Unit - 2 Angle Modulation , Phase & Frequency Modulation

Question 1 Define angle modulation. Ans.

Angle modulation is the process in which the angle of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping constant amplitude of the carrier wave. The angle modulated wave is mathematically expressed as, s(t )  Ac cos 2fc t  (t ) where s (t ) is angle modulated wave, Ac is peak amplitude of the carrier, f c is carrier frequency and (t ) is instantaneous phase deviation (time varying phase). There are two types of angle modulation. 1.

Frequency Modulation (FM)

2.

Phase Modulation (PM)

Question 2 Write the advantages of angle modulation over analog modulation. Ans.

Advantage of angle modulation over analog modulation are : 1.

Noise reduction.

2.

Improved system fidelity (the fidelity is the ability of a receiver to reproduce all the modulating frequencies equally).

3.

More efficient use of power.

Question 3 Give applications of angle modulation. Ans.

Applications of angle modulation are : 1.

Radio broadcasting

2.

TV sound transmission

3.

Two way mobile radio

4.

Cellular radio

5.

Microwave communication

6.

Satellite communication

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Question 4 Why angle modulation is a non linear modulation? Ans.

In a non linear modulation scheme, new frequency spectrum components are generated that are not present in the modulating signal. Hence, angle modulation (both FM and PM) is a non linear modulation.

Question 5 Explain frequency modulation and phase modulation? Ans.

[CSVTU May 2008]

Frequency Modulation : Frequency modulation is defined as the modulation in which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal keeping its (carrier) amplitude and phase constant. Time domain equation of frequency modulated wave is given by,

s(t )  Ac cos (t )  Ac cos 2f c t  2k f  m(t ) dt  where k f is in Hz/volt. Phase Modulation : Phase modulation is defined as the modulation in which the phase of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) amplitude and frequency constant. Time domain equation of phase modulated wave is given by, sPM (t )  Ac cos (t )  Ac cos  2f c t  k p m(t )   Ac cos  2f c t  .cos(2f m t ) 

In a non linear modulation scheme, new frequency spectrum components are generated that are not present in the modulating signal. Hence, angle modulation (both FM and PM) is a non linear modulation. Question 6 What is frequency modulation? Derive the time domain equation for FM wave in terms of modulation index. Ans.

Frequency Modulation : Frequency modulation is defined as the modulation in which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal keeping its (carrier) amplitude and phase constant. Time domain equation of frequency modulated wave is given by,

s(t )  Ac cos (t )  Ac cos 2f c t  2k f  m(t ) dt  where k f is in Hz/volt.

s(t )  Ac cos (t )  Ac cos 2f c t  k f  m(t ) dt  where k f is in rad/volt-sec. For m(t )  Am cos(2f mt ) , s(t )  Ac cos 2fc t   sin(2f mt )

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Angle Modulation

If m(t ) is the message signal, then in an FM system, the instantaneous frequency deviation from the carrier frequency is proportional to the message signal.   fi (t )  fc   k f m(t )  fi (t )  fc   m(t ) Where, f c is frequency of the unmodulated carrier and k f is frequency sensitivity of the modulator expressed in hertz per volt. fi (t )  f c  k f m(t ) For m(t )  Am cos(2f mt ) , Frequency deviation, f  k f m(t ) max  k f Am

fi (max)  fc  k f m(t ) max  fc  f

…..(i)

fi (min)  fc  k f m(t ) min  fc  f

…..(ii)

On adding (i) and (ii), Carrier frequency, f c 

f i (max)  f i (min)

2 On subtracting (ii) from (i), Carrier swing, 2f  f i (max)  f i (min) Frequency deviation, f 

f i (max)  f i (min)

2 So, Carrier swing  2  Frequency deviation The instantaneous angular frequency of the FM signal is given by, fi (t )  f c  k f m(t ) where k f is in Hz/volt

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Multiplying both sides by 2 , 2fi (t )  2f c  2k f m(t ) i (t )  c  k f m(t ) where k f is in rad/volt-sec

Where, i (t ) 

d (t ) rad  dt sec

So, (t )   i (t ) dt  2 f i (t ) dt  2   f c  k f m(t )  dt (t )  2f c t  2k f  m(t ) dt

Also, (t )  c t  (t ) So, k f m(t ) 

(t )  2k f  m(t ) dt



1 d (t ) where k f is in Hz/volt 2 dt

Maximum frequency deviation, f  k f m ( t ) max 

1 d ( t ) 2 dt max

Time domain equation of frequency modulated wave is given by,

sFM (t )  Ac cos ( t )  Ac cos  2fc t  2k f  m( t ) dt  Let m(t )  Am cos(2f mt ) then,

Am

 m(t )dt  2f

sin(2f m t )

m

k f Am     A s(t )  Ac cos  2f c t  2k f  m sin(2f m t )   Ac cos 2f c t  sin(2f mt )  2  f f m m     f k f Am   mf Modulation index,   fm fm 

sFM (t )  Ac cos 2fc t   sin(2fm t )

Question 7 Define : (i) Modulation index in FM Ans.

[CSVTU May 2011] (ii) Frequency deviation in FM.

Frequency deviation in FM : Frequency deviation f represents the maximum departure of the instantaneous frequency fi (t ) of the FM wave from the carrier frequency f c . Its unit is Hz.

f  fi (t )  f c

max



1 d (t ) 2 dt max

Frequency deviation in terms of frequency sensitivity k f is given by, f  k f m(t ) max For m(t )  Am cos(2f mt ) , f  k f Am Frequency deviation in FM is directly proportional to the peak modulating voltage but independent of modulating frequency. Modulation index in FM : Modulation index is defined as the ratio of frequency deviation ' f ' to the modulating frequency ' f m ' . It is unit less.

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Angle Modulation

2-5

k f Am f Frequency deviation    mf fm fm Modulating frequency

Modulation index in FM is directly proportional to the peak modulating voltage but inversely proportional to the modulating frequency. Modulation index in FM : 1. can be greater than 1. 2. decides the bandwidth of the FM wave. 3. decides the number of sideband having significant amplitude. Question 8 Define deviation ratio and percentage modulation. Ans. Deviation ratio is defined as the ratio of the maximum frequency deviation to the maximum modulating frequency. f D  max f m (max) For angle modulation, the percentage modulation is defined as the ratio of actual frequency deviation to maximum allowable frequency deviation. Question 9 Explain phase modulation. Ans. Phase Modulation : Phase modulation is defined as the modulation in which the phase of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) amplitude and frequency constant. If m(t ) is the message signal, then in a PM system, the phase is proportional to the message signal. (t )  m(t )



( t )  k p m ( t )

Where k p is phase sensitivity or phase deviation constant in rad/volt. The angle modulated wave is mathematically expressed as, s(t )  Ac cos 2fc t  (t ) The phase modulated wave is mathematically expressed as, s(t )  Ac cos  2f c t  k p m(t )  where k p is in rad/volt.

Phase deviation or modulation index,   k p m(t ) max  m p Let m(t )  Am cos(2f mt ) then,   k p m(t ) max  k p Am Modulation index,   k p Am  m p Modulation index in PM is directly proportional to the peak modulating voltage but independent of modulating frequency. Frequency deviation, f  m p f m  k p Am f m Frequency deviation in PM is directly proportional to the peak modulating voltage and the modulating frequency. Time domain equation of phase modulated wave is given by,

sPM (t )  Ac cos (t )  Ac cos 2fc t  k p m(t )  Ac cos 2fc t  .cos(2f m t )

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The features of phase modulation are : 1.

The envelope of PM wave is a constant and equal to the amplitude of the unmodulated carrier.

2.

The zero crossings of a PM wave no longer have a perfect regularity in their spacing like AM wave. This is because instantaneous frequency of PM wave is proportional to time derivation of m(t ) .

Question 10 Describe with the help of block diagrams schemes for generating. (1) FM wave using PM

(2) PM wave using FM Or

Relate FM and PM.

[CSVTU May 2012] Or

Illustrate relationship between phase and frequency modulation. [CSVTU Dec 2008] Or State the relationship between phase and frequency modulator. [CSVTU Dec 2007] Or With a block diagram explain how a phase modulator generates frequency modulation and a frequency modulator generates phase modulation.

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Ans.

1.

Angle Modulation

2-7

Generation of FM (Frequency Modulator) using PM (Phase Modulator)

FM can be generated by first integrating m(t ) and then using the result as the input to a phase modulator as shown in above figure. t s(t )  Ac cos  2f c t  k p  m() d     

Substituting 2k f  k p , where k f is in Hz/volt and k p is in rad/volt.

 s(t )  Ac cos  2f c t  2k f  m() d   t



2.



Generation of PM (Phase Modulator) using FM (Frequency Modulator)

The PM signal can be generated by first differentiating m(t ) and then using the result as the input to a frequency modulator as shown in above figure. t d   s(t )  Ac cos 2f c t  2k f  m(t ) dt    dt  

s(t )  Ac cos  2f c t  2k f m(t )  Substituting 2k f  k p , where k f is in Hz/volt and k p is in rad/volt.

 s(t )  Ac cos  2f c t  k p m(t )  Question 11



Determine the power of any frequency modulated signal A cos c t  K f  m ( t ) . dt Ans.

Power of any sinusoidal signal is given by, Power 

Amplitude 2 A2  2 2



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Question 12 Determine the instantaneous frequency in hertz of the following angle modulated signals at t  0 .   (1) 10cos  200t   (2) 10 cos  20t  t 2  6  (3) cos(200t )cos(5sin2t )  sin(200t )sin(5sin2t )

Sol.

  (1) 10cos  200t   6 

s(t )  Ac cos (t )

i (t ) 

d (t )  200 dt



(t )  200t 



fi (t )  100 Hz

 6

Instantaneous frequency at t  0 is fi (t ) t  0  100 Hz

Ans.

(2) 10 cos  20t  t 2 

s(t )  Ac cos (t )

i (t ) 

d (t )  20  2t dt



(t )  20t  t 2



fi  (10  t ) Hz

Instantaneous frequency at t  0 is fi (t ) t  0  10 Hz

Ans.

(3) cos(200t )cos(5sin2t )  sin(200t )sin(5sin2t )

cos A cos B  sin A sin B  cos( A  B)

s(t )  cos(200t  5sin 2t ) s(t )  Ac cos (t )



(t )  200t  5sin 2t

d (t )  200  10 cos 2t  2(100  5cos 2t ) dt fi (t )  (100  5cos 2t ) Hz

i (t ) 

Instantaneous frequency at t  0 is fi (t ) t  0  95 Hz

Ans.

Question 13 An angle modulated signal is given by s(t )  5cos(12000t ) for 0  t  1 . Let the carrier frequency be 10000 rad/sec. (1) If s(t ) is an FM signal with k f  500 rad/sec-volt, determine the modulating signal m(t ) over the interval 0  t  1 . (2) If s(t ) is a PM signal with k p  500 rad/volt, determine m(t ) over 0  t  1 . Sol.

Angle modulated signal , s(t )  5cos (12000t ) Carrier frequency, c  10000 rad/sec (1) The FM wave is mathetatically expressed as,

s(t )  Ac cos c t  k f  m(t ) dt  where k f is in rad/volt-sec

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Angle Modulation

s(t )  5cos(12000t )  5cos(10000t  2000t ) On comparing, 2000t  k f  m(t ) dt  500  m(t ) dt  Modulating signal, m(t )  4

Ans.

(2) The PM wave is mathematically expressed as, s(t )  Ac cos c t  k p m(t )  where k p is in rad/volt

s(t )  5cos(12000t )  5cos(10000t  2000t ) On comparing, 2000t  k p m(t )  500m(t )  Modulating signal, m(t )  4t

Ans.

Question 14 An angle-modulated signal is described by s( t )  10cos  2(106 )t  0.1sin(103 )t  (a) Considering s(t ) as a PM signal with k p  10 , find m(t ) . (b) Considering s(t ) as an FM signal with k f  10 , find m(t ) . Sol.

(a) An angle modulated signal in PM is given by, sPM (t )  Ac cos c t  k p m(t )  Given : s (t )  10 cos  2(106 )t  0.1sin(103 ) t  On comparing, k p m(t )  0.1sin(103 )t

m(t )  

0.1 0.1 sin(103 )t  sin(103 )t kp 10

m(t )  0.01sin(103 )t

Ans.

(b) An angle modulated signal in FM is given by, sFM (t )  Ac cos c t  k f  m(t ) dt    Given : s (t )  10 cos  2(106 )t  0.1sin(103 ) t  On comparing, k f  m(t ) dt  0.1sin(103 ) t d 0.1sin(103 )t   0.11000 cos(103 )t dt  100 100 m(t )  cos(103 )t  cos(103 )t kf 10

k f m(t ) 



m(t )  10 cos(103 )t

Ans.

Question 15 An FM signal which is modulated by a 3 kHz sine wave reaches a maximum frequency of 100.02 MHz and a minimum frequency of 99.98 MHz. Find the following : (1) Carrier swing (2) Carrier frequency (3) Frequency deviation of the signal (4) Modulation index of the FM signal Sol. Modulating frequency, f m  3 kHz The highest and lowest frequencies attained by the FM signal are fi (max)  100.02 MHz f i (min)  99.98 MHz

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(1) Carrier swing, C.S .  2f  f i (max)  f i (min)  100.02  99.98  0.04 MHz (2) Carrier frequency, f c 

f i (max)  f i (min) 2

(3) Frequency deviation, f  

f 



100.02  99.98  100 MHz 2

f i (max)  f i (min) 2

Ans.

C.S . 2



100.02  99.98 0.04   0.02 MHz 2 2

(4) Modulation index of FM signal,   m f 

Ans.

Ans.

f 0.02 MHz   6.67 fm 3 kHz

Ans.

Question 16 The instantaneous frequency of a single tone FM signal varies between 95.95 MHz and 96.05 MHz. The modulating signal frequency is 2 kHz. Determine : (a) the carrier frequency. (b) the maximum frequency deviation. (c) the modulation index. Sol.

Modulating frequency, f m  2 kHz The highest and lowest frequencies attained by the FM signal are fi (max)  96.05 MHz f i (min)  95.95 MHz f i (max)  f i (min)

96.05  95.95  96 MHz 2 2 f i (max)  f i (min) 96.05  95.95   50 kHz (b) Frequency deviation, f  2 2 f 50   25 (c) Modulation index of FM signal,   fm 2 (a) Carrier frequency, f c 



Ans. Ans. Ans.

Question 17 In an FM system, a 7 kHz modulating (or baseband) signal modulates 107.6 MHz carrier wave so that the frequency deviation is 50 kHz. Find (1) carrier swing in the FM signal and modulating index m f (2) the highest and lowest frequencies attained by the FM signal. Sol.

Modulating frequency, f m  7 kHz

Frequency deviation, f  50 kHz Carrier frequency, fc  107.6 MHz

(1) Carrier swing in FM,  2f  2  50  103  100 kHz Modulation index in FM, m f 

f 50  10   7.143 fm 7  103

Ans.

3

Ans.

(2) The upper or highest frequency attained by FM signal is given by f c  f  107.6  106  50  103  107.65 MHz

The lower or lowest frequency attained by FM signal is given by f c  f  107.6  106  50  103  107.55 MHz

Ans. Ans.

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Question 18 A single tone FM is represented by the voltage equation as :

v ( t )  12sin  6  108 t  5sin1250t  Determine the following : (i) Carrier frequency (iii) The modulation index Sol.

(ii) Modulating frequency (iv) Maximum deviation

Given : v(t )  12sin 6 108 t  5sin1250t  The expression of FM wave is given by, s(t )  Ac cos(c t   sin mt ) On comparing, Ac  12 V , c  6  108 rad/sec ,  m  1250 rad/sec Modulation index,   5

Ans.

c 6 10   95.5 MHz 2 2  1250 Modulating frequency, f m  m   199 Hz 2 2 Frequency deviation, f  f m  5 199  995 Hz Carrier frequency, fc 

8

Ans. Ans. Ans.

Question 19 An FM wave is given by s( t )  20sin  6  108 t  7 sin1250t  . Determine (i) The carrier and modulating frequencies, the modulation index, and the maximum deviation. (ii) Power dissipated by this FM wave in a 100 ohm resistor. Sol.

Given : s(t )  20sin 6 108 t  7 sin1250t  The expression of FM wave is given by, s(t )  Ac cos(c t   sin mt ) On comparing, Ac  20 V , c  6  108 rad/sec ,  m  1250 rad/sec Modulation index,   7

Ans.

c 6 10   95.5 MHz 2 2  1250 Modulating frequency, f m  m   199 Hz 2 2 Frequency deviation, f  f m  7 199  1393 Hz Carrier frequency, fc 

8

Power dissipation in 100  resistor, P 

Ac2 202  2 W 2 R 2 100

Ans. Ans. Ans. Ans.

Question 20 An FM wave is represented by the voltage equation v ( t )  20sin(5  108 t  4sin1450 t ) . Find : (1) The carrier and modulating frequencies. (2) The modulation index and the minimum deviation of the FM wave. (3) Power dissipated by this FM wave in a 20  resistor.

Analog Communication Sol.

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Given : s (t )  20sin 5 108 t  4sin1450t  The expression of FM wave is given by, s(t )  Ac cos(c t   sin mt ) On comparing, Ac  20 V , c  5  108 rad/sec ,  m  1450 rad/sec

c 5 108   79.6 MHz 2 2  1450 Modulating frequency, f m  m   230.8 Hz 2 2 (2) Modulation index,   4 (1) Carrier frequency, fc 

Ans. Ans. Ans.

Frequency deviation, f  f m  4  230.8  923.2 Hz (3) Power dissipation in 20  resistor, P 

Ans.

2

2 c

A 20   10 W 2 R 2  20

Ans.

Question 21 The equation of an angle modulated voltage is v ( t )  10sin 108 t  3sin104 t  . What form of angle modulation is this? Calculate the carrier and modulating frequencies, the modulation index and deviation and power dissipated in a 100  resistor. [CSVTU May 2011] Sol.

Given : v(t )  10sin 10 t  3sin10 t  An angle modulated signal is given by, v(t )  Ac cos(c t   sin mt ) 8

4

This being a single-tone angle-modulated signal, unless m(t) is specified, we cannot determine whether it is an FM or PM signal. On comparing, Ac  10 V , c  108 rad/sec ,  m  104 rad/sec Modulation index,   3

Ans.

c 10   15.9 MHz 2 2  104 Modulating frequency, f m  m   1.59 kHz 2 2 Frequency deviation, f  f m  3 1.59  4.77 kHz 8

Carrier frequency, f c 

Power dissipation in 100  resistor, P  Question 22 Given the

angle

modulated

2 c

Ans. Ans. Ans. 2

A 10   0.5 W 2 R 2 100

signal

s(t )  10cos c t  2sin(2000t ) ,

Ans.

where

c  2 107 rad/sec. Find

Sol.

(a) the average transmitted power. (b) the peak phase deviation. (c) the peak frequency deviation. Is this an FM or PM signal? Explain. Given : s(t )  10cos c t  2sin(2000t ) An angle modulated signal is given by, (t )  Ac cos c t  sin mt   Ac cos c t  (t )

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Angle Modulation

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On comparing, Ac  10 V , m  2000 rad/ sec and   2 Modulating frequency, f m 

m 2000   1000 Hz 2 2

(a) Average transmitted power, P 

Ac2 102   50 W 2 2

Ans.

(b) Peak phase deviation,   (t ) max  2sin 2000t max  2 rad

Ans.

(c) Peak frequency deviation, f  f m  2 1000  2 kHz

Ans.

