a network are equal then the network is said to be symmetrical network. 10. Draw a symmetrical network in ‘T’ and Π sections.

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11. Give the characteristic impedance of low pass constant – K filter The characteristic impedance is given as Z OT =

(Z

2 1

+ Z1 Z 2 )

Z OT = L / C 1 − ω 2CL / 4 12. Give the formula to calculate the cutoff frequency for low pass constant –k f c = 1/ π LC 13. What are the constant – k filters? A constant – k filters is a ‘T’ or ‘π ’ network in which the series and shunt impedance Z1 and Zz are connected by the relation. Z1 – Z2 = R2 K. Where ‘Rk’ is a real constant i.e., a resistant independent of frequency. 14. Why constant – k filters are also called as prototype filter? Constant – k filters are also called as prototype filter because other complex networks can be derived from it. VEL TECH


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15. For a low pass filter what is the condition for which the characteristic impedance Z0 is real? The characteristic impedance Z0 is real if (ω

2

LC/4)<1

16. Give the expression for the cutoff frequency of constant – k high pass filter. f c = 1/ 4π LC 17. What are the phase shift and attenuation of constant –k low pass filter?

(

)

−1 The phase shift of constant – k low pass filer is given as, β = 2sin ω / ωC radians −1 The attenuation of constant low pass filter is given as α = 2 cos ( ω / ω c ) nepers

18. What are the phase shift and attenuation of constant –k high pass filter? The phase shift of constant – k high pass filter is given as β = 2 sin-1 ( ω c2/ω 2) radians. The attenuation of constant high pass filter is given as α = 2 cos −1 ( ω c / ω ) nepers 2

19. For a high pass filter what is the conditions for which the characteristic impedance ZOT is real and imaginary. For a high pass filter, the characteristic impedance ZOT is real if (1/4ω

2

LC ) < 1

And Imaginary if (1/4ω 2 LC) >1 20. Give the characteristic impedance of high pass filter constant – k filter. The characteristic impedance of high pass filter constant- k filter is given as

Z OT = L / C 1 − ( 1/ 4ω 2 LC )

21. Draw the T- type and π type low pass constant – k filter. The prototype low pass filters are as shown below.

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22. Draw the T- type and Π type high pass constant –k filter

23. What are the advantages of m- derived filters? The advantages of m – derived filters are, 1. A sharper cutoff characteristic with steeper rise at fc, the slope of the rise being adjustable by fixing the distance between fc and f. 2. ‘Zo’ of the filter will be more uniform within the pass and when m – derived half section having m = 0.6 are connected at the ends. 3. m- derived filters makes it possible to construct composite filters to have any desired attenuation characteristics. 24. What is composite filter? A filter designed using one or more prototype constant – k filters and m – derived filters to have an attenuation between low pass and high filters is called as composite filters. 25. Mention the different sections of a composite filters? [email protected] The different sections of a composite filters are 1. One or more prototype constant filters are 2. One or more m – derived sections 3. Two terminating m – derived half sections with m = 0.6 VEL TECH


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26. What are the drawbacks of constant – k filters (or) what are the disadvantages of constant – k filters? 1. The attenuation does not increase rapidly beyond cutoff frequencies. 2. Characteristic impedance varies widely in the transmission or pass band from the derived value. 27. Why m–derived half section is used as terminating section? Two m – derived half section with m = 0.6 is used as a terminating section to give constant input and output impedance. 28. Distinguish between low pass and high pass filter. S. No 1

2

Low Pass Filter

High Pass filter

This filter passes the frequencies without attenuation upto a cutoff frequency fc and attenuates all other frequencies greater than fc. An ideal low pass filter is as shown below

It transmits frequencies above a designed cutoff frequency but attenuates frequencies below this An ideal high pass filter is as shown below

29. Differentiate between band pass filter and band elimination filter S.No 1

2.

Bandpass filter This filter passes the frequencies between two designated cutoff frequencies and attenuates all other frequencies An ideal band pass filter is as shown below

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Band elimination filter It transmits all frequencies while attenuates a band of frequency. An ideal band elimination filter is as shown below.


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30. Draw a block diagram of a composite filter?

31. Draw the T-type and Π type low pass m – derived filter The m – derived low pass filters are as shown below.

32. Draw the type and Π type high pass m- derived filter The m- derived high pass filters are as shown below VEL TECH


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33. Give the value of ‘m’ in m – derived low pass and high pass filters. The value of ‘m’ in the case of low pass filter is given as below m = The value of ‘m’ in the case of high pass filter is as below m =

1 − ( fc / f ∞ )

1 − ( f ∞ / fc )

2

2

PART – B 1. Define the characteristic impedance of symmetrical networks. When Z1 = Z2 or the two series arms of a T network are equal, or Za = Zc and the shunt arms of a π network are equal, the networks are said to be symmetrical. Filter networks are ordinarily set up as symmetrical sections, basically of the T or π type, such as shown at (b) and (d), Fig.

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Figure: The T and π sections as derived from unsymmetrical L sections, showing notation used in symmetrical network analysis. Attention is called to the peculiarities of notation employed on the variouos arms. This peculiarity is largely dictated by custom, arising from the fact that both T and π networks can be considered as built of unsymmetrical L half sections, connected together in one fashion for the T network, and oppositely for the π network as at (a) and (c), Fig. A series connection of several T or π networks leads to socalled “ladder networks,” which are indistinguishable one from the other except for the end or terminating L half sections, as can be seen in fig. For a symmetrical network the image impedances Z1i and Z2i, are equal to each other, and the image impedance is then called the characteristic impedance or the iterative impedance, Z0. That is, if a symmetrical T network is terminated in Z0, its input impedance will also be Z0, or its impedance transformation ratio is unity. The term iterative impedance is apparent if the terminating impedance Z0 is considered as the input impedance of a chain of similar networks, in which case Z0 is iterated at the input to each network.

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Figure: (a) Ladder network made from T section; (b) ladder network built from π sections. The parallel shunt arms will be combined. The value of Z0 for a symmetrical network can be easily determined. For the T network of Fig. (a), terminated in an impedance Z0, the input impedance is Z1in =

Z1 Z 2 ( Z1 / 2 + Z0 ) + 2 Z1 / 2 + Z2 + Z0

It can be assumed that if Z0 is properly chosen in terms of the network arms, it should be possible to make Z1in equal to Z0. Requiring this equality gives Z0 =

Z12 / 4 + Z1Z2 + Z2Z0 + Z1Z0 / 2

Z02 =

Z1 / 2 + Z2 + Z0 Z12 4

+ Z1Z2

For the symmetrical T section, then, Z0T =

 Z  + Z1Z2 = Z1Z2  1+ 1  4 4Z2  

Z12

becomes the characteristic impedance. This result could also have been immediately obtained from eq. and for the image

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Figure: Determination of Z0: (a) for a T section; (b) for a π section impedance of a T section, by using the values of the arms of Fig. Similarly, for the π section of Fig. (b) the input impedance is

Z1in

  2Z2Z0    Z1 +    2Z2  2Z 2 + Z0    = 2Z2Z0 Z1 + + 2Z2 2Z 2 + Z0

Requiring that Z1in = Z0 leads to Z0π =

Z1Z2 1+ Z1 / 4Z2

which is the characteristic impedance of the symmetrical π symmetrical π section. Certain information concerning networks was developed from measurements of Z∞ and Z∞ . If these measurements are made on the T section of (a), Fig. exclusive of the load Z0, then Z1 + Z2 2 Z Z Z /2 Z1∞ = Z∞ = 1 + 1 2 2 Z1 / 2 + Z2 Z2 Z∞ Z∞ = 1 + Z1Z2 = Z0T 2 4 Z1∞ = Z∞ =

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Similar work for the π section leads to Z∞ Z∞ =

4Z22 Z1 Z1 + 4Z2

= Z0π2

Therefore, for a symmetrical network, Z0 = Z ∞ Z∞ This result could have been directly obtained from the image impedance relations of section. It is a valuable relationship, since it supplies an easy experimental means of determining the Z0 of any symmetrical network. 2. Explain the current and voltage ratios as exponentials; the propagation constant. Under the assumption of equal input and output impedances, which may now be interpreted as a Z0 termination on the network, the absolute value of the ratio of input current to output current of a given symmetrical network was defined as an exponential function,* for the purpose of simplifying network calculations. obviously, the magnitude ratio does not express the * In the general case of unsymmetrical 4-terminal networks, terminated on an image basis, it is customary to define a transfer constant θ , by input volt-amperes E1I 1 = E2I 2 output volt-amperes 1 EI θ = ln 1 1 2 E2I 2 ε 2θ = Or

where θ is in general a complex number. For symmetrical networks Z 14 = Z2 = Z0, and with a termination of Z0, the above discussion follows, with γ customarily replacing θ and implying symmetry and Z0 termination. complete network performance, the phase angle between the currents being needed as well. the use of the exponential can be extended to include the phasor current ratio if it be defined that, under the condition of Z0 termination, I1 = εγ I2 where γ is a complex number defined as γ = α + jβ

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Hence


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I1 = ε γ = εα+ jβ I2

To illustrate further, if I 1 / I 2 = A β ,then A = I 1 / I 2 = εα β = ε jβ Since the input and output impedances are equal under the Z 0 termination, it is also true that V1 = εγ V2 The term γ has been given the name propagation constant. The exponent α is known as the attenuation constant, since it determines the magnitude ratio between input and output quantities, or the attenuation produced in passing through the network. The units of α are nepers. The exponent β is the phase constant as it determines the phase angle between input and output quantities, or the shift in phase introduced by the network. The units of β are radians. If a number of sections all having a common Z 0 value are cascaded, the ratio of currents is I1 I2 I 3 I × × × .. = 1 I 2 I 3 I4 In from which ε γ 1 × ε γ 2 × εγ 3 × ... = εγn and taking the natural logarithm, γ 1 + γ 2 + γ 3 + ... = γ n Thus the over-all propagation constant is equal to the sum of the individual propagation constants. 3. Explain the properties of symmetrical networks. Use of the definition of γ , and the introduction of ε Z0 terminated network, leads to further useful results.

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γ

as the current ratio for a

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Figure: Symmetrical network with generator and load. In Fig. the T network is considered equivalent to any connected symmetrical network, and is terminated in a load Z0. The mesh equation are Z E = I 1  1 + Z2  − I 2Z2  2  Z 0 = −I 1Z2 + I 2  1 + Z2 + Z0   2  The current ratio for the two meshes, which is equal to ε obtained from the second equation as I 1 Z1 / 2 + Z2 + Z0 = = εγ I2 Z2 After thus introducing ε

γ

γ

by definition, can be

, the above may be written

Z0 = Z2 ( ε γ − 1) −

Z1 2

From Eq. for the characteristic impedance, Z02 =

Z12 4

+ Z1Z2

If Z0 is eliminated by use of Eq. in Eq. there results Z2 ( ε γ − 1) − Z1εγ = 0 2

ε 2γ − 2ε γ + 1 =

Z1 γ ε Z2

ε γ + ε− γ Z = 1+ 1 2 2Z2 Z cosh γ = 1+ 1 2Z2 Equation and its other derived forms will be of considerable value in the study of VEL TECH


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filters. By use of the identify, Eq., that cosh2 γ − sinh2 γ = 1 it is possible to write Z0 Z2 Combining Eqs. and leads to sinh γ =

Z0 Z1 / 2 + Z2 By use of Eq. it is possible to write tanh γ =

 γ 1 Z1 sinh = − 1  1+ 2 2 2Z2  =

Z1 4Z2

an expression which will serve to predict filter performance. The propagation constant γ can be related to the network parameters by use of Eq. for Z0T, in Eq. as 2

 Z  Z Z ε = 1+ 1 +  1  + 1 2Z2  2Z2  Z2 Taking the natural logarithm γ

2   Z1  Z1  Z1  γ = ln 1+ +   +  2Z2 2Z2  Z2    

For a network of pure reactances this is not difficult to compute. For an impedance it may be noted that the logarithm of a complex quantity B α = ln B + jα. The input impedance of any T network, terminated in any impedance ZR, may also be written in terms of hyperbolic functions of γ . Writing Zin = Z11 −

Z122

Z22 and substituting the required mesh relations from Fig. with Z0 replaced by ZR, then VEL TECH


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Z 22 Z1 + Z2 = 2 Z1 / 2 + Z2 + ZR

Z12 / 4 + Z1Z2 + ( Z1 / 2 + Z2 ) ZR Z1 / 2 + Z2 + ZR

Use of Eqs. and leads to Zin =

Z02 + ZR Z0 / tanh γ Z0 / tanh γ + ZR

 Z cosh γ + Z0 sinh γ  = Z0  R   Z0 cosh γ + ZR sinh γ  This is the input impedance of a symmetrical T network terminated in a load Z R, in terms of the propagation constant and Z0 of the network. For a short-circuited network ZR = 0. The input impedance is then Z∝ where, from the above equation, Z∞ = Z0 tanh γ For an open circuit ZR = ∝ in the limit, and Z∝ is then limZZR∞→∞ =

Z0 tanh γ

From these two equations it can be seen that tanh γ = and

Zsc Zoc

Z0 = Z ∞ Z ∞

which has already been proved from the properties of the characteristic impedance. In chapter 1, open-circuit and short-circuit measurements were used to describe the performance of a network. in this chapter, two new parameters, the characteristic impedance Z0, and the propagation constant γ , have been introduced, and the properties of the network have been developed in terms of these new parameters. The last few equations are relations between the two sets of parameters. 4. Explain the constant-k low-pass filter. If Z1 and Z2 of a reactance network are unlike reactance arms, then VEL TECH


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Z1Z2 = k2 where k is a constant independent of frequency. Networks or filter sections for which this relation holds are called constant-k filters. As a special case, let Z1 = jωL and Z2 = − j/ ωC, then the product Z1Z2 =

L = R k2 C

The term Rk is used since k must be real if Z1 and Z2 are of opposite type. A T section so designed would appear as at (a), Fig.

Figure: (a) Low-pass filter section; (b) reactance curves demonstrating that (a) is a low-pass section or has a pass band between Z1 = 0 and Z1 = -4Z2. The reactances of Z1 and 4Z2 will vary with frequency as sketched at (b), Fig. The curve representing -4Z2 may be drawn and compared with the curve for Z1. It has been shown by Eq. that a pass band starts at the frequency at which Z1 = 0 and runs to the frequency at which Z1 = 4Z2. Thus the reactance curves show that a pass band starts at f = 0 and continues to some higher frequency fc. All frequencies above fc lie in a stop, or attenuation, band. Thus the network is called a low-pass filter. The cutoff frequency fc may be readily determined, since at that point Z1 = −4Z2 , jωcL =

4j ωcC

1 LC This expression may be used to develop certain relations applicable to the lowpass network. then sinh γ / 2 may be evaluated as fc =

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and in view of Eq. this is γ f sinh = j 2 fc Then if the frequency f is in the pass band or f/fc < 1, so that -1 < Z1/4Z2 < 0, then f β=2sin-1    fc 

f < 1, α =0, fc

Figure: Variation of α and β with frequency for the low-pass section whereas if frequency f is in the attention band or f/fc > 1, so that Z1/4Z2 < - 1, then f f > 1, α =2cosh-1   , f  fc 

β=π

thereby allowing determination of α and β . The variation of α and β is plotted in Fig. as a function of f/fc. This method shows that the attenuation α is zero throughout the pass band but rises gradually from the cutoff frequency at f/f c = 1.0 to a value of ∝ at infinite frequency. The phase shift β is zero at zero frequency and increases gradually through the pass band, reaching π at fc and remaining at π for all higher frequencies. The characteristic impedance of a T section was obtained as  Z  Z0T = Z1Z2  1+ 1  4Z 2   which becomes VEL TECH

Z0T =

L  ω2LC   1−  C 4  VEL TECH MULTI TECH TECH HIGHTECH 22

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for the low-pass constant-k section under discussion. by use of Eq. the characteristic impedance of a low-pass filter may be stated as

Z0T

L f = 1−   C   fc  = Rk

f 1−    fc 

2

  

2

in accordance with the definition of Rk in Eq. Values of Z0T/Rk are plotted against f/fc in Fig. It may be seen that cutoff, then becomes imaginary in the attenuation band, rising to infinite reactance at infinite frequency. A low-pass filter may be designed from a knowledge of the cutoff frequency desired and the load resistance to be supplied. It is desirable that the Z0 in the pass band match the load; but because of the nature of the Z 0 curve in Fig., this result can occur at only one frequency. This match may be arranged to occur at any frequency which it is desired to favor by an impedance match. For reasons which will appear in section, the load is chosen as R = R k = L C , which will favor zero frequency for a low-pass filter. The design of a low-pass filter may be readily carried out. From

Figure: Variation of Z0T / R k with frequency for the low-pass section. the relation that at cutoff Z1 = −4Z2 VEL TECH


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it is seen that ωeL =

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4 ω0C

Using the cutoff frequency equation changes this to π2fc2LC = 1 and use of the relation R = L C gives for the value of the shunt capacitance arm C=

1 πfcR

By similar methods the inductance for Z1 is obtained as L=

R πfc

Since the design is based on an impedance match at zero frequency only, power transfer only, power transfer to a matched load will drop at higher pass-band frequencies. This condition may be undesirable in certain applications, and a remedy will be discussed in section. A network such as is described here is called a prototype section. It may be employed when a sharp cutoff is not required, although cutoff may be sharpened by using a number of such networks in cascade. This is not usually an economic use of circuit elements, and introduces excessive losses over other available methods of raising the attenuation near the cutoff frequency. 5. Explain the constant-k high-pass filter. If the positions of inductance and capacitance are interchanged to make Z1 = − j/ ωC and Z2 = jωL , then Z1Z2 will still be given by Z1Z2 = k2 and the filter design obtained will be of the constant-k type. The T section will then appear as in (a), Fig. The reactances of Z1 and Z2 are sketched as functions of frequency in (b), and Z1

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Figure: (a) High-pass filter section; (b) reactance curves demonstrating that (a) is a high-pass section or has a pass band between Z1 = 0 and Z1 = -4Z2. Is compared with -4Z2, showing a cutoff frequency at the point at which Z1 equals -4Z2, with a pass band from that frequency to infinity where Z 1 = 0. The network is thus a high-pass filter. All frequencies below fc lie in an attenuation, or stop, band. The cutoff frequency is determined as the frequency at which Z1 = −4Z2 , or −j = − j4ω0L, 4ωc2LC = 1 ω0C 1 fc = 4π LC Using the above expression −j γ Z1 1 f sinh = = − 2 = = −j c 2 4Z2 4ω LC 2ω LC f The region in which fc / f < 1 is a pass band, so that the variation of γ inside and outside the pass band will be indentical with the values for the low-pass filter, and the curves of fig. will apply if the abscissa be considered as calibrated in terms of f c/f, except that the phase angle β will be negative, changing from 0 at infinite frequency or fc/f = 0, to -π at cutoff or fc/f = 1. The high-pass filter may be designed by again choosing a resistive load R equal to Rk such that L R = Rk = C From the relation that at cutoff Z1 = −4Z2 it was shown that 4ωc2LC = 1 and again L/C = R2, so that the value of the capacitance for Z1, the series element, is VEL TECH


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1 4πfcR

It should be noted that since Z1/2 is the value of each series arm, the capacity use din each series Z1/2 element should be 2C. By similar methods the value for the inductance for Z2, the shunt arm, is R L= 4πfc The characteristic impedance for the high-pass filter may be transformed to Z0T = R k

f  1−  c  f

2

6. Explain the m-derived T section filter. The constant-k prototype filter section, though simple, has two major disadvantages. The attenuation does not rise very rapidly at cutoff, so that frequencies just outside the pass band are not appreciably attenuated with respect to frequencies just inside the pass band. Also, the characteristic impedance varies widely over the pass band, so that a satisfactory impedance match is not possible. In cases where an impedance match is not important, the attenuation may be built up near cutoff by cascading or connecting a number of constant-k sections in series. It is more economical to attempt to raise the attenuation near cutoff by other means. Consider first the circuit of (a), Fig. The reactance curves sketched at (b) show that this circuit is alow pass filter. However, it can be seen that the shunt arm is a series circuit resonant at a frequency above fc. At this resonant frequency the shunt arm appears as a short circuit on the network, or the attenuation becomes infinite. This frequency of infinite or high attenuation is called f∝, will always be higher in value than fc. If, then, f∝ can be chosen arbitrarily close to fc, the attenuation near cutoff may be made high.

Figure: (a) Derivation of a low-pass section having a sharp cutoff action; (b) reactance curves for (a). VEL TECH


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The attenuation above f∝ will fall to low values, so that if high attenuation is desired over the whole attenuation band, it is necessary to use a section such as in Fig. for high attenuation near cutoff, in series with a prototype section to provide high attenuation at frequencies well removed from cutoff. For satisfactory matching of several such types of filters in series, it is necessary that the Z0 of all be identical at all points in the pass band. They will consequently also all have the same pass band. The network of fig. may be derived by assuming that Z'1 = mZ1 the primes indicating the derived section. It is then necessary to find the value for Z 2’ ' such that Z0 = Z0. Setting the characteristic impedances equal, Z 0 ' = Z0

( mZ1 ) 4 Z'2 =

2

+ mZ1Z2 ' =

Z12 + Z1Z2 4

Z 2 1− m2 + Z1 m 4m

It then appears that the shunt arm Z2 ' consists of two impedances in series, as shown in Fig. As required, the characteristic impedance and fc remain equal to those of the T section prototype containing Z1 and Z2 values.

Figure: The m-derived low-pass filter. Since m is arbitrary, it is possible to design an infinite variety of filter networks meeting the required conditions on Z0 and fc. However, Z2 will be opposite in sign to Z1, and it is desired that this relation continue in the two series impedances given by Eq. for the Z2 ' arm. Equation then indicates that (1-m2)/4m must be positive, forcing the terms 1 – m2 and m always to be positive. Thus m must always be chosen so that <1 VEL TECH


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Filter sections obtained in this manner are called m-derived sections. The shunt arm is to be chosen so that it is resonant at some frequency f ∝ above fc. This means that at the resonant frequency and for the low-pass filter 1 1− m2 = 2πf∞ L 2πf∞ mC 4m 1 f∞ = π ( 1− m2 ) LC Since the cutoff frequency for the low-pass filter is 1 fc = π LC the frequency of infinite attenuation will be f∞ = from which

fc 1− m2

m = 1− ( fc − f∞ )

2

This equation determines the m to be used for a particular f ∝. Similar relations for the high-pass filter can be derived as f∞ = fc 1− m2 and

m = 1− ( f∞ / fc )

2

The m-derived section is designed following the design of the prototype T section. The use of a prototype and one or more m-derived sections in series results in a composite filter. If a sharp cutoff is desired, an m-derived section may be used with f∝ near fc, followed by as many m-derived sections as desired to place frequencies of high attenuation where needed to suppress various signal components or to produce a high attenuation over the entire attenuation band. The variation of attenuation over the attenuation band for a low-pass m-derived section in the stop band is dependent on the sign of the reactances or α = 2cosh−1 fc < f < f∞

Z1 Z1 or α =2sinh-1 4Z2 4Z2 f∞ < f

For Z1 = jωL and Z2 = − j/ ωC for the prototype, then VEL TECH


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Z1 mωL = 4Z2 41/ mωC − ωL ( 1− m2 ) / 4m   so that for fc < f < f∞ α = 2cosh−1

mf / fc 1− f 2 / f∞ 2

Figure: Variation of attenuation for the prototype and m-derived sections, and the composite result of the two in series. and for f∝ < f mf / f α = 2sinh−1 2 2c f / f∞ − 1 The value of α may be determined from the expression. Figure is a plot of a against f/fc for m=0.6, which gives a value of f ∝ equal to 1.25 times the cutoff frequency fc. The great increase in sharpness of cutoff for the m-derived section over the prototype is apparent. The higher attenuation over the whole attenuation band obtained by use of a prototype section and an m-derived section in series as a composite filter is also readily seen. Again following the procedure of section, the phase shift constant β may be determined, in the pass band, from β = 2sin−1 = 2sin−1

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Z1 4Z2 mf / fc

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In the attenuation band, up to f∝, β has the value π . Above f∝ the value of β drop to zero, because the shunt arm becomes inductive above resonance. The phase shift of the m-derived section is plotted as a function of f/fc in fig.

Figure: Variation of phase shift β , for the m-derived filter This material demonstrates the ability of the m-derived section to overcome the lack of a sharp cutoff in the simple prototype filter. Although it may be note that the sharpness of cutoff increases for small values of m, the attenuation beyond the point of peak attenuation becomes smaller for small m. This emphasizes the necessity of supplementing the m-derived section with a prototype section in series to raise the attenuation for frequencies well removed from cutoff. 7. Explain the derived π section filter. An m-derived π section may also be obtained. The characteristic impedance of the π section is Z1Z2 Z0π = Z1Z2 ( 1+ Z1 / 4Z2 ) The characteristic impedances of the prototype and m-derived sections are to be equal so that they may be joined without mismatch. By use of the transformation for the shunt arm, Z2 ' =

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Z2 m


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Figure: (a) Usual symmetrical π section; (b) the m-derived π filter. it is possible to equal the characteristic impedances as Z1 'Z2 / m Z1Z2 = Z1 ( Z2 / m) ( 1+ Z1 'm/ 4Z2 ) Z1Z2 ( 1+ Z1 / 4Z2 ) from which Z1 ' =

1 1 1 + 4m mZ1 Z2 1− m2

It is apparent that the series arm Z1 ' is represented by two impedances in parallel, 2 one being mZ1, the other being 4m/ ( 1− m ) Z2 in value. Equations and thus give the values to be used in designing the m-derived π section. The circuit is drawn in fig. 8. Explain the Band-pass filters. Occasionally it is desirable to pass a band of frequencies and to attenuate frequencies on both sides of the pass band. The action might be thought of as that of low-pass and high-pass filters in series, in which the cutoff frequency of the low-pass filter is above the cutoff frequency of the high-pass filter, the overlap thus allowing only a band of frequencies to pass. Although such a design would function, it is more economical to combine the low-and high-pass functions into a single filter section.

Figure: (a) Band-pass filter network; (b) reactance curves showing VEL TECH


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possibility of two bands. Consider the circuit of (a), Fig. with a series-resonant series arm and an antiresonant shunt arm. In general, the reactance curves show that two pass bands might exist. If, however, the antiresonant frequency of the shunt arm is made to correspond to the resonant frequency of the series arm, the reactance curves become as shown in fig. and only one pass band appears. For this condition of equal resonant frequencies,

Figure: Reactance curves for the band-pass network when resonant and antiresonant frequencies are properly adjusted. ω02L 1C1 = 1 = ω02L 2C2 L 1C1 = L 2C2

or

The impedances of the arms are 2  1  ( ω L 1C1 − 1) Z1 = j ωL 1 − =j ωC1  ωC1  jωL 2 ( − j/ ωC2 ) jωL 2 Z1 = = j( ωL 2 − 1/ ωC2 ) 1− ω2L 2C2

That a network such as (a), Fig. is still a constant-k filter is easily shown as Z1Z2 = −

L 2 ( ω2L 2C1 − 1)

C1 ( 1− ω2L 2C2 )

and if L 1C1 = L 2C2 , then Z1Z2 =

L 2 L1 = = R k2 C1 C 2

Thus the previously developed theory still applies VEL TECH


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At the cutoff frequencies, Z1 = −4Z2 Multiplying by Z1 gives Z12 = −4Z1Z2 = −4Rk 2 from which the value of Z1 at the cutoff frequencies is obtained as Z1 = ± j2R k so that

Z1 at lower cutoff f1 = - Z1 at upper cutoff f2

The reactance of the series arm at the cutoff frequencies then can be written by use of the above as 1 1 − ω1L 1 = ω2L 1 − ω1C1 ω2C1 ω 1− ω12L 1C1 = 1 ( ω22L1C1 − 1) ω2 Now from eq. 1 L 1C1 = 2 ω0 so that eq. may be written as  f12 f1  f22 1− 2 =  2 − 1 f0 f2  f0  2 f0 ( f1 + f2 ) = f1f2 ( f1 + f2 ) f = f1f2 or the frequency of resonance of the individual arms should be the geometric mean of the two frequencies of cutoff. If the filter is terminated in a load R = Rk, as is customary, then the values of the circuit components can be determined in terms of R and the cutoff frequencies f1 and f2. At the lower cutoff frequency,

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1 − ω1L 1 = 2R ω1C1 1−

f12 = 4πRf1C1 f02

In view of Eq., the expression for C1 becomes C1 =

f2 − f1 4πRf1f2

It follows, then, from eqs. and that L1 =

R π ( f2 − f1 )

From equation it is possible to obtain the values for the shunt arm as R ( f2 − f1 ) 4πf1f2 L 1 C 2 = 12 = R πR ( f2 − f1 ) L 2 = C1R 2 =

This completes the design of the prototype band-pass filter.

Figure: m-derived band-pass section An m-derived band-pass section is also possible. Use of the transformation relation developed in section leads to a network of the form of fig. The shunt arm then consists of series-resonant and antiresonant circuits in series. Plotting reactance curves for these two circuits and adding to obtain the reactance variation of the shunt arm, Z2 of the filter, gives the dashed curve of fig. The antiresonant frequency of the arm as a whole must, by previous reasoning, be f0 of the filter. The reactance curve for Z2 then shows that the shunt arm becomes resonant at a frequency below f 0 and VEL TECH


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again at a frequency above f0.

Figure: Reactance curves for the shunt arm of the m-derived band-pass section. At these frequencies the network is short-circuited, and thus they are frequencies of high attenuation, f∝. These frequencies of high attenuation are placed on each side of the pass band, and the m-derived section may be used to increase the attenuation near cutoff, as for the high-or low-pass cases. At one f∝, the reactances Xr and Xar are equal and opposite, so that j ( jω∞ L 2 / m) ( − j/ ω∞ mC2 )  1− m2  jω∞  =  L1 − 2 ω∞  4m/ ( 1− m )  C1 j( 1− / ω∞ mC 2 − ω∞ L 2 / m)  4m  ω 2L C 1− m2 ω∞ 2L 1C1 − 1) = 2∞ 2 1 ( 4 ω∞ L 2C 2 − 1 In view of the fact that L 1C1 = L 2C2 =

1 L ω02

Equation becomes 2

1− m2  f∞ 2  2 2  2 − 1 = 4π f∞ L 2C1 4  f0  The term L2C1 can be evaluated as a function of f1 and f2 from Equations and f − f ( f − f ) R  ( f − f ) L 2C1 = 2 1  2 1  = 2 21 4 4πRf1f2  4πf1f2  16π f0

2

Equation then reduces to VEL TECH


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( 1− m ) ( f 2

∞

( f2 − f1 )

f∞ 2 −

1− m2

2

− f1f2 ) = f∞ 2 ( f2 − f1 ) 2

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2

− f1f2 = 0

Solving for the values of the frequencies of peak attenuation, f∞ =

f2 − f1

( f2 − f1 )

±

2 1− m2

2

4( 1− m2 )

+ f1f2

It is apparent that the radical is larger than ( f2 − f1 ) 2 1− m2 , and thus one root would appear as a negative frequency that has no physical significance here. Thus the expression for f∝ should be reversed so that the two frequencies of peak attenuation are f∞1 = f∞ 2 =

( f2 − f1 )

2

4( 1− m

2

( f2 − f1 )

)

+ f1f2 −

2

4( 1− m

2

)

+ f1f2 +

f2 − f1 2 1− m2 f2 − f1

2 1− m2 Equation may be solved to determine the value of m, giving f ( f − f )  m = 1−  ∞ 2 2 1   f∞ − f1f2  =

2

1− ( f∞ 2 − f12 ) ( f∞ 2 − f22 ) f∞ 2 − f1f2

The value of m may be chosen to place either one of the two frequencies of peak attenuation at a desired point, the other frequency of peak attenuation then being definitely fixed. That this is true may be chosen by forming the product for f∞1f∞ 2 from Equations and: f∞1f∞ 2 =

f22 + 2f1 + f12 − 4m2f1f2 − f22 + 2f1f2 − f12 = f1 f2 4( 1− m2 ) f∞1f∞ 2 = f0

Thus f0 is the geometric mean of the frequencies of peak attenuation and, by Equation, of the cutoff frequencies as well. If m is selected to place f 01 at a desired point, then by Equation, f∞ 2 and f∞ 2 by the use of two m’s or an mm’-derived filter, as shown by Zobel (Reference 2). VEL TECH


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An m-derived T section, rearranged as a π , may be split into two half sections and used as terminating half sections. If m is given the value 0.6, then satisfactory impedance matching conditions are maintained over the pass band. This usage follows the previously developed theory for low-or high-pass sections. 9. Explain the band-elimination filters. If the series and parallel-tuned arms of the band-pass filter are interchanged, the result is the band-elimination filter of (a), fig. That this circuit does eliminate or attenuate a given frequency band is shown by the reactance curves for Z1 and -4Z2 at (b). The action may be thought of as that of a low-pass filter in parallel with a highpass section, in which the cut-off frequency of the low-pass filter is below that of the high-pass filter.

Figure: (a) Band-elimination filter; (b) reactance showing action of bandelimination section. As for the band-pass filter, the series and shunt arms are made antiresonant and resonant at the same frequency f0. Again, it is possible to show that R k2 =

L1 L2 = C 2 C1

f0 = f1f2 and that At the cutoff frequencies, Z1 = −4Z2 , Z1Z2 = 4Z2 2 = Rk 2 jR Z2 = ± k 2 If the filter is terminated in a load R = Rk, then at the lower cutoff frequency,  1  jR Z2 = j − ω1L 2  =  ω1C 2  2 VEL TECH


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1 Since L 2C 2 = ω 2 , 0 1−

f 12 f02

C2 =

= πf1C 2R

1  f2 − f1    πR  f1f2 

In view of the fact that f0 = f1f2 =

then

L2 =

1 2π L 2C 2

R 4π ( f2 − f1 )

By use of Equation, the values for the series arm are obtained as R ( f2 − f1 ) πf1f2 1 C1 = 4πR ( f2 − f1 ) L1 =

Section of the m-derived form may also be obtained. 10. Explain the crystal filters. The lattice structure can also be shown to have filter properties. Considering the network of fig.

