Singnal and System

Communication Systems (Vol-2)

Communication Systems

Kuestions
Type 1 – Probability Theory, Random Variables and Processes .................................. 2
Type 2 – Autocorrelation and Power Spectral Density .................................................. 5
Type 3 – Properties of White noise ........................................................................................ 9
Type 4 – Amplitude Modulation and Demodulation ................................................... 11
Type 5 – Angle Modulation and Demodulation ............................................................. 14
Type 6 – Superheterodyne Receivers ................................................................................. 18
Type 7 – Entropy and Mutual information ....................................................................... 20
Type 8 – Channel Types and Capacity ............................................................................... 21
Type 9 – Pulse Modulation methods-PCM, DPCM ........................................................ 23
Type 10 – Baseband Shaping for Data Transmission .................................................... 26
Type 11 – Digital Modulation methods-ASK, FSK, PSK AND QAM ............................ 28
Type 12 – SNR and BER for Digital Modulation Schemes............................................ 30
Type 13 – MAP and ML decoding and Matched Filter Receiver ................................ 32
Type 14 – Basics of TDMA, FDMA and CDMA ................................................................... 35
Solution ......................................................................................................................................... 37

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 Communication Systems (Vol-2)

Kuestions

Type 1 – Probability Theory, Random Variables and Processes

For Concept, refer to Communication K-Notes, Random Process and Noise

 Common Mistake / Point to remember

• Special care has to be taken while using error function and complementary error function.
• erf(x)+erfc(x)=1
• Normal Gaussian pdf has mean=0, variance=1.

Sample Problem
Let X1, X2 and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. The

probability P X1 , + X 2  X 3  is.
Solution: 0.16

X1 + X2  X3

X1 + X2 − X3  0

 
So, P X1 + X 2 − X 3  0 = P Y  0  
and Y is another random variable whose probability density function is given by convolution of X 1. X2, and X3.

)  Y2 dy
2

(
So, P Y  0 =
−1
0
Y 
3
P ( Y  0 ) =   = 0.16
 6  −1

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 Communication Systems (Vol-2)

Sample Problem
− (x − 4)2
1
The PDF of a Gaussian random variable X is given by Px (x) = e 18
.The probability of the event {X = 4} is
3 2
1 1
(a) (b)
2 3 2
1
(c) 0 (d)
4

Solution: Option (C) is correct

Probability of a Gaussian random variable is defined for an interval and not at a point, So at X = 4, it is zero.

Sample Problem
1 1
Let U and V be two independent zero mean Gaussian random variables of variances and respectively. The
4 9
probability P (3 V > 2U) is
(a) 4/9 (b) 1/2
(c) 2/3 (d) 5/9

Solution: Option (B) is correct

The probability
P ( 3 V − 2 U) = P ( 3 V − 2 U  0 ) = P ( W  0 )
Where W = 3 V – 2 U
U and V are independent random variables and can be expressed in terms of mean and variance as shown below:
 1
U = N  0, 
 4
 1
V = N  0, 
 9
W=3V–2U
 1 1
 W = N  0,9  + 4   = N ( 0,2.7 )
 4 9
Hence W is Gaussian variable with 0 mean having pdf curve as shown below:

1
 P (W  0) = = Area under the curve from 0 to  .
2

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Problems

( ) ( )
01. A probability density function is given by p x = K exp −x 2 2 ; −  x   . The value of K should be

( 2 )
(a) 1 (b) (2 )
(c) (1 2  ) (
(d) 1  2 )
( )
02. If X is a standardized Gaussian Random Variable, the probability P X  2 is given by

(a) 0.5Q ( 2) (b) 0.5erfc 2 ( )

(c) 0.5erfc ( 2 ) (d) 0.5Q 2 ( )
03. A signal m(t) of bandwidth ‘B’ Hz is Gaussian distributed with zero mean and variance 
The probability that the random variable m  −2 is
(a) Q(2) (b) Q(-2)
(c) 1-Q(2) (d) 1+Q(2)

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 Communication Systems (Vol-2)

Type 2 – Autocorrelation and Power Spectral Density

For Concept, refer to Communication K-Notes, Random Process and Noise

 Common Mistake / Point to remember

• A common mistake is made while using transfer function of low pass filter and high pass filter.
• Common mistake is made while taking expectation operation, do consider that particular random variable.
• Fourier transform of Autocorrelation function can never be negative or complex.

Sample Problem

The auto-correlation function of an energy signal has

(a) no symmetry
(b) conjugate symmetry
(c) odd symmetry
(d) even symmetry

Solution: Option (B) & (D) is correct

Auto-correlation function of energy signal has conjugate symmetric
R x (  ) = R *x ( − )
If the function is real, then the autocorrelation function has even symmetry
R x (  ) = R x ( − )

Sample Problem

X   
n = n =
n n =−
is an independent and identically distributed (i.i.d.) random process Xn equally likely to be + 1 or – 1 Yn is
n =−

 
n =
another random process obtained as Yn = Xn + 0.5 Xn−1 . The autocorrelation function of Yn denoted by R y k  , is
n =−

(a) (b)

(c) (d)

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 Communication Systems (Vol-2)

Solution: Option (B) is correct

R y (K ) = R y (n,n + K ) = E  Y n Y n + k  

Given
1
P  x (n) = 1  =
2
1
P  x (n) = −1  =
2
Y[n] = X [n] + 0.5 [n]
R y (K ) = E  x n + 0.5x n − 1   x n + k  + 0.5x n + k − 1  

= E  x n .x n + k   + 0.5E  x n x n + k − 1   + 05E  x n − 1  x n + k   + 0.25E  x n − 1  x n + k − 1  
       
( ) ( ) (
= R x K + 0.5R x K − 1 + 0.5R x K + 1 + 0.25R x K ) ( )
( ) ( )
= 1.25R x K + 0.5R x K − 1 + 0.5R x R + 1( )
= 1.25R x (K ) + 0.5R (K − 1 ) + 0.5R (K + 1 )
x x

Now if K = 0
1 1
R x ( 0 ) = E  x 2 n  = 1  +1 = 1
2 2
K0
R x (K ) = E  x (n) x (n − k )  = E  x (n)  E  x (n − k )  = 0
 1 1 
 ()
Because  E  x n  = 1  − 1  = 0 
 2 2
 
R y ( 0 ) = 1.25R x ( 0 ) + 0.5R x ( −1 ) + 0.5R x (1 ) = 1.25

Similarly,
R y (1 ) = 0.5
R y ( −1 ) = 0.5

Sample Problem

X (t) is a stationary process with the power spectral density Sx (f) > 0 for all f. The process is passed through a system
shown below.

Let SY (f) be the power spectral density of Y(t). Which one of the following statements is correct?
(a) SY (f) > 0 for all f
(b) SY (f) = 0 for |f| > 1 kHz
(c) SY (f) = 0 for f = nf0, f0 = 2 kHz, n any integer
(d) SY (f) = 0 for f = (2n + 1) f0, f0 = 1 kHz, n any integer

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 Communication Systems (Vol-2)

Solution: Option (D) is correct

(
y1 ( t ) = x ( t ) + x t − 0.5  10 −3 )
 j2 f ( 0.510−3 ) 
y1 ( f ) = X ( f ) 1 + e 

Y1 ( f )
H1 ( f ) =
−3
= 1 + e− jf 10
X (f )
− j( f 10−3 ) 
H1 ( f ) = H1 ( f ) .H2 ( f ) = ( j2f ) 1 + e 


( )
H ( f ) = 4 2 f 2 2 + 2cos f  10 −3  = 82 f 2 1 + cos f  10 −3 
2

 
() ( )
S x ( f ) = 82 f 2 1 + cos f  10 −3  S x ( f )
2
Sy(f) = H f
 
For f = (2n + 1)f0; with f0 = 1 kHz
f  10−3 is an odd multiple of 
Sy (f) = 0

Problems

() ( )
01. A signal having an auto correlation function R  = 100 sinc 100  is pass through an ideal LPF with pass band
magnitude of 1 and cut-off at 25Hz. The normalized energy of the output signal in joules is
(a) 25 (b) 50
(c) 100 (d) 75

02. Consider a random process x t = 100 cos 50 t +  () ( )

(
‘  ’ is a RV uniform distributed in the interval −,  , the PSD is )
( ) (
(a) 25 f − 25 + 25 f + 25 )
(b) 50 ( f − 50 ) + 50 ( f + 50 )

(c) 25 ( f − 50 ) + 25 ( f + 50 )

(d) 50 ( f − 25 ) + 50 ( f + 25 )

03. The impulse response of a (LTI system) is given by

h ( t ) =  ( t ) − e−tu ( t )
Where  is a positive constant if the input to the filter is White Gaussian noise with the PSD N0 / 2 W/Hz. The ACF of the
output is

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 Communication Systems (Vol-2)

()
(a) 0.5N0   t − 0.5e−t 

(b) 0.5N   ( t ) − e  −t

 0 
(c) 0.5N   ( t ) − 0.5e −t

 0 
(d) 0.5N   ( t ) − 0.5e 
− t

 0


Statements for linked answer Q4 and Q5 is given below

Consider a random process Y t = X t cos 2fc t +  () () ( )
()
Where X t is a WSS random process having ACF R x (  ) and PSD S ( f )
x
 is a random variable uniformly distributed in
the interval 0 to 2 ( )
X ( t ) and  are independent

04. The ACF of the random process Y t is ()

()
(a) 0.5R x  cos2fc t ()
(b) 0.5R x  cos 4 fc t

(c) 0.25R (  ) cos2f t

x c
(d) 1.5R x (  ) cos2f t
c

05. The PSD of the random process Y t is ()

( )
(a) 0.5 S X f − fc + S X f + fc  ( )
( )
(b) 0.25 S X f − fc + S X f + fc  ( )
( ) ( )
(c) 0.125 S X f − fc + S X f + fc 

(d) 0.215 S X ( f − f ) + S ( f + f )

c X c

06. X(t) and Y(t) are two independent stationary process of means 2 and 3 respectively. Then, the magnitude spectrum of
R xy (  ) * e−2  .u (  ) is
(a) Double sided exponential with respect tot frequency
(b) Constant with respect to frequency
(c) Linear with respect to frequency
(d) Gaussian with maximum at zero respect to frequency

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 Communication Systems (Vol-2)

Type 3 – Properties of White noise

For Concept, refer to Communication K-Notes, Random Process and Noise

 Common Mistake / Point to remember

• Total energy of a signal is value of autocorrelation function at t=0
• DC energy is value of energy spectral density at f=0
• AC Energy=Total Energy-DC Energy

Sample Problem

Noise with double-sided power spectral density of K over all frequencies is passed through a RC low pass filter with 3 dB

cut-off frequency of fc. The noise power at the filter output is

(a) K (b) Kfc

(c) kfc (d) 

