ANSWERS - IES MasterAfter passing through frequency doubler, NBFM becomes, = c c m m f A cos 2 2 f t sin2 f t f = c c m m 2 f A cos 2 2f t sin2 f t f Thus, new carrier frequency =

ESE-2018 PRELIMS TEST SERIESDate: 19th November, 2017

ANSWERS

1. (d)

2. (a)

3. (d)

4. (b)

5. (a)

6. (c)

7. (b)

8. (d)

9. (c)

10. (b)

11. (c)

12. (c)

13. (b)

14. (a)

15. (a)

16. (b)

17. (a)

18. (c)

19. (a)

20. (b)

21. (d)

22. (c)

23. (b)

24. (c)

25. (b)

26. (a)

27. (b)

28. (c)

29. (a)

30. (c)

31. (c)

32. (d)

33. (c)

34. (c)

35. (b)

36. (d)

37. (b)

38. (c)

39. (d)

40. (a)

41. (d)

42. (c)

43. (c)

44. (c)

45. (d)

46. (d)

47. (a)

48. (a)

49. (d)

50. (d)

51. (b)

52. (a)

53. (d)

54. (c)

55. (a)

56. (a)

57. (c)

58. (c)

59. (c)

60. (d)

61. (b)

62. (c)

63. (d)

64. (d)

65. (b)

66. (b)

67. (b)

68. (d)

69. (c)

70. (c)

71. (a)

72. (b)

73. (b)

74. (a)

75. (b)

76. (d)

77. (d)

78. (c)

79. (a)

80. (b)

81. (a)

82. (b)

83. (a)

84. (a)

85. (a)

86. (d)

87. (c)

88. (a)

89. (c)

90. (d)

91. (c)

92. (d)

93. (c)

94. (b)

95. (c)

96. (b)

97. (b)

98. (b)

99. (a)

100. (b)

101. (b)

102. (b)

103. (b)

104. (c)

105. (a)

106. (a)

107. (d)

108. (d)

109. (a)

110. (c)

111. (d)

112. (a)

113. (d)

114. (c)

115. (c)

116. (a)

117. (d)

118. (b)

119. (d)

120. (c)

121. (a)

122. (b)

123. (c)

124. (d)

125. (d)

126. (c)

127. (a)

128. (d)

129. (a)

130. (b)

131. (a)

132. (b)

133. (c)

134. (a)

135. (b)

136. (b)

137. (a)

138. (a)

139. (a)

140. (a)

141. (a)

142. (c)

143. (a)

144. (b)

145. (c)

146. (a)

147. (d)

148. (a)

149. (a)

150. (a)

IES M

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Mobile : E-mail: 8010009955, 9711853908 [email protected], [email protected]. office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064

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1. (d)

x t = 1 2x t jx t

x(t) = cj tRe x t e

= cj t

1 2Re x t jx t e

=

1 c 2 c

2 c 1 c

x t cos t x t sin tRe

j x t cos t x sin t

= 1 c 2 cx t cos t x t sin t

2. (a)RLC = 400×103×100×10–12 = 4×10–5 s

2

Lm

1 1 mR Cm

25

m

1 1 0.754 106.28f 0.75

5

m10 0.6614f 3510Hz

4 6.28 0.75

mf 3510Hz

max. frequency = 3510 Hz

3. (d)

Given,

xc(t) = m c5 1 2cos t cos t

This is an over-modulated AM signal. Hence,the envelope will be distorted and anenvelope detector cannot be used.

Let us check whether a square-law detectorcan be used.

yc(t) = Output of the square-law device

= 2cax t

= 2m ma 25 1 4cos t 4cos t

c1 1 cos2 t2

= m m25 5050cos t 1 cos2 t2 2

c25 cos2 t2

c m50cos2 t cos t

2m c50cos t cos2 t a

If the dc component is blocked by acoupling condenser and the high frequencycomponents are removed by using an LPFof cutoff frequency fm after the square-lawdevice, the final output will be z(t) =

ma.0.5cos t , which is proportional to themodulating signal.

Hence, a square-law detector can be used.

Now, let us check whether a synchronousdemodulator can be used. Recall that insynchronous demodulation, we first multiplythe received modulated signal by the locallygenerated carrier signal and then pass theproduct through an LPF having a cutofffrequency of W Hz, the bandwidth of themodulating signal.

2c c m mx t cos t 5 1 2cos t cos t

m c5 1 2cos t 1 2cos t2

c m5 5 cos2 t 5cos t2 2

c m c m5 5cos 2 t cos 2 t4 4

the output of the LPF = z(t) =

m55cos t2

The dc component, i.e., 5/2, can be rejectedby using a coupling condenser, and theoutput will then be only the message signal.