This being a single-tone angle-modulated signal, unless m(t) is specified, we cannot determine whether it is an FM or PM signal. Question 23 Find the expression for FM wave if a carrier wave of frequency 1 MHz and amplitude 3 volts is frequency modulated by a sinusoidal modulating signal frequency 500 Hz and of peak amplitude 1 volt. Sol.

Carrier frequency and amplitude, f c  1 MHz and Ac  3 V Modulating signal frequency and amplitude, f m  500 Hz and Am  1 V The frequency modulated wave is mathematically expressed as,

s(t )  Ac cos 2f c t  2k f  m(t ) dt  where k f in Hz/volt. k f Am   s(t )  Ac cos  2f c t  sin(2f m t )  f m   

s(t )  3cos 2106 t  2k f  cos(2 500t ) dt  kf   s(t )  3cos  2106 t  sin(2 500t )  500  

Ans.

Question 24 A carrier which attains a peak voltage of 5 volts has a frequency of 100 MHz. This carrier is frequency modulated by a sinusoidal waveform of frequency 2 kHz to such an extent that the frequency deviation from the carrier frequency is 75 kHz. The modulated waveform passes through zero and is increasing at time t = 0. Write the expression for the modulated FM wave. [CSVTU May 2010] Sol.

Carrier frequency and amplitude, fc  100 MHz and Ac  5 V Modulating frequency, f m  2 kHz

Frequency deviation, f = 75 kHz

Because the frequency modulated carrier waveform passes through zero and is increasing at t  0 , therefore, the FM signal must be a sine wave (signal). The frequency modulated wave is mathematically expressed as,

s(t )  Ac sin 2fc t   sin(2f mt )

FM Modulation index,  

f 75  103   37.5 fm 2  103

…..(i)

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Substituting all the values in equation (1),

s(t )  5sin  2108 t  37.5sin(2 2 103 t )  

s(t )  5sin  2 108 t  37.5sin(4103 t ) 

Ans.

Question 25 A carrier which attains a peak voltage of 45 volts has frequency of 100 MHz. This carrier is frequency modulated by a sinusoidal waveform of frequency 2 kHz to such extent that the frequency deviation from the carrier frequency is 75 kHz. The modulated waveform passes through zero and is increasing at time t  0 . Write an expression for the modulated carrier waveform. Sol.

Carrier frequency and amplitude, fc  100 MHz and Ac  45 V Frequency deviation, f = 75 kHz

Modulating frequency, f m  2 kHz

Because the frequency modulated carrier waveform passes through zero and is increasing at t  0 , therefore, the FM signal must be a sine wave (signal). The frequency modulated wave is mathematically expressed as,

s(t )  Ac sin 2fc t   sin(2f mt ) FM Modulation index,  

…..(i)

f 75  10   37.5 fm 2  103 3

Substituting all the values in equation (1),

s(t )  45sin  2108 t  37.5sin(2 2 103 t )  

s(t )  45sin  2108 t  37.5sin(4103 t ) 

Ans.

Question 26 Consider an angle-modulated signal s( t )  10cos (108 )t  10sin 2(103 )t  Determine the maximum phase deviation and the maximum frequency deviation. Sol.

Given : s(t )  10 cos (108 )t  10sin 2(103 )t  Comparing the given s (t ) with standard FM wave equation, s(t )  Ac cos (t ) So, (t )  c t  (t )  (108 )t  10sin 2(103 )t where (t )  10sin 2(103 )t The maximum phase deviation is given by,   (t ) max 

  10sin 2(103 )t

Now,

max

 10 rad

d (t )  10( 2)(103 ) cos 2(103 )t dt

The maximum frequency deviation is given by, f  f 

Ans.

1 10  2 103 cos 2(103 )t  10 kHz max 2

1 d (t ) 2 dt max Ans.

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Angle Modulation

2 - 15

Question 27 A single-tone modulating signal cos(15  103 t ) frequency modulates a carrier of 10 MHz and produces a frequency deviation of 75 kHz. Find (i) the modulation index and (ii) phase deviation produced in the FM wave. (iii) If another modulating signal produces a modulation index of 100 while maintaining the same deviation, find the frequency and amplitude of the modulating signal, assuming k f  15 kHz per volt. Sol.

Frequency deviation, f = 75 kHz

Carrier frequency, f c  10 MHz

Modulating signal, m(t )  cos(15 10 t )  cos(2 7.5  103 t )  cos(2f m t ) 3

Modulating frequency, f m  7.5 kHz (1) FM modulation index,  

f 75 103   10 f m 7.5 103

Ans.

(2) The FM wave is given by, FM  A cos(c t   sin mt ) for m(t )  Am cos(m t ) The instantaneous total phase shift is given by, i (t )  c t   sin mt

i  c t  (t )



(t )   sin mt  10sin mt

Phase deviation,   (t ) max  10sin mt max  10 rad

Ans.

Frequency deviation, f = 75 kHz

(3) Modulation index,   100 Frequency sensitivity, k f  15 kHz/volt



f fm



Modulating signal frequency, f m 

fm 

f 

75 103  750 Hz 100

Ans.

Frequency deviation, f  k f m(t ) max  k f Am  Modulating signal amplitude, Am 

f 75  5 V k f 15

Ans.

Question 28 A carrier wave of frequency 1 MHz and amplitude 3 volts is frequency modulated (FM) by a sinusoidal modulating signal frequency of 500 Hz and peak amplitude 1 volt. As a consequence, the frequency deviation is 1 kHz. The level of the modulating waveform (signal) is changed to 5 volt peak and the modulating frequency is changed to 2 kHz. Write the expression for the new modulated waveform. Sol.

Carrier frequency and amplitude, f c  1 MHz and Ac  3 V Modulating signal frequency and amplitude, f m  500 Hz and Am  1 V Frequency deviation, f = 1 kHz

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The frequency modulated wave is mathematically expressed as, k f Am   s(t )  Ac cos  2f c t  sin(2f m t )  where k f is in Hz/volt. fm  

  f s(t )  Ac cos  2f c t   sin(2f m t )   Ac cos 2f c t  sin(2f mt )  fm  



k f Am fm



f fm

kf 



f Am

1000  1 kHz/volt 1 Now, for the second case, Am '  5 volt and f m '  2 kHz Frequency sensitivity, k f 

k f Am '

Modulation index,  ' 

fm '



103  5  2.5 2  103

The new FM signal can be expressed by, s(t )  Ac 'cos 2fct   'sin(2f m ' t ) Substituting all the values, s(t )  3cos  2106 t  2.5sin(2 2 103 t )  

s(t )  3cos  2106 t  2.5sin(4103 t ) 

Ans.

Question 29 In an FM system, when modulating frequency is 500 Hz and modulating voltage is 2.5 V, frequency deviation produced is 5 kHz. If modulating voltage is now increased to 7.5 V, calculate the new value of frequency deviation. If audio frequency voltage is increased to 10 V and modulating frequency drops to 250 Hz. What is frequency deviation? Calculate modulation index for each case. Sol.

For Am  2.5 V and f m  500 Hz : Frequency deviation, f  5 kHz f  k f Am

Modulation index,  



kf 

f 5000   10 fm 500

f 5   2 kHz/Volt Am 2.5 Ans.

For Am '  7.5 V and f m '  500 Hz : Frequency deviation, f '  k f Am '  2  7.5  15 kHz Modulation index,  ' 

f ' 15000   30 fm ' 500

Ans. Ans.

For Am "  10 V and f m "  250 Hz : Frequency deviation, f "  k f Am "  2 10  20 kHz Modulation index,  " 

f " 20000   80 fm " 250

Ans. Ans.

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Angle Modulation

2 - 17

Question 30 The modulating signal in an FM wave is 500 Hz with amplitude 3.2 Volts and frequency deviation is 6.4 kHz. If the audio frequency voltage is now increased to 8.4 Volts, determine the new frequency deviation and modulation index. If the audio frequency voltage is raised to 20 Volts while the audio frequency is dropped to 200 Hz, determine the frequency deviation and modulation index. Sol.

For Am  3.2 V and f m  500 Hz : Frequency deviation, f  6.4 kHz f  k f Am

Modulation index,  

kf 



f 6.4   2 kHz/Volt Am 3.2

f 6400   12.8 fm 500

For Am '  8.4 V and f m '  500 Hz : Frequency deviation, f '  k f Am '  2  8.4  16.8 kHz Modulation index,  ' 

f ' 16800   33.6 fm ' 500

Ans. Ans.

For Am "  20 V and f m "  200 Hz : Frequency deviation, f "  k f Am "  2  20  40 kHz Modulation index,  " 

f " 40000   200 fm " 200

Ans. Ans.

Question 31 A 25 MHz carrier is modulated by 400 Hz audio frequency. If the carrier voltage is 4 V and maximum deviation of 10 kHz. Write down equations for : (i) FM wave, and

(ii) PM wave

If the modulating frequency is changed to 2 kHz. Write new equations for : (i) FM wave, and Sol.

(ii) PM wave

Carrier frequency and amplitude, f c  25 MHz and Ac  4 V Modulating signal frequency, f m  400 Hz Frequency deviation, f  10 kHz Modulation index in FM,  

f 10000   25 fm 400

Modulation index in PM, m p    25 Frequency modulated wave, sFM (t )  Ac cos 2fct   sin(2f mt ) 

sFM (t )  4 cos  2 25 106 t  25sin(2 400t ) 

Ans.

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Phase modulated wave, sPM (t )  Ac cos  2f c t  m p sin(2f m t )  

sPM (t )  4 cos  2 25 106 t  25sin(2 400t ) 

Ans.

New modulating signal frequency, f m '  2 kHz In PM, with a change in the modulating frequency the modulation index  changes as  depends inversely on modulating frequency. ' New modulation index in FM,  

f 10000  5 f m ' 2000

Frequency modulated wave, sFM (t )  Ac cos 2fct   sin(2f mt ) 

sFM (t )  4 cos  2 25 106 t  5sin(2 2000t ) 

Ans.

In PM, with a change in the modulating frequency the modulating index m p remains unchanged as m p is independent of modulating frequency. Phase modulated wave, sPM (t )  Ac cos  2f c t  m p sin(2f m t )  

sPM (t )  4 cos  2 25 106 t  25sin(2 400t ) 

Ans.

Question 32 Consider the frequency modulated signal

10cos  2 105 t  5sin(2  1500t )  7.5sin(2  1000t ) with f c  105 Hz. Find the modulation index. Sol.

Given : sFM (t )  10 cos  2105 t  5sin(21500t )  7.5sin(21000t )  The FM wave is mathematically expressed as,

sFM (t )  Ac cos c t  2k f  m(t ) dt  where k f is in Hz/Volt. On comparing, 2k f  m(t ) dt  5sin(21500t )  7.5sin(21000t ) 2k f m(t )  5  21500  cos(2 1500t )  7.5  2 1000 cos(2 1000t ) k f m(t )  7500 cos(21500t )  7500 cos(21000t )

Frequency deviation, f  k f m(t ) max  7500  7500  15 kHz Modulating frequencies, f m1  1500 Hz and f m 2  1000 Hz Maximum modulating frequency, f m  1500 Hz  1.5 kHz Modulation index,  

f 15   10 f m 1.5

Ans.

Question 33 A baseband signal m(t ) modulates a carrier to produce the following angle modulated signal s( t )  Ac cos  2  108 t  k P m( t )  , where m(t ) is shown in the below figure. Determine the value of k P so that the peak frequency deviation of the carrier is 100 kHz.

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Sol.

Angle Modulation

2 - 19

Given : s(t )  Ac cos  2108 t  k P m(t )  as an angle modulated signal. Since, phase deviation (t )  kP m(t ) i.e., proportional to message signal, therefore, it is a phase modulated signal. Now, instantaneous frequency is given by, i  c  k P

dm(t ) dt

Then, maximum frequency deviation is given by, i  c

max

 kP

dm(t ) dt max

Peak frequency deviation  100 kHz  100 103 Hz

100 103  2  kP

dm(t ) dt max

From the given figure,

…..(i)

dm(t ) 12.5  (5) 17.5    2.5 103 dt max 5  (2) ms 7 ms

Then, from equation (i), kP 

100 103  2  80 2.5 103

Ans.

Question 34 Given a signal s(t )  cos(2fc t )  0.2cos(2f m t )sin(2fc t ) (i) Prove that s(t ) is a combination of AM-FM signal. (ii) Draw the phasor diagram at t  0 . Or If v(t )  cos(c t )  0.2cos(m t )sin(c t ) , show that v(t ) is a combination of AM-FM Sol.

signal. (i) Given : s(t )  cos(2fc t )  0.2cos(2f mt )sin(2f c t ) Representation of quadrature signal is given by,   B  R  A cos   B sin   A2  B 2 cos   tan 1     A   The given signal s (t ) can be modified in the following form : 1/ 2

s(t )  1  {0.2cos(2f mt )}2 

cos  2f c t  tan 1  0.2cos(2f mt ) 

For small value of  , tan 1   From binomial expansion : (1  x) n  1  nx for small value of x  0.04  s(t )  1  cos 2 (2f mt )  cos  2f c t  0.2cos(2f mt )  2  

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1  cos 2 2  1  cos(4f m t )  s (t )  1  0.02   cos  2f c t  0.2 cos(2f m t )  2   

cos 2  

s(t )  1  0.01  0.01cos(4fmt ) cos 2fc t  0.2cos(2fmt ) s(t )

1  0.01cos(4fmt ) cos 2fct  0.2cos(2fmt )

From above equation, we can see that the amplitude as well as the instantaneous phase angle of the given signal changes in accordance with the modulating or message signal. Hence, the given signal s (t ) is a combination of AM-FM signal. (ii) For the construction of phasor diagram, we can express the signal s (t ) as under :

s(t )  cos(2fc t )  0.2cos(2f mt )sin(2f c t ) 2sin A cos B  sin( A  B)  sin( A  B)

s(t )  cos(2fct )  0.1sin 2( fc  f m )t   0.1sin 2( fc  f m )t  Phasor diagram for above equation has been shown in below figure.

Question 35 An angle-modulated signal is given by s( t )  5cos  2(106 )t  0.2cos 200t  Can you identity whether s(t ) is a PM or an FM signal? Sol.

Given : s (t )  5cos  2(106 )t  0.2 cos 200t 

…..(i)

For angle modulation, the modulated carrier wave is represented by,

s(t )  Ac cos ct  (t )

…..(ii)

In case of PM, (t )  kP m(t ) Where m(t ) is the message signal and kP is phase deviation constant in rad/volt. On comparing equation (1) and (2), k p m(t )  (t )  0.2 cos 200t Phase of carrier signal varies with message signal. Hence s (t ) is a PM signal. In case of FM, s(t )  Ac cos  2f c t  k f  m(t ) dt    Where k f is frequency deviation constant in rad/volt-sec.

…..(iii)

On comparing equation (i) and (iii), k f  m(t ) dt  0.2 cos 200t k f m(t ) 

d 0.2 cos 200t    0.2  200 sin 200t dt

From equation (iii), (t )  2f c t  k f  m(t ) dt

…..(iv)

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Instantaneous angular frequency is given by, i (t ) 

d (t )  2f c  k f m(t ) dt

From equation (iv), i (t )  2fc  0.2  200 sin 200t So, the instantaneous angular frequency of carrier signal varies with message signal. Hence s (t ) is an FM signal. Hence s (t ) can be either PM or an FM signal.

Ans.

Question 36 (i) Show with a phasor diagram that the signal s(t ) expressed by

s( t )  cos(2  106 t )  0.02cos  2(106  103 )t  represents a carrier wave which is modulated both in amplitude and frequency. (ii) Represent s(t ) in the form

s( t )  1  m cos(2  106 t )  cos  2  106 t   sin(2  103 t )  Find the value of m and  . Also, write an expression for instantaneous frequency as a function of time t. Verify that both amplitude and frequency vary approximately sinusoidally with frequency 1 kHz. Sol.

(i) The phasor diagram of the signal s(t )  cos(2106 t )  0.02 cos  2(106  103 )t  has been shown in below figure.

It may be easily observed from above figure that the resultant amplitude differs from the carrier amplitude and the resultant is not in phase with the carrier. Hence, the given signal is modulated both in amplitude and phase. Proved (ii) s(t )  cos(2106 t )  0.02 cos  2(106  103 )t 

cos( A  B)  cos A cos B  sin A sin B s(t )  cos(2106 t )  0.02 cos(2106 t ) cos(2103 t )  0.02sin(2106 t )sin(2 103 t )

s(t )  1  0.02 cos(2103 t )  cos(2106 t )  0.02sin(2106 t )sin(2 103 t ) Representation of quadrature signal is given by,

R  A cos   B sin  



  B  A2  B 2 cos   tan 1     A   2

s(t )  1  0.02cos(2103 t )   0.02sin(2103 t ) 



2 1/ 2



 0.02sin(2103 t )  cos 2106 t  tan 1  1  0.02cos(2103 t )  

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

2

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s(t )  1  0.02cos(2103 t )   0.02sin(2103 t ) 



2 1/ 2



 0.02sin(2103 t )  cos 2106 t  tan 1  1  0.02cos(2103 t )  

Let s (t )  x(t ) cos 2106 t  



2

x(t )  1  0.02cos(2103 t )   0.02sin(2103 t ) 



2 1/ 2

1/ 2

x(t )  1  0.0004cos 2 (2103 t )  0.04cos(2103 t )  0.0004sin 2 (2103 t )  1/ 2

x(t )  1  0.0004  0.04cos(2103 t ) 

1/ 2

1  0.04cos(2103 t )  From binomial expansion : (1  x) n  1  nx for small value of x

 0.04  x(t )  1  cos(2103 t )  1  0.02cos(2103 t )  2   3 0.02sin(2 10 t )   tan 1 tan 1 0.02sin(2 103 t ) 1  0.02 cos(2103 t ) 1  0.02 cos(2 103 t )

For small value of  , tan 1   

s(t )  1  0.02 cos(2103 t )  cos 2106 t  0.02sin(2103 t ) 

Given : s(t )  1  m cos(2 103 t )  cos  2 106 t   sin(2 103 t )  On comparing, m  0.02 and   0.02

Ans.

The instantaneous frequency is given by, 1 d  1 d 6 3 fi  (t )    210 t  0.02sin(210 t )  2  dt  2 dt 

fi (t )  106  20 cos(2 103 t )

Ans.

It can be observed that the amplitude and instantaneous frequency of modulated signal vary with cos(2  103 t ) i.e. with frequency 1 kHz. Question 37 Draw the phasor diagram for NBFM signal. Compare it with the phasor diagram of an AM signal. [CSVTU May 2010] Or Explain the differences between narrow band FM and AM with the help of phasor diagram. Sol. Frequency Modulation : Frequency modulation is defined as the modulation in which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal keeping its (carrier) amplitude and phase constant. The expression of FM signal in terms of modulation index is given by, s(t )  Ac cos 2fc t  sin 2fmt 

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Angle Modulation

cos( A  B)  cos A cos B  sin A sin B s(t )  Ac cos(2fc t ) cos( sin 2f mt )  Ac sin(2f c t )sin( sin 2f mt ) Mathematical identity : 1. cos   1 for small value of  . 2. sin    for small value of  . For NBFM   1 sNBFM (t )  Ac cos(2fc t )  Ac sin(2f c t )sin(2f mt )

sin A sin B  cos( A  B)  cos( A  B)

Ac  A cos 2( f c  f m )t  c cos 2( f c  f m )t …..(i) 2 2 The spectrum of NBFM consists of the carrier, upper sideband and lower sideband. Single sided spectrum of NBFM : sNBFM (t )  Ac cos(2f c t ) 

Bandwidth of NBFM is 2 f m . Phasor diagram of NBFM :

The single sided spectrum and phasor diagram for single-tone AM is given below : Am Am s AM (t )  Ac cos 2f c t  c a cos 2( f c  f m )t  c a cos 2( f c  f m )t …..(ii) 2 2 Single sided spectrum of AM :

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Phasor diagram of AM :

From equation (1) and (2) LSB of NBFM is 1800 phase shift with respect to LSB of AM. From the phasor diagram of NBFM, we see that the resultant of the phasors corresponding to the two sidebands is perpendicular to the carrier phasor. Since   1 , the resultant NBFM phasor has approximately the same amplitude as the carrier phasor but is out of phase with it. From the phasor diagram of AM, we see that the resultant of the phasors corresponding to the two sidebands is always in phase with the carrier phasor. Hence, the resultant AM phasor will have an amplitude different from that of the carrier phasor but is always in phase with it. Question 38 Derive an expression for sideband terms produced in the WBFM? Write down its effects on calculation of significant bandwidth. [CSVTU Dec 2008] Or Prove that in frequency modulation, there are produced sideband terms extending theoretically upto infinity. Ans.

