UNIT 1. Amplitude modulation Topics : 1. Review of spectral characteristics of periodic and non periodic signals 2. Principles of AM 3. Generation and detection of AM 4. Generation and detection of DSBSC 5. Generation and detection of SSB 6. Generation and detection of VSB 7. Comparison of Amplitude modulation systems 8. Frequency translation 9. Frequency Division Multiplexing
1
Definition for a Signal: A signal is defined as a physical quantity that varies with time, space or any other independent variable or variables. It contains some information. Eg. Electric voltage or current Radio signal, TV Signal, Computer Signal, etc., Signal can be represented either using Time domain approach or Frequency domain approach. •
In time domain representation of signals the amplitude is represented on one axis & the time is represented on the other axis.
•
In frequency domain representation the amplitude or power is shown on one axis and frequency is displayed on the other. It specifies the relative amplitudes of the freq component.
•
Any signal can be represented on either way.
•
Signals can be broadly classified into Continuous Time signal (CT) and Discrete Time Signal (DT).
•
A continuous time signal varies in its amplitudes continuously with time.
•
A DT signal is one in which the amplitude of the signal is discrete with respect to time.
•
A continuous time signal is called as Analog Signal.
•
A discrete time signal is known ad Digital Signal.
Characteristics of a Signal : The parameters such as amplitude, frequency & phase of the signal are known as characteristics of the Signal.
2
3
Energy and power signals: Power Signal : •
A signal is called a power signal if its “average normalized power” is non zero and finite .It has been observed that almost all the periodic signals are power signals.
Energy signals: •
A signal having a finite non zero total normalized energy is called as an energy signal. It is observed that almost all the non periodic signals defined over a finite period , are energy signals. As these signals are defined over a finite period, they are called as time limited signals
Average Normalized Power: Average normalized power
The above definition can be generalized for a complex signal X(t) as
4
For periodic signal with a period To the equation 1 and 2 get modified as
For a complexPeriodic signalx ( t) the average normalized power is given by,
Energy: The total normalized energy for a "real" .signal x ( t) is given by,
……5 However if the signal is complex then the expression for total normalized energy is given by
A Review of Fourier Series and Fourier Transform •
In the field of communication engineering we need to analyze a given signal.
•
To do so we have to express the signal in its frequency domain
•
The translation of a signal from time domain to frequency domain is obtained by using the tools such as Fourier series' and Fourier transform
5
Fourier Series: •
Sine waves and cosine waves are the basic building functions for any periodic signal.
•
That means any periodic signal basically consists of sine waves having different amplitudes, of different frequencies and having different relative phase shifts.
•
Fourier.series represents a periodic waveform in the form of sum of infinite number of sine and cosine terms. It isa representation of the signal in a time domain series form.
•
Fourier series is a "tool" used to analyze any periodic signal. After the "analysis” we obtain the following information. about the signal :
(i) All the frequency components present in the signal (ii) Their amplitudes and (iii) The relative phase difference between these frequency components. All the "frequency components" are nothing else but .sine waves at those frequencies Exponential Fourier Series [OR Complex Exponential Fourier Series : •
Substituting the sine and cosine functions in terms of exponential function in the expression for the quadrature fourier series, we can obtain another type of Fourier series called the exponential Fourier series.
•
A periodic signal x ( t) is expressed in the exponential Fourier series form as follows :
6
Amplitude and Phase spectrum: •
The amplitude spectrum of the signal X(t) is denoted by,
The Phase of the spectrum is denoted by,
•
The amplitude spectrum is a symmetric or even function That means ICn I = IC_n I. But the phase spectrum is an asymmetric or odd function. That means arg ( Cn ) = - arg ( C_n ).
Fourier Transform : • A Fourier transform is the limiting case of Fourier series. It is used for the analysis of non-periodic signals. • The Fourier transform of a signal x ( t) is defined as follows.
……………5 This equation is known as analysis equation. Inverse Fourier Transform: •
The signal X(t) can be obtained back from fourier transform X(f) by using the inverse fourier transform. The inverse transform (IFT) is defined as follows:
7
……..6 Amplitude and Phase spectrums: • The amplitude and phase spectrums are continuous rather than being discrete in nature. Out of them, the amplitude spectrum of a real valued function X(t) exhibits an even symmetry and •
The phase spectrum has an odd symmetry. X(f)=X(-f)
…………….6
……….7
Properties of Fourier Transform:
8
Problems: Ex.1 Obtain the Fourier transform of a rectangular pulse of duration T and amplitude A as shown in the fig
The rectangular puse shown can be expressed mathematically as
This is also known as gate function Therefore the Fourier transform will be
9
………….1 As per the Euler’s theorem
Applying this in equation (1) we get …………….2 Multiply and divide the RHS of equation 1 by T to get
…………….3 In the above equation
…………4
Thus the rectangular pulse transform into a sinc function.
10
Amplitude spectrum: The amplitude spectrum of the rectangular function is as shown in fig below As, sinc(0) = 1 Therefore AT sinc(0) = AT The sinc function will have zero value for the following values of “fT”:
The phase spectrum has not been shown as it has zer value for all the values of f , To absorb negative values of X(f) in the phase shift: The negative amplitude of the amplitude spectrum has been absorbed by introducing a ±180º phase shift as shown in the figure below
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EX. 2: For the sinc function shown in the fig obtain the fourier transform and plot the spectrum. Soln: The sinc signal shown in the figure can be expressed as X(t)= Asinc(2wt) To evaluate the fourier transform of this function we are going to apply the duality and time domain properties of the Fourier transform.Refering to the example 1, where we have obtained
The Fourier transform of a rectangular pulse of amplitude A and duration T, as,
Using the duality property we can write that
Comparing equation (3) with the RHS of Equation (1) which states the expression for X(t), as
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•
Thus a sinc pulse in the time domain is transformed into a rectangular pulse in the frequency domain. The spectrum of the sinc pulse is shown in the figure.
Fourier Transform for the Periodic signals: •
Sometimes it is essential to obtain the FT of periodic signals .For example, the sampling theorem
•
FT of a periodic signal is given by
Where Cn is the Fourier coefficient given by
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Fourier Transforms of Standard signals: The FT of Some signals is given in the following table.
Sr.No Signal
Mathematical
Fourier transform
representation 8
Cosine signal
X(t) = cos(2πfct)
9
Sine signal
X(t) = sin(2πfct)
10
Signum function
X(t) = sgn(t)
11
Unit step
X(t) = u(t)
X(f)=1/jπf
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Power and energy theorem: The theorems are: 1. Parseval’s power theorem 2. Rayleigh’s energy theorem. Parseval’s Power theorem: This theorem relates the average power of a periodic signal to its “fourier series” coefficients. This theorem states that the total average power of a periodic signal X(t) is equal to the sum od the average powers of the individual fourier coefficients i.e Cn Average Power of X(t)=(Power of C1) + (Power of C2) +……. Or total average power
Rayleigh’s energy theorem: This is analogous to the Parseval’s power theorem.It states that the total energy of the sinal X(t) is equal to the sum of energies of the individual spectral components in the frequency domain. The total normalized energy of a signal X(t) is given by,
The normalized energy:
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Definition for Communication: •
Electronic Communication System – defined as the whole mechanism of sending and receiving as well as processing of information electronically from source to destination.
•
Example – Radiotelephony, broadcasting, point-to-point, mobile communications, computer communications, radar and satellite systems.
Introduction to Communication: •
Communication – Basic process of exchanging information from one location (source) to destination (receiving end).
•
Refers
–
process
of
sending,
receiving
and
processing
of
information/signal/input from one point to another point. NEED FOR COMMUNICATION: •
Interaction purposes – enables people to interact in a timely fashion on a global level in social, political, economic and scientific areas, through telephones, electronic-mail and video conference.
•
Transfer Information – Tx in the form of audio, video, texts, computer data
and
picture
through
facsimile,
telegraph
or
telex
and
internet.Broadcasting – Broadcast information to masses, through radio, television or teletext
Terms Related To Communications: •
Message – physical manifestation produced by the information source and then converted to electrical signal before transmission by the transducer in the transmitter.
•
Transducer – Device that converts one form of energy into another form. 16
•
Input Transducer – placed at the transmitter which convert an input message into an electrical signal. Example – Microphone which converts sound energy to electrical energy
Output Transducer – placed at the receiver which converts the electrical signal into the original message. Example – Loudspeaker which converts electrical energy into sound energy Signal – electrical voltage or current which varies with time and is used to carry message or information from one point to another
Elements of a Communication System The basic elements are : Source, Transmitter, Channel, Receiver and Destination
Function of each Element •
The message produced by the information source is not electrical in nature; It may be a voice signal, picture signal etc. Hence a transducer is required to convert the original physical message into a time varying electrical signal.
•
These signals are called as Base band signals or Message Signals or modulating Signals.
•
Transmitter comprises of electrical and electronic components that converts the message signal into a suitable form for propagating over the communication medium. .
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•
Channel provides the connection between sources and destinations. It can be of many forms like coaxial cable, microware link, Radio leave link or Optical fiber.
•
The receiver extracts the message signal from the degraded version of transmitted signal.
•
At the destination, another transducer is used to convert the electrical signal into the appropriate message.
Base band transmission: •
Baseband signal is the information either in a digital or analogue form.
•
Transmission of original information whether analogue or digital, directly into transmission medium is called baseband transmission.
•
Example: intercom (figure below)
Baseband signal is not suitable for long distance communication because of the following reasons: •
Hardware limitations •
Requires very long antenna
•
Baseband signal is an audio signal of low frequency. For example voice, range of frequency is 0.3 kHz to 3.4 kHz. The length of the antenna required to transmit any signal at least 1/10 of its wavelength (λ). Therefore, L = 100km (impossible!)
•
Interference with other waves
Simultaneous transmission of audio signals will cause interference with each other. This is due to audio signals having the same frequency range and receiver stations cannot distinguish the signals. In Order to overcome this disadvantage of Baseband transmission we go for Modulation where the low frequency message signal is converted to high frequency signal.
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Modulation & Need for modulation Modulation is the process of changing the characteristics of a high frequency carrier signal in proportion with the instantaneous value of the modulating signal. Need for modulation 1.
Antenna Height: For communication the antennas are needed to transmit & receive the signals. The antenna radiates effectively when its height is of the order of wavelength of the signal to be transmitted. χ c Antenna Height = ––– = ––– 2 2f f → frequency of the signal. c → Velocity of the light. For a low frequency signal say 1KHz. 3 x 108 o Antenna height = ––––––––– = 150 km 2 x 1 x 103 Where as for a high frequency signal say, 3 x 108 o 1MHz Antenna height = ––––––––– = 150 m 2 x 1 x 103 This height is practically achievable.Hence modulation is needed .
2.
Narrow Banding: Say we are transmitting a baseband signal in the range of 50Hz to 10KHz. The ratio of highest frequency to lowest freq. will be 103 x 103 / 50 = 200. It we design an antenna for 50 Hz. It will be too large for 10KHz. Once we modulate this signal range using a 1MHz carrier. The ratio of highest freq. will be (104 + 106) / 106 + 50 = 1.01/1 ≈ 1. Same antenna holds good for the entire range of signal.
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3.
Multiplexing : If more than one signal uses a single channel then modulation may be used to translate different signal to different spectral location, thus, enabling the severer to select the desired signal.
4.
To overcome equipment limitations: When ever we are designing the equipment, it will be designed for fixed range of freq. By modulation we can make any signal to pass through the same equipment.
5.
Modulation to reduce noise and interference: It is not possible to eliminate noise and interference in communication system. But can be minimized using modulation technique.
Amplitude Modulation: It is the process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal (Information, Message). Though there are 3 types of Am Double side band full carriers is the most commonly used and is also know as conventional Am or simply Am.
Fig 1. a) Message signal b) carrier signal c) Amplitude modulated signal. 20
Principles of Amplitude Modulation: AM envelope and the equation: • Consider a message signal em= Em sin 2πfmt. •
Let the carrier Signal be ec= Ec sin 2πfct.
•
According to the definition the Amplitude modulated wave can be indicated as Eam=(Ec+Em sin 2πfmt.) sin 2πfct The shape of the modulated wave is called the Am envelop.
Depth of Modulation/Modulation Index: Coefficient of modulation and percent modulation: If is defined as the ratio of maximum amplitude of the message signal to the maximum amplitude of the carrier signal. Em m = ––––– Ec Percent modulation is indicated ad M Em M = ––––– x 100 Ec Relationship b/w m, Em & Ec
or
M = m x 100
From the figure below.
Em =
½ (Emax - Emin)
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EC = ½ (Emax + Emin)
Emax - Emin ½ (Emax - Emin) M = –––––––––––––– x 100 = –––––––––– ½ (Emax + Emin) Emax + Emin •
x 100
Based on the modulation index modulation can be either, (i). Critical Modulation (ii). Over Modulation (iii). Under Modulation
•
When Em = Ec modulation goes to 100% this situation is known as critical modulation.
•
Em < Ec leads to under modulation.
•
Em > Ec leads to over modulation.
22
AM frequency Spectrum & Bandwidth : •
Am envelope is a complex wave made up of a dc, voltage, the carrier freq. & the sum (fc + fm) and difference (fc - fm) frequencies. (Cross Products)
•
The sum and difference frequencies are displaced from the carrier frequency by an amount equal to the modulating signal.
•
The effect of modulation is to translate the modulating signal in the frequency domain.
•
The figure 2 shown the frequency spectrum of Am.
•
It extends from fc -fm (max) to fc + fm (max).
•
The band of frequencies b/w fc and fc - fm (max) is called lower side band [LSB] and any frequency within this band is called lower side frequency [LSF].
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•
The band of frequencies b/w fc and fc + fm (max) is called upper side band [VSB] and any frequency within this and is called upper side frequency [USF]
Bandwidth of AM. The Bandwidth of Am wave is equal to the difference b/w the highest upper side frequency and lowest lower side frequency. B
=
fc + fm (max) – [fc - fm (max)]
=
fc + fm (max) - fc + fm (max)
BW =
2fm (max)
Ex. 1 : For an Am DSBFC modulator with a carrier frequency fc = 100 KHz and maximum modulating signal frequency fm (max) = 5 KHz, determine. a. Frequency limits for upper and lower sidebands. b. Bandwidth. c. Upper and lower side frequencies produced when modulating a signal in a signal frequency 3 KHz tone. d. Draw the freq spectrum. Soln : fc =
100 KHz
fm =
5 KHz
a. Lower side band Freq.
= f c - fm
= 95 KHz
Upper side band Freq.
= f c + fm
= 150 KHz
b. B.W
= 2 fm (max) = 10 KHz.
For a 3 KHz msg signal. LSF
= 100 k - 3KHz
→ 97 KHz
USF
= 100 k + 3KHz
→ 103 KHz
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AM Voltage Distribution : •
The carrier signal can be described as , Vc cts = Ec sin 2πfct
•
Vc cts →
Time varying voltage of carrier.
Ec
→
Peak carrier amplitude.
Fc
→
Carrier freq in
The modulation signal can be expressed as, Eam
= Vam cts = (Ec + Em sin 2πfmt) sin 2πfct = Ec sin 2πfct + Em sin 2πfmt sin 2πnfct Em Em = Ec sin 2πfct – –––– cos 2π (fc+fm) t + –––– cos 2π (fc+fm) t 2 2
Eam
Em Em = Ec sin 2πfct – –––– cos 2π (fc+fm) t + –––– cos 2π (fc+fm) t 2 2
Ec sin 2πfct
→
Carrier signal (volts)
-mEc –––– Cosπ (fc + fm) t. 2
→
Upper side frequency signal (volts).
→
Lower side frequency signal (volts).
+mEc Cosπ (fc - fm) t. 2 V (max) = •
Ec + Ec/2 + Ec/2 = 2Ec
V (min) = Ec - Ec/2 - Ec/2 = OV The peak change in the amplitude of the o/p wave is the sum of the voltages from the upper and lower side frequencies. Hence Em = Eusf + Elsf If, Eusf = Elsf then EUSF = ELSF = Em / 2 = ¼ (Emax + Emin)
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Ex. 2 : One i/p to a conventional Am modulator is 500 KHz carrier with an amplitude of 20 Vp. The second i/p is a 10 KHz modulating signal that is sufficient to cause a change in the o/p wave of ± 7.5vp Determine. a. Upper and Lower side frequencies. b. Modulation coefficient and percent modulation c. Peak amplitude of the modulated carrier and the upper and lower side frequency voltages. d. Maximum and minimum amplitudes of the envelope. e. Expression for the modulated wave. f. Draw the o/p spectrum. g. Sketch the o/p envelop. Soln : a. fusf
=
500 KHz + 10 KHz =
510 KHz
Flsf
=
500 KHz - 10 KHz =
490 KHz
b. m = Em / Ec = 7.5 / 20 = 0.375 M = 0.375 x 100 = 37.5% c. Ec = 20 VP EVSF = ELSF = mEc / 2 = 0.375 x 20 /2 = 3.75 Vp d. Vmax
=
Em + Ec
=
27.5 Vp
Vmin
=
Ec - Em
=
12.5 Vp
e. Vam lts = 20 sin 2π 500 kt – 3.75 cos (2π 510 kt) + 3.75 cos 2π 490 kt Am power Calculation : The power of the carrier is
Pc
Ec2 Pc = ––– 2R → Carrier power.
Ec
→
Peak carrier Voltage.
26
R
→
Load resistance
PUSB = PLSB =
PUSB = PLSB
m2Ec2 (mEc/2)2 ––––––– = –––––– 2R 8R
Ec2 m2 = ––––– =–––––– 4 2R m2 = –––– PC 4
Pt
= PUSB + PLSB + PC
Pt
m2Pc m 2 Pc = Pc + ––––––– + –––––– 4 4
Pt
m2 = Pc 1 + –––– 2
Ex. 3 : An Am transmitter operating at a carrier freq of 1 MHz and a modulating frequency of 5 KHz and modulated at 60% depth delivers a carrier power of 6kw into 50Ω. a. Obtain the total average power of the modulated signal in DB watts & dsm. b. Obtain the RMS voltage of the modulated signal. Soln : a. Pt
=
Pc (1+m2 /2)
=
6 ( 1 + 0.36 / 2) = 10.8kw = 40.334 dB.
b. The Modulated signal power.
27
V2(rms) =
2 x RL x P Signal. =
2.5 x 10.8 x 103
=
1.028 kV.
Ex. 4 : For a Am DSBFC wave with a peak unmodulated carrier voltage V1 = 10Vp, a load resistance Rc =10Ω. and modulation coefficient m=1, determine a. Power of the carrier and the upper and lower sidebands b. Total solebard power c. Total power of the modulated wave. d. Draw the power spectrum. e. Repeat the above for m=0.5. Soln : a. Pc
102 100 Ec2 = –––– = –––––– = –––––– =5w 2R 2(10) 20
PUSB =
PLSB
m2Pc = –––––– = ¼ x 5 = 1.25w 4
b. Total sideband power = PUSB + PLSB = 1.25 x 2 = 2.50w c. The total power in the modulated wave is found by, Pt
=
Pc( 1+ m2/2) = 7.5ω
d. The power spectrum is given by. Am Current Calculations It2R It 2 m2 Pt ––– = –––– = ––– = 1 + –––– Pc Ic2R Ic 2 2 Pt
→
Total transmit Power
Pc
→
Carrier Power
It
→
Total Current 28
Ic
→
Carrier Current
m
→
Modulation index
R
→
Antenna Resistance
It 2
=
Ic2 (1+m2/2)
It 2
=
Ic2 √1+m2/2)
Modulation by a complex information signal : •
If the modulating signal contains two frequencies say fm1 & fm2, the modulated wave will contain the carrier and two sets of side frequencies. Spaced symmetrically about the carrier. Vam (t) = sin 2π fct + ½ cos 2π (fc - fm1) t - ½cos 2π (fc+ fm1) t + ½cos 2π (fc- fm2) t - ½cos 2π (fc+ fm2) t
•
The coefficient of modulation or modulation index is given by mt
=
√m12 + m22
•
In general for n different signals mt = √m12 + m22 + . . . mn2
•
Pusbt
= Plsst = Pcmt2 / 4 •
Pt
= Pc (1+mt2 / 2)
Ex. 5 : Find the power in each side band of a DSBSC signal with the carrier signal at 1 MHz and of a peak signal voltage on 100V modulated simultaneously by 3 different signals. The freq. of the modulating signal are 2KHz, 3KHz and 5KHz respectively and Peak modulating vaoltages are 10V, 20V and 30V respectively. Assume a load resistance of 100Ω. Soln : Given fm1 = 2KHz; fm2 = 3KHz ; fm3 = 5KHz Vm1 = 10V; Vm2 = 20V ; Vm3 = 30V Vc = 100V 29
m1 = 10/100 = 0.1 m2 = 20/100 = 0.2 m3 = 30/100 = 03 mt = m12 + m22 + m32 = 0.374 PSB = Pc x m2/4 = Vc2 (Peak) / 2RL x m2/4 1002 x 0.3742 = –––––––––––– = 1.75ω 2 x 100 x 4
AM modulating Circuits •
Based on the location in the transmitter Am modulating circuits are classified as a. Low level AM modulator b. High level AM modulator Difference b/w low level and High level AM mod.
1.
2.
Low level AM modulator
High level AM modulator
Modulation takes place prior to 1.
Modulation takes place in the
the final stage of the transmitter
final element of final stage.
Less modulating signal power 2.
More modulating signal power
is required
is required.
30
Low level AM modulator
•
Class A amplifier can perform amplitude modulation.
•
Amplifier must have 2 inputs one for the carrier signal & second for modulating signal.
•
Without modulating signal the carrier signal is just amplified.
•
The carrier is applied the base and the modulating signal to the emitter. Hence it is also called as Emitter Modulation.
•
The modulating signal varies the gain of the amplifier at a sinusoidal rate equal to the frequency of the modulating signal.
•
The depth of modulation achieved is proportional to the amplitude of the modulating signal.
•
The voltage gain for an emitter modulator is expressed as, Av = Aq (1 + m sin 2πfmt) Av
→ Amplifier voltage gain with modulation.
Aq
→ Amplifier quiescent (without modulation) voltage gain.
sin 2πfmt varies form + 1 to -1 Hence,
Av = Aq (1±m)
Av max =
2Aq,
when m = +0
Av min
0,
when m = -1
=
31
Advantages of Low level modulation: 1. Less modulating signal power is required to obtain high percentage modulation. 2. Modulating circuit is designed for low power. Disadvantage of Low level modulation Amplifiers following modulator stage must be linear. At high operating powers linear amplifiers are very inefficient. High power AM modulator / Medium Power AM modulator.
•
The class C amplifier is used. It operates nonlinear and is capable of nonlinear mixing (modulation).
•
This is known as collector modulator because modulating signal is applied to the collector.
•
When the amplitude of the carrier exceeds the barrier potential (0.7V) Q1 turns on collector current flows.
•
When carrier voltage drops below 0.7V Q1 turns off and collector current cases.
•
The corresponding current and voltage waveforms are shown. 32
•
When the modulating signal is applied it adds up with the Vcc and gets submitted from Vcc producing an Am o/p.
Advantages of high level modulators: •
There is no constraint of linear operation on amplifiers preceding modulator stage.
•
Power efficiency is good
Disadvantages of high level modulators: •
High modulating power is required.
•
Final modulating signal amplifier has to supply all the sideband power.
AM transmitters Low level Transmitter
•
Modulating signal is got from an acoustical transducer (Converts voice or music). Such as microphone a magnetic tape, CD or a phonograph record.
•
Bufferamplifier : Class a linear voltage amplifier. It raises the amplitude of the source.
•
Driver-amplifier The information signal to an adequate level to sufficiently drive the modulator.
•
Buffer amplifier: It is a low gain, high input impedance linear amplifier.
•
Matching Network: matches the o/p impedance of the final power amplifier to the transmission line and antenna.
Application: used in low capacity system such as wireless intercoms, Remote control units, pagers, shot range talkie. 33
High level Transmitters:
•
Here modulating signal power should be higher than the low level.
•
The amplification takes place prior to modulation.
•
Used for long distance communications.
AM Detector circuits :
34
•
Figure shows the circuit and the i/p and o/p waveforms.
•
Initially at time t=0 the diode remains off and the capacitor is completely discharges Vc = OV.
•
The diode remains off until the i/p voltage exceeds the barrier voltage.
•
When Vin reaches the Barrier voltage the diode turns on, the diode current flows, charging the capacitor. To the max amplitude.
•
When i/p voltage begins to decrease, the diode turns off. The capacitor begins to discharge through the resistor bur RC time constant is made sufficiently long so that the capacitor cannot discharge as rapidly as Vin is decreasing.
•
The diode remains off until the next i/p cycle when Vin goes 0.3V more +ve than Vc, the diode turns on the capacitor charges to the maximum value.
•
This sequence repeats itself on each successive positive peak of Vin and the capacitor voltage follows the +ve peak of Vin. Hence it is called as peak detector.
•
The output waveform resembles the shape of the i/p envelope. Hence the name shape detector.
•
The o/p wave form has a high frequency ripple that is equal to the carrier freq. This is due to the diode turning on during the positive peak of the envelope.
•
This can be removed by the audio amplifiers.
Amplitude Modulation Classification:
•
•
Double side band full Carrier (DSBFC)
•
Double side Band suppressed Carrier (DSBSC)
•
Single Side Band Suppressed Carrier (SSBSC)
•
Vestigial side Band (VSB)
In the frequency spectrum, the carrier frequency is not carrying any information hence can be suppressed and is called Double side band with suppressed carrier ( DSBSC) or simply Double side Band(DSB)
35
•
Both the sidebands are carrying the same information hence only one sideband is sufficient to convey the message. So we can suppress one sideband and transmit the other - Called as the Single side band with suppressed carrier (SSBSC) or single side band (SSB).
•
In some cases only a portion (vestige) of the modulated signal will be carrying the information hence we can transmit only that portion and suppress the rest and this type is called as vestigial side band (VSB).
DSBSC system:
It is a technique where it is transmitting both the sidebands without the carrier (carrier is being suppressed/cut) Characteristics: Power content less Same bandwidth Disadvantages - receiver is complex and expensive SSB System
Improved DSBSC and standard AM, which waste power and occupy large bandwidth • SSB is a process of transmitting one of the sidebands of the standard AM by suppressing the carrier and one of the sidebands Advantages: Saving power Reduce BW by 50% Increase efficiency, increase SNR Disadvantages:Complex circuits for frequency stability •
36
Generation and Demodulation of DSBSC : A circuit that produces a double sideband suppressed carrier signal is balanced modulator. Balanced modulator / Balanced lattice Modulator.