Figure: Lattice filter section

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( Z / 2 + 2Z2 ) Z∞ = 1 2( Z1 / 2 + 2Z2 ) 2

Z1 + Z2 4 Z1Z 2 Z1Z2 Z∞ = + Z1 / 2 + 2Z2 Z1 / 2 + 2Z2 Z1Z2 = Z1 / 4 + Z2 =

The characteristic impedance of the fig. Lattice filter section. Lattice section then is Z0L = Z∞ Zsc = Z1Z2 Thus if the section elements are reactive, Z0L is real, or a pass band exists for frequencies for which Z1 and Z2 are of opposite sign. Over ranges where Z1 and Z2 have the same sign, an attenuation band exists. Propagation can be investigated further by noting that tanh γ =

 Zsc Z1  1 =   Zoc Z2  1+ Z1 / 4Z2 

It may be noted that Z0L depends on the product of Z1 and Z2, whereas γ depends on the ratio of Z1 to Z2. This feature permits somewhat greater versatility in design of the lattice section over the T or π section, especially for filters in which certain of the elements are constructed of piezoelectric crystals. These crystals have a resonant frequency of mechanical vibration dependent on certain of their dimensions; and because of the very high equivalent Q of the crystals, it is possible to make very narrow band filters and filters in which the attenuation rises very rapidly at cutoff. The equivalent electric circuit of a quartz mechanical-filter crystal is shown in fig. (a), which shows a possibility of both resonance and antiresonance occurring. The inductance Lx is very large, being in henrys for crystals resonating near 500 kc, so that while Rx may approximate a few hundred or few thousand ohms, the effective Q may be in the range of 10,000 to 30,000. Considering the properties of resonant circuits, such as Q would provide a band width of 20 to 50 cycles at 500 kc.

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Figure: (a) Equivalent electrical circuit for a piezoelectric crystal; (b) reactance curves for the circuit of (a). The resistance of the crystal is due largely to mechanical damping introduced by the electrodes and by the surrounding atmosphere. By placing a crystal in an evacuated container, the value of Q can be notably increased. The electrodes are normally electroplated onto the crystal faces and need not introduce much damping. Capacitance Cs is the equivalent series capacitance of the crystal forming a resonant by the crystal electrodes. The values of Cs, and Cp are such that Cp >> Cs, so that resonant and anti-resonant frequencies of the circuit lie very close together, differing by a fraction of 1 per cent of the resonant frequency. The reactance curve sketch of fig. (b) shows the resonant frequency below the antiresonant one. By placing adjustable capacitors in parallel with the crystal, Cp can be increased, resulting in the antiresonant frequency being moved closer to the resonant point. Since the crystal represents either a resonant or antiresonant circuit, it may be used to replace the normal elements of the band-pass or band-elimination filter. as previously shown for band-pass action, the resonant frequency of one arm must equal the antiresonant frequency of the other arm.

Figure: (a) Circuit of a lattice crystal filter with series inductors and parallel capacitors; (b) the electrical equivalent of (a). The pass band with crystal elements will then be found to extend from the VEL TECH


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lowest crystal resonant frequency to the highest crystal antiresonant frequency, or a width of pass band equal to twice the separation of the resonant and antiresonant frequencies of one crystal. This range will result in a pass band a fraction of 1 per cent wide. The band width can be reduced by putting adjustable capacitors in parallel with the crystal, furnishing a means of adjustment of the width of the pass band. By the addition of coils in series with the crystals the pass bands may be widened. Since the added coils have Q values very much below those of the crystals, there will be some loss in sharpness at cutoff. A circuit including series coils is shown in fig. (a), with its equivalent drawn at (b). The reactance curves for the A and B portions of this circuit are drawn in fig. (a), which shows how the resonances and antiresonances are arranged. The presence of the series coil adds an additional resonance, and the pass band exists from the lowest resonance of one crystal to the highest resonance of the other. If f1 and f2 are the frequencies of resonance of one of the circuits and fR is that of the antiresonance, then C f1,2 = fR 1m s Cp The separation of f1 and f2 represents two-thirds of the pass band and is seen to depend on the C s / C p ratio. Since C s / C p may be of the order of 0.01, it can be seen that the separation of f1 and f2 be of the order of 0.10 fR, or 10 per cent of the resonant frequency. By placing coils in series with the crystals, it has been possible to widen the pass band considerably. By adjustment of Cp it is then possible to narrow the band to any desired amount.

Figure: (a) Reactacne curves for the circuit of fig. (a); (b) attenuation curves for that circuit. Thus the use of coils permits the bands to be widened to pass speech frequencies, and crystal filters are quite generally used to separate the various VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 41

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channels in carrier telephone circuits, in the range above 50 kilocycles. 11. Design a composite low pass filter to meet the following specifications. The fitler is to be terminated in 500 ohms resistance and it is to have a cutoff frequency of 1000 Hz with very high attenuation at 1065 and 1250 Hz. Solution: Given R= 500 ; fc = 1000 Hz; f∞ = 1065 Hz. f∞2 = 1250 Hz. i) Design of low pass constant – kT section R 500 = = 0.159 H ∏ fc ∏×1000 1 1 C= = = 0.63µ f ∏ fc.R ∏×1000 × 500 I=

The assembly of this filter is a shown below with inductance L/2 in each series arm. ii. Design of m – derived low pass T- section (Ref fig (b) a) for fω 2 = 1065 2

2

 fc  1000  m1 = 1 −   = 1 −  = m1 = 0.344 1065   f ∞t 

The components of m- derived lowpass T – section filter for m = 0.344 are

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mL 0.344 × 1.59 = = 0.0273H 2 2 mc = 0.344 × 0.636 = 0.218µ f − ( 0.344 ) 1 − m2 .L = × 0.159 4m 4 ( 0.344 ) 2

= 0.102 H The assembly of this m – derived lowpass t – section filter is as shown below 2

2

 fc  1000  m2 = 1 −   = − 1250   f∞ 2  The components of m – derived low pass T – section filter for m = 0.6 are mL 0.6 × 0.159 = = 0.048 H 2 2 mc = 0.6 × 0.636 = 0.3816 µ f 1 − m 2 1 − ( 0.6 ) ⊥ × 0.159 4m 4 ( 0.6 ) 2

= 0.0424 H The assembly of the m – derived low pass T – section filter for m = 0.6 as shown below.

Assembling all the three sections we will get the desired composite filter.

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In the 3rd section i.e., for m = 0.6, the filter is divided into sections so these value get changed. 1 − m2 2× .L = ( 0.0424 ) × 2 = 0.0848 4m mc 0.218µ f and = = 0.109µ f 2 2 Combine the elements with ever possible. The series inductors may be added and the resulting final design I as shown below. 12. i. Describe a prototype T section band stop filter. ii. Determine the formulae required for designing band stop filter. iii. Explain the advantages of m – derived band stop fitler. i. Band stop filter: A band stop or bad elimination filter attenuates a certain range of frequencies and passes all other frequencies there fore a band stop action may be thought of as that of a low pass filter in parallel with a high pass filter in which the cut off frequency of low pass filter is below that of high pass filter. ii.

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1 L1C1

L1C1 = L2 C2 Z1 = jω L1 +

1 jωC

= j ( ω L1 − 1/ ωC1 ) = j ( ω 2 L1C1 − 1/ ωC1 )

jω L2 ( 1/ jωC2 ) jω L2 = jω L2 + 1/ jωC2 1 − ω 2 L2 C2 If the filter is to be constant K type Z1Z2 = RK2 =

 ω 2 L1C1 − 1  jω L2 j= = RK 2  2  ωC1  1 − ω L2 C2 L1C1 = L2 C2 L2 L1 = − RK 2 C1 C2 Z1 = 1; Z1 = −4 Z 2 L1Z 2 Z12 = −4 Z1 Z2 = −4 RK 2 Z1 at lower cut off fi = Z1 at upper cut off f  −1  1 + jω L1 = −1  2 + jω2 L1  jω1C1  jω C1  1 −1 − ω1 L1 = + ω2 L1 ω1C1 ω2 C1

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ω1 2 ( ω2 L1C1 − 1) ω2

1 ω02

ω12 ω1  ω22  1− 2 =  − 1 ω0 ω2  ω02  f2 f f2  1 − 12 = 1  22 − 1 f0 f 0  f0 

f 02 ( f1 + f2 ) + f1 f2 ( f1 + f2 ) fr =

f2 f2

Z1 = −2 JRk 1 + jω1 L1 = −2 jRk jω1C1  −1  j + ω1 L1  = 2 jRk  ω1C1  2 1 − ω1 C1 L1 = 2 Rk ωC1

ω12 = 2 Rk 2 ∏ f1C1 ω02 f 2 f1 = 4 ∏ Rk f1C fr

1−

C1 =

f 2 − f1 4 ∏ Rk f1 f 2

L1C1 = L1 =

1 1 = 2 2 ω0 4 ∏ f1 f2

Rk ∏ ( f 2 f1 )

L2 = C1 Rk2 = C2 =

Rk ( f 2 − f1 ) 4 ∏ f2 f2

L1 1 = 2 Rk π Rk ( f 2 − f1 )

iii) Advantage of m – derived filters VEL TECH


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1. The attenuation does not increase rapidly beyond the cut off frequencies. 2. Characteristic impedance varies widely in the transmission or pass band, from the desired value the design impedance Rk. 13. For a given T – section low pass filter, determine the cut-ff frequency and normal characteristic impedance Re. Solution: Given : L= 80 mH ; c = 0.02 µf 1 ∏ LC Cut off frequency fc =

fc =

1 ∏

( 80 ×10−3 ) ( 0.02 ×106 )

Hz

f c = 7.962 × 103 Hz L 80 ×10−3 R0 = = C 0.02 × 10−6 R0 = 2 ×103 ohms 14. Design a constant – k low pass T and II – sections fitlers having cut-off frequency = 3000 Hz and nominal characteristic impedance R0= 600 ohm. Solution: Given fc = 3000 Hz; Ro = 600 L=

R0 600 × 1 = ∏ f c ∏×3000 × 5

L = 63.68mH 1 1 C= = ∏ R0 f c ∏×600 × 3000 C = 0.1753µ f Hence the require T and II- section lowpass filter are 63.68 mH

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15. Design a constant – k low pass filter having Fc= 2000 Hz and nominal characteristic impedance R0=600Ω. Also find the frequency at which this filter offers attenuation of 19.1 dB. Solution: Given: fc = 2000 Hz: R0 = 600 ohm ; α = attenuation = 19.1 dB. R 600 L= 0 = ∏ f c 600 × ∏×2000 L = 95.54mH 1 1 C= = F ∏ R0 f c ∏×600 × 2000 C = 0.263µ F Attenuation of 19.1 dB is expressed in nepers as 19.1 α= = 2.2 nepers 8.686

α = 2 cosh -1 ( f / fc ) f 2.2 f = cosh : = 1.6685 fc 2 fc ∴ F = 1.6685 × 2000 f = 3337 Hz 16. Design a T- section constant – K high pass filter having cut off frequency of 10KHz and nominal characteristic resistance of Ro=600 om. Find i) its characteristic impedance and phase constants at 25 KHz and ii) Attenuation 5KHz. Solution: Given Fc = 10 KHz ; R0 = 600 ohm R0 600 L= = 4 ∏ f c 4 ∏×10 ×103 L = 4.77 mH VEL TECH


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1 1 = F 4 ∏ .R0 . f c 4 ∏×10 ×103 × 600

C = 0.0132663µ F Each capacitor in the series arm T – section is 2C= 0.02652 µF and the inductor in shunt arm is L=4.777 mH. At f= 25 KHz 2

Z OT

 fc  = R0 1 −    f 

Z OT

 10  = 600 1 −    25 

2

Z OT = 545Ω

β = 2 sin -1 ( f c / f ) = 2sin − ( 10 / 25 ) β = 47.2 or β =47.52 × ∏ /180=0.0824 radians In the attenuation band, α is given as,

α = 2 cosh -1 ( f c / f ) nepers = 2 cosh-1 ( 10 / 5 ) nepers α = 2.6 nepers 17. Determine a prototype band pass fitler section having cut off frequencies of 2000 Hz and 500 Hz and nominal characteristic impedance of 600 ohms. Solution: R0 600 L1 = = ∏ ( f 2 − f1 ) ∏ ( 5000 − 2000 ) L1 = 63.68mH L1 = 31.84mH 2 ( 5000 − 2000 ) f 2 − f1 C1 = F= F 4 ∏ Ra , f1 , f 2 4 ∏×600 × 5000 × 2000 C = 0.0381µ F ICI = 0.0762 µ F

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L2 =


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R0 ( f 2 − f1 ) 600 × ( 5000 − 2000 ) = H 4 ∏ .R0 . f c 4 ∏×5000 × 2000

L2 = 14.33mH C2 =

1 1 = ∏ R0 ( f 2 − f1 ) ∏×600 × ( 5000 − 6000 )

C2 = 0.1769 µ F Hence the band pass filter is as shown below.

UNIT – II PART – A 1. What is transmission line? Energy can be transmitted either by the radiation of free electromagnetic waves as in the radio or it can be constrained to move or carried in various conductor arrangement known as transmission line. It is a conductive method of guiding electrical energy from one place to another. 2. What are lumped parameters and distributed parameters? The parameters which are physically separable and can be shown to be at one place in the circuit in the lumped form are lumped parameters. The parameters which are not physical be separable and are distributed all over the length of the circuit like transmission line are called distributed parameters. 3. State the properties of infinite line. 1. No waves will ever reach receiving end hence there is no reflection. 2. The Zo of the sending end decides the current flowing when voltage is applied VEL TECH


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ZR has no effect on the sending current. 4. What is short line? The short line means a practical line with finite length and the word short does not reflect anything about the actual length of the line. 5. Sketch the group of Zo agains w. R+jwL G+jwC Zo =

w → 0, R L Zo → and w → ∞Z0 → G C R Practically is alwas higher then G when

L C

6. What is called an infinite line? The analysis of the transmission of the electric waves along any uniform and symmetrical transmission line can be done in-terms of the result existing for an imaginary line of infinite length, having electrical constants per unit length identical to that of the line under consideration. 7. Discuss the importance of smooth line. A line terminated in its characteristic impedance Ro is called properly terminated line which acts as a smooth line. Because of proper termination, there is no mismatch of impedance. Hence no reflection takes place. Thus no standing waves are produced. Then the maximum power transfer from generator to load is possible. 8. Draw the equivalent circuit of a unit length of transmission line.

9. Define characteristic impedance. VEL TECH


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When a finite transmission line is terminated with ZO and the input impedance is also ZO, then ZO is known as characteristic impedance. i.e the input impedance of an infinite line is characteristic impedance of line, 10. Define return loss. Return loss is defined as the ratio of power at the receiving end due to incident wave to power due to reflected wave by the load. Returns loss = 20 log

Z R + Zo Z R − Zo

db

11. Define reflection factor? Reflection factor is defined as the ratio which indicates the change in current in the load due to reflection at the mismatched junction. 2 Z R Zo K= Z R + Zo 12. Express reflection factor in terms of impedance. Reflection factor =

2 Z1Z 2 Z1 + Z 2

where Z1 and Z2 are the impedance seen looking both ways at any junction. 13. Find out the value of Reflection coefficient (K) for the following; i) ii) iii)

Properly matched condition - |K| = 1 Short circuited line - |K| = 0 Open circuited line -|K| = 0

14. What is Campell’s equation? It makes possible the calculation of the effects of loading coils in reducing attenuation and distortion. Zc CoshN λ 1 = sinh N λ + CoshN λ 2 Zo when ZC → loading coil impedance Zo → Characteristic impedance N → Length of Transmission line section λ→ Propagation constant 1 λ → Modified propagation constant. VEL TECH


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15. What are the basic application of transmission lines? i. ii. iii.

It is used to transmit energy It can be used a circuit elements like L>C. and resonant circuits. It can be used as filters, transformers, measuring devices.

16. State the condition for minimum attenuation with L and C variable. CR G LG With ‘C; variable , C = R In general, RC = LG 17. Give the types of transmission line With ‘L’ variable , L =

The commonly used transmission lines are: i) Open wire line. ii) Co-axial line iii) Strip line iv) Waveguides v) Optical fibers. 18. Write short notes on transmission line. The open wire transmission line consists of two conductors spaced at a certain distance apart. The spacing between the conductors is large in comparison to the diameter of the line conductors. As a result these lines can operate at higher frequencies. This type of line is used in transmission of electric power, telegraphy and telephony, The conductor of open wire line is as shown below.

19. Give the advantages of open wire transmission line. VEL TECH


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The advantages of open wire line are: (i) (ii) (iii)

Simple to construct and low cost. Insulation between the line and conductor is air so, the dielectric loss is extremely small. It is balanced with respect to ground

20. What are the disadvantages of open wire transmission line? (i)

These lines are unsuitable for use at frequencies above 100MHz. Because energy loss takes place due to radiation. 21. Write short notes on co-axial line (or) Coaxial cable. To avoid radiation losses taking place in open wire lines at high frequencies a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. 22. Give the advantages and disadvantages of co-axial transmission line. Advantages: (i) The electric and magnetic fields are confined with in the outer conductor so the radiation losses are eliminated. (ii) It provides outer shielding from outer interfacing signals. (iii) The co-axial line can be used up to the frequency range of about 1 GHz for transmission of signals. Disadvantages: (i) (ii)

They are costlier than open wire line. Beyond 1GHz these cables cannot be used because losses in the dielectric increases with frequency.

23. What do you mean by Waveguide? Wave guides are hollow conducting tubes of uniform cross section used for U. H. F. Transmission by continuous reflection from the inner walls of the guide. 24. Give the application of microwaves (i)

Satellite communication (ii) Telemetry (iii) Transmission of video signals (iv) Microwave oven.

25. What are the advantages and disadvantages of Waveguide? The advantages of waveguides are: VEL TECH


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i) In wave guides, no power is lost through radiation because the electric and magnetic fields are confined to the space within the guides. ii) The dielectric loss is negligible. iii) Frequencies of the wave higher than 3GHz can be easily transmitted iv) Several modes of electromagnetic waves can be propagated with in a single Waveguide. The Disadvantages of wave guides are: (i) (ii)

Cost of the wave guide is so high The wave guide walls should be specially plated to reduce resistance to avoid skin effect and power loss.

26. What is an optical fiber? An optical fiber is a dielectric wave guide that operates at optical frequencies. It confines electromagnetic energy in the form of light within its surfaces and guides the light in a direction parallel to its axis. 27. What are the advantages and disadvantages of optical fiber? Advantages: (i) (ii) (iii) (iv)

Low transmission loss and very high band width Small size and weight No radio frequency and electromagnetic interference Ruggedness and flexibility

Disadvantages: (i) (ii)

It is difficult to run cables where the bending occurs. Different specialized techniques have to be followed to join ends of two cables.

28. What are the four important parameters of a transmission line? The main four parameters of transmission line are Resistance Inductance Capacitance Conductance

-

R L C G

29. Define propagation constant of uniform line. The propagation constant per unit length of a uniform line is defined as the natural VEL TECH


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logarithm of steady state vector ratio of current or voltage at any point, to that at a point unit distance further from source, when the line is infinitely long 30. When a transmission line is said to be uniform? A line is said to be uniform, when the primary constants R, L, C and G are uniformly distributed along the entire length of transmission. 31. Name the secondary constants of transmission line. Characteristic impedance Zo and Propagation constant P (or) Secondary constants of transmission line

γ

are the

32. What is the value of characteristic impedance of open-wire line? The characteristic impedance of open-wire line is S Zo = 276 log 10 ohms r Where ’S ‘ is the spacing between two wires-centre to centre ‘r’ is radius of either of the wire. 33. What is the value of characteristic impedance of coaxial cable? The characteristic impedance of co-axial cable Is Zo = 138log10

D ohms d

Where ‘D’ is inner diameter of outer conductor ‘d’ is diameter of inner conductor 34. Define velocity of propagation. Velocity of propagation is defined as the velocity with which a signal of single frequency propagates along the line at a particular frequency ‘f’ . It is denoted as Vp and its unit is km/sec. 35. Define group velocity. Group velocity is defined as the velocity of envelope of a complex signal. (or) It is a velocity with which a signal produced by variation of a steady-state wave or by introduction of group frequencies. It is denoted as Vg.

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The process of achieving the condition RC = LG either by artificially increasing ‘L’ or decreasing ‘C’ is called loading of a line. 37. What are the advantages of lumped loading? The advantages of lumped loading are • There is no practical limit to the value by which the inductance can be increased. • Cost is small • Hysterisis and eddy current losses are small. 38. What are the disadvantages of reflection? 1. Reduction in efficiency, 2. If the attenuation is not large, then the reflected wave appears as echo at the sending end. 3. The part of received energy is rejected by the load, hence output reduces. These are the disadvantages of reflection. 39. What is frequency distortion? A complex applied voltage, such as a voice voltage containing many frequency will not have all frequencies transmitted with equal attenuation of the received wave form will not be identical with the input wave form at the sending end. This variation is called frequency distortion. 40. What is phase or Delay Distortion? All frequencies applied to a transmission line will not have the same time of transmission, some frequencies being delayed more than others. For an applied voicevoltage wave, the received wave will not be identical with the input wave form at the sending end, since some components will be delays more than the other. This phenomenon is called Delay distortion. 41. Define Reflection co-efficient. The ratio of amplitudes of the reflected of incident voltage waves at the receiving end of the line is called reflection co-efficient. K=

Reflected voltage at load Incident voltage at load

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It is defined as number of repers or decibels by which the current in the load is changed by the insertion. 43. What is return loss? Return loss is defined as the ratio of power at receiving end due to incident wave and power due to reflected wave in the load. 2 + 2o Reflection Loss = R 2R − 2o It is the reciprocal of Reflection co-efficient 44. Write down the expression for transfer impedance ZT =

Es 2R + 2o γl = e + Ke−γl IR 2

(

)

ZT = zR cos h γl+zo sin h γl 45. What do you mean by Lumped circuits? The network where in the resistance, inductance and capacitance are individually concentrated or lumped at discrete points in the circuit is called Lumped circuit. 46. Write short notes on co-axial cable. One conductor is a hollow tube, the second conductor being located inside of co-axial with the tube.

PART – B 1. Derive the General solution a transmission line. A transmission line is a circuit with distributed parameters hence the method of analyzing such circuit is different than the method of analysis of a circuit with lumped parameters. It is seen that the current and voltage varies from point to point along VEL TECH


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the transmission line. The general solution of a transmission line includes the expressions for current and voltage at any point along a line of any length having uniformly distributed constants. The various notations used in this derivation are, R = Series resistance, ohms per unit length, including both the wires. L = Series inductance, henrys per unit length. C = Capacitance between the conductors, farads per unit length. G = Shunt leakage conductance between the conductors, mhos per unit length. ω L= Series reactance per unit length. ω C = Shunt susceptance in mhos per unit length Z = R + J ω L = Series impedance in ohms per unit length. Y = G + j ω C = shunt admittance in mhos per unit length. S = Distance out to point of consideration, measured from receiving end. I = Current in the line at any point. E= Voltage between the conductors at any point. l = Length of the line. The transmission line of length l can be considered to be made up of infinitesimal T section. One such section of length ds is shown in the Fig.4.17. It carries a current I. The point under consideration is at a distance a from the receiving end. The length of section is ds hence its series impedance is Zds and shunt admittance is Yds. The current is I and voltage is E at this section. The elemental voltage drop in the length ds is dE = I Zds dE ∴ = I Z ………(1) ds the leakage current flowing through shunt admittance from one conductor to other is given by dI = EY ds ∴

dI = EY ds

…….(2)

Differentiating equation (1) and (2) with respect to S we get VEL TECH


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d 2E dI =Z 2 ds ds and d 1I dE =Y 2 ds ds This is because both E and I are function of S. d 2E = ZEY ..........(3) ds 2 IZ and d 2I = YIZ .....(4) ds 2 The equations (3) and (4) are the second order differential equations describing the transmission line having distributed constants, all along its length. It is necessary to solve these equations to obtain the expression of E and I. Replace the operation d/dS by m hence we get. (m2 –ZY) E = 0 but E ≠ 0 m = ± ZY …….. (5) Same result is true for the current equation. So, there exists two solutions for positive sign of m and negative sign of m. The general solution of the equations for E and I are, E = Ae ZYS + Be − ZYS ..........(6) I = (e ZYS + 1)e − ZYS ...........(7) Where A, B, C and arbitrary constants of integration. It is now necessary to obtain the values of A, ,B, C and D. As distance is measured from the receiving end S = 0 indicates the receiving end E = IER and I = IR at S = 0 Substituting in the solution, VEL TECH


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ER = A + B ….. (8(a)) IR = C + D …..(9(b)) Same condition can be used in the equations obtained by differentiating the equations (6) and (7) with respect to S. dE = A ZY e ZY s + B( − ZY )e − ZY s ds and dI = C ZY e ZYs + D(− ZY )e − ZYs ds But dE dI = IZand = Ey ds ds ∴ IZ = A ZY e Zys − B (− Zy )e − Zys ...(9) andEY = C ZY e ZYs − D( − ZY )e − ZYs ..(10) I=

A B ZY e Zys − ( − ZY )e − ZYs Z Z

i.e.I = A

andE = C

Y Y e ZYS − B e − ZYS ....(11) Z Z Z Z e ZYS − D − ZYS .....(12) Y Y

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Now use S = 0 , E = ER and I - IR ∴ IR = A

Y Y −B ............(13( a) Z Z

andER = C

Y Y −D .........(13(b) Z Z

The equation 8a, 13a, 13b are to be solved simultaneously to obtain the values of the constants A, B, C and D. Now while solving these equations use the results, ZR =

ER R + jω L Z andZo = = IR G + jω C Y

Hence the various constants obtained, after solving the equations simultaneously are, A=

ER I R + 2 2

B=

ER I R − 2 2

Z ER  Z O  = 1 +  .......(14) Y 2  ZR  Z ER  Z O  = 1 −  ........(15) Y 2  ZR 

C=

I R ER + 2 2

Y IR = Z 2

 ZR 1 +  ZO

D=

I R ER − 2 2

Y IR = Z 2

 ZR  1 −  ............(17)  ZO 

  .........(16) 

Hence the general solution of the differential equation is,  Z O   ZO  1 +  e ZYS + 1 −  ZR  Z R 

E=

ER 2

E=

 ZR I R  Z R   1 +  e ZYS +  1 − 2  Z O   ZO

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   e ZYS  ..(18)      e − ZYS  ..(19)  


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Taking LCM as ZR and taking


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Z R + ZO out from equation (18) ZR

 E ( Z + ZO )  E ( Z − ZO ) I = R R e ZYs R R e − ZYs  ..(20) 2Z R 2 Z R ( ZR + ZO )   Taking LCM as Zo and taking

Z R + ZO out from equation (19) ZR

 I ( Z + ZO )  ( Z − ZO ) I = R R e ZYS R e − ZY S  ..(21) 2ZO ( Z R + ZO )   The negative sign is used to convert Zo – ZR to ZR - Zo The equation (20) and (21) is the general solution of a transmission line. Another way of representing the equation is E=

ER  ( Z R + ZO ) e ZYs + ( ZR − ZO )e ZY s  2Z R 

E=

ER  Z Re ZYs + ZOe ZYs + ZRe − ZYs −  2Z R 

Z Oe − ZYs ER E = Z R hence R = I R IR ZR But

 e ZYs + e − ZYs  ∴ E = ER  + 2  

 e ZYs + e − ZYs  I R ZO   ......(22) 2  

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 e ZYs + e − ZYs  I = IR  + 2  

and

ER ZO

 e ZYs + e − ZYs    .............(23) 2  

e ZYs + e − ZYs = COSh ZYsand 2 But e ZYs + e − ZYs = Sinh ZYs 2 ∴ E = ERCosh

(

)

ZYs + 1R ZO Sinh

(

)

ZYs ....(24)

and I I R Cosh

(

)

ZYs ER / ZO sinh

(

)

ZYs ..........(25)

The equation (24) and (25) give the values of E and I at any point along the length of the line. Important Note: The similar equations can be obtained in terms of sending and voltage Es and Is. If X is the distance measured down the line from the sending end then, X=1–s And the equation (24) and (25) get transferred in term Es and Is as E = ES Cosh I = I S Cosh

(

(

)

ZYx + I S ZO Sinh

)

(

ZY x + ES / ZO Sinh

ZYx

(

)

ZYx

)

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I = IR cosh ( γ s) +ER /Zo sin h ( γ s) And if sending end parameters are known and X is distance measure from the receiving end then, E = Es cosh ( γ x ) + Is Zo sinh ( γ x ) I = Is cosh ( γ x ) +Es/Zo sinh( γ x) Any set of equations can be used to solve the problems depending on the values given. 2. Explain the physical significance of General solution. From the qeuesed solutions, the sending end current can be obtained by substituting S = I measured from the receiving end. Es = ER cosh ( γ Is = IR [cos h ( γ

I ) + IR Zo sinh ( γ I ) ………(1) I) + ER /Zo sin h ( γ I ) ……..(2)

Now ZR = ER / IR ∴ Is = [ IR cos h ( γ I) + ZR /Zo IR sin h ( γ I ) ] ∴ Is = [ IR cos h ( γ I )+ ZR / Zo sinh ( γ I ) ] ……(3) Now if the line is terminated in its characteristic impedance Zo then, Is = [ IR cosh ( γ I ) + sinh ( γ I ) ]…. As ZR = Zo Is / IR = [ cosh ( γ I ) + sinh ( γ I ) ]= eW …….(4) This is the equation which is already derived for the line terminated in Zo. Using ER = IR ZR in equation (1), Es = ZR IR cosh ( γ I ) + IR Zo sinh ( γ I ) Es = IR [ (ZR cos h ( γ I ) + Zo sinh ( γ I ) ]…..(5) Dividing (5) by (3), I [ Z cosh(γ I ) + ZO sinh(γ I ) ] Es = R R I s I R [ cosh(γ I ) + ZR / ZO sinh(γ I )] But Es / Is = Zs

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∴ ZS =


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Z O [ Z R cosh(γ I ) + ZO sinh(γ I ) ]

[ ZO cosh(γ I ) + Z R sinh(γ I )]

When the line is terminated in Zo then ZR = Zo SO substituting in equation (6) we get Zs = Zo This shows that for a line terminated in its impedance is also its characteristic impedance.

characteristic impedance, it input

Now consider an infinite line I → ∞ Using this in equation (6) we get, ZS =

Z O [ Z R + ZO tanh(γ I ) ]

[ ZO + Z R tanh(γ I )]

and tanh(γ I ) → IasI → ∞ ∴ Z S = ZO ........(8) This shows that finite line terminated in its characteristic impedance behaves as an infinite line, to the sending end generator. Thus the equations for Ex and Ix are applicable for the finite line terminated in Zo. The equations are reproduced here for the convenience of the reader. Ex = Es e

–yx

and Ix = Is e

–yx

If in practice instruments are connected along the line then the instruments will show the magnitude Es e –yx and Is e –yx while the phase angles cannot be obtained. If the graph for Ex or Ix is plotted against x then it can be shown. This is the physical significance of the general solution of a transmission line . Its use will be more clear by studying the various cases of the line. 3. State and explain different types of distortions in line. When the received signal is not the exact replication of the transmitted signal then the signal is said to be distorted. There exists some kind of distortion in the signal. There are three types of distortions present in the transmitted wave along the transmission line. 1. Due to variation of characteristic impedance Zo with frequency. 2. Frequency distortion due to the variation of attenuation constant α with frequency. 3. Phase distortion due to the variation of phase constant with frequency. Distortion due to Zo varying with Frequency: VEL TECH


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The characteristic impedance Zo of the line varies with the frequency while the line is terminated in an impedance with does not vary with frequency in similar fashion as that of Zo. This causes the distortion. The power is absorbed at certain frequencies while its gets reflected for certain frequencies. So there exists the selective power absorption, due to this type of distortion. It is known that, ZO =

R + jω L = G + jω C

R + (1 + jω L / R ) G (1 + jω C / G )

If for the line, the condition LG += CR is satisfied the L/R = C/G and hence (1 + J ω L / R) = (1 + J ω C / G ) ∴ Z O = R / G < Oo = L / C < OoΩ For such a line Zo does not vary with frequency ω and it is purely resistive in nature. Such a line can be easily and correctly terminated in an impedance which matches with Zo at the frequencies for such a line Z R = R / Gor L / G . This eliminates the distortion and hence selective power absorption. 4. Write brief notes on lumped loading. In this type of loading the inductors are introduced in lumps at the uniform distances in the line. Such inductors are called lumped inductors. The inductors are introduced in the limbs to keep the line as balanced circuit. The lumped inductors are in the form of coils called loading coils.

The lumped loading is preferred for the open wire lines and cables for the transmission improvement. The loading coil design is very much important in this method. The core of the coil is usually Toro dial in shape and made if dimensions, very low eddy current losses and negligible external field which restricts the interference with neighboring circuits. The loading coil is wound of the largest gauge of wire consistent with small size. Each winding is divided into equal parts, so that exactly half the inductance can be VEL TECH


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inserted in to each leg of the circuit. These are built into steel sizes to accommodate one or more coils. The pots protect the coils from external magnetic fields, weather and mechanical damage. The fig. shows the construction of loading coils. White installing the coils, the care must be taken so that the circuit balance is maintained. No winding is reversed. If winding is, it will neutralized the inductance of other winding reduced the overall inductance. In the case of lumped loading. The line behaves properly provided spacing is uniform and loading is balanced, up to a certain frequency called cutoff frequency of the line. Upto this frequency, the added inductance behaves as if it is distributed uniformly along the line. But above this cut-off frequency the attenuation constant increases rapidly. The line acts as low pass fitter. The graph of ∝ against the frequency called the attenuation frequency characteristics of the line show in the fig. It can be see that for continuous loading the attenuation is independent if frequency while for lumped loading it increases rapidly after the cut-off frequency. If the loading section distance is d than keeping inductance LS of the loading coil constant, cut off frequency is found to be proportional to the *. Hence to get the higher cut off frequency, small lumped inductance must be used at smaller distances. 5. Write short notes on different types of transmission lines. Transmission is the process it transmitting some signals from one place to another. Here it can be data as in the case of transmission of computer data along telephone lies or it can be audit/video signals from radio or television broad cost. Electrical energy can be transmitted from one point to another by one of the two methods namely 1. By radiation of electromagnetic waves through free space. 2. By use of electrical conductor arrangement known as transmission line. “Transmission line is a conductive method of guiding electrical energy from one place to another”. In communication these lines are used as link between transmitter and receiver. Transmission lines may be grouped as lumped lines or distributed lines while in distributed liens, the electrical parameters like inductance, capacitance, resistance and conductance are distributed uniformly across the entire line length white in lumped lines these parameters are jumped at intervals along the line. The commonly used transmission lines are 1. Open wire line. 2. Coaxial line. 3. Strip line. 4. Wave guides 5. Optical fibers. VEL TECH


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1. Open wire line: These lines consist of two conductors spaced at a certain distance apart. This type of line is used in transmission of electric power, telegraphy and telephony signals. The construction of open wire line is as shown below. These lines have distributed sonic resistance and inductance, shunt capacitance and conductance. The electric and magnetic fields existing in the lines are shown below. The spacing between the conductors is large in comparison to the diameter of the line conductors. As a result these lines can operate at high voltage. However when operating at higher frequency the larger spacing proves to be a disadvantages because radiation of energy from the open wire line takes place. However the radiation loss can be minimized by reducing the spacing between wires. The primary constants of a open wire transmission line are, resistance, inductance, capacitance and conductance per unit length can be given as, R = 2p /r π 2 ohms. L = µ ο / π log e (S / r ) Henries. Where, P – Specific resistance r – radios S – Spacing between conductors. Advantages of open wire lines: 1. Simple to construct and low cost 2. Insulation between the line conductors is air. extremely small 3. It is balanced with respect to ground.