The expression of FM signal in terms of modulation index is given by,

s(t )  Ac cos c t   sin mt 

…..(i)

Using Euler’s identity : e j  cos   j sin , we can express s (t ) as :

s(t )  Ac Re e jc t  e j sin( m t ) 

…..(ii)

Where Re() denotes the real part of () . Equation (ii) may be rewritten as :

s(t )  Ac Re e j ( c  sin m )t 

…..(iii)

Let us consider the second exponential term on the right hand of equation (ii) and denote it by v(t ) i.e. v(t )  e j sin(mt ) …..(iv) Where v(t ) is periodic with period T, where T 

2 . Hence, v(t ) can be expressed as m

a Fourier series. We choose the exponential Fourier series representation. Hence :

v(t )  e j sin( mt ) 



Ve

n  

n

jnm t

…..(v)

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where Vn is the exponential Fourier series coefficient given by : T /2

Vn 

T /2

1 1 v(t ) e  jnm t dt  e j sin( mt ) e  jnmt dt  T T / 2 T T/ 2

Put m t  x, then dt 

…..(vi)

1 T dx  dx . Substituting these in equation (vi), we obtain : m 2



Vn 

1 j   sin x  nx  dx e 2 

…..(vii)

The integral in equation (vii) has a standard solution given by J n () , which is referred to as the Bessel function of the first kind of order n and argument  . In the context of FM, n and  refer to the sideband number and modulation index, respectively. Thus : Vn  J n () and equation (v) becomes :

v(t )  e j sin mt 



J

n 

n

() e jnmt

…..(viii)

Substituting equation (viii) in equation (iii), we obtain       s (t )  Ac Re e jc t   J n ()e jnmt   Ac Re   J n ()e j ( c  nm )t  n     n   Some of the properties of the Bessel function, J n () are :

1.

It is a real value function. 

3.

J

n 

2 n

2.

…..(ix)

J n ()  (1) n J  n ()

()  1

Using property 1, equation (ix) can be written as

s(t )  Ac s(t )  Ac 1.



J

n 

n

() cos (c  nm )t 

n

() cos  2( fc  nf m )t 



J

n 

…..(x)

s (t ) contains an infinite number of components of the form

f c  nf m

where

n  0, 1,  2,..... so theoretically bandwidth is infinite. 2.

The power of s (t ) is distributed among the components. The power carried by the component f c  nf m is Ac2 J n2 () / 2 . The power is  Ac2 J n2 () / 2  Ac2 / 2 , which is reasonable since the overall FM signal has a constant amplitude Ac .

3.

Most of the power is carried by components

f c  nf m for n    1 . (Since

 J   1, we know that most of the power is carried by a finite number of 2 n

components. The value of   1 is determined empirically). 4.

For small  , using the approximations for the Bessel functions, we have

Ac cos 2( f c  f m )t  cos 2( f c  f m )t 2  s(t )  Ac cos(2fc t )  Ac sin(2f mt )sin(2f c t ) s(t )  Ac cos(2f c t ) 

Which is narrowband FM approximation.

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Equation (x) can be expanded to yield : s(t )  Ac J 0 () cos(c t )  Ac  J1 () cos (c  m )t  J 1 () cos (c  m )t  Ac  J 2 () cos (c  2m )t  J 2 () cos (c  2m )t  ......

Using property 2 of the Bessel function, we note that, J n ()  J  n () for even value of n, and J  n ()   J n () for odd values of n, we obtain : s (t )  Ac J 0 () cos(c t )  Ac J1 () cos (c  m )t  cos (c  m )t  Ac J 2 () cos (c  2m )t  cos (c  2m )t 

…..(xi)

Ac J 3 () cos (c  3m )t  cos (c  3m )t  ..... A J () A J12 () Ac2 J 21 () Ac2 J 22 () Ac2 J 22 ()      ...... 2 2 2 2 2 A2 A2     c  J 02 ()  J12 ()  J 21 ()  J 22 ()  J 22 ()  ......  c   J n2 ()  2 2  n  

PWBFM 

PWBFM

2 c

2 0

2 c



Using property 3,

J

n 

2 n

()  1

Transmitted power, PWBFM 

Ac2 2

Carrier power, PC 

Ac2 J 02 () 2

Ac2 J12 () Ac2 J 21 ()   Ac2 J12 () 2 2 A2 J 2 () Ac2 J 22 () Power in second order sideband, PSB 2  c 2   Ac2 J 22 () 2 2

Power in first order sideband, PSB1 

Concept of Bandwidth in FM : 1. Bandwidth of an FM wave is defined as the frequency difference between the highest pair of sidebands. 2. Ideally the bandwidth of FM is infinite, because its spectrum consists of infinite number of upper and lower sidebands.

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3.

But practically it depends on the number of significant sidebands which is n   1 .

4.

The number of sidebands having significant amplitudes will increase with increase in the value of modulation index  .

5.

The simplest method to calculate the bandwidth is as follows : B.W.  2 f m  Number of significant sidebands  2nf m

6.

With increase in modulation index, the number of significant sidebands increases. This will increase the bandwidth. Carson's rule (rule of thumb) states that the bandwidth of FM wave is twice the sum of the deviation and the highest modulating frequency. B.W.  2  f  f m (max) 

Note : The Carson's rule gives correct results if the modulation index is greater than 6. Value of the Bessel function values J n ( ) for various order n and integral values of  n

  0.2

  0.5

  1.0

2

0 1 2 3 4 5 6 7 8 9 10 11 12

0.990 0.100 0.005

0.938 0.242 0.031 0.003

0.765 0.440 0.115 0.020 0.002

0.224 0.577 0.353 0.129 0.034 0.007 0.001

5

8

  10

0.178

0.172

 0.246

 0.328

0.235  0.113  0.291  0.105

0.043 0.255 0.058  0.220  0.234  0.014

0.047 0.365 0.391 0.261 0.131 0.053 0.018 0.006 0.001

0.186 0.338 0.321 0.223 0.126 0.061 0.026

0.217 0.318 0.292 0.207 0.123 0.063

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The Bessel functions J n ( ) plotted as a function of  FOR n = 0, 1, 2,……..5.

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Question 39 Define Carson’s rule.

[CSVTU Dec 2010] Or

Ans.

Explain Carson’s rule of Bandwidth. [CSVTU Dec 2007] Carson's rule provides a thumb formula to calculate the bandwidth of a single-tone wideband FM. Carson's rule (rule of thumb) states that the bandwidth of FM wave is twice the sum of the deviation and the highest modulating frequency.

B.W.  2  f  f m (max)  The Carson's rule gives correct results if the modulation index is greater than 6. Number of sideband terms, n    1  f  B.W.  2nf m  2(  1) f m  2   1 f m  2  f  f m   fm  For NBFM signal, B.W.  2 f m

Question 40 What is the difference between Narrowband FM and Wideband FM. Ans. Difference between Wideband and Narrowband FM : S. 1. 2. 3. 4. 5. 6. 7.

Wideband FM Modulation index is greater than 1. Frequency deviation is 75 kHz. Modulating frequency range from 30 Hz to 15 kHz. Bandwidth is large about 15 times higher than bandwidth of NBFM.

Narrowband FM Modulation index is less than 1. Frequency deviation is 5 kHz. Modulating frequency range 3 kHz. Bandwidth = 2 f m . Bandwidth is small approximately same as that of AM. Less suppression of noise. Used in mobile communication. Pre-emphasis and De-emphasis needed.

Noise is more suppressed. Used in entertainment broadcasting. Pre-emphasis and De-emphasis needed.

Question 41 What is the power of WBFM signal? Ans.

The expression of WBFM is given by, s(t ) 



 AJ

n  

Power of WBFM signal is given by, P 

Ac2 2



Using property of Bessel functions

J

n 

2 n

n

2 n

()



J

n 

() cos 2( f c  nf m )t

c

()  1 in equation (ii), P 

…..(i) …..(ii)

Ac2 2

Question 42 FM is constant bandwidth system, justify. Ans. Consider an FM system with frequency deviation f  75 kHz, carrier frequency

fc  100 MHz and maximum modulating frequency f m  1.5 kHz .

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The bandwidth using Carson’s rule is given by, BW= 2(f  f m )  2(75  1.5)  153 kHz Now keeping everything constant except f m  15 kHz New BW = 2(f  f m )  2(75  15)  180 kHz % change in modulating frequency  % change in bandwidth = 

15  1.5  100 = 900% 1.5

180  153  100  17.6 % 153

The percentage change in the bandwidth is only 17.6% corresponding to the ten time increases in the modulating frequency. This is why FM is called as a constant bandwidth system. Question 43 A 10 MHz sinusoidal carrier is frequency modulated by a unit amplitude sinusoid of frequency 1 kHz. The frequency modulation sensitivity k f  10 radians/sec-Volt. (i) What is the modulation index? (ii) Is this NBFM or WBFM? Why? (iii) What is the bandwidth of the transmitted signal? Sol.

(i) Given : Am  1 Volt , f m  1 kHz and k f  10 radians/sec-volt = Modulation index in FM,   

5 Hz/Volt . 

f k f Am  where k f is in Hz/Volt. fm fm

5 1 1    1000 200

(ii) Since  

Ans.

1  1, the signal is NBFM. 200

Ans.

(iii) NBFM signal has a bandwidth of 2 f m as long as  is very small (NBFM). Hence, the bandwidth of the transmitted signal is 2 kHz.

Ans.

Question 44 An angle-modulated signal with carrier frequency c  2  106 rad/sec is described by the equation, EM (t )  10cos c t  0.1sin 2000t  (i) Find the power of the modulated signal. (ii) Find the frequency deviation f . (iii) Find the phase deviation  . (iv) Estimate the bandwidth of  EM (t ) . Sol.

Given : EM (t )  10cos c t  0.1sin 2000t  An angle-modulated signal is given by,

(t )  Ac cos c t   sin mt   Ac cos c t  (t )

[CSVTU May 2010, Dec 2007]

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On comparing, Ac  10 V ,  m  2000 rad/sec ,   0.1 and (t )  0.1sin 2000t Ac2 102   50 W 2 2 (2) Frequency deviation, f  f m  0.11000  100 Hz

(1) Power of the modulated signal, P 

(3) Phase deviation,   (t ) max  0.1sin 2000t max  0.1 rad .

Ans. Ans. Ans.

(4) Bandwidth of angle modulated signal, B  2 f m for   1 

B  2 1000  2 kHz

Ans.

Question 45 A carrier is frequency modulated with a sinusoidal signal of 2 kHz, resulting in maximum frequency deviation of 5 kHz. [CSVTU May 2009] (i) Find the bandwidth of a modulated signal. (ii) The amplitude of the modulating sinusoid is increased by a factor of 3, and its frequency is lowered to 1 kHz. Find the maximum frequency deviation and the bandwidth of the new modulated signal. Sol.

Modulating frequency, f m  2 kHz

Frequency deviation, f  5 kHz

(1) Bandwidth of the FM signal, BW  2(f  f m )  2(5  2)  14 kHz

Ans.

(2) When the amplitude of the modulating signal is tripled, then the frequency deviation becomes three times. f  k f Am  f  Am New frequency deviation, f  3  5  15 kHz

Ans.

New modulating frequency, f m  1 kHz New bandwidth of the FM signal, BW  2(f  f m )  2(15  1)  32 kHz

Ans.

Question 46 Given an angle-modulated signal xc (t )  10cos c t  3sin m t  Assume PM and f m  1 kHz. Calculate the modulation index and the bandwidth when (i) f m is doubled and (ii) f m is decreased by one-half. Sol.

Given : xPM (t )  10cos c t  3sin mt   10cos c t  kP Am sin mt  The phase modulated wave is mathematically expressed as,

s(t )  Ac cos c t  k p m(t )   Ac cos c t  Am sin m t  where k p is in rad/volt. Modulation index in PM, m p  k P Am  3 It may be observed that the value of m p is independent of f m . For f m  1 kHz , Bandwidth  2(m p  1) f m  2(3  1)  1  8 kHz (1) When f m is doubled, f m  2 kHz Bandwidth  2(m p  1) f m  2(3  1)  2  16 kHz

Ans.

(2) When f m is decreased by one-half, f m  0.5 kHz Bandwidth  2(m p  1) f m  2(3  1)  0.5  4 kHz

Ans.

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Question 47 An angle-modulated signal is given as, s(t )  20cos c t  4sin m t  Assuming as a PM signal and f m  2 kHz, calculate the modulation index and bandwidth when (i) f m is increased two times Sol.

(ii) f m is decreased by half.

Given : s(t )  20cos c t  4sin mt   20cos c t  kP Am sin mt  The phase modulated wave is mathematically expressed as,

s(t )  Ac cos c t  k p m(t )   Ac cos c t  Am sin m t  where k p is in rad/volt. Modulation index in PM, m p  k P Am  4 It may be observed that the value of m p is independent of f m . For f m  2 kHz , Bandwidth  2(m p  1) f m  2(4  1)  2  20 kHz (1) When f m is doubled, f m  4 kHz Bandwidth  2(m p  1) f m  2(4  1)  4  40 kHz

Ans.

(2) When f m is decreased by one-half, f m  1 kHz Bandwidth  2(m p  1) f m  2(4  1) 1  10 kHz

Ans.

Question 48 Consider a modulating signal m( t )  2sin(2 103 t ) which is used to modulate a carrier of frequency 106 Hz. Find the bandwidth for PM if modulating frequency is doubled and amplitude of modulating signal is halved. Use modulation index as 10 for PM. Sol.

Modulating signal , m(t )  2sin(2103 t )  Am sin(2f m t ) Modulating frequency and amplitude, f m  1 kHz and Am  2 V The PM wave is mathematically expressed as, sPM (t )  Ac cos c t  k p Am sin m t  Modulating index, m p  k p Am m p  10 



k p Am  k p  2  10

k p  5 rad/Volt

When f m is doubled and Am is halved, f m '  2 kHz and Am '  1 V Modulation index, m p '  k p Am '  5 1  5 Bandwidth  2(m p  1) f m '  2(5  1)  2  24 kHz

Ans.

Question 49 The maximum deviation allowed in an FM broadcast system is 75 kHz. If the modulating signal is a single-tone sinusoid of 10 kHz, find the BW of FM signal. What will be the percentage change in the bandwidth, if modulating frequency is doubled? Determine the bandwidth when modulating signal amplitude is also doubled. [CSVTU May 2012]

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Frequency deviation, f  75 kHz Modulation index, f m  10 kHz Bandwidth of FM  2(f  f m )  2(75  10)  170 kHz

Ans.

When modulating frequency is doubled, the bandwidth increases, BW  2(75  20)  190 kHz % change in BW =

190  170  100  11.76 % 170

Ans.

When modulating signal’s amplitude is doubled, the frequency deviation becomes 2  75  150 kHz .



f  k f Am

f  Am

With modulating frequency doubled if modulating signal’s amplitude is doubled, then f '  150 kHz and f m '  20 kHz . 

BW  2(150  20)  340 kHz

Note : % change 

Ans.

New value  Old value 100% Old value

Question 50 A carrier signal c( t )  10cos(2  106 t ) is angle modulated (FM and PM) by the single tone signal m( t )  5cos(2  103 t ) . Let k f  2000 Hz/Volt and k p  3 rad/Volt. (1) Using Carson’s formula, compute the bandwidth of the resulting FM and PM signals. (2) What is the effect on the bandwidth in the two cases if f m is increased by a factor of 3? Sol.

Modulating signal, m(t )  5cos(2 103 t )  Am cos(2f m t ) On comparing, Am  5 V and f m  1 kHz (1) FM system : k f  2 kHz/Volt Modulation index,  

k f Am fm



25  10 1

Band width  2(  1) f m  2(10  1) 1  22 kHz

Ans.

PM system : k p  3 rad/volt Modulation index, m p  k p Am  3  5  15 Band width  2(m p  1) f m  2 (15  1)  1  32 kHz (2)

f m '  3 f m  3 1  3 kHz FM system : Modulation index,  ' 

k f Am fm '



2  5 10  3 3

 10  Band width  2( ' 1) f m '  2   1  3  26 kHz 3 

Ans.

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PM system : Modulation index is independent of f m and remain 15. Bandwidth  2(m p  1) f m '  2(15  1)  3  96 kHz

Ans.

Question 51 A modulating signal 5cos 2 15  103 t , angle modulates a carrier A cos c t . (i) Find the modulation index and the bandwidth for (a) the FM system (b) the PM system. (ii) Determine the change in the bandwidth and the modulation index for both FM and PM, if f m is reduced to 5 kHz. Assume k p  k f  15 kHz/Volt. [CSVTU Dec 2009, Dec 2008] Sol.

Modulating signal, m(t )  5cos 215 10 t  Am cos(2f m t ) 3

On comparing, Am  5 V and f m  15 kHz (1) FM system : Frequency deviation, f  k f Am  15 10  5  75 kHz 3

Modulation index,  

f 75  5 f m 15

Bandwidth  2(f  f m )  2(75  15)  180 kHz

Ans. Ans.

PM system : Frequency deviation, f  k p Am fm  15 103  5 15 103  1125 MHz Bandwidth = 2(f  f m )  2(1125  0.015)  2250.03 MHz

Ans.

The bandwidth is quite large as compared to FM. This is because the modulation index in PM is quite large. Modulation index, mp  k p Am  5 15 103  75000

Ans.

The tremendous value of the modulation index m p produce a large number of sidebands, and this is the reason for a large bandwidth. (2) New modulating frequency, f m  5 kHz FM system : The new frequency deviation in FM (f  k f Am ) is independent of f m and remains 75 kHz. New modulation index,  

f 75   15 fm 5

New bandwidth  2(f  f m )  2(75  5)  160 kHz Initially, BW was 180 kHz and modulation index 5 for f m  15 kHz . 160  180  100  11.11% 180 15  5 % change in modulation index  100  200 % 5

% change in BW 

Ans. Ans.

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Angle Modulation

Thus, in FM the modulation index changes considerably with a change in the modulating frequency, but the bandwidth changes only slightly. PM system : Frequency deviation (f ) in PM is dependent on f m and is given by,

f  k p Am fm  15 103  5  5 103  375 MHz Hence, BW  2(f  f m )  2(375  0.005)  750.01 MHz The new modulation index in PM is independent of f m and is given by,

mp  k p Am  15 103  5  75000 Initially, BW was 2250.03 MHz and modulation index 75000 for f m  15 kHz . 750.01  2250.03 Ans. 100   66.67 % 2250.03 75000  75000 % change in modulation index  Ans.  100  0 % 75000 Thus, in PM, the bandwidth changes considerably with a change in the modulating frequency but the modulation index m p remains unchanged.