37
•
Has two i/p and the carrier and the modulating signal.
•
The amplitude of carrier must be sufficiently greater than the amplitude of the modulating signal to ensure the on and off condition of 4 diodes.
•
When modulating signal is +ve. The carrier signal is as shown then the diodes D1 & D2 are forward biased (on) D3 & D4 are reverse biased (off)
•
When the modulating signal is Zero & the carrier signal is in the positive half cycle, the Diode D1 & D2 are forward biased and D3 & D4 are reverse biased.
38
•
The current divides equally in the upper and lower portions of the primary winding of T2. This produces a magnetic field which is equal in magnitude but in opposite dissection in the upper and lower portions of primary winding hence they cancel each other.
•
IIIrly when the carrier signal is in the negative half cycle the diodes D3 & D4 are forward biased and D1 & D2 are reverse biased.
•
Here also magnetic fields in primary winding are equal and opposite canceling each other. Thus the carrier is suppressed.
•
The Diodes D1 to D4 are the electronic switches that control whether the modulating signal is passed from T1 to T2 as it is or with 1800 phase shift.
•
When the carrier is in +ve half cycle the message signal is transferred across the closed switches to T2 without a phase reversal.
•
When the carrier is in -ve half cycle the message signal is transferred across the closed switches to T2 with a 1800 phase reversal.
Coherent detection of DSBSC waves: The coherent detector for the DSBSC signal as shown in the figure.
•
DSBSC wave S(t) is applied to a Product modulator in which it is multiplied with the locally generated carrier
•
The locally generated carrier is exactly coherent in frequency and phase with the original carrier. Hence the method is called as coherent detection or synchronous detection.
•
The output of the product modulator is applied to the low pass filter which eliminates the unwanted frequency components to produce the original message signal.
39
Costas loop detection for DSBSC modulated wave:
•
It consists of 2 coherent detectors with same input(the DSBSC).
•
The local oscillator signals for individual detectors will be in phase and quadrature with respect to each other.
•
The local oscillator is adjusted to the carrier frequency.
•
The detector in the upper path is known as in-phase coherent detector or I-channel .
•
The detector in the lower path is known as Quadrature –phase coherent detector of Q-channel.
•
These two detectors are coupled together to form negative feedback system (to maintain local oscillator in synchronous with the carrier wave).
•
If the local oscillator signal is same as that of the carrier, I-channel output contains the desired demodulated output and Q-channel output is zero.
•
Else Q-channel contains the output.
•
Assume that the local oscillator drifts in phase by a small value φ radians, the output of I-channel will remain same but the Q-channel will produce a small voltage as its output, which is proportional to sinφ=φ. 40
•
The outputs of I and Q channel are combined in the phase discriminator, which is a multiplier followed by a low pass filter.
•
The phase discriminator produces a DC voltage proportional to the error φ. The dc voltage is applied to the input of the VCO to correct the frequency and phase of the VCO automatically to reduce the phase.
Single Sideband signal: Since both the sidebands are carrying the same information it is sufficient to transmit only one side band. Such a transmission is called as Single side band transmission.Figure below shows the SSBSC system
Advantages of SSB: •
Since only single sideband is transmitted the bandwidth of the transmitter and channel is only fm.This is half of the BW of DSBFC system.
•
The power of the suppressed sideband is saved.
•
The effect of noise at the receiver is reduced.
•
Fading effect which arises because of the interference of carrier and two sidebands is removed in SSB.
41
Generation of SSB: Suppression of unwanted sideband: •
The balanced modulator produces DSB o/p. This DSB signal contains both the sidebands.
•
This is given to sideband suppression filter to remove unwanted sidebands.
•
The filter must have a flat pass band and extremely high attenuation outside the passband.
Phase shift method to generate SSB :
•
Consists of 2 Balanced modulators1 and M2.
•
One Modulator say M1 has the direct modulating signal and 900 phase shifted carrier signal
•
Hence the o/p of modulator M1 is given as x = cos wc - (wm + 90) – cos wc + wm + 90
---- (1)
42
•
The modulator M2 gets the i/p of direct carrier signal and 900 phase shifted message signal, the o/p of M2 is given by, y = cos ( wc - 90 ) - wm – cos (wc + 90) + wm
---- (2)
At the summer we will have, x + y = cos (wc - wm) - 90 – cos (wc + wm) + 90 cos (wc - wm) + 90 – cos (wc + wm) + 90 x + y = - 2cos (wc + wm) + 90 Thus one side band is totally eliminated while retaining the other. Vestigial Sideband ,Transmission: Definition : One of the sideband is partially suppressed and vestige (portion) of the other sideband is transmitted, This vestige (portion) compensates the suppression of the sideband. It is called vestigial sideband transmission. Generation and demodulation of VSB: •
Fig.shows the generation of VSB. The product modulator generators DSBSC signal from the message and carrier.
•
The bandpass filter is designed in such a way that. it suppresses one side band partially and passes a portion (vestige) of other sideband.
•
The output of the bandpass filter is VSB signal.
43
Magnitude Response of VSB Filter Fig. shows the magnitude response of VSB filter.
•
Here observe that fc to fc+W is USB. It's portion from fc to fc +fv is suppressed partially. fc to fc - W is LSB. It's portion from fc -fv to fc is transmitted as vestige.
•
Observe that H(fc)=1/2. And the frequency response fc-fv<=H(f) <= fc+fv exhibits odd symmetry. The sum of any two frequency components in, the range. Is
equal to unity. i.e H(f-fc) + H(f+fc) = 1 •
Phase response is linear.
Transmission bandwidth From Fig. the transmission bandwidth of VSB modulation is, Br = fc +W
44
Applications and Advantages of VSB Advantages: 1. Low frequencies, near fc are, transmitted without any attenuation. 2. Bandwidth is reduced compared to DSB. Applications: VSB is mainly used for TV transmission, since low frequencies near fc represent significant picture details. They are unaffected due toVSB. Frequency translation: •
In communication system the frequency of the modulated wave is translated either upward or downward. So that the modulated wave occupies a new slot in the frequency spectrum.
•
Mixer is a device which carries out the frequency translation of the modulated wave.
•
The operation of frequency translation is called as mixing or heterodyning.
•
The mixing is alinear operation in which it preserves the relation between the sidebands of the incoming signal with the carrier.
45
Fig a) Spectrum of modulated signal at the input of the mixer b)Spectrum of modulated signal at the output of the mixer .
Frequency division multiplexing: Channel 1
Channel 2
Channel 3
Channel 4
Source Source Source Source Information Information Information Information 100 KHz 105 KHZ 110 KHz 115 KHz 120 KHz Fig 1 •
FDM is analog multiplexing scheme.
•
A familiar example is the commercial Am Broadcast band. Which occupies a freq spectrum from 535 KHz to 1605 KHz.
•
Each Broadcast station carries an information signal that occupies a Bandwidth b/w 0Hz to 5 KHz.
46
•
If the information is transmitted with original freq spectrum, it will be impossible to separate one stations transmission from the other.
•
Each station amplitude modulates a different carrier frequency and produces a 10 KHz signal.
•
With FDM each narrow band channel in converted to a different location in the total frequency spectrum.
•
The channels are staked on top of one another in frequency domain.
•
In the figure, a simple FDM system where four 5 KHz channel are frequency division multiplexed into a single 20 KHz channel (Combined).
•
Channel 1 amplitude modulates a 100 KHz carrier in a Balanced Modulator. The o/p of Balanced modulator is 10 KHz DSBSC. This is passed through BPF where it is converted into a SSB signal for this it will be
•
100 KHz - 105 KHz. IIIrly for Channel 2 → 105 KHz to 110 KHz
•
,,
,,
→ 110 KHz to 115 KHz
,,
,,
→ 115 KHz to 120 KHz
The combined o/p spectrum by combining the o/p and from the four BFP is shown which has a total Bw of 20 KHz. With each channel Bw = 5 KHz.
47
The operation can be explained with the above block diagram •
The input message signals from various sources are passé through the LPF to obtain a band limited signal .
•
Then the output of this is Amplitude modulated using different carrier frequencies in the modulator.
•
The output of the modulator is a DSBSC signal. This is passed through a BPF to obtain a SSB signal.
•
The output of n different BPF s are multiplexed and passed through a single channel.
•
At the receiver The original signal is recovered by using the same carriers used in the transmitter
Application of FDM 1. Commercial FM 2. Television broad casting. 3. High-volume telephone. 4. Data communication System. 5. Cable Television.
48
TUTORIAL Problem : 1. An Audio frequency signal 10 sin 2π x 500 t is used to amplitude modulate a carrier signal of 50 sin 2π x 105t calculate. 1. Modulation index 2. Sideband frequencies 3. Amplitude of each sideband frequencies 4. Bandwidth required 5. Total Power delivered to the load of 600 Ω Soln : 1. 0.2 or 20% 2. wm = 2π x 500
→
fm = 500 Hz
wc = 2π x 105
→
f c = 105 Hz or 100 KHz.
fusb
= fc + fm
= 100 KHz + 500 Hz
= 100.5 KHz
flsb
= fc - fm
= 100 KHz - 500 Hz
= 99.5 KHz
1. Amplitude of side bands = mEc / 2 = 0.2 x 50 / 2 = 5 V 2. Bandwidth of AM = 2 fm = 2 x 500 Hz = 1 KHz. 3. Total Power delivered to the load Ptotal
Ec2 m2 502 = ––––– 1 + ––– = –––––––– 2R 2 2 x 600
Ptotal
= 2.125 watts
(0.2)2 1 + –––––– 2
Problem : 2. A 400w carrier is modulated to a depth of 80% calculate the total power in the modulated wave. Soln : Pc = 400w PTotal
m = 0.8
= Pc ( 1 + m2 /2)
= 400 ( 1 + (0.8)2 / 2 )
49
= 528 W. Problem : 3. A broad cast transmitter radiates 20 Kw when the modulation % is 75 calculate carrier power & Power of each side bard. Soln : PTotal = 20,000W Ptotal
&
m = 0.75
m2 = Pc 1+ ––– 2
(0.75)2 20,000 = Pc 1+ ––––––– 2 Pc
=
Ptotal
m2 = Pc 1+ ––– 2
PSB
= Pcm2 / 4
PSB
15.6 Kw =
Pc + Pcm2 /2
= 2.2 kw.
Problem : 4. The total antenna current of an Am transmitter is 5A. If modulation index is 0.6, Calculate the carrier current in antenna. Soln : I Total
=
Ic
M2 1+ –––– => 2
5 = Ic
(0.6)2 1 + ––––– 2
Ic = 4.6A
Problem :
50
5. Calculate the total modulation index if the carrier wave is amplitude modulated by three modulating signals with modulation Indies of 0.6, 0.3 & 0.4 respectively. Soln : Mt
Mt
=
√ m12 + m22 + m32
=
√ 0.62 + 0.32 + 0.42
=
0.781
Problem : 6. An Am transmitter radiator 10 kw with the carrier and 11.8 kw when the carrier is sinusoidally modulated calculate the modulation index. If another sinewave corresponding to 30% modulation, is transmitted simultaneously determine the total radiated power. . M
M
=
PTotal 2 –––––– - 1 Pc
=
11.8 2 –––––– - 1 10
=
0.6
m1 = 0.6 the another signal modulates 30% Hence m2 = 0.3. The combined total modulation index due to two signal.
Soln : : Mt
=
√ m12 + m22
=
√ 0.62 + 0.32
P Total = Pc
mt2 1 + –––– 2
(0.67)2 = 10 1+ ––––– 2
51
Problem : 7. A complex modulating waveform consisting of a sin wave of amplitude 3V and frequency 1 KHz
plus a cosine wave of amplitude 5 V and freq 3 KHz.
Amplitude modulates a 500 KHz and 10V peak carrier voltage plot the spectrum of modulated wave and determine the average power when the modulated wave is fed into 50Ω load. em
= 3 sin (2π x 1000) t + 5 cos (2π x 3000) t
ec
=
10 sin (2π x 500 x 103) t
Soln : m1
=
3/10 = 0.3 ;
mt
=
√ m12 + m22
=
0.58
m2
= 5/10 = 0.5
Problem : 1. A telephone transmitter using Am has unmodulated carrier o/p power of 20kw and can be modulated to a max depth of 80% by a sinusoidal voltage without causing a overloading find the value to which unmodulated carrier power may be increased without resulting is overloading if the maximum permitted modulating index is restricted to 60% determine the carrier swing the maximum and minimum frequencies altained, and the modulating index of the Fm signal generated by frequency modulation at 101.6 MHz carrier with an 8 KHz sine wave causing a frequency deviation of 24 KHz. Given : fm
=
8 KHz
fc
=
101.6 MHz
∆f
=
40 KHz
Mf
=
∆f ––––– Fm
40 KHz = –––––––– = 8 KHz
5
52
For modulation index 5, n = 8 Max freq =
f c + n fm
= 101.6 MHz + 8 x 8 KHz = Min freq =
101.664 MHz fc + 8 fm
=
101.6MHz - 8 x 8 KHz
=
101.664 MHx
Min freq =
Max freq - Min freq
=
101.664 MHz - 101.536 MHz
=
128 KHz.
53
UNIT 2 ANGLE MODULATION SYSTEM TOPICS:
1. PHASE AND FREQUENCY MODULATION 2. SINGLE TONE FM 3. NARROW BAND AND WIDE BAND FM 4. TRANSMISSION BANDWIDTH 5. GENERATION AND DEMODULATION OF FM SIGNAL
54
Angle Modulation Definition We know that amplitude, frequency or phase of the carrier can be varied by the modulating signal. Amplitude is varied 'in AM. When frequency or phase of the carrier is varied by the modulating signal, then it is called angle modulation, There are two types of angle modulation. 1. Frequency Modulation : When frequency of the carrier varies as per amplitude variations of modulating signal, then it is called Frequency Modulation (FM). Amplitude of the modulated carrier remains constant. 2. Phase Modulation : When phase of the carrier varies as per amplitude variations of modulating signal, then it is called Phase Modulation (PM). Amplitude of the modulated carrier remains constant, Frequency Modulation: The frequency of the high frequency carrier signal is carried in accordance with the modulating signal. Vm
= Em sin 2πfmt
Vc
= Ec sin 2πfct
Vfm
= Ec sin (2πfct + mf sin 2πfmt)
Vfm
= Ec sin (wct + mf sin wmt)
mf
→ modulating index of fm
Relationship/Difference between FM and PM: •
The basic difference between FM and PM lies in which property of the carrier isdirectly varied by modulating signal. Note that when frequency of the carrier varies,phase of the carrier also varies and viceversa.
•
But if frequency is varied directly, thenit .is called FM.,
•
And if phase is varied. directly, then it is called PM.
55
The instantaneous phase deviation is denoted by θ (t). It is the instantaneous change in phase of the carrier with respect to reference phase. The instantaneous phase of the carrier is precise phase of the carrier at a given instant .It is mathematically expressed as, Instantaneous phase = ...................(1) Here θ(t) is the instantaneous phase deviation and ωc is the carrier frequency. Now the instantaneous frequency deviation is defined as ................(2) Definition for instantaneous frequency deviation: It is the instantaneous change in carrier frequency. It is equal to the rate at which instantaneous phase deviation takes place. Definition of instantaneous frequency:It is the frequency of the carrier t a given instant of time. It is given as
...................(3) Instantaneous phase deivation θ(t) is proportional to modulating signal voltage
-....................(4) Where K is the deviation sensitivity of phase Similarly the instatneous frequency deviation is proportional to modulating Signal voltage. ........................(5) Where k1 is the deviation sensitivity of frequency. From equation (2)
56
We have
.......................(6) Let the modulating signal be given as
Using the equation in equation (6)
.......................(7) The angle modulated wave is mathematically expressed as .......................(8) Using the value of θ(t) in the above equation from equation (7)
....(9) Similarly using the value of θ(t) from equation (5) in equation (8) we get ..........(10)
57
FM and PM waveforms:
From the above waveform we can note the following •
For FM signal maximum frequency" deviation takes Place when Modulating signal is at positive .and negative peaks.
•
For PM signal the maximum frequency. deviation takes place near zero crossings of the modulating signal.
•
Both FM anid PM waveforms are identical except the phase shift.
58
Definition of Modulation index of PM and FM The modulation index of PM is given as M=KEm For FM It is the ratio of maximum frequency deviation (δ) to the modulating frequency (fm). Maximum frequency deviation δ mf = ––––––––––––––––––––––––––––– = –––– Modulating frequency fm •
The maximum frequency deviation is the shift from centre frequency fc when the amplitude of message is maximum. ∆ f = K1 Em (Hz) K1 = Deviation sensitivity.
The Bandwidth of FM: •
By Carson’s rule the Bandwidth needed by fm is given as, Bω =
2 (δ + fmmax)
δ
Maximum frequency Deviation
→
fmmax is Maximum modulating frequency. Deviation Ratio : The modulation index corresponding to maximum modulating frequency is called deviation ratio. Deviation Ratio =
Maximum frequency deviation ––––––––––––––––––––––––––––– Maximum modulating frequency
Frequency Spectrum of angle modualted wave : •
FM and PM analysis is quite complicated. It is derived with the help of Bessel function. Efm = Ec sin (wct + mf cos wmt) Using Bessel function this can be expanded as,
59
Efm = A {Jomf sin wct + J1mf [sin (wc + wm)t – sin (wc - wm)t] + J2mf [sin (wc + 2wm)t – sin (wc - 2wm)t] + J3mf [sin (wc + 3wm)t – sin (wc - 3wm)t] + J4mf [sin (wc + 4wm)t – sin (wc - 4wm)t] …..} J0, J1, J2, J3 …. Are Bessel functions. The value of this depends on modulation index mf.
From the figure the Bandwidth of FM is given by B = fc + nfm – fc + nfm BW = 2nfm Bessel Function Table : m
J0
J1
J2
J3
J4
J5
J6
J7
J8
J9
J10
0
1
-
-
-
-
-
-
-
-
-
-
0.25
0.98
0.12
-
-
-
-
-
-
-
-
-
0.5
0.94
0.24
0.03 -
-
-
-
-
-
-
-
1
0.77
0.44
0.11 0.02
-
-
-
-
-
-
-
1.5
0.51
0.56
0.23 0.06
0.01
-
-
-
-
-
-
2
0.22
0.58
0.35 0.13
0.03
-
-
-
-
-
-
0.50
0.45 0.22
0.07
0.02
-
-
-
-
-
0.34
0.49 0.31
0.13
0.04
0.01
-
-
-
-
0.36 0.43
0.28
0.13
0.05
0.02
-
-
-
0.05 0.31
0.39
0.25
0.13
0.05
0.02
-
-
2.5 3 4 5
0.05 0.26 0.04 0.18
0.07 0.33
60
Problems: Ex.1 : Determine the Bω occupied by a sinusoidally frequency modulated carrier for which the modulation index is 2.4 & maximum frequency deviation is 15 KHz. Soln : mf
=
2.4
∆f
=
15KHz
=
∆f ––––––––– => fm (max)
mf
15 KHz –––––––––– 2.4
fm(max) =
∆f ––––– = mf
Bω
=
2 (∆f + fm(max))
=
42.5 KHz.
=
6250 Hz
Given fm modulator with following parameter, K1
=
1'5 KHz / V Fc - 500 KHz
Vm
=
2 sin 2π 2kt
Detersive modulation index, o/p spectrum, change the signal amplitude to 4 Vp repeat 1. Ex.2 : For an fm modulator with a peak frequency deviation ∆f = 10 KHz, a modulating signal frequency fm = 10 KHz, Vc = 10 V and a 500 KHz carrier is applied Determine. a. Achial minimum Bw from Bessel table. b. Bandwidth using casson’s rule c. Modulation Index. Soln : a.
∆f mf = ––––––– fm (max) 61
10 KHz mf = ––––––– = 1 10 KHz From the Bessel table, for the modulation index = 1, The n value is 3. Bw = 2 (3 x fm) = 2 ( 3 x 10 ) b.
= 60 KHz.
Bw using Carson’s rule. Bw
= 2 ( ∆f + fm ) = 2 (10 k + 10 k) = 40 KHz.
Ex.3 : In a fm system the frequency deviation is 1 KHz / V. A sinusoidal modulating voltage of amplitude 15 & frequency 3 KHz is applied find, a. The maximum frequency deviation b. Modulation index.
Soln : ∆f
=
1 KHz / V
Modulating voltage is 15V at 3 KHz. Maximum frequency deviation
m
Af = ––––– fm
m
=
=
1 KHz / V. 15V
=
15 KHz.
15 KHz = –––––––– = 5 3KHz
5
Ex.4 : Determine the peak frequency deviation ∆f and modulation index (m) for an Fm modulator with a deviation sensitivity K1 = 5 KHz / V and a modulating Signal Vm(t) = 2 cos (2π 2000t) ∆f
=
KIVm
62
=
5 KHz –––––– x 2 V = 10KHz. V
∆f 10KHz = –––– = –––––––– = 2. fm 5KHz Classification of FM: 1. Narrowband FM 2. Wide band FM Narrow band FM: When the modulation index is less than I, it is called narrowband FM. The FM Equation given by eq. 9 can also be expressed as,
…………………(1)
(2) For narrowband FM, the modulation index, m is very small therefore following approximations can be considered.
Using this in equation (2)
Expanding
This equation gives the spectrum of narrowband FM. Observe that there is carrier frequency fc, upper sideband (fc + fm) and lower sideband (fc - fm).
63
Wide band FM If the modulation index is higher than 10 it is called as wide band FM
then the above equation becomes
The above integral is known as the nth order Bessel function of the first kind. It is given as
64
Using the value of Cn in equation (2)
Using the value x(t) in equation (1)
………………(3) The Fourier transform of the above equation becomes
This equation shows that there are infinite number of components located fc±fm, fc±2fm,fc ±3fm……………. Comparison between narrowband and wideband FM Sr.no
Narrow band FM
Wide band FM
1
Modulation index is < 1
Modulation index > 10
2 3
Spectrum contains 2 sidebands and Spectrum carrier
4
BW=2fm
5
It is used for mobile communication
cont6ains
infinite
number of sidebands and carrier
It is used for broadcasting and entertainment
6
Maximum deviation =75Hz
Maximum deviation = 5 Hz
7
Range of modulating frequency
Range of modulating frequency
30Hz to 15 Kz
30Hz to 3 Kz
65
Comparison between FM and PM
1
2 3 4 5
Frequency modulation
Phase modulation
The maximum frequency deviation depends upon amplitude of modulating voltage and modulating frequency Frequency of the carrier is modulated by modulating signal Modulation index is increased as modulation frequency is reduced and vice versa Noise immunity is bette than AM and PM FM is widely used
The maximum phase deviation depends only upon the amplitude of modulating voltage Phase of the carrier is modulated by modulating signal Modulation index remains same if modulating frequency is changed Noise immunity is better than AM but worse than FM PM is used in some mobile systems.
Generation of FM waves: Fm Modulators: There are 2 types of FM modulators. 4. Direct Method 5. Indirect Method
Direct FM Modulators : In this type the frequency of the carrier is varied derectly by the modulating signal. Indirect FM Modulators: In this type FM is obtained by phase modulatiuon of the carrier.
66
Generation of Narrow band FM:
Direct FM reactance modulator.
•
It
behaves as reactance across terminal A-B.
67
•
The terminal A-B of the circuit may be connected across the tuned circuit of the oscillator to get fm o/p.
•
The varying voltage (modulating voltage) V, across the terminals A-B changes the reactance of FET.
•
This charge in reactance can be inductive or capacitive.
•
Neglecting the gate current, let the current through C & R be I1.
•
At the carrier freq. the reactance of C is much larger than V R & I1 = –––––––––– R + 1/ jwc Jwc >> R I1 = jwcV From the Circuit, Vg = I1R
= jwcrv
Id = gmvgs = gmVg Id = jwcRgm V From the circuit impedance of the FET is, V Z = –––– Id
V 1 = ––––––––––– = ––––––––––– jwCR gm V jw (gmCR)
1 = ––––––––– jw (Ceq)
•
The impedance of FET is capacitive.
•
By carrying the modulating voltage across FET, the operating paint gm can be varied and hence Ceq.
•
This change in the capacitance will change the frequency of the oscillator.
Frequency Modulation using Varactor diode. 68
•
We know that the junction capacitance of the varactor diode changes as the reverse bias across it is varied.
•
L1 & C1 forms the tank circuit of the carrier oscillator.
•
The capacitance of the varactor diode depends on the fixed bias set by R1 & R2 & AF modulating signal.
•
Either R1 or R2 is made variable.
•
The radio frequency choke [RFC] has high reactance at the carrier frequency to prevent carrier signal from getting into the modulating signal.
•
At +ve going modulating signal adds to the reverse bias applied to the varactor diode D, which decreases its capacitance & increases the carrier frequency.
•
A –ve going modulating signal subtracts from the bias, increasing the capacitance, which decreases the carrier frequency.
Direct Fm Transmitters : 69
•
Fig. shows the FM Crosby transmitter with an AFC loop. (Automatic frequency correction loop).
•
The Frequency modulator can be either a reactance modulator or voltage controlled oscillator.
•
The carrier freq is 5.1MHz. which multiplies by 18 in three steps to produce a final frequency of 91.8 MHz.
•
When the frequency modulated carrier is multiplied, its frequency & phase deviations are also multiplied.
•
The rate at which the carrier is deviated is unaffected by the multiplication process. Hence the modulation index is multiplied.
•
When an angle modulated carrier is heterodyned with another freq in a non linear mixer, the carrier can either be up converted or down converted.
AFC loop : 70
•
The purpose of the AFC loop is the achieve near crystal stability of the transmit carrier freq. without using a crystal in the carrier oscillator.
•
The cassier frequency is mixed with a local oscillator freq and then down converted in freq. & the fed to a frequency discriminator.
•
Frequency discriminator is a device whose o/p voltage is proportional to difference b/w i/p freq and its resonant freq.
•
Discriminator responds to low freq changes in the carrier center freq because of master oscillator freq drift.
•
When the discriminator responds to frequency deviation, the feedback loop would cancel the deviation and this remove the modulation.