Solution: Option (C) is correct

1 1
H( f ) = =
1 + j2fRC f
1+ j
fc
fc2
H( f ) =
2

f 2 + fc2
fc2
()
2
Output PSD = H f .Input PSD = .K
f 2 + fc2
 
fc2
Output noise power =  ( output PSD )df
−
=K f
−
2
+ fc2
df = Kfc (By substitution f = fc tan  )

Sample Problem

A white noise process x (t) with two-sided power spectral density 1 × 10–10 W/Hz is input to a filter whose magnitude
squared response is shown below:

The power of the output process y (t) is given by

(a) 5 × 10–7 W (b) 1 × 10–6 W
(a) 2 × 10–6 W (a) 1 × 10–5 W

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 Communication Systems (Vol-2)

Solution: Option (B) is correct

PSD of white noise = 1 × 10–10 W/Hz  k ( )
PSD of output
G0 ( f ) = H ( f ) .Gi ( f ) = k. H ( f )
2 2

Output noise power

+ f0
 1 
 G ( f ) df = k   area under H( f ) curve  = k  2  bh 
2
N0 = 0
− f0  b 
K f0 × 1 = 1 × 10–10 × 10 × 103 = 10–6 W

Problems

01. Zero mean Gaussian noise of variance N is applied to a half wave rectifier. The mean squared value of the rectified
output will be
(a) Zero (b) N/2
(c) N (d) N
2

02. Zero mean noise is passed through an ideal low pass filter with cut-off at W Hz. The filter output will be uncorrelated at
time instants spaced
(a) 1 ω sec apart
2
(b) m sec apart m = 1,2,3.......
2
(c) 1 m sec apart m = 1,2,3.......
4

( )
2
(d) 1 m sec apart m = 1,2,3.......
2

03. White Gaussian noise is passed through a linear narrow band filter. The probability density function of the envelope of
the noise at the filter output is:-
(a) Uniform (b) Poisson
(c) Gaussian (d) Raleigh

04. For a narrow band noise with Gaussian quadrature components, the probability density function of its envelope will be
(a) Uniform (b) Gaussian
(c) Exponential (d) Rayleigh

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 Communication Systems (Vol-2)

Type 4 – Amplitude Modulation and Demodulation

For Concept, refer to Communication K-Notes, Analog Communication

 Common Mistake / Point to remember

• A common mistake is made while writing complex baseband representation of Bandpass signal.
• Product in time domain is equivalent to convolution in frequency domain.
• Student often make mistake in taking dB value.

Sample Problem

Which of the following demodulators (s) can be used for demodulating the signal
x(t) = 5 (1 + 2 cos 200 t) cos 20000 t
(a) Envelope demodulator (b) Square-law demodulator
(c) Synchronous demodulator (d) None of the above

Solution: Option (C) is correct

Given that
x(t) = 5 (1 + 2 cos 200t) cos 20000 t ---(i)
The standard equation for AM signal is
X AM ( t ) = A c (1 + mcos mt ) cos c t
By comparing the equation (i) and equation (ii),
We have m = 2
Since the modulation index is more than 1 here, so it is the case of over modulation. When the modulation index of AM
wave is more than 1 (over modulation) then the detection is possible only with synchronous modulator only. Such signals
can not be detected with envelope detector.

Sample Problem

A 4 GHz carrier is DSB-SC modulated by a lowpass message signal with maximum frequency of 2 MHz. The resultant signal
is to be ideally sampled. The minimum frequency of the sampling impulse train should be
(a) 4 MHz (b) 8 MHz
(c) 8 GHz (d) 8.004 GHz

Solution: Option (B) is correct

fc = 4G Hz = 4000 MHz
fm = 2 MHz
fH = fc + fm = 4000 + 2 = 4002 MHz
fL = fc − fm = 4000 − 2 = 3998 MHz
2fH
fs =
K
fH 4002 MHz
and K = = = 1000.5 1000
fH − fL 4 MHz
2  4002 MHz
fs = = 8.004 MHz 8 MHz
1000

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 Communication Systems (Vol-2)

Sample Problem

The diagonal clipping in Amplitude Demodulation (using envelope detector) can be avoided if RC time-constant of the
envelope detector satisfies the following condition, (here W is message bandwidth and  is carrier frequency both in
rad/sec)
1 1
(a) RC  (b) RC 
W W
1 1
(c) RC  (d) RC 
 
Solution: Option (A) is correct
For envelope Detector
1 1
 RC 
fc fm
1
To prevent Diagonal Clipping RC 
W

Problems

01. A 1MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100 sec . Which of the
following frequencies will NOT be present in the modulated signal?
(a) 990KHz (b) 1010KHz
(c) 1020KHz (d) 1030KHz

02. A square law demodulator is used to demodulate an AM signal. The ratio between wanted component to unwanted
component, with a the modulation index of 0.5 is
(a) 4 (b) 5
(c) 2 (d) 8

03. The signal to noise ratio at the input of a AM receiver is 10dB. Calculate the signal to noise ratio at the output of the
receiver if the modulation index is ‘1’
(a) 7.67 (b) 8.67
(c) 6.67 (d) 5.67

Statement for linked answer Q4 & Q5 is given below

( )
An AM signal 10 1 + 0.8 cos25000 t cos2 10 t is passed through tuned circuit. The resultant signal is transmitted
6

through a channel. The gain of the tuned circuit at 1MHz is 0.8 & at 1MHz + 5KHz is 0.5

04. The modulation index of the AM signal at the output of the tuned circuit is
(a) 0.4 (b) 0.5
(c) 0.8 (d) 1

05. The transmitted power is

(a) 18W (b) 9W
(c) 36W (d) 66W

06. A sinusoidal modulated AM signal is applied to a square law device of characteristics V0 = Vi with modulation index
2

m. The ratio of second harmonic to first harmonic at the output is

(a) m 2 (b) m 4
(c) m 5 (d) m 6

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 Communication Systems (Vol-2)

() ()
07. Consider a message signal m t = e− tu t and the carrier is AC cos2fC t .The carrier in modulated by the message
signal to generate a DSB signal. Which of the following demodulator is used to get back the message?
(a) Envelope detector (b) Synchronous detector
(c) Both ‘A’ and ‘B’ (d) None

08. An SSB signal is generated by modulating 800KHz carrier by the signal

m ( t ) = 2sin2000 t, A C = 100
The time domain equation of the SSB signal is (consider USB transmission)
(a) 50sin2000tcos2fc t + 50cos21000tsin2fct
(b) 100sin21000tsin2fc t + 100cos21000tcos2fct
(c) 50sin21000tcos2fc t + 50cos21000tsin2fct
(d) 100sin21000tcos2fc t − 100cos21000tsin2fct

09. A tone signal cos2fmt is used to DSB and SSB modulators respectively having same carrier frequency
fc = 2000 KHz and carrier amplitudes A d and A s respectively. in order for the modulated signals to have equal average
powers. The ratio of the carrier amplitudes Ad / As is
(a) 2 (b) 0.707
(c) 1 (d) 1.404

10. A communication system operates in the presence of while noise in the presence of with a two sided power spectral
density of 0.25  10−1.4 W / Hz with total path losses of 100 dB . The input signal is band limited to 10 KHz. The signal to
noise ratio required at the output of the receiver is 40 dB
(
The minimum power required to transmit if the, modulation used is AM  = 7.707 is )
(a) 25 KW (b) 200 W
(c) 10 KW (d) 1404 W

() ()
11. A signal m(t) whose spectrum is shown in figure ‘a’ is generated by using the signal m1 t and m2 t . M1 f and ()
M2 ( f ) are shown in the figure ‘b’ and ‘c’

The signal m(t) can be represented mathematically as

() ()
(a) m1 t + m2 t cos23t () ()
(b) m1 t + 2m2 t cos23t

(c) 2m ( t ) + m ( t ) cos23t
1 2
(d) 2m ( t ) + 2m ( t ) cos23t
1 2

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 Communication Systems (Vol-2)

Type 5 – Angle Modulation and Demodulation

For Concept, refer to Communication K-Notes, Analog Communication

 Common Mistake / Point to remember

• Generally Students make mistake in calculating Phase Deviation and Frequency deviation for Phase
Modulation and Frequency Modulation and Thus Errors are made in calculating Bandwidth.
• Calculation mistakes are made while calculating frequency deviation when frequency modulation by
indirect method is done.

Sample Problem

The signal m(t) as shown is applied both to a phase modulator (with kp as the phase constant) and a frequency modulator
with (kf as the frequency constant) having the same carrier frequency

The ratio kp/kf (in rad/Hz) for the same maximum phase deviation is
(a) 8 (b) 4
(a) 2 (d) 

Solution: Option (B) is correct

For phase modulator
 ( t ) = 2fc t + k pm ( t )
Maximum phase deviation is
( )
D max
= k p max m ( t )  = 2k p …(1)
For frequency modulator
 ( t ) = 2fc t + 2k f  m ( t ) dt
t

(' ) = 2k f t   m ( t ) dt 
t
D max  0 max

( ' ) = 2k f   m ( t ) dt 
2
D max
 0 

(' ) = 2k f   2 dt 
2
D max
 0 
(' )D max
= 8k f …(2)

given
(' )D max
= ( D )max

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8k f = 2kp
kp
= 4
kf

Sample Problem

A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with carrier frequency fc1 = 106 Hz. The

resulting signal is then passed through a Narrow-Band Frequency Modulator with carrier frequency fc2 = 109 Hz.

The bandwidth of the output would be

(a) 4 × 104 Hz (b) 2 × 106 Hz

(c) 2 × 109 Hz (d) 2 × 1010 Hz

Solution: Option (B) is correct

 fm = 106 Hz
(as 10 k is small in comparison to 106)
B.W. = 2fm = 2  106 Hz

Problems

01. In an FM system, a carrier of 100MHz is modulated by a sinusoidal signal of 5KHz. The bandwidth of Carson’s
()
approximation 1MHz. If y t = (Modulated waveform)3, than by using Carson’s approximation, the bandwidth of y t ()
around 300MHz and the spacing of spectral components are respectively
(a) 3MHz, 5KHZ (b) 1MHZ, 15KHz
(c) 3MHZ, 15KHz (d) 1MHZ, 5KHz

02. Consider an FM system with frequency deviation of 75KHz and fm = 15KHz No = 10 Watt / Hz . Calculate the SNR
−9

at the input of the receiver so that the output signal to noise ratio is 40dB
(a) 7.7dB (b) 8.7dB
(c) 6.7dB (d) 5.7dB

03. In FM the rate of frequency deviation is proportional to

(a) The impedance of the antenna
(b) The power output of the transmitted
(c) The amplitude of the modulating signal
(d) The frequency of the modulating signal

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 Communication Systems (Vol-2)

04. Consider the angle modulated signal

S ( t ) = cos ( 200 t + 0.4 sin10 t )
The frequency sensitivity of the modulator is 4Hz/ V. The message signal is
(a) sin10t (b) 0.4sin10t
(c) 0.5cos10t (d) 0.5sin10t