% change in BW 

Note : % change 

New value  Old value 100% Old value

Question 52 A bandwidth rule for FM signal is sometimes used as B  (2   1) f m . What fraction of the signal power is included in that frequency band? Consider   1 and 10. Sol.

Given : Bandwidth for space communication system, B  (2   1) f m The bandwidth for FM signal can be calculated on the basis of 98% power requirement given by Carson’s rule as BW  2(  1) f m …..(i) The fraction of the signal power included in the frequency band B is (2  1) f m B  98  B  0.98 P   0.98   2(  1) f m BW  100  BW For   1 , P 

(2 1)  1 3  0.98   0.98  73.5% 2(1  1) 4

For   10 , P 

(2 10)  1 21  0.98   0.98  93.54% 2(10  1) 22

…..(ii) Ans. Ans.

Question 53 A carrier A cos c t is modulated by a signal 2cos104.2t  5cos103.2t  3cos104.4t . Find the bandwidth of the FM signal by using Carson’s rule. Assume k f  15  103 Hz per volt. Also find the modulation index m f . Ans.

Modulating signal, m(t )  2cos104.2t  5cos103.2t  3cos104.4t Modulating frequencies, f m1  10 kHz , f m 2  1 kHz and f m3  20 kHz Maximum modulating frequency, f m  20 kHz

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Frequency deviation, f  k f m(t ) max

f  15 103 2cos104.2t  5cos103.2t  3cos104.4t

max

 15 103 2  5  3  150 kHz

Bandwidth  2(f  f m )  2(150  20)  340 kHz Modulation index,  

Ans.

f 150   7.5 fm 20

Ans.

Question 54 An angle modulated signal with carrier frequency f c  2  106 Hz is given by

s( t )  cos 2  2  106 t  30sin150 t  40cos150t  (1) Determine the maximum frequency deviation (2) Find maximum phase deviation, and (3) Find the bandwidth of s(t ) . Sol.

Given : s(t )  cos 2  2 106 t  30sin150t  40 cos150t 

s(t )  cos  2 2  106 t  2 30sin150t  2 40 cos150t  Modulating frequency, f m 

150 Hz 2

The FM wave is mathematically expressed as, s(t )  Ac cos c t  (t ) On comparing, (t )  2 30sin150t  2 40cos150t The maximum frequency deviation is given by, f 

1 d (t ) 2 dt max

d (t )  2 150  30 cos150t  2 150  40sin150t dt

f 

1 d 1 (t )  2150  30cos150t  2150  40sin150t max 2 dt 2  max

f  150  30cos150t 150  40sin150t max  4500cos150t  6000sin150t max If R  A cos   B sin  then, R  A2  B2

f  45002  60002  7500 Hz  Frequency deviation, f  7.5 kHz

Ans.

The maximum phase deviation is given by,   (t ) max

  2 30sin150t  2 40cos150t max  2 30sin150t  40cos150t max   2 302  402  2 50  100  Phase deviation,   100 rad

The bandwidth of angle modulated signal is given by, 150   BW  2(f  f m )  2  7500    15.04 kHz 2  

Ans.

Ans.

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Question 55 A device with input x(t ) and output y(t ) is characterized by y( t )  x 2 ( t ) . An FM signal with f  90 kHz and f m  5 kHz is applied to this device. Find the Sol.

bandwidth of output signal. x(t ) is an FM signal.

x(t )  Ac cos 2fc t   sin 2f mt 

y(t )  x2 (t )  Ac2 cos2 2fc t   sin 2fmt  cos 2 A 

1  cos 2 A 2

Ac2 A2 A2 1  cos  4f c t  2 sin 2f m t   c  c cos  4f c t  2 sin 2f m t  2 2 2 New modulation index,  '  2



y (t ) 

New modulation frequency, f m '  f m New frequency deviation, f '   ' f m '  2 f m  2f New band width  2(f '  f m ')  2(2f  f m )  2(2  90  5)  370 kHz

Ans.

Question 56 Consider a single-tone FM signal with a peak frequency deviation of 5 kHz and frequency deviation constant k f  10

4

Hz/volt.

Let the modulating signal

frequency be 1 kHz and the carrier frequency be 100 kHz. (1) Find the peak and RMS value of the modulating signal voltage. (2) Obtain the FM modulation index  .

Sol.

(3) What is the FM transmission bandwidth? (4) How many sideband terms are present in the FM spectrum and what is the separation between the adjacent sideband terms? Frequency deviation, f  5 kHz Carrier frequency, f c  100 kHz Modulating frequency, f m  1 kHz

k f  104 Hz/volt  2104 rad/sec-volt

(1) Frequency deviation is given by, f  k f m(t ) max  k f Am 5 kHz  104  Am



Am  0.5 V

RMS modulating signal voltage is given by, A 0.5  0.354 V  Am rms  m  2 2 f 5000  5 (2) Modulation index in FM,   f m 1000

Ans. Ans.

(3) FM transmission bandwidth by using Carson’s formula is given by, BW  2(  1) f m  2(5  1) 1  12 kHz

Ans.

(4) The number of sideband terms n is given by, n    1  5  1  6

Ans.

The separation between the sideband terms is f m  1 kHz

Ans.

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Question 57 An angle modulated signal with carrier frequency c  2  105 is described by the equation  AM (t )  10cos(c t  5sin 3000t  10sin 2000t ) . Find the following values : (i) Modulating signal bandwidth (iii) Frequency deviation (v) Phase deviation Sol.

(ii) Power of the modulated signal (iv) Deviation ratio (vi) Estimate the bandwidth of the modulated wave Given :  AM (t )  10cos (c t  5sin 3000t  10sin 2000t ) The standard expression for FM is given by, s(t )  Ac cos c t  (t ) (i) Modulating signal bandwidth 3000 f m 2  1000 Hz f m1  Hz 2 Modulating signal bandwidth, f m (max)  1000 Hz (ii) Power of the modulated signal A2 102 P c   50 W 2 2 (iii) Frequency deviation (t )  5sin 3000t  10sin 2000t

Ans.

Ans.

d (t )  15000 cos 3000t  20000 cos 2000t dt 1 d 1 f  (t )  15000cos3000t  20000 cos 2000t max 2 dt 2  max 7500  10000  12.4 kHz  (iv) Phase deviation f 

Ans.

  (t ) max  5sin3000t  10sin 2000t max  5  10  15 rad

Ans.

(v) Deviation ratio f 12400  = = 12.4 f m (max) 1000 (vi) Bandwidth of the modulated wave BW  2(  1) f m  2(12.4  1) 1000  26.8 kHz Question 58 For the message signal m(t ) shown in the figure.

Ans.

Ans.

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Angle Modulation

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(a) If m(t ) is frequency modulated on a carrier with frequency 106 Hz, with a frequency deviation constant k f  5 Hz/V, what is the maximum instantaneous frequency of the modulated signal? (b) If m(t ) is phase modulated with phase-deviation constant k p  3 rad/V, what is the maximum instantaneous frequency on the modulated signal? Sol.

Carrier frequency, f c  106 Hz (a) From the figure, m(t ) max  105 The instantaneous frequency is given by, fi (t )  f c  k f m(t )

max  fi (t )  max  f c  k f m(t )   106  5 105   1.5 MHz

Ans.

(b) The phase deviation of the phase modulated signal is (t )  k p m(t ) and the instantaneous frequency is given by, fi (t )  fc  

 fi (t )max   fc  

kp d 1 d (t )  f c  m(t ) 2 dt 2 dt

kp d  m(t )  2 dt  max

From figure

d 105 m(t )   105 dt 1 max

 fi max  106 

3  10  1.048 MHz 2

Ans.

Question 59 Let m1 ( t ) and m2 ( t ) be two message signals, and let xc1 ( t ) and xc 2 (t ) be the modulated signals corresponding to m1 ( t ) and m2 ( t ) , respectively. (i) Prove that if the modulation is DSB(AM), then m1 (t )  m2 (t ) will produce a modulated signal equal to xc1 (t )  xc 2 (t ) . (This is why AM is sometimes referred to as a linear Modulation.) (ii) Prove that if the modulation is PM, then the modulated signal produced by m1 (t )  m2 (t ) will not be xc1 (t )  xc 2 (t ) , that is, superposition does not apply to angle-modulated signal. (This is why angle modulation is sometimes referred to as a nonlinear modulation.) Sol.

(i) DSB (AM) For modulating signal m1 (t ) , modulated signal is xc1 (t )  m1 (t ) cos c t For modulating signal m2 (t ) , modulated signal is xc 2 (t )  m2 (t ) cos c t For m1 (t )  m2 (t ) , modulated signal is xc (t )  m1 (t )  m2 (t ) cos ct

xc (t )  m1 (t ) cos c t  m2 (t ) cos c t  xc1 (t )  xc 2 (t ) Therefore, DSB (AM) modulation is a linear modulation.

Proved.

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(ii) PM For modulating signal m1 (t ) , modulated signal is xc1 (t )  A cos c t  kP m1 (t ) For modulating signal m2 (t ) , modulated signal is xc 2 (t )  A cos c t  kP m2 (t ) For m1 (t )  m2 (t ) , modulated signal is xc (t )  A cos c t  k P  m1 (t )  m2 (t ) 

xc (t )  xc1 (t )  xc 2 (t ) Therefore, PM is a nonlinear modulation.

Proved.

Question 60 Consider the signal cos c t  (t ) where ( t ) is a square wave taking on the value



2  sec . every fc 3

[CSVTU Dec 2011]

(i) Sketch cos c t  (t ) (ii) Plot the phase as a function of time. (iii) Plot the frequency as a function of time. Ans.

From above figure 0  t 

2    ; (t )  then s(t )  cos  c t   fc 3 3 

2 4    t  ; (t )   then s(t )  cos  c t   fc fc 3 3 

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(i) Sketching of cos c t  (t )

(ii) Plot of phase as a function of time.

Angle Modulation

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(iii) Plot of frequency as a function of time.

Question 61 Consider a carrier signal c( t )  50cos(2 .108 t ) frequency modulated by a message signal m(t )  5 sin c(500t ) . Let the modulation index be 10.

Sol.

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

Write the expression for the modulated wave. Obtain the power in the modulated signal. Obtain the maximum frequency deviation of the modulated signal. What is the transmission bandwidth of the modulated signal? The modulated signal in FM is given by, t   rad …..(i) s(t )  Ac cos c t  k f  m() d   where k f is in Volt-sec    From the concept of Fourier transform :  t  FT Arect    A sin c  f     t  FT Let A  1 ,   500 then rect    500sin c  500 f   500  FT FT x( f ) X ( f ) then X (t )  Apply duality property : x(t ) 

f   f  FT 500sin c  500t   rect    rect  500   500    1  f   f  FT 5sin c  500t   rect    0.01rect  500  100 500    

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This corresponds to a rectangular spectrum with bandwidth of 250 Hz or B  2 250 rad/sec . The modulation index is given by,  



kf 5 2 250

k f m(t ) max B



k f 5sin c  500t  max B

 k f  1000 rad/sec-volt

 10

Carrier signal, c(t )  50 cos(2 .108 t )  Ac cos c t From equation (1), m()  5sinc 500  , k f  1000 rad/sec-volt t   s(t )  50cos  2 .108 t  1000 5  sin c  500  d     



t   s(t )  50cos 2 .108 t  5000  sin c  500  d     

Ans.

(b) The (normalized) power in the modulated signal is given by, Ac2 502   1250 W 2 2 (c) Maximum frequency deviation of the modulated signal is given by, P

Ans.

f   f m  B  10  250 Hz  2.5 kHz

Ans.

(d) The transmission bandwidth by using Carson’s formula is given by, BFM  2(  1) B  2(10  1)250  5500 Hz

Ans.

Question 62 An FM signal s( t )  20cos  2  106 t  10sin(2  103 t )  is radiated to free space and antenna resistance is 50  . Calculate the carrier and total radiated power if

J 0 (10)  0.25 . Sol.

s (t )  20 cos  2106 t  10sin(2103 t )   Ac cos  2f c t   sin(2f m t ) 

Modulation index,   10 Carrier power, PC  

Resistance, R  50 

Ac2 J 02 () 202  J 02 (10) 202  0.252   2R 2  50 2  50

PC  0.25 W

Radiated power, PWBFM 

Ans. 2 c

2

A 20  4 W 2R 2  50

Ans.

Question 63 An FM transmitter radiates 100 W when no modulation occurs. The carrier is now frequency modulated by a sinusoidal signal and  is adjusted such that amplitude of first order sideband is zero in spectrum index. Given : J 0 (0)  1, J 0 (2.4)  0, J 0 (3.8)   0.41, J 1 (2.4)  0.52, J 1 (3.8)  0, J 1 (5.1)  0.33, J 2 (2.4)  0.42, J 2 (3.8)  0.41, J 2 (5.1)  0

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Calculate : (a) Power in carrier frequency. (b) Remaining sideband power. (c) Power in second order sideband. Sol.

Power with no modulation, PC  100 

Ac2 2

Ac2 2

Ac  14.14 V



Power in first order sideband is zero i.e. PSB1  Ac2 J12 ()  0 From the given data, J1 (3.8)  0



  3.8

(a) Power in carrier frequency PC 

Ac2 J 02 () 14.142  J 02 (3.8) 14.142  ( 0.41) 2    16.8 W 2 2 2

Ans.

(b) Remaining sideband power Total power, PT 

Ac2 14.142   99.96 W 2 2

Sideband power = Total power – Carrier power 

PSB  PT  PC  99.96  16.8  83.16 W

Ans.

(c) Power in second order sideband PSB 2  Ac2 J 22 ()  14.142  J 22 (3.8)  14.142  0.412  33.61 W

Ans.

Question 64 Determine the relative power of carrier wave and sideband frequency when   0.2 for a 10 kW FM transmitter. Sol.

Ac2  10000 W 2 From table of Bessel function, J 0 (0.2)  0.99

Transmitted power, PT 

Power of carrier wave, PC  

Ac2 J 02 ()  10000  J 02 (0.2)  10000  (0.99) 2 2

PC  9.8 kW

Ans.

Power of sideband frequency, PSB  PT  PC  (10  9.8) kW = 200 W Power of each sideband frequency, PSB1  PSB 2  100 W

Ans.

Question 65 A carrier is angle modulated by two sinusoidal modulating waveform simultaneously so that v(t )  A cos(c t  1 sin 1 t  2 sin 2 t ) Show that this waveform has sidebands separated from the carrier not only multiples of 1 and of

 2 but also has sideband as well at separations of multiples of 1  2 and 1  2 . [CSVTU Dec 2011]

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Sol.

2 - 45

Angle Modulation

Given : v(t )  A cos(c t  1 sin 1t  2 sin 2t ) The given signal v(t ) can be expressed as under : v(t )  Re  Ae jc t e

j 1 sin 1t 2 sin 2 t 

  A Re e jc t e j1 sin 1t e j2 sin 2t    

…..(i)

Where Re(.) denotes the real part of (.). 

e j1 sin 1t 

J

n 

e j2 sin 2t 

n

(1 ) e jn1t



J

m 

m

(2 ) e jm2t

…..(ii) …..(iii)

Substituting equations (ii) and (iii) in equation (i) and taking the real part, we get 

v(t )  A 



J

n  m 

n

(1 ) J m (2 ) cos(c  n1  m2 ) t

…..(iv)

From equation (4) it may be observed that the spectrum of v(t ) consist of 1.

sideband lines at c  n1 due to one tone alone.

2.

sideband lines at c  m2 due to the other tone alone.

3.

sideband lines at c  n1  m2 because of the nonlinear property of angle modulation.

This shows that the waveform v(t ) has sidebands as multiples of 1  2 and 1  2 . Question 66 An FM carrier is sinusoidally modulated. For what values of  does all the power lie in the sideband (i.e. no power in the carrier)? Sol.

The wide band FM waveform is given by, s(t )  Ac cos[c t   sin mt ] where   1

cos( A  B)  cos A cos B  sin A sin B

s(t )  Ac cos(c t )cos( sin mt )  sin(ct )sin( sin mt )

cos( sin mt )  J 0 ()  2 J 2 ()cos 2mt  2 J 4 () cos 4mt  .......

sin(  sin mt )  2 J1 ()sin mt  2 J 3 ()sin 3mt  ....... s(t )  Ac J 0 ()cos c t  Ac 2 J 2 ()cos c t cos 2 mt  Ac 2 J 4 ()cos c t cos 4 mt  .......  Ac 2 J1 ()sin c t sin mt  Ac J 3 ()sin c t sin 3mt....... If all power lie in the sideband i.e. power of carrier should be zero i.e., J 0 ()  0

The values of  at which J 0 ()  0 are   2.4, 5.5, 8.5, 11.8 . These values are referred as magic values.

Ans.

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Question 67 Consider the angle-modulated waveform cos(c t  6sin m t ) i.e.   6, so that the waveform may be approximated by a carrier and seven pairs of sidebands. In a system in which the carrier phasor  0 is at rest, determine the phasor  1 ,  2 , etc., representing the first, second, etc., sideband pairs at a time when sin m t  1 .

For

this time draw a phasor diagram showing each phasor  i and the resultant phasor. Given J 0 (6)  0.15, J1 (6)   0.28, J 2 (6)   0.24, J 3 (6)  0.11,

J 4 (6)  0.36, J 5 (6)  0.36, J 6 (6)  0.25, J 7 (6)  0.13 . Sol.

Given : s(t )  cos(c t  6sin mt )

cos( A  B)  cos A cos B  sin A sin B s(t )  cos(c t ) cos(6sin mt )  sin(c t )sin(6sin mt ) cos(6sin mt )  J 0 (6)  2 J 2 (6) cos 2mt  2 J 4 (6) cos 4mt  2 J 6 (6) cos 6mt  ....... sin(6sin mt )  2 J1 (6)sin mt  2 J 3 (6)sin 3mt  2 J 5 (6)sin 5mt  2 J 7 (6)sin 7mt  .......

sin m t  1 At m t 



m t 

 2

 , 2

0  J 0 (6) cos c t  0.15cos c t (0 is at rest)

 1  2 J1 (6)sin mt  2 J1 (6)sin    2   0.28 1   0.56 2   2  2 J 2 (6) cos 2mt  2 J 2 (6) cos  2    2   0.24  1  0.48 2    3  2 J 3 (6)sin 3mt  2 J 3 (6)sin  3    2  0.11 1   0.22  2

  4  2 J 4 (6) cos 4mt  2 J 4 (6) cos  4    2  0.36 1  0.72 2    5  2 J 5 (6)sin 5mt  2 J 5 (6)sin  5    2  0.36 1  0.72  2   6  2 J 6 (6) cos 6mt  2 J 6 (6) cos  6    2  0.25  1   0.50  2   7  2 J 7 (6)sin 7mt  2 J 7 (6)sin  7    2  0.13  1   0.26 2 

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Phasor diagram at m t 

Angle Modulation

2 - 47

 2

Steps for drawing phasor diagram : (1) Odd phasors are perpendicular to even phasors. (2) An even phasor with positive value will point right while even phasor with negative value will point left (  2 ,  4 being positive are pointing right while 6 being negative is pointing left). (3) An odd phasor with positive value will point up while odd phasor with negative value will point down ( 1 , 3 , 7 being negative are pointing down while 5 being positive is pointing up). Question 68 Consider the angle-modulated waveform cos(c t  6sin m t ) i.e.   6, so that the waveform may be approximated by a carrier and seven pairs of sidebands. In a system in which the carrier phasor  0 is at rest, determine the phasor  1 ,  2 , etc., representing the first, second, etc., sideband pairs at a time when sin m t  1 .

For

this time draw a phasor diagram showing each phasor  i and the resultant phasor. Given J 0 (6)  0.1506, J1 (6)   0.2767, J 2 (6)   0.22429, J 3 (6)  0.1148,

J 4 (6)  0.3516, J 5 (6)  0.3621, J 6 (6)  0.2458 . Sol.