•
The dc correction voltage is added to the modulating signal to automatically
adjust
the
master
oscillator’s
centre
frequency
to
compensate for low freq drift. Ex. : A crossby direct Fm transmitter model is used. The total freq multiplication is 20 & Transmit carrier frequency ft = 88.8 MHz. determine 1. Master oscillator center frequency. 2. Frequency deviation at the o/p of the modulator for a freq deviation of 75 KHz at the antenna. 3. Deviation ratio at the o/p of the modulator for a maximum modulating signal freq fm = 15 KHz. 4. Deviation ratio at the Antenna. Ft Fc = ––––––––– =
88.8 MHz ––––––––––– = 4.43 MHz 71
N1N2N3
20
∆ Ft Af = ––––––––– = N1N2N3
75 KHz ––––––––––– = 3750 Hz 20
∆ f max Dr = ––––––––– = f m (max)
3750 –––––––– = 15 k
0.25
Dr = 0.25 x 20 = 5. PLL Direct FM transmitter:
•
Fig shows a wide band FM transmitter.
•
The VCO o/p freq is divided by N & fed bark to the PLL phase comparator, where it is compared to a stable reference freq.
•
The phase comparator generator a correction voltage that is proportional to the difference b/w the 2 frequencies.
•
The correction voltage is added to the modulating signal & applied to the VCO i/p.
•
The correction voltage adjusts the VCO centre freq to its proper value.
•
The LPF prevents the changes in the VCO o/p frequency due to the modulating signal from being converted to a voltage & fed back to VCO.
72
•
The LPF also prevents the loop from locking onto a side frequency.
Indirect Fm transmitter
•
Here the modulating signal directly deviates the phase of the carrier, which indirectly changes the frequency.
•
The carrier source is a crystal oscillator hence stability can be achieved without a AFC.
•
A carrier is phase shifted to 900 & fed to the Balanced modulator. Where it is mixed with the i/p modulating signal.
•
The o/p of balanced modulator is DSBSC.
73
•
The o/p of Balanced modulator is combined with original carrier in the combining N/W. to produce a low index, phase modulated wavefrom.
•
Fig (b) shows phasor of original carrier, modulating signal and the resultant Vector.
•
Fig (b) shows the phasors for the side freq. components of the suppressed carrier wave. As suppressed carrier is out of phase with Vc, the upper & Lower side bands combine to produce Vm – 90o with Vc.
•
The phase modulated signal is obtained by vector addition of carrier and modulating signal.
•
Modulating signal vector adds to the carrier OA with 900 phase Shift.
•
The resultant phase modulated vector is OB with phase shift θ.
•
This works only if both have the same frequency. The means carrier & modulating signal should have same frequency. Under this condition phase modulation produces FM o/p.
FM Demodulators / Detectors FM demodulator must satisfy the following requirements •
It must convert the frequency variations into amplitude variations
•
This conversion must be linear and efficient.
•
The demodulator circuits must be insensitive to amplitude changes.
•
It should not be too critical in its adjustment and operation. Types of FM demodulator: •
Round Travis Detector or Balanced discriminator.
•
Foster – Seley Discriminator or Phase discriminator.
•
Ratio Detector.
Slope Detector / Round Travis Detector :
74
•
Consists of 2 identical circuit connected back to back.
•
FM signal is applied to the tuned LC circuit.
•
Two tuned LC circuits are connected in series.
•
The inductance of the secondary tuned LC circuit is coupled with the inductance of the primary LC circuit this forms a tuned transformer.
•
Upper tuned circuit is T1 & lower tuned circuit is T2.
•
I/P side LC is tuned to be T1 is tuned to fc + δf - max freq fm. T2 is tuned to fc - δf - max freq fm.
•
Secondary of T1 & T2 are connected to diodes D1 & D2 with RC loads.
•
The total o/p is equal to difference b/w Vo1 & Vo2.
75
•
When i/p freq is fc, both T1 & T2 produce the same voltage hence o/p = 0
•
When i/p freq is fc + δf, the upper circuit T1 produces maximum voltage since it is tuned to this freq. Hence this produces maximum votalge. V01 is high compared to V02. Vout = V01 - V02 is positive for fc + δf.
•
When i/p freq is fc - δf. T2 produces maximum signal since it is tuned to it. But T1 produces minimum voltage. Hence o/p Volt = V01 – V02 is negative. Thus we get a modulating signal.
Foster - Seeley Discriminator :
76
•
The primary voltage is coupled through C3 & RFC to the centre tap on the secondary.
•
The capacitor C3 passes all the frequencies of Fm. The voltage V1 is generated across RFC.
•
RFC offers high impedance to frequencies of Fm.
77
•
The voltage V1 thus appears across centre tap of secondary and ground also.
•
The voltage of secondary is V2 & equally divided across upper half & lower half of the secondary.
•
In the figure the voltage across diode D1 is VDI = V1 + 0.5 V2 and that across D2 is VD2 = V1 + 0.5 V2 .
•
The o/p of upper rectifier is V01 and lower rectifier is V02.
•
The net o/p V0 = V01 – V02 ≅ V0 = | VD1 | - | VD2 |
•
At carrier frequency VD1 x VD2 are equal hence the net o/p of the discriminator will be zero.
•
When the i/p frequency increases above fc the phase shift b/w V1 & V2 reduces | VD1 | > | VD2 | hence V01 = | VD1 | - | VD2 | will be +ve.
•
When the i/p frequency reduces below fc then | VD1 | > | VD2 | hence V01 = | VD1 | - | VD2 | will be –ve.
Ratio detector :
Ratio detector can be obtained by sight modifications in the foster-Seeley discriminator. Fig shows the circuit diagram of ratio detector. As shown in the diagram the diode D2 is reversed, and output is taken from different points. In the above circuit the regular conversion from frequency to phase shift and phase shift to amplitude takes place as in faster–Seeley discriminator. The polarity of voltage in the lower capacitor is reversed. Hence the voltages V01 and V02 across two capacitors add. (Note that these voltages subtract in faster-seely circuit). And 78
we know that when V01 increases, V02 decreases and vice-versa as we have seen in faster-Seeley circuit. Since, V0' is sum of V01 and V02, it remains constant. From the circuit of Fig we can write two equations for the output voltage V0 (Note that V0 is the net output voltage and taken across points A and B). The First equation will be, V0 = ½ V'0 – V02 V0 = - ½ V'0 – V01
and
adding the above two equations, 2 V0 = V01 - V02 V0
= ½ (V01 - V02)
Since V01 ≈ | VD1| and | V02 | above equation will be, V0 = ½ ( | VD1 | - | VD2 |) •
Here VD1 & VD2 are obtained as discussed earlier in foster seeley circuit.
•
From the equation we know that the output of ratio detector is half compared to that of Foster-Seeley circuit
•
As frequency increases above fc' | VD1 | > | VD2| hence o/p is +ve.
•
III
rly
as frequency decreases below fc = | VD2 | > | VD1|, hence o/p is –
ve. Advantage : •
Reduced fluctuations in the o/p voltage compared to foster seeley circuit
PLL Demodulator circuit .
79
•
Fig. shows the block diagram of PLL FM demodulator.
•
The output frequency of VCO is equal to the frequency of unmodulated carrier.
•
The phase detector generates the voltage which is proportional to difference between the FM signal and VCO output.
•
This voltage is filtered and amplified. It is the required modulating voltage.
•
Here frequency correction is not required in VCO since it is already done at transmitter
Comparison of FM and AM Amplitude Modulation 1.
Frequency Modulation
Amplitude of the carrier is varied Frequency of the carrier is according to amplitude of modulating varied according to amplitude of signal.
2.
the modulating signal.
Am has poor fidelity due to narrow bandwidth. Most of the power is in carrier hence less efficient. Noise interference is more. Adjacent Chennai Interference is present Am Broadcast operates in MF and HF range In Am only carrier and two sidebands are present The transmission equipment is simple. Transmitted power varies according to modulation index.
Since the bandwidth is large, fidelity is better. 3. All the transmitted power is useful. 4. Noise interference is minimum. 5. Adjacent Chennai Interference is avoided due to guard bands. 6. FM Broadcost operates in VHF and UHF range. 7. Infinite number of sidebands are present 8. The transmission equipment is complex. 9. Transmitted power remains constant irrespective of modulation index. 10. Depth of modulation have limitation. Depth of modulation have no It can not be increased above I. limitation. It can be increased by increasing frequency deviation.
80
TUTORIAL Problem : 1. For an Fm modulator with a peak frequency deviation ∆f = 10 KHz, a modulating signal frequency fm = 10 KHz, Vc = 10V & 500 KHz carrier, Draw the frequency spectrum. 2. For an Fm Modulator with a modulation index of m = 1 a modulating signal Vm sin 2π1000t and unmodulated carrier Vc (t) = 10 sin (2π500kt) determine a. Number of sets of significant side freq. b. Their amplitudes. c. Then draw the freq spectrum showing their relative amplitudes. Relative amp J0 (m) = 0.77
7.7
J1 (m) = 0.44
4.4
J2(m) = 0.11
1.1
J3 (m) = 0.02
0.2V
81
UNIT 3 NOISE THEORY
TOPICS 1. Review of Probability 2. Random Variables 3. Random Process 4. Gaussian Process 5. Noise introduction and types 6. Thermal noise 7. White noise 8. Narrow band noise 9. Noise temperature 10. Noise Figure
82
Review of probability: •
Probability theory is the phenomenon that is modeled explicitly or implicitly on an experiment with an out come that is subjected to chance.
•
If the out come is repeated ,the outcome can differ because of the random
•
phenomenon or chance mechanism-refered to as Random Experiment
Features of Random Experiment: •
The Experiment is repeatable under identical conditions.
•
On any trial of the experiment, the out come is unpredictable.
•
For large number of trials of the experiment, the outcomes exhibit statistical regularity.
Axioms of Probability: •
Random experiment and its possible out come are studied in terms of a Space and its points.
•
Say for Kth outcome of an experiment, the sample point is denoted as Sk.
•
The total sample points corresponding to all possible outcomes of the experiment are called the sample space. Denoted by S.
•
Event corresponds to either single sample point or set of sample points.
•
Entire sample space is called sure event.
•
Null set is the impossible event.
•
Single sample point is called elementary event.
•
Eg: consider the throwing of dice. We know that there are six possible outcomes. [1, 2, 3, 4, 5, 6]. Hence the sample space is 6.If we consider the event which is described as “an eve number” then the sample space is 3, [2, 4,6].
•
Let A denote the possible out come of the random experiment, Suppose that event A occurs Nn(A) times.Nn(A)/n is assigned to the event A. This ratio is called as the relative frequency. The relative frequency is a nonnegative number and is less than or equal to 1. 83
•
If the event A occurs in non of the trial then Nn(A)/n=0.
•
If the event A occurs in all the n trials then Nn(A)/n=1
Definition of probability: Probability of an event A is defined as
•
A sample space S of elementary events(outcomes)
•
A class of event that are the subset of S.
•
A probability measure P(.) assigned to each event A in the class, which has the following property
1. P(S)=1 2. 3. If A+B is the union of two mutually exclusive events in the class then P (A+B) =P (A) +P (B) •
Properties 1, 2 and 3 are known as the axioms of probability.
•
Axiom 1 states that the probability of a sure event is unity.
•
Axiom 2 states that the probability of an event is non negative real number that is less that or equal to unity.
•
Axiom 3 states that the probability of the union of two mutually exclusive events is the sum of probabilities of the individual events. These 3 axioms are sufficient to deal with experiments with finite sample space.
84
Basic Properties of Probability: Property 1
Where
is the complement of event A
The use of this property helps us investigate the non occurrence of an event. Proof: The sample space is expressed as the union of two mutually exclusive events A and The use of Axioms 1 and 3 yields Property 2: If mutually exclusive events A1,A2,……….Am exhaustic property A1+A2+……….Am=S then P(A1)+P(A2)+………+P(Am)=1 Property 3: When event A and B are not mutually, then the probability of the union event “A or B” equals P(A+B)=P(A)+P(B)-P(AB) Where P (AB) is the probability of the joint event “A and B” The Probability P (AB) is called a joint probability. It has the following relative frequency interpretation.
Where Nn(AB) is the number of times the events A and B occur simultaneously in the trials of the experiment. Conditional Probability: Consider an experiment involving a pair of events A and B . •
Let P(B/A) denote the probability of event B, given the event A has occurred, The probability P(B/A) is called the conditional probability of B given A.
•
The conditional probability P(B/A) is defined by the equation P (B/A) =P (AB)/P (A)-------------------------1
85
Where P (AB) is the joint probability. Rewriting the equation (1) We have P (AB) =P (B/A) P (A) --------------2 It is apparent that we may also write P (AB) =P (A/B) P (B) -------3 Thus we may state that the joint probability of two events may be expressed as the product of the conditional probability of one event given the other, and the elementary probability of the other. •
Using eqn 3 in eqn 1 we have P(B/A)=P(A/B) P(B)/P(A)- this relation is a special form of Baye’s rule
•
If the conditional probability P(B/A) is simply equal to the elementary probability of occurrence of event B that is, P(B/A) = P (B) Then P(AB) = P (A) P (B)
Random variables: •
This term is used to signify a rule by which a real number is assigned to each possible out come of an experiment.
•
Random variables are of two types . Discrete Random variables and continuous random variables.
Discrete random variables: •
We know that a random variable associated with an experiment is a rule or relationsh ip X(s) that assigns a real number X to every sample point “s”.
•
If the sample space contains a countable number of sample points the X(s) will be discrete random variable. A discrete random variable will thus have a countable numb er of distinct values.
86
Continuous random variable: •
A continuous random variable is not restricted to finite numberof distinct values
•
A continuous random variable has an “uncountable” number of possible values.
•
Such a random variable is defined for systems which generate an infinite number of outputs within a finite period of time.
NOISE: Definition The noise is defined as the unwanted signal that tends to interfere with the required signal. This gives rise to audible or visual disturbances in the communication systems and errors in the digital communication. Sources of Noise: The noise arises from different types of sources. They are 1. Natural noise sources 2. Man made noise sources 3. Internal noise sources Types of noise
87
External Noise: This noise is caused by the sources external to receiving system. They are mainly categorized into natural and man made noise. The natural noises are Atmospheric and Extraterrestrial noises. Atmospheric noise: This is also known as static noise. This noise is mainly due to lightening. This source causes electrical disturbance and results in a transient electrical signal that generates harmonic energy that can travel extremely long distance. Atmospheric noise generally affects the reception at frequencies less than 30MHz. Extraterrestrial noise: This noise is mainly due to electronic storms, solar flares and radiation in space. This is again divided into two types 1. Solar noise 2. Cosmic noise Solar Noise: Primary source of solar noise is sun. Sun radiates in wide range of spectrum, these noise radiation vary with time. The noise radiation from sun varies with temperature changes on its surface. The temperature changes follow a cycle of 11 years, hence the periods of great electrical disturbances also repeats after every 11 years. Cosmic noise: It comes from stars. This is identical to the noise radiated by the sun because stars are also large hot bodies. This noise is also called as Black body noise. It is distributed uniformly over the entire sky.
88
Man made noise: It’s produced by the make and break process in a current carrying circuit. Such as automotive ignition systems, electric motors and generator brush contact. Electrical equipment when abruptly switched ON or OFF produces transients that create noise. Fluorescent light also produce noise. Internal Noise: Some noises are generated internally in the circuit... Electronic components such as resistors, diodes and transistors produce noise as their characteristics vary with temperature. Though this is a low level noise, it can interfere with weak signal. As this noise obeys certain laws, it is possible to design the equipment in which the effects are minimized. The internal noise is classified as, 1. Thermal noise 2. Shot noise 3. Partition noise 4. Low frequency or flicker noise 5. High frequency or transit time noise. Thermal Noise or Johnson noise •
The free electrons within a conductor are always in random motion. This random motion is due to the thermal energy received by them.
•
The distribution of these free electrons within a conductor at a given instant of time is not uniform.
•
It is possible that an excess number of electrons may appear at one end or the other of the conductor.
•
The average voltage resulting from this non-uniform distribution is zero but the average power is not zero. As this power results from the thermal energy, it is called as the “Thermal noise Power”.
•
The average thermal noise power is given by, Pn = k TB watts
89
Where,
k
=
Boltzmann’s constant = 1.38 x 10-23
Joules/Kelvin. B = Banwidth of the noise spectrum (Hz). T = Temperature of the conductor, oKelvin. The Equation indicates that a conductor operated at a finite temperature
•
can work as a generator of electrical energy. The thermal noise power Pn is proportional to the noise bandwidth and
•
conductor temperature. Problem: Ex.1 : A receiver has a noise power bandwidth of 12 kHz. A resistor which matches with the receiver input impedance is connected across the antenna terminals. What is the noise power contributed by this resistor in the receiver bandwidth? Assume Temperature to be 30o C. Soln : Given :
B = 12 kHz, T = 30oC = 30 + 273 = 303o K
The noise power contributed by the resistance is given by, Pn
=
k TB
=
1.38 x 10-23 x 303 x 12 x 103
=
5.01768 x 10-17 W
Equivalent Circuit: •
As the Conductor is generating electrical noise energy, it must have an equivalent voltage or equivalent current generator circuit which will represent the noise source.
90
(a)
(b) Fig 3.1 (a) and (b)
•
The voltage generator and the current generator equivalent circuits are as shown in Fig.3.1 (a) and (b) respectively. Consider the circuit shown in Fig.3.1 (a)
•
Assume that the rms source voltage is Es. Then the maximum power is delivered to the load resistance RL when RL = Rs. Es 2 PL max. =
–––4Rs
•
Applying the same logic to a conductor which is generating noise. We can write the expression for the noise power Pn as, En2 Pn=
––– 4R
Where, R En •
= Resistance of the conductor and = RMS noise voltage.
But referring to Equation Es 2 Pn = kTB
∴ PL max. =
––– 4Rs
∴
En
= √ 4k TBR
This is the expression for the rms value of the thermal noise voltage.
91
•
The current equivalent circuit for the thermal noise is as shown in Fig.3.1 (b). It can be obtained by using the same logic as that for the voltage equivalent circuit. The rms noise current is given by. In
= √ 4 G k TB 1
Where,
G
= Conductance = ----R
Shot Noise: •
The shot noise is produced due to shot effect. Due to the shot effect, shot noise is produced in all the amplifying divices rather in all the active devices.
•
The shot noise is produced due to the random variations in the arrival of electrons (or holes) at the output electrode of an amplifying device.
•
Therefore it appears as a randomly varying noise current superimposed on the output. The shot noise “sounds” like a shower of lead shots falling on a metal sheet.
•
The shot noise has a uniform spectral density like thermal noise. The exact formula for the shot noise can be obtained only for diodes. For all other devices and approximate equation is stated. The mean square shot noise current for a diode is given as, I2
=
2(I + 2 Io )q B amperes2
I
=
Direct current across the junction (in amp.)
I0
=
Reverse saturation current (in amp.)
q
=
Electron charges = 1.6 x 10-19 C.
B
=
Effective noise bandwidth in Hz.
n
Where,
For the amplifying devices the shot noise is : (i)
Inversely proportional to the transconductance of the device.
(ii)
Directly proportional to the output current.
92
•
Shot noise is present in the devices such as diodes, transistors, photo detectors etc. because of the discrete nature of current flow in these devices.
•
Consider a photo detector circuit. Every time when the light is incident on it, the cathode emits an electron. This generates a current pulse.
•
These electrons are being randomly emitted at time instants donoted by τk where
•
-∞ < k < ∞.
If we assume that such a random emission is going on for a long time, then the photo current flowing through the photodetector can be imagined to be an infinite sum of current pulses.
•
This can be expressed mathematically as follows : ∞ X(t)
=
Σ
h ( t - τk )
k = -∞ •
In Equation above, h ( t - τk ) . . . h(t - τ 1) + h(t - τ0) + h(t - τ 1) + ... Represents the current pulse generated at instance τk, and the process X(t) is called as shot noise.
•
The number of electrons N (t), which are emitted in the time interval (0,t) constitute a discrete random process.
•
The value of this process increases by 1 every time electron is emitted. A sample function of such a random process is shown in Fig.3.2
93
Fig 3.2 •
Let the number of electrons emitted between t and t+t0 be ν. Then the mean value of number of number of electrons is given by Mean E [ν] = λ t0
•
λ is a constant and called as the “rate” of the process.
•
So the total number of electrons emitted in the interval t to t + t0 i.e. ν is given by, ν = N ( t + t0 ) - N (t)
•
ν in the above expression will follow a Poisson distribution with a mean value of λ t0.
•
The probability of emitting k number of electrons in the interval (t, t + t0) is given by the poisons distribution as (λt0)k P( ν = k ) =
––––– e -λt0
k = 0, 1 . . . .
.. 1
k! •
After the detailed statistical analysis, the first two moments of the process X(t) are given by ∞ 94
•
The mean of X(t)
= mX = λ ∫ h (t) dt
.. 2
-∞ where λ = rate and h (t) = waveform of the current pulse. •
The auto covariance function of X (t) is given by ∞ CX (τ) = λ
∫ h (t) h (t + τ ) dt
.. 3
-∞ This expression is also called as Cambell’s theorem. Partition Noise •
Partition noise is generated when the current gets divided between two or more paths.
•
It is generated due to the random fluctuations in the division. Therefore the partition noise in a transistor will be higher than that in a diode.
•
The devices like gallium arsenide FET draw almost zero gate bias current, hence keeping the partition noise to its minimum value.
Low Frequency or Flicker Noise •
The flicker noise will appear at frequencies below a few kilohertz. It is sometimes called as “1/f” noise.
•
In the semiconductor devices, the flicker noise is generated dud to the fluctuations in the carrier desity.
•
These fluctuations in the carrier density will cause the fluctuations in the conductivity of the material. This will produce a fluctuating voltage drop when a direct current flows through a device. This fluctuating voltage is called as flicker noise voltage.
•
The mean square value of flicker noise voltage is proportional to the square of direct current flowing through the device.
95
Problem: Ex.2 : A 600Ω resistor is connected across the 600Ω antenna input of a radio receiver. The bandwidth of the radio receiver is 20 kHz and the resistor is at room.Temperature of 27oC. Calculate the noise power and the noise voltage applied at the input of the receiver. Soln : Given : R1 = 600Ω, R2 = 600Ω ∴
Reff = 600 || 600 = 300Ω B = 20 kHz, T = 27oC or 300oK
Noise Power : Pn = KTB = 1.38 x 10-23 x 300 x 20 x 103 ∴
Pn = 8.28 x 10-17 W
Noise Voltage at the receiver input : To calculate the noise voltage at the receiver input, we have to consider the effective resistors are in parallel, Reff = 600 || 600 = 300Ω ∴ Noise voltage En ∴
En
=
√ 4 k TBR
=
√ 4 x 1.38 x 10-23 x 300 x 20 x 103 x 300
=
3.15 x 10-7 volts or 0.3152 µ V
High Frequency or Transit Time Noise: •
If the time taken by an electron to travel from the emitter to the collector of a transistor becomes comparable to the period of the signal which is being amplified then the transit time effect takes place.
•
This effect is observed at very high frequencies. Due to the transit time effect some of the carriers may diffuse back to the emitter.
•
This gives rise to an input admittance, the conductance component of which increases with frequency.
96
The minute currents induced in the input of the device by the random
•
fluctuations in the output current, will create random noise at high frequencies. Once this noise appears, it goes on increasing with frequency at a rate of
•
6 dB per octave. Ex.3 : Calculate the noise voltage at the input of a receiver’s RF amplifier, using a device that has a 100Ω equivalent noise resistance and a 200 input resistor. The B.W. of the amplifier is 1M the temperature is 20oC. Boltzmann’s constant = 1.38 x 10-23 J/K. Soln : Given : Ri = 200Ω, Rn = 100Ω, B = 1 MHz, T= 25oC = 298oK Noise voltage at the input of the RF amplifier is,
∴
Vn
=
√ 4 kTBReq
...(1) ...(2)
Here,
Req
=
Ri + Rn = 100 + 200 = 300Ω
∴
Vn
=
√ 4 x 1.38 x 10-23 x 298 x 1 x 106 x 300
∴
Vn
=
2.22 µ V
...(3)
Ex.4 : An amplifier has a band width of 4 MHz with 10 k Ω as the input resistor. Calculate the rms noise voltage at the input to this amplifier if the room temperature is 25oC . Soln : Given : B = 4 MHz, Ri = 10 k Ω, T = 25oC = 298oK RMS noise voltage Vn
= √ 4 kTBR = √4 x 1.38 x 10-23 x 298 x 4x 106 x 10 x 103
Vn
=
25.65 µ V
97
Ex.5: Calculate the thermal noise power available from any resistor at room temperature (290o K) for a B.W. of 2 MHz. Also calculate the corresponding noise voltage given that R = 100Ω. Soln : (i)
Thermal noise power Pn
= kTB = 1.38 x 10-23 x 290 x 2 x 106 = 8 x 10-15 watt
(ii)
Noise voltage Vn
= √ Pn R
=
√8 x 10-15 x 100 = 0.894 µ V
Ex.6 : Calculate the thermal noise power available from any resistor at room temperature (290o K) for a B.W. of 2 MHz. Also calculate the corresponding noise voltage given that R = 100 Ω.Given A=60 and B=20 KHz Soln : (i)
Thermal noise power Pn
= kTB = 1.38 x 10-23 x 290 x 2 x 106 = 8 x 10-15 watt
(ii)
Noise voltage Vn
= √ Pn R
=
√8 x 10-15 x 100 = 0.894 µ V
Ex.7 : The noise output of a resistor is amplifies by a noiseless amplifier having a gain of 60 and a bandwidth of 20 kHz. A meter connected to output of the amplifier reads one millivolt (RMS). Refer the given figure i)
If the bandwidth of the amplifier is now reduced to 5 kHz, its gain remaining constant, what does the meter read?
98
ii)
If
the
resistor
is
operated
at
80oC,
what
is
its
resistance?
Soln : (i)
From the Fig. we can write, 1mV En
=
––––– 60
=
1.66 x 10-5 volts
But,
En
=
√ 4 k TBR
∴
En2
=
4 k TBR
….(1) ... (2)
The bandwidth B = 20 kHz (1.66 x 10-5)2 ∴
kTR
=
––––––––––– =
3.47 x 10-15
... (3)
4 x 20 x 103 (ii)
Now the gain of the amplifier is constant i.e. A = 60 and the bandwidth is changed. ∴
B
=
5 kHz.