As a result the electric loss is

Disadvantages: 1. These lines are unsuitable for use at frequencies above 100 Mhz because energy loss takes place due to radiation. 2. Coaxial lines (or) Coaxial cables: The avoid radiation losses taking place in open wire lines at high frequencies, a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. The construction of coaxial cable, the electric magnetic field existing in a coaxial cable s as shown above and below. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 69

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A typical air inner copper conductor held in position by insulating discs

Flexible cables have polythene dielectric and outer conductor in the form of copper braided or flexibility. It ‘D’ and ‘d’ are diameters of the outer and inner conductor, and also µ and ε be the permeability and permittivity of the insulting medium, then the characteristics impedance P of the cable is given by Z o = 138 µ r / ∈ r log10 (d / d ) the velocity of propagation V = 3*108 / µ r ∈ r Advantages: 1. The electric and magnetic fields are continued with in the conductor there by eliminating radiation losses. 2. It provides effective shielding from outer interfering signals. Disadvantages: 1. They are costlier then open wire lines. 2. Beyond 1Ghz these cables cannot be used because losses in the dielectric increases with frequency. STRIP LINES VEL TECH


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A form of line using finite plates and an inverting dielectric medium is turned as strip line. Strip lines are two types: 1. Trip late 2. Micro strip. .Trip late Line: This line resembles as a co-axial line in which side conductors have been removed. The energy propagation in this line is in the form of TEM waves, provided. The distance b/w the centre and outer plate is smaller compared with the wave length of the signal Disadvantages: 1. It is costlier and it also requires a manufacturing skill. Micro strip line: The micro strip line has a narrow conductor supported by a dielectric. The bottom conducting plates serve as earth plate. The micro strip line is widely used in microwave integrated circuits these components find wide application in couplers circulators and receiver. Micro strip lines are fabricated on fiber glass or polystyrenes printed circuit boards as about 1.5mm thickness with copper strips. 3mm wide.

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Wave guides: Wave guides are hollow conducting takes a uniform gross section used for U. H. F. transform by continuous reflection from the inner walls of the guide. Wave guide are used to minimize losses and for high power transmission at microwave freq. The shape of the wave guide may be rectangle or cylindrical thro which electromagnetic waves are propagated. The propagation takes place in open wire and co-axial lines propagation takes place in the form d transverse electric (TE) and transverse magnetic ™ waves.

Rectangular waveguide

Elliptical waveguide.

Advantages: 1. In wave guide no power is cost throughout, because she elastic and magnetic fields are confined to the space with in the guides. 2. The dielectric loss is negligible. 3. Several modes of electro magnetic waves can be propagated with in a single wave guide. 4. Frequency of the wave higher than 39 hz can be easily transmitted. Disadvantages: 1. Cost of the wave guide is very high. 2. Wave guide walls should be specially plated to reduce resistance to avoid skin effect and power less. Optical fibers: VEL TECH


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The entrance appearance of optical fibers are similar to C0-axial cables. Here the copper cores are replaced by highly put e – glass (or) silica which is used to carry modulated light energy similar to micro wave energy. An optical fiber is a dielectric wave guide that operated at optical frequency. It confines electromagnetic energy in the form a light to within its surfaces and guide the light in a direction 11 d to the axis. The structure of a time cable is show below.

Advantages: 1.Low transmission loss and high band width. 2. Electrical isolation is there. 3. No radio frequency and electromagnetic item. 4. High degree of data securing is afforded. 5. Small sue and weight. Disadvantages: 1. The cost of fiber able is very high (Rs.7000/- per meter). 2. It is difficult to run the cables where the bending occurs. 3. Different specializes technique have to the followed to join ends of two cables. 6. Explain about transmission line parameter: Transmission line parameter: In a open wire line, when a current is passed through it, a magnetic fields are produced around the conductors and the voltage drop occurs along the line similarly when a voltage is passed through a open wire line, an electric field is produced b/w the two conductor. The magnetic field proportional to the usual indicate that the line L as series inductance ‘L’ and the voltage drop indicate the presented series resistance ‘R’ Similarly the electric field proportion to the voltage indicates that the line contains VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 73

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shunt capacitance ‘C’ and this capacitance is new useless (or) perfect so, the line also contains conductance ‘G’ These four parameter R, L, C and G are distributed along the whole length of the line. The four line parameters, R, L, C. and G are termed as primary customs of a transmission line. They are defined Resistance ‘R’ Resistance ‘R’ is designed as loop resistance per unit length of line. They it is the sun of resistance of both the wire for unit line length. It is ohm/kn. Inductance ‘L’ Inductance t ‘L’ is defined as loop inductance per unit length of line. Thus it is the sum of inductance of both wire for unit length. Its unit is HCGrier /km. Conductance ‘G’ Conductance ‘G’ is define as shunt conductance b/w the two wires per unit length of line. Its unit is mhos / km. Capacitance: Capacitance is defined as shunt capacitance b/w the two wipes per unit length: It’s unit is farad / km Impedance ‘Z’ The series impedance of a transmission line per unit length is given as , Z=R+jωL, Where, R – line resistance, and j ω L – Line reactance. Admittance: ‘Y’: The shunt admittance. ‘Y’ of a transmission line per unit length is given at y = G’ + j ω c. Where, G – line conductance and j ω C – line susceptance. Characteristic Impedance ‘Zo’: VEL TECH


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Characteristic impedance is defines as the ratio of square root of impedance to admittances. Zo = Z / Y Z o = ( R + jω L) /(G + jω C ) Propagation constant ’Y’: Propagation constant is defined at the product of square root of impedance and admittance. r = ZY = ( R + jω L)(G + jω L) These characteristic impedance ‘Zo’ and ‘r’ are called as secondary constants of transmission line. In addition to primary and secondary constants of a transmission line there are 3 more units of transmission line theory. Wave length: It is defined as the distance that the wave travels along the line in order that the total shift is 2 π radiance. It is denoted by ‘ λ ‘ and its unit is meter. λ =2π / β Group Velocity: The group velocity is defined as the velocity of the develop is a complete signal. It is denoted at √g. g = (ω 2 − ω1 ) /( β 2 − β1 ) Velocity of propagation: It is defined as the velocity with which a signal of single frequency propagation along the line at a particular frequency ‘f’. √g = λ f √g = 2 ∏ f/ β

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7. Derive the expression for telephone cables: Telephone cables (or) telex lines are used as low frequencies transmission line. A telephone cable is formed by two wines insulated from each other by a layer oil impregnated paper and then twisted in pains. A large no of each pairs form an underground cable. Such transmission lines are called as telephone cables. At low frequencies, the series inductance reactance is quite negligible as negligible as compared to line resistance ‘R’. Similarly line conductance is also vary small as compared to susceptance substituting the condition in general equation for propagation constant (γ ) and the characteristic input (Zo) given.

γ = XY γ =

( R + jω L)(G + jω c)..........(1) subG = 0; L = 0

γ = γ =

jω RC

ω Rc 45o..........(2) α + j β = ω RC cos 45o + j ω RC sin 45o

ω RC / 2 + j ω RC / 2 γ = α = ω RC / 2............(3)

β = ω RC / 2............(4) Characteristic impedance Z o = ( R + jω L)(G + jω L) Z 0 = R / jω C Z 0 = R / ω C < 45oand ............(5) Velocity of propagation Vp is given as γ p 4= ω / β VEL TECH


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ω RC / 2 ⇒ V p = 2ω / RC ...........(6)

From the above equation its observed that α and √ are independent on frequency. This the higher frequency are attenuated move and travel faster than the lower frequency resulting in frequency and delay distortion. Hence large distortion occurs at higher frequencies in a telephone lines.

8. Explain insertion loss in detail and derive the expression for the same. Insertion loss occurs due to insertion of a network or a line in between source and load.

If input impedance Zs is not equal to generator impedance Zg, then reflection loss occurs at terminals 1-11. If ZR is not equal to Zo, then second reflection loss occurs at terminals 2-21. The overall effect of insertion of a line is to change the current through the load and hence power delivered to load is less compared too power delivered to load when it was directly connected to generator. Thus insertion loss of a line or a network is defined as the number of ropers or dB by which the current in the load is changed by insertion of a line or a network between the load and the source. Consider the circuit in which generator of impedance Zg is connected to a load of impedance ZR. I R1 =

E → (1) Zg + ZR

The line is inserted between load and generator. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 77

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Let Zs be input impedance of a line which different than Zg, IS =

E → (2) Z g + ZS

We know that input impedance of line is  eγ l + ke−γ l  Z S = ZO  γ l → (3) −γ l   e − ke  sub(3)in(2) IS =

E  eγ l + ke−γ l  Z g + ZO  γ l −γ l   e − ke 

∴ Is =

γl

Z g (e − ke

−γ l

E ) + Zo (eγ l + ke−γ l )

we know that Is =

I R ( Z R + Zo ) γ l (e − ke −γ l ) → (4) 2Z o

This equation is obtained from general solution of line by substituting s = l 2ZO I S → (5) ( Z R + ZO )  eγ l − ke −γ l 

IR =

2 Z o E (eγ l − ke −γ l ) = → (6) Z g (eγ l − ke −γ l ) + Z 0 (eγ l + ke−γ l ) ( Z R + Z o )(eγ l − ke −γ l ) Z R − Zo → (7) Z R + Zo

we know that ref. 10eff. K=

∴ IR =

  ( Z R + Z o ) Z o γel  

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2Z 0 E Z R − Z0 + Z R + Zo

 lγ e 

−

 l Z+ g e 

γ

Z R − Z R

−Z o +Z o

 e−l  

γ

  


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I R=


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2Z o E  ( Z R + Z o ) Z o eγ l 

∴ IR =

 Z − Zo  − + R  Zo e  Z R + Zo 

lγ

+Z g e l γ

Z −Z o − R Z R +Z o

 −l Z g e 

γ

  

2Z 0 E → (8) ( Z R + ZO )( Zo + Z g )eγ l + ( ZR − Zo )e−γ l

Insertionloss =

I R1 IR

E Z g + ZR I ∴ I .loss = = 2Z o E IR γl ( Z R + Z o )( Z 0 + Z g )e + ( ZR _ Zo )( Zo − Zg )e −γ l 1 R

γl −γ l I R1 ( Z R + Z o )( Zo + Z g )e + ( ZR − Zo )( Zo − Zg )e = → (9) IR 2 Zo ( Zg + zR )

The length of line is usually very large hence e −γ l → 0 ∴2nd term in numerator can be neglected. ∴

γl I R1 ( Z R + Z o )( Z 0 + Z g )e = → (10) IR 2Zo ( Z g + Z R )

=

( Z R + Z o )( Zo + Z g )eα l e j β l 2Z 0 ( Z g + Z R ) (Q γ = α + j β )

But insertion loss has too be calculated as a function of ratio of current magnitudes β and hence e j l can be neglected. Z R + Z o Zo + Z g eα l I R2 ∴ = → (11) IR 2 | Zo || Z g + ZR | Xand ÷ by 2 Z g Z R ,

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I R1 2 Z g Z R | Z R + ZO || ZO + Zg | e = IR 4 Z g Z R | ZO || Zg + Z R | ∴

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I R1 | Z g + ZO || Z R + ZO | 2 Zg ZR α l = e → (12) IR 2 Z g ZO 2 Z R ZO | Zg + ZR |

All the terms on RHS are reflection factors. 2 Z g Zo Let Ks = = reflection factor at source side →(13) | Z g + Zo | kR =

2 Z R Zo = reflection factor at load side →(14) | Z R + Zo |

k SR =

2 Zg ZR | Zg + ZR |

= refl factor for direct connection →(15)

eα l indicates loss in the line. k I R1 ∴ = SR eα l IR kS k R Insertion loss = (or)

 I R1  1 1 1 = lu + lu − lu + α l  Nepers. I R  kS kR kSR 

  1 1 1 + 0.4343α l  dB Insertion loss = 20 log + log − log kS kR kSR   The term corresponding to kSR is negative. It is the loss if generator and load would have been directly connected. It is not related to insertion hence it is subtracted from overall loss. 9. Prove that (i ) Z 0 = Z oc. Z SC (ii ) tanh γ =

Z sc Z oc

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We know that finite line terminated in Zo behaves as an infinite line, hence Zo must be Zo The i/p impedance Z in of equivalent T network is Z  Z  Z in = 1 +  Z 2 ||  1 + Zo   2   2  Z  z2  1 + Z 0  Z 2  Z in = 1 +  2 Z + Z1 + Z 2 0 2 ButZ in = Z o Z  Z2  1 + Zo  Z 2  Z0 = 1 +  2 Z + Z1 + Z 2 o 2 Z Z     Z  2 Z o  Z 2 + 1 Z0  = Z1  Z2 + 1 + Zo  + 2 Z2  1 + Z0  2  2     2  2 Z ∴ 2 Z o 2 = 2 Z1 Z 2 + 1 2 2 Z ⇒ Z o 2 = 1 + Z1 Z 2 4 ∴ Zo =

Z12 + Z1Z 2 4

to obtain Z1 and Z2 we open circuit and short circuit the network In open circuit the line is kept open and input impedance is measured. Z Z oc = 1 + Z 2 2 In short circuit the second end of line is short ed and input impedance is measured.

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Z1  Z  +  Z 2 || 1  2  2 Z1 Z2 Z = 1+ 2 2 Z1 + Z 2 2 Z sc =

Z12 Z1 Z 2 + + Z1Z 2 2 = 4 Z1 + Z2 2 Z2 ∴ Z sc = o c Zo Z o 2 = Z oc. Z sc ⇒ Zo = Zsc .Zoc (iii)

we know that

eγ = 1 +

Z1 Z o + 2Z 2 Z 2

subZ o Z1 Z12 Z e = 1+ + + 1 2 2Z 2 4Z2 Z2 γ

2

 Z  Z Z ∴ e = 1 + 1 +  1  + 1 → (1) 2Z 2  2Z 2  Z2 γ

mathematically 2

 Z  Z Z e ; 1 + 1 −  1  + 1 → (2) 2Z 2  2Z 2  Z2 (1) = (2) Z eγ + e −γ = 2 + 1 Z2 −γ

∴

eγ + e −γ Z = 1+ 1 2 2Z 2

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Z1 2Z 2

Noweγ = coshγ + sinhγ  Z  ∴ (cosh γ + sinh γ ) − cosh γ = eγ − 1 + 1   2 Z 2.  Z  Z Z  ∴ sinh γ = 1 + 1 + o − 1 + 1  2 Z 2 Z 2  2 Z2 

γ =

Zo Z2

Zo Z0 sinh γ Z2 Sinh tanh γ = = = Z cosh γ 1 + Z1 Z2 + 1 2Z 2 2 ButZ o = Z sc .Zoc andZ 2 +

∴ tanh γ =

Z1 = Z 0c 2

Z sc Z oc Z oc

⇒ tanh γ =

Z sc Z oc

10. Write a note on Reflection factor and same.

Reflection loss and derive the

Reflection occurs due to improper termination at the receiving end. This concept can be extended to the function of any two impedances . Let a source of voltage Es and impedance Z1 is connected to a load of impedance Z 2 . If Z 2 is not equal to Z1 , reflection of energy takes place resulting in a change in the ratio of V to current and alteration in the distribution of energy between the Electric and magnetic field. The energy transferred to Z 2 is less than that with impedance matching. A reflection is said to have occurred. The magnitude of this loss can be computed by taking the ratio of current actually flowing into the load to the load to the current that would have flown if the impedance is were matched. The matching of impedance is called as image matching and can be obtained on a live by connecting a transformer. According to the transformer theory,

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z2 z1

→


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(1)

For matching, the magnitude of Z1 can be made equal to Z 2 by choosing proper transformer ratio. The current which flows through the generator is I1 I1 =

E 2 Z1

→ (2)

1 The current I 2 (Under image matching condition) which flows through the load

I 21 =

Z1 Z2

• I1

I 21 =

Z1 Z2

•

I 21 =

E Z1Z 2

E 2 Z1 → (3) I2.

E → (4) Z1 + Z 2 The ratio of the current actually flowing into the load to that current flowing under image condition Without image matching, the current flowing through the load is

is

∴ I2 =

2 Z1Z 2 I2 = = R → (5) 1 I2 Z1 Z2

This is called as Reflection factor (R) Reflection loss It is defined as the no.of repers or dB by which the current in the load under image matched condition would exceed the current actually flowing in the load. Then The reflection loss in repers is  Z + Z2  Rloss = lu  1  in lepers  2 Z1 + Z 2   Z + Z2  Rloss = 20 log  1  − dB  2 Z1 + Z 2  VEL TECH


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11. Explain the Reflection of a line not terminated by Z0. Referring to Equation     ( Z − Z R ) − zys  IR I= ( Z 0 + Z R )  e ZYS + o e  2ZO (Z R + Zo )   E=

 (Z − Zo ) − ER ( Z R + Z o ) e ZYS + R e 2 ZR (Z R + Zo ) 

zys

There are current and voltage relationships derived for the lines which are terminated in Z o But if a line is not terminated in Z o (or) it is joined to same impedance other than Z o then part of the wave is reflected back phenomenon exists for a line which is not terminated in Z o . Reflection is maximum when the line oCi, eZR= α Reflection is maximum S c i eZ R =O Reflection is zero when Z R = Zo S → distance measured from the receiving end and treated positive. When Z R ≠ Zo 1) One part varying exponentially with positive S 2) One part varying exponentially with negative S E=

ER ( Z R + Z o )  Z − Zo )  e zys +  R e 2 ys 2Z R  ZR 

ER ( Z R + Z o ) 2 ZR ⇒E=

ER ( Z R + Z o ) E e zys + R ( Z R − Z o )e − 2 ys 2Z o 2 ZR

III rly I =

IR ( Z R + Z o ) 85 I R ( ZR − Z 0 ) 85 e − e 2Z o 2 Zo

The first component of E or I which varies exponentially with TS is called incident wave which flows from the sending end to the receiving end. S=O at the receiving and maximum m(s=l) at the sending end. Thus as incident wave travels from the sending end to the receiving end, its amplification decreases. A wave which flows from sending end to the receiving end, with decreasing amplitude is the incident wave. VEL TECH


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The second component of voltage or current which travels from the receiving end to the sending end which varies e85 and its amplitude decreases as its progress towards the sending end. The total instantaneous voltage at any point on the line is the vector sum of incident and the reflected currents are in out of phase with each other. If z R = Z o it can be seen that the reflected wave is absent end there is not reflection. Such a line is uniform and there is no discontinuity existing to send the reflected wave back along the line. Similarly along the line and the energy is absorbed wave. Such a finite line terminated in Z o without having any reflection is called a smooth line. 12. Derive an expression for input impedance and transfer impedance and of transmission line terminated by an impedance. From the general solution Zinτ Z s Z τ

Z o ( Z R wsh(81) + Zo Sinh(81) ) ( Z o wsh(81) + ZR Sinh(81) )

Dividing by ( Z R cosh (81) )both Nr and Dr. We get 1+ Zsτ Z o

Zo Jauh(81) ZR

Zo + Jauh(81) ZR

 Z + Z o Jauh(81)  (or ) zsτ  R  Z  Z o + Z R Jauh(81)  or equ (3) jauh (81)

e81 − e −81 e81 + e −81

 Z + Z o (e81 − e −81 / e81 − e−81 )  ∴ Z sτ Z o  R 81 −81 81 −81   Z o + Z R (e + e / e + e )  Equ. (3) and (4) are the I/P Impedance of a live terminated by an impedance. Let ZT =

Es = Transfer impedance of a live IR

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I Rτ cosh(81) − Iτ Now

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Es Sinh(81) Z o xIR

Is Z cosh (81) − T sinh(81) IR Zo

while ERτ Es cosh (81) − Zo sinh(81)

τ Es cosh(81) I s Zo sinh(81) − ER Is =

Es cosh(81) − ER Z o sinh(81)

Sub in equ (5) 1τ = 1τ

cosh(81) IR

 Es cosh(81) − ER  ZT sinh(81)  −  Zo sinh(81)  Zo

cosh 2 (81) − Z R cosh(81) − ZT sinh2 (81) Z o sinh(81)

Z o sinh(81)τ ZT cosh(81) − ZR cosh(81) − ZT sinh 2 (81)

τ ZT (cosh 2 (81) − Sinh2 (81)) − ZR cosh(81)

∴

τ ZT (1) − Z R cosh(81) ZTτ Z R cosh(81) + Zo sinh(81)  e81 + e −81   e81 + e−81  ZT τ Z R   + Zo   2 2    

This is the reg. transfer impedence. 13. Derive an expression for the Input impedance of a lossless line. I/P impedance of a lossless live of any length or obtained from equ. Z in = Z o = Zo

Z R + Z o tanh(81) Z o + Z R tanh(81)

______(1)

for loss live α = o, therefore Pτ j β will become jB only Hence,

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Z R + Z o tanh( j β l ) _____(2) Z o Z R tanh( j β l )

But tanh j β lτ tan β l Therefore  Z + jZ o tan β l  Zin = Z o  R   Z o + Z R tan( j β l )  2π l Since β = λ 2π l    Z R + jZ o tan λ  Zin = Z o  ______(3) 2π l   Z o + Z R tan  λ   Again for a lossless live, the resistive component of the live i.e., R and G will be equal to zero P = α + j β = ( R + jwl )(G + jwc) Thus

O + jβ =

jwLxjwc

j β = jw LC

β = w LC ⇒ β = w LC

(or)

If f is the frequency of operation and terminating impedance is a pure resistance RR Equ (3) will become

Z in =

RR + jZ o tan 2π f LCl Z o + JRR tan 2π f LCl

Therefore, I/P Impedance of a lossy and lossless line. 14. Derive the equation for T and π section equivalent to lines.

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Z10c = Z1 + Z3 Z15c = Z1 +

Z2Z3 Z2 + Z3

Z1 = Z10c = Z3 Z2 = Z20c − Z3  Z Z  Z3 = Z2 + Z3  Z1 + Z3 − Z1 + 2 3  Z2 + Z 3    Z ( Z +Z ) −Z Z  3 2 3 = Z2 + Z3  3 2   Z2 + Z3  

(

)

(

)

Z3 = Z3Z2 + Z32 − Z2Z3 Z3 = Z3 ∴ Z3 = Z20c ( 210c − Z15c ) The input impedance of oc and SC lines are,  γl  ZR Zo  −γl  e + e  Z + Z Zo R o    Z10c = = Zo   γl  Z Z  −γl  tan h γl  e −  R o  e   ZR + Z o    for oc, ZR=∞

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 eγl + e−γl ∴ 210c = Zo  γl −γl e +e

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 eγl + e−γl  III ly Z13C = 20tan hγl=20 γl −γl   e +e  since the line is symmetrical Z10c=Z20c ∴ Z3 =

 Z o  Zo − Zo tan hrl   tan h γl  tan h γl 

Zo  Zo − Zo tan2hγl  =   tan h γl  tan h γl  Zo = 1− tan2 h γl tan h γl Zo Zo 1 = sec2 hγl = × coshγl × tan h γl sin h γl cos h hγl Zo = sin h γl  eγl + e−γl  Z1 = Z2 = Z10c − Z3 = Zo  γl −γl   = Z3 e +e 

(

)(

 eγl + e−γl eγl + e−γl 2 Z 3 = Zo  2  eγl + e−γl 

( (

)

(e

  

γl

+e

−γl

)( e

γl

+ e−γl

)

) )

 eγl + e−γl  −1  eγl + e−γl   

Zo2 

Z3 =

=

(

)  − Zo2 ( eγl + e−γl ) ( eγl + e−γl )

(

)

 eγl + e−γl + 2 − e2γl − e−2γl − 2     eγl + e−γl 2  

Zo2 

(

)

  4 2    z2 = z  =  o o  2 γl −γl   γl −γl  e −e   e −e    VEL TECH

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)


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 eγl + e−γl  2 Z10c − Z2 = Zo  γl − γl −γl −γl  e + e  e +e  γl  e + e−γl − 2 = Z o  γl  e + e−γl     γl −γl / 2 2 e − e = Zo  γl / 2 −γl / 2 γl / 2 −γl / 2  e −e e +e 

(

(

)

(

)(

)

)

   

 eγl / 2 − e−γl / 2  = Zo  γl / 2 −γl / 2  −e e  z1 = z2 = zo tan hγl/2 ∴ Zo tan hγl/2 Zo tan hγl/2

π - Section equivalent :-

ZA =

Z1Z2 + Z2Z3 + Z3Z1 Z (Z + Zc ) Z10c = A B Z2 Z A + Z B + Zc

ZB =

Z1Z2 + Z2Z3 + Z3Z1 Z (Z + Z B ) Z20c = c A Z3 Z A + Z B + Zc

ZC =

Z1Z2 + Z2Z3 + Z3Z1 ZA Z B Z15c = Z1 ZA + Z B

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ZA=Zc=


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z20C + z15c

Z20c − Z20c − Z20c ( Z10c − Z15c )

 eγl + e−γl   eγl + e−γl  × = Z o  γl Z  e + e−γl  o  eγl + e−γl    γ l −γ e +e l  2Z = Z o  γl − γl o−γl −γl   e −e  e −e

=

=

= Zo2  eγl + e−γl  2Z Zo  γl − γl o−γl −γl   e −e  e −e

(

Zo eγl / 2 − e−γl / 2

γl / 2

( eγl / 2 − e−γl / 2 )

Z A = Zc = ZB =

)(e

Zo tan hγl

− e−γl / 2

)

2

Zo tan hγl

Z 2oc.Z15c

Z20c ( Z10c − Z15c )

Zo2 Z2 sin hrl = = o = Zo sin hγl Zo / sin hγl Zo

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UNIT – III THE LINE AT RADIO FREQUENCIES PART – A 1. What is dissipation line? A line for which the effect of resistance ‘R’ is completely neglected is called dissipation less line. Dissipation less there is used for transmission of power at high frequency in which losses are neglected completely. 2. What is the nature and value of Zo for dissipation less line? For dissipation less line, Zo is purely resistive and is given by L Z o = Ro = C 3. What is the value of ∝ and β for a dissipation less line? The value of attenuation constant ‘ ∝ ‘ for a dissipation less line is zero and the value of phase constant ‘ β ‘ for a dissipation less line is ω LC radian / m. 4. What are nodes and antinodes on a line? Nodes are points of zero voltage or current in a standing wave system and Antinodes are points of maximum voltage or current in a standing wave system.

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5. Draw the graph between standing wave ratio (s) and reflection co efficient (k)

6. Give the expression for skin depth. Skin depth or nominal depth of penetration is given by l S= meters ∏ f /m where ‘p’ is the resistivity of conductor in Ω /m ‘f’ is the frequency in HZ ‘m’ is absolute magnetic permeability of conductor in H/M 7. What are the advantages of dissipation less line? The advantages of dissipation less line are • The line acts as a smooth line • No reflection takes place at receiving end • Standing waves are not produces. VEL TECH


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8. Calculate the reflection co efficient if VSWR of the line is 1.5. 1+ | K | 1− | K | 1+ | K | 1.5 = 1− | K | VSWR = 1+ | K |= 1.5 − 1.5 | K | 0.5 =| K | +1.5 | K | 0.5 =| K | 2.5 ∴ | K |= 0.2 9. Define standing wave ratio. The ratio of maximum and minimum magnitude of current or voltage on a line having standing waves is called standing wave ratio. i.eSWR =

| Vmax | | I max | = | Vmin | | I miin |

10. What are standing waves? When a transmission line is not terminated in its characteristic impedance, the traveling electromagnetic wave from generator at sending end is reflected completely or partially at the terminating end. The combination of incident and reflected wave gives rise to standing waves of current and voltage with definite maxima and minima along the line. 11. Give the relationship between VSWR and reflection coefficient for a transmission line The relationship between VSWR(S), and reflection coefficient ’K’ is S=

1+ | K | 1− | K |

12. Define reflection coefficient. Reflection coefficient is defined as ratio of reflected voltage or current to the incident voltage or current. It is denoted as ‘K’. V I K = r (or ) K = − r Vi Ii The current ratio is negative because the reflected current suffer a 180° phase shift at VEL TECH


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the receiving end while the reflected voltage does not. 13. Give the velocity of propagation of open wire dissipation less line. The velocity of propagation for open wire dissipation less line is given by V = 3X108 m/sec 14. List out the features of power frequency line. •

• •

Power transmission lines are electrically short in length, with the length not λ exceeding 10 The power efficiency of power transmission line is very high as compared to other energy sources. Power transmission lines are operated at constant output voltage.

15. Write the expression for input impedance of RF line. The input impedance of dissipation less RF line is given by  Z R + jRo tan β s  Z in = Ro    Ro + jZ R tan β s  where Ro is characteristic impedance ZR is terminating impedance β is phase constant ‘s’ is length of the line. 16. Give the expression for input impedance of short circuited line. The input impedance of short circuited line is  2π s  Z sc = jRo tan    λ  where λ is wavelength s is length of the line Ro is characteristic impedance 17. Give the expression for input impedance of open circuited line. The input impedance of open circuited line is  2Π S  Z oc = − jRo cot    λ  where λ is wave length s is length of the line VEL TECH


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Ro is characteristic impedance 18. Define dissipation factor. Dissipation factor is defined as the ratio of energy dissipated to energy stored in dielectric per cycle. 19. Name the device used for measuring standing wave. The derive used for measuring standing wave is directional coupler. 20. What is the maximum resistive input impedance of a dissipation less line? The maximum resistive input impedance of a dissipation less line is Rmax = S Ro Where S is standing wave ratio and Ro is characteristic impedance. 21. What is the minimum resistive input impedance of a dissipation less line? The minimum resistive input impedance of dissipation less line is R Rmin = s S where ‘S’ is standing wave ratio Ro is characteristic impedance 22. A 50 Ω line is terminated in load ZR = (90 + j60) Ω . Determine VSWR due to this load. S=

1+ | K | 1− | K |

whereK = ∴s =

Z R − ZO 90 + j 60 − 50 = = Z R + ZO 90 + j 60 + 50

1+ 1−

23. A lossless line of 300 Ω characteristic impedance is terminated in a pure resistance of 200 Ω Find the value of SWR? Zo = 300 Ω VEL TECH

; ZR =200 Ω VEL TECH MULTI TECH TECH HIGHTECH 97

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1+ | K | 1− | K | Z R − ZO 200 − 300 = Z R + ZO 200 + 300 100 =− 500 k = −0.2

whereK =

1 + 0.2 1.2 = 1 − 0.2 0.8 S = 1.5

∴S =

24. Sketch the standing waves on a dissipation less line terminated in a load not equal to Ro.

25. Sketch the standing waves on a line having open-or- short circuit termination.

26. What is directional coupler? Directional coupler is a device which is used to measure standing waves. It consists of Coaxial transmission line having two small holes in the outer sheath 1 spaced by wavelength clamped over these holes is a small section of line, 4 terminated in its Ro value at both ends to prevent reflection.

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27. State the use of half wave line. The expression for the input impedance of the line is given as ZS = ZR. Thus the line represent repeats its terminating impedance. Hence it is operated as one to one transformer. The main application of a half wave line is to connect a load to a source where both of them cannot be made adjacent. 28. What are the uses of quarter wave line? The expression for input impedance of quarter wave line is given by Zs =

RO 2 ZR

This equation is similar to the equation for impedance matching using transformer. Hence the quarter wave line is considered as transformer to match impedance of ZR & ZS. It is used as an impedance matching section. It is used to couple a transmission line to a resistive load such as antenna. 29. What do you mean by reflection loss? When there is mismatch b/w the line and load, the reflection takes place. Because of this the energy delivered to the load by the line is less it composition with the power delivered to the load by a properly terminated line. This loss in power is called reflection loss. 30. Explain how smith chart can be used as an admittance chart. If the smith chart is to be used for admittance, the ‘ri’ axis becomes ‘gi’ axis, while’Xi’ axis becomes ‘bi’ axis. Then above real axis, the susceptance is inductive which is negative. The extreme left point on the real axis represents zero conductance while the extreme right point on the real axis represent infinite conductance. 31. What is the practical value of SWR we can achieve by double stub matching? VEL TECH


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SWR = 1.2 32. What is smith chart? Smith chart is an impedance (or) admittance chart which is used to calculate all the parameters of transmission line. It consists of two sets of circles. 33. What is stub matching? A section of transmission line is used as a matching by inserting then b/w load and source. This is called sub matching. 34. What are the advantages of stub matching? 1. The length and characteristic impedance of the line remain unaltered. 2. From mechanical stand point, adjustable susceptance are added in shunt with the line. 35. Give type of stub matching. 1. Single stub matching 2. Double stub matching. 36. Name the impedance transformer that are used at higher frequencies? At higher frequency the impedance transformers consists as a section of transmission line in various arrangements as listed below. 1. Quarter wave transformer (impedance inverter) 2. Stub matching. 1. Single stub matching 2. Double stub matching. 37. What are the advantage and disadvantages as quarter transformer? Advantages: 1. It is very useful device because d its simplicity. 2. Its behaviour can be easily calculated. Disadvantages: 1. It is so sensitive to change in freq. 38. What do you mean by impedance circle diagram? VEL TECH


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If impedance are plotted in the form of R – X diagram it turns that for a loss less line terminated in some fixed impedance ZF, the locus of the i/p impedance. Zin as electrical length pl is varied as a circle. This circle diagram is known as impedance circle diagram. 39. Draw the family of constant S – Circle diagram.