Hence, either a square-law detector, or asynchronous detector, may be used, butnot the envelope detector.

4. (b)

Amax = 12V, Amin = 4V

Modulation index,

12 4 8 1 0.512 4 16 2

Unmodulated carrier amplitude

max min

cA A 12 4A 8V

2 25. (a)

We know that bandwidth occupied by anAM signal is equal to twice the highestaudio frequency in its modulating signal.

Bandwidth required for each station = 2×5KHz = 10 KHz

Bandwidth available = 1.5 MHz – 1.0 MHz

= 500 KHz

Number of stations that can be

accomodated = 500 50.10

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6. (c)

For conventional AM

Antenna current,

2

tI 12

For DSB-SC

It (Antenna current)

2

2For SSB-SC

It (Antenna current)

2

4 2

Thus, In SSB-SC antenna current is directlyproportional to modulation index, so ifmodulation index is halved, antenna currentwill also halved.

7. (b)

8. (d)

For negative peak clipping occurs,

ac

dc

RmR , where m = modulation index

It is given that ac load resistance is very

much smaller than the dc load So, ac

dc

RR

becomes negligible, this results negativepeak clipping.

9. (c)

The phase deviation t produced by themodulating signal

3pk x t 4 3sin2 2 10 t

3t 12sin4 10 t

If the modulated signal = xc(t) =

c cA cos t t , the instantaneous

frequency fi is given by

fi = c c1 d 1 dt f t

2 dt 2 dt

fi = 3c

1 df 12sin 4 10 t2 dt

= 3 3cf 24 10 cos 4 10 y

peak frequency deviation of the carrier is

3f 24 10 24 kHz

10. (b)

Messagesignal

PMsignaldifferent-iator FM

modulator

11. (c)

Modulation index for phase modulatingsignal,

P pK max x t

p

So, modulation index also double.

12. (c)

Average power of FM signal is = 2cA

2RWhere, Ac is amplitude of carrier wave.

So, Average power of FM signal only dependupon amplitude of carrier.

13. (b)

General expression of NBFM

=

f mc c m

m

K AA cos 2 f t sin2 f tf

=

c c mm

fA cos 2 f t sin2 f tf

After passing through frequency doubler,NBFM becomes,

=

c c mm

fA cos 2 2 f t sin2 f tf

=

c c mm

2 fA cos 2 2f t sin2 f tf

Thus, new carrier frequency = 2fc

and new frequency deviation = 2 f

14. (a)

Given, fc = 12 MHz, f = 3.2 KHz

Filter

Oscillator

f =10 MHzo

Centered atdifference frequency

Outputf =12 MHzc

General expression of FM signal

f mFM c c m

m

K AS t A cos 2 f t sin2 f tf

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After passing through mixer,

f mFM c c m o

m

K AS t A cos 2 f t sin2 f tK 2 f tf

But filter passes only difference of frequency.

FM c c m om

fS t A cos 2 f t sin2 f t 2 f tf

max c of f f f

min c of f f f

Thus new carrier frequency

= fc – fo = 2 MHz

but f is unchanged.

15. (a)

16. (b)

Foster-seeley responds to both amplitudeand frequency variation.

17. (a)

For WBFM signal, a/c to carson’s law,

bandwidth is given as = m2 1 f

mBW f

For WBFM signal bandwidth is linearly variedwith fm.

18. (c)

For good image rejection a high value of IFis required and that for good sensitivity andselectivity, a low value of IF required. Hencethe choice of IF value is generally based ona compromise between these conflictingrequirements. How ever these problem maybe solved by the use of double hetrodyne,or double conversion receivers that can givegood image rejection as well as goodselectivity. Use a high first IF to get goodimage rejection and a low second IF to getgood selectivity.

19. (a)

In an FM receiver the amplitude variationsof the received FM signal, caused by noiseetc are removed by using amplitude limiters.Amplitude limiting action may be obtainedin an IF stage by including back to backconnected diode in the input tuned circuitof the IF amplifier.

20. (b)

All FM communication system use pre-emphasis at the transmitter and de-emphasis at the receiver, to improve SNRat the desitnation. Pre-emphasis consits ofboosting the high frequency components ofthe message signal before modulation andde-emphasis attenuates the high frequencycomponents of the message signal obtainedin the receiver at the output of thediscriminator.