En
=
√ 4 k TBR x B
En2
=
8.33 x 10-6 volt
∴ The meter reading will be
= √4x3.47x10-15x5x103 ... (4)
=
A x En = 60 x 8.33 x 10-6
=
0.5 mV
… (5)
99
The operating temperature is T = 80oC = 353oK
(iii)
we know that kTR
=
3.47 x 10-15 3.47 x 10-15
∴
R
=
–––––––––––––– 1.38 x 10-23 x 353
∴
R
=
712.77 k Ω
… (6)
Ex. 8: An amplifier operating over the frequency range from 3 to 10 MHz has a 20 k Ω input resistance. What is the r.m.s. noise voltage at the input to this amplifier at room temperature? Soln : Given : RMS noise voltage Vn Vn
= √ 4 kTBR =
√4x1.38x 10-23x(27+273)x(10-3)x106x20x103
=
48.14 µ V
. . . Ans
NOISE CALCULATIONS (THERMAL NOISE) Resistors in Series •
The resistors act as the sources of thermal noise. So let us see the effect of connecting tow noise sources i.e. two resistors in series with each other. The situation is as shown in Figs.3.3. (a) and (b). The two resistances R1 and R2 are replaced by their voltage source equivalent circuits.
•
Since the two resistors are in series they can be replaced by a resistance R = R1 + R2. The noise voltage generated by the resistor R is given
Fig 3.3 (a) and (b)
100
by, En2
∴ •
En
=
4 k TBR
=
4 k TB (R1+R2) 2
=
4 k TB R1 + 4 k TB R2
2
=
En1 + En2
=
√ En12 + En22
Since the two resistors are in series they can be replaced by a resistance R = R1 + R2. The noise voltage generated by the resistor R is given by,
•
And the effective
En = √ En12 + En22+ En32 + . . .
resistance
R is given as, R = R1 + R2 + R3 + . . . Thus the equivalent noise voltage of the series combination of resistors is equal to the vector sum of the noise voltages produced by the individual resistors. Resistors in Parallel: •
For the parallel connection of the resistors R1 and R2 the current equivalent circuit’s equivalent circuits are to be used. The equivalent circuits are as shown in the Fig.3.4(a) and Fig.3.4(b).
•
As the two conductances G1 and G2 are in parallel, the effective value of conductance is given by, GP = G1 + G2 . . . . . 101
The noise current generated by the conductance GP is given as,
•
In2 = 4 GP k TB = 4(G1 + G2) k TB = 4 G1 k TB + 4 G2 k TB ∴ In 2 = I n1 2 + I n2 2 . . . The same logic can be extended for “n” number of resistors connected in
•
parallel, to give In2 = 4 (G1 + G2 + G3 + . . . + Gn) k TB If it is required to obtain the voltage generator equivalent circuit for parallel
•
connection of the resistors then the equivalent parallel resistance RP is given as,
and
1 RP = –– + R1 2 En = 4 RP
1 1 1 –– + –– + ….. + –– …. R2 R3 Rn k TB . . .
Where RP is the parallel combination of resistors. Problem: Ex.9 : Two resistors 20 kΩ and 50 kΩ are at room temperature (290oK). Calculate for the bandwidth of 100 kHz, and thermal noise for the following conditions : (i)
For each resistor
(ii)
(iii)
For two resistors in parallel
For to resistors in series
Soln : Given : k = T= ∴ kT= (a)
1.38 x 10-23 Joules / Kelvin. 290oK, B = 100 kHz. 4 x 10-21 W/Hz at 290oK
For the 20 k Ω resistor: En12 = 4 k TBR = 4 x 4 x 10-21 x 100 x 103 x 20 x 103 x 290 = ∴
En22 = 5.66 µ V
32 x 10-12 V2, . . . (1)
102
(b)
For the 50 k Ω resistor : En22 = 4 x 4 x 10-21 x 100 x 103 x 50 x 103 x 290 ∴
(c)
En22 = 8.95 µ V
. . . (2)
For the two resistor in series: Effective resistance R = R1 + R2 = 70 k Ω Hence,
(d)
En = √ (5.66)2 + (8.95)2 = 10.59 µ V
. . . Ans.
For the two resistor in Parallel: Effective resistance RP = 20 x 50/20+50=100/7 k Ω = 14.28 k Ω Hence,
En = √ 4 k TBRP
=
√ 4x4x10-
21
x290x100x103x14.28x103 ∴
Ex.10 :
En = 4.78µ V
An amplifier operating over a frequency range from 17 to 19MHz
has a input resistance of 5 kΩ. What is the rms thermal noise voltage at the input of this amplifier? Assume the operating temperature to be 27oC . Soln : Given : k =
1.38 x 10-23 Joules / Kelvin.
T =
27oC, = 27 + 273 = 300oK
B =
19 - 17 = 2 MHz
R =
5 kΩ
∴ RMS thermal noise at the input of the amplifier is given by, En = √4 x 1.38 x 10-23 x 300 x 2 x 106 x 5 x 103 = 12.86 µ V Practically, the signals applied at the input of this amplifier for amplification must be much larger than the rms noise voltage. Otherwise the signal will be masked by the noise.
103
Ex.11 :Two resistors 10 kΩ and 25 kΩ are at room temperature (290oK). for the bandwidth of 150 kHz, calculate thermal noise for each resistor, if two resistors are in series and if two resistors are in Parallel. Soln : Given :
R1 = 10 kΩ, R2 = 25 kΩ, k = 1.38 x 10-12 J/K kJ = 4 x 10-21 W/Hz at 290 K,
To find :
B = 150 kHz.
(i)
thermal noise for 10 kΩ resistor.
(ii)
thermal noise for 25 kΩ resistor.
(iii)
if two resistances are in series.
(iv)
thermal noise when the two resistances are in parallel.
Step I : For 10 kΩ Ω resistor : En1
= √4 kTBR
En1
= √ 4 x 4 x 10-21 x 290 x 150 x 103 x 10 x 103
= 83.42 µ V
Step II : For 25 kΩ Ω resistor : En1
= √4 kTBR
En1
= √ 4 x 4 x 10-21 x 290 x 150 x 103 x 25 x 103 = 131.9 µ V
Step III : For Rs = (10+25) = 35 kΩ Ω resistor : En
= √4 kTBRs
En
= √ 4 x 4 x 10-21 x 150 x 103 x 290 x 35 x 103 = 156.07 µ V
Ω : Step IV: For RP = 10k || 25k = 7.14 kΩ En
= √4 kTBR
En
= √ 4 x 4 x 10-21 x 290 x 150 x 103 x 7.14 x 103 = 70.49 µ V
104
White Gaussian Noise : •
“White” noise is the noise whose power spectral density is uniform over the entire frequency range of interest, as shown in Fig.3.5
Fig 3.5
•
The white noise contains all the frequency components in equal proportion. This is analogous with white light which is a superposition of all visible spectral components.
•
The white noise has a Gaussian distribution. That means the PDF of white noise has the shape of gaussian PDF. Therefore it is called as gaussian noise.
Power spectral density of white noise : As shown in Fig 3.5 the power spectral density (psd) of a white noise is given by,
No Sn ( f ) = ––– 2
This equation shows that the power spectral density of white noise is independent of frequency. As N0 is constant, the psd is uniform over the entire 105
frequency range including the positive as well as the negative frequencies. N0 in above Equation is defined as : No = K Te Where,
K = Boltzmann’s constant and Te = Equivalent noise temperature of the system.
Example of white noise: The best example of white noise is the thermal or Johnson noise. Problem Ex.12 :
Derive and plot the auto correlation function of a white gaussian
noise which has a power spectral density of N0 / 2 . Soln : We know that the power spectral density and auto correlation function form a Fourier transform pair. That means. F R(τ) ↔ S(f) i.e.
Ft {R (τ )} = S ( f )
OR
R (τ ) = IFT { S ( f ) }
Therefore for the white noise the autocorrelation function R (τ) is obtained as follows : R (τ ) = IFT { Sn ( f ) } = IFT { No /2} N0 ∴
R (τ ) = ––– δ (τ ) 2
This is the expression for the autocorrelation function of white noise. The autocorrelation function can be plotted as shown in Fig.
106
Note : As shown in Fig. the autocorrelation function of white noise is a delta function weighted by a factor (N0 / 2) at τ = 0. The meaning of this is that any two samples of white noise are totally uncorrelated. Equivalent Noise Resistance (Rn) : •
The noise generated by an active or passive device is represented by means of a “fictitious” resistance Rn.
•
The resistance Rn is assumed to generate the noise at room temperature, which is equal to the noise generated by the device.
•
Rn is called as the equivalent noise resistance and the device is then assumed to be noiseless.
•
The equivalent noise resistance is a part of the manufacturers data. The concept of equivalent noise resistance will be clear by looking at Fig. 3.6
107
Fig 3.6 i)
The amplifier shown in the Fig.3.6 has a specified noise resistance Ri.
ii)
The equivalent mean square thermal noise voltage, which appears at the input of the amplifier is, Vn2 = 4 (Rn + Ri) k To B
-----1
Where, T0 = room temperature iii)
From equation 1 it is clear that Ri and Rn are in series with each other. As shown in Fig 3.6, the amplifier is now assumed to be noiseless.
108
iv) It is important to note that Rn is not a real resistance. Hence the actual input resistance of the amplifier is not changed. It remains the same. Problem: Ex.13: Calculate the noise voltage at the input of an RF amplifier which has an equivalent noise resistance of 100Ω and input resistance of 300Ω. The noise equivalent bandwidth is 5 MHz and the temperature is 17oC. Soln : Noise voltage at the input Vn2 = 4 k T0 B (Ri + Rn) ∴
Vn2 = 4 x 1.38 x 10-23 x 290 x 5 x 106 x 400
∴
Vn2 = 3.2 x 10-11
∴
Vn = 5.62 x 10-6 volt or 5.65 µV
Noise due to Several Amplifiers in Cascade: •
In the receivers a number of multiplying stages are connected in cascade. e.g. the RF amplifier and mixer stages are connected in cascade.
•
To find the combined effect of such a connection on the receiver noise.
Consider the two stage amplifier shown in Fig.3.7. Let the gains of the
two
stages be A1 and A2 respectively. The first stage has a total input noise resistance R1, the second stage has total input noise resistance R2 and the output resistance is R3. The rms noise voltage at the output due to R3 is given by,
109
Fig 3.7 Noise of several amplifiers in cascade Vn3 •
=
√ 4 k TB R3
. . ..1
The rms noise voltage Vn3 is produced due to the rms noise voltage Vn3 at the input of the second stage such that, Vn3
=
A2 x V′n3
. . .2
OR
•
Vn3 V′n3 = ––– . . ..3 A2 We cab assume that the rms noise voltage V′n3 is generated by another resistor R′3 at the input of second stage, with R3 absent on the output side.
•
∴
V′n3
=
But
V′n3
=
√ 4 k TB R3 √ 4 k TB R3 Vn3 ––– = –––––––––– A2 A2
Comparing the two equations we get, R3 R′3 = ––– A2
...4
. . ..5
. . . .5
110
•
This equation shows that when the noise resistance is to be transferred from the output of a stage to its input, it must be divided by square of the voltage gain of the stage.
Therefore the equivalent noise resistance at the input of the second stage is given by. R3 R′eq = R2 + R3 = R2 + ––– . . . ..7 A22 • Similarly we can transfer the resistance R′eq from the output to the input of the first stage. The value of R′eq when transferred to the input of the first stage is given by, R′eq R2 R3 R′2 = ––––– = –––– + ––––– . . . .8 2 2 2 2 A1 A1 A2 A1 • Hence the equivalent noise resistance at the input of the first stage is given as, = R1 + R2 Req R2 R3 = R1 + ––––– + –––––––– ...9 2 2 2 A1 A1 A2 The same concept is applicable to n number of stages connected in cascade. Problems: •
Ex.14 :
Two amplifiers are connected in cascade. The first amplifier has a
voltage gain of 10, input resistance of 600Ω, equivalent noise resistance of 1500Ω and output resistance of 25kΩ. These values for the second amplifier are 20, 80kΩ, 10kΩ and 2MΩ respectively. Calculate the equivalent input noise resistance of the two stage amplifier. Soln : From the data,
and Therefore,
R1 = 600 + 1500 = 2100Ω
. . . (1)
25 x 80 R2 = ––––––– + 10 = 29 kΩ 25 + 80
. . . (2)
R3 = 2 MΩ Req = R1 +
R3 R2 ––––– + –––––––– A12 A12 A22
. . . (3)
111
Therefore,
Req
29 x 103 2 x 106 = 2100 + ––––––– + –––––––– 102 x 202 102
Req = 2440 Ω
. . . (4)
This is the equivalent resistance at the input of the cascade configuration. Ex.15 :The first stage of the two-stage amplifier has a voltage gain of 5, a 300Ω input resistor, a 800Ω equivalent noise resistance and a 20 k Ω output resistor. For the second stage these values are 10, 40kΩ, 5kΩ and 500kΩ respectively. Calculate the equivalent input noise resistor of this two-stage amplifier. Soln : From the data,
R1 = 300 + 800 = 1100Ω R2
and ∴
Therefore,
20 x 40 = ––––––– + 5 = 18.33 kΩ 20 + 40
. . . (1) . . . (2)
R3 = 500 MΩ (given) Req =
Req
R1 +
R2 R3 ––––– + –––––––– A12 A22 A12
. . . (3)
18.33 x 103 500 x 103 = 1100 + ––––––––– + ––––––––– (5)2 (5)2 x (10)2
Req = 2033.3 Ω
. . . (4)
112
Signal to Noise Ratio : •
In the communication systems the comparison of signal power with the noise power at the same point is important, to ensure that the noise at that point is not excessively large.
•
It is defined as the ratio of signal power to the same point. ∴
S Ps ––– = ––– N Pn Ps = Signal Power
Where
Pn = •
. . . .1
Noise Power at the same point.
The signal to noise ratio is normally expressed in dB and the typical values of S/N ratio range from about 10 dB to 90 dB. Higher the value of S/N ratio betters the system performance in presence of noise. S/N (dB) = 10 log10 (Ps/Pn)
•
. . . .2
The powers can be expressed in terms of signal and noise voltages as follows : Ps
•
•
Vs2 = –––– R
Pn
Where Vs = Signal Voltage and Vn = Noise voltage. Vs2 Vs2/R Ps = –––––– = –––– 2 Vn /R Vn2 . . . .3 The signal to noise ratio in dB is given by, (S/N)dB = =
•
and
Vn2 = –––– R
10 log10 [ Vs / Vn ]2 20 log10 [ Vs / Vn ]
. . . .4
All the possible efforts are made to keep the signal to noise ratio as high as possible, under all the operating conditions.
•
Although the signal to noise ratio is a fundamental characteristics of a communication system it is often different to measure. In practice instead of measuring S/N, another ratio called (S + N) / N is measured.
113
SINAD : This is another variation of signal to noise ratio. SINAD stands for signal noise and distortion and it is defined as, SINAD
S+N+D = ––––––––– N+D
Where S = Signal, N = Noise and D = Distortion. SINAD is generally used in the specifications of FM receiver. Problem: Ex.16 :A receiver produces a noise power of 100 mW when no signal is present at its input. The receiver output power is equal to 3W with the signal applied at its input. Calculate the (S + N) / N ratio as a power ratio and in dB. Soln : (S + N) 3 + 0.1 The power ratio is –––––– = ––––––– = 31 N 0.1 In dB it is ( S + N ) / N (dB) = 10 log10 31 = 14.91 dB
1. 2.
Noise Factor : •
The noise factor (F) of an amplifier or any network is defined in terms of the signal to noise ratio at the input and the output of the system. It is defined as, S/ N ratio at the input F = ––––––––––––––––––– S/ N ratio at the output Pno Psi = –––– x –––– Pni Pso
. . ..1
. . ..2
Where Psi and Pni = Signal and noise power at the input and Pso and Pno = Signal and noise power at the output
114
•
The temperature to calculate the noise power is assumed to be the room temperature.
•
The S/N at the input will always be greater than that at the output. This is due to the noise added by the amplifier. Therefore the noise factor is the means to measure the amount of noise added and it will always be greater than one. The ideal value of the noise factor is unity.
•
The noise factor F is sometimes frequency dependent. Then its value determined at one frequency is known as the spot noise factor and the frequency must be stated along with spot noise factor.
Noise output Power in Terms of F : •
The available power gain is defined as, Pso G = ––– Psi
•
•
Substitute this value in Equation we get, Pno F = –––––– G Pni
. . . ..5
As we already know, Pni = k TB
•
. . . .4
Therefore the noise power at the amplifier output is given by, Pno = F G P ni
•
. . ..3
. . . .6
But the noise factor is determined at room temperature. Therefore substitute To in the equation for Pni.
•
∴
Pni = k T0 B
and
Pno = F G k T0 B
. . . .7
Thus with increase in the noise factor F, the noise power at the output will increase. Higher the noise factor value is more will be noise contributed by the amplifier.
Noise factor in Terms of Rn : •
The S/N ratio at the output of an amplifier is given as 115
(S/N) out =
Vs 2 –––––––––––––––– 4 k To B ( Rp + Rn )
...1
Where, RP = parallel combination of amplifier input resistance Ri and the source resistance Rs. Rs Ri i.e. Rp = ––––––––– . . .2 ( Rs + Ri ) •
The S/N ratio at the input of the amplifier is given by, Vs2 (S/N) in = ––––––––– 4 k To Rp
. . .3
Note here that the effective noise resistance Rn is not included. •
Therefore the noise factor is given as, (S/N) in F = –––––––– (S/N) out
•
Substituting the values, ∴ F =
Rp + Rn ––––––– Rp
...4
If the amplifier does not produce any noise i.e. it is an ideal amplifier then Rn = 0 and the noise factor from Equation 4 will be unity Noise Figure : •
Noise factor is expressed in decibels it is called noise figure. Noise Figure = FdB = 10 log10 F
•
Substituting the expression for the noise factor we get, Noise figure = 10 log10
S/N at the input –––––––––––––––– S/N at the output
= 10 log10 (S/N)I – 10 log10 (S/N)o ∴ Noise figure FdB = (S/N)I dB – (S/N)o dB •
The ideal value of noise figure is 0 dB. 116
Measure to improve the noise figure: To improve the noise figure of a receiver, the devices used for the
•
amplifiers and mixer stages must produce low noise. The diodes and FETs are therefore preferred. The receiver can be
•
operated at low temperatures. This is done in satellite low noise receivers. Use high gain amplifiers. All this will collectively improve the noise figure of the receiver. Problems: Ex.17 :The signal power and noise power measured at the input of an amplifier are 150 µ W and 1.5 µ W respectively. If the signal power at the output 1.5 W and noise power is 40 mW, calculate the amplifier noise factor and noise figure. Soln : Psi = 150 µ W, Pni = 1.5µ W, Pso = 1.5 W, Pno = 40 mW. 1.
Noise factor
Psi Pno = –––– x –––– Pso Pni 150 x 10-6 40 x 10-3 = –––––––––– x –––––––– = 2.666 1.5 x 10-6 1.5
2.
Noise figure
.
= 10log10 (Noise Factor) =
10log10 (2.667) = 4.26 dB
Ex.18 :The signal to noise ratio at the input of an amplifier is 40 dB. If the noise figure of an amplifier is 20 dB. Calculate the signal to noise ratio (in dB) at the amplifier output. Soln : ∴
Noise Figure (dB)
= (S/N)i (dB) – (S/N)0 dB
(S/N)o (dB)
= (S/N)I (dB) – Noise figure (dB) =
40 - 20 = 20 dB
117
For an ideal amplifier Rn = 0, F = 1 then the noise figure would be zero dB. Amplifier Noise Contribution in Terms of F (Pna) : •
The total noise referred to the input of the amplifier is given by, Pni(total)
Pno = ––––– G
…………….1
Where, G = Power gain of the amplifier. Pno = Noise power at the output of amplifier. Psi Pno Pno = –––– x –––– = ––––– Pni Pso G Pno
But F
Pno ∴ Pni = ––––– FG
…3
•
Let us find expression for total input noise power Pni.
•
From Equation 1 we can write, Pno = FG Pni But
•
Pni
=
... 2
... 4
k TB = k To B
This is because the noise factor F is calculated at the room temperature To. ∴ Pno
•
= FG k To B
…5
∴ Pno/ G = FG k To B
…6
We have defined Pni (total) in Equation 1 as Pno/ G ∴ Pni (total)
•
= F k To B
…7
Out of this total input noise power, the input source contribution is only k ToB and the remaining is contributed by the amplifier. ∴ Pni (total)
= Pni + Pna
…8
Where, Pna = Noise contributed by the amplifier. ∴ Pna =
Pni (total) - Pni = F k To B – k To B 118
Pna = •
(F - 1) k To B
…9
This is the expression for the noise contributed by an amplifier. For an ideal amplifier F = 1 and Pna = 0.
Problems: Ex.19 :The RF amplifier working at 10 MHz has a bandwidth of 200 kHz. The input resistance of the stage is 1 kΩ with an equivalent shot noise resistance of 2 kΩ. The receiver is connected to a 10 micro volt, 75Ω source, i) Equivalent noise voltage
ii) Noise figure of the stage.
Soln : Refer to the above Fig = (S/N)i (dB) – (S/N)0 dB
Noise Figure (dB) ∴
(S/N)o (dB)
= (S/N)I (dB) – Noise figure (dB) =
i)
40 - 20 = 20 dB
Equivalent noise voltage, En = [ 4kTB (Rp + Rn)]1/2
… (1)
Rs Ri 75 x 1000 Here Rp = ––––––– = ––––––––– = 69.8Ω ≈ 70Ω … (2) 119
Rs+Ri and Rn
=
1075
2 k Ω . . . shot noise equivalent resistance.
Assume T = 3000K. ∴
En = [ 4 x 1.38 x 10-23 x 300 x 200 x 103 x (2070)]1/2 = 2.61 µV
ii)
… (3)
Noise factor is give as,
∴
F
Rp + Rn = ––––––––– Rp
F
70 + 2000 = ––––––––– = 29.57 70
∴ Noise figure
… (4)
= FdB = 10log10 (29.57) =
14.7 dB
… (5)
Noise Temperature : •
The concept of noise factor or noise figure is not always the most convenient way of measuring noise.
•
Another way to represent the noise is by means of the equivalent noise temperature.
•
The equivalent noise temperature is used in dealing with the UHF and microwave low noise antennas, receivers or devices.
Definition : The equivalent noise temperature of the system is defined as the temperature at which the noisy resistor has to be maintained so that by connecting this resistor to the input of a noiseless version of the system, it will produce the same amount of noise power at the system output as that produced by the actual system. Equivalent Noise Temperature Teq at Amplifier Input : •
We know that the noise at the input of the amplifier input is given as,
120
Pna = (F - 1) k To B This is the noise contributed by the amplifier. This noise power can be
•
alternatively represented by some fictitious temperature Teq such that, k Teq B = (F-1) k T0 B
… .1
Thus the equivalent noise temperature of the amplifier is given by,
•
Teq = (F - 1) To This equations shows that Teq is just an alternative measure of F.
•
Problems: Ex.20 :An amplifier has a noise figure of 3 dB. Calculate its equivalent noise temperature. Soln : 1.
Calculate the noise figure. = 10log10 (Fratio)
FdB
∴ Noise Factor =
Antilog ( FdB / 10 ) = Antilog (0.3)
∴ Noise Factor = 2. 2.
Teq
= (F - 1) T0 =
∴ Teq
(2-1) x 300 = 300oK … by assuming To = 300oK
= 300oK
Noise Factor of amplifiers in Cascade (Friiss Formula) •
In practice the filters or amplifiers are not used in isolated manner. They are used in the cascaded manner.
•
So it is necessary to see the effect of cascading on the noise factor and noise temperature. The overall noise factor of such cascade connection can be determined as follows : Fig 3.8 shows two amplifiers connected in cascade.
121
Fig 3.8 Two amplifiers connected in cascade •
Let the power gains of the two amplifiers be G1 and G2 respectively and let their noise factors be F1 and F2 respectively.
•
The total noise power at the input of the first amplifier is given as. Pni (total) = F1 k T0 B
•
... .1
The total noise power at the output of amplifier I will be the addition of two terms. ∴ Noise input to amplifier 2 = G1F1k To B +(F2-1) k To B
….2
The first term represents the amplified noise power (by G1) and the second term represents the noise contributed by the second amplifier. •
The noise power at the output of the second amplifier is G2 times the input noise power to amplifier 2.
•
∴ Pno
= G2 x (Noise input to amplifier 2)
∴ Pno
= G1G2F1k ToB + G2(F2 – 1) k T0 B
The overall gain of the cascade connection is given by, G
•
….3
= G1G2
….4
The overall noise factor F is defined as follows : F
Pno = –––––––––
….5 122
G1 G2 Pni Here, Pni = Noise power supplied by the input source = k To B •
Substituting the values of Pno and Pni we get, G1G2F1k T0B + G2(F2-1) k To B = ––––––––––––––––––––––––––––– G1 G 2 k T o B
F
(F2 - 1) = F1 + –––––––– ….6 G1 The same logic can be extended for more number of amplifiers connected
∴ •
F
in cascade. Then the expression for overall noise factor F would be, ∴ •
F
(F2 - 1) = F1 + ––––––– + G1
(F3 - 1) (F4 - 1) –––––––– + –––––––– + …. G1 G2 G1 G2 G3
This formula is known as the Friiss formula.
Conclusion : •
This equation indicates that in the cascade configuration, the first stage is the most important stage in deciding the overall noise factor.
•
This is because due to presence of terms G1, G1 G2, ….. in the denominators of second, third terms respectively, they become negligibly small as compared to the first term. ∴
F
≈ F1
Problems: Ex.21 :In a radio receiver an RF amplifier and a mixer are connected in cascade as shown in Fig.below. The amplifier has a noise figure of 10 dB and the power gain of 15 dB. The noise figure of the mixer stage is 20 dB. Calculate the overall noise figure referred to the input.
123
Soln : It is given that : F2 = 20 dB, F1 = 10 dB and G1 = 15 dB. However to use the Friiss formula, these should be converted into power ratios. ∴ and
F2 = 20 dB = 100
: 1 Power ratio [dB = 10 log10 F2]
F1 = 10 dB = 10
: 1 Power ratio [dB = 10 log10 F2]
G1 = 15 dB = 31.62
: 1 Power ratio [dB = 10 log10 F2]
Using the Friiss Formula, (F2 - 1) F
∴
(100 – 1)
= F1 + –––––––– = 10 + –––––––––– = 13.13 G1
31.62
This is the overall noise factor, convert it to overall noise figure as, FdB
=
Noise figure =
10log10(13.13) 11.18 dB.