40. Explain the direction of movement towards generator or load in circle diagram. In circle diagram, the movement in the clockwise corresponds to transverse from the load towards the generator and the movement in the anti clockwise direction corresponds to transverse from the generator towards the load. 41. What are the advantages and disadvantages * circle diagram. Advantages: 1. It is very useful in calculating line impedance and admittances. Disadvantages: 1. ‘S’ and ‘pl’ circles are not concentric making interpolation difficult. 2. Only limited range impedance value can be contained in nc chart. 42. What are the differences between circle diagram and smith chart? The basic difference between circle diagram and smith charts are: 1. In circle diagram, the resistance component * an impedance represented in rectangular form . 2. In smith chart, the resistive component ‘p’ and resistive component ’x’ and impedance are refines in circular form. VEL TECH


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43. Draw the family of constant – R circles in smith chart.

44. Draw the family of constant x – circle in smith chart.

45. Give the properties or smith chart: The properties of smith chart are: 1. Normalizing impedance. 2. Plotting as an impedance 3. Determination of K in magnitude and direction 4. Determination of swr 5. Movement along the periphery of the chart. 46. Give some applications of smith chart: 1. 2. 3. 4.

Smith chart can be used as an admittance, diagram Used for convey r + aj impedance into admittance. Any value of input impedance can be easily determined. Smith chart can also be used to determined load and impedance

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5. The input impedance and admittance of shot circuited line and open circuited line can be easily calculated. 47. Draw a diagram showing how a quarter wave transformer can be used for matching two lines.

48. What are the disadvantages of single stub marks? 1. The single stub matching system is useful only for fixed frequencies. 2. Final adjustment of the sub has to the moved along the line shifts. 49. Why short circuit stub is used in single stub matching? The short circuit is invariably used because. 1. It radiates ups power and 2. It effective length may be varied by means of a shorting bar which normally takes the shapes of shorting plugs. 50. A loss less line has a characteristic impedance of 400 Ω . Determine the stands wave ratio with the following receiving and impedance Z1 = 70 + 0.0Ω Solution: K=

Z O − Z R Z R − ZO Z O + Z R Z R + ZO

ZR = 70 ; Zo = 400 k=

70 − 400 ⇒ −33 / 47 70 + 400

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∴| k |= 33 / 47 1+ | k | 1 + 33 / 47 s= = = 5.71 1− | k | 1 − 33 / 47 51. Give the formula to calculate the position and length of a short circuited stub, The position of the stub can be calculated using the formula.

λ tan −1 Z O Z 2Π λ 20 S= tan −1 2Π 2R

1=

52. What is dissipation less line? A line for which the effect of resistance R is completely neglected is called dissipation less line. 53. What is the nature of value of Zo for the dissipation less line? For the dissipation less line, Zo is purely resistive of given by Zo = 1Zo =

L C

54. What is the range of values of standing wave ratio? The range of values of standing wave ratio is theoretically 1 to ∞. 55. Determine K of a line for with ZR=200u, Zo=692 −12o

( (

) )

o 200 − ( 676.8 − j143.8) ZR − Zo 200 − 692 −12 K= = = ZR + Zo 200 + 692 −12o 200 + ( 676 − j143.8)

489.4 −162.91o K= 888.51 −9.31o K=0.55 −153.6

56. What is the need for stub matching in transmission lines? When line at high frequency is terminated into its characteristic impedance Ro, then the line operates as smooth line. Under this conditions, losses are absent, hence VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 104

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maximum power is delivered with increased performance. But practically Ro of the line termination are not matching. So to provide impedance matching between line of its termination, stub matching is used. 57. Why are short circuited stubs preferred over open circuited stub? A high frequencies, open circuited stubs radiated some energy which is not the case with short circuited stub. Hence over open circuited stubs, short circuited stubs are preferred. 58. What are the advantages of dissipation less line? i) ii) iii)

The line acts as a smooth line No reflection takes place at the receiving and The standing waves are not produced.

59. If using of line is 1.5 then calculate its reflection co-efficient . VSWR =

1+ 1K1 = 1.5 1− 1K1

1+1K1=1.5-1.51K1 2.51K1=1.5-1=.5 1K1=0.2. 60. Give the expression for L & C for open-wire line at high frequency? h0 d d l n henrys / m(or ) L = 9.21×10−7 log henrys / m 2∏ a a 12.07 C= µµ f / m d ln a L=

61. Give the expression for L & C for coaxial line at high frequency L = 2 × 10−7 l n C=

b henrys/m a

2∏∈ farads / m b ln a

PART – B

1. Write short notes on reflection losses on unmatched line. If a line is not matched to its load then the energy delivered by the line of the load VEL TECH


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is less than the energy delivered by the matched line to load. Due to this unmatched system reflected waves and standing waves are produced. The voltage at a maximum voltage point is due to the in phase sum of the incident and reflected waves. In measurement of power and impedance on a Tx line we found that Emax =

I R Z R + RO (1| k | .........(1) 2

Now this equation can be writes as Emax | =| Ei | + | Er |=

I R Z R + RO (1+ | k | ....(2) 2

The minimum voltage is due to the difference of the incident and reflected waves and it is given as , Emax =| Ei | − | Er |=

I R Z R + RO (1− | k | ....(3) 2

Hence the standing wave ratio is S=

Emax | Ei | + | Er | = ..........(4) Emin | Ei | − | Er |

The total power transmitted along the line and delivered to the load is given as P= =

| Emax | . | Emin | ( | Ei | + | Er |) (| Ei | − | Er |) = Ro Ro

| Ei |2 − | Er |2 ..........(5) Ro

From the above expression we can recognize the transmitted power as the difference of two powers. One power Pi being transmitted in the incident wave and the other power Pr traveling back in the reflected wave. The ratio of the power ‘P’ delivered to the load to the power transmitted by the incident wave is P Pi − Pr | Ei |2 − | Er |2 E = = = 1− r 2 Pi Pi | Ei | Ei

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2


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= 1− | K |2 2

 S − 1 = 1−   S + 1  4S = ............(6) ( S + 1) 2 Now the ratio power absorbed by the load to the power transmitted is plotted as a function of ‘S’ as shown in fig.

2. Explain Eighth wave line and half wave line. Eighth Wave Line: The input impedance of a line of length s = λ /8 is  Z + jRO tan β s  Z S = RO  R  → (1)  RO + jZ R tan β s   Z + jRO tan(Π / 4)  Z S = RO  R  → (2) R + jZ tan( Π / 4 R  O  Whereβ = 2π / λ 2Π λ . =Π/4 λ 8  Z + jRO  Z S = RO  R  → (3)  RO + jZ R  If the line is terminated in a pure resistance RR, then  R + jRO  Z S = RO  R  → (4)  RO + jRR  In equation (4) the numerator and denominator have identical magnitudes so equ ($) becomes Zs = Ro ……(5)

β=

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Thus an eighth – wave line can be used to transform any resistance to an impedance with a magnitude equal to Ro of the line. (or) to obtain a magnitude match between a resistance of any value and a source of R0 internal resistance. Half – Wave Line: When a length of line having s = λ / 2 is used, the input impedance is  Z + jRO tan Π  Z S = RO  R  → (1) Since tan Π = 0  RO + jZ R tan Π  Z S = Z R ..........(2) Thus a half wave length of line may be considered as a one to one transformer. *It has greater utility in connecting a load to a source in cases where the load and source cannot be made adjacent. *A group of capacitors may be placed in parallel by connecting them with sections of line n half waves in length. As a result insulators on a high frequency line should not be spaced at half wave intervals, since their effect would then be cumulative, lowering the insulation resistance of the line. 3. Explain the principle and application of Quarter wave transformer for impedance matching (or) what are the features of a Quarter wave transformer? Quarter wave line Impedance Matching: The expression for the input impedance of dissipation less line is given as  Z + jRO tan β s  Z S = RO  SR  → (1) R + jZ tan β s R  O  Equ (1) is rearranged as  ZR   tan β s + jRO  Z S = RO   → (2)  RO + jZ  R  tan β s  For a Quarter wave line , s = λ / 4 VEL TECH


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∴ equ (2) becomes  ZR   tan Π / 2 + jRO  Z S = RO   RO  + jZ R   tan Π / 2 

R 20 ZS = .............(3) ZR Since tan Π /2 = infinity i.e. the input impedance of the line is equal to the square of Ro of the line divided by the load impedance. A Quarter wave line acts as a transformer to match a load of ZR ohms. Such a match can be obtained if the characteristic impedance Ro of the matching Quarter wave section of line is chosen as, R0 =

Z S Z R .........(4)

A Quarter wave line may be considered as an impedance inverter, here it transforms a low impedance into a high impedance and vice versa. An application of the Quarter wave matching section is to couple a transmission line to a resistive load such as a antenna. The Quarter wave matching section is designed to have a characteristic impedance Ro chosen that the antenna resistance RA is transformed to a value equal to the characteristic, impedance Ro of the Tx line The characteristic impedance Ro of the matching section should be Ro = RA RO ...........(5) VEL TECH


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The transformers also a single frequency or narrow band device. The bandwidth may be increased by using two or more Quarter wave sections in series each accomplishing part of the total transformation. A Quarter wave transformer may also be used if the load is not pure resistance. It should then be connected between points corresponding to Imax or Emin at which places the transmission line has resistive impedances given by Ro /s or s/Ro. For step down in impedances from the line value of Ro, the matching transformer characteristic impedance should be,] RO = RO .RO / S = RO / S Another application of the short circuited Quarter wave line is as an insulator to support an open wire line or the center conductor of a coaxial line. This application is illustrated is as shown below. These lines are sometimes referred as copper insulator. 4. A lossless line having Ro = 300 ohms is terminated by a load resistance of 78 ohm. The frequency of operations is 40MHz. What type of single stub will be required to provide impedance when placed nearest to the load? Calculate its length and find its location. Solution : The reflection coefficient is given by K=

Z R − ZO Z R − RO 78 − 300 = = k = −0.587 Z R + ZO Z R + R 78 + 300

∴| k |= 0.587 The first voltage minimum occurs at y2 = λ / 2, where 300 λ= = 7.5Meters 40 The location of the stub nearest to the load is given by cos −1 | K | λ cos−1 0.587 7.5 d= . = . Π 4 Π 4 d = 0.563 meters Hence the stub is located at a distance of Y1 = 3.75 – 0.563 Y1 = 3.187 meters The susceptance of the line at the location of the stub will be positive. Hence a VEL TECH


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short circuited stub will be needed to provide the impedance match. The length of this short circuited stub is given by  1− | K |2  λ L= tan −1   2Π  2 | K |  L=

 1 − (0.587)  7.5 −1 7.5 tan −1  tan (0.689) = 2Π  2 × 0.587  2Π

L = 0.72meter 5. A load of ZR = 140 ohms is to be connected to a line RO = 100 ohms by a quarter wave matching transformer. a. Find ZO of the matching transformer, b. What is the ‘S’ for the transformer? c. If the input voltage to the line is 100v. Find the load voltage. Solution a ) Z O = Z R .RO = 14,100 b) S = Z R / ZO = 140 /100 Z O = 118.322OHMS . S = 1.1832 c)Vmin = 100v;Vmax = 1.1832 ×100 Vload = 118.32volts d) The location of Vmax on the N4 line is located at the load. 6. A 150 ohm transmission line is terminated by a load of 200 – j300. Determine the location and length of a short circuited matching stub. Solution: Given Zo = 150 ohm ZR = ZL = 200 – j 300 ohm The single stub is constructed as follows

a) Calculate

Zr =

Z R 200 − 300 = ZO 150

Z r = 1.33 − j 2.00

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b) Mark this point as ‘A’ in the smith chart. c) Draw the SWR circle through ‘A’ the ‘S’ value is 4.9 d)Extend the radial line from point ‘A’ through the center to the other side of the SWR circle and mark the point as ‘B’ and to the outer perimeter as point ‘C’ at 0.0057 λ The point ‘B’ is the normalized load admittance (0.23 +j 0.36) equal to the reciprocal of the normalized load impedance at point A (1.33 – j 200) d) Find the intersection of SWR circle and G = 1 circle mark it as point ‘D’ e) Extend the radial line from the centre to ‘D’ continue the line to the perimeter ‘E’ at 0.183 λ The distance from load to the junction of the transmission line and the stub is I1 = 0.183 λ - 0.057 λ ; I1= 0.126 λ f) Final step is the determination of the length of the stub. The susceptance at point ‘D’ is 0 = j 1.8. So the susceptance contributed by the stub must be j 1.8. Point ‘F’ corresponds to this for a short circuited line at 0.33 λ . The length of the matching stub is from 0.25 λ to 0.33 λ i.e., (0.33 – 0.25 ) λ = 0.08λ

= 12

7. Explain in details the constant ‘R’ circles and constant ’X’ circles in a smith chart. A modified form of a circle diagram for the dissipation less line is the smith chart developed by P.H smith. This chart consists of two circles. 1. ‘R’ circles and 2. ‘X’ circles question)

(For R circles & X circles diagram refer part A

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Z R − Z O Z R / ZO − 1 = Z R − Z O Z R / ZO + 1

Zr −1 WhereZ r = Z R / Zo Zr + 1

Zr – Normalized terminating impedance 1+ K Hence Zr = 1− K Since ‘Zr” and ‘K’ are complete quantities we have R + jx =

1 + K r + jK x 1 − ( K r + jK x )

Rationalizing right hand side we get, 1 − K r 2 − K x2 + 2 jK R + jx = (1 − K r ) 2 + K x2 Equating real and imaginary parts R=

1 − K r2 − K x2 ..............(1) (1 − K r ) 2 + K x2

2K x ..............(2) (1 − K r ) 2 + K x2 When equation (1) and (2) are solved, we get two sets of circles. Equation (1) will yield a family of circle called R – circles and equation (2) will yield a family of circles called x – circles. X=

The constant ‘R’ circles: Consider equ (1) and cross multiply R=

1 − K r2 − K x2 (1 − K r ) 2 + K x2

We get

R (1 + K r2 − 2 K r + K x ) = 1 − Kr2 − Kx2 R + K r2 R − 2 RKr + RK x2 − 1 + Kr2 + Kr2 = 0

K r2 ( R + 1) + Kr2 ( R + 1) − 2 Kr R = 1 − R

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 2 K r .R  ( R = 1)  K r2 + K x2 − = 1− R ( R + 1)   2 K r .R 1 − R K r2 + K x2 − = R +1 1+ R Adding

R2 on both sides to make it a perfect square we have . (1 + R ) 2

K r2 + K x2 −

2 K r .R R2 (1 − R ) R2 + = + R + 1 (1 + R ) 2 (1 + R) (1 + R )2 2

R  1− R R  K x2 +  K r − + +  1 + R  1 + R (1 + R) 2  2

2

2

R   1   K x2 +  K r − − ............(3) 1 + R  1 + R  

This equation represents the family of circles on the reflection coefficient plane, these circles are called constant –R circles having radius 1/1 + R and centre (R / 1 + R, O ). These circles have their centers on the positive Kr axis and are contained in the region ‘0’ to ‘1’ as shown in the figure below. R = o corresponds to a circle with center (0,0( on the plane K – plane. This circle forms the periphery of the smith chart. All constant ‘R’ circles touch the point ( 1, 0 ) Including that R = ∞ which is the same as the point itself. Constant X – circles: Consider the equation (2) and cross multiplying, we get 2K x (1 − K r2 ) + K 2 = X 2K x (1 − K r2 ) + K x2 = =0 X Adding (1/x2) on both sides in order to make the ‘Kx” terms a perfect square we get, ( K r − 1) 2 + ( K x − 1/ X )2 = (1/ X ....................(4) Equation (4) represents another family of circles called constant – X circles with centre, (1, 1/x ) and radius (1/x on the k – plane as shown in figure below. ‘X’ being the reactance can be positive or negative whenever ‘X’ is positive the circle lies above the horizontal line. On the other hand when ‘X’ is negative the circle lies below the real axis Kx = 0. When X=0 , the circle degenerates into a straight line because straight line is VEL TECH


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circle whose radius is infinity and for X = 0, the radius 1/x will be infinity. All the circles touch the point (1, 0 ) 8. Derive an expression for the position of attachment and length of short circuited stub will remove the standing wave on a large potion of a transmission line. Consider a transmission line having a characteristic admittance y o terminated in a pure conductance yR as shown in the figure. Since we connect stub in parallel with the main line it is easier to deal with the admittance as they can be added up. We know that yR is different from yo standing waves are set up. When we along the line from the load towards the source (generator), the input admittance will be varying for a maximum conductance through a parallel combination of conductance and inductance a minimum conductance and so on this cycle repeats for every λ / 2. When the line is traversed from the point of maximum (or minimum) conductance to that of minimum (or maximum) conductance, there will be a point at which the real part of the admittance is equal to the characteristic admittance. If a suitable susceptance, obtained by using an appropriate length of a short circuited or open circuited line called stub is added in shunt at this point so as to obtain and 0 resonance with the susceptance already existing, then up to that point matching has been achieved. The input impedance of a transmission line at any point is given as

[ Z R + jZO tan β s ] Z in = ZO

Z R + jZO tan β s ..........(1) Z O + jZ R tan β s

Convert impedance to admittance  Y + jYO tan β s  Yin = YO  R   YO + jYR tan β s   Y Y + j tan β s  Yin =  O r  ..........(2)  YO 1 + jYR tan β s  YR Where Yr = (Normalized load admittance) Yo

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Yin =


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YIN (Normalized input admittance ………..(3) Y0

Rationalizing equ (2) Yin = =

Yr + j tan β s (1 − jYr tan β s) 1 + jYr tan β s (1 − jYr tan β s)

Yr (1 + tan 2 β s ) + j (1 − Yr2 ) tan β s (1 + Yr2 tan 2 β s )

For on reflection Yin = 1 Thus, the stub has too be located at a point where the real part is equal to unity. Yr + (1 + tan 2 β s ) =1 1 + Yr tan 2 β s tan 2 β s (Yr − Yr2 ) = 1 − Yr Yr tan 2 β s = 1 tan 2 β s = 1/ Yr tan β s =

Yo .................(4) YR

This equation gives the location of the stub ‘S’ and can further simplified as

β s = tan −1 Yo / YR 2Π .s = tan −1 Z R / Z o λ λ ∴s = .tan −1 Z R / Zo .................(5) 2Π

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bs (1 − Yr 2 ) tan β s = ............(6) Yo (1 − Yr2 ) tan β s Substitute equation (4) in (6) 2 2 bs (1 − Yr / Yo ) Yo / YR = .................(7) Yo 1 + Yr2 / Yo2 .Yo / YR

=

(1 − Yr2 / Yo2 ) Yo / YR = (1 − YR / Yo ) Yo / YR 1 + YR / Yo

bs =

(Yo − YR ) Yo Yo YR

Advantages of stub matching: 1. It radiates less power. 2. Its effective length may be varied means of a shorting the bars. Disadvantages: 1. Single stub matching is used only for a fixed frequency because as the frequency changes, the location of the stub will too be changed. 2. For final adjustment the stub has to be moved along the line slightly. This is possibly only in open wire line and on co – axial single stub matching may become inaccurate in practice. 9. Determine the maximum value of conductance that can be matched by a double stub tuner with one stub at the load and the other stub at 3/8 back from the load. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 117

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In order to overcome the two disadvantages of the single stub matching. Two short circuit stubs are connected. The lengths of these stubs are adjustable but the positions are fixed.

Let the first stub whose length is It 1 be located at the point ‘A’ at the distance of 'S1’ from load end then the normalized input admittance at that point will be Y Y + j tan β S1 YA = A = r YO 1 + jYr tan β S1 YA = =

Yr + j tan β S1 1 − jyr tan β S1 × 1 + jyr tan β S1 1 − jyr s tan β S1

Yr (1 + tan 2 β S1 ) + j (1 − yr2 ) tan β S1 1 + yr2 tan 2 β S1

y A = s A + jbA Where, Yr Sec 2 β S1 (1 − Yr2 ) tan β S1 SA = ; bA = 1 + Yr2 tan 2 β S1 1 + Yr2 tan2 β S1 When a stub having a susceptance value is altered b1 is added at this point, the new admittance value will be, y A = sA + jbA Since only the susceptance value is altered by the addition of the stub the conductance part remains unchanged The stub length at ‘B’ is adjusted such that the zero value yA is equal to ‘I’ How the location of the stub can be encountered in practice. • The distance s1 can never be more than or equal to λ / 2. VEL TECH


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The distance chosen will be either λ / v or 3 λ / 8. Since matching is obtained between the point ‘B’ and generator, we have reflection loss occurring to the right of ‘B’ due to mismatch.

In order to avoid this loss, sometimes the first stub is located at the load itself. In common practice the distance s1 is of the order of 0.1λ to 0.15λ 10. Give the method of constructing single stub matching using Smith Chart. A single Stub matching is constructing by following the procedure as below. Step 1 : Calculate the normalized impedance of admittance. Step 2 : Mark this point as ‘A’ in the smith chart, Step 3 : Draw the SWR circle through ‘A’ Step 4 : Extend the radial line from point ‘A’ through the centre to the other side of the SWR . Circle and mark this point as ‘B’ Step 5 : Extend the ‘AB’ radial line towards outer perimeter and mark it as ‘C’ Step 6 : Find the intersection of SWR circle and G = 1 circle mark it as ‘D’ Step 7 : Extend the radial line from the center to ‘D’ and continue the line to the perimeter and mark it as ‘E’ Step 8 : The difference between the points ‘C’ and ‘E’ gives the distance of the stub to load. Step 9 : Find the susceptance at the point ‘D’ to cancel this susceptance mark it in opposite direction and mark it as ‘F’. Step 10 : Find the difference between the values of 0.25 and the value at ‘F’ point. This gives the length of the stub. This is the procedure to construct a single stub matching using smith chart. 11. A 9/16λ long lossless line has Zs/Ro = 1.5 + j0.9. Where Zs is the impedance and Ro is the characteristic resistance. Find the load impedance normalized to R0 and also the standing wave ratio. Solution The given values are, Zo = 1.5 + j 0.0 Ro S = 9 /16λ Load impedance may be calculated using the formula.

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Z O Z R + jRO tan βS = RO RO + jZR tan β s  2π 9  Z R / RO + j tan  .  ZS  λ 16  = Z RO  2π 9  1 + j R tan  .  R  λ 16   9π  Z R / RO + j tan    8  1.5 + j 0.9 = Z  9π  1 + j R tan   R  8  (1.5 + j 0.9)1 +

jZ R 9π .tan = Z R / RO + j tan(9π / 8) RO 8

  jZ (1.5 + j 0.9)1 + 1 + R .(0.4142)  = Z R / RO + j (0.4142) RO   (0.5 + j 0.9) + Z R / RO (0.37278 − j 0.6213 − 1) = j (0.4142) Z R j 0.4142 − (1.5 + j 0.9) = RO −0.6272 + j 0.6213 =

j 0.4142 − (1.5 + j 0.9) −1.5 − j 0.4858 = −0.6272 + j 0.6213 −0.6272 + j 0.6213

ZR = 0.8198 + j1.5867 RO K=

Z R − RO Z R / RO − 1 = Z R + RO Z R / RO + 1

k=

0.8198 + j1.5867 − 1 = 0.8198 + j1.5857 + 1

Reflection coefficient

−0.180 + j1.5867 = 0.37569 + j 0.544 1.8198 + j1.58 | k |= 0.3757 Standing wave ratio, S = VEL TECH

1+ | K | 1 + 0.3757 = 1− | K | 1 − 0.3757 VEL TECH MULTI TECH TECH HIGHTECH 120

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S = 2.20 12. If a lossless line has Ro=200 ohms. What length of the line will be required to obtain at the input of an inductance of 8 micro hertz at frequency of 70 mHz with far end short circuited? Repeat the calculation of open circuited received end. Solution The input impedance of a short circuit 4d line is a pure reactance is given by, Z ac = jR0 tan(2π / λ ).S Substitute the values  2π s  J 2π × 70 ×106 × 8 ×10 −6 = j 200 tan    λ  s

 2π s  15π × 70 tan   = 200  λ  2π s = tan −1 17.6 λ 86.75 × π = radians 180 S = 0.241λ The impedance of an open circuited line is given by  2π s  Z oc = jRo cot    λ  Substitute the values,  2π s  J 2π × 70 ×106 × 80 ×106 = − j 200 cot    λ   2π  −2π × 560 cot  = −17.6 = 200  λ  cot ( π − 2π s / λ ) = 17.6

π − 2π s / λ = cos −1 17.6

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3.25 × π radians 180 π (1 − 2s / λ ) = 3.25 × π /180 =

−2 s 3.25 = 1− ; s = 0.981.λ / 2 λ 180 ∴ s = 0.4905λ 13) Explain the terms standing waves , Nodes, standing wave Ratio If the voltage magnitudes are measured along the length of a line terminated in a load other than Ro the plotted values will appear as

Figure: Resistive load of value not equal to RO In the case of either OC(or)SC lines current magnitudes will be same except there will be a λ / ∆ shift of maxima and minima.

Nodes – Minimum (i.e) are points of zero voltage (or)zero currents. Antinodes – maxima (i.e) are points of maxima (voltage or current). For a open circuited like the voltage modes occur at a distance of λ / 4,3λ / 4,5λ / 4.... current mode occurs at a distance of o, λ / 2, λ / 2, λ ..... For a short circuited line, this modal pts get shifted by a distance of λ / 4 and voltage modes occur 0. λ / 2, λ..... and current modes occurs at λ / 4,3λ / 4,5λ / 4......

The ratio of maximum to minimum magnitude of voltage or current in a line having standing waves is called Standing Wave Ratio. VEL TECH


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Scors =


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E max Im ax = E min Im in

14) Derive the interrelation between reflection co-efficient and standing wave Ratio(SWR). From the SWR it is clear that the points of voltage maxima occur at points where the incident and reflected waves are in phase with each other. / E max/ =| Ei | + | Er | _______(1) The voltage minimum occurs at pts. Where with each other. /Emin/=/Ei/-/Er/________(2) From S = |

E max | Ei | + | Er | |= E min | Ei | − | Er |

÷ by Ei on both sides | Ei | Ei | + | Er | Ei | 1+ | k | = | Ei | Ei | − | Er | Ei | 1− | k | 1+ | k | S= (∴ k = Er | Ei ) _____(4) 1− | k | 1− | k | ( s ) = 1+ | k | S=

1k1 =

S −1 S +1

/K/=

⇒

= 1k1=

E ma −1 E min E max + 1 (or ) E min

Im ax −1 Im in Im ax +1 Im in

E max Im ax − Im in or E min Im ax + Im in

15) Draw the phaser diagram for the i/pimpedance on the line and concenent on it . (or) Derive the I/Pimpedance of an dissipationless line.

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2π s 2π s + jI R Ro sin E λ λ Zs = s = 2π s ER 2π s Is I R cos +j sin λ Ro λ ER cos

 ER cos β s +  Zs =  IR cos β + 

IR  Ro sin β s  ER   ER j sin β s  Ro I R  j

divide by c os β s  1+ ER   Zs = I R 1 +  divide by

 I R Ro janβ s  ER  ER j janβ s   Ro IR j

  Ro tanβ s   1+ j ZR  Zs = ZR  ZR 1 + j janβ s    Ro  Z + jRo tan β s  Z s = Ro  R   Ro + jzR tan β s 

__________(2)

Another form of I/P Impedance Equ.

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ER ( Z R + Z o ) β s (ej + ke = j β s) Es 2 zR zs = = Is  zR + Zo  β s I  (ej − ke − j β s )  2 Ro  = R0

 ej β s + ke − j β s   ej β s − ke − j β s   

ej β s =

COS β s + j sin β s

/ ej β s / tan −1

= COS 2 β s + sin 2 | β sτ | = 1

sin β s = tan −1 tan β sτβ cos β s

φ angle of reflection coefficient 1 β s + 1k1 φ − β s   ________(3) = Ro   1 β s − 1k1 φβ s    1 + 1k1 φ − 2 β s  Ro   1 − 1k1 φ − 2 β s  I/P Impedance is maximum at a distance of

φ = 2 β s ________(1) φ i.e. s = 2β 1 + 1k1  zs (max)τ Zo  = szo 1 − 1k1  zs (max) = szo If we travel a distance of λ / 4 from pt. where impedance is maximum, we get a point of minimum impedance. ∴ i / p impedance is min if,

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φ 2π + λ / 4 (∴λ = ) 2β β φ π s= + ) 2β β φ +π ∴s = 2β 1 + 1k1 −π  1 − 1k1  ∴ Z min = Z o  = Zo   1 + 1k1  1 − 1k1 −π  S=

∴ Zmin =

zo s

16) Derive the I/P Impedance expression for a lossless line terminated one a)short circuit b)open circuit a)Short circuit I/P impedance of a dissipation less live is  Z + jRo tan β s  Z s = Ro  R   Ro + jZR tan β s  for a short circuit live ZR = 0, so that

ZSC = Ro

 jRo tan β s    Ro  

ZSC = JRo tan β s As Z is purely reactive or imaginary then Let Z s = jλ s  2π s  ∴ jxs = jRo tan    λ  xs 2π s = tan Ro λ

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Xs Z x is called as normalized impedance. The variation of sc = with length of Ro Ro Ro line S may be as for In this

SJ ;

Xs =∂ Zo

λ Xs ; =∞ 4 Zo λ X SJ ; s = o 2 Zo 3λ X s SJ ; =∞ 4 Zo X S J λ; s = ∂ Zo SJ

For an open-circuited live, Z R = ∞ Ro   1 + j Z tan β S  R  Z oc = R∂  R  o + j tan β S   Z R 

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  1 Z oc = Ro    j tan β S   −j  2π S Z oc = Ro  = − jRo cot  X  tan β S  Xs 2π S = − j cot Ro λ Similarly for open circuit. S = 0,

λ λ λ Xs , ,3 , = ∞, o, ∞, o 4 2 4 Zo

Sc and Oc impedance is purely reactance value. For the first quarter wavelength, shoot ckt, line acts as inductance and next wavelength it acts as capacitance. However, the curves are for the ideal dissipatedness live. In a practical live there will 1o be a small resistance component of impedance indicating source power loss; and P zero or infinite impedance are never achieved the actual values treading to minima and maxima. 1/ R = 1.25+ j.25. Find the length of location of single Go stub funec short circuited. 17. A load of admittance

The normalized load admittance is given by YR = 1.25+ j0.25 Go VEL TECH


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1/ 2R = 1.25+ j.25 1/ R o Ro R o 2R − R o 2R = = R 2R 2R + R o 1− o 2R 1−

1- ( 1.25+j0.25) = 1+ ( 1.25+j.25) -.25-j.25 = -2.25+j.25 0.3535-135o = 2.26386.34o =0.1561-141.34o =0.1561-2.466o Calculating value of cos-1(1K1) Cos-1(1k1)=cos-1 (0.1561)=1.414 Calculation for length of location of stub:Case(1) S1 =

φ + π − cos−1 ( 1k1) 2β

φ + π − cos−1 ( 1k1) = 2π 2   λ  -2.466+π-1.414 = .λI − 0.0587λ 4π

Length of the stub, VEL TECH


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L=

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 1− 1k12  λ  tan−1  2π  21k1 

2  λ −1  1− ( .3535)  = tan  2( 0.3535)  2π  

=0.1469λ Case (2) S1 = =

φ + π − cos−1 ( 1k1) .λ 4π −2.466 + 3.142 + 1.414 .λ 4π

S1 = 0.1662λ length of the stub, L=

 1− 1k12  λ  tan−1  2π  −21k1 

 1− ( 0.3535) 2  λ  = tan−1   −2( 0.3535)  2π  

λ = tan−1 ( -1.3231) 2π

λ =  π − tan−1 ( 1.013231)  2π L = 0.353λ

18. A loss less RF line has Zo of 600 Ω of is connected to a resistive load of 75Ω . Find the position of length of short circulated stub of same VEL TECH


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construction as line which would enable the main length of a line to be correctly terminated at 150MHz. Given ,

f=150MHz Ro=600Ω ZR=75Ω

Finding λ , f.λ =c C 3× 102 λ= = = 2m f 150× 106 The reflection co-efficient is given by, K=

ZR − Zo ZR − R o = Z R + Zo Z R + R o

K=

75− 600 −525 = = −0.777 75+ 600 675

K = 0.7777 πc = 0.7777 180o Case(1) S1 =

φ + π − cos−1 ( 1k1) 2β

β=

2π λ

φ + π − cos−1 ( 1k1) S1 = .λ 4π π+π-cos-1 ( 0.7777) = .2 4π =0.8918M The length of the stub is given by L=

 1− 1k12  λ  tan−1  2π  21k1 

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 1− ( 0.7777) 2  2  L= tan−1   2( 0.7777)  2π   =0.1222m Case (2) S1 =

φ + π + cos−1 ( 1k1) 2β

π+π + cos-1 ( 0.7777) = .λ 4× π =1.108m The length of the stub is given by 2   1− 1k12  λ λ −1 −1  1.( 0.7777)    L= tan = tan  −2( 0.7777)  2π  −21k1  2π  

λ 1 = tan−1 ( −0.4041) = ( π − 0.384) 2π π = 0.8777m Selecting a point located nearest to the load. Hence the stub location nearest to the load is calculated in case1. The stub must be located at a distance 0.8918m from the load of the length of the stub required is 0.1222m. 19. Design a quarter wave transformer to match a load of 200Ω to a source resistance of 500Ω . Operating frequency is 200 MHz. For a quarter wave transformer, the input impedance is given by. Zin = 2s =

R o2 2R

The source impedance Zs = 500 Ω Load impedance =2Ω = 200Ω

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500 =


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R o2 200

R o2 = 500× 200 R o2 = 100000 R o = 316.22Ω f = 200MHz wave length f.λ =c λ=

C 3× 108 = = 1.5m f 200× 106

∴ The length of quarter wave line is given by S=

λ 1.5 = = 0.375m 4 4

C-s=λ /4=.345m. 20. A loss less transmission line with Zo = 75Ω of electrical length l=0.3λ is terminated with load impedance of 2Ω =(40+j20)Ω . Determine the reflection co-efficient at load, SWR of line, input impedance of the line. Solution: Given Zo=Ro=75Ω Zr=(40+J20)Ω Reflection co-efficient is given by

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ZR − R o ZR + R o

( 40+j20) − 75 ( 40+ j20) + 75

=

-35+j20 = 115+ j20 40.31129.74o = 116.7219.86o =0.345319.88o Standing wave ratio is S=

1+ 1k1 1− 1k1

1+0.3453 = 1− 0.3453 =2.0548 Input impedance of the line is given by   2πs    ZR + jR o tan  λ    Zin = Ro   R o + j2R tan ( 2πs/ λ )      2π× 0.3λ     ( 40 + j20) + j75tan  λ   =75  2 π× 0.3 λ   75+ j( 40 + j20) tan   λ     =75

 40 + j20 + j( −230.82)     75+ ( − j123.1) + ( 61.55) 

 40 − j210.82  = 75   136.55− j123.1

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 214.58 −79.25o  = 75 o  183.84 −42.03 

(

= 75 1.167 −37.222o

)

= 69.7 − j52.95Ω 21. A line with zero dissipation has R= 0.006Ω /m, L=2.5µ H/m d C=4.45PF/m. If the line is operated at 10 MHz, find i) Ro,ii)α iii) β iv) ν v) λ Given R=0.006Ω /m L=2.5 × 10-6 H/m C= 4.45 PF/m F=10MHz At f=10MHz, WL=2π fL=2× π fL=2× π × 10× 10-6 × 2.5 × 10-6 =15.708Ω ∴ WL > > R at 10 MHz So according to standard assumption for the dissipation less line, we can neglect R. i) Characteristic impedance Zo = R o =

L 2.5× 10−6 = = 749.53Ω C 4.45× 10−12

ii) Propagation constant ν = α + jβ = 0 + jw LC

(

γ = α + jβ = 0 + j 2π× 10× 10−6

)

2.5× 10−64.45× 10−12

γ = α + jβ = 0 + j0.2095/ m ∴ α =0 β=0.2093 rad/m iii) velocity of propagation 1 1 V= = = 2.998× 108 m/ sec −6 −12 LC 2.5× 10 × 4.45× 10 iv) wave length is given by 2π 2π γ= = = 29.13m β 0.2095

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PART – A 1. What are guided waves? Give examples. The electromagnetic waves that are guided along or over conducting or dielectric surface are called guided waves. Examples: Parallel wire and transmission lines. 2. What is cut-off frequency? The frequency (fc) at which the wave motion ceases, is called the cut-off frequency of the wave guide. 3. Give the expression for guide wavelength when the wave transmitted in between two parallel plates.