05. An angle modulated deviation is given

(
S ( t ) = cos 2 2  106 t + 30sin150t + 40cos150t 
  )
The maximum frequency and phase deviations are
(a) 2.5KHz, 200  (b) 4.5KHz,100 
(c) 7.5KHz, 100  (d) 17.5KHz, 200 

( )
06. The percentage of the total power at 10MHz in the spectrum is  J0 4 = −0.4 
(a) 4% (b) 16%
(c) 2% (d) 5%

Statement for linked answer Question 07 & 08 is given below

A sinusoidal signal Am cos 4103 t is applied to an FM modulator
07. Starting near about zero Am , is gradually increased and when Am = 2V ,it has been found that the carrier component
goes to zero for the first time. The frequency sensitivity of the modulator is
(a) 1.2KHz/V (b) 2.4KHz/V
(c) 4.8KHz/V (d) None

08. Keeping Am at 2V, frequency fm is decreased until the carrier component goes to zero for the second time. The value of
fm for this to happen is
(a) 672Hz (b) 772Hz
(c) 872Hz (d) 1744Hz

09. The smallest value of modulation index in frequency modulation such that all the modulated power is contained in the
side bands is
(a) 0.4 (b) 1.4
(c) 2.4 (d) 8.6

10. A communication system operates in the presence of while noise in the presence of with a two sided power spectral
density of 0.25  10−1.4 W / Hz with total path losses of 100 dB . The input signal is band limited to 10 KHz. The signal to
noise ratio required at the output of the receiver is 40 dB
(
The minimum power required to transmit if the, modulation used is FM f = 30 KHz is )
(a) 270 W (b) 707 W
(c) 370 W (d) 125 W

11. The carrier signal C ( t ) = A cos2106 t ()

angle modulated by the sinusoidal signal m t = 2cos2000 t ,

KP = 1.5 rad / V,KF = 3000 Hz / V

The modulation indices of PM and FM signal and the corresponding bandwidths are
(a) 3, 6, 8000 Hz & 14000 Hz
(c) 3, 3, 8000 Hz & 14000 Hz
(c) 3, 6, 8000 & 1400 Hz
(d) 3, 6, 800 Hz & 14000 Hz

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 Communication Systems (Vol-2)

12. A signal is given as I/P to square law modulator as shown in Figure Below. Output of square law modulator m(t) is
Frequency modulated.

Let ‘ fc ’ bet he carrier frequency of the modulator when m ( t ) = 0 . If s(t) can be written in the form
s ( t ) = A c cos 2fc t −  sin22fmt 
The value of ‘  ’ is
K f Am2 K f Am2
(a) (b)
2f
2
m
2fm
K f Am2 K f Am2
(c) (d)
4f
2
m
4fm

( )
13. A sinusoidal Base band signal 2cos 2  12  103 t ,frequency modulates a carrier of peak 10V, and 100MHz
frequency, using a Modulator of frequency sensitivity of 12KHz/Volt. As per Carson’s rule, the amplitude of the lowest
frequency of the resulting FM signal (in terms of Bessel Coefficients) is
(a) 20J2 2 ( ) ( )
(b) 10J2 2

(c) 10J3 (3) (d) 20J ( 3 )

3

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 Communication Systems (Vol-2)

Type 6 – Superheterodyne Receivers

For Concept, refer to Communication K-Notes, Analog Communication

 Common Mistake / Point to remember

• Calculation mistakes are done in calculating Image frequency.

Sample Problem

In a superheterodyne AM receiver, the image channel selectivity determined by

(a) The preselector and RF stages
(b) The preselector RF and IF stages
(c) The IF stages
(d) all the stages

Solution: Option (A) is correct

The image rejection should be achieved before IF stage because ones it enters into IF amplifier it becomes impossible to
remove it from wanted signal. So image channel selectivity depends upon preselector and RF amplifiers only. The IF
amplifiers helps in rejection of adjacent channel frequency and not image frequency.

Sample Problem

A superheterodyne radio receiver with an intermediate frequency of 455 kHz is tuned to a station operating at 1200 kHz.
The associated image frequency is ___ kHz.

Solution: 2110
fsi = fs + 2 IF
fsi = 1200 + 2 ( 455 )

fsi = 2110 kHz

Problems

01. A super heterodyne receiver is to operate in the frequency range 550 KHz – 1650 KHz. With the intermediate frequency
(
of 450 KHz, Let R = Cmax / Cmin denote the required local oscillator and I denote the image frequency in KHz ) of the

incoming signal. If the receiver is tuned to 70 KHz then

(a) R = 4.41, I=1600KHz (b) R=2.10, I=1150 KHz
(c) R=3.0, I=1600 KHz (d) R=9.0, I=1150 KHz

02. A super heterodyne receiver having no RF amplifier is tuned to 500KHz. The IF is 465KHz. The ‘Q’ of the tuned circuit is
50. The image station is attenuated by the
(a) -42dB (b) 42dB
(c) -21dB (d) +21dB

03. Determine the improvement in the output SNR that if the modulation index is increased from 0.3 to 0.7
(a) 6.6dB (b) 7dB
(c) 7.6dB (d) 8dB

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 Communication Systems (Vol-2)

Statement for linked answer 04 & 05 is given below

A super heterodyne Fm receiver operates in the frequency range 88-108MHz. The IF any local oscillator frequencies are
related that fif  fL0 . We require that the image frequency fall outside the range 88-108MHz

04. The minimum require fif is

(a) 10.7MHz (b) 455KHz
(c) 20MHz (d) 10MHz

05. The range of variation in local oscillator frequency is

(a) 98.7MHz to 118.7MHz
(b) 98MHz to 118MHz
(c) 88.455MHz to 108.455MHz
(d) 108MHz to 128MHz

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 Communication Systems (Vol-2)

Type 7 – Entropy and Mutual information

For Concept, refer to Communication K-Notes, Information Theory

 Common Mistake / Point to remember

• Students make mistake in calculating baud rate.
• In Shannon-Hartley Law formula SNR is not in dB and it is log to the base 2.

Sample Problem

An image uses 512 × 512 picture elements. Each of the picture elements can take any of the 8 distinguishable intensity
levels. The maximum entropy in the above image will be
(a) 2097152 bits
(b) 786432 bits
(c) 648 bits
(d) 144 bits

Solution: Option (B) is correct

n = log2 L
n = log2 8 = 3
Maximum entropy = 512 × 512 × n = 512 × 512 × 3 = 786432

Problems

01. A television picture may be considered as composed of approximately 3,00,000 picture elements. Each of these
elements can assume 10 distinguishable levels for proper constant. Assume that for any picture element, the 10 brightness
levels are equally probable. There are 30 picture frames being transmitted for second
To transmit this information, the channel capacity required is
(a) 30mbps (b) 60mbps
(c) 40mbps (d) 50mbps

02. A telephone channel has a BW of 3KHz and a SNR of 30dB connected to a teletype machine having 32 different
symbols. The symbol rate required for error free
(a) 1800 symbols/ sec (b) 3000 symbols/ sec
(c) 5000 symbols/ sec (d) 6000 symbols/ sec

03. A memory less source has the alphabet [-5, -3, -1, 0, 1, 3, 5] with corresponding probabilities [0.05, 0.1, 0.1, 0.15, 0.05,
0.25, 0.3]
The entropy of the source is
(a) 1.25 bits/symbol (b) 2.25 bits/symbol
(c) 1.5282 bits/symbol (d) 2.5282 bits/symbol

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 Communication Systems (Vol-2)

Type 8 – Channel Types and Capacity

For Concept, refer to Communication K-Notes, Information Theory

 Common Mistake / Point to remember

• For loseless Channel, channel matrix has only one non zero element in each column
• For Deterministic Channel, channel matrix has only one non zero element in each row
• A noiseless channel is both loseless and deterministic.

Sample Problem

A communication channel with AWGN operating at a signal to noise ratio SNR >> 1 and bandwidth B has capacity C 1. If
the SNR is doubled keeping B constant, the resulting capacity C2 is given by
(a) C2  2C1 (b) C2  C1 + B

(c) C2  C1 + 2B (d) C2  C1 + 0.3B

Solution: Option (B) is correct

SNR  1
 S S
C1 = B log2  1 +   B log2  
 N  N
When SNR is doubled
 2S  S
C'  B log2   = B log2   + B log22 = C1 + B
N N =B
= C1

Sample Problem
The capacity of Binary Symmetric Channel (BSC) with cross-over probability 0.5 is.

Solution: 0
Channel capacity of BSC is
C = Plog2 P + (1 − P ) log2 (1 − P ) + 1

C = 0.5log2 0.5 + 0.5log2 0.5 + 1

C=0
It is the case of channel with independent input and output, hence C = 0.

Problems

01. The channel capacity of an ideal channel having infinite bandwidth is

(a) Infinite
(b) Depends on SNR
(c) 2.44 S / N0 (where N0 is the two sided spectral density)
(d) 1.44 S / N0 (Where N0 is the two sided spectral density )

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 Communication Systems (Vol-2)

02. The noise characteristic of a Binary Symmetric communication channel is given as

The capacity above channel is

(a) 1 bit/ message

(b) Infinity

(c) Zero

(d) 0.5 bits/message

03. A communication channel with additive while Gaussian noise, has a bandwidth of 4KHz and SNR of 15. Its channel
capacity is
(a) 1.6 kbps (b) 16 kbps
(c) 32 kbps (d) 256 kbps

04. Assertion A The channel capacity of an infinite bandwidth channel is finite.

Reason R : Signal power is limited but noise power is not
A. Both A and R are true and R is the correct explanation of A.
B. Both A and R are true but R is NOT the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.

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 Communication Systems (Vol-2)

Type 9 – Pulse Modulation methods-PCM, DPCM

For Concept, refer to Communication K-Notes, Digital Communication

 Common Mistake / Point to remember

• For PCM, SNR=1.76 + 6.02n in dB; where n is the number of bits.
• Use general mean and variance calculation formula (as in random variables) when noise given is not
uniform.
• Students often make mistake in using  and f.

Sample Problem

Companding in PCM system lead to improved signal to quantization noise ratio.

This improvement is for
(a) lower frequency components only
(b) higher frequency components only
(c) lower amplitudes only
(d) higher amplitudes only
Solution: Option (C) is correct
Companding results in making SNR uniform, throughout the signal, irrespective of amplitude levels. Since, in uniform
quantization, step size is same, the quantization noise power is uniform, throughout the signal.
Thus, higher amplitudes of signal will have better SNR than the lower amplitudes.
Hence, companding is used for improving SNR at lower amplitudes.

Sample Problem

The line code that has zero dc component for pulse transmission of random binary data is
(a) more-return to zero (NRZ) (b) return to zero (RZ)
(c) alternate mark inversion (AMl) (d) none of the above

Solution: Option (C) is correct

Alternate mark inversion (AMl) code has zero dc component for pulse transmission of random binary data.