Equivalent Noise Temperature of Amplifiers in Cascade : •
The Friiss formula derived for the overall noise factor can be written in terms of the overall noise temperature as follows : We know that the Friiss formula is, F
• •
(F2 - 1) (F3 - 1) = F1 + ––––––– + ––––––– + …. G1 G1 G 2
….9
Subtract 1 from both the sided to get,
124
(F – 1)
(F3 - 1) (F2 - 1) = (F1 – 1) + –––––––– + ––––––– + …. G1 G1 G 2
Teq We can substitute ( F-1) = ––– the T0
•
….10
similarly making the substitutions on
right hand side of the equation above, Teq Teq1 Teq2 Teq3 ––– = –––– + ––––––– + –––––––– + . . . T0 G1 T 0 G1 G2 T 0 T0
∴
Teq = Teq1
Teq2 Teq3 + ––––– + –––––– + . . . G1 G1 G2
….11
….12
Where Teq1, Teq2, etc, are the noise temperatures of amplifiers 1, 2 etc. Problems: Ex.22 :A radio receiver with 10 kHz bandwidth has a noise figure of 30 dB. Determine the signal power required at the input of receiver to achieve input SNR of 30 dB. Soln : (i)
Noise figure, FdB ∴
30
∴ Noise Factor F (ii)
= 10 log F, = 10 log F = 1000
Noise Power at input = Pni = Fk TB = 1000x1.35x10-23x(273+27)x10x103 =
4.14 x 10-14
(iii) input signal to noise ratio, SNR dB = 10 log [Psi/ Pni] Psi ∴
30
= 10 log
–––––––––– 4.14 x 10-14
∴
Psi
= 4.14 x 10-12
125
Ex.23 :If each stage has a gain of 10 dB and noise figure of 10 dB determine the overall noise figure of a two stage cascaded amplifier. Soln : The gain ∴
G
= 10 log10 10
G
= 10 (for each stage)
And noise figure FdB = 10 log10 F Where
F
∴ Noise factor
=
Noise factor
=
F = 10 (for each stage)
F2 - 1 ∴ Over all noise factor, F = F1 + ––––––– G1 ∴
F
∴
F
10 - 1 = 10 + ––––––– 10 = 10.9
∴ Over all noise factor, FdB Ex.24 :
… Friiss formula
= 10 log10 (10.9) = 10.37 dB
An amplifier with 10 dB noise figure and 4 dB power gain is
cascaded with a second amplifier which has a 10 dB power gain. What is the overall noise figure and power gain?. Soln : The given system is an shown in Fig .
Step 1 :
Convert dB into equivalent power ratios : F1 = 10 dB
∴ F1 = 10
F2 = 10 dB
∴ F2 = 10
G1 = 4 dB
∴ 4 dB = 10log G1
∴ G1 = 2.5
126
Step 2 :
∴
Calculate the overall noise factor : F
(F2 - 1) (10 – 1) = F1 + –––––––– = 10 + –––––––– 2.5 G1
F
= 13.6
Step 3 :
Overall noise figure and power gain :
Noise figure = Overall power gain
10 log (13.6) = 11.33 dB
=
log (G1G2)
=
1.39 dB
= log (2.5 x 10)
Ex.25: A mixer stage has a noise figure of 20 dB and it is preceded by an another amplifier with a noise figure of 9 dB and an available power gain of 15 dB. Calculate the overall noise figure referred to the input. Soln : The system is an shown in Fig .
Step 1 :
Convert dB into equivalent power ratios :
F1 = 9 dB
∴9
F2 = 20 dB
∴ 20 = 10 log F2
∴ F2 = Antilog (2) = 100
G1 = 15 dB
∴ 15
∴ G2 = Antilog (1.5) = 31.62
Step 2 :
∴
= 10 log F1 = 10 log G1
∴ F1 = Antilog (0.9) = 7.94
Calculate the overall noise factor: F
(F2 - 1) (100 – 1) = F1 + –––––––– = 7.94 + –––––––– 31.62 G1
F
= 11.07
127
Step 3 :
Calculate the overall noise figure : :
The overall noise Figure
=
10 log F
=
10 log10 (11.07) = 10.44 dB … Ans.
Ex.26:The noise figure of a receiver is 20 dB and it is fed by a low noise amplifier which has a gain of 40 dB and noise temperature of 800K. Calculate the overall noise temperature of the receiver system and the noise temperature of the receiver. Soln : Receiver noise figure
Given :
Amplifier Gain
F2 = 20 dB
G1 = 40 dB
Amplifier Noise temperature Te1 Step 1 :
80o K
Convert dB into ratios : [F2] dB =
20 dB
∴ F2 =
100
Similarly [G1] dB
=
∴ G1 = Step 2 :
=
= 10 log F1
10 log G1 = 40 dB 10,000
Obtain Te2 :
Noise temperature of the receiver is related to its noise factor as follows : Te2 = (F2 – 1) T0 But
F2
= 20 dB
… Given
i.e. F2 = 100
Assume T0 = 300o K ∴ Step 3 :
Te2
=
99 x 300oK = 29700oK
Calculate overall noise temperature :
∴
Te
Te2 = Te1 + ––––– G1
Te
29700 = 80 + –––––––– 10000 = 80 + 2.97 = 82.970 K 128
Noise in Reactive Circuits : •
In communication circuits we come across many circuits which contain reactive components such as inductors and capacitors.
•
These components are used to form the tuned circuits. Refer Fig.3.8 in which a resistance R is connected in parallel with a a parallel tuned circuit.
•
An ideal tuned circuit is theoretically noiseless. Hence it should not have any effect on the noise generated by the resistor R of Fig.3.9.
•
But tuned circuit is a frequency selective network. Hence it will attenuate both, the signal and the noise in the same proportion, above and below the resonant frequency.
•
In other words, the ideal tuned circuit limits the bandwidth of the noise source by attenuating noise outside the passband.
129
•
Now consider the practical turned circuit of Fig.3.10. Here the inductor L of the tuned circuit possesses a small resistance Rs which produces noise. Hence the practical tuned circuit is not noiseless.
Noise voltage across the capacitor : •
Refering Fig.3.10 to calculate the noise voltage across C. i.e. at the input of the amplifier. The noise current flowing in the RLC circuit is given by, In =
Vn ––– Z
…1
Where Vn = Noise voltage and Z = Rs + j (XL – XC) •
At resonant frequency XL = XC
•
Vn ––– Rs The noise voltage across C is given by, ∴
In =
∴
Vnc = In Xc
But ∴
•
=
Vn Xc –––– Rs
…2
Xc = Q Rs at resonance. Vn x QRs Vnc = ––––––––– = VnQ Rs
…3
Taking eh square of both the sides we get, Vnc 2 = Q2Vn2 = Q2 [4 k T B Rs ] …. Since Vn = √ 4k TB Rs But √ Rp = Q2 Rs Where Rp = equivalent parallel resistance at resonance ∴
Vnc2
= 4kTBRP
∴
Vnc
= √ 4kTBRP
…4
130
Problem: Ex.26 :A parallel tuned circuit has a resonant frequency fr = 10 MHz lts Q = 20 and the value of capacitor is 10 pF. If the ambient temperature is 17oC calculate the noise voltage across the parallel tuned cicuit. Soln : Step 1 :
Calculate the bandwidth :
Bandwidth B = Fr / Q = 10 MHz / 20 = 50 kHz. Calculate Rs and Rp :
Step 2 :
Q
1 = ––––– ωr CRs
∴
Rs
1 1 = ––––––– = ––––––––––––––––––––––––––– Qωr C 20 x 2π x 10 x 106 x 10 x 10-12
∴
Rs
= 79.57 Ω
Rp
= Q2 Rs = (20)2 x 79.57 = 31.83kΩ
Step 3 :
Calculate the noise Voltage : ∴
Vnc
= √ 4kTBRP = √ 4 x 1.38 x 10-23 x 290 x 500 x 103 x 31.83 x 103
∴
Vnc
=
1.596 x 10-5 volts or 15.96 µ V
...
Ex. : Calculate the R.M.S. value of noise voltage at 170C development across the network shownFig..
131
Soln : The R.M.S. noise voltage across the capacitor is given by, Vnc
=
= Vnc =
kT ––– C 1.38 x 10-23 x 290 ––––––––––––––– 4 x 10-12 3.163 x 10-15 V or 31.63 µ V
Ex.27 :A parallel tuned circuit, having Q of 20 is resonated to 200 MHz with a 10 pF capacitor. If this circuit is maintained at 17oC, what is the equivalent noise voltage? Soln : Given : Q = 20, Fr = 200 MHz, C = 10 pF, T = 290oK Step 1 : Calculate the value of resistance R of the tuned circuit. We know that the quality factor of resonant circuit is given by : ωr L 1 Q = –––––– = ––––––– R ωr C R
… (1)
Where ωr = 2π fr and fr = Resonant frequency. 1 1 ∴ R = –––––––– = ––––––––––––––––––––––––––– 2π x 200 x 106 x 10 x10-12 x 20 ωr C Q
132
R
∴ Step 2 :
=
3.978Ω
Calculate the equivalent noise voltage : ∴
En
= √ 4k TB Q2 R
Assuming the band width B = 10 kHz we get En
= √ 4 x 1.38 x 10-23 x 290 x 10 x 103 x (20)2 x 3.98 =
50.48 µ V
Superposition of Noises : •
In the earlier sections we have represented noise n (t) in the Fourier series form. That means noise is represented as “Superposition” of individual noise components.
•
However the frequency ranges of
these components were not
overlapping. •
It can be proved that even though the frequency ranges are overlapping then also the total noise power is superposition of powers of individual spectral components. Let us prove it as follows : i)
Suppose that we have two noise processes n1 (t) and n2 (t) whose ranges overlap fully or partially.
ii)
The total noise due to these two components is given by, n(t) = n1(t) + n2(t) . . . . . by superposition
iii)
Then the normalized power P12 of the sum n1(t) + n2(t) is given by
∴ iv)
P12
= [n(t)]2 = [n1(t) + n2(t)]2
P12
= [n1(t)]2 + [n1(t)]2 + 2 n1(t) n2(t)
….1
P12
=
…..2
P1 + P2 + 2 n1(t) n2(t)
In the above expression P1 and P2 are the normalized power of n1 (t) and n2 (t) respectively and the third term is the expected value of the product. The expected value of the product is nothing but “cross-correlation” of the processes n1(t) and n2(t). As these
133
processes are uncorrelated the cross-correlation will be zero and the equation .2 can be simplifies as : P12 = P1 + P2
…3
Conclusion : From the above expression we can draw a conclusion that the total normalized noise power can be obtained by superposition of the powers of individual noise components. For k number of noise components the expression is, P = P1 + P 2 + P 3 + . . . + Pk Effects of Linear Filtering on Noise : •
We have already discussed that the thermal noise has a uniform power spectral density upto the frequency of the order of 1013 Hz.
•
Shot noise also has a wide frequency range. Other noise sources also have very wide spectral ranges.
•
Let us consider the “white noise” for understanding the effect of filtering on noise.
•
The white noise has a uniform power spectral density over the entire frequency range of interest.
•
Filters are connected in order to reduce the noise power. Generally these filters are narrowband filters which are designed to pass a specific range of frequencies.
•
In order to minimize the noise power that is presented to the demodulator of a receiver, we introduce a filter before the demodulator as shown in Fig.3.11
134
Fig 3.11 A filter connected before a demodulator to reduce the noise Types of Filter: In this section we are going to see the effect of following filters on noise. •
R-C low pass filter.
•
Ideal low pass filter.
•
Ideal band pass filter.
Effect of R-C Low Pass Filter : •
The R-C low pass filter and variation of its transfer function with frequency are as shown in Fig.3.12(a) and (b) respectively.
Fig 3.12 (a) RC low pass filter
135
Fig 3.12 (b) Transfer function of RC low pass filter •
|H(f)| is the transfer function of the filter and fc is its cut-off frequency. The transfer function of the R-C low pass filter is given by, H (f)
1 = –––––––– 1 + jf / fe
….1
Assumptions : 1. Let the input noise signal be ni (t) and the output noise signal be no (t). 2. Let the psd of input noise signal be Si (f) and the psd of the output noise signal be So (f). Let us now obtain the expression for the average noise power at the output of the R-C filter. Average noise power at the Filter output: •
We know that the relation between the psd of input and output is given by, So ( f ) = | H (f) |2 Si (f)
•
….2
Assume that the input noise is AWGN (additive White Gaussian Noise) then the psd at the filter input is given by,
•
No Si ( f ) = ––– ….3 2 The magnitude of the transfer function of the filter can be obtained from Equation 1 2 ….4 | H (f) | = ––––––––––– 1 + ( f / fe )2
136
•
Substitute Equations .3 and 4 into Equation 2 to get, S0 (f)
•
1 No = –––– . ––––––––– 2 1 + ( f / fe )2
….5
Thus we have obtained the power spectral density of the output noise signal.
•
The next step is to obtain the average output noise power from the psd. It is done as follows : ∞ P
=
∫ S (f) df
...6
-∞ •
This is the general expression relating the average power and psd. Using this expression we can obtain the output noise power as : P
∞ = ∫ S (f) df = -∞
1 ∞ No ∫ –– ––––––––– df -∞ 1 1 + ( f / fe)
Substitute f / fe = x in the above expression. ∴ fe d x = df • The limits of integration will remain unchanged. ∞ 1 No P = –– fe ∫ –––––– dx 2 -∞ (1 + x2) 1 But ∫ ––––– dx = π (1 + x2)
•
No = –– fe π 2 Thus the average noise power at the output of an R-C filter is given by, π No fe ∴ Pno = –––––– 2 (R-C low pass filter) ….7 This equation shows that the output noise power can be reduced by reducing the value of filter cut-off frequency fc. ∴
•
•
P
137
•
Power spectral Density : Let the psd of the output noise be denoted by S0 (f). Let us obtain the expression for the same.
•
Referring to Equation .5 we can write that the psd of output noise is No / 2 So (f) = –––––––––– 1 + (f / fe)2
•
But the cut-off frequency fe of the RC low pass filter is given by, fe
•
1 = –––––– 2π RC
Substituting this we get No/ 2 So (f) = –––––––––––– 1 + (2π f RC)2
•
… .8
This is the required expression for the psd of ouput noise and it is plotted in Fig.3.13
Fig 3.13 PSD of output noise
Autocorrelation function of Output noise : •
The autocorrelation function of output noise is denoted by RN (τ) and it can be obtained by taking the inverse Fourier transform of PSD So (f). 138
∴
•
RN (τ)
= IFT [So (f)]
No / 2 = IFT –––––––––– 1+ (2π f RC)2 Substitute RC = 1/α to get No / 2 α2 No / 2 = IFT ––––––––––– = IFT –––––––––– 1+ (2π f / α )2 α2 + (2π f)2
•
2α Substitute α2 = α x α = –––– in the numerator to get 2 RC [(No/2) 2α / 2 RC] No 2α RN (τ) = IFT ––––––––––––––– = ––––– IFT ––––––––––– 4RC [α2 + ((2π f)2] [α2 + ((2π f)2]
•
F 2α We know that e -α | (τ) | ↔ ––––––––– α2 + ((2π f)2
∴
No RN (τ) = ––––– . e-α | (τ) | 4RC No = ––––– . e- | (τ) | / RC 4RC
•
…9
This is the required expression for autocorrelation function RN(τ) and it is plotted in Fig. 3.14.
139
Fig 3.14 Autocorrelation function of output noise Rectangular (Ideal) Low Pass Filter : •
The transfer function of an ideal (rectangular) low pass filter is given by: H (f)
= 1, for - B ≤ f ≤ B = 0, elsewhere
….10
Expression for the output noise Power : •
The noise at the input is assumed to be AWGN. Therefore the psd of the input noise is given by, No Si (f) = ––– 2
•
……11
We know that the relation between the input and output spectral densities is given by, So (f) = | H (f) |2 x Si (f) Substituting Equation (11) into the above expression we get, No So (f) = | H (f) | –––– 2 2
….12
140
•
The average noise power at the filter output is given by, Pno
∞ = ∫ So (f) df -∞ No ∞ 2 = ∫ | H (f) | x ––– df -∞ 2
Pno •
=
B No = ∫ ––– x (1)2 df -B 2
No B
Thus the average output noise power of an ideal low pass filter is given by, Pno = No B
•
… Ideal low pass filter
….13
This equation shows that the output noise power can be reduced, by reducing the bandwidth of the ideal low pass filter.
Noise Power Spectral Density : •
Refer to Equation .12 to write the expression for output noise power spectral density as No So (f) = | H (f) |2 ––– 2
•
But | H (f) | = 1 for –B ≤ f ≤ B No ∴ So (f) = –––– 2 = 0
•
….. –B ≤ f ≤ B
….. else where
… 14
The output noise power spectral density is graphically shown in Fig. 3.15
141
Fig 3.15 Output noise power spectral density. Autocorrelation function of output noise : •
Let us now obtain the expression for the autocorrelation function of output noise of the ideal low pass filter.
•
Autocorrelation of the output noise be denoted by RN (τ). It can be obtained by taking the inverse Fourier transform of the power spectral density. ∴ RN (τ)
= IFT [ S0 (f) B No = ∫ ––– -B 2
e
i2πF τ
No df = –––––––––– [ei2πF τ ]B -B j x 2x 2 πτ
No ei2πF τ - ei2πF τ B = ––– –––––––––––– 2 2j No ∴ RN (τ) •
= ––––– sin (2 π τ B) 2πτ Multiply and divide by B to get,
….15
No B ∴ RN (τ)
= –––––– sin (2 π τ B) 2πτ B = No B sinc ( 2τ B)
….16
142
•
This sinc function will have a maximum value of NoB at f = 0 and it will pass through zero at 2 τ B = ± 1, ± 2 . . .
•
So the RN (τ) passes through zero at 1 ∴
τ
= ± –––– ± –––– . . . 2B’
•
2 ….17
2B’
The autocorrelation is plotted in Fig.3.16
Fig 3.16 Auto correlation function of the out put Noise Bandwidth : •
Assume that a white noise is present at the input of a receiver. Let the filter have a transfer function H (f) as shown in Fig.3.17
Fig 3.17 Noise bandwidth of a filter •
This filter is being used to reduce the noise power actually passed on to the receiver.
•
Now contemplate an ideal (rectangular) filter as shown by the dotted plot in Fig.3.17 . The center frequency of this ideal filter also is f0.
143
•
Let the bandwidth “BN” of the ideal filter be adjusted in such a way that the noise output power of the ideal filter is exactly equal to the noise output power of a real R-C filter.
•
Then BN is called as the noise bandwidth of the real filter. Thus the noise bandwidth “BN” is defined as the bandwidth of an ideal
(rectangular) filter which passes the same noise power as does the real filter. Noise bandwidth of an R-C low pass filter : •
Let us calculate the noise bandwidth BN of an R-C low pass filter. The transfer function of an R-C low pass filter is given by 1 H (f)
= –––––––– 1 + j f / fc
For this filter H (f) = 1 at f = 0. •
The noise output power of such a filter is given by Equation π No fc = –––––––
Pno
2 •
In presence of the same AWGN noise, the noise output power of an ideal low pass filter with a bandwidth B will be : Pno
•
= No B
According to the definition of noise bandwidth , if B = BN then output noise power of ideal and real low pass filters should be identical. π No fc ∴
–––––– =
NoBN
2 π ∴ BN = –– fc 2
R-C low pass filer.
144
•
Thus the noise bandwidth of an R-C low pass filter is π/2 times or 1.57 times its 3-dB bandwidth fc.
Narrow-Band Noise : •
The preprocessing of received signal is performed in every communication receiver.
•
A narrow band filter is generally employed for the purpose of preprocessing.
•
Bandwidth of the narrow band filter is just large enough to pass the signals only but the bandwidth is not very large as to allow excessive noise to pass through.
•
The noise at the output of such a filter is called as the narrow band noise.
(b)Sample function of narrow band noise Fig 3.18 •
Fig.3.18(a) shows the power spectral density of the narrow band noise. It shows that the spectral components of narrow band noise are concentrated about some midband frequency ± fe.
•
Narrow band noise is a random process, the sample function of which is denoted by n (t) and shown in Fig. 3.18(b). 145
•
Fig.3.18(b) shows that the sample function n(t) of a narrow band noise is similar to a sinewave. Its frequency is fc and it undergoes slow variations in both amplitude and phase.
Power spectral density SN (f):
Fig 3.19 •
Refer Fig.3.19 Let the noise produced at the output of a narrow band filter be n (t) due to application of a white Gaussian noise having a sample function w (t).
•
The white Gaussian noise applied at the filter input has a zero mean and unit power spectral density.
•
w (t) is the sample function of the input noise process w (t) and n (t) is the sample function of the output noise process N (t).
•
Let the transfer function of the filter be H (f). So the PSD of n(t) is given by, SN (f) = | H (f) |2 x PSD of w (t) ∴
SN (f) = | H (f) |2 x 1 = | H (f) |2
….1
Representing the Narrow band Noise in terms of in Phase and Quadrature Components : •
Let n+ (t) denote the preen elope and n (t) denote the complex envelope of the narrow band noise n (t).
146
•
Let the power spectrum of n (t) be centered about the frequency fc. Then we can write that n+ (t) = n (t) + j n (t)
...2
Where n (t) = Hillbert transform of n (t) as defined in the previous section. •
And n (t) = n+ (t) e –j2πfct
. . . .3
When n (t) is the complex envelope. •
The complex envelope may be expressed in terms of the in phase component nI (t) and the quadrature component nQ (t) of the narrow band noise as follows :
n (t) = n+ (t) + j nQ (t) •
. . . .4
Combining the Equation (2) through Equation.(4), we get, nI (t) = n (t) cos (2πfct) + n (t) sin (2πfct)
. . . .5
and nQ (t) = n (t) cos (2πfct) - n (t) sin (2πfct) •
. . ..6
From Equations .(5) and (6) by eliminating n (t) we can express n(t) in terms of the in phase and quadrature components as follows : N (t) = nI(t) cos (2πfct) – nQ(t) sin (2πfct)
. . . .7
White Noise Passed Through a Ideal Bandpass Filter : •
Let the input to an ideal bandpass filter be a white Gaussian noise of zero mean and power spectral density of No/2.
Fig 3.20 •
The transfer function of an ideal bandpass filter is as shown in Fig.3.21
147
Fig 3.21 •
The transfer of the ideal bandpass filter is given by, H (f)
•
= 1,
for | fc | - B to | fc | + B
= 0,
elsewhere
… .9
The midband frequency of the filter is fc and bandwidth is 2B.
Power spectral density of filtered noise : •
Assuming that the noise at filter input to be AWGN, the psd of the input noise is given by,
•
No Si (f) = ––– 2 The psd of the output noise is given by,
… .10
So (f) = | H (f) |2 . Si (f) •
But H (f) = 1 in the passband of the filter No ∴
So (f) = Si (f) = ––– for | fc | - B to | fc | + B
… .11
2
148
Unit 4 Performance of CW modulation Systems Topics 1. Superheterodyne receiver and its characteristics. 2. Signal to noise Ratio. 3. Noise in DSBSC system using coherent detection. 4. Noise in AM system using envelope detection and FM system. 5. FM threshold effect. 6. Pre-emphasis and De-emphasis in FM. 7. Comparison of Performances.
149
Super Heterodyne Radio receiver: Receiver Parameters 1.
Selectivity
2.
Bandwidth - Improvement
3.
Sensitivity
4.
Dynamic Range
5.
Fidelity
6.
Insertion loss
7.
Noise Temperature
8.
equivalent Noise Temp.
2. Selectivity : It measures the ability of the receiver to accept a given band of frequencies and reject all others. The Bw should be narrow for better selectivity. 3. Bandwidth Improvement : Thermal noise is directly proportional to the Bw. Reducing the Bw, the noise can be reduced. The Noise reduction ratio achieved by reducing the Bw is called Bandwidth improvement and is mathematically expressed as
BI
=
BRF = –––––––– BIF Bandwidth Improvement
BRF
=
RF Bandwidth
BIF
=
IF Bandwidth
BI
150
The corresponding reduction in the noise figure due to the reduction in Bw is called noise figure improvement. NF Improvement =
10 log BI
Ex. 6 : Determine the improvement in the noise figure for a receiver with a RF sand width equal to 200 KHz and an IF Bw = 10 KHz. Soln :
BI
200 KHz = ––––––––– = 20 10KHz
NF Improvement =
10 log 20
= 13 dB.
4. Sensitivity : The ability of the receiver to detect the weak signal, amplify there and produce a usable demodulated information signal incalled the sensitivity. 5. Dynamic Range : It is the i/p power range over which the receiver is useful. 6. Fidelity : It is a measure of the ability of a communications system to produce, at the o/p of the receiver, an exact replica of the original information. Any frequency, phase or amplitude variations that are present in the demodulated waveform that were not in the original information signal are considered as distortion. 7. Insertion Loss : 151
It is the ratio of the output power of a filter to the i/p power. Pout Il (ds) = ––––– Pin
Super Heterodyne Receiver : •
Was developed near the end of world was I.
•
It has its gain, Selectivity and Sensitivity characteristics superior to the other configurations.
•
Heterodyne means to mix to frequencies together in a non linear device. or
•
To translate one frequency to another using non linear mixing.
•
It has 5 sections 1. 2. 3. 4. 5.
RF section Mixer / Converter Section The IF section Audio detector Section Audio amplifier section
1. RF Section : •
Consists of Preselector and an amplifier stage. They
can be either
separate circuit or combined. •
Preselector is a Band pass filter, which provides the initial Band limiting to prevent unwanted radio frequency called the image frequency.
152
•
RF amplifiers can be one or more or not any depending on the desired sensitivity.
•
The advantages of RF amplifier in a receiver are, 1.
Greater gain, thus better sensitivity.
2.
Better selectivity
3.
Better signal to noise ratio.
2. Mixer / Converter section : •
It has a local Oscillator. It is called the first detector.
•
It is a nonlinear device and its purpose it to convert radio frequency into intermediate frequency.
•
Heterodyning takes place here.
•
Through the frequencies of the modulated signal is charged. The envelope remains the same.
3. IF Section : •
If consists of series of IF amplifiers and Band pass filters and is called.
•
Receiver gain and selectivity is achieved in IF section.
•
The If is always lower in frequency than RF.
4. Detector Section : •
Converts the IF signal back to the original source information.
•
It is generally called the audio detector or the second detector.
•
The detector can be as simple as a signal diode or as complex as a phase locked loop or balanced demodulator.
5. Audio amplifier Section : •
The audio section comprises several cascaded audio complifiers and one or more speakers.
•
The number of amplifiers depends on
153
Receiver Model: •
For carrying out the noise related analysis, we have to use the receiver model shown in Fig.4.1
Fig 4.1 Noisy receiver model •
S (t) is the incoming modulated signal and the ω (t) denotes the front end receiver noise.