The guide wavelength

λc =

2π  mπ  ω µε −    a 

2

2

4. What is TEM wave or principal wave? TEM wave is a special type of TM wave in which an electric field E along the direction of propagation is also zero. [OR] The transverse electromagnetic (TEM) waves are waves in which both electric and magnetic fields are transverse entirely but have no components of E z and Hz. it is referred to as principal wave. 5. Mention the characteristic of TEM waves. It is a special type of TM wave. It does not have either Ez or Hz component. It velocity is independent of frequency. Its cut-off frequency is zero. 6. Define attenuation factor. Attenuation factor α =

power lost /unit length 2 × power transmitted

7. Distinguish TE and TM waves. VEL TECH


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Electric field E is entire Magnetic field H is entirely transverse. transverse. It has z component of magnetic field Hz. It has z component of electric field Ez (direction of propagation) It has no z component of electric field Ez It has no component of magnetic field Hz. (Ez = 0) (Hz= 0) 8. Define wave impedance. Wave impedance is defined as the ratio of electric to magnetic field strength Z xy+ =

Ex Hy

Z xy− = −

in the positive direction

Ex in the negative direction Hy

9. What are the characteristics of TEM waves? TEM wave is a special type of transverse magnetic wave in which the electric field E along the direction of propagation is also zero. 10. Give some examples of guided waves. 1. Electromagnetic waves along ordinary parallel wire 2. Wave in wave guides 3. Wave guided along the earth surface from a radio parameter to the receives. 11. What do you mean by cutoff – frequency? Cut – off frequency can be defined as the frequency at which the propagation constant changes form being real to imaginary. fc =

m 2a M ∑

PART – B 1. Derive the field components of the wave propagating between parallel plates. Consider an electromagnetic wave propagating between a pair of parallel perfectly conducting planes of infinite in the y and z directions as shown in Fig 2.1.

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Fig Parallel conducting guides Maxwell’s equations will be solved to determine the electromagnetic field configurations in the rectangular region. Maxwell’s equations for a non-conducting rectangular region and given as

∇ × H = − jωε E ∇ × E = − jωµ H ax ∂ ∇× H = ∂x Hx

ay ∂ ∂y Hy

 ∂H z ∂H z = ax  − ∂z  ∂x

az ∂ ∂z Hz  ∂H x ∂H z  −  + ay  ∂x   ∂z

 ∂H y ∂H x    + ax  ∂x − ∂y    

= jωε  a x Ex + a y E y + a z Ez  Equating x ,y and z components on both sides,  ∂H z ∂H y − = jωε Ex  ∂y ∂z   ∂H x ∂H z − = jωε E y  ∂z ∂x  ∂H y ∂H x  − = jωε Ez  ∂z ∂y  VEL TECH


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ax ∂ Similarly, ∇ × H = ∂x Ex  ∂e ∂E = ax  z − Y ∂z  ∂y

ay ∂ ∂y Ey

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az ∂ ∂z Ez

  ∂Ex ∂Ez −  + ay  ∂x  ∂z 

 ∂E y ∂Ex    + ax  ∂x − ∂y    

= jωµ  a x H x + a y H y + a z H z  Equating x, y and z components on both sides

 ∂Ez ∂E y − = jωµ H x  ∂y ∂z   ∂Ex ∂Ez − = jωµ H y  ∂z ∂x  ∂E y ∂Ex  − = jωε H z  ∂X ∂y  The wave equation is given by ∇2 E = γ 2 E ∇2 H = γ 2 H Where γ 2 =(σ +jωε ) (jωµ ) For a non-conducting medium, it becomes ∇ 2 E = −ω 2 µε E ∇ 2 H = −ω 2 µε H  ∂ 2 E ∂2 E ∂2 E + 2 + 2 = −ω 2 µε E  2 ∂x ∂y ∂z   2 2 2 ∂ H ∂ H ∂ H 2 + 2 + 2 = −ω µε H   ∂x 2 ∂y ∂z

.....(3)

It is assumed that the propagation is in the z direction an the variation of field components in this z direction may be expressed in the form e-yz, Where VEL TECH

γ is propagation constant γ =α +jβ VEL TECH MULTI TECH TECH HIGHTECH 139

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If α =0, wave propagates without attenuation. If α is real i.e. β =0, there is no wave motion but only an exponential decrease in amplitude. Let

H y = H y0e − yz ∂H y

Similarly, Let

= −γ H y0e − yz = −γ H y ∂z ∂H x = −γ H x ∂z E y = E y0 e − yz ∂E y

Similarly

= −γ E y ∂z ∂E x = −γ Ex ∂z

There is no variation in the direction i.e., derivative of y is zero substituting the values of z derivatives and y derivatives in the equation (1), (2) and (3).  γ H y = − jωε Ex   ∂H z  −γ H x − = jωε Ey  .....(4) ∂x  ∂H y  = jωε Ez  ∂x   γ E y = − jωµ H x   ∂Ez  −γ Ex − = − jωε H y  .....(5) ∂x  ∂Ey  = − jωε H z  ∂x  2  ∂ E + γ 2 E = −ω 2 µε E  2  ∂x .....(6)  2 ∂ H 2 2  + γ H = −ω µε H  ∂x 2 ∂2E ∂2 H 2 Where = γ E and = γ 2H ∂x 2 ∂x2 Solving the equation (4) and (5) , the fields Hx, Hy, Ex and E To solve Hx’ VEL TECH


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can be found out.

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∂H z = jωε Ey ∂x γ E y = − jωµ H x −γ Hx −

From the above equation . −γ E y Hx = jωµ Ey =

1 jωε

∂H z   γ H x + ∂x 

Substituting the value of Ey in the above equation. −γ 1 − jωµ jωε

 1  ∂H z  − jωε  γ H x + ∂x   −γ  ∂H z  Hx = − γ H x + jωµ  ∂x  Hx =

  

 γ2  −γ  ∂H  H x = 1 + 2  = 2  z   ω µε  ω µε  ∂x  ∂H z H x ω 2 µε + γ 2  = −γ ∂x Hx =

−γ ∂H Z 2 ω µε + γ ∂x 2

where h 2 = γ 2 + ω 2 µε To solve h y' ∂Ez = jωε E y ∂x γ H y = jωε Ex From the above equations, jωε Hy = Ex γ

γ Ex +

Ex =

1 ∂E  jωε H y − z   γ  ∂x 

Substituting the value of Ex in the above equation. Hy =

jωε 1  ∂E  .  jωε H y − z  γ γ  ∂x 

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Hy =


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−ω 2 µε jωε ∂E2 Hy − 2 2 γ γ ∂x

 ω 2 µε  jωε ∂Ez H y 1 + 2  = 2 γ  γ ∂x  ∂E H y ( y 2 + ω 2 µε ) = jωε z ∂x Hy =

∂Ex − jωε 2 (γ + ω µε ) ∂x 2

h 2 = γ 2 + ω 2 µε Hy =

− jωε ∂E2 h 2 ∂x

To solve Ex,

γ Ex , ∂E z = jωε H y ∂x jωε Hy = Ex γ Substituting the value of H y in the above equation,

γ Ex

 jωε  ∂E z = jωε  Ex  ∂x  γ  −ωµε = Ex γ ω 2 µε ∂E γ Ex + Ex = z γ ∂x

γ Ex

 ω 2 µε  ∂E Ex γ + =− 2  γ  ∂x  ∂E Ex [γ 2 + ω 2 µε ] = −γ 2 ∂x Ex =

−γ ∂Ez h 2 ∂x

To solve Ey, VEL TECH


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∂H z = − jωε Ey ∂x  −γ 2  ∂H z Ey  − jωε  = − ∂x  jωµ 

γ Hx =

E y [γ 2 + ω 2 µε ] = jωµ

Ey =

∂H z ∂x

jωµ ∂H z h 2 ∂x

h 2 = γ 2 + ω 2 µε Where The components of electric and magnetic field strength (E x,Ey,Hx and Hy) are expressed in terms of Ez and Hz. It is observed that there must be a z components of either e or H; otherwise all the components would be zero. Although in general case Ez and Hz may be present at the same time, it is convenient to divide the solutions into two sets. In the first case, there is a component of E in the direction of propagation (Ez), but no component of H in this direction . Such waves are called E or Transverse magnetic ™ waves. In the second case, there is a component of H in the direction of propagation (Hz), but no component of E in this direction. Such waves are called H waves or transverse Electric (TE) waves. 2. Derive the electro Magnetic field for TE waves. TRANSVERSE ELECTRIC WAVES: Transverse electric (TE) waves are waves in which the electric field strength E is entirely transverse. It has a magnetic field strength Hz in the direction of propagation and no component of electric field Ez in the same direction (Ez=0) Substituting the value of Ez =0 in the following equations.

γ ∂E − jωε ∂E2 and Hy = 2 2 h ∂x h ∂x E x = 0 and H y = 0

Ex = Then

The wave equation for the components Ey

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∂2 E + γ 2 EY = −ω 2 µε EY 2 ∂x ∂2E = −ω 2 µε E y − γ 2 E y 2 ∂x =-(ω 2 µε + γ 2 ) E y But

h 2 = y 2 + ω 2 µε ∂2 Ey ∂x

2

+ h2 E y = 0

This is a differential equation of simple harmonic motion. The solution of his equation is given by Ey=C1 sin hx + C2 cos hx Where C1 and C2 are arbitrary constants. 0 − yz If Ey is expressed in time and direction ( E y = E y e ) then the solution becomes

Ey=(C1 sin hx + C2 cos hx)e-yz The arbitrary constants C1 and C2 are determined from the boundary conditions. The tangential components of E is zero at the surface of conductors for all values of z. Ey=0 at x=0 Ey= 0 at x=a Applying the first boundary condition (x=0) 0=0+C2 C2=0 Then

Ey=C1 sin hx e-yz

Applying the second boundary condition (x=a)

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Where Therefore, ∴

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 mπ  − yz Ey=C1Sin  xe  a  ∂E y mπ  mπ  − yz = C1 cos  xe ∂x a  a 

Equations (5) are γ E y = − jωµ H x ∂E y ∂x

= − jωε H z

γ Ey jωµ Substituting the value of E y in the above equation: From the first equation , H x = −

Hx =

−γ  mπ C1 sin  jωµ  a

 x  e −γ x 

From the second equation , Hz = −

1 ∂E y jωε ∂x

Substituting the value of Ey, in the above equation. =

-mπ  mπ C1Cos  jωµ a  a

Hz =

 x  e − yz 

-mπ  mπ C1Cos  jωµ a  a

 x  e− yz 

The field strengths for TE waves between parallel planes are     −γ  mπ  − yz  Hx = C1 Sin  xe  jωµ  a    −mπ  mπ  − yz  Hz = C1Sin  xe  jωµ a  a  

 mπ E y = C1Sin   a

 x  e− yz 

.......(7)

Each value of m specifies a particular field of configuration or mode and the wave associated with integer m is designated as TEm0.wave or TEm0 mode. The second subscript refers to another integer which varies with y. VEL TECH


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If m =0 , then all the fields becomes zero, Ey=0,Hx=0, Hz=0. Therefore, the lowest value of m =1. The lowest order mode is TE10. This is called dominant mode in TE waves. The propagation constant γ =α +jβ . attenuation, α =0, only phase shift exists.

If

the

wave

propagates

without

γ =jβ Then the fields strengths for TE waves.  mπ  − j β z E y = C1Sin  xe  a  −β  mπ  − j β z Hx = C1Sin  xe jωµ  a  Hz =

jmπ  mπ C1Cos  ωµ a  a

 x  e− jβ z 

The field distributions for TE10 mode between parallel planes are shown in fig.

Fig. Electric and magnetic fields between parallel planes for the TE10. 3. Derive the Electromagnetic fields expression for TM waves. TRANSVERSE MAGNETIC WAVES: Transverse magnetic (TM) waves are in which the magnetic field strength H is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 146

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entirely traverse. It has an electric field strength Ez in the direction of propagation and no component of magnetic field Hz in the same direction (Hz=0) Substituting the value of Hz =0 in the following equations. −γ ∂H z jωµ ∂H z and E y = 2 2 h ∂x h ∂x Then Hx = 0 and E y = 0 Hx =

[ Q Hz = 0]

The wave equation for the component Hy ∂2 H y ∂x

2

+ γ 2 H y = ω 2 µε H y ∂2H y ∂x 2

But

= −(ω 2 µε + γ 2 ) H y

h 2 = γ 2 + ω 2 µε ∂2 H y ∂x

2

+ h2 H y = 0

This is also a differential equation of simple harmonic motion. The solution of this equation is H y = C3 sinh x +C4 cosh x Where C3 and C4 are arbitrary constants. If Hy is expressed in time and direction , then the solution becomes. H y = (C3 sinh x +C4 cosh x)e − yz The boundary conditions cannot be applied directly to Hy, to determine the arbitrary constants C3 and C4 because the tangential component of H I not zero at the surface of a conductor . However Ez can be obtained in terms of Hz.

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∂H Y = jωε Ez [eqn.(4)] ∂x 1 ∂H y Ez = jωε ∂x h = [C3 cos h x − C4 sinh x]e − yz jωε Applying the first boundary condition (E z =0 at x=0) C3 = 0 Then Ez =

−h C4 sin hx e-yz jωε

Applying the second boundary condition (Ez=0 at x=a) mπ a where m is a mode m=1,2,3......... mπ  mπ  − yz Therefore, Ez = − C4 sin  xe jωε a  a  jmπ  mπ  − yz =− C4 sin  xe ωε a  a  h=

 mπ H y = C4 cos   a

 x  e − yz 

But , γ H y = jωε Ex Ez = =

γ Hy jωε γ  mπ C4 cos  jωε  a

 x  e − yz 

The field strengths for TM waves between parallel planes are  mπ  − yz H y = C4  xe  a  γ  mπ Ex = C4 cos  jωε  a

 x  e − yz 

jmπ  mπ C4 sin  ωε a  a

 x  e − yz 

Ez =

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The transverse magnetic wave associated with the integer m is designated as TMm0 wave or TMm0 mode. If m=0 all the fields will not be equal to zero i.e., Ex and Hy exist and only Ez=0. In the case of TM wave there is a possibility of m=0 If the wave propagates without attenuation (α =0), the propagation constant become γ =jβ . The field strengths for TM waves between parallel conducting planes are:  mπ H y = C4 cos   a

 x  e− jβ z 

β  mπ  − j β z C4 cos  xe ωε  a  jmπ  mπ  − j β z Ex = C4 cos  xe ωε  a  Ex =

The field distributions for TM10 wave between parallel planes shown in fig 2.3 Fig The TM10 wave between parallel planes. 4. Describe the field expression for TEM wave guided by parallel conducting plane. TRANSVERSE ELECTROMAGNETIC WAVES: It is a special type of transverse magnetic wave in which electric field E along the direction of propagation is also zero. The transverse electromagnetic (TEM) waves in which both electric and magnetic fields are transverse entirely VEL TECH


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but has not component of Ez and Hz. It is referred to as principal waves. The field strength for TM waves are:  mπ  − j β z H y = C4 cos  xe  a  β  mπ  − j β z Ex = C4 cos  xe ωε  a  jmπ  mπ  − j β z C4 sin  xe ωε A  a  for TEM waves E=0 and the minimum value of m=0

Ez =

H y = C4 e − j β z

β C4 e − j β z ωε Ez = 0 Ex =

The fields are not only entirely transverse , but they are constant in amplitude between the parallel planes. Characteristics: For, lowest value m=0 and dielectric is air. Propagation constant

γ = 0 − ω 2 µ 0ε 0 =jω µ 0ε 0

β = ω µ 0ε 0 Velocity v=

ω 1 = =c β µ 0ε 0

2π c = β f Unlike TE and TM waves, the velocity of TEM wave independent of frequency and has the value c=3 x 108 m/sec. Wavelength λ =

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Fig. The TEM wave between parallel planes The cut-off frequency for TEM wave is zero fc =

m =0 2a µε

(m=0)

This means that for TEM wave, all frequencies down to zero can propagate along the guide, the ratio of E to H between the parallel planes for a traveling wave is E = H

µ0 ε0

The fields distribution are shown. 5. Describe the different of various of propagation between 2 plates and prove that Vg Vp=C2. VELOCITY OF PROPAGATION: The velocity with which the energy propagates along a guide is called group velocity dω dβ If the frequency spread of the group is small enough vg =

dω dβ

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may be considered to be a constant through the group. It is always less then the free space velocity c. Phase velocity is defined as the velocity of propagation of equiphase surfaces along a guide. It is denoted by dω vp = dβ It is always greater then the free space velocity c. The phase-shift is given by  mπ  β = ω µε −    a 

2

2

 mπ  Squaring on both sidesβ 2 = ω 2 µε −    a  Differentiate 2β d β = 2ω d ωµε − 0 dω β 1 = − d β ωµε  ω   β  µε  

2

v2 vp dω vg = group velocity dβ ω phase velocity Where v p = β 1 v= free space velocity µε The product of group velocity and phase velocity is the square of free space velocity vgv p v = 2 vg =

=c 2

vg v p = c 2 6. Obtain the attenuation factor in parallel plane. The field strengths between parallel conducting planes for TE, TM and TEM waves have been obtained without any loss. In actual wave guides, there are some losses. These losses will modify the field strength by the introduction of multiplying α factor e- z. The attenuation factor α that is caused by losses in the walls of the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 152

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guide is determined as follows. The voltage and current phasors in waveguide are V = V0 e −α z e− j β z I = I0 e −α z e− j β z The average power transmitted is 1 Real part of [ VI*] 2 1 = Real part of  V0 e −α z e− j β z I 0 e−α z e j β z  2 1 = Real part of  V0 I 0*  e −2α z 2

Wav =

Where I* is the complex conjugate of I I * = I 0*e −α z e j β z The rate of decrease of transmitted power −∂Wav 1  = 2α  Re al  V0 I*0  e −2α z  ∂z 2  = 2α Wav This is the power lost per unit length or power dissipated per unit length. ∴

Power lost/unit length 2α Wav = = 2α power transmitted Wav

The attenuation factor α is α=

Power lost/unit length 2 × power transmitted

This is the attenuation factor for more general case of guided wave transmission. The α can be determined for TE, TM and TEM waves. 7. Derive an expression for attenuation Factor for TE Waves. The electric and magnetic field strengths between perfectly conduction parallel planes for TE waves are

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 mπ  − j β z E y = C1 sin  xe  α  Hx =

−β  mπ C1 sin  ωµ  α

Hx =

jmπ  mπ C1 cos  ωµα  α

 x  e− jβ z   x  e− jβ z 

The amplitude of linear current density in the conduction planes will be equal to the tangential component of H at x=0 and x = a, J zy = H z =

mπ C1 ωµα

The power loss in each conduction plane is

1 2 J zy Rs 2

Where Rs is the surface resistance. Rs =

ωµ 2α

The power loss in two conducting (upper and lower) planes m2π 2C12 ωµ / 2α 1 2 2 × J zy Rs = 2 ω 2 µ 2α 2 The power transmitted in the z direction is 1 Re al  E × H *  2 1 Power transmitted/unit area = = −  E y H x  2 β c12  mπ  = sin 2  x 2ωµ  a  =

Power transmitted in z direction for a guide 1 metre wide with a spacing between conductors of metres is

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x=a

β c12 2  mπ ∫x=0 2ωµ sin  a

β c2   mπ x  dx = 1 sin2  2ωµ   a

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 x  dx  a

  mπ   sin 2  x 2  β c1 x a    −  = mπ 2ωµ  2  4   a 0 2 βc a = 1 4ωµ Power transmitted/unit length = =

α=

power lost/unit length 2 × power transmitted

m 2π 2C 2 ωµ / 2σ ω 2 µ 2a2 = β C 2a 2× 1 4ωµ

The attenuation factor

α=

β C12 a 4ωµ

2m 2π 2 ωµ / 2σ β ωµ a 3

mπ  Substitute the value of β = ω µε −    a 

2

2

α=

2m 2π 2 ωµ / 2σ  mπ  ωµ a 3 ω 2 µε −    a 

2

The attenuation factor α decreases from infinity at cut-off frequency to a low value at higher frequency.

α=

 mπ  2   a 

2

ωµ 2σ

 mπ  ωµ a ω µε −    a 

2

2

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2

=

 mπ  2  Rs  a   mπ  ωµ a ω 2 µε −    a 

2

where Rs =

 mπ  At cut- off frequency, ω µε −    a 

ωµ 2σ

2

2

2

 mπ  2 2 ω µε −   = ω µε − ω c µε  a  2

ω  = ω µε 1 −  c  ω   f  = ω µε 1 −  c   f  Substituting this value,

α=

2

2

2ω c2 µε Rs  f  ω µα µε 1 −  c   f 

2

2

=

ω  2 Rs  c  ω 

2

f  µ a 1−  c  ε  f 

α=

f  2 Rs  c   f 

2

2

 f  a η 1−  c   f 

2

 Q η = 

µ  ε

8. Derive on Expression for Attenuation Factor for TM Waves. The Expressions for E and H for the Transverse magnetic waves between perfectly conducting parallel planes are

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 mπ  − j β z H y =C4 cos  xe  a  β  mπ  − j β z Ey = C4 cos  xe ωε  a  Ez =-

mπ  mπ C4 sin  jωε a  a

 x  e− jβ z 

The amplitude of the current density in each plane is J = C4 The power loss per unit length in each conducting plane is 1 2 1 j Rs = C42 Rs 2 2 ω µm Rs = 2σ m The power transmitted down the guide per unit are 1 Re al (E × H*) 2 1 = (E x H y ) 2 1  βC  mπ   mπ   =  4 cos  x  C4 cos  x  2  ωε  a   a  1 β C42  mπ  = cos 2  x 2 ωε  a  =

Power transmitted in the z direction for a guide 1 metre wide with a spacing between conductors of ‘s’ metre is 1 β C42 1 β C42  2  mπ cos  x  dx = a 2 x∫=0 ωε 4 ωε  a  Power lost/ unit lengh Attenuation factor (α ) = 2 × power transmitted a

C 24

α= 2

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VEL TECH

VEL TECH MULTI TECH TECH HIGHTECH 2ωε =

But

VEL

ω µm 2σ m aβ

 mπ  β = ω µε −    a 

2

2

The attenuation factor for TM waves is

αTM =

2ω ωµ m / 2σ m  mπ  a ω 2 µε −    a 

2

The attenuation factor for TE waves is 2m 2π 2

αTE = ωµ a Dividing α

TE

by α

α TE α TM

3

ωµ m 2σ m

 mπ  ω µε −    a 

2

2

TM

m 2π 2  mπ  ωµ a 3  a  = = 2 ωε ω µε a

2

2

 mπ    a  =  2 ( 2π f ) µε 2

But

m 1   2a µε =  2 f m 1 fc = . 2a µε f c2 α TE = α TM f 2 2

αTE VEL TECH

 f  =  c  αTM  f  VEL TECH MULTI TECH TECH HIGHTECH 158

VEL

VEL TECH


VEL

The attenuation factor for TM waves is

αTM =

2 ωε R s  mπ  a ω 2 µε −    a 

2

2

 mπ  2 substituting the value of   = ω c µε a  

αTM = =

2 ωε R s a ω 2 µε − ω c2 µε 2 ωε R s ω  aω µε 1 −  c  ω  2R s

=

f  µ a 1−  c  ε  f 

2

 µ ∴η =  ε 

2R s

=

 f  a η 1−  c   f  Rs =

π f µm σm

=

π fc µm σm

Where

2 ∴αTM =

2 or α TM =

2

π f µm σm aη

2

f fc f fc

f  1−  c   f 

2

π fc µm σm

2  f c    fc   a η   1 −     f    f  

VEL TECH


VEL

VEL TECH


The minimum value of α  f    , to zero.  fc 

TM

VEL

is determined by equating derivative of α with respect to

π fc µm σm K= aη f x= fc 2

Let

α =K

x 1 1−    x

=K

2

x3 x2 −1 1

 x3  2 =K  2   x −1  Differentiating with respect to ‘x’,  2 2  3 dα K  3 x ( x − 1) − x (2 x)  = 1 2   dx ( x 2 − 1)  x3  2   2 2   x −1    K x 2 − 1  3x 4 − 3 x 2 − 2 x4  =  2 x 3  ( x 2 − 1) 2   dα =0 Equating dx   K x 2 − 1  x 4 − 3 x2  =0 2 x 3  ( x 2 − 1) 2     4 2   x − 3x  = 0  ( x 2 − 1) 2    x4 = 3 x2 Dividing by x2, x2 = 3

VEL TECH


VEL

VEL TECH


∴

VEL

x= ± 3 f x= fc

But

f =± 3 fc Take only positive value of frequency f = 3 fc The attenuation α TM reaches a minimum value at a frequency of 3 times the cut-off frequency and then increases with frequency. After substituting this value,

α min =

2 Rs  1  η a 1-    3

2

=

2.5R s ηa

9. Define Wave Impedance. Obtain the wave impedance expressions for TE, TEM and TM waves. WAVE IMPEDANCES In transmission-line theory power is propagated along one axis only, and only one impedance constant is involved. However, in the three dimensional wave propagation power may be transmitted along three axes of the coordinate system and consequently three impedance constants must be defined. Wave impedances are defined by the following ratios of electric to magnetic field strengths for the positive directions of the coordinates. Z xy + =

Ex ; Hy

Z yx + = −

Ey Hx

Z yz + =

Ey Hz

; Z zy + = −

;

Zzx + =

Ez Hx

E Ez ; Zxz + = − x Hy Hz

The wave impedances for the negative directions of the coordinates are

VEL TECH


VEL

VEL TECH

Z xy− = − Z yx− =


Ex ; Hy

Ey 1+ x

Z yz− = − Z zy− =

;

Ey Hz

Zzx− = −

;

Ez ; Hy

Zxz− =

VEL

Ez Hx

Ex Hz

For TE waves, the wave impedance is given by Z yx+ = −

Ey Hx

jωµ jβ ωµ = β

=

Z yx+ =

ωµ  mπ  ω µε −    a 

2

2

=

ωµ  mπ  ω µε 1 −  / ϖ µε   a 

µ ε

=

Z yx+ =

 mπ Q ω c = a µε 

1 f  1−  c   f 

2

2

  

η f  1−  c   f 

2

2

 mπ  + At cut-off frequency ω 2 µε =   , wave impedance Z yx becomes infinity. At  a  very high frequency (greater than cut-off frequency) wave impedance becomes, + yx

Z = =

ωµ ω 2 µε

2   mπ   2 Q ω µε >>     a   

µ µ = =η ε µε

VEL TECH


VEL

VEL TECH


VEL

At ω >> ω c, the wave impedance is equal to the intrinsic impedance. =η

Z+yx

For TM waves, the wave impedance is given by  mπ  ω 2 µε −   Ex β  a  + Z xy = = = H y ωε ωε  mπ  1−  / ω µε  a   ωε

ω µε =

ω  1−  c  ω 

µ = ε

f  Z = η 1−  c   f 

2

2

2

2

+ xy

2  2  mπ   At cut-off frequency ω µε =    , the wave impedance becomes zero.  a   

At very high frequency (greater than cut-off frequency), the wave impedance becomes

ω 2 µε ωε µ = =η ε Z xy+ = η Z xy+ =

For TEM wave, the wave impedance is Z xy+ =

Ex Hy

β ωε µo = εo =

=η VEL TECH

o

intrinsic impedance VEL TECH MULTI TECH TECH HIGHTECH 163

VEL

VEL TECH


VEL

Z xy+ = ηo The wave impedances for TE, TM and TEM waves between parallel planes are shown as functions of frequency in figure.

Wave impedance versus frequency characteristics of waves between parallel conducting plane 10. A parallel plane wave guide consists of two sheets of good conductor separated by 10 cm. Find the propagation constant at frequencies of 100 MHz and 10 GHz, when the wave guide is operated in TE10 mode. Does the propagation take place in each case. Given: TE10 mode : m = 1, n = 0 a = 10 cm = 0.1 m f = 100 MHz, 10 GHz For free space µ = µ o and ε = ε

o

1 = 3 × 108 m / sec µ oε o Propagation constant is given by Free space velocity c =

2

 mπ  2 γ=   − ω µε  a  2

 π   2π f  =   −  a  c  2

1 2f  =π   −  a  c  VEL TECH

2

[Q ω = 2π f ]

2


VEL

VEL TECH


VEL

For f = 100 MHz  2 × 100 ×106  1 =π −   (0.1) 2  3 × 108 

2

2

2 = π 100 −   3 = 31.346 Nepers / metre Here propagation constant γ has real value. i.e., γ = α Hence no propagation takes place at 100 MHz. For f = 10 GHz  2 × 10 ×109  1 γ =π −   (0.1) 2  3 ×108   200  = π 100 −    3  = j 207.07

2

2

Here γ has imaginary value i.e., γ = jβ . Hence propagation takes place at 10 GHz. 11. A pair of perfectly conducting planes are separated 8 cm in air. For frequency of 5000 MHz with the TM1 mode excited find the following. (i) cut-off frequency (ii) characteristic impedance (iii) attenuation constant for f = 0.95 fc (iv) phase shift (v) phase velocity and group velocity (vi) wavelength measured along the guiding walls Given : TM1 mode : m = 1 1 εo = F /m a = 8 cm = 0.08 m 36π ×109 µ o = 4π ×10−7 H / m f = 5000 MHz (i) Cut-off frequency:

VEL TECH


VEL

VEL TECH


VEL

2

 mπ  2 γ=   − ω µε a    mπ ω c µε =   a mπ ωc = . a 2

  

2

1 µε

m v 2a

fc =

[Q v =

1 , ω = 2π f ] µε

But v = c = 3 × 108 m/sec 1 × 3 × 108 fc = 2 × 0.08 = 18.75 × 108 Hz = 1.875 GHz (ii) Characteristic impedance:

µo = 120π εo η = 120π ohms or 377 ohms

z=

(iii) Propagation constant becomes attenuation constant (i.e., real value) if the operating frequency is less than the cut-off frequency. f = 0.95 fc 2

 mπ  2 γ =α =   − ω µε  a  2

 mπ   2π f  =   −   a   v  f = 0.95 f c

2

Cut-off wave length λc =

VEL TECH

v 2a = fc m


VEL

VEL TECH


VEL

2

m  mπ   α=   −  2π 0.95  2a   a   2

mπ   mπ   =   −  0.95  a   a   =

2

mπ 1 − (0.95) 2 a

= 39.27 0.0975 = 12.26 Nepers / m (iv) Phase shift:  mπ  β = ω 2 µε −    a 

2

f = 5000 MHz 2

 2π f   mπ  β=   −   v   a 

2

 2 × 5000 ×106   1  =π  −  3 ×108    0.08 

2

2

 100  =π   − 156.25  3  = 97.08 (v) Phase velocity:

ω β 2π f = β

vp =

or

2π × 5000 ×106 = 97.08 = 3.236 ×108 m / sec

VEL TECH


VEL

VEL TECH vp =

=


VEL

c f  1−  c   f 

2

3 ×108  1.875 × 109  1−  6   5000 × 10 

2

= 3.236 ×108 m / sec c2 Group velocity vg = vp (3 ×108 ) 2 = 3.236 ×108 = 2.78 m / sec. (vi) Wave guide wavelength: λ λg = 2 λ  1−    λc  But λ =

3 × 108 = 0.06 m 50 ×108

12. For a frequency of 6000 MHz and plane separation = 7 cm, find the following for the TE1 mode. (i) cut-off frequency (ii) angle of incidence of the planes (iii) phase velocity and group velocity is it possible to propagate TE3 mode? Given : TE1 mode : m = 1, a = 0.07 m f = 6000 MHz (i) Cut-off:

VEL TECH


VEL

VEL TECH  mπ  ω c 2 µε =    a  m 1 fc = . 2a µε


VEL


VEL

2

m v 2a 1 = × 3 ×108 2 × 0.07 f c = 21.4286 ×108 Hz (ii) Phase velocity =

vp =

=

c f  1−  c   f 

2

3 × 108  21.4286 ×108  1−   6  6000 ×10 

= 3.2118 × 108 m / sec Group velocity vg =

c2 vp

(3 ×108 ) 2 3.2118 × 108 = 2.802 m / sec. =

(iii) Angle of incidence c cos θ c 3 ×108 cos θ = = v p 3.2118 ×108 vp =

θ = cos −1 (0.934) = 20.92°

VEL TECH

VEL TECH


VEL

2

 mπ  2 γ=   − ω µε a   2

 mπ   2π f  =   −   a   v  2

2

2

m 2f  =π   −  a  v  2

6  3   2 × 6000 ×10  =π  −    3 ×108  0.07   

2

= π 1836.735 − 16000 = π 236.735 = 48.337 Nepers / m Propagation constant has real value is v = α . Propagation is not possible for TE3 mode. 13. Consider a parallel plate wave guide with plate separation 20 cm with the TE10 mode excited at 1 GHz. Find the propagation constant, the cut-off frequency and guide wavelength assuming ε r = 4 for medium of propagation in the guide. Given: TE10 mode : m = 1, n = 0, a = 0.2 m f = 1 GHz Propagation constant 2

 mπ  2 γ=   − ω µε  a  2

9  1   2 × 10 × 2  =π  −    8  0.2   3 × 10 

 Q v = 

1 = µε

2

 1  µ oε oε r 

= j 39.3 radians / m Cut-off Frequency

VEL TECH


VEL

VEL TECH fc = =

m 2a


VEL

1 µε

m c 2a 4

1 c = µε εr

3 ×108 2 × 0.2 × 2 = 375 MHz =

Guide wavelength

λg =

λ 2

λ 1−    λc  2a λc = = 2 × 0.2 m = 0.4 m 3 × 108 λ= 1×109 = 0.3 m

λg =

0.3  0.3  1−    0.4 

2

= 0.16m 14. A 4 GHz signal is propagated in a rectangular wave guide with internal dimensions of 2.5 × 5 cm. Assuming the dominant mode, calculate: (i) cut-off wavelength (ii) guide wavelength (iii) group velocity (iv) phase velocity and (v) wave impedance Given : TE10 mode : m = 1, n = 0 f = 1 GHz a = 0.05 m b = 0.025 m (i) Cut-off wave length

VEL TECH


VEL

VEL TECH


VEL


VEL

2a m = 2 × 0.05 = 0.10 m

λc =

(ii) Guide wave length :

λg =

λ λ 1−    λc 

2

c f 3 × 108 = = 0.075 m 4 × 109 0.075 λg = = 0.1134 m 2 0.075   1−    0.1 

λ=

(iii) Group velocity λ  vg =   c  λg     0.075  8 =  3 × 10 0.1134   = 1.984 × 108 m / sec. (iv) Phase velocity

VEL TECH

VEL TECH

=

VEL

c

vp =

fc =


f  1−  c   f  mv = 2a

2

m .c 2a

1 × 3 ×108 0.1

f c = 30 × 108 Hz vp =

3 × 108  30 ×108  1−  9   4 ×10 

2

= 4.5356 ×108 m / sec. or

 λg  vp =   c λ   0.1134  8 =  × 3 × 10  0.075  = 4.5356 × 108 m / sec. (iv)

Wave impedance z0 zTE = 2 λ 1−    λc 

Intrinsic impedance z0 = 120π ohms zTE =

120 π

 0.075  1−    0.1  = 570 ohms.