Sample Problem

The minimum step-size requ0ired for a Delta-Modulator operating at 32K samples/sec to track the signal (here u(t) is the
unit-step function) X (t) = 125t (u(t) – u(t – 1)) + (250 – 125t) (u(t – 1) – u(t – 2)) so that slope-overload is avoided, would be
(a) 2–10 (b) 2–8
(c) 2–6 (d) 2–4
Solution: Option (B) is correct
To avoid slope overload

 m(t)
Ts
.32  1024  125
.215 = 125
27
  15
2
  2−8

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 Communication Systems (Vol-2)

Problems

01. A random variable and uniform density in the interval 0 to 1 is quantized as follows
If 0  X  0.3,X1 = 0
If 0.3  X  1,X1 = 0.7
Where X 1 is the quantized value of X?
The root mean square value of the quantization noise is
(a) 0.573 (b) 0.176
(c) 2.205 (d) 0.266

( )
02. A base band signal m(t) is uniformly distributed between −2V, +2V and is quantized into 20 levels. If ‘S’ is the signal
power and Nq is the quantization noise power, the ratio of peak S/Nq to average S/Nq for the quantized signal is
(a) 3 (b) 5
(c) 8 (d) 10

03. Consider a quantizer with transfer characteristics as shown in figure

The input to the quantizer is a random variable whose PDF is

f ( x ) = Ae
−x
for −4  X  4
= 0 Otherwise
The signal to quantization noise ratio is
(a) 5.8dB (b) 6.8dB
(c) 7.8dB (d) 8.8dB

04. Consider a speech signal with maximum frequency of 3.4KHz and a maximum amplitude of 1Volts. The speech signal is
applied to a delta modulator whose bit rate is set at 20kbps. The minimum step size for the modulator to avoid slope
overload error is
(a) 1V (b) 2V
(c) 3V (d) 4V

05. Consider a PCM system with sinusoidal input. Calculate the minimum number of Quantization levels required so that
the maximum Quantization error is 0.1% of the peak signal amplitude?
(a) 9 (b) 10
(c) 512 (d) 1024

Data for Question number 06 and 07 are given below. Solve the problem and choose correct answer:
In a delta modulation system the input is sinusoidal signal. The step size is 0.628V.
The bit rate is 10kbps

06. If the signal frequency is 1 KHz the peak amplitude of the signal to prevent distortion is
(a) 1V (b) 2V
(c) 3V (d) 4V

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 Communication Systems (Vol-2)

07. The granular noise occurs when

(a) Am = 2V fm = 100Hz
(b) Am = 5V fm = 200Hz
(c) Am = 10V fm = 1KHz
(d) Am = 20V fm = 50Hz

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 Communication Systems (Vol-2)

Type 10 – Baseband Shaping for Data Transmission

For Concept, refer to Communication K-Notes, Digital Communication

 Common Mistake / Point to remember

Bit Rate
• Baud Rate = ; M is no. of symbols.
Log2 M
• ISI factor i.e Excess Bandwidth has to be taken into account when it is explicitly mentioned.

Sample Problem

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75%
excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is
(a) 1750 (b) 2625
(c) 4000 (d) 5250

Solution: Option (C) is correct

Rb
BW= (1 +  )
2
Rb
3500 = (1 + 0.75)
2
Rb = 4000bits / sec
Symbol rate=Bit rate=4000 symbols/sec

Sample Problem

The raised cosine pulse p(t) is used for zero |S| in digital communications. The expression for p(t) with unity roll-off factor is
given by
sin4  W t 1
p(t) = . The value of p(t) at t = is
4 Wt (1 − 16W 2 t2 ) 4W
(a) –0.5 (b) 0
(c) 0.5 (d) 

Solution: Option (C) is correct

sin 4  Wt
p(t) =
4  Wt(116 W 2 t2 )
1
 1  sin 4  W 
p 4W
=
 4W  4  W  1  1 − 16W 2 1 
 
4W  16W 2 

As it comes in the form of 0/0 so applying LH rule

d
sin 4  Wt
 1  dt cos 4  Wt
P  = =
 4W  d 1 − 48 W 2 t2
4  Wt(1 − 16W2 t2 )
dt

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 Communication Systems (Vol-2)

1
Putting t =
4
cos 
P(t) = = 0.5
1−3

Problems

() ()
01. A base band binary communication system uses the line encoding formats S1 t and S2 t defined as

Si ( t ) = A for 0  t  T To represent a Binary 1 and

2
= -A for T tT
2
S 2 ( t ) = A for 0  t  T to represent a binary ‘0’. Which of the following can be a basis function for the corresponding
constellation diagram?
(a) 1 T  u ( t ) − 2u ( t − T 2 ) + u ( t − T ) 
 
(b)  A T  u ( t ) − 2u ( t − T 2 ) + u ( t − T ) 
 
() ( ) (
(c)  A T  u t − 2u t − T 2 + u t − T 
  )
() ( ) (
(d)  T  u t − 2u t − T 2 + u t − T 
  )

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 Communication Systems (Vol-2)

Type 11 – Digital Modulation methods-ASK, FSK, PSK AND QAM

For Concept, refer to Communication K-Notes, Digital Communication Techniques

 Common Mistake / Point to remember

• Probability of error can be estimated from Constellation Diagram by calculating dmin ie minimum distance
between any two symbols.

Sample Problem

In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by since waves
of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of
(a) 45  sec (b) 200  sec
(c) 50  sec (d) 250  sec

Solution: Option (B) is correct

For orthogonality in Tb duration there should be integral multiple of cycles.
1
Tb = = 100 s
0
10 K
1
Tb = = 40 s
1
25 K
 200 s is the integral multiple of both.

(Tb0
and Tb
1
)
Sample Problem

An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with
bandwidth 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess
bandwidth. The minimum value of M is _________.

Solution: 16
Rb
B= (1 +  )
log2 M
200  2
100 =
log2 M
log2 M = 4
So, M = 16

Sample Problem

The bit rate of digital communication system is R kbits/s. The modulation used is 32-QAM. The minimum bandwidth
required for |S| free transmission is
(a) R/10 Hz (b) R/10 kHz
(c) R/5 Hz (d) R/5 kHz

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 Communication Systems (Vol-2)

Solution: Option (C) is correct

Rb
BW of M-QAM= (1 +  )
log2 M
For ISI free Transmission α=0
Rb Rb
BW= =
log2 32 5

Problems

01. Bit stream 011101 is applied to a DPSK modulator, the phase shifts of the carrier are
(a)  0  0 0  (b)  0 0   0
(c) 0    0  (d)  0 0 0  0

() ()
02. A digital communication system uses the signals S t t = A = 10mV for 0  t  0.1msec and S2 t = − A = −10mV for

( )
to represent a binary ‘1’ and ‘0’ respectively. This data, modulates a carrier 2cos 2  5  10 t using ASK. If this signal is
6

to have the best noise performance, the minimum value of ‘A (in mV) should be
(a) 15 (b) 20
(c) 25 (d) 30

03. Local oscillator is not required at the receiver for

(a) DSB (b) ASK
(c) FSK (d) DPSK

04. Frequency shift keying is used mostly in

(a) Radio transmission (b) Telegraphy
(c) Telephony (d) None of these

05. If carrier modulated by a digital bit stream had one of the possible phase of 0, 90, 180 and 270 degrees, then the
modulated is called
(a) BPSK (b) QPSK
(c) QAM (d) MSK

06. The bit rate of a digital communication system is 34 M bit s . The modulation scheme is QPSK. The band rate of the
system is
(a) 68 M bit s (b) 34 M bit s
(c) 17 M bit s (d) 8.5 M bit s

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 Communication Systems (Vol-2)

Type 12 – SNR and BER for Digital Modulation Schemes

For Concept, refer to Communication K-Notes, Digital Communication Techniques

 Common Mistake / Point to remember

• Be careful while Calculating Probability of error in terms of Q function or Error function or complementary
error function

Sample Problem

Let Q(  ) be the BER of a BPSK system over an AWGN channel with two-sided noise power density N0/2. The parameter
 is a function of bit energy and noise power spectral density.
A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the
figure. The BPSK demodulator receives the sum of outputs of both the channels.

If the BER of this system is Q(b  ) , then the value of b is __________.

Solution: 1.414
BER for the given system is Q( 2  )
Hence b = 2 = 1.414

Sample Problem

Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1 (t) =  cos 2f1t
and s2 (t) =  cos 2f2t , where  = 4 mV. Assume an AWGn channel with two-sided noise power spectral density

N0 1

−
u2

2
= 0.5  10 −12 W/Hz. Using an optimal receiver and the relation Q(v) =
2
e
v
2
du, the bit error probability for a

data rate of 500 kbps is

(a) Q(2) (b) Q(2 2)
(c) Q(4) (d) Q(4 2)
Solution: Option (C) is correct
1  E 
Pe = erfc  b

2  2N 0 
Rb = 500  103 bps
N0
= 0.5  10 −12
2
 = 4  10−3 V

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 Communication Systems (Vol-2)

1
Then, Tb = = 2  10 −6 sec
Rb
2  10 −6  16  10 −6
Eb = = 16  10−12
2
1  16  10 −12 
Hence Pe = erfc  
2  2  10
−12

1
Pe = erfc  8 
2  
1  8 2
Pe = erfc  
2  2 

1  16 
Pe = erfc  
2  2 
1  4 
Pe = erfc  
2  2
 1  x  
Pe = Q(4)  erfc   = Q(x)
 2  2 

Problems

01. A polar binary signal is a +A Volts or –A Volts pulse during the interval (0, T)= AWGN with PSD of

(N 2 ) = 10
0
−5
W Hz is added to the signal. The bit rate of the system is 7.26kbps. Calculate the value of ‘A so that the

Pe = 10−4

Q ( x ) = 10 −4 when x=3.71
(a) 1V (b) 2V
(c) 3V (d) 4V

02. A NRZ signal is a +1V or -1V pulse during the interval (0, T). Additive while noise with two sided power spectral density
10−3 W Hz is added to the signal. Determine the bitrates that can be sent with a bit error probability of 10−4
Q ( x ) = 10 −4 When X=3.71

1
Where Q x = ( )  exp ( −y )
2 dy
2

2 X

(a) 2.5kbps (b) 5kbps

(c) 6.26kbps (d) 7.26kbps

03. In digital communication system, transmission of successive bits through a noisy channel are assumed to be
independent events with error probability p. The probability of at most one error in the transmission of an 8-bit sequence
is
( ) ( ) + 8p (1 − p )
8 7
(a) 7 1 − p 8 +p 8 (b) 1 − p

( ) + (1 − p ) (d) (1 − p ) + p (1 − p )
8 7 8 7
(c) 1 − p

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Type 13 – MAP and ML decoding and Matched Filter Receiver

For Concept, refer to Communication K-Notes, Digital Communication & Digital Communication Techniques

 Common Mistake / Point to remember

• Be careful while calculating impulse response for Complex signal in case of matched filter.

Sample Problem

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this
matched filter?