•
The received signal is therefore the addition of s (t) and ω (t) as shown in Fig.4.1 the receiver has to work on this signal.
•
The combined signal s (t) + ω (t) is applied at the input of a bandpass filter. This is not the actual filter. It represents the combined filtering action of the tuned amplifiers used in the receiver prior to demodulator.
•
The bandwidth of this filter is just sufficient to pass the signal s (t) without any distortion.
•
The type of demodulator in Fig.4.1 depends on the type of modulation used.
•
Noise ω (t) is assumed to be on additive white Gaussian noise. The power spectral density of this noise is No / 2 and it is constant, independent of frequency.
154
•
No is the average noise power per unit bandwidth and it is measured at the front end of the receiver.
•
The bandwidth of the bandpass filter in Fig.4.1 is exactly equal to the transmission bandwidth BT of the modulated signal s (t). This filter is an ideal filter. The midband frequency is equal to fc.
•
Since the midband frequency of the BPF is fc we can model the power spectral density SN (f) of the noise n (t) due to passing of ω (t) through the ideal BPF as shown in Fig.4.1
•
Generally fc >> BT. Hence we can treat the filtered noise n (t) as a narrow band noise. So it is possible to represent n (t) in the canonical form as follows. n(t) =
nI ( t ) cos ( 2π fc t ) – nQ ( t ) sin ( 2π fc t )
Where nI (t) is the in phase noise component and nQ (t) is the quadrature noise component. •
At the output of the filter, the signal present is x ( t ) which is given by, X(t) = s(t)+n(t) This signal is then applied to the demodulator input.
•
Out of them s ( t ) is dependent on the type of modulation and the average noise power at the demodulator input is equal to area under the curve of PSD SN ( f ).
•
The average noise power can be obtained from Fig.below as the area under SN (f) curve. It is given by, Average noise Power
=
No 2 x ––– x BT = No BT 2
155
Signal to noise ratio at the demodulator input: •
The signal to noise ratio at the input of the demodulator can be defined as,
(SNR)I
=
Average power of s (t) ––––––––––––––––––––––––––––– Average power of filtered noise n (t)
Output signal to noise ratio: •
This is more useful measure of noise performance. It is denoted by SNRo and defined as the average power of the demodulated message signal to the average power of the noise measured at the output of the demodulator.
Factors affecting SNRo: Following are the factors which affect the SNRo •
Type of modulation used at the transmitter.
•
Type of demodulation used at the receiver.
Channel Signal to Noise Ratio (SNRc): •
The channel signal to noise ratio is denoted by SNRc and it is defined as the ratio of average power of the modulated signal s(t) to the average 156
power of noise in the message bandwidth W, both measured at the receiver input.
∴ SNRc •
=
Average signal power at receiver input –––––––––––––––––––––––––––––– Average noise power at receiver input
We can view this ratio as the signal to noise ratio which results from the transmission of the baseband signal (i.e. the message signal m(t) ) without modulation. This is illustrated in Fig.4.2
Fig 4.2 •
Here we assume the following two points. 1. The message power at the input of the LPF is equal to the average power of the modulated signal. 2. The bandwidth of LPF is W Hz. Therefore it will pass only the message signal and rejects all other signals and noise components outside this band.
Figure of Merit: •
In order to compare different CW modulation systems, we have to normalize the receiver performance by dividing the output signal to noise ratio (SNRo) by the channel signal to noise ratio (SNRc).
•
Hence the figure of merit for a receiver is defined as, Figure of merit
=
(SNRo) –––––– (SNRc)
157
•
Figure of merit can be less than greater than or equal to 1 depending on the type of modulation.
•
The figure of merit should be as high as possible because higher value of Figure of merit indicates better noise performance of the receiver
Noise in DSBSC receivers using coherent detector: The model of DSB-SC receiver using coherent detection is shown in Fig.4.3
Fig 4.3 Model of DSBSC receiver using coherent detection •
For achieving the coherent detection, we have to multiply signal x (t) by a locally generated carrier cos (2π fc t) and then pass the product through a low pass filter.
•
It is assumed that the local oscillator has been synchronized in phase and frequency with the oscillator at the transmitter.
•
Let the DSBSC component of the filtered signal x (t) be expressed as follows. s(t) =
C Ac cos ( 2π fc t ) m ( t )
----1
158
Where Ac cos (2π fc t) represents the carrier, m (t) is the message signal, and C is a system dependent scaling factor. •
Let m (t) is the sample function of a stationary process having a zero mean. Let the power spectral density of m (t) be SM (f) which extends only upto W Hz (since W is the message bandwidth).
•
Hence the average power P of the message signal can be obtained by calculating the total area under the power spectral density curve. That means, W P
=
∫ SM (f) df
-----2
-W •
Average power of the DSB-SC modulated signals s (t) is given by
C2 Ac2
P / 2. •
The noise spectral density of the additive white Gaussian noise is
No /
2. If the message bandwidth is No then the noise power in the message bandwidth W is equal to WNo. •
Hence the channel signal to noise ratio is given by, (SNR)c
=
Average signal power at receiver input –––––––––––––––––––––––––––––– Average noise power at receiver input
=
C2 Ac2 P / 2 ––––––––– WNo
Here C2 is a constant which is included in order to make the ratio dimensionless. SNRo of the DSB-SC Receiver: •
To obtain the signal to noise ratio at the output of the system, use the narrow band representation of filtered noise n (t).
•
The total signal at the input of the coherent detector (i.e. at the output of the band pass filter) is given by, x(t) =
s(t) + n(t)
∴ x ( t ) = C Ac cos (2π fc t) m(t) + nI(t) cos (2π fc t) – nQ(t) cos (2π fc t) 159
Where nI (t) and nQ (t) are the in phase and quadrature components Of noise n (t) •
This signal is multiplied with the local oscillator in the product modulator. So the product modulator output is given by, v(t) =
x ( t ) . cos (2π fc t)
= C Ac cos2 (2π fc t) m ( t ) + nI ( t ) cos2 (2π fc t) – nQ ( t ) sin (2π fc t) cos (2π fc t) But cos (2π fc t)
=
1 – 2
and sin (2π fc t) cos (2π fc t)
=
1 – sin (4π fc t) 2
2
1 + – cos (4π fc t) 2
Hence v(t) =
1 1 1 – C Ac m ( t ) + – nI ( t ) + – cos (4π fc t) [C Ac m ( t ) + nI ( t ) ] 2 2 2
1 - – Ac nQ ( t ) sin (4π fc t) 2 • The low pass filter after the product modulator removes all the high frequency components. Hence the receiver output is given by, 1 1 y(t) = – C Ac m ( t ) + – nI ( t ) 2 2 From the above equation we can conclude that: 1. At the receiver output we obtain the sum of the message signal m (t) and the in phase component of noise i.e. nI (t). 2. The coherent receiver rejects the quadrature component of noise n (t), completely. 3. The output is independent of the signal-to-noise ratio.
160
Signal to noise ratio at the output. •
1 The message component in the receiver output is – C Ac m (t). 2 Therefore the average power associated with the message component 1 is – C2 Ac2 P where P is the average power of the message signal m(t). 2
•
For DSB-SC modulation, the transmission bandwidth BT = 2W. Hence the bandwidth of bypass filter also is equal to 2W.
•
Hence the average power of the filtered noise n (t) is 2 WNo as discussed earlier.
•
From Equation, the noise component at the receiver output is nI (t) / 2. Hence the average power of the noise at the receiver output is.
1 – 2 •
2
1 x 2 W No = – W No 2
The SNRo for the DSB'SC receiver using coherent detection is therefore given by,
•
C2 Ac2 P / 4 ––––––––– = SNRo = W No / 2 Figure of merit is given by,
C2 Ac2 P –––––– 2 W No
SNRo –––– SNRc Substituting the expressions for SNRo and SNRc we get Figure of merit
Figure of merit
=
=
C2 Ac2 P / 2 WNo ––––––––––––– C2 Ac2 P / 2 WNo
=1
161
Noise in SSB Receivers: •
Consider the detection of SSB signal using the coherent detection. Let only the LSB is transmitted. Hence the SSB signal is mathematically expressed as, s(t) =
1 1 Λ – C Ac cos (2 π fc t) m ( t ) + – C Ac sin (2 π fc t) m ( t) 2 2
Λ Where m ( t ) is represents the Hilbert transform of m ( t ). •
Λ The Hilbert transform m ( t ) is obtained by passing the message m ( t ) Λ o through 90 phase shifter. Hence m ( t ) and m ( t ) will always be orthogonal to each other.
•
Λ Hence if m ( t ) has zero mean then m ( t ) and m ( t ) are uncorrelated and their power spectral densities can be added.
•
Λ m ( t ) is obtained by passing m ( t ) through a linear filter with a transfer function – j sgn ( f ). The squared magnitude of this transfer function is unity for all the values of f.
•
Λ Hence m ( t ) and m ( t ) will have the same power spectral density.
•
So we can use the same procedure as that used in the previous section to find the in-phase and quadrature components of s ( t ). The average power contributed by these components is C2 Ac2 P / 8 each.
•
Where P is the average power of m ( t ). The average power associated with s ( t ) is,
•
C2 Ac2 P 1 C2 Ac2 P –––––– + –––––– = – C2 Ac2 P 8 8 4 Note that this power is exactly half of the power in the DSB-FC signal.
•
The average noise power in the message bandwidth W is WNo. This is same as that in the DSBSC receiver.
162
•
Therefore the channel signal to noise ratio is given by, SNRc =
•
C2 Ac2 P –––––– 4 WNo
Refer Fig (a). Which shows that the bandwidth of the SSBSC system is BT = W and the midband frequency is offset from the carrier frequency fc by W / 2.
•
From Fig (a). we can write the expression for n ( t ) as follows. n ( t ) = nI ( t ) cos 2π
•
W fc - –– t - nQ ( t ) sin 2π 2
W fc - –– 2
t
We apply the combination of SSBSC signal s ( t ) and the noise n ( t ) at the input of the receiver. The output of the coherent detector is then given by, 1 1 1 y ( t ) = – C Ac m ( t ) + – nI ( t ) cos ( π W t) + – nQ ( t ) sin ( π W t) 4 2 2
Λ • Note that the m ( t ) term has been eliminated from the output. But the quadrature component of noise i.e. nQ ( t ) appears in the output. Output signal power: •
In the expression for y (t) the message signal component is
1
/
4 C Ac m (t). Hence the average power in the recovered message signal is given by C2 Ac2 P / 16. Output Noise Power: •
The noise component in the receiver output is, 1 – nI ( t ) cos ( π W t) + nQ ( t ) cos ( π W t) 2
•
To calculate the power associated with it notes the following.
1. The power spectral density of nI (t) and nQ (t) is same and it is as s hown in Fig (b). 2. The cos ( π W t) is independent of nI ( t ) and nQ ( t ). Hence the power spectral density of n'I ( t ) = nI ( t ) cos ( π W t) is obtained as follows : 163
•
Shift SNI ( f ) to the left by W/2.
•
Then shift it to right by W/2.
•
Add the shifted spectra and divide it by 4. The PSD of n'I ( t ) is S'NI ( f ) and it is shown in Fig(c).
3. The PSD of n'Q ( t ) = nQ ( t ) sin ( π W t) can be obtained in a similar manner. Note that PSD of n'Q ( t ) is S'NQ ( f ) and it is equal to S'NI ( f ) as shown in Fig (c) •
From Fig (c) it is evident that the average power of the noise component n'I ( t ) or n'Q ( t ) is WNo / 2. Hence from the equation from y ( t ), the average output noise power is WNo / 4.
Output Signal to Noise Ratio (SNRo): •
Therefore the output signal to noise ratio SNRo of the SSB receiver is given by, SNRo = ∴
SNRo =
Figure of Merit: • •
C2 A c 2 P –––––– x 4 16 WNo
The figure of merit of SSB receiver is given by, Figure of merit =
• •
Signal power –––––––––– = Noise power C2 Ac2 P –––––– 4 WNo
SNRo –––– SNRo
=
C2 Ac2 P / 4 WNo ––––––––––––– C2 Ac2 P / 4 WNo
∴ Figure of merit = 1 Note that C2 factor gets cancelled out. The expression of figure of merit shows that if the average modulated ( or transmitted ) signal power and the average noise power in the message bandwidth is same as that of DSBSC receiver then the output SNR and figure of merit of SSB coherent receiver is same as that of the DSBSC receiver.
Noise Performance of AM & FM Receivers Noise in AM Receiver (Using Envelope Detection): •
Let us now analyze the AM receiver using envelope detector. 164
•
In the AM signal, we have carrier and both the sidebands. The transmitted AM wave is expressed mathematically as, S ( t ) = Ec [1 + mx (t) ] cos (2 π f c t)
Hence Ec cos (2 π f c t) is the carrier wave, x (t) is the message signal, and m is the modulation index. •
The average power of the carrier component is given by, Pc = Ec 2 / 2
•
The information bearing component in Equation is given by, mEc x m (t) cos (2 π f c t). The total average power associated with it is
•
Ec2 m2 P/2 where P is the average power of the message signal x ( t ). Em2
•
The average power of the full AM signal s ( t ) is there fore equal to Ec2 1 + m 2 P 2 Em2
•
The DSB-SC system has the message bandwidth of WNo. The channel signal to noise ratio of AM is therefore given by, Ec2 (1 + m2 P) Em2 ––––––––––– 2WNo
•
Now let us calculate the output signal-to-noise ratio (SNRo). For this let us represent the filtered noise n (t) in terms of its in phase and quadrature components nI (t) and nQ (t).
•
The filtered signal f (t) applied at the envelope detector in the receiver model as shown in Fig.4.4
AM signal S(t)
Σ
Band-pass filter
f (t)
Envelope detector
Output signal y(t)
Noise ω(t) Fig 4.4Noisy model of AM receiver 165
f ( t)
=s(t)=n(t) = [Ec + Ec mx (t) + nI ( t ) ] cos (2 π f c t) – nQ ( t ) sin (2 π f c t)
Resultant y (t) nQ(t)
Ec [1 + mx (t)]
nI(t)
Fig 4.5 Phasor diagram of AM wave plus narrow band nois •
The components which comprise the signal f ( t ) are s ( t ) and
n(t)
and they are represented by a phasor diagram as shown in Fig. 4.5. •
From the phasor diagram of fig (a) the receiver output y ( t ) can be obtained as, follows, y(t)
•
=
envelope of f ( t )
=
{ Ec2 [1 + mx (t) + nI ( t ) ]2 + nQ2 ( t )}1/2
This is the output of an ideal detector. Note that we have not considered the phase because an ideal envelope detector is completely insensitive to phase.
•
If the average carrier power is large as compared to the average noise power then the expression for y ( t ) can be approximated as follows : y(t) = Ec + Ec mx (t) + nI (t) Noise Message signal DC term
•
The DC term in the above expression can be neglected because it has no relation with the message signal. 166
•
The remaining terms in y (t) are similar to those in the DSB-SC receiver using coherent detection.
•
Hence the output signal to noise ratio can be approximately given by,
SNRo = •
Ec2 m2P Em2 –––––––– 2WNo
This expression for SNRo is valid if and only if the following two conditions are satisfied. 1. If the average carrier power is higher than the noise power. 2. The percentage modulation is adjusted to be less than or equal to 100%.
Figure of merit: The figure of merit for the A.M. receiver is given by, (SNR)o Figure of merit = ––––– = (SNR)c
=
Ec2 m2P 2 WNo –––––––– x ––––––––––– 2 WNo Em2 Ec2 (1 + m2 P) Em2
( m / Em )2 P ––––––––––––– 1 + ( m / Em )2 P
This is less than 1 Note: The figure of merit of DSBSC and SSB receivers with coherent detection is 1. But that of an AM receiver with envelope detection is less than 1. That means the noise performance of the AM receiver is poor as compared to that of the SSB or DSB-SC receovers. The inferior noise performance of the AM receivers is due to the transmission of carrier as a part of AM signal. Problem: Ex.1. Calculate the figure of merit of an AM receiver operating on single tone AM.
167
Soln: •
The message signal x ( t ) is given by, x ( t ) = Em cos (2 π fm t)
•
The AM signal corresponding to this signal is given by, s ( t ) = Ec [ 1 +m cos (2 π fm t)] cos (2 π fc t)
•
The average power of the modulating signal x ( t ) is given by, P
•
= ½ m 2 Ec 2
Hence output signal to noise ratio is, SNRo =
Ec2 (m / Em )2 x P ––––––––––––– 2 WNo
Substitue P = Em2 /2 ∴
•
Em2 –– 2
=
m2 Ec2 ––––– 4WNo
Channel signal to noise ratio is given by,
∴
•
SNRo =
Ec2 m2 ––––––– x 2 WNo Ec2
SNRc =
Ec2 [1 + (m / Em )2 x P] ––––––––––––––––– 2 WNo
SNRo =
Ec2 ––––– 2 WNo
x 1+
m2 Em2 ––––– = 2 Em2
Ec2 (1 + m2/2) ––––––––––– 2WNo
Hence the figure the merit is given by, Figure of merit
(SNR)o = ––––– (SNR)c
=
=
m2 Ec2 ––––– 4 WNo
4WNo x ––––––––– Ec2 [2 + m2]
m2 ––––– 2 + m2
where m represents the modulation index or percent modulation. Conclution:
168
1. m = 1 corresponds to 100% modulation. The figure of merit at m = 1 is given by, Figure of merit =
1 –––– 2+1
=
1 – 3
That means if all other parameters are identical then the average power that should be transmitted by an AM transmitter should be three times as large as the power transmitted by the suppressed carrier system with coherent detection in order to obtain the same quality of noise performance. 2. The figure of merit and hence the noise performance of AM system is dependent on the value of the modulation index m. 3. The figure of merit increases as the value of m is increased. So we get better noise performance with increase in the value of modulation index. Threshold Effect in AM Receivers: •
If the carrier to noise ratio is small as compared to unity (1), then the effect of noise dominates the signal and the performance of AM receiver (with envelope detector) will degrade to a great extent.
•
Under such working conditions it is more convenient to represent the narrowband noise n ( t ) in terms of its envelope r ( t ) and ψ ( t ) phase as follows, n(t) =
r ( t ) cos [ 2 π fc t + ψ ( t )]
Resultant y (t) Ec [1 + mx(t)] ψ (t) r (t) Fig 4.6 Phasor diagram for AM wave plus narrowband noise for low carrier to noise ratio
169
•
This expression is represented in the phasor form in fig 4.6. This phasor diagram corresponds to the detector input F ( t ) = s ( t ) + n ( t ).
•
In fig.(b) the noise has been used as reference because now noise is the dominant signal.
•
To the noise r ( t ) phasor is added to the signal term Ec [1 + mx ( t )]. The angle between them is ψ ( t ).
•
From fig.(b) we can directly obtain the approximate value of detector output y ( t ) is as follows: y ( t ) = r ( t ) + Ec cos [ψ ( t )] + mEc x (t) cos [ψ ( t )]
Conclusion : 1. The expression for y ( t ) indicates that when the carrier to noise ratio is small, the detector output does not contain any term which is directly proportional to message signal x ( t ). 2. The last term in Equation contains the message signal x ( t ) multiplied by the noise term cos [ψ ( t )]. 3. But the phase ψ ( t ) of the narrow band noise n ( t ) is uniformly distributed over 2π radians. Hence there will be a complete loss of information and there will not be any message term present in the detector output y ( t ). Threshold effect : •
The loss of message signal in the output of the envelope detector due to low carrier-to-noise ratio is called as threshold effect.
•
The meaning of word threshold is the value of carrier to noise ratio below which the performance of a deteriorates much more rapidly than the proportion decided by the carrier to noise ratio.
•
Every nonlinear detector such as envelope detector exhibits the threshold effect. But threshold effect does not arise in the coherent detector.
Noise in FM Receivers :
170
•
Now let us carry out the noise analysis of a frequency modulation (FM) system. The receiver model of an FM receiver is shown in Fig. 4.7
•
The noise ω (t) in fig. is a white Gaussian noise with zero mean value. Its power spectral density is No /2.
FM signal s(t) + Σ
Band-pass x(t) filter
Limiter
Discrimintor signal
+
v(t)
Noise ω(t)
Output signal
Baseband LPF
Fig 4.7 Noisy model of an FM receiver •
s ( t ) represents the received FM signal having a carrier frequency fc and transmission bandwidth BT.
•
We assume that almost all the transmitted power lies inside the frequency band fc + (BT / 2).
•
Bandpass filter : The bandpass filter of fig. has a midfrequency fc and bandwidth BT. This is necessary in order to pass the FM signal without any distrotion.
•
Noise : Since BT is small as compared to fc, it is possible for us to use the narrowband representation for n(t) . Not that n (t) represents the filtered version of the received noise ω(t). We can represent n ( t ) in the form of in phase and quadrature components nI ( t ) and nQ( t ) respectively.
•
Limiter : In FM, only the carrier frequency is changed and the amplitude of the FM wave is supposed to remain constant. But due to noise added to the FM wave, its amplitude changes. To avoid this from happening an amplitude limiter is connected after the bandpass filter as shown in fig.
•
The limiter actually clips the modulated waveform at the output of bandpass filter. Due to clipping the FM wave gets converted into a rectangular wave as shown in the Fig.4.8
171
FM wave
Limiter
Clipped FM wave
Fig 4.8 Action of the amplitude limiter •
The rectangular wave obtained at the output of the limiter is again rounded off by another bandpass filter which is an integral part of the limiter (not shown in fig.)
•
This is essential order to suppress any harmonics of fc generated in the process of clipping.
•
Thus at the output of the limiter will be sinusoidal FM wave, the amplitude of which is practically independent of the carrier at receiver input.
Discriminator : •
The FM discriminator consists of two parts namely a slope network having a purely reactive transfer function and an envelope detector.
•
Both these parts are implemented as integral parts of a single unit.
Post detection low pass filter : •
In Fig., a post detection low pass filter is connected after the discriminator. It is named as baseband low pass filter.
•
The bandwidth of this filter is just large enough to pass the highest frequency component of the message signal.
•
This filter will remove all the out of band frequency components and noise, present in the discriminator output. This is necessary to minimize the effects of noise on the output.
Analysis of Noise Performance of FM system: •
The output of the bandpass filter shown in Fig 4.9 consists of the filtered component n ( t ) of the input noise ω ( t ).
172
Fig 4.9 Noise model of FM receiver drawn partially •
The filtered noise n (t) can be expressed in terms of the inphase and quadrature component as follows: n ( t ) = nI ( t ) cos ( 2π fc t) - nQ ( t ) sin ( 2π fc t)
•
It is possible to express n ( t ) in terms of its envelope and phase as follows : n ( t ) = r ( t ) cos [ 2π fc t + ψ ( t )] Where the envelope is given by, r ( t ) = [ nI2 ( t ) + nQ2 ( t ) ]1/2 And the phase is given by, ψ ( t ) = tan
•
-1
nQ ( t ) –––––– nI ( t )
Note that the envelope r (t ) has a Rayleigh distribution and phase ψ ( t ) is distributed uniformly over 2π radians.
•
The FM wave at the input is given by, s ( t ) = Ec cos [ 2π fc t + 2π kf m ( t ) dt ] Where Ec = Carrier amplitude, fc = Carrier frequency and kf = Frequency sensitivity and m ( t ) = message signal
•
Let 2 π kf m ( t ) dt = φ ( t ). Hence the expression for FM wave is givenby, s ( t ) = Ec cos [ 2π fc t + φ ( t )]
•
The noisy signal at the bandpass filter output is given by, x(t) = s(t)+n(t) = Ec cos [ 2π fc t + φ ( t )] + r ( t ) cos [ 2π fc t + ψ ( t )]
173
x (t) can be represented by a phasor diagram as shown in Fig.4.10 Resultant x (t) r (t) θ(t) - φ ( t )
ψ (t) – φ (t)
Ec Fig 4.10 Phasor diagram of FM wave plus narrow band noise for high carrier to noise ratio •
Note that signal term Ec is being used as a reference. The resultant phasor x (t) has a phase θ(t). It can be obtained from the phasor diagram of fig. as, θ(t) - φ ( t )
= tan
-1
r (t ) sin [ψ (t) – φ (t)] –––––––––––––––––––– Ec + r ( t ) cos [ψ (t) – φ (t)] r (t ) sin [ψ (t) – φ (t)]
∴ θ(t)
•
= φ ( t ) + tan-1
––––––––––––––––––––– Ec + r ( t ) cos [ψ (t) – φ (t)]
We are not interested in the envelope variations of x (t) because they are removed by the limiter. But our focus of attention is on finding the error in the instantaneous frequency of the carrier wave due to filtered noise n ( t ).
Expression for noise in the detector output: •
If the discriminator is ideal, then it acts as a pure differentiator. Hence its output is proportional to θ` ( t ) / 2π where θ` ( t ) is the time derivative of input θ(t).
•
To simplify the analysis assume that the carrier to noise ratio at the discriminator input is large, compared to unity.
•
Also assume that the expression for θ(t) can be simplified as follows : θ(t) But
= φ(t)+
r(t) –––– Ec
sin [ψ (t) – φ (t)]
φ ( t ) = 2π kf m ( t ) dt 174
∴ θ(t) = 2π kf m ( t ) dt + r ( t ) sin [ψ (t) – φ (t)] Ec •
The out put of an ideal discriminator is given by, 1 v ( t ) = –– 2π =
dθ ( t ) ––––– dt
1 d 2π dt
∴ v(t)
=
2π kf m ( t ) dt + r ( t ) sin [ψ (t) – φ (t)] Ec kf m ( t ) + nd ( t )
Where nd ( t ) represents the noise term and it is given by, 1 nd ( t ) = –––– 2π Ec
d –– { r ( t ) sin [ψ (t) – φ (t)] } dt
Conclusion: From the expression of v ( t ), we conclude that if the carrier to noise ratio is high, then the discriminator output v ( t ) consists of the original message term m ( t ) with a multiplying factor kf ,plus the additional noise component nd ( t ). Simplification of the expression for nd ( t ) : •
If the carrier to noise ratio is high, then we can assume that the phase difference ψ (t) – φ (t) is uniformly distributed over 2π radians.
•
With assumption, the noise nd ( t ) at the output of the discriminator will be independent of the modulating signal and it will depend only on the characteristics of the carrier and narrowband noise.
•
Hence the expression for nd ( t ) can be simplified as follows : 1 nd ( t ) = –––– 2π Ec
•
d –– { r ( t ) sin [ψ (t) ] } dt
But r ( t ) sin [ ψ ( t ) ] = nQ ( t ) i.e. the quadrature component of the filtered noise n ( t ).