2

15. An uniform plane wave at 2.45 GHz is transmitted through a medium having σ =2.17 s/m ∑=47∑ o µ =µ o. Find the complex propagation constant, phase velocity and the wave impedance of the medium. If the electric field mag is 10V/m, find the time average power flow per unit area. VEL TECH


VEL

VEL TECH


VEL

Given : F=2.45GHz=2.45× 109Hz W=2π f=2π × 2.45× 109=15.394× 109 σ =2.17s/m ∑=47∑o=47× 8.854× 10-12=416.14× 10-12 F/m µ =µ o=4π × 10-7 H/m γ = jωµ ( σ + jω∑ ) = jωµσ-ω2µ ∑ = -123922.51+j41977.94 =361.580.63o γ =58.85+j356.65 α =58.85N/m β =356.65 rad/m Vp=ω /β =4316× 108m/s ωµ wave impedance Z = β

=54.24Ω jωµ o σ + jω∑ Time average power flow / unit area 1 Pav= EH 2 1 = × 10× 192=96w/m2 2

Intrinsic impedance η =

16. What are the characteristics of TE and TM waves? (1) Propagation constant

VEL TECH


VEL

VEL TECH


VEL

h2 = γ 2 + ω2µ 2 ∑ γ 2 = h2 − ω2µ 2 ∑ r = h2 − ω2µ 2 ∑ mπ sub h= a 2

mπ  2 r =   −ω µ∑  a  γ = α + jβ 2

 mπ  ,∴ γ at lower freq. ω µ ∑ is <  becomes real with value of α , and β =0. ∴ there   a  is only attenuation, without any propagation. 2  mπ  , 2 At higher frequencies value of ω µ ∑ becomes greater than  making   a  γ .imaginary. mπ 2 2 For f
2

mπ  2 o =   −ω µ∑  a  mπ  ω2µ ∑ =    a  ωc µ ∑ =

mπ a

2πfc µ ∑ = fc =

2

mπ a

m 2a µ ∑

VEL TECH


VEL

VEL TECH


VEL

(iii) Phase – constant :2

mπ  sub ωc2µ ∑ =   in eqn of γ =  a 

2

 mπ  − ω2µ ∑    a 

γ = ωc2µ ∑ −ωc2µ ∑

(

γ = µ ∑ − ωc2 − ω2 γ = j µ∑

)

ω2 − ωc2

γ = j µ ∑ f 2 − fc2 = jβ β = 2π µ ∑ f 2 − fc2 (iv) cut-off wave length λc =

V fc

fc =

m  1 mv  =  but v=  2a µ ∑  µ ∑ 2a 

λc =

2a µ ∑ j1 × m µ∑

λc =

2a m

(v) Group wavelength:-

VEL TECH


VEL

VEL TECH v=

ω = fλ β

λ=

ω 2πf/ = β f βf/

λg =


VEL

2π mπ  ω2µ ∑ −    a 

2

2

mπ  2 sub   = ωc µ ∑ a   2π

λg =

ω µ ∑ − 1− =

ωc2 ω2

2π

ωc2 2πf 1− 2 ω

V f λg = ωc2 1− 2 ω vi) Velocity of propagation:ω ω V= = mπ 2 β ω2µ ∑ −    a  mπ  2 2 sub  = ωc µ ∑ a   ω

V=

ω µ ∑ 1− V=

−ωc2 ω2

Vo f 1−  c  f

2

UNIT – V VEL TECH


VEL

VEL TECH


VEL

PART – A 1. Write down the Maxwell’s equation for the loss-free region within the guide? ∂Ez + γ EV = − jw µ Hz ∂y ∂Ez ∂H z + γ E x = jw µ Hv + γ H x = − jw ∈ EV , ∂y ∂x ∂H y −∂H x ∂E y −∂E x = jw ∈ Ez , = − jw µ H z ∂x ∂y ∂x ∂y

∂Hz + γ Hv = jω ε Ex,

2. What is the wave equation for E2 and H2? 2 ∂ 2 E z ∂ 2 Ez + + γ E z = −ω 2 µε Ez 2 2 ∂x ∂y 2 ∂ 2H z ∂ 2 Hz + + γ H z = −ω 2 µε Hz 2 2 ∂x ∂y

3. In waves between parallel plates what are the classification by field configurations? 1. Transverse Magnetic Waves (TM) 2. Transverse Electric waves (TE) 4. For Rectangular guide shown in figure what is the boundary condition?

Ex = Ez = 0 At y = 0, and y = b Ev = Ez = 0 at x = 0 and x = a. 5. Write down the propagation constant for a rectangular guide for TM waves. VEL TECH


VEL

VEL TECH


VEL

ν = G 2 − ω 2 µε =

A 2 + B 2 − ω 2 µε 2

2

 mπ   nπ  =  + − ω 2 µε    a   b  6. What is the propagation constant for an ordinary transmission line?

γ =α + jβ α → Attenuation constant β → Phase constant. 7. What is the expression of attenuation constant, propagation constant for a perfectly conducting wall?

α = 0 such that ω > ω 2

 mπ   nπ  = ω µε −   −   a   b 

β

c

2

2

2

8. The Cut-off frequency that propagation will not occur, is fc =

1 2π µε

2

 mπ   nπ  =   +   a   b 

2

 mπ   nπ  =   +   a   b 

1 The value of ω c = µε

is

the

frequency

below

which

wave

2

9. The cur-off wavelength, that is the cutoff wavelength which wave propagation will not occur is 2 λ0 = 2 2 m n   a  +b     10. What is dominant wave? The wave which has the lowest cut-off frequency is called the dominant wave. 11. For TE10 wave, what is cut-off frequency? Cut-off frequency is that frequency for which the corresponding half wavelength is equal to the width of the guide, the cut-off frequency is independent of the VEL TECH


VEL

VEL TECH


VEL

dimension b. 12. What is a waveguide? A hallow conducting metallic tube of uniform section (rectangular or circular) is used for propagating electromagnetic waves. Waves that are guided along the surface (walls) of the tube is called waveguide. 13. What do you mean by propagation of waveguide? Propagation of waveguide can be considered as a phenomenon in which waves are reflected from wall to wall and hence pass down the waveguide in a zig-zag fashion. 14. Write down the maxwell’s equation for non-conducting medium. ∇xH = jωξ E ax ay ∂ ∂ ∇xH = ∂x ∂y Hx Hy

[∴ σ =0] az ∂ ∂z Hz

= jω ε ( a xEx+ a yEy+ a zEz]

15. Write down the wave equation for rectangular waveguide. ∂ 2 E z ∂ 2 Ez + + γ 2E z = −ω 2 µε Ez 2 2 ∂x ∂y 16. What is Dominant mode? The lowest mode for TE wave is TE10 (m = 1, n = 0) whereas the lowest mode for TM wave is TM11 (m = 1, n=1). This wave has the lowest cut-off frequency. Hence the TE10 mode is the dominant mode of a rectangular waveguide. 17. What is wave impedance? The wave impedances defined as the ratio of electric field intensity to magnetic field intensity are E E E + + Z xy = x ; Zyx = y ; Zzx+ = z Hy Hx Hx 18. What are the wave impedance for different modes? fc For TM, z = η 1 −   f  VEL TECH

2


VEL

VEL TECH

For TE,

z=


VEL

η  fc  1−   f 

For TEM, z = η =

2

µ ε

19. What is characteristic impedance? For transmission lines the integrated characteristic impedance Zo can be defined as in terms of the voltage current or interms of the power transmitted for a given voltage or given current. V i.e., Zo (V, I) = I 20. What are the sources of attenuation in wave guide? Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. 21. What is attenuation constant in propagation? It is given by Power lost per unit length α= 2 x power transmitted. 22. How will you calculate power? The power transmitted is obtained by integrating the axial component of the pointing vector through the cross-section of the guide the pointing vector Pz is given by Pt = ½ |Etrans||Htrans| 23. What is attenuation constant in terms of power? The attenuation constant is

α=

Power lost 2 x power transmitted.

24. What is attenuation constant for TM11 mode?

VEL TECH


VEL

VEL TECH

α=


VEL

a   b 2Rs  2 + 2  b  a 2

 fc   1 1  ηab 1 −    2 , 2   f  a b 

α in (dB| m)

25. Draw the variation in attenuation with frequency due to wall losses in a rectangular waveguide?

TM11 TE10F 26. Write the field expression for rectangular TM waves. − j βc B cos Bx sin Ay h2 − j βc = A sin Bx cos Ay h2 jωε c = 2 A sin Bx cos Ay h − jωε c = B cos Bx sin Ay h2

Ex0 = Ey0 Hx0 Hy0

27. Write the field expression of TE wave guide. jβ CB sin Bx cos Ay h2 jβ = 2 CA cos Bx sin Ay h

Hx0 = Hy0

Hz0 = C cos Ay cos Bx jωµ CA cos Bx sin Ay h2 − jωµ Ey0 = CB sin Bx cos Ay h2 28. What are the types of waveguides? Ex0 =

1. Rectangular waveguide 2. Circular waveguide VEL TECH


VEL

VEL TECH


VEL

3. Elliptical waveguide 29. Mention the application of waveguides. The waveguides are employed for transmission of energy at every high frequencies where the attenuation caused by waveguide is smaller. Waveguide are used in microwave transmission circular waveguides used as attenuation and phase shifters. 30. Why are rectangular waveguides preferred over circular waveguides? Rectangular waveguides are preferred over circular waveguides because of the following reasons. 1. Rectangular waveguide is smaller in size than a circular waveguide of the same operating frequency. 2. It does not maintain its polarization through the circular waveguide. 31. Why is rectangular or circular form used as waveguides? Waveguides usually take the form of rectangular or circular cylinders because of its simpler form in use and less expensive to manufacture. 32. For an air filled copper x-band waveguide with dimension a = 2.286 cms and b = 1.016 cms determine the cut off frequencies for TE11 and TM11 modes? a = 2.28 6 cm = 2.286 x 10-2 m b = 1.016 cm = 1.016 x 10-2 m For TE11 mode, m = 1, n = 1 Cut-off frequency, fc = 2

=

c m n  + 2  a   b 

2

1 2π µε

 mπ   nπ   a  + b     

2

2

2

2

3 × 108  102   102  =   +  2  2.286   1.016 

FC = 16.156 GHz. The cut of frequency for TE11 mode is same as that of TM11. 33. What is an evanescent mode? When the operating frequency is lower than the cut-off frequency the propagation constant becomes real i.e., Y = α . The wave cannot be propagated. This non-propagating mode is known as evanescent mode. VEL TECH


VEL

VEL TECH


VEL

34. Which are the non-zero field components for the TE10 mode in a rectangular waveguide? Hx, Hz and Ey 35. Which are the non-zero field components for the TM11 mode in a rectangular waveguide? Ex, Hy Ey and Ez

Wave impedance, η

36. Draw a neat sketch showing the variation of wave impedance with frequency for TE and TM waves in a waveguide.

Region of number TE fc 37. What is the cut-off wavelength and cut-off frequency of the TE10 mode in a rectangular waveguide? Propagation TM Cut-off wavelength, λ c = 22 C Cut-off frequency, fc = 22 38. What is the cut-off wavelength and cut-off frequency of the TM 11 mode in a rectangular waveguide? 2 Cut-off wavelength, λ

c

Cut-off frequency, fc =

2

=

 1  1 a +b     2

1 2 µε 2

2

2

 1  1 a +b    

ν  1  1 = + 2  a   b 

2

39. What is the wave impedance of TEM waves in waveguide? VEL TECH


VEL

VEL TECH


VEL

Wave impedance of TEM becomes the intrinsic impedance of the medium Z =η =

µ ε

40. Write down the expression for wave impedance of TE mode? Z=

η f  1−  c  f 

2

41. Write down the expression for wave impedance of TM mode? f  Z = η 1−  c  f 

2

42. Write down the expression for phase velocity in a waveguide?

ν =

ω 2

2

 mπ   nπ  ω µε −   −   a   b  2

43. Define character is the impedance in a waveguide. For transmission lines the integrated characteristic impedance can be defined as in terms of the voltage current ratio or in terms power transmitted for a given voltage or a given current. V i.e., Zo (V, I) = I 2w Zo ( w , I ) = II * VV * Zo ( w , I ) = 2W Where V and I are peak phasors. W is the power transmitted. * indicates complex conjugate. 44. What are the types of power loss in waveguides? There are two types of power loss occurs in waveguides 1. Loss due to attenuation of signals below cut-off frequency. 2. Loss due to dissipation with in the waveguide walls and the dielectric with in the waveguide. 45. The larger dimension of the cross section of a rectangular waveguide is VEL TECH


VEL

VEL TECH


VEL

2cm. Find the cut off frequency and wavelength for the domain and TE mode. The given data are a = 2 cm. We know the dominant TE mode is TE10 mode, thus m = 1 and n = 0. The cut-off frequency of rectangular waveguides is 2

2

v m n vm Since n = 0 fc = +  =   2  a  a 2 a 3 × 108  1  1.5 × 108 = = = 7.5 × 109 = 7.5GHz   2  0.02  0.02 2π Cutoff wavelength λ

λc =

c

=

2 ( 0.02 ) 1

2

2

 mπ   nπ   a  − b     

=

2a m since n = 0

= 0.04mts

46. Explain why TEM waves are not possible in rectangular waveguide. (Apr. 2004.) Since TEM wave does not have actual component of a E or H. propagated with in a single conductor wave guide.

it cannot be

There fore for a displacement current the guide requires an axial component of E, which is not present in TEM waves. 47. A rectangular wave guide has the following dimensions l=2.54cm b=1.27cm, wave guide thickness = 0.127 cm. Calculate the cut off frequency for TE11 mode. 2

2

C  m π   nπ  + 2π  a   b  48. Explain why TM01 and TM10 modes in a rectangular wave guide do not exist. (May 2006) fc =

For the modes TM01 and TM10, Fe > f Where f → frequency of the wave to be propagated ∴ The dominant mode in TM mode VEL TECH


VEL

VEL TECH


VEL

is TM11 49. Define the difference between the wave impedance and the intrinsic impedance. The intrinsic impedance η is given by the ratio between the permeability and the permittivity. M For free space, η = ; 377Ω for cu. ε The wave impedance is the radio between the electrostatic energy and the magnetic field energy. ∴Z =

Ey Hx

=

Ex Hy

50. Define surface impedance. The surface impedance is defined by the conductivity. Rs =

wµ 2σ

51. Give the attenuation factor for TM waves

α=

Power lost / length 2 × Power transmitted/length

where the power loss in the 4 walls of the guide is the sum of losses in x=0 and y=0. 52. Which mode is the dominant mode in a rectangular wave?

(Nov 2005)

The dominant modes are, TE10 and TM11 But the lowest mode is TE10 53. Show the excitation method of TE11 and TM11 modes is a rectangular wave guide.

VEL TECH


VEL

VEL TECH


Fig. (a) TE11

VEL

(b) TM11

54. Give the attenuation factor for TM wave is a rectangular wave guide. (Apr. 2005)

α TM

  2Rs  = ηab   1    a2 

where

Rs =

a b  a 2 + b2    1  fc  + 2  1−   b  c

2

wµ 2σ

55. What is a guided wavelength?

λg =

λ 2

λ  1−    λc  c where λ = and λc = cut off wave length. f 56. Mention the applications of circular waveguides. Circular waveguides are used as attenuators and phase shifters. 57. Which mode in a circular waveguides has alternation effect decreasing with increasing in frequency? TE01 58. Mention the dominant modes in rectangular and circular waveguides. For a rectangular waveguide, the dominant mode TE10, For a circular waveguide VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 188

VEL TECH


VEL

the dominant mode TE11. 59. Write down the expression for cut-off frequency in a circular waveguide. fc =

hnm 2π µε

where hnm =

( ha ) nm a

60. Calculate the cut off frequency of copper tube with 3 cm diameter inside with air filled, in TE11 mode. fc =

hnm 2π µ m

h11 =

; a=

3 × 10 −2 m 2

( ha ) 11

a 3.85 × 2 = 3 × 10−2 = 2.566 x 102

[ (ha)11 = 3.85]

2.566 × 102 × 3 × 108 2π = 12.25 GHz

fc =

 1  Q µε = c   

61. Determine the cut-off frequency of a circular waveguide with a diameter of 2.36 cms operating in the dominant mode. 2.36 × 10−2 m. 2 Dominant mode is TE11, ( ha ) nm hnm = a ( ha ) 11 = a 3.85 × 2 × 102 = 2.36 = 3.263 x 102 a=

fc =

hnm 2π µε  1 8 Q µε = 3 × 10    VEL TECH MULTI TECH TECH HIGHTECH 189

3.263 × 102 = × 3 × 108 2π VEL TECH

VEL

VEL TECH


VEL

fc = 15.58 GHz. 62. Why is TM01 mode prepared to the TE01 mode in a circular waveguide? TM01 mode is preferred to the TE01 mode, since it requires a smaller diameter for the same cut-off wavelength. 63. Define Q of a waveguide. Quality factor Q is given by Q = ω =

energy stored/unit length energy lost / unit length / second.

64. Give the relation b/w quality factor and attenuation factor of a waveguide? Q=

ω 2Vg 2

65. Which of the following wave guide is easier to manufacture? Circular ⇒ Rectangular ⇒ Elliptical ⇒ none Ans (a) circular ⇒

66. Which of the following waveguide / transmission line would offer the maximum attenuation? a) Coaxial b) Rectangular waveguide c) Circular waveguide. Ans: (a) coaxial. 67. A circular waveguide will behave like a) Low pass filter b) Band pass filter c) High pass filter d) Non of the above Ans: (c) High pass filter. 68. What are the uses of circular wave guides? Circular waveguides are used as attenuators and phase shifters. 69. Draw the variation of attenuation as a function of frequency for different VEL TECH


VEL

VEL TECH


VEL

Attenuator

modes.

TE11

TM01

TE11

70. What is the wave impedance of a circular waveguide? The wave impedance of a circular waveguide is the ratio of the resultant Frequency transverse electric fields to the transverse magnetic field. For TM waves E 2TM = H 71. What is Bessel function? In solving for the electromagnetic fields within guides of circular cross section, a differential equation known as Bessel’s Equation is encountered. 1   ρ ∞ r 2 P = C1  ( −1) ∑ 2 r =0 ( r !)

2r

This series is convergent for all values of p1 either real complex. It is called Bessel function of first kind of order zero d is denoted by J0 (P). 72. What the cut-off freq or critical frequency below which the transmission of a wave will not occur? hnm fc = 2π µε where hnm =

( ha ) nm a

73. What is the expression for TM waves in circular guides? H20 = Cn Jn (hp) cos n φ VEL TECH


VEL

VEL TECH


VEL

− j βCn Jn (hp) cos n φ h jn βCn Hφ 0 = Jn (hp) sin n φ h2 p ωµ 0 Fp0 = H β p −ωµ 0 Eφ0 = Hp β Hp0 =

74. What are the boundry conditions for TM waves in circular guides? The boundry conditions to be met for TM waves are that Eφ = q at P = a. 75. What is the wave impedances at a point? + Z xy =

+ Z yx =

Ex Hy

−E y Hx

Zy+z =

Ey

Zzx+ =

Hz

Z zy+ = −

Ez Hy

Ez Hx

+ Zxz =

−E x Hz

76. What is the work impedence for waves guided by transmission lines / wave guides? E x2 + Ey2 Etrans = Z2 = Htrans H x2 + H y2 Zz = ZPφ = -Zφ p = Zz ( TM ) =

β ωε

µ w 2  1−  c 2  ε  w 

ω 2  = η 1-  c 2   ω  β = DE 77. Draw the characteristic between wave impedance and frequency. TE

Wave impedance TM

VEL TECH

VEL TECH MULTI TECH TECH HIGHTECH 192 fc

f

VEL

VEL TECH


VEL

78. Define wave impedance. The wave impedances are defined by the following rations of electric to magnetic field strengths for the positive direction as well as negative direction of the co-ordinates. Z +xy =

Ex Hy

Z -yx =

Ey Hx

Z +yz =

Ey Hx

Z -zy =

Ez Hy

Z +zx =

Ez Hx

Z -xz =

Ex Hz

79. Write condition for minimum attenuation for TM waves. dα =0 f   d   fc  α TM =

2wμ wε . σ βa =

2wμ . σ

2πfε 2

mπ  ωμε-   .a  a 

simplifying we get, f= 3fc

80. Write the wave impedances for TE, TM Q TEM at cut off frequency. * For TE waves at fc 1 ZTE =α For TM waves at fc 1 ZTM = 0 For TEM waves at fc 1 ZTEM = η 81. Relationship between α TM VEL TECH

and α TE:


VEL

VEL TECH


VEL

αTE  fc2 =  α TM f 2 The ratio between the attenuation factors of TM and TE waves is given by the ratio between the cutoff frequency and the wave frequency. 82. Give the equation for the power loss at the magnetic field. Power loss = I2R = (JYZ)2 . Rs = (Hy)2 . Rs (Hy)

= C4

\ Power loss=C24

ωμ 2σ

PART – B 1. Write the expression for transverse magnetic waves in rectangular wave guides. The wave equations are partial differential equations that can be solved by the usual technique of assuming a product, solution. This procedure leads to two ordinary differential the solutions of which are known. Nothing that Ez (x, y, z) = Ez0 (x, y) e-y2 Ez0 = xy Where X is a function of x alone, and y is a function of y alone. y

2 d 2x d 2y + x + y xy = −ω 2 µε xy 2 2 dx dy

2

Putting h2 = y + ω VEL TECH

2

µε

as before VEL TECH MULTI TECH TECH HIGHTECH 194

VEL

VEL TECH

y


VEL

d 2x dy + x 2 + h 2 xy = 0 2 dx dy Divide by XY 1 d 2x 1 d2y 2 + h = − X dx 2 y dy 2

The above equation equates a function of x alone to a function of y alone. The only way in which such a relation can hold for all values of x and y is due to have each of these function equal to some constant. 1 d 2x + h 2 = A2 2 X dx 1 d 2y = − A2 2 y dy The solution of equation is X = c1 cos Bx + c2 sin Bx B2 = h2 – A2 The above solution is y = c3 cos Ay + c4 sin Ay This gives E20 = xy = c1c3 cos Bx cos Ay+c1c4 cos Bx sin Ay + c2c3 sin Bx cos Ay + c2 c4 sin Bx sin Ay. The constants c1, c2, c3, c4, A and B must now be selected to fit the boundary condition. E2 0 = 0

when x = 0, x = a, y = 0 y = b.

If X = 0 the general expression. E20 = c1 c3 cos Ay + c1 c4 sin Ay E20 = c2 c3 sin Bx cos Ay + c2 c4 sin Bx sin Ay When y = 0 reduces to E20 = c2 c3 sin Bx For this to be zero for all values of x if it possible to have either c 2 (or) c3 equal VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 195

VEL TECH


VEL

to zero (assuming B ≠ 0) putting c2 =0 would make E20 identically zero. So instead c3 will be put equal to zero. E20 = c2 c3 sin Bx sin Ay In addition to the amplitude constant c = c2 c4 If x = a E20 = c sin Ba sin Ay. In order for this to vanish for all values of y (and assuming A ≠ 0) (Because A = 0 would make E20 identically zero) The constant B must have B=

mπ when m = 1, 2, 3 a

Again if y = b; E20 = c sin

A=

mπ x sin Ab a

nπ where n = 1, 2, 3. b

Therefore the final expression for E20 is = c sin

y = jβ

mπ nπ y x sin a b

Ey0

B=

H z0 =

jωε c A sin Bx cos Ay h2

H y0 =

− jωε c B cos Bx sin Ay h2

mπ nπ and A = a b

VEL TECH

− j βc B cos Bx sin Ay h2 − j βc = A sin Bx cos Ay h2

Ex0 =


VEL

VEL TECH


VEL

These expressions show how each of the components of electric and magnetic field strengths varies with x and y. The variation with time and along the axis of the guide, that is M the z direction, is shown by putting back intro each of these expressions the factor ejwt- xz and then taking the real port. In the division of the fields it was found necessary to restrict the constants A and B to the values given by expressions. A2 + B2 = h2 h2 = y 2 + ω µ ε y = h 2 − ω 2 µε = A2 + B 2 − ω 2 µε 2

2

 mπ   nπ  =  + − ω 2 µε    a   b  The above equation defines the propagation constant for a rectangular guide for TM waves. For low frequencies where ω 2 µ ε is small y will be a real number. This propagation constant met with in ordinary transmission-line theory is a complex number. That is y = a + j β where a is the mathematics constant and β is the phase shift constant. If y is real, β must be zero and there can be no phase shift along the tube 2

 mπ   nπ  β = ω µε −   −   a   b 

2

2

This means there can be no wave motion along the tube for low frequencies y = jβ . The attenuation constant a is zero for all frequencies such that w > wc.

ωc =

fc =

2

µε

2

 m π   nπ   a  + b     

1 2π µε

λc =

2

 mπ   nπ   a  + b     

1

2

2 2

m n   a  +b    

VEL TECH

2


VEL

VEL TECH


VEL

fcλ c = v0 v=

ω = β

ω 2

2

 mπ   nπ  ω µε −   −   a   b  2

This last expressions indicates than the velocity of propagation of the wave in the guide is greater than the phase velocity in free space. Since the wave length in the guide is given is given by L = v

f

it will be longer

than the corresponding free space wave length. 2π

L=

2

2

 mπ   nπ  ω µε −   −   a   b  In the above expressions the only restriction on m and n is that they be integers. However from the above equations it is seen that is either m or n is zero fields will all be identically zero. Therefore the lowest possible values for the either m or n is unity. The lowest cut off of frequency will occur for m = n = 1 This frequency TM waves which can be propagated through the guide. This particular wave is called the TM11 wave for obvious reasons. High order waves require higher frequencies in order to be propagated along a guide of given dimensions. 2

2. Derive the Expression for Transverse Electric waves in Rectangular waveguides. The wave equation, in a rectangular waveguide is given by ∂ H z ∂ 2 Hz + + γ 2H z = −ω 2 µε Hz 2 2 ∂x ∂y The solution of the equation is, 2

Hz (x, y) = Hz0 (x, y) e-vz Hz0 (x, y) = xy Let Where x → function of x only. Y → function of y only. Substituting the value of H2 in wave equation,

VEL TECH


VEL

VEL TECH


y

d 2x d 2y + x + γ 2 xy = −ω 2 µε xy 2 2 dx dy

y

d 2x d 2y + x + h 2 xy = 0 dx 2 dy 2 where G2 = γ + ω

2

VEL

µ ε

Dividing by xy 1 d 2x 1 d2y + + h2 = 0 2 2 X dx y dy 1 d 2x 1 d2y 2 +h = − X dx 2 y dy 2 The equation relates a function of x alone to a function of y alone and this can be equated to a constant. 1 d 2x + h 2 = A2 X dx 2 1 d 2x + h 2 − A2 = 0 2 X dx let 1 d 2x + B2 = o 2 X dx The solution of this equation is, X = C1 cos Bx + C2 sin Bx Similarly −

1 d 2y = A2 2 y dy

1 d 2y + A2 = 0 2 y dy The solution of this equation is y = C3 cos Ay + C4 sin Ay But H20 = xy = (c1 cos Bx + c2 sin Bx) (c3 cos Ay + c4 sin Ay) VEL TECH


VEL

VEL TECH


VEL

= c1 c3 cos Ay cos Bx + c2 c3 cos Ay sin Bx + c1 c4 cos Bx sin Ay + c2 c4 s in Ay sin Bx It is known that, Ex = −

γ 2 ∂E2 jωµ ∂H z − 2 h 2 ∂x h ∂y

For TE waves Ez = 0 − jωµ ∂H z h 2 ∂y − jωµ = [ -c1 c3 A sin Ay cos Bx – c2 c3 A sin Ay sin Bx + c1 c4 A cos Bx cos Ay + c2 h2 c4 A cos Ay sin Bx] Applying Boundary conditions. Ex =

\

E2 = 0, when y = 0, y = b If y = 0, the general solution is Ex =

− jωµ (c1 c4 A cos Bx + c2 c4 A sin Bx ] = 0 h2

For Ex = 0, C4 = 0 ( c4 is common) Then the general solution is Ex =

− jωµ [ - c1 c3 A sin Ay sin Bx – c2 c3 A sin Ay Sin Bx] h2

If y = b,

Ex = 0

For Ex = 0, it is possible either B = 0 or A =

nπ , b

It B = 0, the above solution is identically new, so it is bitter to select A =

nπ . b

The general solution is jωµ [ c1 c3 Ay cos Bx + c2 c3 A sin Ay + Sin Bx] h2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 200

Ex0 =

VEL

VEL TECH


VEL

Similarly for Ey Ey = −

γ 2 ∂E2 jωµ ∂H z + 2 h 2 ∂y h ∂x

jωµ ∂H z h 2 ∂x

=

jωµ [ -c1 c3 B cos Ay Sin Bx + c2 c3 cos Ay cos Bx -c1 c4 B sin Bx sin Ay + c2 c4 B sin B h2 sin Ay cos Bx] =

Applying boundry conditions, Ey = 0; Ey0 =

x = 0 and x = a

jωµ [c2 c3 B cos Ay+c2 c3 B cos Ay + c2 c4 B Sin Ay] h2

For Ey0 =0, c2 = 0 Then the general solution is Ey0 =

jωµ [- c1 c3 B cos Ay sin Bx – c1 c4 B sin Bx Sin Ay] h2

If x = a, then Ey0 = 0 jωµ Bc1 sin Bx [c3 cos Ay + c4 Sin Ay] h2 mπ For Ey0 = 0, either A = 0 or B = a Ey0 =

Since A = 0, will make Ey identically zero, it is better to take B = − jωµ [ c1 c3 B sin Bx cos Ay + c1 c4 B sin Bx Sin Ay] h2 jωµ = 2 [c1 c3 A sin Ay cos Bx + c2 c3 A sin Ay Sin Bx] h

mπ a

Ey0 = Ex0

Substituting the value of c2 = c4 = 0 VEL TECH


VEL

VEL TECH


VEL

jωµ c1 c3 A sin A cos Bx h2 − jωµ Ey0 = c1 c3 B sin Bx cos Ay h2 Let C = c1 c2 Ex0 =

Ex0 =

jωµ CA sin A cos Bx h2

Ey0 =

− jωµ CB sin Bx cos Ay h2

Where A =

nπ mπ ,B= b a

Similarly for Hx0 −γ h2 −γ = 2 h

H xo =

∂H z jωε ∂Ez + 2 ∂x h ∂y ∂H z ∂x

for propagation γ = jβ , H xo = but Ey =

− j β ∂H z h 2 ∂x jωµ ∂H z h 2 ∂x

∂H z h2 = ⋅ Ey ∂x j ∂ωµ

Substituting the value of H xo = =

− j β h2 ⋅ E oy 2 h jωµ

∂H z in the above Hx0 equation. ∂x

−β 0 E ωµ y

Substituting the value of Ey0 in the above Hx0 equation.