[CSVTU Dec 2010, Dec 2007]

Given : s(t )  cos(c t  6sin mt )

cos( A  B)  cos A cos B  sin A sin B s(t )  cos(c t ) cos(6sin mt )  sin(c t )sin(6sin mt ) cos(6sin mt )  J 0 (6)  2 J 2 (6) cos 2mt  2 J 4 (6) cos 4mt  2 J 6 (6) cos 6mt  ....... sin(6sin mt )  2 J1 (6)sin mt  2 J 3 (6)sin 3mt  2 J 5 (6)sin 5mt  2 J 7 (6)sin 7mt  .......

sin m t  1



m t 

 2

 , 2 0  J 0 (6) cos c t  0.1506cos c t (0 is at rest)

At m t 

  1  2 J1 (6)sin mt  2 J1 (6)sin    2   0.2767 1   0.5534 2

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  2  2 J 2 (6) cos 2mt  2 J 2 (6) cos  2    2   0.22429  1  0.44858 2 

  3  2 J 3 (6)sin 3mt  2 J 3 (6)sin  3    2  0.1148  1   0.2296  2   4  2 J 4 (6) cos 4mt  2 J 4 (6) cos  4    2  0.3516 1  0.7032 2    5  2 J 5 (6)sin 5mt  2 J 5 (6)sin  5    2  0.36211  0.7242  2

  6  2 J 6 (6) cos 6mt  2 J 6 (6) cos  6    2  0.2458  1   0.4916  2  Phasor diagram at m t  2

Steps for drawing phasor diagram : (1) Odd phasors are perpendicular to even phasors. (2) An even phasor with positive value will point right while even phasor with negative value will point left (  2 ,  4 being positive are pointing right while 6 being negative is pointing left). (3) An odd phasor with positive value will point up while odd phasor with negative value will point down ( 1 , 3 being negative are pointing down while 5 being positive is pointing up). Question 69 Consider the angle-modulated waveform cos(c t  2sin m t ) i.e.   2, so that the waveform may be approximated by a carrier and three pairs of sidebands. In a coordinate system in which the carrier phasor  0 is at rest, determine the phasors

 1 ,  2 and  3 , representing respectively the first, second and third sideband pairs. Draw diagrams combining the four phasors for the cases m t  0,  / 4,  / 2, 3 / 4 and  . For each case calculate the magnitude of the resultant phasor. Given :

J 0 (2)  0.22, J1 (2)  0.58, J 2 (2)  0.35, J 3 (2)  0.125 .

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Sol.

2 - 49

Angle Modulation

Given : s(t )  cos (c t  2 sin mt )

cos( A  B)  cos A cos B  sin A sin B s(t )  cos(c t ) cos(2sin mt )  sin(c t )sin(2sin mt ) cos(2sin mt )  J 0 (2)  2 J 2 (2) cos 2mt  2 J 4 (2) cos 4mt  2 J 6 (2) cos 6mt  ....... sin(2sin mt )  2 J1 (2)sin mt  2 J 3 (2)sin 3mt  2 J 5 (2)sin 5mt  2 J 7 (2)sin 7mt  ....... Phasor 0 is given by, 0  J 0 (2) cos c t Phasor 1 is given by, 1  2 J1 (2)sin mt Phasor  2 is given by, 2  2 J 2 (2) cos 2mt Phasor 3 is given by, 3  2 J 3 (2)sin 3mt Case 1 : m t  0

0  J 0 (2) cos c t  0.22 (0 is at rest) 1  2 J1 (2)sin mt  2 J1 (2)sin(0)  2  0.58  0  0 2  2 J 2 (2) cos 2mt  2 J 2 (2) cos(0)  2  0.35 1  0.70

3  2 J 3 (2)sin 3mt  2 J 3 (2)sin(0)  2  0.125  0  0 Phasor diagram at mt  0

Resultant phasor  AC  0.22  0.70  0.92  Case 2 :  m t  4 0  J 0 (2) cos c t  0.22 (0 is at rest)

Ans.

  1  2 J1 (2)sin mt  2 J1 (2)sin    2  0.58  0.707  0.82 4   2  2 J 2 (2) cos 2mt  2 J 2 (2) cos  2    2  0.35  0  0 4    3  2 J 3 (2)sin 3mt  2 J 3 (2)sin  3    2  0.125  0.707  0.176  4  Phasor diagram at m t  4

2 2 Resultant phasor  AD  0.22  (0.82  0.176)  1.02

Ans.

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 2 0  J 0 (2) cos c t  0.22 (0 is at rest)

Case 3 :  m t 

  1  2 J1 (2)sin mt  2 J1 (2)sin    2  0.58 1  1.16 2   2  2 J 2 (2) cos 2mt  2 J 2 (2) cos  2    2  0.35  1   0.70  2   3  2 J 3 (2)sin 3mt  2 J 3 (2)sin  3    2  0.125  1   0.250  2  Phasor diagram at m t  2

2 2 Resultant phasor  AE  (0.70  0.22)  (1.16  0.25)  1.028

Ans.

3 4 0  J 0 (2) cos c t  0.22 (0 is at rest)

Case 4 : m t 

 3  1  2 J1 (2)sin mt  2 J1 (2)sin    2  0.58  0.707  0.82  4   3  2  2 J 2 (2) cos 2mt  2 J 2 (2) cos  2    2  0.35  0  0 4    3  3  2 J 3 (2)sin 3mt  2 J 3 (2)sin  3    2  0.125  0.707  0.176 4   3 Phasor diagram at m t  4

2 2 Resultant phasor  AD  0.22  (0.82  0.176)  1.02

Ans.

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Angle Modulation

2 - 51

Case 5 : m t  

0  J 0 (2) cos c t  0.22 (0 is at rest) 1  2 J1 (2)sin mt  2 J1 (2)sin()  2  0.58  0  0 2  2 J 2 (2) cos 2mt  2 J 2 (2) cos(2)  2  0.35 1  0.70 3  2 J 3 (2)sin 3mt  2 J 3 (2)sin(3)  2  0.125  0  0 Phasor diagram at mt  

Resultant phasor  AC  0.22  0.70  0.92 Ans. Question 70 Write the difference between Frequency and Phase modulation. Or Distinguish between PM & FM based on four major points. [CSVTU Dec 2009] Ans. Frequency and Phase modulation : S.

FM

PM

1.

s(t )  Ac cos c t  2k f  m(t ) dt 

s(t )  Ac cos c t  k p m(t ) 

2.

Frequency deviation is proportional to modulating voltage. f  k f Am

Phase deviation is proportional to the modulating voltage.   k p Am

3.

Associated with the changes in f c , there is

4.

some phase change. Modulation index m f is proportional to

Associated with the changes in phase, there is some change in f c .

the modulating voltage as well as the modulating frequency f m .

mf 

k f Am

Modulation index m p is proportional only to the modulating voltage and independent of modulating frequency. m p  k p Am

fm

5.

In FM the frequency deviation is proportional to the modulating voltage only. f  k f Am

In PM the frequency deviation is proportional to both the modulating voltage and modulating frequency. f  k p Am f m

6.

It is possible to receive FM on a PM receiver.

It is possible to receive PM on a FM receiver.

7.

Noise immunity is better than AM and PM.

Noise immunity is better than AM but worse than FM.

8.

Amplitude of the FM wave is constant.

Amplitude of the PM wave is constant.

9.

Signal to noise ratio is better than that of PM.

Signal to noise ratio is inferior to that in FM.

10.

FM is widely used.

PM is used in some mobile system.

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2 - 52

Question 71 Write the difference between Frequency and Amplitude modulation. Ans.

Frequency and Amplitude modulation :

S.

FM

AM

1.

Amplitude of FM wave is constant. It is independent of the modulation index.

Amplitude of AM wave will change with the modulating voltage.

2.

Transmitted power remains constant. It is independent of modulation index.

Transmitted power is dependent on the modulation index.

3.

All the transmitted power is useful.

Carrier power and one sideband power are useless.

4.

FM receivers are immune to noise.

AM receivers are not immune to noise.

5.

Bandwidth  2(  1) f m . The bandwidth

Bandwidth  2 f m . Bandwidth is not

depends on modulation index.

dependent on the modulation index.

6.

BW is large, hence wide channel is required.

BW is much less than FM.

7.

Space wave is used for propagation. So radius of transmission is limited to line of sight.

Ground wave and sky wave propagation is used. Therefore larger area is covered than FM.

8.

It is possible to operate several transmitters on same frequency.

Not possible to operate more channels on the same frequency.

9.

FM transmission and reception equipment are more complex.

AM equipment is less complex.

10.

The number of sidebands having significant amplitudes depends on modulation index m f .

Number of sidebands in AM will be constant and equal to 2.

11.

The information is contained in the frequency variation of the carrier.

The information is contained in the amplitude variation of the carrier.

12.

Applications : Radio, TV broadcasting, police wireless, point to point communications.

Applications : Radio and TV broadcasting.

Question 72 Compare AM and FM on following points :

Ans.

(i) Power requirement

(ii) Bandwidth

(iii) Noise

(iv) Carrier frequency

Difference between AM and FM : 1.

Carrier frequency : The FM range is 88 - 108 MHz (with broadcast frequencies, or stations, assigned between 88.1 and 107.9 MHz every 0.2 MHz). The AM range is 535 - 1605 kHz (stations are assigned between 540 and 1600 kHz every 10 kHz).

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Angle Modulation

2.

Power requirement : The amplitude of FM wave remains constant. As the power of signal depends upon amplitude of signal, power of FM also remain constant but in case of AM the amplitude cannot be constant. If we increase the value of modulation index in case of AM power requirement increases.

3.

Bandwidth : High quality frequency modulation produces many side frequencies, thus necessitating a much wider bandwidth than is necessary for AM transmission. Each station in the commercial AM radio band is assigned 10 kHz of bandwidth, whereas in the commercial FM broadcast band, 200 kHz is assigned to each station.

4.

Noise : FM systems are much less susceptible to noise than AM systems, since most noise will be additive and amplitude variations in the received signal will be ignored by the receiver.

Question 73 Write down advantages and disadvantages of FM over AM signals. Or Basically what disadvantage of amplitude modulation does frequency modulation cope up with? [CSVTU May 2009] Ans.

Advantages of FM over AM : 1.

An FM system has a much larger allowable range of modulating signal amplitudes because there is less restriction on maximum value of frequency deviation than in AM where the modulation factor is limited to ma  1 .

2.

The FM transmitter is more efficient since the amplitude of an FM wave is constant, as opposed to an AM transmitter where each of the stages must be able to cope with a range of power levels as the carrier and /or sidebands vary between zero and full power.

3.

FM systems are much less susceptible to noise since most noise will be additive and amplitude variations in the received signal will be ignored by the receiver.

4.

The capture effect allows an FM receiver to suppress the weaker of two cochannel signals simultaneously present at its front end.

5.

It is possible to reduce noise still further by increasing deviation whereas this feature is not in AM.

6.

According to CCIR as there is ‘guard band’ between FM stations, so, there is less adjacent-channel interference than in AM.

7.

In AM systems only one third of the total power is carried by the side bands and two third is wasted in the carrier signal transmission. While in FM signal most of the energy is carried in its sidebands. For large modulation index carrier power becomes less while sideband increases.

Disadvantages of FM over AM : 1.

The main disadvantage of FM is the much wider bandwidth required to achieve the signal-to-noise ratio improvement.

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2.

Since FM requires a wider bandwidth, higher frequencies must be used at which communications range is normally limited to line-of-sight.

3.

The capture effect may also be a disadvantage when a receiver is near the edge of the service area and may be captured by an unwanted signal or by noise.

4.

The modulation and demodulation equipment of FM system is more complex than AM system; due to complex circuitry FM systems tend to be costlier than AM ones.

5.

For FM mobile communication it becomes very difficult to receive the signal as reception is limited to line of sight (area of reception is much smaller).

Question 74 Why FM is known as excellent modulation scheme as far analog modulation is concerned? [CSVTU Dec 2009] Ans.

1.

FM systems are much less susceptible to noise.

2.

It is possible to reduce noise still further by increasing deviation whereas this feature is not in AM.

3.

There is less adjacent-channel interference in FM than in AM.

4.

The FM transmitter is more efficient.

Question 75 Compare FM with SSB-SC signal. Ans.

1.

Amplitude of FM wave is constant while amplitude of AM SSB-SC wave will change with the modulating voltage.

2.

Transmitted power in FM remains constant while transmitted power in SSB-SC is dependent on the modulation index.

3.

Bandwidth is large in FM while bandwidth is much less in SSB-SC.

4.

FM accommodates lager bandwidth for better signal/noise ratio whereas SSB limits the bandwidth to accommodates more channels in specified bandwidth.

5.

SC (suppressed carrier) has same purpose in both to limit transmitter power.

6.

FM receivers are immune to noise. SSB-SC receivers are not immune to noise.

7.

FM receivers are easy to implement compare to SSB-SC.

Question 76 Why FM is widely used in microwave radio and satellite communication system? [CSVTU May 2008] Or Explain the nonlinear effects in FM systems. Ans.

There are two basic forms of nonlinearity to consider : 1.

The nonlinearly is said to be strong when it is introduced intentionally and in a controlled manner for some specific applications. Examples of strong nonlinearity included square law modulators, limiters and frequency multipliers.

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2.

The nonlinearity is said to be weak when a linear performance is desired, but nonlinearities of a parasitic nature arise due to imperfections. The effect of such weak nonlinearities is to limit the useful signal levels in a system and thereby become an important design consideration. Consider a communication channel, the transfer characteristics of which is defined by the, non-linear input-output relation. v0 (t )  a1vi (t )  a2 vi2 (t )  a3vi3 (t ) …..(i) where vi (t ) , and v0 (t ) are the input and output signals, respectively and a1 , a2 and

a3 are constant. t

vi (t )  Ac cos 2fct  (t ) where (t )  2k f  m(t ) dt 0

From equation (1), v0 (t )  a1 Ac cos 2fct  (t )  a2 Ac2 cos2 2fct  (t )  a3 Ac3 cos3 2fct  (t )

1 3   a2 Ac2   a1 Ac  a3 Ac3  cos  2f c (t )  (t )   2 4   …..(ii) 1 1 2 3 a2 Ac cos  4f c t  2(t )   a3 Ac cos 6f c t  3(t )  2 4 Thus the channel output consists of a dc component and three frequency modulated signals with carrier frequency f c , 2 f c and 3 f c . v0 (t ) 

To extract the desired FM signal from the channel output v0 (t ) that is, the particular component with carrier frequency f c , it is necessary to separate the FM signal with this carrier frequency 2 f c . Let f denote the frequency deviation of the incoming FM signal vi (t ) and W denote the highest frequency component of the message signal

m(t ) . Then, applying Carson’s rule and noting that the frequency deviation about the second harmonic of the carrier frequency is doubled, we find that the necessary condition for separating the desired FM signal with the carrier frequency f c from that with the carrier frequency 2 f c is

2 fc  (2f  W )  fc  f  W



f c  3f  2W

Thus by using a BPF of mid-band frequency f c and band-width 2 f c  2W , the channel output is reduced to 3   …..(iii) v10 (t )   a1 Ac  a3 Ac3  cos  2f c (t )  (t ) 4   From equation (iii), we see that the only effect of passing an FM signal through a channel with amplitude nonlinearities, followed by appropriate filtering is simply to modify its amplitude. That is unlike amplitude modulation, frequency modulation is not affected by distortion produced by transmission through a channel with amplitude nonlinearities. It is for this reason that we find frequency modulation widely used in microwave radio and satellite communication systems. It permits the use of highly nonlinear amplifiers and power transmitters, which are particularly important to producing a maximum power at radio frequency.

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A common type of phase nonlinearity that is encountered in microwave radio system is known as AM to PM conversion. An FM system is extremely sensitive to phasor nonlinearities. When an FM wave is transmitted through a microwave radio link, it picks up spurious amplitude variations due to noise and interference during the course of transmission and when such an FM wave is passed through a repeater with AM to PM conversion, the output will contain unwanted phase modulation and resultant distortion. It is therefore important to keep the AM to PM conversion at a low level.

FM Generation Question 77 Explain direct method of generation of FM wave. Or Explain varactor diode method for FM generation and derive equation for instantaneous frequency of oscillation. Ans.

Direct method or parameter variation method of FM generation : An FM signal may be generated directly by using a varactor diode. A varactor diode is a semiconductor diode with a small junction capacitance that varies in a definite manner as a function of its reverse bias voltage. The figure below shows the symbol of a varactor diode which is similar to that of p-n diode but with an additional parallel line showing the capacitance of the diode.

If the varactor diode is reversed biased suitably, it will exhibit a definite capacitance value. Now, if we superimpose the modulating signal variation onto the reverse bias voltage, the varactor diode capacitance will vary as a function of the signal across it. In the varactor diode FM generator, we have a conventional tuned circuit oscillator with tank circuit capacitance C in parallel with inductance L which generates carrier signal. The frequency of oscillation of the carrier generation is governed by the expression c 

1 LC

or f c 

1 2 LC

The carrier frequency c can be made to vary in accordance with the baseband or modulating signal m(t ) if L or C is varied according to m(t ) . An oscillator circuit whose frequency is controlled by a modulating voltage is called voltage controlled oscillator (VCO). The frequency of VCO is varied according to modulating signal simply by putting a shunt voltage variable capacitor, called varactor or varicap, with its tuned circuit. The varactor diode is connected in shunt with the tuned circuit of the carrier oscillator.

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In varactor diode FM generation arrangement, the capacitor C is made much smaller than the varactor diode capacitance C d so that the radio frequency (RF) voltage from oscillator across the diode is small as compared to reverse bias d.c. voltage across the varactor diode. In addition to this the reactance of the capacitor C at the highest modulating frequency is made large enough compared to resistor R so that, the shunting of the baseband or modulating signal through the tuned circuit may be checked. Mathematical Analysis : k The capacitance C d of the varactor diode is expressed as Cd   k (vD )1/ 2 vD Here, k is a constant of proportionality and vD is the total instantaneous voltage across the varactor diode. vD  V0  m(t ) The oscillation frequency is given as c 

1

or f c 

1

LC 2 LC The varactor diode capacitance comes in parallel with the oscillator tank capacitance C. The total capacitance of the oscillator tank circuit will be C0  Cd and thus the

instantaneous frequency of oscillation i is expressed as i 

1 L0 (C0  Cd )

or fi 

1 2 L0 (C0  Cd )

In the above equation, substituting the value of C d ,

i 

i 

1 L0  C0  kvD 1/ 2 



or fi 

1

L0 C0  k V0  m(t )

1/ 2



1 2 L0  C0  kvD 1/ 2  or fi 



1

2 L0 C0  k V0  m(t )

1/ 2



The instantaneous frequency i of FM signal depends upon vD which in turn depends upon the value of the modulating signal m(t ) . Thus, the instantaneous oscillator frequency i also depends upon the baseband or modulating signal m(t ) and hence frequency modulation is generated.

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Question 78 Write the drawbacks of direct method for FM generation. Ans.