21. (d)

In a superhetrodyne receiver the differencebetween, local oscillator frequency (fo) andcarrier frequency (fc) should be equal to fif(intermediate frequency) of the receiver. Thusfo may be greater or less than fc. But it isalways arranged to be greater than fc asotherwise, the tunning capacitor rangerequired will be for greater than what weobtained in practise.

22. (c)

The phenomenon of desired signal fs beingreceived at two different dial setting of thereceiver, is known as double spotting. Thiscause is poor image rejection.

23. (b)

24. (c)

In double hetrodyne receiver two IF amplifieris used. The first IF is chosen high to getgood image rejection and second IF ischosen low to get good selectivity andsensitivity.

25. (b)

1

2 5x

F (x)x

For 2 x 5

Fx(x) = 1 x 23

For x > 5Fx(x) = 1

For x < 2

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Fx(x) = 0

P(2.5 < x < 4)= Fx(4) – Fx(2.5)

= 2 0.53 3

= 1.5 0.53

26. (a)

Let x is a random variable which is uni-formally distributed in 3 and 6. Then theirpdf will be

f (x)x

1/3

3x

0 6

E[x] =

xxf x dx

= 6 62

33

1 1xdx x 4.53 6

E[x2] =

2

xx f x dx

= 6 62 3

33

1 1x dx x 213 9

Variance, 22 2E x E x

= 21 – 4.52 = 0.75

27. (b)

Figure of merit (FOM) of an AM system

=2 2

2 2

m x

1 m x(i) m = 1, i.e., 100% modulation

FOM = 2

2

x .1 x

Since nothing has been

mentioned about the average power of themodulating signal, if a single-tone is

assumed, 2x 1/2

m 1

1/ 2 1 2 1FOM1 1/ 2 2 3 3

28. (c)

PSD of noise

/2

f

Noise power = PSD × BW

= 2

2=

since, PSD is two sided.

29. (a)

We know that in case of phase modulation

p p m mt K x t K E cos t

Peak frequency deviation produced by this

phase modulation = p m mmax

d t K Edt

If mf is the modulation index for FM

p m m

f p mm

K Em K E

In case of PM, the ‘figure of merit’ is givenby

(FOM)PM = 2 2pK x t

=2p2

pE

K2

= 2fm

230. (c)

Noise figure =1

figure of merit

Conventional AM has FOM =

2

2 2

SSB-SC has FOM = 1

FM signal has FOM = 232

31. (c)

Antialiasing filter is a low pass filter usedto band limit the signal prior to sampler.

32. (d)

33. (c)

Noise performance of PAM signal is at least3 dB inferior from direct base band analogmessage signal transmission.

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34. (c)

In comparison with FDM, the TDM hardwareis much simpler. It has several otheradvantages over FDM, such as its ability toeasily handle base band signals havingwidely different band widths and its relativerobustness with regards to short-term fading.

35. (b)

From 2Y ,x

we have

2dx 2 2,xdy yy

and 2

Y X ii i

dy y 1,f y f xdx 2 dy dx

Y X X2dx 2 2f y f x fdy yy

36. (d)

For this problem, it is easiest to work withthe expectation operator. The mean functionof the output is

E[Y(t)] = 2 + E[X(t)] = 2

The autocorrelation of the output is

YR t, = E 2 X t 2 X t

= E 4 2X t 2X t X t X t

= 4 2E X t 2E X t

E X t X t

= X4 R

We see that YR t, only depends on the

time difference . Thus Y(t) is wide sensestationary.

37. (b)

For option (A) modulation index is 30 1.520

For option (B) modulation index is 16 420 5

which is less then 1 so this signal can berecovered properly.

For option (C) signal is DSB-SC somessage can not be recovered by envelopdetection method.

38. (c)

The auto-correlation function of x(t)

T 2

x TT 2

1R lim x t x t dtT

= T 2

T

1lim 1 dtT

X T

1 T 1R limT 2 2

Thus X X1S f FT R f2

39. (d)

Consider the AM signal

m ct A 1 mcos t cos t

Let mt A 1 mcos t denote the

corresponding envelope.

max A 1 m 125 A 75

min A 1 m 25

m mt 75 1 0.667cos t cos t

40. (a)

41. (d)

i t 50 t 2 t

id t 3100 50 t

dt 2

i 100 75 t

if 50 37.5 t Hz

42. (c)

By Carson’s rule:

max mB 2 f f

3 3max200 10 2 f 52 10

3maxf 100 52 10 48kHz

But, max ff k max m t

3max

3f

f 48 10max m t 12Voltsk 4 10

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43. (c)

LPF OutputFMSignal

VCO

The frequency interval in which phase ofthe FM signal is approximated to phase ofVCO is called as captured mode of PLL.