VEL TECH


VEL

VEL TECH =


VEL

− β  − jωµ CB sin Bx cos Ay] ωµ  h 2

Similarly for Hyo

γ ∂H z jωε ∂Ez − 2 h 2 ∂y h ∂x −γ ∂H z = 2 h ∂y

H yo = −

[∴ E2 = 0]

For propagation γ = jβ H y0 =

But

Ex =

− j β ∂H z h 2 ∂y − jωµ ∂H z h 2 ∂x

∂H z −h2 = ⋅ Ex ∂y jωµ Substituting this value of

H yo = =

∂H z in the above Hy0 equation. ∂y

− j β ( −h2 ) o ⋅ Ex h 2 jωµ

β 0 E ωµ x Substituting the value of Ex in the above equation Hy0.

Hy0 =

β  jωµ CA sin Ay cos Bx] ωµ  h 2

Hy0 =

jβ CA sin Ay cos Bx h2 H20 = xy = c1 c3 B cos Ay cos Bx + c2 c3 cos Ay sin Bx +c1 c4 cos Bx sin Ay + c2 c4 sin Ay

VEL TECH


VEL

VEL TECH


VEL

sin Bx. But

c2 = c4 = 0 Hz0 = c1 c3 cos Ay cos Bx C = c1 c3 Hz0 = C cos Ay cos Bx The field equation of TE waves are as follows, jβ CB sin Bx cos Ay h2 jβ Hy0 = 2 CA cos Bx sin Ay h 0 Hz = C cos Ay cos Bx jωµ Ex0 = 2 CA cos Bx sin Ay h − jωµ Ey0 = CB sin Bx cos Ay h2 nπ mπ Where A = ,B= b a Hx0 =

For TE waves the equation for β , fc, λ those of TM waves. 2

2

2

2

c

iv and λ are found to be identical to

 mπ   nπ  β = ω µε −   −   a   b  2

ωc =

fc =

1

µε 1

2π µε

 mπ   nπ   a  + b      2

 mπ   nπ   a  + b     

2

The corresponding cut off wavelength is 2 λc = 2 2 m n  a  +b     The velocity of propagation. ω v= β VEL TECH


VEL

VEL TECH =


VEL

ω 2

 mπ   nπ  ω µε −   −   a   b 

2

2

2π

λ=

2

2

 mπ   nπ  ω µε −   −   a   b  3. Explain about dominant mode in rectangular waveguide. 2

The lowest mode for TE wave is TE10 (m = 1, n =0) whereas the lowest mode for TM waves is TM11 (m = 1, n = 1). This wave has the lowest cut-off frequency. Hence the TE10 mode is the dominant mode of a rectangular wave guide. Because the TE 10 mode has the lowest attenuation of all modes in a rectangular wave guide and its electric field is definitely polarized in one direction every where. For TE10 mode m = 1, n = 0 2

 mπ   nπ  h=   −   a   b 

2

2

π  =   +0 a =π a The field expressions are jβ CB sin Bx cos Ay h2

Hx0 =

jβ

H

0 x

( a)

= π =

Hy0 = 0

(

2

( )

(

(

)

C π a sin π a x

j β ac sin π a x π

Hz0 = C cos π a x

)

)

Ex0 = 0

VEL TECH


VEL

VEL TECH − jωµ

E

0 y

=

( a)

= π


( )

(

C π a sin π a x

2

(

− j µwac sin π a x π

VEL

)

)

The instantaneous field expressions for the dominant TE10 mode are obtained by the β phasor expressions in above equations with ej(wt - z) and the taking the real part of the product. Hx0 (x, y, z, t) =

(

)

− β ac sin π a x sin (wt - β z) π

(

)

Hz0 (x, y, z, t) = C cos π a x cos (wt - β z) Ey0 (x, y, z, t) =

(

)

w µa C sin π a x sin (wt - β z) π

Ex0 = Hy0 = 0 For TE10 mode m = 1, n = 0 1

ωc =

µε

= C.π fc = =c

2

a

π    2π µε  a  1

2a

where C =

λc =

π  a  

1

µ0ε 0

2  1 a  

- 3 x 108 m/sec

2

λc = 2a. VEL TECH


VEL

VEL TECH


VEL

For TE10 mode the cutoff wave length is equal to twice the width of the guide. Its cut off frequency is independent of the dimension ‘b’ the field configurations for the lower order TE waves in rectangular wave guider.

TE10 Wave Electric and magnetic field configuration for the lower order mode in a rectangular wave guide. The surface current density on surface waveguide walls is given by Js = an x H At t = 0 When x = 0 Js = −ay Hz VEL TECH


VEL

VEL TECH


VEL

Js (x = 0) = −ay C cos (0 - β z) When x = a J s = ay H z Js (x = a) = ay C cos β

z

When y = 0 Js = ax Hz −ay Hx

(

)

( )

β ac Js (y = 0) = ax C cos π a x cos β z - az sin π a x sin β z π When y = b Js (y = 0) = J3 (y = b) The surface currents on side walls at x = 0 and at y = b are selected the below figure.

Surface currents on wave guide walls for TE10 mode in rectangular wave guide. 4. Explain Wave Impedance. In a Cartesian coordinate s/m three wave impedances (impedance constants) must be defined. The wave impedances defines as the ratio of electric field intensity to magnetic field intensity are E E E + + Z xy = x ; Zyx = y ; Zzx+ = z Hy Hx Hx VEL TECH


VEL

VEL TECH


VEL

The wave impedance in the opposite directions are the negative of those given above. (i.e.) E E E − − Z xy = − x ; Zyx = − y ; Zzx− = − z Hy Hx Hx For waves guided by transmission liens or wave guides, the wave impedance which is seen in the direction of propagation z is given by Zz = zxy = Zyz Zz =

E x2 + Ey2 H x2 + Hy2

For TM waves in a rectangular waveguide. ZTM =

E x −E y β = = Hy Hx ωε

It is known that, the propagation constant is y = h 2 − ω 2 µε At cut-off frequency fc, λ = 0 h 2 − ω c2 µε = 0 h2 – wc2 µ ε w c2 = h

2

µε

= 0, wc2 µ ε h2 fc2 = 2 ( 2π ) µε

= h2

The cutoff frequency is h fc = 2π µε For propagation λ must be imaginary

λ = jβ

λ = h2 − w 2 µε

β = w 2 µε − h2 VEL TECH


VEL

VEL TECH


VEL

= w 2 µε − w c2 µε  wc  =w µε 1 −   w 

2

2

f  = w µε 1 −  c  f  The wave impedance of TM waves Z+ M = p

ωε

f  w µε 1 −  c  f  = ωε = µ ZTM

f  1−  c  ε f 

2

2

f  = η 1−  c  f 

2

Where η is a characteristic impedance

η= µ

ε

The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always less than the intrinsic impedance of the dielectric medium. When the operating frequency is lower than the cut-off frequency. The propagation constant is real. W < wc

λ =a

= h2 − w 2 µε

= w c2 µε − w c2 µε =w c

(or)

w  µε 1 −    wc 

f  a =h 1 −    fc 

VEL TECH

2

2


VEL

VEL TECH


VEL

For a given mode the waves with frequencies lower than the cutoff frequency cannot be propagated i.e. waves are attenuated as e-y2 = e-az with z. It propagates if the operating frequency is greater than the cutoff frequency. Therefore a waveguide exhibits the property of a high pass filter. The wave impedance in a non propagation mode is ZTM = λ

jωε

f  h 1−    fc  = jωε

2

f  − jh = 1−   ωε  fc  for f < fc

2

wave impedance is purely reactive, indicating that there is no power flow for f < fc For TE waves in a rectangular wave guide ZTE = =

E x −E y = Hy Hx

ωµ β For propagation Y must be imaginary

y = j β y = j β = h2 − w 2 µε

β = w 2 µε − h2 = w 2 µε − w c2 µε  wc  =w µε 1 −   w  f  = w µε 1 −  c  f 

2

2

The wave impedance of TE waves

VEL TECH


VEL

VEL TECH


VEL

µ ZTE = ω β =

ωµ

f  w µε 1 −  c  f 

µ =

2

ε 2

f  1−  c  f  η ZTE = 2 f  1−  c  f  Where η is the characteristic impedance

η= µ

ε

The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always larger than the intrinsic impedance of the dielectric medium. When the operating frequency is lower than the cut-off frequency. The propagation codes does not take place. i.e, propagation constant becomes real. f  y = h 1−    fc 

2

The wave impedance in a non-propagating mode ZTE = =

jωµ λ jωµ

f  h 1−    fc 

2

f
Wave impedance is purely reactive, indicating that there is no power flow for f < fc. For TEM waves in a rectangular wave guide. VEL TECH


VEL

VEL TECH


VEL

Wave impedance ZTEM =

−E x jωµ = Hy λ

( or ) ZTEM = λ

jωε

The propagation constant y = a + jβ = jβ = jω µε Substituting in wave impedance equation ZTEM =

( or ) ZTEM =

jω µε = jωε jωµ jω µε

=

µ ε µ ε

Wave impedances of TEM waves is the characteristic impedance of any medium ZTEM = η The variation of wave impedance with frequency is below. Wave impedance versus frequency characteristics of waves between parallel conducting plane. The wave impedance for different are given below. Mod e TM TE

TEM

VEL TECH

Wave Impedance 2

f  Z = η = 1−  c  f  η 2 Z=  fc  1−   f  z =η =

µ ε


VEL

VEL TECH


VEL

The phase velocity in a waveguide is given by

γ =

ω = β

ω f  w µε 1 −  c  f 

2

f  where β =w µε 1 −  c  f  1 r = 2  fc  µε 1 −   f 

2

Vo

V =

f  1−  c  f 

2

1

Where Vo =

µε

The wave length in the waveguide is V λ= = f

Vo

f

f  1−  c  f 

2

2

2 λ  f Since c =  o  f  λc  ho h= 2  λo  1−    λc 

λ=

λo λc λc2 − λ02

or

λ2 =

λo2λc2 λc2 − λ02

VEL TECH


VEL

VEL TECH


VEL

λ 2 ( λc2 − λo2 ) = λo2 λc2 λ 2λc2 − λo2 λc2 = λo2 λc2 λ 2λc2 = λo2 λ 2 + λo2 λc2 λo2 ( λ 2 + λc2 ) = λ 2 λc2 λ02 = λ0 =

λ 2 λc2 λ 2 + λc2 λλc λ + λc

5. Explain the Excitation methods for various modes? In order to launch a particular mode, a type of probe is chosen which will produce lines of E and H that are roughly parallel to the lines of E and H for that mode possible methods for feeding rectangular waveguides are shown. In figure the probe is parallel to that y – axis and so produces lines of E in the y direction and lines of H which the in the xz plane. This is the perfect field configuration for the TE10 mode. In Figure the parallel produces fed with opposite phase tend to set up the TE20 mode in figure the probes which are parallel to the z-axis produces electric field liens in the xy plane for TE11 mode.

In figure the probe parallel to the z-axis produce magnetic field lines in the xy plane. This is the perfect field configuration for the TM11 mode. It is possible for several modes to exist simultaneously in waveguides, if the VEL TECH


VEL

VEL TECH


VEL

frequency is above cut-off for these particular modes. However the waveguide dimensions are often chosen so that only the dominant mode can exist. 6. Derive the equation for Alternation in waveguides. For the practical consideration we assumed that the waveguides has infinite conductivity that the waveguides has infinite conductivity thus there is the no loss, but impractical the conductivity is not infinite, but some high value, due to this the is some less occurs in the waveguide, which is known as “waveguide attenuation”. Attenuation due to waveguide walls can be defines as

α=

1  Power lost in guide walls  2  Power transmitted 

TM Waves The current induced in the waveguide alls depend on the magnitude of the Hx and Hy at the surfaces these the power lost and be written as b a 2  1 2 P = Js Rs = Rs  ∫ Hx dx + ∫ Hy2 dy  2 0 0 

Where Js2 = H, n$ = H x2 i + Hy2 j = linear current density per meter length per conducting wall. Rs = Surface impedance =

ωµm 2σ m

b  ωµ m  a 2 2 ∴P =  ∫ H x dx + ∫ Hy dy  2σ m  0 0 

Substitute the value of Hx and Hy integrate and simplify we get, =

ωµ m  v 2f 2c 2  m2a m2 b   +    2σ m  8η 2fc4  b a 

Power transmitted,

VEL TECH


VEL

VEL TECH Pρ =


VEL

1 E x H y + Ey Hx  2

a b 1  mπ c  mπ x  2  nπ   =  2 ωεβ ∫ ∫ cos   sin  a y   dxdy 2  aω c µε  a    0 0

1 nπ c  mπ x   nπ y  + ωεβ ∫ ∫ sin2  cos2   dxdy 2 2 bω c µε  a   a  0 0 a b

abc 2  f  PT =   8η  fc 

2

f  1−  c  f 

2

watt

 ωµ m  v 2f 2c 2  n2 a m 2 b    +  2 4   a  1  2σ m  8η fc   b α=   2 2 abc 2  f   fc     1−     8η  fc  f    2

2

n m + 2 ωµ m v f  b   a  α= N /m 2 2σ m 2η fc2 f   1−  c  f  for TE waves 1 2 1 2 The power lost = ∫ Jsx Rs dx = ∫ Jsz Rs dz 2 2 R = s 2

2 b a 2 β 2  mπ  2 2  mπ x   2  mπ x   ∫ A cos   dx + ∫ h 4  a  A sin  a   dx  a        0 0

 π A2 β 2  mπ  2 a  + 4   A . 2  for y 2 plane. 2 h a     2 2 2 R  aA β  mπ  2 a  = s + 4 A .  for y 2 plane 2 2  2 h  a   =

Rs 2

2 2 Rs A2  β 2  mπ   nπ   + Total power lost = = ( a + b ) + 4   2  h  a   a   Power transmitted PT = ½ [Ex Hz + Ey Hx]

VEL TECH


VEL

VEL TECH


VEL

 nπ 2 a b  2  mπ x  2  nπ cos sin y      ∫ ∫  a   a  dxdy  b  0 0  2 a b  mπ   2  mπ x  2  nπ y  +  sin cos dxdy   ∫∫  a   a       a  0 0 

1 A2ωµβ PJ = 2 h4

ab 2 ωµβ abA2 f  = A 2 = ωµω µε 1 +  c  8 ω c µε 8 f  2

2

2

ab µ 2  fc  f  = A   1+  c  8 ε f  f  1 Power lost  α−  2  Power transmitted   R A2 = s  2

2 2  β 2  mπ   nπ     + ( a + b ) + 4    h  a   a      

After simplifications,

πµ m α= δm

 2a  fc 2  1 +  2 b  f   nepers/m  fc   aη  1 −  c  f

7. Explain the characteristics of TE & TM waves. The propagation characteristics of TE & TM waves are obtained as follows, From the above analysis we got 2

 mπ   nπ  h = p + ω µω = A + B =      a   b  2

2

2

2

2

2

2

2

 mπ   nπ  or P =  − ω 2 µω     a   b  2

2

or

2

 mπ   nπ  P=     − ωµΣ  a   b 

We know ‘P’ is Q complex number I e

VEL TECH


VEL

VEL TECH


VEL

P = α + jβ 2

2

 mπ   nπ  ∴P = α + jβ  − ω 2 µΣ     a   b  2

 mπ   nπ  At low frequencies ω µΣ <<   +   a   b 

2

2

Thus ‘P’ becomes real i.e. p = α , the wave is hence the wave cannot propagate. 2

2

mπ   nπ  Hence α =  + − ω 2Σ    a   b  2

 mπ   nπ  At high frequencies ω µΣ >>      a   b 

2

2

Thus ‘p’ becomes purely imaginary [i.e. α = 0] hence the wave propagates  mπ 2  nπ 2  j β = j ω µΣ −   +    a   b   2

At the transition ‘p’ becomes zero, the frequency at which ‘p’ becomes just zero, is defined as cut-off frequency’ At F = fc: P = 0 Hence the equation (1) becomes 2 2  mπ   nπ  2 0=   b  − ω µΣ a     2

2

 mπ   nπ  or ω c2 µΣ =      a   b  2 2 1  mπ   nπ   or ωc2 =   µΣ  a   b   or

fc =

or

f=

2

v  m n  + 2   a   b   2

Where v =

1

=

2

 mπ   nπ      2π µΣ  a   b  1

2

   

1

= 3 × 108 m / sec

µΣ 4π × 10 × 8.854 × 10 The cutoff wavelength is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 219 −7

−12

VEL

VEL TECH

λc =


v = fc v 2

v  m  2  n 2        a   b  

VEL

2ab

=

( mb )

2

+ ( na )

2

Guide wavelength: It can be defined as the distance traveled by the wave in order to undergo a phase shift of 2π radians. It is denoted by,

λg =

2π = β

2π  mπ 2  ~ π 2  ω 2 µΣ −   +    a   b  

Practically the guide wavelength is different from free space wavelength. 2

W.K.T.

2π

ω  ω µΣ 1 −  c  ω  2π v v λ = = = 2 2 2  fc   fc  λ  ω 1-   f 1+   1−   f  f   λc 

Thus

W.K.T.

2

 mπ   nπ  ω µΣ =   +   a   b  2π λg = = ω 2 µΣ − ω c2 µΣ 2 c

λcα

1 1 , λα & v = fc f

∴ λc =

1

µΣ

λ λ  1+    λc 

2

squaring on both sides we get, 2

λ  λ    = 1 −    λc   λg  1 1 1 = 2+ 2 2 λ λc λg VEL TECH

2


VEL

VEL TECH


VEL

Phase Velocity:W.K.T. The wave propagates in the waveguide Wavelength λg is greater than the free space wavelength λ , thus the velocity of propagation is defined as the rate at which the wavelength changes its phase in terms of λg v p = λg f = =

2π f .λg 2π

=

2π f ω = (2π / λg ) β

ω  mπ 2  nπ   ω µΣ −       a   b  

ω

=

ω µΣ − ωc2 µΣ 2

2

2

2

 mπ   nπ  ω µΣ =   +  Since at cut-off frequency  a   b  2 c

vp = Then vp =

ω   ωc  ω µΣ 1 −     ω  v  fc  1−   c

v

=

  fc 2  1 −      f  

= velocity of light

2

The phase velocity is the velocity of TE & TM waves. Group velocity It can be defined as vg

=

dω dP

  mπ  2  nπ 2  β = ω µΣ −   +    a   b     2

W.K.T.

= ω 2 µΣ − ω 2 µΣ = µΣ(ω 2 − ω 2c ) Differentiate the above equation w.r.t. ‘w’ we get,

VEL TECH


VEL

VEL TECH


VEL

µΣ

dβ 2ωµΣ = = dω 2 (ω 2 − ωc2 )µΣ

2

 ωc  1−    ω 

2

 fc  1−   2  dω  fc   f   or = = v 1−   dβ  µΣ c    or

λ  dω vg = = v 1−   λ  dβ  y

2

8. For an a/r filled copper x –brand wave side with dimensions a=2.286cm, b=1.016cm determine the cut-off frequency of the first four propagating modes. What is the alteration for metre length of the guide when operating at the frequency of 10GHZ? Given a =2.286cm; b=1.016cm F = 10GHZ length = 1m TE10 ,TE01,TE11,TE2 2

2

 m π   nπ  ωc 2µ ∈=   +   a   b  2

2

 m π   nπ  ωc =   +  M∈  a   b  1

2

c m n fc = + 2  a   b 

2

for TE10 mode: m=1, n=0 c fc = 2a 3 × 108 = 2 × 2.286 × 10 −2 = 6.56 GHZ

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for TE01 mode : m = 0, n = 1 fc =

c 2b

3 × 108 = = 14.76 GHZ 2 × 1.016 × 10−2 for TE11 mode m=1, n=1 2

c  1  1 fc=   +   2a b

2

2

2

3 × 108  102   102  =   +  2  2.286   1.016  = 16.156 GHZ for TE02 mode: m = 0, n = 2 fc =

c2 2b

c 3 × 108 = = = 29.53 GH2 b 1.016 × 10−2 Pr opagation constant, 2

2

 m π   nπ  γ=  +   − ω2M ∈   a   b  2

2

2

 m   n   2f  =   +     a  b  c 

If the operating frequency is less than the cut-off frequency alternation takes place i.e., propagation does not take place. For TE01, TE11, TE02 modes propagation will not take place i.e., propagation constant γ = α . For TE11 mode, m =1, n = 1 VEL TECH


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2

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2

 102   102  ν=π   +   2.286   1.016 

 2 × 10 × 109  .  8  3 × 10 

ν = α = 265.77 Nepers in Alternation = αl = 337.4 × 1 = 337.4 Nepers for TE01 mode; m=0, n=1 2

2

 102   2 × 10 × 109  γ=π   −  8  1.016   3 × 10  α = 227.5 Nepers / m Alternation = αl = 227.5 × 1 =227.5Nepers 2

2

 2 × 102   2 × 10 × 109  γ=π   −  8  1.016   3 × 10  α = 581.88 Nepers / m

Alternation = αl = 581.88 × 1 = 581.88 Nepers 9. A rectangular waveguide has cross-section dimensions a = 7cm & b = 4cm. Determine all the modes which will propagate through the waveguide at frequency of 6 GHZ Given a = 7cm = 7 x 10-2m b = 4cm = 4 x 10-2m f = 6GHZ The cut-off wavelength 2 λc = 2 2 m n +  a  b     I m ethod, λo = c f 3 × 108 = = 5cm 6 × 109

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for TE10 mode : m = 1,n = 0 λ c = 2a = 2 × 7=14cm If λ c>λ o then the propagation takes place the cut off wavelength .λ c, should be greater then the minimum wavelength (λ o) for propagation, since λ c,λ o propagation is possible for TE10 mode, for TE01 mode : m = 0;n = 1 λ c = 2b = 2 × 4=8cm λ c > λo ,propagation is possible for TE01 mode, for TE11 mode; m=1, n=1 2

2

 102   102  γ = π  +  − 1600  7   4  = 87.228 rad / m propagation is possible for TE11 mode for TE zo mode m = 2, n = 0  2 × 102  γ=π   − 1600  7  = j87.95rad / m Propagation is possible for TE20 mode For TE02 mode:  2 × 102  γ=π   − 1600 = 94.24  4  γ Becomes real value i.e., γ = α (∴ β =0) Propagation will not take place in this made ∴ TE10, TE01, TE11, TE20 modes will propagate VEL TECH


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10. A rectangular waveguide measures 3 x 5 cm in equally of ahs a 10GHz signal propagated in it. Calculate the cut off wavelength the guide wavelength of the characteristic wave impedance for the TE10 mode. Given a = 5 cm = 5 x 10-2 cm b = 3 cm = 3 x 10-2 cm f = 10 GHz, TE10, m = 1, n = 0 Cut – off wavelength 2a λC = m =2 × 5 × 10 −2 = 10 × 10-2m Cut – off frequency c fc = λC 3 × 108 = 10 × 10 −2 = 3GHz Guide wavelength λ0 λg = 2  fe  1=   f  But λ 0 = c f 8 3 × 10 = 10 × 109 = 3cm or 3 × 10-2m λg =

3  3  1−    10 

2

= 3.145cm / 2.145 × 10−2 m Characteristic wave impedance

VEL TECH


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VEL TECH 2z =

=


η f  1−  c  f  120π

 3  1−    10  = 395.19Ω

VEL

( Q η=120πΩ )

2

2

11. Design a rectangular waveguide with the following specifications a) At a 7.5 GHz the guide wave length for TE10 mode is 90% of the cut off wave length. B) TE30 & TE12 wave the same cut off frequency. Given f = 7.5GHz λg = 0.9 λe λ0 = c

f 3 × 108 = = 4cm 7.5 × 109 λ0

λg =

λ  1+  0   λC  2

 λg    =  λc 

2

1 2

λ  1−  0   λc  cross multiplying 2

 λs   λg    −  =1  λ 0   λc  2

2

 λg   λg    = 1+    λ0   λc  = 1+(0.9) 2 = 1.81 λg = 1.81λ0 = 1.81× 4 = 7.24 cm VEL TECH


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But

TE10 :

λg = 0.9 λc λ λc = g 0.9 7.29 = = 8.04 cm 0.9 2a λc = m λc =2a λ a = c 2 8.04 a = 2 a = 4.02 cm. 2a m λm a= c 2 a = 12.067cm

λc =

for TE 30 :

2

for TE12 : m = 1, n = 2

λc =

2 2

 λs   λg    = 1+    λo   λo 

2

4 m n   +  = 2  a   b  λc

m n   +   a  b = 0.0619 − 0.00687 = 0.055 A b2 = 0.055 b = 8.5cm a = 12.067 cm 2

VEL

2

( 2/b )

2

=

4 1 − 2 (8.04) ( 12.069 ) 2

2

= 1+(0.9) 2 = 1.81 λg = 1.81 > 0 = 1081× 4 =7.24cm =7.24cm VEL TECH


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VEL TECH


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λ9 = 0.9 λc 7.24 = = 8.04cm But 0.9 2a λc = m For TE10: λc = 2a a = λc / 2 8.04 2 a = 4.02cm a=

For TE30: 2a m > xm a= 2 a = 12.067cm.

λc =

For TE12:

M=1, n=2. 2

λc =

2

m n      a  b 2

2

2

m n 2   +  = 4/λ a b     4 1 2 −  = 2 2  b  (8.0 ~) (12.067) = 0.0619-0.00687 = 0.055 b2 =

4 0.055

b = 8.5 cm a =12.067 cm. VEL TECH


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VEL TECH


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12. An air filled hollow rectangular conducting waveguide has cross section dimensions of 8x10 cm. How many TE modes will this waveguide transmit at frequencies below 4 GHz? How these mode are designated & What are their cut-off frequencies? Given a = 0.1m, b = 0.08m, f = 4 GHz λ = c/ f =

3 ×108 = 0.075m. 4 ×109

The cut-off wavelength

λc =

=

2 2

m n      a  b 2

2

2

 m  n       0.1   0.08 

2

Propagation constant, 2

2

2

2

 mπ   nπ  2 ν=   +  − ω µΣ  a   b  2

 mπ   nπ   2π f  =   +  −   a   b   c  2

=π

2

2

m n 2f    +  −   a  b  c 

f 4 ×109 40 = = c 3 ×108 3 2

2

2

 m   n   2 × 40  γ =π   +  +   0.1   0.08   3  Let m = 1, n = 0

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VEL TECH


 1   80  γ =π   +0−   0.1   3

VEL

2

= j77.66rad/metre 2

 1   80  m=0, n=1 γ = π 0+  −   0.08   3 

2

= j74 rad/m m=1, n=1, 2

 1   80  γ =π 0 +   +   0.08   3 

2

= j74rad/m m=1, n=1, 2

2

2

 1   1   80  γ =π   +   −   0.1   0.08   3  =j64 rad/m. m = 2, n = 0,

2

 2   80  γ =π   +0−   0.1   3

2

= j 55.4 rad/m m=0, n=2 2

 2   80  γ =π 0+   +   0.08   3 

2

=j39.1 rad/m m=1, n=2 2

2

2

 1   2   80  γ =π   +   −   1   0.8   3  VEL TECH


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VEL TECH


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= 11.708 nepers/m m=2, n=2 2

2

 2   2   80  γ =π   +  −   −1   0.08   3 

It propagation constant γ is imaginary, propagation will take place. For TE10, TE01, TE11, TE20, TE02, TE21 modes, γ is imaginary. These are the nodes will be nodes will be propagated. The corresponding wavelength for each mode is given by. 2

TE10 :

c m n fc =   +  2  a  b

2

2

3 × 108  1  fc =   +0 2 1 = 1.5 GHz. 3 ×108  1  0+  2  0.08  = 1.875 GHz

TE01:

2

TE11: TE20:

2

fc =

3 × 108  1   1  =     2  0.1   0.08  = 2.5 GHz

2

3 ×108  2   +0 2  0.1  = 3 GHz

fc =

2

TE02:

3 ×108  2  fc = 0+  2  0.08 

TE21:

3 ×108  2   1  fc =     2  0.1   0.08  = 3.457 GHz

2

2

13. A rectangular air filled copper waveguide with dimension 2cm x 1cm VEL TECH


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cross-section & 300m length is operation at a cuttz with a dominant mode find cut-off frequency, guide wavelength, phase velocity, characteristic impedance and atomization. Assume σ = 5.8 × 107 for copper . Given a = 2cm = 2 ×10−2 m b = 1cm = 1× 10−2 m f = 9 ×109 Hz l = 30cm = 30 ×10−2 m c λ= f =

3 ×108 = 3.33 × 10−2 m 9 ×109

The dominant mode is TE10, Cut-off frequency c c fc = = λ 2a 3 ×108 = 2 × 2 × 10−2 f c = 7.5GHz Guide wavelength

λg =

=

Phase velocity VP =

λ  fc  1−    f 

2

3.33 ×10

−2

 7.5  1−    a 

2

λ = 6.02 ×10−2 m

c  fc  1−    f 

2

VEL TECH


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VEL TECH

VEL TECH MULTI TECH TECH HIGHTECH =

VEL

3 ×108  7.5  1−    a 

VP = 5.43 ×108 m / sec Characteristic impedance 7 ZTE = 2  fc  1−    f  =

ZTE

120π

 7.5  1−    9  = 682ohms.

2

Surface resistance Rs = =

π fµ σ π × 9 ×109 × 4π ×10−7 5 : 8 × 10−7

Rs = 2.475 × 10−2 ohms. Alteration constant  2b  fc  2  Rs  1 +     a  c   α=  2  fc  µo b 1−   Σo  f  2  1  7.5   2.47 ×10-2 × 1× 2 × ×    2  9    =

 7.5  1×10 ×120π 1 −    9  = 0.02 Nepeirs/m

2

−2

Total Alteration

α l = 0.02 × 0.30 = 0.006nepers. 14. What are the reasons for impossibilities of TEM mode in a rectangular VEL TECH


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waveguide? TEM mode cannot propagate in a rectangular waveguide can be proved by the following argument: 1) By definition, TEM wave means that there is no variation of electric & magnetic fields in the transverse plane. If this is to be satisfied than the transferees field components Ex, Ey, Hxy & Hy have to be constant w.r.t x & y. if this so, then this will violate the boundary conditions E tan = 0< H normal =o. if the boundary conditions are to be satisfied for a rectangular waveguide, then this wave cannot be a TEM wave. 2) In the expression for the field components, if we put E 2=H2=0 then all other field components Ex, Ey, Hx & Hy will also be zero as could be seen from equation. X 11 = kx 2 X y11 = ky 2 y ∂ ∂ & are not zero for a ∂x ∂y rectangular waveguide. Hence no TEM wave propagation can take for a rectangular waveguide.

∴ There is no wave propagation if both E2 & H2=0 since

3) Let us assume that a TEM wave exist inside a rectangular waveguide which is a single conductor system. Existence of TEM wave means the magnetic field must the entirely in the transverse plane. For a magnetic field, div H = 0 that is the magnetic field lines must form closed loops in the x-y transverse plane inside the waveguide. If use apply amperes circuit law tot his magnetic field, the lines of this magnetic field aloud these closed path us must be equal to the current enclosed in the axial direction current or a displacement orient in the 2-direction existence of such displacement current will require an axial component of electric field E2 B~A if E2 is present then this wave cannot a TEM wave. Further, if instead of displacement current conduction current exist, then there should be a centre conductor to provide return path, which is not the case in a rectangular waveguide. This argument holds good for any ingle conductor waveguide. There no TEM wave can exist inside a single conductor waveguide. 15. A x-band waveguide which is over filled has inner dimensions of a =2.286 cm and b=1.016 cm. Calculate the cut-off frequencies of the following modes. TE10, TE20, Tm11, TM21 and TM12. Also find our which of the modes will propagate along the waveguide and which of them will evince when the signal frequency is 10 GHz? The cut-off frequency for a rectangular waveguide equation.