1
∴ nd ( t ) = –––– 2π Ec
d nQ ( t ) –––––– dt
175
•
This expression shows that the additive noise nd ( t ) which appears in the discriminator output is dependent only on carrier amplitude Ec and the quadrature component nQ ( t ) of the narrow band noise n ( t )
Output Signal to Noise Ratio : •
It is the ratio of average output signal power to the average output noise power.
•
Equation states that the discriminator output is given by v ( t ) = kf m ( t ) + nd ( t )
•
Hence the message component in the output of the discriminator output and hence in the low pass filter output is equal to kf m ( t ).
•
So the average signal power is equal to kf2 P, where P represents the average power of message signal m ( t ). ∴ Average signal power = kf2 P
•
Now let us obtain the average noise power at the output of the discriminator. We know that the noise in the discriminator output i.e. nd ( t ) is proportional to derivative of nQ ( t ). F
•
dx ( t ) We also know that –––– dt
•
Hence we can obtain the noise process nd ( t ) by passing nQ ( t ) through
j 2π f X ( f )
a linear filter with a transfer function equal to, j 2π f jf –––– = – 2π Ec Ec •
Therefore the relation between the power spectral density of nd ( t ) i.e. SNd ( f ) and power spectral density of the quadrature component
nQ ( t )
i.e. SNQ ( f ) is
176
SNd ( f ) •
=
f2 – SNQ ( f ) Ec 2
Assume that the bandpass filter in the FM receiver model has an ideal frequency response, with midband frequency fc and the bandwidth Bf .
•
Therefore the quadrature component nQ ( t ) or n ( t ) exhibits the ideal low pass characteristics as shown in Fig 4.11(a).
•
The power spectral density of nd ( t ) is shown in Fig 4.11 (b)and given athematically as
SNd ( f )
=
No f2 ––– Ec2
= 0 •
BT ….. | f | < – 2 ….. elsewhere
In the noisy FM receiver model a low pass filter is connected at the output of the frequency discriminator. The bandwidth of this filter is equal to the message bandwidth W.
•
If the type of FM is wideband FM, then W < < BT / 2 where BT is the transmission bandwidth of FM signal.
•
This filter will reject all the out of band components ( components outside the band – W < f < W ) of nd ( t ). Hence the power spectral density of the output noise no ( t ) appearing at the receiver output is given by,
SNo ( f )
=
No f2 ––– Ec2
= 0 •
….. | f | < W ….. elsewhere
The power spectral density SNo ( f ) has been shown in Fig 4.11©.
177
Fig 4.11 (a) (b) and (c) (a) PSD of nQ ( t ) (b) PSD of noise nd ( t )at the discriminator output (c) PSD of no ( t ) at the receiver output •
The average noise power in the output is obtained by integrating the power spectral density from -W to + W as follows : ∴ Average output noise power = SNo ( f ) df =
No f2 df –– Ec 2
=
2 No ––– W 3 3 Ec2
=
No –– Ec2
f3 – 3
W
-W
Note : Equation indicates that the average output noise is power is inversely proportional to average carrier power E / 2. Therefore with increase in carrier power the output noise will reduce •
We have obtained the values of average output signal power and average noise output power. So the output signal to noise ratio is given by, Average output signal power kf2P SNRo = ––––––––––––––––––––––– = ––––––––––– Average output noise power 2 No W 3 / 3 Ec2 3 Ec2 kf2P = ––––––– 2 No W 3
178
Channel Signal - to - Noise Ratio : •
The average power in the FM wave s ( t ) is Ec2 / 2.
•
The average noise power in the message bandwidth is WNo . Hence the channel signal to noise ratio is given by, SNRC =
Ec 2 –––– 2WNo
Figure of Merit : •
Figure of merit
SNRo = –––– SNRC =
•
=
3 Ec2 kf2P –––––– 2 No W 3
2 WNo x –––– Ec2
3 kf2P ––––– W2
But deviation ratio D = ∆f / W. So it is proportional to the ratio
kf
P1/2 / W. Hence figure of merit α D2 . Conclusion : So the figure of merit of a wideband FM is proportional to the square of deviation ratio. •
But in wideband FM the transmission bandwidth BT is approximately proportional to the deviation ratio.
•
Hence for a wideband FM with high carrier-to-noise, if we increase the transmission bandwidth, then the output signal to noise ratio and figure of merit increases in the square proportion.
•
Thus in FM system the noise performance improves with increase in bandwidth.
Noise Performance of a Single Tone FM : •
The modulating signal is sinusoidal for the single tone FM. The frequency modulated wave with single tone FM is as follows : s ( t ) = Ec cos 2π fc t + ∆f sin ( 2π fm t ) fm
179
•
Hence we can substitute ∆f sin ( 2π fm t ) = 2π kf m ( t ) dt fm Differentiate both the sides with respect to time to get, ∆f 2π fm cos ( 2π fm t ) fm m(t)
∴ •
= 2π kf m ( t ) = ∆f cos ( 2π fm t ) kf
This shows that the amplitude of message signal m ( t ) is ∆f / kf . So the average power of the message signal developed across a 1Ω resistor is P
•
= ( ∆f )2 2 kf2
The output signal to noise ratio for the output signal to noise ratio is given by, SNRo =
But
2
kf P
=
∴ SNRo
Where mf
=
3 Ec2 kf2P ––––––– 2 No W 3 ( Χ )2 –––– 2 =
3 Ec2 ( ∆f )2 –––––––– 4 No W 3
=
3 Ec2 ( mf )2 –––––––– 4 No W
∆f –– W
is the modulation index.
Figure of merit : • •
3 Ec2 ( mf )2 2 WNo 3 SNRo Figure of merit = –––– = ––––––––– x ––––– = – (mf)2 4 No W Ec2 2 SNRc Thus the figure of merit is directly proportional to the square of modulation
index.
180
Comparison of AM and FM: •
Assume that the modulation index for both the system is adjusted to 1.
•
Hence as proved earlier the figure of merit of AM system will be 1/3. But the figure of FM system will be 3/2. Moreover we can increase the value of modulation index mf of FM beyond 1. This will increase the value of figure of merit further.
•
Therefore the noise performance of FM system will always be better than that of AM.
FM Threshold Effect : •
We have derived the expression for output signal to noise ratio of FM receiver as, SNRo
=
3 Ec2 kf2P ––––––– 2 No W 3
•
But this expression is valid only if the carrier to noise ratio is high.
•
However practically it is observed that as the input noise increases, and the carrier to noise ratio decreases, the FM receiver "breaks".
•
First of all individual clicks are heard in the receiver output. This is the indication of the receiver begins to break.
•
If the carrier to noise ratio reduces further, then the clicks merge into a crackling or sputtering sound.
•
The expression for SNRo begins to fail as the breaking point is approached.
•
The failure of the expression for SNRo is evident when it starts predicting higher values of SNRo than the actual ones.
•
This phenomenon is called as threshold effect. The threshold is defined as the minimum carrier to noise ratio below which the SNRo deteriorates very fast.
181
Analysis of FM Threshold Effect : •
Assume first that no signal is present, and the carrier wave is unmodulated. Then the frequency discriminator output is given by, x ( t ) = [ Ec + nf ( t ) ] cos ( 2π fc t ) – nQ ( t ) sin (2π fc t)
•
This expression can be represented in the phasor form as shown in Fig.
•
As the amplitude and phase of nI ( t ) and nQ ( t ) change with time, the point "A" fig. will move in a random manner around point "B".
•
If carrier to noise ratio is large then Ec is large as compared to r ( t ) and so the moving point A will be close to point B. So angle θ ( t ) is approximately given by, nQ ( t ) ––––– Ec
θ(t) = •
But if the carrier to noise ratio is small, then r ( t ) will be large as compared to Ec and the moving point A will sometimes sweep around the origin and θ ( t ) increases or decreases by 2π radians.
•
From the phasor diagram of x ( t ) we can obtain the condition for the clicks to occur. A positive going click will occur when the envelope r ( t ) and phase ψ ( t ) of the filtered noise n ( t ) satisfy the following condition :
•
r(t) > Ec ψ(t) < π < ψ(t)+dψ(t) dψ(t) –––––– > 0 dt The conditions for a negative going click are as follows : r(t) ψ(t) dψ(t) –––––– dt
> Ec > -π > ψ(t)+dψ(t) <
0
182
FM Threshold Reduction: •
In some applications such as space communication using FM, we have to reduce the noise threshold.
•
This is necessary to operate the FM receiver with minimum possible signal power.
•
Threshold reduction can be achieved by using an FM demodulator with a negative feedback. Such a demodulator is called as FMFB demodulator. Another way to reduce the threshold is by using a phase-locked loop demodulator.
PLL Demodulator: •
The PLL demodulator is also called as extended threshold demodulator. The principle of such a demodulator is illustrated in Fig.
•
The amount of threshold extension shown in Fig. is measured with respect to a standard frequency discriminator which does not use negative feedback.
FMFB Demodulator : •
Fig. shows the block diagram of FMFB demodulator.
•
The instantaneous frequency of VCO is controlled by the demodulator output.
•
Assume that VCO is taken out from the circuit and the feedback loop is broken. Also assume that a wideband FM signal is applied at the input of the demodulator and another FM signal having a modulation index slightly smaller than that of the first one is applied to the other input of the mixer.
183
•
The output of mixer will consist of only the difference component because the sum frequency component is removed by the bandpass filter.
•
The mixer output will contain an FM wave with a smaller modulation index since the modulation indices of the two inputs to mixer would subtract.
•
This FM is passed through a bandpass filter. It is now evident that the second wideband FM applied to the mixer can be obtained by feeding the output of frequency discriminator back to VCO.
•
It can be proved that the signal to noise ratio of an FMFB receiver is same as that of a conventional FM receiver, if the carrier to noise ratio is large enough.
•
The question is that how dew the threshold extension take place? The process of threshold extension takes place as follows:
•
An FMFB receiver is a tracking filter which is capable of tracking only the slowly varying frequency of a wideband FM.
•
Therefore it responds only to a narrow band of noise which is centered about the instantaneous carrier frequency. The noise bandwidth to which the FMFB receiver responds corresponds to the band of noise tracked by the VCO .
•
Hence the FMFB receiver can obtain a threshold extension of the order of 5 to 7 dB which is a very good improvement.
•
The PLL demodulator also is a tracking filter. But the improvement in threshold extension is typically 2 to 3 dB which is not as high as that of FMFB receiver.
Pre-emphasis and De-emphasis : Pre-emphasis : •
It has been proved that in FM, the noise has a greater effect on the higher modulating frequencies.
•
This effect can be reduced by increasing the value of modulation index ( mf ) for higher modulating frequencies ( fm ). This can be done by 184
increasing the deviation " δ " and δ can be increased by increasing the amplitude of modulating signal at higher modulating frequencies. •
Thus if we "boost" the amplitude of higher frequency modulating signals artificially then it will be possible to improve the noise immunity at higher modulating frequencies. The artificial boosting of higher modulating frequencies
•
is
called
as
pre-emphasis.
Boosting of higher frequency modulating signal is achieved by using the pre-emphasis circuit of Fig(a).
•
The modulating AF signal is passed through a high pass RC filter, before applying it to the FM modulator.
•
As fm increases, reactance of C decreases and modulating voltage applied to FM modulator goes on increasing. The frequency response characteristics of the RC high pass network is shown in Fig(b).
•
The boosting is done according to this prearranged curve.
•
The amount of pre-emphasis in US FM transmission and sound transmission in TV has been standardized at 75 µsec.
•
The pre-emphasis circuit is basically a high pass filter. The pre-emphasis is carried out at the transmitter. The corner frequency for the RC highpass network is 2,122 Hz as shown in Fig.
185
De-emphasis : •
The artificial boosting given to the higher modulating frequencies in the process of pre-emphasis is nullified or compensated at the receiver by a process called "De-emphasis".
•
The artificially boosted high frequency signals are brought to their original amplitude using the de-emphasis circuit.
•
The 75 µsec de-emphasis circuit is standard and it is as shown in Fig. It shows that it is a low pass filter. 75 µsec de-emphasis corresponds to a frequency response curve that is 3 dB down at a frequency whose RC time constant is 75 µsec.
•
1 1 i.e. f = ––––– = ––––––––––– 2π RC 2π x 75 x 10-6 ∴ f = 2,122 Hz. The demodulated FM is applied to the De-emphasis circuit with increase in fm the reactance of C goes on decreasing and the output of deemphasis circuit will also reduce as shown in Fig.
FM versus AM : •
In general FM is considered to be superior to AM.
•
This is because FM has some significant advantages over AM as shown in Table. Advantages and disadvantages of FM compared to AM
Sr.No.
Advantages
Disadvantages
1.
Better noise immunity.
More bandwidth required.
2.
Better transmitter efficiency.
It is more complex and costly.
3.
Rejection of interference due to capture effect.
186
Advantages of FM over AM : •
FM has a better noise immunity as compared to that of AM for the reasons already discussed.
•
Another advantage of FM is that the interfering signals on the same frequency are effectively rejected.
•
When two or more FM signals occur simultaneously on the same frequency, the stronger signal of the two will capture the channel and will completely eliminate the weak signal. This is known as the capture effect in FM.
•
On the contrary when two AM signals occupy the same frequency, both of them will be heard. The weaker of the two will be heard in the background.
•
The third advantage of FM is its high transmission efficiency.
•
This is due to the use of highly efficient class C amplifiers used in the FM transmitters. It is possible to use the class C amplifiers for FM signal is a constant amplitude signal.
•
We can not use class C amplifiers for the AM transmitters because AM signal continuously varies in its amplitude.
•
Instead the less efficient linear amplifiers (class A, AB, B) are required to be used for AM transmission.
Disadvantages of FM as Compared to AM : •
The biggest disadvantage of FM over AM is that its bandwidth is too large. FM signal needs a much larger bandwidth as compared to that of AM to transmit the same information.
•
The bandwidth requirement of FM is reduced by using the narrow band FM (NBFM).
•
Another disadvantage of FM is that the circuits used for modulation and demodulation is too complex as compared to those used for AM.
187
Comparison performance:
188
UNIT V INFORMATION THEORY TOPICS 1. DISCRETE MESSAGES AND INFORMATION CONTENT. 2. AMOUNT OF INFORMATION. 3. AVERAGE INFORMATION, ENTROPY. 4. INFORMATION RATE. 5. SOURCE CODING. 6. SHANNON – FANO CODING. 7. HUFFMAN CODING. 8. LEMPEL ZIV CODING. 9. SHANNON’S CAPACITY AND CODING THEOREM. 10. MUTUAL INFORMATION. 11. CHANNEL CAPACITY.
189
Scope of Information Theory 1. Determine the irreducible limit below which a signal cannot be compressed. 2. Deduce the ultimate transmission rate for reliable communication over a noisy channel. 3. Define Channel Capacity - the intrinsic ability of a channel to convey information. Communication system The function of any communication system is to convey the information from source to destination. Discrete message Message which is selected from a finite number of predetermined messages. During one time one message is transmitted. During the next time interval the next from the set is transmitted. Memory source A source with memory for which each symbol depends on the previous symbols. Memoryless source Memoryless in the sense that the symbol emitted at any time is independent of previous choices. Probabilistic experiment involves the observation of the output emitted by a discrete source during every unit of time. The source output is modeled as a discrete random variable, S,which takes on symbols form a fixed finite alphabet. S = s0, s1, s2, · · · , sk-1 with probabilities P(S = sk) = pk, k = 0, 1, · · · ,K – 1 We assume that the symbols emitted by the source during successive signaling intervals are statistically independent. A source having the properties is
190
described is called discrete memoryless source, memoryless in the sense that the symbol emitted at any time is independent of previous choices. Source alphabet The set of source symbols is called the source alphabet. Symbols or letters The element of the set is called symbol. Uncertainty The amount of information contained in each symbols is closely related to its uncertainty or surprise. If an event is certain (that is no surprise, of probability 1) it conveys zero information. We can define the amount of information contained in each symbols. I(sk) = Here, generally use log2 since in digital communications we will be talking about bits. The above expression also tells us that when there is more uncertainty(less probability) of the symbol being occurred then it conveys more information. Unit of the information The unit of the information depends on the base of the logarithmic function. UNIT
b VALUE
Binit
2
Decit (OR) Hartley
10
Natural unit(nat)
e
When pk = ½ ,we have I(sk) = 1 bit. Some properties of information are summarized here: 1. For certain event i.e, pk = 1 the information it conveys is zero, I(sk) = 0. Absolutely certain of the outcome of an event. 2. For the events 0 ≤ pk ≤ 1 the information is always I(sk) ≥0.
191
Either provides some or no information, but never brings about a loss of information. 3. If for two events pk > pi, the information content is always I(sk) < I(si). The less probable the event is, the more information we gain when it occurs. 4. I(sksi) = I(sk)+I(si) if sk and si are statistically independent. Proof: p(sj, sk) = p(sj) p(sk) if sk and si are statistically independent. I (sj, sk) = log (1 / p(sj, sk)) = log (1 / p(sj) p(sk)) = log (1 / p(sj)) + = log (1 / p(sk)) = I(sj) I(sk) Average information or entropy The amount of information I(sk) produced by the source during an arbitrary signalling interval depends on the symbol sk emitted by the source at that time. Indeed, I(sk) is a discrete random variable that takes on the values I(s0), I(s1), · · · , I(sK-1) with probabilities p0, p1, · · · , pK-1 respectively. The mean of I(sk) over the source alphabet S is given by
The important quantity H(φ) is called the entropy of a discrete memoryless source with source alphabet φ. It is a measure of the average information content per source symbol. Note the entropy H(φ) depends only on the probabilities of the symbols in the alphabet φ of the source.
192
Problem 1. A DMS has four symbols S1 , S2, S3 S4 with probabilities 0.40, 0.30, 0.20, 0.10 a. Calculate H(φ). b. Find the amount of information contained in the message S1S2S3 S4 and S4S3S3 S2 , and compare with the H(φ). Solution a.
H(φ) = ∑ Pk log2(1/ Pk) = - 0.4 log2 0.4 - 0.3 log2 0.3 - 0.2log2 0.2 - 0.1log20.1 = 1.85 b/symbol
b.
P(S1S2S3 S4 ) = (0.4)(0.3)(0.2)(0.1) = 0.0096 I(S1S2S3 S4 )= - log2 (0.0096) = 0.60 b/symbol I(S1S2S3 S4 ) < 7.4 ( 4 H(φ) ) P(S4S3S3 S2) = (0.1)(0.2)2 (0.3) = 0.0012 I(S4S3S3 S2) = -log2(0.0012) =9.70 b/ symbol I(S4S3S3 S2) > 7.4 ( 4 H(φ) )
Some properties of entropy The entropy H(φ) of a discrete memoryless source is bounded as follows: 0 ≤ H(φ) ≤ log2(K) where K is the radix (number of symbols)of the alphabet S of the source. Furthermore, we may make two statements: 1. H(φ) = 0, if and only if the probability pk = 1 for some k, and the remaining probabilities in the set are all zero; this lower bound on entropy corresponds to no uncertainty. 2. H(φ) = log2(K), if and only if pk = 1/ K for all k; this upper bound on entropy corresponds to maximum uncertainty. Proof: H(φ) ≥0. Since each probability pk is less than or equal to unity, it follows that each term Pk
is always nonnegative. So H(φ) ≥0.
193
The term Pk
is zero if, and only if, pk = 0 or 1. That is pk =1 for
some k and all the rest are zero. H(φ) ≤ log2(K) To prove this upper bound , we make use of a property of the natural logarithm.
To proceed with this proof, consider any two probability distributions {p0, p1, · · · , pk-1 } and {q0, q1, q2, · · · , qk-1 } on the alphabet φ = {s0, s1, s2, · · · , sk-1 } of a discrete memoryless source. Then changing to the natural logarithm, we may write
Hence, using the inequality, we get
We thus have the fundamental inequality
Where the equality holds only if pk = qk for all k. Suppose we next put
194
So, Thus H(φ) is always less than or equal to log2k. This equality holds only if the symbols are equiprobable. Entropy of a binary memoryless channel Consider a discrete memoryless binary source shown defined on the alphabet φ = {0, 1}. Let the probabilities of symbols 0 and 1 be p0 and 1- p0 respectively. The entropy of this channel is given by
From which we observe the following: 1. When p0=0, the entropy H(φ) =0. 2. When p0=1, the entropy H(φ)=1. 3. The entropy attains its maximum value, Hmax = 1 bit, p0= p1= 1/2., that is, symbols 1 and 0 are equally probable.
Figure - Entropy of a discrete memoryless binary source Information rate If the source of the message generates messages at the rate of r messages per second, then the information rate is defined to be R= rH = average number of bits of information / second. Example problem 195
An analog signal is bandlimited to B Hz,sampled at the nyquist rate, and the
samples
are
quantized
into
four
levels.
The
quantization
levels
Q1,Q2,Q3,Q4(messages ) are assumed independent and occur with probabilities P1=P4=1/8 and P2=P3=3/8. Find the information rate of the source. Solution The average information H is H=
+ +3/8
=1/8
+ +3/8
+ +1/8
= 1.8 bits / message The information rate R is R= rH =2B(1.8) = 3.6 bits /s. Shannon source coding theorem An important problem in communication is the efficient representation of data generated by a discrete source. The process by which this representation is accomplished is called source encoding. The device that performs that representation is called a source encoder. Variable length code If some source symbols are known to be more probable than others, then the source code is generated by assigning short code to frequent source symbols, and long code to rare source symbols. EX : Morse code, in which the letters and alphabets are encoded into streams of marks and spaces, denoted as dots “.” And dashes “-“. Our primary interest is in the development of an efficient source encoder that satisfies two functional requirements: 1. The code words produced by the encoder are in binary form. 2. The source code is uniquely decodable, so that the original source sequence can be reconstructed perfectly from the encoded binary sequence. We define the average code word length, L, of the source encoder as L=
196
In physical terms, the parameter L represents the average number of bits per source symbol used in the source encoding process. Let Lmin denote the minimum possible value of L. We then define the coding efficiency of the source encoder as η = Lmin / L The source encoder is said to be efficient when η approaches unity. Source coding theorem: Given a discrete memoryless source of entropyH(φ) , the average codeword length L for any distortionless source encoding scheme is bounded as L ≥ H(φ) According to the source-coding theorem, the entropy H(φ)represents a fundamental limit on the average number of bits per source symbol necessary to represent a discrete memoryless source in that it can be made as small as, but no smaller than, the entropy H(φ).Thus with Lmin = H(φ), we may rewrite the efficiency of a source encoder in terms of the entropy H(φ)as η = H(φ)/ L Data Compaction: 1. Removal of redundant information prior to transmission. 2. Lossless data compaction – no information is lost. 3. A source code which represents the output of a discrete memoryless source should be uniquely decodable. Prefix Coding Consider a discrete memoryless source of alphabet {s0, s1, s2, · · · , sk-1} and statistics {p0, p1, p2, · · · , pk-1}. For each finite sequence of symbols emitted by the source, the corresponding sequence of code words is different from the sequence of code words corresponding to any other source sequence. For the above mentioned symbol, let the code word be denoted by {mk0, mk1, mk2, · · · , mkn-1} – the element are 0s and 1s. n - denotes the code word length
197
Prefix condition The initial part of the code word is represented by the elements {mk0, mk1, mk2, · · · , mki}Any sequence made up of the initial part of the code word is called prefix. Prefix code 1. The Prefix Code is variable length source coding scheme where no code is the prefix of any other code. 2. The prefix code is a uniquely decodable code. 3. But, the converse is not true i.e., all uniquely decodable codes may not be prefix codes.
From 1 we see that Code I is not a prefix code. Code II is a prefix code. Code III is also uniquely decodable but not a prefix code. Prefix codes also satisfies Kraft-McMillan inequality which is given by
Code I violates the Kraft – McMillan inequality. Both codes II and III satisfies the Kraft – McMillan inequality, but only code II is a prefix code.
198
Decoding procedure 1. The source decoder simply starts at the beginning of the sequence and decodes one codeword at the time. 2. The decoder always starts at the initial state of the tree. 3. The received bit moves the decoder to the terminal state if it is 0,or else to next decision point if it is 1.
Given a discrete memoryless source of entropy H(φ), a prefix code can be constructed with an average code-word length l, which is bounded as follows: H(φ)≤L≤ H(φ)+1 The left hand side of the above equation, the equality is satisfied owing to the condition that, any symbol sk is emitted with the probability
where, lk is the length of the codeword assigned to the symbol sk. Shannon – fano coding Procedure 1. List the symbols in order of decreasing probability. 2. Partition the set into two sets that are as close to equiprobable as possible, and assign 0 to the upper set and 1 to the lower set.