The main advantage of the direct FM generation is the simplicity of the modulators and their low cost. Disadvantages of direct method : 1. In the direct method of FM generation, we have to use the LC oscillator. The LC oscillator frequency is not stable enough. This means that the carrier generation cannot be of high stability which is a necessary requirement. 2. It is not possible to use such oscillators for the communication or broadcast purpose. Therefore we have to use the automatic frequency control scheme in which we can use the crystal oscillator to control the carrier frequency. 3. The non-linearity of the varactor diode produces a frequency variation due to harmonics of the modulating or baseband signal and therefore the FM signal is distorted. Question 79 At low carrier frequencies, it may be possible to generator an FM signal by varying the capacitance of a parallel resonant circuit. Prove that the output xc ( t ) of the tuned circuit shown in below figure is an FM signal if the capacitance has a time dependence of the form C (t )  C0  km (t ) and

Sol.

k m ( t )  1 C0

The effective tank circuit capacitance is related to modulating signal as C (t )  C0  km (t ) If m(t )  0 , then C (t )  C0 and the oscillator frequency is equal to the carrier frequency according to the expression c 

1 LC0

If we assume that km(t ) is small and slowly varying, then the output frequency the oscillator will be given by

i 

Because

1 LC (t )



1 L C0  km(t )



1   k LC0 1  m(t )   C0 



  k 1  m(t )  LC0  C0  1

k m(t )  1 , the following approximation can be used C0

1  1  (1  z)1/ 2  1    z  1  z 2 2  

1/ 2

i of

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

i 

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 km(t )   km(t )  c k m(t )  c  k f m(t ) 1    c 1    c  2C0  2C0  2C0 LC0   1

Where k f 

c k in rad/sec-volt. 2C0

Thus, xc (t ) is an FM signal.

Proved

Question 80 The figure shows that a voltage variable capacitor in a reversed-biased pn junction diode whose capacitance is related to the reverse biasing voltage v by





Cv  100 / 1  2v pF . The capacitance C0  200 pF and L is adjusted for resonance at 5 MHz when a fixed reverse voltage v  4 Volts is applied to the capacitor C v . The modulating voltage is m( t )  4  0.045sin 2  103 t . If the oscillator amplitude is 1 volt, write expression for the angle output waveform which appears across the tank circuit. [CSVTU Dec 2011, Dec 2008]

Sol.

Given : Cv 

Cv  Cv 

100 1012

F and v  4  0.045sin 2103 t

1  2v 100  1012

1  2  4  0.045sin 2 103 t  100 1012 9  0.09sin 2103 t



100  1012



1  8  0.09sin 2 103 t

100 1012 3 1  0.01sin 2103 t

1/ 2 100 1012 100 1012 1  0.01sin 2103 t   3 3 1 1 1/ 2 1  z   1    z  1  z 2  2

Cv 

At t  0 , Cv0 

 0.01 3  1  2 sin 210 t   

100 1012 F 3

100 1012 1  0.005sin 2103 t   Cv 0 1  0.005sin 2103 t  3 Cv  Cv0  0.005 Cv0 sin 2103 t  Cv0  Cv Cv 

The resonance frequency is given by, f 

1 2 L(C0  Cv )

 Cv f   1   2 L(C 0  Cv0  Cv ) 2 L(C0  Cv0 )  C0  Cv0 1

1

   

1/ 2

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f 

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  1  Cv  1 1     2 L(C0  Cv0 )   2  (C0  Cv0 )  2 L(C0  Cv0 ) 1

  Cv 1    2(C0  Cv0 ) 

  Cv 1 f  f 0 1   where f 0  2 L(C0  Cv )  2(C0  Cv0 )  0

With a fixed reverse voltage V  4 Volts and with C0  200 pF , Cv 0 

100  1012 F , 3

Cv  0.005 Cv0 sin 2103 t and f0  5 MHz

f 0 Cv f  f0   5  106  2(C0  Cv0 )

100 1012 3 100   2  200 1012  1012  3  

5  106  0.005sin 2103 t 

f  5  106  1785.7 sin 2 103 t 2f    2 5  106  2 1785.7 sin 2 103 t

Integrating  to obtain angle modulated output waveform,     dt    2 5  106 dt  21785.7 sin 2 103 t  dt

  2 5 106 t 1.7857cos 2103 t The angle modulated output waveform is expressed as, s(t )  Ac cos  Carrier oscillator amplitude is 1 Volt i.e. Ac  1 V 

s( t )  cos  2  5  106 t  1.7857 cos 2  103 t 

Ans.

Question 81 Define frequency multiplier circuit. Ans. A frequency multiplier is an amplifier whose output signal frequency is an integer multiple of the input frequency. A frequency multiplier is a combination of a nonlinear element and a band-pass filter. The process of frequency multiplication performed by the multiplier under consideration is one in which a periodic signal of frequency f serves to generate a second periodic signal of frequency nf , with n an integer.

Circuit such as in above figure is commonly used for multiplication by factors from about 2 to 5. Where higher orders of multiplications are required, multipliers may be cascaded. Multipliers of order n1 , n2 , n3 ....... yield an overall multiplication of order

n1n2 n3 ....... .

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Angle Modulation

Question 82 Draw the block diagram for conversion from NBFM to WBFM. Ans. Consider single tone NBFM, SNBFM (t )  Ac cos 2fc t   sin 2fmt  where   1 and f c is a stable sub carrier frequency. When applied to a frequency multiplier, the NBFM is converted to a WBFM signal given by, SWBFM (t )  Ac cos  n  2f c t   sin 2f m t  

SWBFM (t )  Ac cos 2(nfc )t  (n)sin 2f mt   Ac cos 2fc ' t  'sin 2f mt 

Where fc '  nf c is the final carrier frequency and  '  n is the final modulation index. Note that f m is unaffected by frequency multiplication. The maximum frequency deviation of NBFM also gets multiplied by a factor of n since f '   '/ f m  nf .

Question 83 Explain with the help of block diagram of Armstrong modulation system. [CSVTU May 2011] Or Explain Armstrong method of frequency modulation with a suitable diagram. [CSVTU May 2009, May 2008] Or Draw & explain the block diagram of an Armstrong system of generating an FM signal using multipliers to increase the frequency deviation. [CSVTU Dec 2008] Or Explain the generation of FM signals using Indirect method. What is the need of multipliers in it? Ans. Indirect method or Armstrong method of FM generation : In this method, an NBFM signal is generated first using an integrator and a phase modulator. The NBFM signal is then converted to WBFM using a frequency multiplier. A frequency multiplier is an amplifier whose output signal frequency is an integer multiple of the input frequency. The expression of FM signal in terms of modulation index is given by, s(t )  Ac cos 2fc t  sin 2fmt  where m(t )  cos 2f mt

cos( A  B)  cos A cos B  sin A sin B

s(t )  cos(2fct )cos  sin 2f mt   sin(2fct )sin  sin 2f mt 

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Mathematical identity : 1.

cos   1 for small value of 

2.

sin    for small value of 

For NBFM :   1

SNBFM (t )  cos(2fc t )  sin(2fc t )sin(2f mt )  cos(c t )  sin(c t )sin(mt )

The block diagram of below figure represents an Armstrong FM transmitter system (indirect-method) which supplies a signal whose carrier is at 96 MHz. It allows direct phase modulation (PM) of the carrier before multiplication. The multiplication process is performed in several stages in order to increase the carrier frequency as well as frequency deviation to the assigned value. The desired multiplication factors for carrier frequency and frequency deviation are not the same. Hence multiplication is done in stages.

Initially, the carrier frequency f c and the deviation f generated by NBFM, has the following typical values : fc  200 kHz, f  25 Hz which corresponds to   0.5 and

f m  50 Hz . The multiplication factor is determined by the lower frequency limit of the baseband spectrum which is (25Hz -15kHz) .

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The final carrier frequency and the deviation desired at the transmitter output is as follows : fc  96 MHz, f  75 kHz . Therefore, the multiplication factor (one shot) needed for a desired deviation is

75 kHz  3000; whereas, the multiplication factor 25 Hz

needed to achieve the desired carrier frequency is

96 MHz  480 only. Thus, if we 200 kHz

adjust the multiplication factor of 3000 for achieving the desired deviation, the final carrier frequency generated will be 600 MHz, which is much higher than the assigned value of 96 MHz and also outside the 88-108 MHz FM band. However, by performing multiplication in two stages, we can achieve both, the carrier frequency f c and deviation f as assigned. The design procedure is as follows : (1) Choose the multiplication factors of multiplier stages so that the net multiplication achieves the desired deviation f .

n1n2  3000

..…(i)

(2) Before second multiplication, the carrier frequency at the output of the first multiplier is shifted downward to (nfc1  f c 2 ) by mixing it with a sinusoidal signal of frequency f c 2 . The shifted frequency is increased n2 times by second multiplier to get the assigned carrier frequency f c  96 MHz . Therefore,

n2  n1 fc1  fc 2   fc Here fc1  200 kHz  0.2 MHz and fc 2  10.8 MHz . Substituting the values,

n2  0.2n1 10.8  96

..…(ii)

Solving equation (i) and (ii), we get n1  64 and n2  48 . For convenience in the design of the multiplier circuits, the values are rounded off, so that the multiplication may be done by factors of 2 and 3, i.e. n1  64  26 and n2  48  3  24 .

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Question 84 Give an example of Armstrong’s method using frequency multipliers which allows direct phase modulation of the carrier before multiplication as m  0.5 , where

f m  50 Hz , f  25 Hz and carrier frequency f c  96 MHz . Sol.

Maximum angular deviation, m  0.5   Modulating frequency, f m  50 Hz Frequency deviation at NBFM, f  25 Hz Carrier frequency at input, fc  200 kHz

(1) The final carrier frequency and the deviation desired at the transmitter output is as follows : fc  96 MHz, f  75 kHz . Therefore, the multiplication factor needed for a desired deviation is 75 kHz / 25 Hz  3000 Choose the multiplication factors of multiplier stages so that the net multiplication achieves the desired deviation f .

n1n2  3000

..…(i)

(2) Before second multiplication, the carrier frequency at the output of the first multiplier is shifted downward to (nfc1  f c 2 ) by mixing it with a sinusoidal signal of frequency f c 2 . The shifted frequency is increased n2 times by second multiplier to get the assigned carrier frequency f c  96 MHz . Therefore,

n2  n1 fc1  fc 2   fc Here fc1  200 kHz  0.2 MHz and fc 2  10.8 MHz . Substituting the values,

n2  0.2n1 10.8  96

..…(ii)

Solving equation (i) and (ii), we get n1  64 and n2  48 . For convenience in the design of the multiplier circuits, the values are rounded off, so that the multiplication may be done by factors of 2 and 3, i.e. n1  64  26 and n2  48  3  24 .

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Question 85 In an Armstrong modulator, the crystal oscillator frequency is 200 kHz. It is desired, in order to avoid distortion to limit the maximum angular deviation to m  0.2 .

Sol.

The system is to accommodate modulation frequencies down to 40 Hz. At the output of the modulator the carrier frequency is to be 108 MHz and the frequency deviation 80 kHz. Select multiplier and mixer oscillator frequencies to accomplish this end. Maximum angular deviation, m  0.2   Carrier frequency at input, fc  200 kHz Carrier frequency at output, fc '  108 MHz Frequency deviation at output, f '  80 kHz Modulating frequency, f m  40 Hz Frequency deviation at NBFM, f  m f m  40  0.2  8 Hz

From above figure, f '  n1n2 f

n1n2 

f ' 80 kHz   10000 f 8 Hz

…..(i)

fc '  (n1 fC  f 0 )n2  108 MHz where fc  200 kHz  0.2 MHz

 From (i), 0.2 10000  f0 n2  108  (0.2n1  f 0 )n2  108

0.2n1n2  f 0 n2  108 f 0 n2  1892

Select n2  200 then f 0  9.46 MHz From (i), n1n2  10000



n1  50

Multipliers are : n1  200, n2  50

Ans.

Mixer oscillator frequency is f 0  9.46 MHz.

Ans.

Question 86 Design an Armstrong indirect FM modulator to generate the FM carrier with carrier frequency of 98.1 MHz and f  75 kHz. A NBFM generator is available at a carrier frequency of 100 kHz and frequency deviation f  10 Hz. The stock room has oscillator with adjustable frequency in the range 10 to 11 MHz. There are frequency doublers, triplers and quintuplers available. [CSVTU May 2012]

Analog Communication Sol.

Doublers : 2

2 - 66 Triplers : 3

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Quintuplers : 5

(Quaduplers : 4)

Carrier frequency at input, f c  100 kHz Carrier frequency at output, fc '  98.1 MHz Frequency deviation at NBFM, f  10 Hz Frequency deviation at output, f '  75 kHz

From above figure, f '  n1n2 f

n1n2 

f ' 75 kHz   7500 f 10 Hz

…..(i)

fc '  (n1 fC  f 0 )n2  98.1 MHz where fc  100 kHz  0.1 MHz

 From (i), 0.1 7500  f0 n2  98.1  (0.1n1  f 0 )n2  98.1

0.1n1n2  f0 n2  98.1 f 0 n2  651.9

Select n2  60  22  3  5 then f0  10.865 MHz [10  11 MHz] From (i), n1n2  7500



n1  125  53

Multipliers are : n1  125, n2  60

Ans.

Mixer oscillator frequency is f 0  10.865 MHz.

Ans.

Question 87 Design an Armstrong indirect FM modulator to generate an FM carrier with a carrier frequency of 96 MHz and f  20 kHz. A narrowband FM generator with

f c  200 kHz and adjustable f in the range of 9 to 10 Hz is available. We also have an oscillator with adjustable frequency in the range of 9 to 10 MHz there is a bandpass filter with any center frequency and only frequency doublers are available. [CSVTU Dec 2009, Dec 2007] Sol.

Carrier frequency at input, fc  200 kHz Carrier frequency at output, f c '  96 MHz Frequency deviation at output, f '  20 kHz Frequency deviation at NBFM, f  9  10 Hz

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Carrier oscillator frequency, f0  9  10 MHz

From above figure, f '  n1n2 f

9 Hz  f  10 Hz

f ' 20000 Hz  n1n2 n1n2



f 



9 Hz 

20000 Hz  10 Hz n1n2

Since only doublers are available, therefore n1n2 has to be power of 2. By trial and error, n1n2  2048 , n1  64, n2  32 .

fc '  (n1 fC  f0 )n2  96 MHz where fc  200 kHz  0.2 MHz (0.2n1  f0 )n2  96 0.2  64  32  f0  32  96

 

0.2n1n2  f0 n2  96 f 0  9.8 MHz

Multipliers are : n1  64 , n2  32

Ans.

Mixer oscillator frequency is f 0  9.8 MHz .

Ans.

Question 88 A block diagram of an indirect (Armstrong) FM transmitter is shown in below figure. Compute the maximum frequency deviation f of the output of the FM transmitter and the carrier frequency f c if f1  200 kHz, f LO  10.8 MHz, f1  25 Hz, n1  64 and n2  48 .

Sol.

Given : f1  200 kHz , f L0  10.8 MHz , From the figure, f  n1n2 f1  64  48  25 Hz  76.8 kHz

Ans.

f 2  n1 f1  64  200 103  12.8 106 Hz  12.8 MHz

23.6 MHz f 3  f 2  f LO  12.8  10.8  106 Hz =   2.0 MHz Thus, when f3  23.6 MHz, then fc  n2 f3  48  23.6  1132.8 MHz

Ans.

When f3  2 MHz then, fc  n2 f3  48  2  96 MHz

Ans.

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Question 89 Given a frequency multiplier circuit and NBFM signal xNBFM (t )  A cos(c t   sin m t ) with   0.5 and f c  200 kHz. Let f m range from 50 Hz to 15 kHz, and let the maximum frequency deviation f at the output be 75 kHz. Determine the required

Sol.

frequency multiplication n and the maximum allowed frequency deviation at the input. Given : xNBFM (t )  A cos(c t   sin mt ) Frequency deviation, f  75 kHz

Carrier frequency, fc  200 kHz

Range of modulating frequency f m : 50 Hz to 15 kHz Modulation index is given by,   min 

f f m(max)



f fm

f 75 103 75 103 and     1500  5 max f m(min) 50 15 103

If 1  0.5 , where 1 is the input  , then the required frequency multiplication will be

n

max 1500   3000 1 0.5

Ans.

The maximum allowed frequency deviation at the input, denoted f1 , will be

f 75 103 Ans.   25 Hz n 3000 Question 90 Assume that the 10.8 MHz signal in below figure is derived from the 200 kHz oscillator by multiplying by 54 and that the 200 kHz oscillator drifts by 0.1 Hz. f1 

(a) Find the drift, in hertz, in the 10.8 MHz signal. (b) Find the drift in the carrier of the resulting FM signal.

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(a) The drift, in hertz, in the 10.8 MHz signal is  0.1 54   5.4 Hz

Ans.

(b) The drift in the carrier of the resulting FM signal is 48 Hz.

Ans.

FM Demodulation Question 91 Explain the method of FM Demodulation. [CSVTU May 2010] Or With the aid of diagram explain the demodulation of FM signals. Ans. FM demodulator or Detectors : The process of getting a modulating or baseband signals from a frequency modulated (FM) signal is called demodulation or detection. The electronic circuits which perform the demodulation process are known as FM demodulators or detectors. The FM demodulator or detector perform the extraction of modulating signal in two steps as follows : 1. It converts the frequency modulated (FM) signal into a corresponding amplitude modulated (AM) signal with the help of frequency dependent circuits, i.e. the circuits whose output voltage depends upon the input frequency. These circuits are generally known as frequency discriminators. 2. The original modulating or baseband signal is recovered from this AM signal with the help of the linear diode detector or envelope detector. A simple R-L circuit may be used as a discriminator, but this circuit has a very poor sensitivity as compared to a tuned L-C circuit. Therefore, L-C circuits are generally used as frequency discriminators. Requirements of FM Demodulator (Detector) : The FM demodulator must satisfy the following requirements : (i) It must convert frequency variations into amplitude variations. (ii) Conversion must be linear and efficient. (iii) The demodulator circuit should be insensitive to amplitude changes. It should respond only to the frequency changes. (iv) It should not be critical in its adjustment and operation. Types of FM Demodulators : FM demodulators are of following types : 1. Slope detector : The principle of operation of slope detectors depends upon the slope of the frequency response characteristics of a frequency selective network. The two main FM detectors, which use detuned resonant circuits come under this category, are as under :

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(i) Single tuned detector circuits or simple slope detector. (ii) Stagger-tuned detector circuits or balanced slope detector. 2.

Phase difference detectors : The following two circuits come under this category :

(i) Foster-Seeley detector (ii) Ratio detector Slope detector :

The circuit uses two tuned circuits which are tuned to two different frequencies called as discriminator. First one is tuned to the incoming FM carrier frequency c whereas the second one is tuned to a frequency slightly different from the carrier frequency c . This circuit converts the FM signal into an AM signal as shown in the slope detector characteristic curve. Another portion of the circuit is envelope detector. The AM signal from the output of the discriminator is applied at the input of envelope detector.

At the output of envelope detector, the original modulating or baseband signal is obtained. Although this circuit is simple and inexpensive, it has following drawbacks.

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(i) The circuit’s non-linear characteristic produces a harmonic distortion. The nonlinearity is obvious from the fact that the slope is not the same at each point of the characteristics. (ii) The circuit does not eliminate the amplitude variations and the output is sensitive to any amplitude variations in the input FM signal which is obviously not a desirable feature. A good discriminator circuit must respond only to frequency variations and not to amplitude variations. Question 92 Explain Balanced slope detector of FM detection. Also show that operation of detector is limited to small duration only. [CSVTU May 2012] Or Explain balanced frequency discriminator method for demodulator of FM signal. [CSVTU May 2008] Ans. Balanced Slope Detector or Stagger Tuned Detector : 1. The balanced slope detector is an improvement over the simple slope detector. This circuit of balanced slope detector consists of two L-C circuits. 2. It consists of two slope detector circuits. These two circuits are connected back to back, to the opposite ends of a center-tapped transformer.

3.

Due to this connection the input fed to these circuits is 1800 out of phase. The primary circuit is tuned to the IF ( f c ) .