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1/ 2

 mπ  2  nπ 2  fcm,n =      2π µε  a   b   For the TE10 mode put m=1 and n=0 in the above equation 1

c π2 c 3 1010 fc / TEVo = × = = × = 6.56GHz 2π 92 2a 2 2.286 For the TE01 mode put m= 0 and n = 1. 1/ 2

π 2  f c / TE01 =  2 2π mε  b  1

1

=

2b µε

=

3 ×1010 2 × 1.016

For the TE20 mode 1/ 2

c  4π  c 2π c f c / TE20 = = = z   2π  a  2π a a 3 ×1010 = = 13.12GHz 2 × 2.286 For the TE11 mode. π π2  2 + 2 b   a 2

c f c / TE11 = 2π

1/ 2

c  a 2 + b2  =  2 2  2 a b 

1/ 2

1/ 2

3 ×1010  (2.286)2 + (1.016)2  = 2  (2.286)2 (1.016)2  1/ 2

3 ×1010  2.5  = 2  2.32 

= 16.16GHz

For the TM11 mode f c / TM 11 will be same as above Fc /TM11 = 16.16 GHz For the TM21 mode 1/ 2

1/ 2

 4 1  = 1.5 ×10  + 2 2   (0.286) (1.06)  =1.5 ×1010 [0.765 + 0.887]1/ 2 = 19.28GHz

3 ×1010  4 1  f c / TM 21 = + 2  a 2 b2 

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10


For the TM12 mode

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1/ 2

3 ×1010  1 4  f c / TM 12 = + 2  a 2 b2 

1/ 2

 1 4  = 1.5 ×10  + 2 2   (2.286) (1.016)  = 1.5 ×1010 (0.191 + 3.875)1/ 2 = 30.24 GHz. 10

Since the signal frequency is 10 GHz only the modes with cutoff frequencies less than 10 GHz will propagate and the others will evanesce. The waves that will propagate is only TE10 mode. The modes that will evanesce are: TE 01, TE20, TE11, TM21 and TM12. 16. A 10GHZ signal is propagated is a dominant mode in a rectangular wave guide if vg is to be 90% of the free space velocity of the light, then what be the breath of the waveguide. Find the characteristic impedance also. Given: Vg C

= 0.9

f = 10GHZ a=? Z TE = ?  fc  Vg = C 1 −   c

( i)

sub. fc= Vg

V 2a

 fc  = 1−   C c

( 0.9 )

2

2

2

 fc  = 1−   c

2

∴ fc = 4.358GHZ for f= 10 GHZ VEL TECH


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VEL TECH

( ii)

fc=


VEL

V = 4.358GHZ 2a

∴ a = 0.0344m and Z TE =

η  fc  1−   b

2

Z TE = 418.868Ω 17. Design a rectangular wave guide with following specifications (a) at .5 GHZ the guide λ g for TE10 is 90% of λ c (b) TE30 & TE12 have same fc. Given: f = 7.5GHZ

λg = 0.9 λc λ

λg =

 λ  1−    λc  2

 λg    = λ

2

1  λ  1−    λc 

2

2

2

2  λg   λg   λ    −    = 1  λ   λ   λc 

2

 λg  2   = 1 + ( 0.9 ) λ

λg2 = 1.8λ 2 c λg = 1.8   f

λg = 0.053m VEL TECH


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VEL TECH

λc =

λg

= 0.9


VEL


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0.058

WKT λc = 2a 0.058=2a ∴ a = 0.029 m

( b)

TE30 = fc=

∴ λc =

3V = fc 2a

3V 2a

2a 3

TE10 & TE30 have sane 0.024 fc & λc

λ c ( 30 ) = 0.059 λc 2 a = 0.08m

∴a = 3

fc =

fc =

3V = 5.08x109 Hz 2a 1 2π µε

2

 m π   nπ   a  + b     

2

fc ( 30 ) = fc ( 12 ) at TE12 ⇒ λ c ( 12) =

2π 2

 m π   nπ   a  + b     

2

sub a=0.08m, we get, b=2.063m VEL TECH

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18. Derive the Equation for Beset function: Beset Function : In solving for the electromagnetic fields with in guides or circular cross section a differential equation known as Bessel’s equation is encountered. The solution of the equation dents to Bessel functions. These functions will be considered briefly in this section in propagation for the following section on circular wave-guides. These same functions can be expelled to appear in any two dimensional problem in which these is circular symmetry. Examples of such problems are the vibrations of a circular membrane. The propagation of waves within a circular cylinder , and the electromagnetic field distribution about an in finitely long wire. The differential equation involved in there problems these the form. d2p 1 dp  n2  + + 1−  p = 0 dp2 e de  p2  Where n is any integer. One solution to this equation can be obtained by assuming a power series solution. p = a0 + a1e + a2p2 + ............. Substitution of this assumed solution back into and equating the coefficients of like power leads to a series solution for the direrential eqauation, for example in me special case when n=0 d2p 1 dp + +p = 0 dp2 p de When the power series is interasted in and the sums of the coefficients of each power of p are equated to zero. The folowing series is obtained. 4 6   1  1  e e 2     2  e 2     p = p1 = C1 1 −   + − + ....... 2 2  2  ( 2!) ( 3!)     2 4 2  P  P −P = C1 1 − 2 + 2 2 2 2 2 + .... 2 ⋅4 2 ⋅4 ⋅6  2  α

2r

(1/ 2e) = C1 ∑ ( −1) (r1 ) T =0 this series is convergent for all values of p, either real or complex. iT is called bissels funtion of the first kind of order zero and is denoted by the symbol. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 240 r

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J0 (p) The zero order refers to the fact that if is the solution case of n=0. The coresponded solution for n=1,2,3…etc are designated. J0 (p), J2 (p), J3 (p) where the subscript n donotes the order of the bessel function since second order differential equation, thermust be two linearly independent solutions for each value of n . The second solution may be obtained ina manner somewhat similar to thaeir used for the first but starting with a slightly different senes that is suitably manipulated to yield a solution. The second soulution is known as bessel’s function of the seond kind, or neuman’s function and is designated by the symbol Nn (p) Where again n indicates the order of the funtion. In the zero order of this solution of the secend kind the following series is obtained 2 P  N0 (p) = ln   + r  Jo(p) π  R  2 α 1 r ( 1/ 2p )  1 1 = − ∑ ( −1) 1 + + + .... +   2 π r =1 r ( r!)  2 3 2r

p = AJ0 (p) + Bw 0 (p) A plot of Jo (p) and No (e) is shown is shown in figure. Because all the neumarn function become infinte at p=0; these second solutions cannot be used for any physical problem in which the origin is included us for example the hellow waveguide problem

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In is apperent the (except near the origin for No (p) ) these curves beat a marked similary to J E ( e) →

2 cos( P − π / 4) πe

No( P ) →

2 sin( P − π / 4) πe

19. Derive the solution of the field equations of cylindrical co-ordinates. The method of solution of the electromagnetic equations for guided of-circular cross section is similar to that followed for rectangular guides however, in order to simplify the applications of the boundary conditions, it is expedient to express the field equations & the wave equations in the cylindrical co-ordinate system.

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In cylindrical co-ordinates in a non-conducting region mass ell’s equations are. ∂Η 2 + γ Ηϕ = jωΣE ρ ρ∂ϕ ∂E2 + γ Eφ = − jωΣEφ ρ∂φ −γ H ρ −

γ Eρ −

∂E2 = jωµ Eφ ∂ρ

∂E2 = jwµ H φ ∂ρ

1  ∂ ( ρ H φ ) ∂H ρ  − = jωΣE2 ρ  ∂ρ ∂φ  1  ∂ ( ρ Eφ ) ∂E ρ  − = − jωµ Hz ρ  ∂ρ ∂φ  These equations can be combined to give h2 H ρ = j

ωΣ ∂E2 ∂Η 2 −γ ρ ∂φ ∂ρ

∂E2 γ ∂Η 2 − − ∂ρ ρ ∂φ ∂E jωµ ∂H 2 h 2 E ρ = −υ 2 − ∂ρ ρ ∂φ h 2 H φ = − jωΣ

H2E ϕ =

γ ∂Σ 2 ∂H 2 + jωµ l ∂φ ∂ρ

h 2 = γ 2 + ω 2 µΣ The wave equation in cylindrical co-ordinates for E2 is ∂ 2 f 2 1 ∂ 2 E2 ∂ 2 E2 1 ∂E2 + 2 + + = ω 2 µΣΕ 2 2 2 2 ∂ρ ρ ∂φ ∂2 ρ ∂ρ proceeding in a manner similar to that followed in the rectangular case, let γ2

E2 = P(P) e (φ ) e VEL TECH

γz

= E2c e


VEL

VEL TECH


Where P(p) is a function of P alone & Q(φ ) is a function of φ expression for E2 in the wave equation gives. Q

VEL alone.

Sub the

2 d 2 P QdP d 2Q + + P + PQγ + ω 2 µΣPQ = 0 2 2 dP Pdl dQ

by PQ, 1 d 2P 1 1 d 2Q + + + + h2 = 0 2 2 2 P dP ρ P QP dQ As before equation above can be broken up in to two ordinary different equations d 2Q = − n 2Q dQ 2 d 2 P 1 dP  2 −n2 + +h dP 2 P dP  P 2

 P = 0 

Where n is a constant. The structure of above equation is, Q = ( An cos φ + b ∩ sin nϕ ) Through by h2, equation (32) is founded in to  d 2P 1 dp n2  + + 1 + p=0 d ( ph) 2 Ph d ( ph)  ( p 2)2  This is a standard form of Bessel’s equation is term of (eh) using only the solution that is finite at (eh)=0, gives P(eh)=Jn(eh) Where Jn (ph) is Bessel’s function of the first kind of order n sub the solution of (3) & (5) in (2 V2

E2 = S-7 ( ( ρ h)( An cos nϕ )e The solution of Hz will have exactly the save form as fel E2 & can ∴ be written V2

H 2 = Jn( Ph)(cn cos nϕ + Dn sin nϕ )e

for Tm waves the remaining beld components can be obtained by inserting (b) into equation (b) for 7E waves (7) must be inserted into the set corresponding to(b). 20. Derive the equation for Tm & TE waves in circular waveguides. VEL TECH


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As in the case of rectangular guides, it is convenient to divide the possible solution for circular guides into transverse magnetic & transverse electric waves for the TM waves H2 is identically zero and the wave equation for E2 is used. The Bonn dry conditions require the E2 must vanish at the surface of the guide ∴ from (6) Jn(ha) = 0 (8) Where a is the radius of the guide. There is an infinite number of possible TM 2 waves corresponding to the infinitive number of roots of (8) As before h2= γ + ω 2 µΣ & in the case of rectangular guides h2 must be less than ω 2 µ 2 for transmission to occur. There extract high frequencies will be required. This in turn means that only the first few roots of (8) will be of practical interest the first few roots are. (ha)01= 2.405 (ha)11 = 3.85 (ha)02= 5.25 – (ha)12 = 7.02 The first subscript refers to the value of n & the second refers to the roots n their order of magnitude. The various Tm waves will be referred to as Tm0, & Tm12 etc. Since γ = h 2 − ω 2 µΣ this gives for β

β

= ω 2 µΣ − h2 nm

mn

the cut-off or critical frequency below which transmission of a wave will not occur is hnm 2π mΣ Where fc =

(ha)nm a The phase velocity is hnm=

V=

ω ω 2 µΣ − h2 nm

from equation (b) the various components of TM waves can be computed of Tm waves can be computed in terms of E2. The expression for Tm waves in circular guides are E20=An Jn (h ρ )cos n ϕ − jAn ω 1Σn ρ Jn((h) sin nϕ ) H °== h2 ρ

VEL TECH


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− jAn ωΣ jn( ρ h) cos nϕ h β El ° = Hϕ ° ωΣ β Eϕ ° = H ρ° ωΣ Hϕ ° =

The variation of each of there field components with time & in the 2 direction are shown by multiplying each of the expressions of (3) by the factor ei (ω t − β z ) & taking the real part. In the original expression (6) for E 2 1 the arbitrary constant bn has been part equal to zero. The relative amplitudes of An & Bn determine the orientation of the field in the guide, & for a circular guide & any particular value of n1 the ϕ =0, can always i.e. oriented to make either An or Bn equal to new. For transverse electric waves E2 is identically zero & H2 is given by equation (6), by substituting (6) into (5), the remaining for Tm waves in circular guides are. H 2 ° = CnJn(hl ) cos nϕ − j β cn Jn '( h ρ ) cos nϕ h φ jo β cn Hϕ ° = 2 Jn(h ρ )sin nϕ h ρ −ωµ Eρ° = H ρ° β Hl ° =

The boundary conditions to be met for Tm waves are that Eφ at ρ = a1 from (b) Eφ is proportional to ∂Hz / ∂ρ & ∴ to Jn’ (h ρ )’ (h ρ ) where the prime denote the derivative write (h ρ ) ∴ for Tm waves the boundary conditions require that Jn −1 (ha) = o & its the roots of (47) which must be determined. The first few of these roots are (ha)101= 3.83 (l a)102 = 7.02

(ha)111= 1.84 (ha)112

= 3.33

The corresponding TE waves are referred to as TE01, TE11 & so on The equations for fc, β , λ , &γ are identical to those for the Tm waves. It is understood, of course that the roots of equation (7) are to be used in connection with TE waves only. The equations shows that the wave having the lowest cut-off frequency is the VEL TECH


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TE11 wave. The wave having the next lowest cut-off frequency is the Tm01. 21. Derive the expression for wave impedance & characteristic impedance. The wave impedances at a point have been defined by equation already. For waves guided by transmission lines 1 wave guider, interest centers on the wave impedance which is seen when looking in the direction of propagation that is along the =axis. −

+ −

Hy

=

−Ey Hx

∴ 2xy = 2yx =

=

Ex 2 + Ey 2 Hx + Hy 2

2

=

β ωΣ

β = 2z ωΣ

The wave impedances looking in the 2-direction are equal & may be put equal to 2 z, where E 2z = trans = H trans

Σx 2 +Σy 2 Hx 2 + Hy 2

is the ratio of the total transverse electric field transverse electric field straits to the total transverse magnetic field strength. A similar inspection of egn (b) for in waves in circular grids shows that for them also 2z=2pφ = -2φ ι

=

β ωΣ

If is seen that for in waves in rectangular a circle girder a indeed in cylindrical grids of any cress-section the wave impedance in the direction of propagation U Constant over the cross section of the guide, & is the same for girder of different shapes reaching that.

β = ω 2 µ − h2 & that the cut-off angular frequency ω c has been defined is that frequency that wakes.

ω c2 ΜΣ = h 2 VEL TECH


VEL

VEL TECH

it follows that


VEL

β = ω M Σ 1 − (ω c 2 / ω 2 )

Then from (3) a (+) the wave impedance is the 2-direction for Tm waves is 2 z (Tm) =

m 1 − (ω c 2 / ω 2 ) Σ

= 7 1-(ω c 2 / ω 2 ) Thus for any cylindrical guide the wave impedance for Tm waves is dependent only on the intrinsic impedance of the dielectric & the ration of the frequency to the cut-off frequency For TE waves the same conclusion can be recharged. found that. Z z (TE ) =

However for TE waves it is

ωµ 7 β 1 − (ω c2 / ω 2 )

for TEM waves between parallel planes or on ordinary parallel wire or co-trig transmission lines the cut-off frequency is zero, d the wave impedance reducer to, Zz (TEM) =7 The dependence of β on the ratic of frequency to cut-off frequency as shown by (3) effects the phase velocity & the wavelength in a corresponding manner thus the phase or wave velocity in a cylindrical guide of any cross section is given by.

γ=

uo ω 1 1 = = = 2 2 β µΣ 1 − (ω c / ω ) 1 − (ω c2 / ω 2 )

Where Vo = 1 µΣ , d µ & Σ are the constants of the dielectric. The wavelength in the guide measured in the direction of propagation, is V 2π 1 λ= = ρ β f µΣ 1 − (ω c 2 / ω 2 ) =

>o

1-(ω c 2 / ω 2 ) Where λ o is the wavelength of a TEM wave of frequency f in a dielectric having the constants µ & Σ . Since ω 2 / ω 2 = λ o 2 / λ c2 it follows that

VEL TECH


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λo λc

λ=

λc2 − λo2

λo = λλc A quantity of great usefulness in correction with ordinary two-conductor transmission liner is the characteristic impeding, 20 of the line for such lines , 20 can be defined in terms of the voltage-current ratio or in terms of the power transmitted for a given voltage or a given current. That is for or infinitely long line. 2ω ν v∗ z o = ∗ ; 20= π 2ω

v Zo = ; I

Where V & I are peak phasors. For ordinary transmission fives these definitions are equivalent but for wave guides they lead to three values that depend upon the guide dimensions in the same way but which differ by a constant. For example consider the three definitions given by (6) for the case of the TE 10 mode in a rectangular grids. The voltage will be taken as the maximum voltage from the lower face of the grids to the upper face this warms at x=9/2 & has a value, Un=-Hx=

− j β ac πx sin π a

The total longitudinal current in the lower face is − j 2a 2 β c I = ∫ J 2 dx = π2 b a

Then the integrated characteristic impedance by the first definition 20(V .I ) =

π bGwm π b 2 = = 2a β 2a

π b7 2a

1 − ( fc 2 / f 2 )

Terms of the second definition, the characteristic impendence for the TE 10 wave in a rectangular guide is found to be. Z o (w. I) =

π 2b π Z z = Z o (u , I ) 8a x

Terms of the third definition the integrated characteristic impedance.

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VEL TECH

Zo(W, V) =


VEL

2b 4 Z z Z o (u , I ) a π

Explain the excitations of modes in circular waveguides:TE modes here no 2 component of an Electric field , & TM modes have no 2 – component of magnetic field. If a device is used in a circular waveguide in such a wary that it excites only a 2 – component of electric field, the wave propagative through the guide will be in the TM mode, or the other hand if a device is placed in a circular waveguide in such a way that it exists only the 2-component of magnetic field, the traveling wave will be in the TE mode. The methods o excitation for various modes in circular waveguides are shown. A common method of excitation of TM modes in a circular waveguide by co-axial line is shown. At the end of the co-axial line a large magnetic field exists in the direction of propagation the magnetic field from the co-axial line will excide the TM modes in the guide however, when the guide is connected to the source by a co-axial a discontinuity problem. At the function will increase the eventually decrease the power transmission, it is after necessary to place a turning device around the function in order suppress the reflection.

22. Given a circular waveguide used for a signal at a frequency of 11GHz propagated in the TE11 mode & the internal diameter is 4-5 cm, Calculate. 1) Cut-off wavelength 2) Group velocity 3) Grid wavelength 4) Phase velocity VEL TECH


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5) Characteristic impedance Given f = 11 GHz d = 4.5 cm, a = 2.25 for TE01, (ha)11 = 1.84 λ = c/f 3 ×108 = 11× 109 = 0.02727 m. i)

cut – off wavelength: 2π a λc = (ha )11 2 π × 2.25 1.84 = 7.68 cm / 0.0768 m.

=

ii)

Guide wavelength:

λg =

=

iii)

 λ  1−    λc 

2

0.02727  0.02727  1-    0.0768 

2

= 0.029 m. Phase velocity : λ  =  g c λ  VP

iv)

A

 0.029  8 =  3 ×10 0.02727  

= 0.029 × 108 m / sec. Group velocity:

VEL TECH


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 λu  =  c λ   g Vg

 0.02727  8 =  3 ×10  0.029  = 2.80u ×108 m / sec

v)

Characteristic impedance: =

Z

=

20 2

λ 1−    λc  120π

 0.2727  1−    0.029 

2

= 1108 ~ .

23. Calculate the cut-off wavelength the guide wavelength & the characteristic wave impedance diameter is ucm for a g GHZ signal propagated in it is the TE11 mode. Given: F = 9 GHZ D = 4cm; a=A/2 =2cm For TE11: (ha)11 = 1.8~ cut-off wavelength 2π a = ( ha ) 11 2π a = 6.8 β cm / 0.068.3m Xc 1.8~ x = c/ f =

=

3 ×102 = 3.33cm / 0.0333m, 9 × 109

Guide wavelength,

VEL TECH


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VEL TECH


λ9 = =

VEL

λ 1 − (λ / λ c ) 2 3.33  Σ.33  1-    6.83 

2

= 3.81cm/ 0.0381 m. Diacharacteristic wave: m Polence = =

η 1 − ( > / > c) 2 120π 2

 3.35  1-    6.83  Zz =435.7 ohms. 24. Determine the cut-of frequencies of the first two propagating modes of circular waveguide with a=0.5 cm & Σ1 = 2.25 if the guide is 50cm in length operating at f=13 GHz determine the attenuation Given : A=0.5cm = 0.5x10-2m Σr = 2.25, µ = µ 0 l = 50cm, = 0.5m f = 13∠Hz Cut-off frequency: (ha )11c Fc = 2π a Fx TE01 mode: (ha) 0.1 = 3.832 3.832 × 3 × 108 for = 2π × 0.5 × 102 =36.6 GHz for TE11 mode: (ha)11=1.8~1 VEL TECH


VEL

VEL TECH

VEL TECH MULTI TECH TECH HIGHTECH Fc=

VEL

1.8 ~ 1× 3 × 108 2π × 0.5 ×10 −2

Propagator constant

γ = hnm − ω 2 µΣ 2

 (ha) nm  =  + ω 2 µΣ   a  (or)

 1 c  = ∴  µΣ Σγ  

TE01

2  (ha )01   2π f Σγ  γ=    − c  a   

2

2 9  3.832   2π × 13 × 10 × 2.25  =  −  2  3 ×108  0.5 ×10   

2

α = 648.5 Nepers / m. Propagation constant γ becomes real value i.e. v= α If the length of the waveguide is 0.5~ than the Alternation α l = 648.5 x 0.5 = 324.26 Nepers/m 25. A TE11 mode is propagating through a circular waveguide. The radius of the guide is 1cm and the guide content an air dielectric a) Determine in the out-off frequency b) Determine the wavelength in the guide for an operating frequency of 3GHz. c) Determine the wave impedance in the guide. For TE11 mode (ha)11 = 1.8~1 A = 5 x 10-2m (a) Cut –off frequency: (ha )11 c = 2π a 1.8 ~ 1× 3 × 103 2π × 5 ×10−2 b) The phase constant in the guide is . =

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β = ω 2 µ − h2 f = 3 × 109

µ = 4π ×10−7 Σ = 8.854 ×10−12 ( ha)11 h= a  1.8 ~ 1  β = (2π × 3 ×109 ) 2 4π × 10−7 × 8.854−12 −  −12   5 ×10  the wavelength in the guide is 2π β 2π λ= 50.9 λ = 12.3 cm.

λ=

c) the wave impedance is 2TE =

ωM β

2π × 3 × 109 × 4π × 10−7 50.9 2TE = 465 ohms. =

26. An air filled circular loan guide having an inner radius of icon is excited in dominant mode at 10 GHz. Find the a) cut off frequency of dominant mode at 10GHz . Find the cut-off frequency: guide wavelength and wave impedance. Find the bandwidth. For operation in dominant mode only. The dominant mode is TE, For TE,, mode (ha)11 = 1.84) F = 10x10-9H2 A = 1x10-2m

a) unit off frequency:

VEL TECH


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VEL TECH


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(ha ), c 2π a 1.841× 3 × 108 = 2 ×1× 10−2

fc =

b) Guide wavelength:

λ

λg =

 fc  1−  2  1  λ = c/ f

2

3 ×102 = 10 ×10−9 = 3 × 10−2 m

λg =

3 × 10−2  8.795  1−    10 

2

= 6.3 ×10−2 m c) wave impedance: zTE =

=

η  fc  1−    f  120π

2

2

 8.795  1−    10  = 792 Bandwidth = cut-off frequency of TM01-cut-off frequency of TE,, ( ha ) 01 c fc of TM 01 = 2π a (ha) 01 = 2.405 fc of TM 01 =

1.405 × 3 ×108 2 × π ×1×10 −3

=11.49 GHz

Band width = 11.49 - 8.795 = 2.695 GHz

RECTANGULAR WAVEGUIDES VEL TECH


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27. Derive the field configuration, cut-off frequency and velocity of propagation for TM waves in rectangular wave guides.

Consider the shape of the rectangular waveguides above with dimensions a and b (assume a > b) and the parameters e and m. For TM waves, Hz =0 and E, should be solved from equation for TM mode; ° 2 xy E 0 + h 2 E 0 = 0 N z z 0 − gz Since Ez (x,y,z) = Ez ( x, y ) e , we get the following equation,

 ∂2  ∂2 + + h 2  Ex0 ( x, y ) = 0  2 2  ∂x ∂y  0 If we use the method of separation of variables, that is Ez ( x, y ) = X ( x ) , Y ( y ) we get 2 d 2 X ( x) 1 d Y ( y) − = + h2 X ( x ) dx 2 Y ( y ) dy2

1

Since the right side contains x terms only and the left side contains y terms only, they are both equal to a constant. Calling that constant as kx2, we get, d 2 X ( x) + k x2 X ( x ) = 0 2 dx 2 d Y ( y) + k y2Y ( y ) = 0 2 dy 2 2 2 Where k y = h + kx

VEL TECH


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Now, we should solve for X and Y from the preceding equations. Also we have the boundary conditions of; Ez0 ( 0, y ) = 0 Ez0 ( a, y ) = 0 Ez0 ( x, 0 ) = 0 Ez0 ( x, b ) = 0 From all these, we conclude that X(x) is in the form of sin k, x where kx = mp/a, m= 1,2,3… Y (y) is in the form of sin kyy, where ky = np/b, n=1,2,3…. 0 So the solution of Ez ( x, y ) is

 mπ Ez0 ( x, y ) = E0 sin   α

  nπ  x  sin  y   b V / m

2 2 2 From k y = h − kz , we have

 mπ h =  a 2

  nπ  +    b 

2

For TM waves, we have H x0 =

jW ε ∂Ez0 h 2 ∂y

H y0 = −

jW ε ∂Ez0 h 2 ∂x

Y ∂Ez0 h 2 ∂x γ ∂Ez0 0 Ey = 2 h ∂y Ex0 = −

From these equations, we get

VEL TECH


VEL

VEL TECH


Ex0 ( x, y ) = −

γ  mπ   mπ   nπ  x  sin  y  E0 cos  2  h  a   a   b 

E y0 ( x, y ) = −

γ  nπ  h2  b

  mπ  Eo sin    a

VEL

  nπ  x  cos  y   b 

H x0 ( x, y ) = −

jwε h2

 nπ   b

  mπ  Eo sin    a

  nπ  x  cos  y   b 

H y0 ( x, y ) = −

jwε h2

 mπ   b

  mπ  Eo cos    a

  nπ  x  sin  y   b 

Where 2

 mπ   nπ  γ = j β = j w µε −   −   a   b 

2

2

Here, m and n represent possible modes and it is designated as the TM mn mode. M denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. When we observe the above equations we see that for TM modes in rectangular waveguides neither m nor n can be zero. This is because of the fact that the field expressions are identically zero if either m or n is zero. Therefore, the lowest mode for rectangular waveguide TM mode is TM11 Here, the cut-off wave number is 2

 mπ   nπ  kc =   +   a   b 

2

and therefore,

β = k 2 − kc2 The cut-off frequency is at the point where g vanishes. Therefore, 1 f = 2 εµ VEL TECH

2

m n   +   a  b

2

( Hz ) VEL TECH MULTI TECH TECH HIGHTECH 259

VEL

VEL TECH


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Since I= u/f, we have the cut-off wavelength 2

λc =

2

m n   +   a  b

2

( m)

At a given operating frequency f, only those frequencies , which have fc < f will propagate. The modes with f < fc will lead to an imaginary b which means that the field components will decay exponentially and will not propagate. Such modes are called cut-ff or evanescent modes. The mode with the lowest cut-off frequency is called the dominant mode. Since TM modes for rectangular waveguides start from TM11 mode, the dominant frequency is

(

1 f c ) 11 = 2 εµ

2

1 1   +  a b

2

( Hz )

The wave impedance is defined as the ratio of the transverse electric and magnetic fields, Therefore, we get from the expressions for Ex and Hy (see the equations above). ZTM =

Ex Y jβ jβ βη = = = ⇒ ZTM = H y jW ε jwε jwε k

The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to

λ=

2π 2π > =λ β k

Which is thus greater than 1, the wavelength of a plane wave in the filling medium. The phase velocity is up =

w w > = β k

1 µε

Which is greater than the speed of light (plane wave) in the filling material Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. The attenuation constant due to the losses in the dielectric can be found as follows: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 260

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2

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2

f  f   f  σ γ = j β = j k − k = jk 1 −  c  = jw 1 −  c  = jw µ ε + 1−  c  jw  f   f   f  2

2 c

28. Derive the field configuration cut-off frequency propagation for TE waves in rectangular wave guide.

and

velocity

of

TE Modes Consider again the rectangular waveguide below the dimensions a and b (assume a>b) and the parameters e and m.

For TE waves Ez =0 and Hz should be solved from equation of TE mode; 0 − gz Since Hz (x,y,z) = H z ( x, y ) e , we get the following equation,

 ∂2  ∂2 + + h 2  H z0 ( x, y ) = 0  2 2  ∂x ∂y  If we use the method of separation of variables, that is Hz0(x,y) =X(x), Y(y) we get, −

2 d 2 X ( x) 1 d Y ( y) = + h2 2 2 X ( x ) dx Y ( y ) dy

1

Since the right side contains x terms only and the left side contains y terms only, they are both equal to a constant. Calling that constant as kx2, we get; d 2 X ( x) + k z2 X ( x ) = 0 2 dx VEL TECH


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VEL TECH


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d 2Y ( y ) + k y2Y ( y ) = 0 dy 2 2 2 2 Where k y = h − kx

Here, we must solve for X and Y from the preceding equations. Aslo we have the following boundary conditions: ∂H z0 = 0 ( Ey = 0) at x=0 dx ∂H z0 = 0 ( Ey = 0) at x=a dx 0 ∂H z = 0 ( Ex = 0 ) at y=0 dy ∂H z0 = 0 ( Ex = 0 ) at y = b dx From al these, we get  mπ H z0 ( x, y ) = H 0 cos   a

  nπ  x  cos  y   b ( A / m )

2 2 2 From k y = h − kz , we have;

2

 mπ   nπ  h =  +   a   b 

2

2

For TE waves, we have

γ H =− 2 h γ H y0 = − 2 h 0 x

∂H z0 ∂x ∂H z0 ∂y

Ex0 =

jwµ ∂H z0 h 2 ∂y

E y0 =

jW µ ∂H z0 h 2 ∂x

From these equations, we obtain VEL TECH


VEL

VEL TECH


Ex0 ( x, y ) =

jwµ  nπ   mπ   nπ  x  sin  y  H 0 cos  2  h  b   a   b 

E y0 ( x, y ) =

jwµ  mπ  h2  b

H x0 ( x, y ) =

γ  mπ  h2  a

  mπ  H 0 sin    a

  nπ  x  cos  y   b 

H y0 ( x, y ) =

γ  nπ  h2  b

  mπ  H 0 cos    a

  nπ  x  sin  y   b 

  mπ  H 0 sin    a

VEL

  nπ  x  cos  y   b 

Where 2

 mπ   nπ  γ = jβ = j w µ s −   −   a   b 

2

2

As explained before, m and n represent possible modes and it is shown as the TE mn mode. ,m denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. Here, the cut-off wave number is 2

 mπ   nπ  kc =   +   a   b 

2

And therefore,

β = k 2 − kc2 The cut-off frequency is at the point where g vanishes, Therefore, fc =

1 2 εµ

2

2

m n   +   ( Hz )  a  b

Since I = u/f, we have the cut-off wavelength

VEL TECH


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λc =

2

m n   +   a  b

2

VEL

( m)

At a given operating frequency f, only those frequencies, which have f>fc will propagate. The modes with fb and so the dominant frequency is

( f c ) 10 =

1 2a µε

( Hz )

The wave impedance is defined as the ratio of the tranverse electric and magnetic fields. Therefore, we get from the expressions for E x and Hy (see the equations above); ZTE =

Ex jwµ jwµ kη = = ⇒ ZTE = Hy γ jβ β

The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to

λg =

2π 2π > =λ β k

Which is thus greater than 1, the wavelength of a plane wave in the filling medium.

The phase velocity is up =

w w > = β k

1 µε

Which is greater than the speed of the plane wave in the filling material The attenuation constant due to the losses in the dielectric is obtained as follows: 2

2

2

f  f   f  σ γ = j β = j k − k = jk 1 −  c  = jw 1 −  c  = jw µ ε + 1−  c  jw  f   f   f  2

VEL TECH

2 c


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After some manipulation, we get cn

αd =

f  2 1−  c   f 

=

2

k 2 tan δ 2β

Example: Consider a length of air-filled copper x-band waveguide, with dimensions a=2.286 cm, b= 1.016 cm. find the cut-off frequencies of the first four propagating modes. Solution: From the formula for the cut-off frequency 1 fc = 2 εµ

2

2

2

2

c m n m n air − filled →   +      +  2  a  b  a  b

( Hz )

B.E./B.TECH. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2008 FIFTH SEMESTER ELECTRONICS AND COMMUNICATION ENGINEERING EC 1305 – TRANSMISSION LINES AND WAVEGUIDES (COMMON TO B.E. (PART – TIME) FOURTH SEMESTER REGULATION 2005) PART - A 1. Briefly discuss the difference between wavelength and period of a sine wave. VEL TECH


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2. Find the attenuation and phase shift constant of a wave propagating along the line whose propagation constant is 1.048× 10-4 ∠ 88.8o. 3. Give the minimum and maximum value of SWR and reflection coefficient. 4. Why is the quarter wave line called as copper insulator? 5. Enumerate the properties of TEM waves between parallel planes of perfect conductors. 6. Plot the frequency – versus – attenuation characteristic curve of TM and TE waves guided between parallel conducting plates. 7. How is the TE10 mode launched or initiated in rectangular wave guide using an open ended coaxial cable? 8. Calculate the cut-off frequency of a rectangular wave guide whose dimensions are ‘a’=2.5cm and ‘b’=1.5cm operating at TE10 mode. 9. Why is the Bessel’s function of the second kind (neumann’s function applicable for the field analysis inside the circular wave guide? 10.

Distinguish between wave guides and cavity resonator.

PART – B 11. (a) Derive the general transmission line equations for voltage and current any point on a line. Or (b)(i) Write a brief notes on frequency and phase distortions. (ii) The characteristic impedance of a 805m-long transmission line 94∠ 23.2oΩ , the attenuation constant is 74.5× 10-6 Np/m and the phase shift constant is 174× 10-6 rad / m at 5KHz. Calculate the line parameters R, L, G and C per meter and the phase velocity on the line.

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12. (a) (i) A 75Ω loss less transmission line is to be matched to a resistive load impedance of ZL=100Ω via a quarter-wave section. Find the characteristic impedance of the quarter wave transformer. (ii) A 50Ω loss less transmission line is terminated in a load impedance of ZL=(25+j50)Ω . Use the SMITH chart to find. (1)Voltage reflection coefficient, (2)VSWR, (3) Input impedance of the line, given that the line is 3.3λ long and (4)Input admittance of the line. Or (b) A 50Ω loss less feeder line is to be matched to an antenna with ZL(75-j20)Ω at 100MHz using SINGLE shorted stub. Calculate the stub length and distance between the antenna and stub using smith chart. 13.

(a)(i) Derive the components of Electric and Magnetic field strength between a pair of parallel perfectly conducting planes of infinite extent in the ‘Y’ and ‘Z’ directions. The planes are separated in X direction by “a” meter. (ii) A parallel perfectly conducting plates are separated by 5cmin air and carries a signal with frequency of 10 GHz in TM11 mode. Find the cut-off frequency and Cut-off wave length. Or (b) (i) Discuss on the characteristics of TE, TM and TEM waves between parallel conducting planes. And also derive the expressions for the cut off frequency and phase velocity from the propagation constant. (ii) Describe the Velocity of propagation of wave between a pair of perfectly conducting plates.

14.

(a) Derive the field configuration, cut of frequency and velocity of propagation for TE waves in rectangular wave guide. Or (b) A TE10 wave at 10 GHz propagates in a X-band copper rectangular wave

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guide whose inner dimensions are ‘a’=2.3cm and ‘b’ =1cm, which is filled with Teflon ε r=2.1, µ r=1. Calculate the cut-off frequency, velocity of propagation, Phase velocity, Phase constant, Guide wave length and Wave impedance. 15.

(a)(i) Derive the expression for TM wave components in circular wave guides using Bessel function. (ii) Write a brief note on excitation of modes in circular wave guides. Or (b) Derive the equation for Q – factor of a rectangular cavity resonator for TE 101 mode. ***************

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Transmission Lines & Wave Guides

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