199
3. Continue the process, each time partitioning the sets with nearly equal probabilities as possible until further partitioning not possible. Example problems 1. A DMS has six symbols S1
,
S2, S3 S4, S5, S6, with corresponding
probabilities 0.30, 0.25, 0.20, 0.12, 0.08, 0.05. construct a Shannon – fano code for S. sk
pk
S1
0.30
S2 S3 S4 S5 S6
Step 1
Step 2
0
0
00
0
1
01
1
0
10
1
1
0
1
1
1
0
1110
1
1
1
1
1111
Step 3
Step 4
Step 5
0.25 0.20 0.12 110
0.08 0.05
200
2. A DMS has six symbols S1
,
S2, S3, S4 with corresponding
probabilities, 1/2 , 1/4, 1/8, 1/8, construct a Shannon – fano code for S. sk
pk
Step 1
S1
1/2
0
Step 2
Step 3
Step 4 0
S2 S3 S4
1/4 1
0
10
1
1
0
110
1
1
1
111
1/8 1/8
3. A DMS has five equally likely symbols S1 , S2, S3 S4, S5 construct a Shannon – fano code for S. sk
pk
Step
Step 2
Step 3
Step 4
1 S1 S2 S3 S4 S5
0.20 0
0
00
0
1
01
1
0
10
1
1
0
110
1
1
1
111
0.20 0.20 0.20 0.20
201
Huffman coding It is to assign to each symbol of an alphabet a sequence of bits roughly equal in length to the amount of information conveyed by the symbol in question. Algorithm 1.The source symbols are listed in order of decreasing probability. The two source symbols of lowest probability are assigned a 0 and a 1. This part of the step is reffered to as a splitting stage. 2. These two source symbols are regarded as being combined into a new source symbol with probability equal to the sum of the original probabilities. The probability of the new symbol is placed in the list in accordance with its value. 3. The procedure is repeated until we are left with a final list of source statistics of only two for which a 0 and a 1 are assigned. Symbol
Stage I
Stage II
S0,
0.4
0.4
0.4
S1
0.2
0.2
0.4
S2
0.2
0.2
0
S3
0.1 0
0.2
1
S4
0.1
Stage III
Stage IV
0.6
0
0
0.4
1
0.2
1
1
The code for each source symbol is found by working backward and tracing the sequence of 0s and 1s assigned to that symbol as well as its successors. Symbol
probability
code word
S0,
0.4
00
S1
0.2
10
S2
0.2
11
S3
0.1
010
S4
0.1
011
202
Drawbacks: 1. Requires proper statistics. 2. Cannot exploit relationships between words, phrases etc., 3. Does not consider redundancy of the language. Lempel-ziv coding 1. Overcomes the drawbacks of Huffman coding 2. It is an adaptive and simple encoding scheme. 3. When applied to English text it achieves 55% in contrast to Huffman coding which achieves only 43%. 4. Encodes patterns in the text This algorithm is accomplished by parsing the source data stream into segments that are the shortest subsequences not encountered previously. Let the input sequence be 000101110010100101......... We assume that 0 and 1 are known and stored in codebook subsequences stored : 0, 1 Data to be parsed: 000101110010100101......... The shortest subsequence of the data stream encountered for the first time and not seen before is 00 subsequences stored: 0, 1, 00 Data to be parsed: 0101110010100101......... The second shortest subsequence not seen before is 01; accordingly, we go on to write Subsequences stored: 0, 1, 00, 01 Data to be parsed: 01110010100101......... We continue in the manner described here until the given data stream has been completely parsed. The code book is shown below:
203
Numerical positions: 1
2
3
4
5
6
7
8
9
subsequences:
1
00
01
011
10
010
100
101
11
12
42
21
41
61
62
0
Numerical Repre sentations: Binary encoded blocks:
0010 0011 1001 0100 1000 1100 1101
Discrete Memoryless Channels Let X and Y be the random variables of symbols at the source and destination respectively. The description of the channel is shown in the Figure
The channel is described by an input alphabet {x0, x1, x2, · · · , xJ-1}, J – input alphabet size an output alphabet {y0, y1, y2, · · · , yK-1}, K – output alphabet size and a set of transition probabilities p(yk/xj) = p(Y= yk / X= xj ) for all j and k channel matrix or transition matrix
P=
p(y0/x0) p(y0/x1) . . . p(y0/xJ-1)
p(y1/x0) p(y1/x1)
……… ………
p(yK-1/x0) p(yK-1/x1)
p(y1/xJ-1)
………
p(yK-1/xJ-1)
204
each row corresponds to a fixed channel input. each column corresponds to a fixed channel output. Input probability distribution p(xj) , j=1,2,……J-1, the event that the channel input X= xj occurs with probability p(xj) = p(X=xj) for all j. joint probability distribution p(xj, yk) = p(X=xj, Y= yk) = p(yk/xj) p(xj) Marginal probability distribution of the output random variable Y is obtained by averaging out the dependence of p(xj, yk) on xj,
Binary Symmetric Channel – A discrete memoryless channel with J = K = 2. – The Channel has two input symbols(x0 = 0, x1 = 1)and two output symbols(y0 = 0, y1 = 1). – The channel is symmetric because the probability of receiving a 1 if a 0 is sent is same as the probability of receiving a 0 if a 1 is sent. – The conditional probability of error is denoted by p. Abinary symmetric channel is shown in Figure and its transition probability matrix is given by
205
Mutual Information If the output Y as the noisy version of the channel input X and H(X) is the uncertainity associated with X, then the uncertainity about X after observing Y , H(X|Y) is given by
The quantity H(X|Y) is called Conditional Entropy. It is the amount of uncertainity about the channel input after the channel output is observed. Since H(X) is the uncertainity in channel input before observing the output, H(X) H(X|Y) represents the uncertainity in channel input that is resolved by observing the channel output. This uncertainity measure is termed as Mutual Information of the channel and is denoted by I(X; Y).
206
Where the H(Y) is the entropy of the channel output and H(Y/X) is the conditional entropy of the channel output given the channel input. Properties of Mutual Information Property 1: The mutual information of a channel is symmetric, that is
I(X; Y) = I(Y;X) Where the mutual information I(X; Y) is a measure of the uncertainty about the channel input that is resolved by observing the channel output, and the mutual information I(Y;X) is a measure of the uncertainty about the channel output that is resolved by sending the channel output. Proof:
Substituting Eq.3 and Eq.10 in Eq.4 and then combining, we obtain
207
From Bayes’ rule for conditional probabilities, we have
Hence, from Eq.11 and Eq.12
Property 2: The mutual is always non-negative, that is I(X; Y) ≥0 Proof: We know,
Substituting Eq. 14 in Eq. 13, we get
Using the following fundamental inequality which we derived discussing the properties of Entropy,
Drawing the similarities between the right hand side of the above inequality and the left hand side of Eq. 13, we can conclude that
With equality if, only if, p(xj |yk) = p(xj)p(yk ) for all j and k. Property 2 states that we cannot lose information, on the average, by observing the output of a channel. Moreover, the mutual information is zero if,
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and only if, the input and output symbols of the channel are statistically independent. Property 3: The mutual information of a channel is related to the joint entropy of the channel input and channel output by where, the joint entropy (X, Y) is defined as
Proof:
Therefore, from Eq. 18 and Eq. 23, we have
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Channel Capacity Channel Capacity, C is defined as ‘the maximum mutual information I(X; Y) in any single use of the channel(i.e., signaling interval), where the maximization is over all possible input probability distributions {p(xj)} on X”
C is measured in bits/channel-use, or bits/transmission. Example: For, the binary symmetric channel discussed previously, I(X; Y) will be maximum when p(x0) = p(x1) = 1/2 . So, we have
Since, we know
Using the probability values in Eq. 3 and Eq. 4 in evaluating Eq. 2, we get
FIGURE-Variation of channel capacity of a binary symmetric channel with transition probability p.
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1. When the channel is noise free, p=0, the channel capacity C attains its maximum value of one bit per channel use. At this value the entropy function attains its minimum value of zero. 2. When the conditional probability p=1/2 due to noise, the channel capacity C attains its minimum value of zero,whereas
the entropy
function attains its maximum value of unity, in such a case the channel is said to be useless. Channel Coding Theorem: Goal: Design of channel coding to increase resistance of a digital communication system to channel noise. Channel coding Mapping of the incoming data sequence into channel input sequence. It is performed in the transmitter by a channel encoder. Channel decoding (inverse mapping) Mapping of the channel output sequence into an output data sequence. It is performed in the receiver by a channel decoder. The channel coding theorem is defined as process to introduce redundancy in order to reconstruct the original source sequence as accurately as possible. 1. Let a discrete memoryless source – with an alphabet φ – with an entropy H(φ) – produce symbols once every Ts seconds 2. Let a discrete memoryless channel – have capacity C – be used once every Tc seconds. 3. Then if,
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There exists a coding scheme for which the source output can be transmitted over the channel and be reconstructed with an arbitrarily small probability of error. The parameter C / Tc is called critical rate. 4. Conversly, if
it is not possible to transmit information over the channel and reconstruct it with an arbitrarily small probability of error. Example: Considering the case of a binary symmetric channel, the source entropy H(Φ) is 1. Hence, from the above equation, we have
But the ratio Tc / Ts equals the code rate, r of the channel encoder.
r≤C Hence, for a binary symmetric channel, if r ≤ C, then there exists a code capable of achieving an arbitrarily low probability of error. Information Capacity Theorem: The Information Capacity Theorem is defined as ‘The information capacity of a continuous channel of bandwidth B hertz, perturbed by additive white Gaussian noise of power spectral Density N0/ 2 and limited in bandwidth to B, is given by
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where P is the average transmitted power. Proof: Assumptions: 1. band-limited, power-limited Gaussian channels. 2. A zero-mean stationary process X(t) that is band-limited to B hertz, sampled at Nyquist rate of 2B samples per second 3. These samples are transmitted in T seconds over a noisy channel, also band-limited to B hertz. The number of samples, K is given by
We refer to Xk as a sample of the transmitted signal. The channel output is mixed with additive white Gaussian noise(AWGN) of zero mean and power spectral density N0/2. The noise is band-limited to B hertz. Let the continuous random variables Yk, k = 1, 2, · · · ,K denote samples of the received signal, as shown by ' The noise sample Nk is Gaussian with zero mean and variance given by
The transmitter power is limited; it is therefore
Now, let I(Xk; Yk) denote the mutual information between Xk and Yk. The capacity of the channel is given by
The mutual information I(Xk; Yk) can be expressed as
This takes the form 213
When a symbol is transmitted from the source, noise is added to it. So, the total power is P + σ2. For the evaluation of the information capacity C, we proceed in three stages: 1. The variance of sample Yk of the received signal equals P + σ2. Hence, the differential entropy of Yk is
2. The variance of the noise sample Nk equals σ2. Hence, the differential entropy of Nk is given by
3. Now, substituting above two equations into
bits per transmission. The number K equals 2BT. Accordingly, we may express the information capacity in the equivalent form:
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUBJECT CODE : EC2252 SEM / YEAR : IV / II SUBJECT NAME : COMMUNICATION THEORY UNIT I – AMPLITUDE MODULATIONS PART A 1. Define amplitude modulation and modulation index. (MAY/JUNE 2007) 2. Related to AM, what is over modulation, under modulation and 100% modulation? 3. Compare low level modulation and high level modulation.( NOV/DEC 2008) 4. Draw the frequency spectrum of VSB, where it is used? 5. Define modulation index of an AM signal. 6. Determine the expression for AM and give its frequency spectrum. (NOV/DEC 2006) 7. Draw the circuit diagram of an envelop detector. 8. What is the mid frequency of IF section of AM receivers and its bandwidth. 9. A transmitter radiates 9 kW without modulation and 10.125 Kw after modulation. 10. Determine depth of modulation. 11. Draw the spectrum of DSB. 12. Define the transmission efficiency of AM signal. 13. Draw the phasor diagram of AM signal. 14. What are the Advantages of SSB. (MAY/JUNE 2007) 15. Give the applications of SSB SC-AM. ( NOV/DEC 2008) 16. Disadvantages of DSB-FC. 17. What are the advantages of superhetrodyne receiver? 18. What are the Advantages of VSB. 19. Distinguish between low level and high level modulator. 20. Define FDM & frequency translation.
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21. Give the parameters of receiver. 22. Define sensitivity and selectivity. 23. Define fidelity. 24. What is meant by image frequency? 25. Define multitone modulation. 26. Suggest one application for AM, SSB, DSB and VSB modulation techniques and justify your answer. ( NOV/DEC 2005) 27. What is the Need for modulation? 28. What are the Applications of AM. 29. What is meant by diagonal clipping and negative peak clipping? 30. Define envelope. 31. Distinguish between linear and non linear modulator. 32. What are the limitations of AM? 33. Draw the envelope of AM. 34. Differentiate phase modulation and frequency modulation. 35. Suggest one application for AM, SSB, DSB and VSB modulation techniques and justify your answer. 36. When a signal m(t) = 3 cos (2p x 103t ) modulates a carrier c(t) = 5 cos (p x 106t), find the modulation index and transmission bandwidth if the modulation is AM. ( NOV/DEC 2005) PART B 1.
Explain the generation of AM signals using square law modulator. (16) . ( NOV/DEC 2005)
2.
Explain the detection of AM signals using envelope detector. (16) . ( NOV/DEC 2005)
3.
Explain about Balanced modulator to generate DSB-SC signal. ` (16)
4.
With a neat block diagram explain the SSB transmissions.
5.
Explain the operation of a ring modulator. State its advantages. ( NOV/DEC 2008)
6.
Explain about coherent detector to detect SSB-SC signal. (16)
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7.
Explain the generation of SSB using balanced modulator. NOV/DEC 2006
8.
Draw the circuit diagram of Ring modulator and explain with its operation?
9.
Discus the coherent detection of DSB-SC modulated wave with a block
diagram of detector and Explain. (16) 10. With a neat block diagram Explain frequency translation. ( NOV/DEC 2008) 11. Explain the working of Superheterodyne receiver with its parameters. (16) 12. Draw the block diagram for the generation and demodulation of a VSB signal and explain the principle of operation. (16) (MAY/JUNE 2007) 13. Write short notes on frequency translation and FDM? (16) 14. Explain about AM transmitters. (16) 15. Define sensitivity, selectivity and image frequency of a Receiver system. (MAY/JUNE 2007) UNIT II – ANGLE MODULATION PART A 1. What do you mean by narrowband and wideband FM? 2. Give the frequency spectrum of narrowband FM? 3. What is diversity reception? ( NOV/DEC 2008) 4. Why Armstrong method is superior to reactance modulator. 5. Define frequency deviation in FM? 6. State Carson’s rule of FM bandwidth? 7. Write thee expression for FM wave and give its frequency spectrum. NOV/DEC 2006. 8. Differentiate between narrow band and wideband FM.? ( NOV/DEC 2005) 9. What are the advantages of FM.? 10. State the disadvantages of FM. ( NOV/DEC 2008) 11. Define PM. 12. What is meant by indirect FM generation? 13. Draw the phasor diagram of narrow band FM. NOV/DEC 2006 14. Write the expression for the spectrum of a single tone FM signal. 15. What are the applications of phase locked loop?
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16. Define modulation index of FM and PM. 17. Differentiate between phase and frequency modulation. (MAY/JUNE 2007) 18. A carrier of frequency 100 MHz is frequency modulated by a signal x(t)=20sin (200¶x103t ). What is the bandwidth of the FM signal if the frequency sensitivity of the modulator is 25 KHz per volt? 19. What is meant by FM Threshold? Explain. . ( NOV/DEC 2005) 20. What is the bandwidth required for an FM wave in which the modulating frequency signal is 2 KHz and the maximum frequency deviation is 12 KHz? 21. A 15 KHz audio signal is frequency modulated with modulation index= 5.Calculate the transmission bandwidth of FM Signal. ( NOV/DEC 2005) 22. Determine and draw the instantaneous frequency of a wave having a total phase angle given by ø(t)= 2000t +sin10t. 23. Draw the block diagram of PLL. (MAY/JUNE 2007) 24. What are the applications of phase locked loop? NOV/DEC 2006. 25. How will you generate message from frequency-modulated signals? 26. What are the types of FM detectors? 27. What are the disadvantages of balanced slope detector?
PART B 1. Explain the indirect method of generation of FM wave and any one method of demodulating an FM wave. (16) NOV/DEC 2006 2. Derive the expression for the frequency modulated signal. Explain what is meant by FM. 3. Explain any two techniques of demodulation of FM. . (16) ( NOV/DEC 2006) 4. Explain the working of the reactance tube modulator and drive an expression to show how the variation of the amplitude of the input signal
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changes the frequency of the output signal of the modulator. (16) (MAY/JUNE 2007) 5. Discuss the effects of nonlinearities in FM. . (8) 6. Discuss in detail FM stereo multiplexing. . (8) 7. (i) Derive the expression for the frequency modulated signal. Explain what is meant by narrow –band FM and wideband FM using the expression NOV/DEC 2006. (ii) Discuss the indirect method of generating a wideband FM signal. 8. Draw the frequency spectrum of FM and explain. Explain how Varactor diode can be used for frequency modulation. . (16) Discuss the indirect method of generating a wide-band FM signal. (8) . ( NOV/DEC 2008) 9. Draw the circuit diagram of Foster-Seelay discriminator and explain its working. (16) 10. Explain the principle of indirect method of generating a wide-band FM signal with a neat
block diagram. (8) ( NOV/DEC 2006)
11. Explain the principle of indirect method of generating a wide-band FM signal with a neat diagram. 12. (ii) Differentiate narrow band and wide band FM. (MAY/JUNE 2007) UNIT III PART A 1. Define probability. 2. What are mutually exclusive events? 3. Define probability density function. 4. Define random variable 5. Define Random process. 6. Give the Laws of probability. 7. Define noise. 8. Give the classification of noise. 9. What are the types of External noise?
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10. What are types of internal noise? 11. Define noise figure. MAY/JUNE 2007, ( NOV/DEC 2006) 12. Define noise factor . ( NOV/DEC 2008), ( NOV/DEC 2006) 13. What is white noise MAY/JUNE 2007, ( NOV/DEC 2007), MAY/JUNE 2007 14. What is thermal noise? Give the expression for the thermal noise voltage across a resistor . ( NOV/DEC 2008), ( NOV/DEC 2006) 15. What is shot noise ( NOV/DEC 2007) 16. Define noise temperature. 17. Define signal to noise ratio. 18. What is narrowband noise? 19. Give the representation of narrowband noise in terms of envelope and phase components. 20. Give the expression for equivalent noise temperature in terms of hypothetical temperature. 21. Give the Friss formula in terms of noise temperature. 22. What is narrowband noise? 23. Define Partition noise. 24. Give the expression for noise voltage when several sources are cascaded. 25. What do you understand by additive white Gaussian Noise? Explain. ( NOV/DEC 2006) 26. Find the thermal noise voltage developed across a resistor of 700ohm. The bandwidth of the measuring instrument is 7MHz and the ambient temperature is 27’C. PART B 1. Derive the effective noise temperature of a cascade amplifier. Explain how the
various noise are generated in the method of representing them.(16)
2. Show how a narrow band noise can be represented as n(t) = nc(t) coswct - ns(t) sinwct where nc(t) and ns(t) are the in-phase and quadrature phase components of noise respectively. ( NOV/DEC 2006)
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3. Derive the Friis formula. Explain noise effect on bandwidth. ( NOV/DEC 2008) 4. What is noise temperature? Deduce the expression for effective noise temperature for a cascaded system. Explain narrow band noise.( NOV/DEC 2008) 5. Explain how the various noises are generated and the method of representing them. .(16) 6. Explain concept of noise equivalent Bandwidth ( NOV/DEC 2007) 7. Write notes on noise temperature and noise figure. (8) 8. What is meant by noise equivalent bandwidth? Illustrate it with a diagram. ( NOV/DEC 2006) 9. What is a narrow band noise ? Discuss the properties of the quadrature components of a narrowband noise. ( NOV/DEC 2006) 10. ( i) Derive the expression for output signaltonoise for a DSBSC receiver using coherent detection. (ii) Write short notes on noise in SSB receivers. ( NOV/DEC 2006) Derive the noise figure for cascade stages. (8) 11. Write short notes on : (i) Shot noise. (4) (ii) Thermal noise. (4) (iii) Noise figure and Noise temperature. (8) MAY/JUNE 2007 12. What is narrowband noise discuss the properties of the quadrature components of a narrowband noise. (8) 13. What is meant by noise equivalent bandwidth? Illustrate it with a diagram(8) 14. Derive the expression for output signal to noise for a DSB-SC receiver using coherent detection. . (16) 15. Write short notes on noise in SSB. (16) 16. Discuss the following: . (16) i) noise equivalent bandwidth (4) ii) narrow band noise (4)
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iii) noise temperature (4) iv) noise spectral density (4) 17. How sine wave plus noise is represented? Obtain the joint PDF of such noise component. (16)
UNIT IV – NOISE PERFORMANCE OF AM & FM RECEIVERS PART A 1. How to achieve threshold reduction in FM receiver? ( NOV/DEC 2007) 2. What is meant by FOM of a receiver? 3. What is extended threshold demodulator? 4. Draw the Phasor representation of FM noise. 5. Define pre-emphasis and de-emphasis. . ( NOV/DEC 2005), ( NOV/DEC 2007), MAY/JUNE 2007 6. What is capture effect in FM? . ( NOV/DEC 2008), MAY/JUNE 2007 7. What is the use of pre-emphasis and de-emphasis in FM systems? MAY/JUNE 2007, ( NOV/DEC 2006) 8. What is the SNR for AM with small noise case? 9. What do you understand by additive white Gaussian Noise? Explain. ( NOV/DEC 2006) 10. What is threshold effect with respect to noise? MAY/JUNE 2007 11. Define SNR. 12. Define CSNR. 13. Write the expression of the SNR for a synchronous detector . ( NOV/DEC 2008) 14. Discuss the factors that influence the choice of intermediate frequency in a radio receiver. 15. What are called extended threshold demodulators? ( NOV/DEC 2006) 16. What is TRF receiver? 17. What are the advantages of superheterodyne receiver over TRF?
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18. What is threshold effect PART B 1. Explain how threshold improvement is done through de-emphasis. ( NOV/DEC 2008) 2. Derive an expression for the output signal-to-noise ratio of an AM DSB-FC system. (8) MAY/JUNE 2007 3. Discuss the noise performance of AM system using envelope detection. (16) 4. Compare the noise performance of AM and FM systems. (16) ( NOV/DEC 2007) 5. Explain the significance of pre-emphasis and de-emphasis in FM system? (8) 6. Derive an expression for the output SNR of an FM receiver and hence obtain the figure of merit. ( NOV/DEC 2006). 7. Derive the noise power spectral density of the FM demodulation and explain its performance with diagram. (16). 8. Draw the block diagram of FM demodulator and explain the effect of noise in detail. 9. Explain the FM threshold effect and capture effect in FM? (16) 10. Discuss the following: (i) Noise equivalent bandwidth [MARKS 4] (ii) Narrow-band noise [MARKS 4] (iii) Noise Temperature [MARKS 4] (iv) Noise Power Spectral Density [MARKS 4] MAY/JUNE 2007 11. Explain the noise in AM receiver using its noisy model block diagram. (ii) What are pre emphasis and de emphasis in FM? Draw suitable circuits and Explain. ( NOV/DEC 2006) 12. Explain the capture effect and FM threshold effect. [MARKS 8](ii) Draw the block
diagram of FM demodulator and explain the effect of noise
in detail and compare the performance of AM and FM in the presence of noise( NOV/DEC 2006)
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UNIT V – INFORMATION THEORY PART A 1. What is entropy? ( NOV/DEC 2005) 2. What is prefix code? ( NOV/DEC 2006) 3. Define information rate. ( NOV/DEC 2007) 4. State any four properties of Entropy. ( NOV/DEC 2008) 5. Give the expressions for channel capacity of a Gaussian channel. ( NOV/DEC 2008) 6. What is channel capacity of binary synchronous channel with error probability of 0.2? 7. State channel coding theorem. 8. Define entropy for a discrete memory less source. 9. What is channel redundancy Write down the formula for the mutual information. 10. When is the average information delivered by a source of alphabet size 2, maximum? 11. Name the source coding techniques. 12. State channel capacity theorem and Explain. ( NOV/DEC 2006) 13. Write down the formula for mutual information. 14. Write the expression for code efficiency in terms of entropy. 15. Define entropy of a discrete memory less source emitting K symbols. ( NOV/DEC 2006) 16. Is the information of a continuous system non negative? If so, why? 17. What is channel capacity of Binary Synchronous Channel with error probability of O.2? . ( NOV/DEC 2007) 18. Explain the significance of the entropy H(X/Y) of a communication system where X is the transmitter and Y is the receiver. An event has six possible outcomes with probabilities ½.1/4,1/8,1/16,1/32,1/32. Find the entropy of the system.
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19. Define Shannon-Hartley theorem. . ( NOV/DEC 2005) PART B 1) Discuss Source coding theorem, give the advantage and disadvantage of channel coding in detail, and discuss the data compaction. (16) 2) Define mutual information. Find the relation between the mutual information and the joint entropy of the channel input and channel output. ( NOV/DEC 2006) 3) Explain in detail Huffman coding algorithm and compare this with the other types of coding. (8) 4) Explain the properties of entropy and with suitable example, explain the entropy of binary memory less source. (8) 5) Give the advantage and disadvantage of channel coding in detail. Discuss the data compaction. MAY/JUNE 2007 6) What is entropy? Explain the important properties of entropy. (8) 7) Five symbols of the alphabet of discrete memory less source and their probabilities are given below. (8) S=[S0,S1,S2,S3,S4] P[S]=[.4,.2,.2,.1,.1] Code the symbols using Huffman coding. 8) Write short notes on Differential entropy, derive the channel capacity theorem and discuss the implications of the information capacity theorem. . (16) MAY/JUNE 2007 9) What do you mean by binary symmetric channel? Derive channel capacity formula for symmetric channel. (8) 10) Construct binary optical code for the following probability symbols using Huffman procedure and calculate entropy of the source, average code Length, efficiency, redundancy and variance? 0.2, 0.18, 0.12, 0.1, 0.1, 0.08, 0.06, 0.06, 0.06, 0.04 (16)
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11) Define mutual information. Find the relation between the mutual information and the joint entropy of the channel input and channel output. Explain the important properties of mutual information. . (16) 12) Derive the expression for channel capacity of a continuous channel. Find also the expression for channel capacity of continuous channel of a infinite bandwidth. Comment on the results. (16) 13) A discrete memory less has a alphabet given below. Compute two different Huffman
codes for this source, hence for each of the two Codes. Find i. The average code-word length. ii.
The variance of the average code-word length over the ensemble of source symbol. ( NOV/DEC 2008)
SYMBOL S0 S1 S2 S3 S4 PROBABILITY 0.55 0.15 0.15 0.10 0.05 14) Show that the joint entropy H(X,Y) = H(X) +H(Y/X) 15) Find the joint entropy and redundancy of a source producing 4 symbols A, B, C and D, which are related with the following probabilities. ( NOV/DEC 2006) iABCD P(i) 0.3 0.2 0.4 0.1 P(j/i) j A B C i A 0 0.8 0.2 B 0.5 0.5 0.0 C 0.5 0.4 0.1 16) What is the capacity of the Discrete memory less channel? (2) (ii) A Discrete memory less channel has the following alphabet with probability of occurrence. Symbol I SO Sl S2 S3 S4 S5 S6 Probability : 0.125 0.0625 0.25 0.0625 0.125 0.125 0.25 Generate the Huffman code, Find average encoded Length, entropy and n. (14) ( NOV/DEC 2007)
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17) Prove that the entropy of a discrete memoryless source IS maximised when the symbols are equiprobable. (6) (ii) A source has five outputs denoted by [ml, m2, m3, m4, ms] with respective probabilities [0.41,0.19,0.16,0.15,0.09]. Determine the code words to represent the source outputs using Shannon-Fano Encoding technique and determine its efficiency. (10) MAY/JUNE 2007 18) (i) Explain any two properties of Binary symmetric channel. (6) (ii) Encode the source symbols with following set of probabilities using Huffman coding. m = {0.4, 0.2, 0.12, 0.08, 0.08, 0.08, 0.04}. (10) B 2174 MAY/JUNE 2007
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