(a) (b)

(c) (d)

Solution: Option (C) is correct

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Sample Problem

A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) =
9/10, then the probability of error for an optimum receiver will be
7 63
(a) (b)
80 80
9 1
(c) (d)
10 10

Solution: Option (B) is correct

For the optimum reception we choose to detect zero at the output irrespective of whatever bit is transmitted. Thus the
1
probability of error will occur when the input bit is one and the probability of occurrence of one is .
10

Problems

01. Consider a matched filter with input signal

S ( t ) = e− t 0  t  T
=0 otherwise

The output of the filter is

(a) e−t (b) −e−t
(c) e−t cosht (d) e− t sinht

02. To obtain 100% modulation of a carrier with a sine wave, the modulating power must equal what minimum percent of
the carrier power?
(a) 10% (b) 15%
(c) 25% (d) 50%

03. The input to a matched filter is

S (t) = A 0  t 1

=A 2t 3

=0 otherwise

The output of the matched filter at t=3 is

(b) 2 A
2
(a) 2
(c) A
2
(d) none

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The data for Question number 04-05 is given below:

The input to a Matched filter is

04. The input response of the Matched filter is

(a) (b)

(c) (d)

05. The output of the Matched filter is

(a) (b)

(c) (d)

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Type 14 – Basics of TDMA, FDMA and CDMA

For Concept, refer to Communication K-Notes, Digital Communication & Digital Communication Techniques

 Common Mistake / Point to remember

• Add synchronisation bits (if given) to calculate bit rate in the formula.

Sample Problem

In a GSM system, 8 channels can co-exist in 200 kHz bandwidth using TDMA, A GSM based cellular operator is allocated 5
MHz bandwidth. Assuming a frequency reuse factor of 1/5, i.e. a five-cell repeat pattern, the maximum number of
simultaneous channels that can exist in one cell is
(a) 200 (b) 40
(c) 25 (d) 5

Solution: Option (B) is correct

Allocated Bandwidth = 5 MHz
1
Frequency reuse factor =
5
1
Bandwidth allocated for 1 cell = 5  = 1 MHz
5
1 MHz
Number of simultaneous channels =  8 = 40
200 kHz

Sample Problem

In a Direct Sequence CDMA system the chip rate is 1.2288 × 106 chips per second. If the processing gain is desired to be
AT LEAST 100, the data rate

(a) must be less than or equal to 12.288 × 103 bits per sec
(b) must be greater than 12.288 × 103 bits per sec
(c) must be exactly equal to 12.288 × 103 bits per sec
(d) can take any value less than 122.88 × 103 bits per sec

Solution: Option (A) is correct

Rc
Processing gain =
Rb
Rc Rc
 100 => R b 
Rb 100
Rb  12.288  103 bits per sec .

Sample Problem

Four message band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing
(TDM). The minimum bandwidth required for transmission of this TDM signal is
(a) W (b) 3 W
(c) 6 W (d) 7 W

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Solution: Option (D) is correct

fs = 2  W = 2 W
1

fs = 2  W = 2 W
2

fs = 2  2W = 4 W
3

fs = 2  3W = 6 W
4

fs = fs + fs + fs + fs
1 2 3 4

fs = 14 W

For minimum bandwidth n = 1

Rb = nfs
Rb = 1  14 W = 14 W
Rb
(B. W)min =
2
14 W
(B. W)min = =7W
2

Problems

01. Six independent low pass signals of bandwidths W, 2W, 4W, 4W, 6W and 6W are to be time division multiplexed on a
common channel using PAM. To achieve this the minimum transmission bandwidth of the channel should ne [Where,
W=730Hz]
(a) 40KHz (b) 35KHz
(c) 42KHz (d) 45KHz

02. A TDM link has 20 signal channel and each channel is sampled 8000 times/sec. Each sample is represented by seven
binary bits and contains an additional dit for synchronization. The total bit rate for the TDM link is
(a) 1180 k bits sec (b) 1280 k bits sec
(c)1180 M bits sec (d) 1280 M bits sec

03. Three signal each band limited to 600Hz, 600Hz and 1200HZ are sampled at Nyqusit rate and transmitted through a
channel using TDM. Each sample is encoded into 12bits. The bit rate of the multiplexed signal is
(a) 57.6mbps (b) 5.76mbps
(c) 5.76kbps (d) 57.6kbps

04. A comparison of frequency division and time division multiplexing system shows that
(a) FDM requires a lower bandwidth but TDM has greater noise immunity
(b) FDM has greater noise immunity and requires lower bandwidth than TDM
(c) FDM requires channel synchronization, while TDM has greater noise immunity
(d) FDM requires more multiplexing while TDM requires band pass filter

05. Assertion A Digital communication needs some synchronization signals

Reason R Bit sync’ signal is required in time division multiplexing to distinguish groups of data
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

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Solutions
Solution

Type 1 – Probability Theory, Random Variables and Processes

01. Ans: (a)
x2

( )
−
Solution: P.d.f → P x = ke 2
, −  x  
P.d.f satisfies the property → area under p.d.f is unity.


  P ( x ) dx = 1
−

 x2
−
 ke
−
2
dx = 1

 1
− x2
ke 2
dx = 1
−

 1
− x2 
e
−
2
dx =
1
= 2

2
1
 2 =
k
1
k=
2

02. Ans: (c)

1
Solution: P(X  m) = Q(m) = erfc( m)
2

03. Ans: (a)

Solution: S/g m(t) → B  w = BHz

mean = m = 0

variance = 2

 −2 − mean   −2 − 0 

P (m  −2 ) = 1 − P (m  −2 ) = 1 − Q   = 1 − Q 
 s tandard deviation    

P (m  −2 ) = 1 – Q (– 2) = Q (2)

Type 2 – Autocorrelation and Power Spectral Density

01. Ans: (b)
Solution: R(z) = 100 sinc [100 τ]
o p PSD = i p PSD  H ( f )
2

y ( f ) = x ( f )  H( f )
2 2 2

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R (  ) ⎯⎯ → X (f )
2
F.T

 25
 y ( f ) = X ( f )  H( f ) =  y (f )
2 2 2 2 25
Ey ( f ) =
−
df =  df = f 
−25
−25

E y ( f ) = 50

02. Ans: (a)

()
Solution: x t = 10 cos 50t +  ( )
 →Rv
R x (  ) = E  x ( t ) x ( t +  ) 

R x (  ) = E 10cos (50 t +  )  10cos 50  t +   +  

  ( )
R x ( z ) = E 100cos (50 t +  ) cos (50 t + 50  +  ) 

R x ( z ) = 100E cos (50 t +  ) cos (50 t + 50 z +  ) 

1
R x ( z ) = 100  E cos (50 ) + cos (100t + 2 + 50 ) 
2
R x ( z ) = 50 cos (50 ) + 50E cos (100 t + 2 + 50  ) 

E cos (100 + 2 + 50 )  = 0

 R x ( z ) = 50 cos (50 )

R x ( z ) ⎯⎯⎯⎯⎯
fourier transform
→P  S  D

P  S  D = F  T 50 cos (50 ) 

50
P  S D =  ( f + 25 ) +  ( f − 25 ) 
2  
P  S  D = 25 ( f + 25 ) + 25 ( f − 25 )

03. Ans: (d)

()
Solution: h t =  t − e u t
−t
() ()

H( t ) = 1 −
( f + )
j2 

H ( f ) = H ( f )  H* ( f )
2

2
 H( f ) = 1 −
2

2 + ( 2f )
2

o / P PSD = i / P PSD  H ( f )
2

N0  2


o / P PSD = 1−
2  2 
 + ( 2f ) 
2


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 
2
P  S  Do/P = 0.5N0 1 − 
 2 + 2f 2 
 ( )
Inverse Fourier transform of P  S  D is autocorrelation function.
  
2
 ACFo/P = Inverse FT 0.5N0  1 − 
  2 
 + ( 2f )  
2
  
R ( z ) = ACFo/P = 0.5N0  ( z ) − 0.5e 
− z

 
 
 1 − z FT  
 e ⎯⎯ → 2 
 2  + ( 2f ) 
2

04. Ans: (a)

() ()
Solution: y t = x t cos 2fc t +  ( )
 →Rv
( )
R y ( z ) = E  y ( t ) y ( t + z )  = E  x ( t ) cos ( 2fc t +  )  x ( t +  ) cos 2fc ( t +  ) +  
 
R y (  ) = E  x ( t )  x ( t +  )  cos ( 2fc t +  ) cos ( 2fc t + 2fc  +  ) 

x ( t ) and  are independent, we can write

R y (  ) = E  x ( t )  x ( t +  )  E cos ( 2fc t +  )  cos ( 2fc t + 2fc  +  ) 

 
1
R y (  ) = R x (  )  E cos ( 2fc  ) + cos ( 4 fc t + 2 + 2fc  ) 
2
1
R y (  ) = R x  E cos ( 2fc  )  + 0
2
Because, E cos ( 4 fc t + 2 + 2fc  )  = 0

1
 R y (  ) = R x (  )  E cos ( 2fc  ) 
2
E cos ( 2fc  )  = cons tant
 

 E cos ( 2fc  )  = cos ( 2fc  )

Hence,
Rx ( )
R y ( ) =  cos ( 2fc  )
2
R y (  ) = 0.5R x (  ) cos ( 2fc  )

05. Ans: (b)

Solution: P  S  D of y(t) is Fourier transform of autocorrelation function R y   of y(t).
  ()
P  S  D y ( t ) ⎯⎯⎯⎯⎯→ R y ( z )
Fourier transform

P  S  D y ( t ) = F  T R y (  )  = F  T 0.5R x (  ) cos ( 2fc  ) 

We charge variable ‘z’ to ‘t’, for case of finding Fourier transform.

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  e j2 fc t + e− j2 fc t 
P  S  D y ( t ) = F  T 0.5R x ( t ) cos ( 2fc t )  = F  T 0.5R x ( t )   
  2
  
R x ( t ) ⎯⎯
F T
→ Sx (t )

R x ( t ) ⎯⎯ → S x ( f − fc )
j2 fc t
e F T

06. Ans: (b)

Solution: x(t) and y(t) are independent
 R xy (  ) = E  x ( t )  y ( t +  )  = E  x ( t )   E  y ( t +  )  = x   y = 2  3 = 6

 R xy (  )  e−2 u (  )
6  e−2 u (  )
Convolution in time domain is multiplication in frequency domain.
1 1 6 ( f )
 R xy (  )  e−2 u (  ) ⎯⎯
F T
→ 6 ( f )  = 6 ( f )  = = 3 ( f )
( 2 + j2f ) 2 + j2f 
f =0
2

Hence, magnitude spectrum is independent of frequency i.e., zero at any frequency other than zero.

Type 3 – Properties of White noise

01. Ans: (d)
Solution: mean = 0, variance = N
1
( )
2
i/P P.d.f Px x = e− x
, −  x   /2N

N  2
 1 
o/P P.d.f Px ( x ) = 2 
2
e− x /2N  ,0  x  
 N  2 

After rectification, x takes only +ve values. o/P P.d.f has been multiplied by 2, so that area under P.d.f = 1.