4.

There are three tuned circuits. Out of them the primary is tuned to IF i.e. f c . The upper tuned circuit of the secondary (T1 ) is tuned above f c by f i.e. its resonant frequency is ( fc  f ) . The lower tuned circuit of the secondary is tuned below f c by f i.e. at ( fc  f ) .

5.

The secondary tuned circuits are connected to a diode detector with an RC load. Therefore, the total output of the balanced slope detector is the sum of the individual outputs of the diode detectors. R1C1 and R2 C2 are the filters used to bypass the RF ripple. V01 and V02 are the output voltages of the two slope detectors. The final output voltage V0 is obtained by taking the subtraction of the individual output voltage, V01 and V02 .

 V0  V01  V02

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Operation of the circuit : 1.

When the input frequency is equal to f c , the voltage applied to the two diodes will be identical. The dc output voltages will also be identical and thus the detector output will be zero.

2.

When the instantaneous frequency to be equal to f c  f . Since T1 is tuned to this frequency, the output of D1 will be quite large and the output of D2 will be very small, since the frequency f c  f is quite a long way from f c  f .

3.

When the input frequency is equal to f c  f , the output of D2 will be a large negative voltage, and that of D1 a small positive voltage.

4.

When the instantaneous frequency is between these two extremes, the output will have some intermediate value. It will then be positive or negative, depending on which side of f c the input frequency happens to lie.

Characteristics of the balanced slope detector : The characteristic of the balanced slope detector is as shown in below figure. Due to the shape it is called as the S-shape characteristics.

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The slope is identical at all the points on the linear portion of the characteristics curve, no harmonic distortion is caused when the operation is restricted to the linear region. Advantages : 1.

This circuit is more efficient than simple slope detector.

2.

It has better linearity than the simple slope detector.

Disadvantages : 1.

Even though linearity is good, it is not good enough.

2.

This circuit is difficult to tune since the three tuned circuits are to be tuned at different frequency, i.e. f c , f c  f and f c  f .

3.

Amplitude limiting is not provided.

Question 93 Describe the operation of Foster-Seeley discriminator with circuit diagram. [CSVTU May 2011] Or Write short note on Foster-Seeley discriminator.

[CSVTU May 2010]

Or Explain the Phase shift discriminator detector of FM demodulation. Ans.

Foster Seeley Discriminator or Phase shift discriminator : 1.

This detector circuit makes use of the frequency sensitive nature of the series resonant circuit and provides a frequency dependent phase shifting of the modulated signal. When the phase shifted signal is added to a portion of the original signal, the resultant signal will have a magnitude varying with frequency.

2.

This circuit provides frequency dependent phase shifting. The output voltage, Vout is taken across a capacitor, Vout and the current I flowing in the circuit are in phase difference of 900 for all frequencies.

3.

At resonant frequency, the series R-L-C circuit behaves as purely resistive circuit in which input voltage Vin and current I will be in phase. Above resonant frequency, the circuit is inductive, and hence current I lags Vin . Below resonant frequency, the circuit is capacitive and current I leads Vin . Thus the circuit provides frequency dependent phase shift.

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Operation of the circuit :

1.

Here, the two tuned circuits primary and secondary are resonant at the same frequency. The circuit depends upon phase shift for its operation and so is commonly called a phase shift discriminator.

2.

The centre of the secondary is connected to the top side of the primary P by the coupling capacitor C p which is almost short at signal frequency. The voltage developed across primary effectively appears across the radio frequency choke [RFC], which is denoted by V3 .

3.

The voltage V1 and V2 across half the secondary windings are 900 out of phase with the primary voltage V3 . The radio frequency voltages applied to the two diodes D1 and D2 are the vector addition of two voltages V1 and V3 [ V1 3 ] and V2 and V3 [ V2 3 ].

4.

At the resonant frequency of the tuned secondary circuit, the secondary voltages V1 and V2 are in quadrature with V3 existing across the primary inductance. However, when the applied frequency is either higher or lower than the resonant frequency of the secondary, the phase position of V1 and V2 relative to V3 will differ from 90 .

5.

The result of this situation is that at resonance the two resultant voltages V1 3 and

V2 3 are equal in amplitude. But at frequencies below resonance, the amplitude of one of these voltages is decreased while that of the other increases. Above resonance, the situation reverses. The amplitudes of the voltage V1 3 and V2 3 will vary with instantaneous frequency.

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The phasor diagrams for different input frequency ranges are shows in below figure.

The two voltages, V1 3 and V2 3 , developed by the discriminator are applied to the two diodes D1 and D2 to produce the output voltage V4  [V13 ]  [V23 ] . Then the output voltage V4 will vary with instantaneous signal frequency.

The variation of output voltage with frequency is known as the discriminator characteristic, which is also called curve for the discriminator. The phase-shift discriminator gives a very linear relation over a range of instantaneous frequencies only slightly less than the frequency separation of the two peaks of S curve. Advantages of phase discriminator : 1. As the circuit uses two tuned circuits which are to be resonated to the same center frequency, the circuit can be easily aligned practically. 2. The circuit has very good linearity between input frequency deviation and corresponding detected output. Disadvantages : The Foster-Seeley discriminator is sensitive to input amplitude variations, and hence must be preceded by the limiter stage.

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Question 94 Explain Ratio detector in FM receiving system. Ans. Ratio Detector : 1. Ratio detector is another frequency demodulator circuit. 2. The ratio detector circuit is similar to Foster seeley discriminator except for, (i) The direction of diode D2 has been reversed in polarity. (ii) A large value capacitor C5 has been included in the circuit. (iii) Method of taking the output voltage Vout .

Operation of the circuit : 1. The primary and secondary circuits are tuned to carrier frequency. C p is the coupling capacitor which is almost short at signal frequency, while L3 is RFC. It develops two output voltages, applied to the two diodes, D1 and D2 . These 2.

voltages are frequency dependent. The capacitor C5 is selected to be large in value. Any amplitude variations due to noise or other interference cause very little effect on the charge of capacitor C5 ,

3.

and the voltage across this long time constant circuit remains essentially constant which is the advantage of the ratio detector. The output voltage is not affected by amplitude variations in the signal, limiter circuit is not essential. The two resistances are chosen to be equal (R), voltage across each of them is one-half the voltage across C5 ; i.e. voltage between Q and S is half the voltage across C5 , which is constant for all frequencies.

4.

5.

At the carrier frequency, since both diodes conduct equally, the voltages across C3 and C4 are equal (at one-half the voltage across C5 ). Also, their voltage drops are equal. This makes point P and Q at the same potential, and the output voltage is zero. As the input frequency changes, so that D2 conducts more than D1 ; VC 4 increases but VC 3 decreases, but in such a way that the sum of these voltages is always constant. The output voltage available is positive. Conversely, when D1 conducts more heavily than D2 ; VC 3 exceeds VC 4 , but their sum remains constant. This makes point P negative with respect to Q. Thus the output voltage swings negative.

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Thus, in the circuit, the sum of the voltages VC 3 and VC 4 remains constant, but their ratio changes, depending on signal frequency. Hence the circuit is called ratio detector.

Advantages of ratio detector : The primary advantage of the ratio detector over the discriminator is that it is insensitive to noise and amplitude variations. This is because of the use of a large capacitor C5 . 1.

The ratio detector provides amplitude limitation; hence additional limiting stage is not required.

2.

Practically the circuit can be tuned to desired frequency, easily.

3.

The circuit provides linearity between detector output and input frequency deviations.

4.

The voltage developed across large capacitor used in circuit is proportional to average signal amplitude and hence can be used for automatic gain control purposes.

Question 95 Explain the need of hard limiter in FM demodulation. Ans.

[CSVTU May 2010]

If the amplitude Ai (t ) of the input signal is not fixed then the demodulator output will respond to the input amplitude variations as well as to frequency deviations. Ordinarily in a frequency modulated communication system the amplitude of the transmitted signal will not be modulated deliberately so that any such modulation which does appear will be due to noise. This is accomplished by passing the incoming signal through a hard limiter as shown in below figure. The purpose of the hard limiter (or comparator) is to insure that variations of amplitude Ai (t ) are removed. Below figure shows the original FM waveform with amplitude variations at the output of the hard limiter. Since the waveform has been reduced to a frequency modulated square wave, a band-pass filter is inserted to extract the first harmonic frequency f 0 . The resulting FM waveform is now applied to the FM demodulator.

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Question 96 Write the comparison of FM Demodulators. Ans.

Comparison of FM Demodulators :

Performance Parameters

Slope detector

Balanced slope detector

Foster-Seeley discriminator

Ratio detector

Tuned circuits used

One

Three

Two

Two

Linearity

Very poor

Poor

Very good

Good

Sensitivity to noise and amplitude variation

Sensitive

Sensitive

Sensitive

Insensitive

Use

Not used in practice

Not used in practice

Used in applications where linearity is critical

Used in applications where linearity is not critical

Output characteristic depends on.

Primary and secondary frequency relationship

Primary and secondary frequency relationship

Primary and secondary phase relation

Primary and secondary phase relation

Question 97 Explain how phase locked loop (PLL) can be used as frequency demodulator. [CSVTU Dec 2007] Ans.

Phase Locked Loop (PLL) : The Phase Locked Loop is a negative feedback system. It can be used to track the phase and the frequency of the carrier component of an incoming signal. Hence it is useful device for the demodulation of angle modulated signals, especially when signal to noise ratio is poor. A PLL has three basic components : 1.

A voltage-controlled oscillator (VCO)

2.

A multiplier serving as a phase detector or a phase comparator.

3.

A loop filter, which is a low-pass filter

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The frequency of VCO can be controlled by the external voltage. In a VCO, the oscillation frequency varies linearly with its input voltage. The output of the multiplier is passed through a low-pass filter (loop filter) and then applied to the input of the VCO. This voltage then changes the frequency of the VCO and keeps the loop in locked condition.

Initially, the control voltage to VCO is zero, the VCO is adjusted so that 1. The frequency of the VCO is exactly made equal to the unmodulated carrier frequency fC . The VCO output has a phase shift of 900 with respect to the unmodulated carrier wave. Suppose that the input signal applied to PLL is an FM wave : s(t )  Ac sin 2fC t  1 (t ) with a modulating signal m(t ) 2.

t

1 (t )  2k f  m(t ) dt 0

If r (t ) denotes the output of VCO, then r (t )  Av cos 2fC t  2 (t ) If v(t ) is the control voltage applied as input voltage to VCO, then t

2 (t )  2kv  v(t )dt 0

where k v is the frequency sensitivity of VCO output measured in Hz/Volt. The incoming FM wave s (t ) and the VCO output r (t ) are the two inputs to multiplier. The output e(t ) which is input to the loop filter, will be

e(t )  km [s(t )][r (t )] where km is multiplier gain.

e(t )  km  AC sin[2fC t  2 (t )] Av cos[2fct  2 (t )] 1 km AC AV sin[4fC t  1 (t )  2 (t )]  sin[1 (t )  2 (t )] 2 2sin A.cos B  sin( A  B)  sin( A  B)

e(t ) 

This is the input to the loop filter, which is a low pass filter which will attenuate the first term. 1 e(t )  km AC Av sin 1 (t )  2 (t ) 2 Then the phase error is e (t )  1 (t )  2 (t ) t

e (t )  1 (t )  2kv  v(t )dt 0



t

2kv  v(t )dt  1 (t )  e (t ) 0

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t

Assuming small error e (t ) , 2kv  v(t )dt  1 (t ) 0

t

For FM wave input : 1 (t )  2k f  m(t ) dt 0



t

t

0

0

2kv  v(t ) dt  2k f  m(t ) dt

Differentiating both sides w.r.t. time t,

kv [v(t )]  k f [m(t )] 



v (t ) 

kf kv

[ m(t )]

v(t )  m(t )

Thus the output v(t ) of the low pass loop filter is proportional to the original modulating signal. The input FM signal is demodulated and at the output of loop filter, we get the modulating signal. In PLL FM demodulator the bandwidth of the incoming FM wave can be much wider than that of the loop filter. Question 98 t   In the WBFM signal s( t )  A cos  2f c t  k  m( ) d   , m(t ) is an ergodic random    M M  1  m  process having a probability density f ( m )   M 2 2 . Obtain an  0 elsewhere expression for G( f ), power spectral density of v(t ) .

Sol.

Since G (v) (with v  f  f c ) is proportional to f (m), we have

G(v)  f (m)

…..(i)

Where  is a constant of proportionality. Since v(t )  km(t ) / 2 ,

kM kM   v  G (v )   M …..(ii) 4 4  0 elsewhere Replacing v by f  fc , and expressing the power spectral density for both positive and negative frequencies (i.e. a two-sided density), we have

   G ( f )   2M  0

fc 

kM kM  f  fc  4 4 elsewhere

…..(iii)

To evaluate  , we note that the power of the FM waveform is A2 / 2 . Hence 

 G( f ) df



fc  ( kM / 4  )



A2 2

 A2 df  for single sided spectrum  2M 2 fc  ( kM / 4  )

…..(iv)

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fc  ( kM / 4  )

2

 A2 df  for double sided spectrum 2M 2 fc  ( kM / 4  )

 M 



kM kM  A2  f   f   c  c 4 4  2 



 kM A2   M 2 2



A2 k

From equation (3) we get,

 A2  G ( f )   2kM  0 

fc 

kM kM  f  fc  4 4 elsewhere

Ans.

Question 99 t   If v ( t )  cos  c t  k  m( ) d   , where has a m( t )    1  m2 / 2 . Calculate the rms frequency deviation. f (m )  e 2

Sol .

probability

density

[CSVTU Dec 2010]

t   1  m2 / 2 Given : v(t )  cos c t  k  m() d   , f (m)  e 2   

Gaussian density function is given by, f (m) 



1  2

e

( m  )2 2 2

…..(i)

Compare equation (1) with f (m) then we get,  2  1,   0 2 2 Mean square value (MSV) is given by, MSV      1

MSV = E  m 2 (t )   1 t

From given expression : (t )  c t  k  m( ) d  

Instantaneous frequency is given by, fi 

fi 

t  1 d 1 d  (t )  c t  k  m() d   2 dt 2 dt   

1 k  2fc  k m(t )  f c  m(t ) 2 2

Frequency deviation is given by, f  fi  f c

E  f   2

k2 k2 2   E m ( t )   42 42 

max



k m(t ) max 2 E  m 2 (t )   1

RMS frequency deviation is given by, f rms  E  f   2



f rms 

k 2

k2 4 2 Ans.

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Question 100 k   Consider the FM signal v ( t )  cos  c t    k cos(k t   k )  k 1  

Let  k k  1 for each k . (a) Find B if B  2(f )1  (f )2  .....  (f )k  (b) Find B if B  2(f )1rms  (f )2rms  .....  (f )krms  Sol.

Given : k k  1

…..(i)

The modulation index in FM is given by,  

f where f is frequency deviation and fm

f m is modulating frequency. So, k 

(f ) k fk

…..(ii)

From equation (i) and (ii),

(f )k 1  2f k  1  (f )k  fk 2

(a) B  2(f )1  (f )2  .....  (f )k  k

B  2 (f )k  2  k  f  2  k  k 1



B

1 2

k 

Ans.

(b) B  2(f )1rms  (f )2rms  .....  (f )krms  k

B  2 (f )krms  2k  f rms k 1

Frequency deviation in FM is given by, f 

f 

1 d 2 dt

k

 k 1

k

cos (k t  k )

1 k k cos (k t  k ) k f k cos (k t  k ) 2

f rms  E  f   2

1  cos (2k t  2k )  E  f k22k cos2 (k t  k )   f k k E   2  

E represents expected value or average value.

1 1 E   2 2 f rms 

f k k

 cos(2k t  2k )  E 0 2  

B  2  k  f rms  2k 

2 From equation (1) we get, B

k 2

f k k 2



2k 2f k k 2k k k  2 2 2 2

Ans.

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Angle Modulation

2 - 83

Question 101 The two independent modulating signals m1 ( t ) and m2 ( t ) are both Gaussian and both of zero mean and variance 1 volts 2 . The modulating signal m1 ( t ) is connected to a source which can be frequency-modulated in such manner that, when m1 ( t )  1 volt (constant), the source frequency, initially 1 MHz, increases by 3 kHz. The modulating signal m2 ( t ) is connected in such a manner that, when m2 ( t )  1 volt (constant), the source frequency decreases by 4 kHz. The carrier amplitude is 2 volts. The two modulating signals are applied simultaneously. Write an expression for the power spectral density of the output of the frequency-modulated source. Sol.

Given : Vc  2 Volts, f c  1 MHz , m1 (t )  1 V , m2 (t )  1 V , f1  3 kHz and f 2  4 kHz The angle modulated waveform in terms of FM is given by, t t   sFM (t )  Vc cos c t  k1  m1 () d   k2  m2 () d      

The frequency deviation for modulating signal m1 (t ) is given by,

f1 

t k m (t ) 1 d k1  m1 () d   1 1 2 dt  2

k1  1 2

3  103 



k1  6103 Hz

The frequency deviation for modulating signal m2 (t ) is given by,

f 2 

t k m (t ) 1 d k2  m2 () d   2 2 2 dt  2

4  103 

 f rms 

2

k2  1 2



k2  8 103 Hz

2 t  1 d  t       E  k1  m1 () d   k2  m2 () d      2 dt        

E  m1 () 2   1

E  m2 () 2   1

E  m1 () 2 m2 () 2   0

since both are independent signals.

 f rms 



1 1  k12  k22  k12 E  m1 ()2   k22 E  m2 ()2   (2)2 (2) 2 

 f rms 



(6π  103 ) 2  (8π  103 ) 2 36  106  64  106   25  106 4 (2π) 2

2

2

f rms  5 kHz

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The power spectral density of the output of the frequency modulated source is given 2  ( f  fc )2    ( f  fc )2 2 1 2( f rms ) e by, G( f )   e 2( frms )   f rms 2π     ( f 10 )    ( f 106 ) 5010 50106 e  e   5 103  2π   6 2



G( f ) 

6 2

1

Ans.

Question 102 k   An FM signal is given by v ( t )  cos  c t    k cos( k 0 t   k )  k 1  

(a) If  k  0 and k = 1, 2 find the maximum frequency deviations. (b) If each k is an independent random variable, uniformly distributed between

  and  , find the rms frequency deviation. (c) Under the condition of (b) calculate the rms phase deviation. Sol.

Given : k  0 and k = 1, 2 (a) v(t )  cos 0t  1 cos 0t  2 cos 20t 

f 

1 d 1 1 cos 0t  2 cos 20t   10 sin 0t  220 sin 20t  2 dt 2

f max 

 1 max  10 sin 0t  22 0 sin 20t   0 max 1 sin 0t  2 2 sin 2 0t  2 2

f max 

0 max 1 sin 0 t  42 sin 0 t cos 0 t  2

f max 

0 max 1 sin 0 t  42 sin 0 t 1  sin 2 0 t    2

sin 2  2sin .cos 

Let x  sin 0t , then f max  f0 max 1 x  42 x 1  x 2    Consider A  1 x  42 x 1  x 2 , 1/ 2 dA 1  1  42 x  1  x 2   2 x  42 1  x 2  0 dx 2

1  42 1  x 2  42 x 2 1  x 2  Multiply by

1/ 2

1  x2 , 1 1  x 2 

1/ 2

82 x 2  42  1 1  x 2 

 42 1  x 2   42 x 2 82 x 2  42   12 1  x 2 



1/ 2

6422 x 4  1622  6422 x 2  12 1  x 2 

2



6422 x 4   12  6422  x 2  1622  0

2 Let x  y , then 6422 y 2   12  6422  y  1622  0

ay 2  by  c  0



y

b  b 2  4ac 2a

GATE ACADEMY PUBLICATIONS ®

x  2

6422  12 



2 1

 6422   4 1622  6422 2

2  6422

 2 2  642  1  sin 0 t  x    f max

Angle Modulation

2 - 85

1/ 2

2 2   6422    6422    12822 



2 1

 2 2  642  1  2    f 0 1 x  42 x 1  x where sin 0 t  x     

1/ 2

2 2   6422    6422    12822 



2 1

Ans. (b) RMS frequency deviation, f rms  E  f 

2

E represents expected value or average value. k 1 d k 1 f  k cos(k 0t  k )  k  k 0  sin(k 0t  k )  2 dt k 1 2 k 1 k

f  k f 0 k  sin(k 0t  k ) k 1

2

k    k  f rms  E k f 0 k  sin(k 0t  k )   k f 0 k E   sin(k 0t  k )  k 1    k 1 

2

 k 1  cos(2k 0t  2k )   f rms  k f 0 k E    2   k 1  Average value of a constant is a constant and that of a sinusoidal signal is zero. 

f rms  k f 0 k

k 2

Ans.