 1 2 
E  x2  =  x2  2  e− x /2N  dx
0  N  2 
x2 Ndt
Put t =  dx =
2N 2Nt
 1  3 
2  2N2 1 2N −1
E  x2  = e t dt = e t dt [ Gamma Function e t dt = (m) ]
−t 2 −t 2 − t m−1

N  2 2N 0  0 0

2N 3 2N 1 1  n = (n − 1 ) n − 1 
E  x 2  = = 
 2  2 2  

N 1 
E  x 2  =    = 
  2 

E  x 2  = N = mean square value of rectifier o/P.

02. Ans: (b)

Solution: Ideal low pass filter through which noise is passed is
Impulse response of the filter, h(t) is given by

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 
h(t) =  H( f ) e
j2 ft
df = e
j2 ft
df
− −

 
1 e j2 t − e j2 t 
h ( t ) = e j2 ft  = 


− t  2j 
 
sin ( 2t )
h(t) =
t
Now, h(t) is zero at
2t = n
n
t = sec, n = 1, 2, 3,
2
Filter output will be uncorrelated at time instants where h(t) = 0.

03. Ans: (c)

Solution: White Gaussian noise when passed through narrow band filter, envelop of noise at the filter output is also
Gaussian.

04. Ans: (b)

Solution: For narrow band noise with Gaussian quadrature. Components, the P.d.f of its envelope is also Gaussian.

Type 4 – Amplitude Modulation and Demodulation

01. Ans: (c)
Solution: fc = 1 MHz = 1000 KHz .
Symmetrical square wave of period 100µ sec.
T = 100 sec fm = 10 KHz
1 2 1 1 
(t) = + cos mt − cos3mt + cos5mt − − − − 
2  3 5 

()
Message (t) t [square wave] when modulates sinusoidal carrier, we have

 ( t )  ( t ) = ( A c cos c t )  ( t )

1 2  1  A 2Ac 2A
= Ac cos c t  + cos mt − cos3mt − − − −   = c cos c t + cos c t cos mt − c cos c t  cos3mt
2   3  2  3
 frequencies terms present in the modulated signal with be c , c + m , c − m , c + 3m ,
c − 3m , c + 5m , c − 5m , c − 5m − − − −

Hence, 1020 KHz will not be present in the modulated signal.

02. Ans: (a)

Solution: AM S/g = A c 1 + mcos mt  cos c t

i/P to square law = Ac cos c t + Acmcos mtcos c t

( )
2
At o/P → A c cos c t + A cmcos mt cos c t

At the o/P we have components,

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A 2c cos2 c t , A2c m2 cos2 mtcos2 c t , 2A2c mcos mtcos2 c t

Out of this, wanted component is one with cos mt and unwanted component is term with cos m t
2

2A2c mcos2 c t 2 2
 ratio = = = =4
A m cos c t
2
c
2 2
m 0.5

03. Ans: (c)

S
Solution:   = 10 dB
 N i/P
m=1
S
  =?
 N o/P
PR S S
=   → receiver → = ? = 10 dB = 10
2N0B N
 i/P N0

PR
 = 10
2N0B
S PR
  = = 20
 N ref N0B
S
  → is the signal to noise ratio at the input of receiver for baseband system considering the same received signal
 N ref
power.
S PR
  =
 N ref N0B
PR → signal power at the receiver input.
N0B → noise power at the receiver input for baseband system.
i.e., noise power in the message signal B  w .
(SNR ) o/P m2
Figure of merit for A  M = =
(SNR ) ref
2 + m2
 m 2
 1
(SNR ) =   ( SNR )ref =  20 = 6.67
3
2+m
o/P 2


04. Ans: (b)

Solution: AM s/g → 10 1 + 0.8 cos2500t  cos210 t
6

Amplitude of carrier = Vc = 10V

After passing through tuned circuit, amplitude of carrier will be
Vc = 10  0.8 = 8 V
Amplitude of message s/g = Vm = 8V
After passing through tuned circuit, amplitude of message s/g will be
Vm = 8  0.5 = 4 V

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Vm 4
 m = =
Vc 8
1
m =
2
m = 0.5

05. Ans: (c)

 m2 
Solution: Pt = Pc 1 + 
 2 
At the o/P of tuned circuit
Vc = 8V
 ( 0.5 )2 
1
 Pt =  ( 8 ) 1 +
2

2  2 
 
Pt = 36watt

06. Ans: (a)

Solution: V0 = Vi
2

A  M s / g → A c 1 + mcos mt  cos c t = Ac cos c t + mAc cos mtcos c t

2
V0 =  A c cos c t + mA cos mt cos c t 

V0 = Ac2 cos2 c t + m2 Ac2 cos2 mtcos2 c t + 2mA2c cos2 c tcos mt

2nd harmonic component → cos2 mt
1st harmonic component → cos mt
m2 A 2c cos2 c t m
Hence, required ratio = =
2mA cos c t
2
c
2
2

07. Ans: (c)

()
Solution: m t = e u t
−t
()
carrier = Ac cos2fc t
Message signal m(t) does not cross x=axis and become negative.
Thus envelope of the modulated signal is still m(t).
Hence, envelope detector can be used to demodulate the S/g.

08. Ans: (b)

Solution: fc = 800 KHz, Ac = 100
m ( t ) = 2sin2000t
m(t ) m (t)
USB =  Ac cos c t  − h  A sin c t 
2 2  c
Where, mh ( t ) → Hilbert transform of m ( t ) .

m ( t ) = 2sin2000t
 
mh ( t ) = 2sin  2000t − 
 2

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mh ( t ) = −2cos2000t

USB =
2sin2000t
100cos2fc t  −
( −2cos2000t ) 100sin2f t 
2 2  c 

200sin2000t cos2fc t 200cos2000 t sin2fct

USB = +
2 2
USB = 100sin2000tcos2fc t + 100cos2000tsin2fct

09. Ans: (b)

Solution: fc = 200 KHz
m ( t ) = cos2fmt
We need, PowerDSB = PowerSSB

()
DSB S/g → m t  A d cos c t 

1
 PowerDSB = m2 ( t )A 2d
2
m ( t ) AS cos c t mh ( t ) A s sin c t
SSB LSB S / g → +
  2 2
1 m ( t ) A s 1 mh ( t )A s
2 2 2 2

 PowerSSB =  + 
2 4 2 4
m2 ( t ) = mh2 ( t )
1 2
 PowerSSB = m ( t )A 2s
4
for PowerDSB = PowerSSB
1 2 1
m ( t )A 2d = m2 ( t )A 2s
2 4
1
A 2d = A 2s
2
Ad 1
= = 0.707
As 2

10. Ans: (a)

Solution: B = 10 KHz
N0
= 0.25  10 −14 w / Hz
2
Total path loss = 100 dB = 1010
S
  = 40dB = 10
4

 N o/P
S
 
m  N 0
2
Figure of merit of AM detector = =
2+m2
S
 
 N ref

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( 0.707 )
2
10 4
=
2 + ( 0.707 ) SNRref
2

PR 10 4
 SNRref = = = 5  10 4
N0B 0.2
PR → received S/g power at the receiver input.
N0B → average noise power in message S/g B  w
PR = 5  104  N0B
= 5  104  2  0.25  10−14  10  103
PR = 2.5  10 −6 w
PR is power at the channel o/P or receiver i/P channel causes power loss of 100 dB i.e. 1010
Power at channel i / P
 power at the channel o/P =
1010
Hence, power at channel i/P = 1010  2.5  10−6
i.e., transmitted power = 25  103 w = 25 kW

11. Ans: (b)

()
Solution: m2 f is shifted to left and right by f = 3 Hz.

Hence, m ( f ) must be multiplied by a carrier of frequency f

2 c
= 3 Hz .

Hence, m ( t ) = m ( t ) + 2m ( t ) cos23t
1 2

1
m2 ( t ) cos2fc t ⎯⎯
F T
→ m2 ( f + fc ) + m2 ( f − fc ) 
2
() ()
Hence, multiplying m2 t cos2fc t by 2 gives shifted version of m2 f without scaling.

Type 5 – Angle Modulation and Demodulation

01. Ans: (a)
Solution: fc = 100 MHz, fm = 5 KHz
B  w = 1 MHz
 B  w = 2 fm + f 
1 MHz = 2 5 KHz + f 
f = 495 KHz

() ( )
3
Now, if y t = modulated waveform
Carrier frequency and frequency deviation both will become 3 times and modulating S/g frequency will remain unchanged.
 fc = 300 MHz
f = 1485 KHz
fm = 5 KHz
New bandwidth = 2  fm + f  = 5 + 1485  = 2.98 MHz  3 MHz

Spacing of spectral components = fm = 5 KHz

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02. Ans: (b)

Solution: f = 75 KHz
fm = 15 KHz
S
Signal Power = PR → Receiver →   = 40 dB = 10 4
 N 0
PR → received signal power
f 75
modulation indexFM = FM = =
fm 15
FM = 5
3 2
Figure of merit for FM = 
2 FM
S
 
 N o/P 3 2
F  o  MFM = = FM = 37.5
S 2
 
N
 ref
S
 
S  N o/P 10
4

  = = = 266.67
 N ref 37.5 37.5
S
  → is signal to noise ratio at the input of receiver for baseband system considering the same received signal power.
 N ref
S PR
  = = 266.67
N
 ref N0
B
PR → signal power at the receiver i/P.
N0B → noise power at the receiver i/P for baseband system.
i.e., noise power in the message S/g B  w
PR = 266.67  N0  B = 266.67  2  10−9  10  103 = 5.33  10−3 w
For FM system,
signal power at the receiver i/P = 5.33 mW
noise power at the receiver i/P[N] is
FM is band pass signal
N = 2N0BFM , BFM → bandwidth of FM signal.
BFM = 2 ( f + fm ) = 2 75 + 15  = 180 KHz
Hence, N = 2  2  10−9  180  103 = 7.2  10−4
S 5.33  10 −3
  = = 7.4167
 N i/P 7.2  10 −4
S
   8.7 dB
 N i/P

03. Ans: (c)

()
Solution: i t = c + k f m t()
= c + k f Am cos mt

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 = k f Am cos mt

d
 = −k f Amm sin m t
dt
d
Hence,  − Am
dt

04. Ans: (c)

() (
Solution: s t = cos 200 t + 0.4 sin10 t )
k f = 4 Hz / volt
i ( t ) = 200t + 0.4 sin10 t
i ( t ) = 200t + 0.4 cos10t
fi ( t ) = 100 + 0.4  5cos10t
fi ( t ) = 100 + 2cos10 t
In general,
fi ( t ) = fc + k f m ( t )
 k f m ( t ) = 2cos10t
2
m(t ) = cos10t
4
m ( t ) = 0.5cos10 t

05. Ans: (c)

()
Solution: i t = 2 2  106 t + 30 sin150t + 40 cos150t 

i ( t ) = 4   106 t + 2 (30 sin150t + 40 cos150t )

 max phase deviation = max 2 ( 30 sin150t + 40 cos150t ) 

= 2 max 30 sin150t + 40 cos150t  = 2 302 + 402 = 2 50 = 100 rad
i ( t ) = 4   106 + 2 30  150 cos150t − 40  150 sin150t 

Max frequency deviation   = max 2  4500 cos150t − 6000 sin150t  

( 4500 ) + ( 6000 )
2 2
 = 2
f = 7500 Hz = 7.5 KHz

06. Ans: (b)

Solution: fc = 10 Hz = 10 MHz
7

f = 4 KHz
 = f = 4
FM wave is also given by

S ( t ) = Ac  J () cos 2 ( f
m c
+ nfm ) t  for n = 0, we have only carrier component of 10 HMz.
n=−

( ) (
i.e., for n= 0, A c J0  cos2fc t = 100  −0.4 cos2fc t )

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1
 (100  0.4 ) = 800 w
2
Power in carrier component of 10 MHz =
2
Power in 10 MHz 800
 % power in 10 MHz component =  100% =  100
Total Power 5000
= 16%

07. Ans: (b)

()
Solution: m t = Am cos 410 t
3

Am = 2v
fm = 2 KHz
When Am = 2 carrier component goes to zero.

S ( t ) = Ac  J () cos 2 ( f
n c
+ nfm ) t 
n=−

Carrier component goes to zero, when Jo  goes to zero( )

Jo (  ) goes to zero for the first time, when  = 2.4
f
 = = 2.4
fm
f = 2.4  2  103
f = 4.8  103
k f Am = 4.8  103
4.8  103
kf = = 2.4 KHz / volt
2

08. Ans: (c)

Solution: Am is kept at 2v fm is decreased until the carrier component goes to zero for the second time.
k f = 2.4 KHz / volt
Am = 2V
f = k f Am = 4.8  103 Hz
( )
Carrier component goes to zero for the second time, when Jo  goes to zero for the second time.

Jo (  ) goes to zero for the second time, when  is around 5.5.

f 4.8  103
 = 5.5 = =
fm fm
4.8  103
fm =
5.5
fm  872.72 Hz

09. Ans: (c)

( )
Solution: Jo  goes to zero for the first time at β around 2.4

10. Ans: (c)

Solution: B = 10 KHz

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N0
= 0.25  10 −14 W / Hz
2
Total path loss = 100 dB = 1010
f = 30 KHz
f 30 KHz
= = =3
fm 10 KHz

(SNR )
o/P 3 2
Figure of merit of FM receiver = = 
(SNR )
ref
2 FM

(SNR ) 3
 ( 3) = 13.5
o/P 2
=
(SNR ) ref
2

(SNR ) 10 4
(SNR ) ref
=
13.5
o/P
=
13.5
PR
(SNR ) ref
=
N0B
= 740.74

PR = N0B  740.74 = 2  0.25  10−14  10  103  740.74

PR = 3.70  10−8 W
 transmitted power = power at channel i/P × 1010
 transmitted power = 3.70  10−8  1010 = 370W

11. Ans: (a)

()
Solution: c t = A cos210 t
6

m ( t ) = 2cos2000 t
kp = 1.5 rad / V
k f = 3000 Hz / V
for P  M ,
i ( t ) = c t + k pm ( t ) .
d
i ( t ) = c + k p m(t) .
dt
 = k p ( 200 )  2
fPM 3 KHz
PM = =
fm 1 KHz
Bandwidth PM = 2 fm + fPM  = 2 103 + 3  103  = 8 KHz
for F  M,
i ( t ) = c + k f m ( t )
f = k f Am
fFM = 6 KHz
fFM 6 KHz
FM = =
fm 1KHz

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BWFM = 2 fm + fFM  = 2 103 + 6  103  = 14 KHz

12. Ans: (d)

Am2 1 − cos2mt 
( )
2
Solution: Am sin mt → →
2
A 2

 m(t) = − cos2mt
m

2
fi ( t ) = fc + k f m ( t )
 A2 
fi ( t ) = fc + k f  − m cos2mt 
 2 
 
k f Am2
fi ( t ) = fc − cos2mt
2
2k f Am2
i ( t ) = c − cos2mt
2
2k f Am2 sin2mt
i ( t ) = c t −
2 2  2fm
k f Am2
i ( t ) = 2fc t − sin22fmt
4fm
 By comparison,
k f Am2
=
2fm

13. Ans: (b)

()
Solution: m t = 2cos 2  12  103 t ( )
Ac = 10v fc = 100 MHz
k f = 12  103 Hz / volt
eqn of FM S/g,

S ( t ) = Ac  J () cos 2 ( f
n c
+ nfm ) t 
n=−

f 24 KHz
= = =2
fm 12 KHz
Lowers frequency of FM signal is
fi = fc − f = fc − k f Am = 100 − 12  2 = 76 KHz
min

 fc − nfm = 76 KHz
n = 2
A c Jn (  ) = 10J2 ( 2 )

Type 6 – Superheterodyne Receivers

01. Ans: (a)
Solution: 550  fs  1650 KHz
I  F = 450 KHz

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Cmax
R=
Cmin
Receiver tuned to 700 KHz
 fs = 700 KHz
fs = fs + 2I  F
i

fs = 700 + 2  450
i

fs ,Image Frequency = 1600 KHz

i

Hence, option (b), (c) are eliminated.

f0 = fs + I  F
f0 = fs + I F
max max

f0 = fs + I F
min min

2
Cmax  f0 
R= =  max 
Cmin  f0min 
2
 2100 
R= 
 1000 
R = 4.41

02. Ans: (d)

Solution: fs = 500 KHz
fi = 465 KHz
Q = 50
fS = fS + 2fi = 500 + 2 × 465 = 1430 KHz
i

fS fS
= i
−
fS fS
i

1430 500
 == − = 2.51
500 1430
 = 1 + Q 2 2
 = 125.52
 dB = 10log10 125.52 
dB  21dB

03 Ans: (a)
(SNR ) o/P m2
Solution: Figure of merit AM = =
(SNR ) ref
2 + m2
Where m is the modulation index
For m=0.3
(SNR ) o/P1
= 0.043 (SNR )ref

( 0.7 )
2

For m = 0.7, Figure of Merit = = 0.19678

2 + ( 0.7 )
2

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(SNR ) o /P2
= 0.19678 (SNR )ref

(SNR ) o/P2
Hence, = 4.5762
(SNR ) o/P1

Hence, improvement in SNR is by, = 10log10  4.5762  = 6.6dB

04. Ans: (d)

Solution: 88  fs  108 MHz
fLO  fIF
fo = fs + I  F
fsi should be outside 88 – 108 MHz range
fsi = fs + 2I  F
When fs = 88 MHz, fsi should be greater than 108 MHz.
108 = 88 + 2I  F
 IFmin = 10 MHz

05. Ans: (b)

Solution: fo = fs + I  F
fo = fs + 10 MHz
when, fs = 88 MHz fo = 98 MHz
fs = 108 MHz fO = 118 MHz
98  fo  118 MHz

Type 7 – Entropy and Mutual information

01. Ans: (a)
Solution: Given → 3  105 element/frame
30 frames/sec
Each of the elements can assume 10 distinguishable levels.
1
Probability of occurrence of each level =
10
Max : Information in each level in bits is
H = log2 10 =3.32 bits/element
 channel capacity in bits/sec is = 3.32  3  105  30 = 29.88  106 bps  30 Mbps

02. Ans: (d)

Solution: B  w = 3 KHz
SNR = 30 dB = 103
Number of different symbols = 32
Channel capacity in bits/sec is
 S
C = Blog2 1 + 
 N 

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C = 3  103 log2 1 + 103 

C = 29.9 kbps
C  30 kbps = 30000 bit / sec

Telephone machine has 32 different symbols

i.e, m = 32
H = log2 m = log2 32 = 5 bits / symbol
30000 bits / sec
Channel capacity in symbols/sec is = = 6000 symbols/sec
5 bits / symbol
Hence, max symbol rate with error free transmission is 6000 symbols/sec.

03. Ans: (d)

Solution: P1 = 0.05, P2 = 0.1, P3 = 0.1, P4 = 0.15, P5 = 0.05, P6 = 0.25, P7 = 0.3
1 
H =  Pk log2  
k  Pk 
 1 
H = 0.05log2  2
 0.05 
 1 
+2  0.1log2  
 0.1 
 1   1   1 
+0.15log2   + 0.25log2   + 0.3log2  
 0.15   0.25   0.3 
H = 0.4322 + 0.6643 + 0.4105 + 0.5 + 0.5211
H = 2.5282 bits/symbol

Type 8 – Channel Types and Capacity

01. Ans: (d)
 S
Solution: C = Blog2 1 + 
 N
 S 
C = Blog2 1 + 
 N0B 
 S 
C = limBlog2 1 + 
 N0B 
B →

 S 
Blog2 1 + 
C = lim  N0B 
B → 1
B
We use L’ Hospital’s rule
  S  −1 
   
1 1 N B 
C = lim    0
B →  S  loge 2  −1 
1 +   
 N0B  B 
S 1
C= 
N0 loge 2

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S
C = 1.44
N0

02. Ans: (c)

Solution: From the transition probability diagram,
y 
P  1  = P1
 x1 
y  y 
 P  2  = 1 − P  1 
 x1   x1 
P1 = 1 − P1
1
 P1 =
2
1
Similarly P2 =
2
1
 error probability Pe  of given BSC is Pe =
2
1  1 
 (Pe ) = Pe log2   + (1 − Pe ) log2  
 Pe   1 − Pe 
 (Pe ) = 1

Hence,
Channel capacity CS  = 1 −  Pe ( ) =1–1=0
03. Ans: (b)
Solution: B = 4 KHz
SNR = 15
C = Blog2 1 + SNR  = 4  103 log2 1 + 15  = 4  103 log2 16 = 4  103  4
C = 16 kbps

04. Ans: (a)

 S
Solution: C = Blog2 1 + 
 N
 S 
C = Blog2 1 +  S → signal power
 N0B 
When B → , C →  
Channel capacity is finite because noise power also increases with increase in channel bandwidth, which leads to finite
channel capacity.

Type 9 – Pulse Modulation methods-PCM, DPCM

01. Ans: (b)
Solution:
0  x  0.3 xq = 0
0.3  x  1 xq = 0.7
Above is case of non-uniform quantization i.e., step size is not constant.