(c) RMS phase deviation, rms  E  

2

2

 k   k  rms  E   k cos (k 0t  k )   k E   cos (k 0t  k )   k 1   k 1  

rms  k

k 2

2

Ans.

Question 103 Derive the bandwidth required for a Gaussian modulated WBFM signal so that the modulation may be recovered without distortion. Ans.

We know that when the carrier was sinusoidally modulated, Carson’s rule B  2(f  f m ) specifies the bandwidth required to transmit enough of the power (98%) so that the modulation may be recovered without distortion. If we consider m(t ) as a Gaussian signal, then power spectral density will be Gaussian.

Analog Communication

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2 - 86

 m2

1

Let m(t ) having density function is given by, f (m) 

e 2 where mean = 0, 2

 2

variance = 2 and   f rms

f ( m) 

 m2

1 f rms 2

e

2( f rms )2



Spectral density, G(v)  f (v) Since,









 f



 v2

 rms

2

e

2( f rms )2

2

Let

v x 2(f rms )2 2 f rms 2





A2 dv  2

x

e (f rms )dx 2x

0







2 0

2v dv  dx 2(f rms ) 2





f rms 2

e

v2 2( f rms )2

A2 2

 G( f ) df   G(v) dv 



G (v ) 





2

A 2

x

f rms 2

dx(f rms )2 2 x (f rms )



 

By using definition of Gamma function,

dv 

 v2





e

2( f rms )2



dv 

A2 2

(f rms )dx 2x



1/ 2  x  x e dx  0

A2 2

n 1  x

e dx  n

0

 1 A2  2  2





A2  2





 

G (v ) 

A2 2  v2

A2 2 2f rms

e

2( f rms )2

Let v  f  f c and G( f )  Two sided power spectral density Then, G( f ) 

  f  fc 

A2 4 2f rms

e 2( frms )

2

2

Where A  Aptitude of the FM waveform, f rms  standard deviation. We considered a rectangular band pass filters, centered at f c which will pass 98% of the power of the FM waveform.

A2 2 2 f rms

 v2

B/2



B/2

e

2( f rms )2

A2 dv  0.98  2



1 2f rms

 v2

B/2



B/2

e

2( f rms )2

dv  0.98

GATE ACADEMY PUBLICATIONS ®

Let

v 2f rms e x

2

2

 v2

B/2

2

For even function,

Angle Modulation

2 - 87

2 f rms

e

2( f rms )2

dv  0.98

0

 x , then dv  2f rms dx B / 2 2f rms





e x dx  0.98 2

Error function, erf (t ) 

t

2

e 

 y2

dy

0

  B Oncomparing, erf    0.98  2 2f rms  B 2 2f rms 



 erf 1 (0.98)

B  2 2(1.645)  f rms

B  4.6 f rms

This B represents the Gaussian bandwidth without distortion. Question 104 Consider that the WBFM signal having the power spectral density of 2  ( f  f c )2    ( f  fc2) 2 A2 2  f  e rms  e 2 frms  is filtered by a Gaussian filter having the G( f )    4f rms 2   band-pass characteristics H ( f )  e  ( f  fc ) 2

2

/ 2 B2

 e  ( f  fc )

2

/ 2 B2

Assume fc  B .

2

(a) Sketch H ( f ) as a function of f. (b) Calculate the 3-dB bandwidth of the filter in terms of B. (c) Find B so that 98 percent of the signal power of the WBFM signal is passed. Sol.

(a)

H ( f )  e  ( f  fc ) 2

2

/ 2 B2

 e  ( f  fc )

At f  0 , H (0)  e fc 2

2

/ 2 B2

At f   f c , H ( f )  e 2

/ 2 B2

 e  fc

At f  f c , H ( f c )  e  0  e 2

2

2

/ 2 B2

 (2 fc )2 / 2 B 2

 ( 2 fc )2 / 2 B 2

e

0

1 0

1

(b) Given H ( f )  S ( f ) is a two sided spectral density function. 2

S (v )  e



v

2

2 B2

where v  f  f c for f  0 and v  f  f c for f  0

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Condition for 3-dB frequency or half-power frequency : VB2

 1 2  e 2B 2

VB2 1 ln     2 2B 2



So, VB  1.177 B 3-dB bandwidth of the filter is B3 dB  2VB  2.354 B

Ans.

(c) The band-pass fitter centered at f c will pass 98% of the power of the FM waveform. B  4.6 f rms Where B is the Gaussian bandwidth without distortion.

B3-dB  2.354B

4.6 f rms  2.354 B



B  1.95 f rms

Ans.

Question 105 Explain stereophonic FM Broadcast system.

[CSVTU May 2012]

Or Write a short note on stereophonic FM transmitter.

[CSVTU Dec 2010]

Or Explain stereophonic FM transmitter. Ans.

[CSVTU Dec 2007]

Stereophonic FM transmitter system : 1.

A stereo FM transmission is a system in which, adequate information is sent to the receiver in order to reproduce the original stereo sound signal.

2.

The stereo FM system is made compatible with the usual mono FM systems, so that it is possible to receive FM stereo transmission even using mono FM receiver.

3.

The block diagram of stereo FM multiplex generator is shown in figure.

Operation of the System : 1.

The two input channels are passed through a matrix which produces two outputs namely, (L + R) the sum component and (L – R), the difference component.

2.

The sum component (L + R) modulates the carrier in the same manner as the signal in the mono FM system. This signal is demodulated and reproduced by a mono FM receiver tuned to stereo FM transmission.

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Angle Modulation

3.

The difference signal (L – R) is applied to a balanced modulator. It will suppress the 38 kHz subcarrier applied to it and produces two sidebands.

4.

If we assume that the frequency range of input signal is 50 Hz to 15 kHz, then the two sidebands produced by the balanced modulator will occupy a frequency range from 23 kHz to 53 kHz as shown in figure.

5.

A 19 kHz subcarrier generator generates a 19 kHz stable subcarrier which is multiplied by two in the frequency doubler and used as a carrier for the balanced modulator, where two sidebands of (L – R) signal are produced.

6.

The sum signal (L + R), the 19 kHz subcarrier, the two sidebands of (L – R) signal produced by the balanced modulator and the output of SCA generator are added in the adder to form a modulating signal.

7.

This modulating signal is then used to modulate a carrier in the frequency modulator to produce FM signal.

8.

A "subsidiary communications authorization" (SCA) signal is an optional signal which may be transmitted along with the other signals as shown in the spectrum of figure. SCA uses a subcarrier at 67 kHz modulated to a depth of  6.5 kHz, by the audio signal.

9.

Frequency modulation is used. The frequency band occupied by the SCA signal is from 59.5 kHz to 74.5 kHz.

10. Some stations provide SCA as a second or medium quality transmission which can be used as background music in stores, restaurants etc. (SCA signals can contain telemetry, facsimile signals etc).

FM Receiver Question 106 Draw the block diagram of FM receiver and explain briefly each block. Ans.

FM Receiver : The function of the FM receiver is to intercept the FM signal incoming from an FM transmitter, and then recover the original modulating signal. Below figure shows the block diagram of an FM broadcast receiver which is similar to an AM superheterodyne receiver.

Analog Communication

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The stages of the block diagram are described as follows : 1.

RF amplifier : This stage amplifies the received radio signal. FM uses radio frequency ranging from 40 MHz to 1 GHz for various applications such as FM broadcasting, television sound transmission, police radio, military systems, etc. The bandwidth needed for an RF amplifier may also be large (150 kHz). The RF amplifier also rejects the image signal, as in AM receivers.

2.

Frequency mixer and Local oscillator : Separate active devices are used for the mixer and local oscillator as the frequency involved is in VHF and UHF range. The IF frequency in an FM receiver is much higher than AM. The FM broadcast receiver use 10.7 MHz as intermediate frequency.

3.

IF amplifier : This stage amplifies the intermediate frequency signals. It comprises multistage double tuned or stagger tuned amplifiers to provide a high gain and a high overall bandwidth (150 kHz). This stage is responsible for sensitivity and selectivity of the receiver.

4.

Limiter : The limiter keeps the IF amplifier output voltage constant to a predetermined value and removes all amplitude fluctuations due to noise, or any other interference. This is essential because the FM detector needs a constant amplitude FM voltage at its input for a satisfactory operation. The limiter may be ignored if the ratio detector has been used. Simple diodes and amplifying devices are used in making a limiter circuit.

5.

FM detector : It recovers the modulating signal from the IF signal. Various types of FM detector circuit have already been discussed. De-emphasis circuit does the inverse job of the pre-emphasis circuit. The high modulating frequency terms boosted by pre-emphasis are brought back to original amplitude level by deemphasis circuit.

6.

AF amplifier and speaker : This stage amplifies the audio frequency (AF) modulating signal recovered by the FM detector. The amplifier has wider bandwidth than that of AM receiver. The loudspeaker converts the electrical signal into sound signal.

Question 107 Difference between FM and AM receiver. Ans.

Difference between FM and AM receiver are : 1.

The operating frequencies in FM are much higher than in AM.

2.

The IF frequency in an FM receiver is much higher than AM.

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Angle Modulation

3.

The FM receivers need the circuits like limiter and de-emphasis.

4.

The FM demodulators are different from AM detectors.

5.

The method to obtain the AGC is different in FM receivers.

Question 108 A carrier signal is frequency modulated by a sinusoidal signal of amplitude A m and frequency fm . In a certain experiment conducted with fm = 1 kHz and increasing A m (starting from 0 V). It was found that the carrier component in the spectrum was reduced to zero for the first time when Am = 2 V . (i) Determine frequency sensitivity of the modulator. (ii) Determine the value of A m for which the carries component will be reduced to zero for the second time.

[CSVTU May 2013]

Sol.

The amplitude of the carrier in FM wave is Ac J 0 () . The means that if we can make

J 0 ()  0 , the carrier gets suppressed in the FM waveform. The typical values of  for which J 0 ()  0 are 2.44, 5.52, 8.65, 11.2 etc. These values are shown in figure. (i) Given : Am  2 Volts; f m  1 kHz

  2.44 since the first time J 0 () is zero 

f k f Am  fm fm

Frequency sensitivity, k f 

f m 2.44  1 103   1.22  103 Hz/V Am 2

Ans.

(ii) Also,   5.52, since for the second time J 0 () is zero. Hence, Am 

f m 5.52 1103   4.52 Volts kf 1.22 103

Ans.

Question 109 In a FM system, when the audio frequency (AF) is 500 Hz and the AF voltage is 2.4 V, the deviation is 4.8 kHz. If the AF voltage is now increased to 7.2 V what is the new deviation? If the AF voltage is raised to 10 V while the AF is dropped to 200 Hz, what is the deviation? Find the modulation index in each case. [CSVTU Dec 2012]

Analog Communication Sol.

2 - 92

GATE ACADEMY PUBLICATIONS ®

For Am  2.4 V and f m  500 Hz : Frequency deviation, f  4.8 kHz

Modulation index,  

kf 



f  k f Am

f 4.8   2 kHz/Volt Am 2.4

f 4800   9.6 fm 500

Ans.

For Am '  7.2 V and f m '  500 Hz : Frequency deviation, f '  k f Am '  2  7.2  14.4 kHz Modulation index,  ' 

Ans.

f ' 14400   28.8 fm ' 500

Ans.

For Am "  10 V and f m "  200 Hz : Frequency deviation, f "  k f Am "  2 10  20 kHz Modulation index,  " 

Q.1

Ans.

f " 20000   100 fm " 200

Ans.

The signal m(t) as shown is applied both to a phase modulator (with k p as the phase constant) and a frequency modulator (with k f as the frequency constant) having the same carrier frequency.

[GATE 2012, IIT-Delhi]

The ratio k p / k f (in rad/Hz) for the same maximum phase deviation is (A) 8 Sol.

(B) 4

(C) 2

Given :

Fig. (a)

(D) 

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Fig. (b)   S FM (t )  Ac cos  2 f c t  2 k f  m( ) d  where k f in Hz/volt 0   t

S PM (t )  Ac cos  2 f c t  k p m(t )  where k p in rad/volt

Given : Same Maximum phase deviation Phase deviation is given by,   (t ) max t

For FM (t )  2k f  m() d  0

For PM (t )  k p m(t )

t  k p  m(t )Max  2 k f   m( ) d  0  Max From Fig. (a) m(t )max  2 t From Fig. (b)   m() d   4  0  max k p (2)  2  k f (4)

kp kf

Ans. Q.2

 4

(B) Consider an angle modulated signal X(t)  6 cos[2106 t  2sin(8000t)  4 cos(8000t)]V . The average power of X(t) is : [GATE 2010, IIT-Guwahati] (D) 28 W

Sol.

Ans. Q.3

(A) 10 W (B) 18 W (C) 20 W Average power of FM is given by V 2 62 P C   18 W 2 2 (B) Consider the frequency modulated signal 10 cos[2105  5sin(21500t)  7.5sin(21000t)] with carrier frequency of 105Hz . The modulation index is : (A) 12.5

(B) 10

(C) 7.5

[GATE 2008, IISc-Bangalore] (D) 5

Analog Communication Sol.

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Modulation index for multi-tone is given by,  

f W

Where f = maximum frequency deviation

W = maximum frequency component Given that : f m1 = 1500 Hz, f m 2  1000 Hz W  f m1  1500 Hz FM (t)  10 cos[2105t  5sin(21500t)  7.5sin(21000t)]   2 105t  5sin(2 1500t)  7.5sin(2 1000t) i 

d  2 105  2 7500 cos(2 1500t)  2 7500cos(21000t) dt

( fi  fc)max  (7500  7500)

f  15000 Hz 

f 15000   10 W 1500

Ans.

(B)

Q.4

The signal cos ct  0.5cos mt sin ct is :

Sol.

[GATE 2008, IISc-Bangalore]

(A) FM only

(B) AM only

(C) Both AM and FM

(D) Neither AM nor FM

s(t )  cos c t  0.5cos mt sin c t s(t )  cos c t  0.25[2sin c t cos mt ] s(t )  cos c t  0.25[sin(c  m )t  sin(c  m )t ]

s(t ) represents AM signal (DSB – FC). If we consider envelope X  A cos   B sin  X 

A2  B 2 cos(  )

  tan 1 B / A

s(t )  12  (0.5cos mt )2 cos[c t  tan 1 (0.5cos mt)]  1  0.25cos2 mt cos[c t  tan 1 ] tan 1 

 (if  is small)

 [1  0.125cos 2 m t ]cos (c t  )

 [1.0625  0.125cos 2t ]cos(c t  0.5cos mt ) 1.0625  0.125cos 2t 1.0625 (since DC power is higher than the AC power)

s(t ) 1.0625cos(c t  0.5cos mt )

s(t ) represents NBFM signal. Ans.

(C)

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Q.5

Sol.

2 - 95

Angle Modulation

A carrier is phase modulated (PM) with frequency deviation of 10 KHz by a single tone frequency of 1 KHz. If the single tone frequency is increased to 2 KHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is : [GATE 2005, IIT-Bombay] (A) 21 KHz (B) 22 KHz (C) 42 KHz (D) 44 KHz Given : f  10KHz, f m  1KHz



f  10 fm

  p    10

f m  2 kHz Bandwidth for PM system is given by BW  2  p  1 f m  2 10  1 2  44 KHz Ans. Q.6

(D) A device with input x(t ) and output y (t ) is characterized y (t )  x 2 (t ) . An FM signal frequency deviation of 90 KHz and modulating signal bandwidth of 5 KHz is applied to this device. The bandwidth of the output signal is : [GATE 2005, IIT-Bombay] (A) 370 KHz (B) 190 KHz (C) 380 KHz (D) 95 KHz

Sol.

Bandwidth of FM is given by BW [ y(t )] = 2(f  f m ) BW = 2(180  5) BW = 370 KHz Ans. (A) Note : When FM signal applied to frequency doubler then frequency deviation doubles but f m Q.7

Sol.

remains the same. Two sinusoidal signals of same amplitude and frequencies 10 KHz and 10.1 KHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is : [GATE 2004, IIT-Delhi] (A) 0.1 KHz sinusoid (B) 20.1 KHz sinusoid (C) a linear function of time (D) a constant s(t)  A cos[210000 t]  A cos[210100 t]

fi  10000 Hz , f 2  10100Hz

T1  100  sec , T2  99  sec s(t ) will be periodic with period LCM of T1 & T2 = 9900  sec

T0  9900  sec

f0  Ans.

(A)

1  101.01Hz  0.1KHz T0

Analog Communication Q.8

2 - 96

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Let m(t )  cos[(4 103 )t ] be the message signal and c(t )  5cos[(2106 )t ] be the carrier, c (t ) and m(t ) are used to generate an FM signal. If the peak frequency deviation of the generated FM is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos[2π(1008  103 t )] in the FM signal (in terms of the Bessel coefficients) is : (B) 5 J 8(3)

(A) 5 J 4 (3) Sol.

[GATE 2003, IIT-Madras] 2

(C)

5 J 8 (4) 2

(D) 5 J 4 (6)

Given : m(t )  cos[(4 103 )t ] , c(t )  5cos[(2106 )t ]

f m  2kHz, fc  1MHz, Vc  5V The expression of WBFM is given by,

X FM  t   Ac



 J    cos  

n 

n

c

 nm  t

…….(i)

Transmission bandwidth of AM = 2 f m

f  3  B.W.  6 f m



f 6 fm

Given : cos[2π(1008  103 t )] c  n m  2 1008  103

2106  n.4103  21008 103 n4 Bessel co-efficient = Vc J n ()  5J 4(6) Ans. Q.9

(D) The scheme shown in below figure is used for the generation of wideband FM from a narrow band FM. The multiplier box multiplies the input frequency by the factor shown. The input x(t ) is a narrow band FM signal of carrier 100 kHz and frequency deviation of 25 Hz . The local oscillator frequency in kHz and the multiplication constant m to achieve an output y (t ) with a carrier of 2.0 MHz and a frequency deviation of 1.0 kHz are respectively :

Sol.

(A) 750,4 Given :

(B) 1000,4

[GATE 2003, IIT-Madras]

(C) 750,8

(D) 1000,8

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Mixer only changes the carrier frequency.

f '''  5  m  f f '''  5  m  25

1000  125m fc " 



m= 8

f c "'  250 kHz 8

From figure f c "  fl  f c '

or

f c "  f c ' fl

250kHz  fl  500kHz

…….(i)

250 kHz  500 kHz  fl

…….(ii)

From equation (i) and (ii), we get

fl  750kHz or 250 kHz Ans.

(C)  



Analog Communication

2 - 98 Space for Notes

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