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→ i → ith step size Pi → Probability that i/p lies in ith interval

1 = 0.3
0.3
P1 =  (1) dx = 0.3
0

2 = 0.7
1
P2 =  (1) dx = 0.7
0.3

1 1 
( 0.3)  0.3 + ( 0.7 )  0.7  = 0.17559
2 2
root mean square value of quantization error =
12
 i2Pi =

12  
rsm value of quantization noise  0.176

02. Ans: (a)

Solution: L = 20
m ( t )max = 2v m ( t )min = −2v

VH − VL 2 − ( −2 ) 4 1
= = = = = 0.2
L 20 20 5
2 ( 0.2 )
2

Nq = = = 3.33  10 −3 W
12 12
(Peak Power / N ) q
=
Peak Power
( Average Power / N ) q
Average Power
P  d  f of message signal m(t)
2
1  x  −2
3
2
1 4
E  x  =  x  dx =
 2
 2
=
−2 4 4 3 3

( ) = (2) = 4
2 2
Peak Power = Peak value

V2
Average Power = mean square value = E  x  = m
2

3
4
Average Power =
3
Peak Power 4
 = =3
Average Power  4 
 
3

03. Ans: (d)

Solution: from the quantizer transfer function,
→ above is uniform quantizer with step size    = 2

2 4
 quantization noise power = Nq = =
12 12
1
Nq =
3
P  d  f of input is

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fx ( x ) = Ae
−x
−4  x  = 0 otherwise
4
−x 1
 Ae
−4
dx =
4
0 4

A   ex dx +  e− x dx  = 1
 −4 0 
 0  e− x  
4

  
A e  + 
x
  =1
 −4
 −1  0 
 
A 1 − e−4 + 1 − e−4  = 1

1
A= = 0.5093
2 1 − e−4 
A  0.51
4
−x
Signal Power = mean square value of x = E  x  = x  Ae dx
2 2

−4
4
E  x2  = 2A  x2e− x dx
0

  x2e− x  4 4 2x  e− x 
 
E  x2  = 2A    − dx 
  −1  0 0 ( −1 ) 
   xe− x  4 4 
 
E  x  = 2A −16e + 2  
   +  e dx  
2 −4 −x

  −1  0 0 
  
 (
E  x 2  = 2A −16e−4 + 2 −4e−4 + 1 − e−4 )

E  x 2  = 2A −16e−4 − 8e−4 + 2 − 2e−4 
E  x 2
 = 2A 2 − 26e  −4

E  x 2  = 2.5257

S E  x  2.5257
2

 = =
Nq Nq 1
3
S
= 7.577
Nq
S
 8.8 dB
Nq
dB

04. Ans: (a)

Solution: fm = 3.4 KHz
A = 1V
Rb = 20 kbps
for delta modulator,
bit rate = sampling frequency = fs

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 fs = 20  103 Hz
Ts = 0.05 msec.
To avoid slope overload error, maximum amplitude should be such that

A
m Tm
when a is maximum, we get minimum step size as
  AmTs
min = A2fmTs
min = 1  2  3.4  103  0.05  10−3
min = 1.06v
min  1v

05. Ans: (d)

Solution: Let sinusoidal i/P with peak value = A
Max quantization error = 0.1%A
 0.1
  A
2 100
A

500
VH − VL A

L 500
 A − ( −A ) 
  A
L 500
2A A

L 500
L  1000
Lmin = 1024

06. Ans: (a)

Solution:  = 0.628
Rb = 10 kbps
fm = 1 KHz
To present slope overload distortion,

A
m TS
fS
A
2fm
In delta modulation,
bit rate =sampling frequency
 fS = 10  103 Hz
fS 0.628  10  103
Amax = =
2fm 2    1  103

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Amax = 0.999
Amax  1V

07. Ans: (a)

Solution: Granular noise occurs when
d 
m(t) 
dt max
TS
if m(t) = Am sin mt

d
m ( t ) = Am m cos mt
dt

 Amm 
TS
  fS
Amfm 
2
Amfm  1000
For option (a) → Amfm = 200 , which is less than 1000. Hence granular noise occurs for option (a).

Type 10 – Baseband Shaping for Data Transmission

01. Ans: (a)
Solution: From the given options, we can see that all the options have same form, they only differ in their amplitude.
 T
i.e., u ( t ) − 2u  t −  + u ( t − T )
 2
Now, a basis function should be orthonormal.
i.e., It’s energy should be 1.
To concert the above function into orthonormal, we divide it by E.
T
2 T
E =  (1 ) dt +  ( −1 ) dt
2 2

0 T
2

T  T
E= + T − 
2  2
T T
E= +
2 2
E=T
 E= T
Hence,
1   T  1   T 
Basis function is = u ( t ) − 2u  t −  + u ( t − T )  = u ( t ) − 2u  t −  + u ( t − T ) 
E  2  T  2 

Type 11 – Digital Modulation methods-ASK, FSK, PSK AND QAM

01. Ans: (a)
Solution: DPSK modulator.
(
initially, b t − Tb = 0 )

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Phase shift 180º 0º 180º 0º 0º 180º

Hence, phase shifts of the carrier is
000

02. Ans: (b)

Solution:
Binary 1 → 10 mV
Binary 0 → -10 mV
P(t) and – P(t) are used for BPSK modulations. Also, BPSK has the best noise performance compared to ASK and FSK.
Hence, for BPSK, Euclidean distance is
d = 2 Eb
0.110−3

 (10  10 )
2
Eb = −3
dt
0

( )
2
Eb = 10  10 −3  0.1  10 −3

Eb = 10 −8 J
Hence, dBPSK = 2  10−4
Now, if we do ASK modulation, then
Binary 0 → No pulse
Binary 1 →
0.110−3
Eb = 
0
A 2dt = A2  10−4

dASK = Eb = A2  10−4 = A  10−2

For both BPSK and ASK to have same noise performance, their Euclidean distance should be equal and for ASK to have
better performance than PSK, dASK  dBPSK .
i.e., dBPSK  dASK
2  10  A  10−2
−4

 Amin = 20mV

03. Ans: (d)

Solution: In DPSK demodulation is done by non-coherent techniques, for which carrier signal is not required.
Hence, local oscillator is not used in DPSK.

04. Ans: (c)

Solution: FSK is used in Digital Telephony

05. Ans: (b)

Solution: Total symbols = 4 [0º, 90º, 180º, 270º]
 QPSK modulation is used.

06. Ans: (c)

Solution: bit rate = 34 M bit/sec.
In QPSK,
M= 4
M = 4 = 2n
n = 2

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bit rate 34
Hence, baud rate = symbol rate = = M bit/sec. = 17 M bit/sec
2 2

Type 12 – SNR and BER for Digital Modulation Schemes

01. Ans: (a)
Solution: Pulse for 1 → +A
Pulse for 0 → -A
No
= 10 −5 w / Hz
2
1
Rb = = 7.26  103 bps
Tb
Pe = 10−4
For polar binary signaling,
 2E 
Pe = Q  b

 No 
 
Eb → avg bit energy.
Ep + Eq
Eb = , Since Ep = Eq
2
Eb = Ep = Eq
Tb

 (A)
2
Eb = dt
0

Eb = A 2 Tb
 2A2 T 
 Pe = Q  b  = 10 −4
 No 
 
2A2 Tb
Hence, = 3.71
No

2  A2
= 3.71
2  10  7.26  103
5

 A  1V

02. Ans: (d)

 2E 
Solution: Pe = Q  b
 = 10−4
 N0 
 
2Eb
 = 3.71
N0
2Eb
= 13.7641
N0
Eb = 13.7641  10−5 J
T

 (1 )
2
Eb = dt = T
0

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 T = 13.7641  10−5
1
bitrate (Rb ) = = 7.2652kbps
Tb

03. Ans: (b)

Solution: Number of bits = 8
Probability of error =P
Probability of at most one error
= Probability of zero error + Probability of one error
= 8 C0P0 (1 − P ) + 8 C1P1 (1 − P ) = (1 − P ) + 8P (1 − P )
8 7 8 7

Type 13 – MAP and ML decoding and Matched Filter Receiver

01. Ans: (d)
Solution: input S t = e ,0  t  T
−t
()
Impulse response → h t = S T − t () ( ) = e− T  et ,0  t  T
y (t) = x (t)  h(t)

y (t) =  x (  ) h ( h −  ) d
−

Since, both x(t) and y(t) are causal,

t t t
 e−2z 
y ( t ) =  e− z   e− T e( )  dz = e− T  et  e2z dz = e− T  et 
t−z

  
0 0  −2  0
 e−2t − 1 
y ( t ) = e − T  et   = e  sinht
−T

 − 2 

02. Ans: (d)

Solution: m = 1
 m2 
Pt = Pc 1 + 
 2 
Pt = Pc + modulating power
m2 P
Modulating power Pc = c [m = 1] = 0.5Pc = 50Pc %
2 2

03. Ans: (b)

Solution:
h ( t ) s ( T − t ) = s (3 − t )
y (t) = h(t)  s (t)
T = 3, width of y(t) = 6
 o / P of matched filter at t = 3 is energy of input S/g.
1 3
 ES ( t ) =  A 2dt +  A 2dt = A2 + A2 = 2A2
0 2

y (t) = 2A 2
t =3

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04. Ans: (a)

Solution: Let the input pulse be P(t)
() (
Impulse response of matched filter is given by h t = P Tb − t )
h ( t ) = P ( −t + Tb )
y(t) = p(t)  h(t)

05. Ans: (d)

Solution:
y(t) = p(t)  h(t)
Convolution of two rectangular pulses with same width is a triangular pulse.

Type 14 – Basics of TDMA, FDMA and CDMA

01. Ans: (b)
Solution:
Bandwidth of signals Sampling frequency
w 2w
2w 4w
4w 8w
4w 8w
6w 12w
6w 12w

We select lowest sampling frequency for commutator.

 fs = 2w
Number of samples of each signal during one rotation of commutator is
1, 2, 4, 4, 6, 6.
 total number of samples = 1 + 2 + 4 + 4 + 6 +6
N = 23
Hence, minimum transmission = 23  fs
Bandwidth = 23  2w = 23 × 2 × 750 = 34500 Hz = 34.5 KHz

02. Ans: (b)

Solution: Number of signal channels = 20
Sampling frequency of each channel = 8 KHz.
Hence, speed of commutator = 8 KHz.
Every channel contributes 1 shape during one rotation of commutator.
Total number of samples in one rotation of commutator = 20 = N.
Number of bits/sample = 7 + 1 for synchronization.
Hence, bit rate = nNfs = [1 + 7] × 20 × 8000 = 1280000 = 1280 K bits/sec.

03. Ans: (d)

Solution: fm = 600 Hz
1

fm = 600 Hz
2

fm = 1200 Hz
3

fm = 2fm = 1200 Hz
1 1

fm = 2fm = 1200 Hz
2 2

fm = 2fm = 2400 Hz
3 3

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We select lowest sampling frequency as speed of commutator.

 fs = 1200 Hz
 number of samples of
Channel 1 = 1
Channel 2 = 1
Channel 3 = 2
N=1+1+2=4
 bitrate = Nnfs = 4 × 12 × 1200 = 57.6 kbps

04. Ans: (a)

Solution:
FDM requires lower bandwidth compared to TDM.
TDM has greater noise immunity than FDM.

05. Ans: (a)

Solution: Digital communication requires proper synchronization which is done using bit synchronization signal

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Singnal and System