44. (c)

Hilbert transform of a signal x(t) is defined

as convolution of x(t) with 1 ,t

and denoted

by x(t).

x t = 1 x tt

If x(t) = t ,

then H.T. of t = 1 1tt t

45. (d)

• The demodulation of the pulse is done easilyusing a low pass filter.

• Amplitude demodulation is done using diodedetector.

• Frequency demodulation is done using ratiodetector.

• Phase demodulation is done using cohrentdetector.

46. (d)

LPF m(t)S (t)DSB Multiplier

LO

SDSB(t) = c cA m t cos2 f t

Local oscillator output = c cA cos 2 f t

No phase synchronization.

Multiplier output = outputDSBS t . LO

= 2c c cA m t cos2 f t cos 2 f t

=

2c

cA m t

cos 4 f t cos2

LPF Output

= 2

cA m tcos

2

cos is constant

So, it will just attenuate (reduce) the output.

47. (a)

Output

1 Amplifierst

G =20dBF =20dB

1

1

G =20dBF =20dB

2

2

Overall noise figure = 2

11

F 1F

G

...(i)

20 dB on linear scale = 100

NF = 99100 100.99

100 101

NF(dB) 20.04dB

48. (a)

90º Phaselead

cos t0

Serialto parallelconverter

QPSKOUTPUT

Localoscillator

sin t0

1 Serial to parallel converter

2 Local oscillator 0cos t

3 90º phase lead

49. (d)

Channel capacity is given by

Cs =

iP xmax I x;y in bits/ symbol

where, I (x;y) is mutual information channelcapacity in bits/second

Cs =

i

s P xR max I x;y

Rs = Transmission rate

Thus, channel capacity depends onmaximum rate of information transmission.

50. (d)

51. (b)

52. (a)

Given, fm = 600 Hz, and fs = 1000 Hz

After sampling, the frequency componentswill be,

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= m s m s mf , f f , 2f f ,....

= 600, 400, 1400, 2400

= ....–2400, –1400, –400, 400, 600,1400, 2400

53. (d)

Shannon Hostley law describes the channelcapacity of communication channel over aspecified band width. So, statement 4 iswrong.

Statement 1, 2 and 3 all about digitalcommunication.

54. (c)

1st case, no. of quantization level, L = 2

Thus no. of bits, n = log22 = 1

BW = nfs = fs

fs = B

2nd case if, L' = 8

then, n' = log28 = 3

BW = 3fs = 3B

55. (a)

Comparison of DM, over PCM

(i) DM transmitter and reciever require verysimple and inexpensive hardware. Althougha higher sampling rate is used in DM, sobit rate is high. Since only one bit is usedso, bandwidth requirement is less in DMcompare to PCM.

(ii) If input message waveform has steelgradients, severe slope overload distortionresults, since step size is fixed.

56. (a)

Properties of auto correlation function.

(i) R 0 R

(ii) R R

(iii)

2x xlim R , mean

57. (c)

Entropy of the source =

3

i 2 ii 1P log P

H(s) = 2 2 2 21 1 3 3 3 3 1 1log log log log8 8 8 8 8 8 8 8

Nyquist rate = 2

Information rate = r H(s) = 2 1.81 =3.62

58. (c)

59. (c)

Companding - to protect the small signalsin PCM from quantizing noise.

TDM - To use only one carrier frequency tohandle different signals.

Source coding - To increase the informationtransmissioin rate.

FDM - To use different frequency band fordifferent signals.

60. (d)

E[XY] = 0, means

X and Y are two orthogonal randomvariables.

If two random variables X and Y areuncorrelated, that means their covarianceis zero.

EXY = 0 = E[XY] – E[X] . E[Y]

E[XY] = E[X] . E[Y]

So, E[XY] = 0 means neither uncorrelatednor independent.

61. (b)

Quantizers - If converts sample of analogsignals into a quantized sample having anamplitude corresponding to the prescribedlevel.

Encoder - Encoder converts quantizedsignal into corresponding digital signal. So,combinedly it converts analog signal intodigital.

62. (c)

Minimum bandwidth required for PAM/TDM= NW

Where, N = no. of messages

and W = Bandwidth of each message signal

Required minimum BW = nB.

63. (d)

Noise power in resistance = KTB

where, T = equivalent temperature

B = bandwidth

K = Boltzman constant

64. (d)

Quadrature carrier multiplexing is a methodwhich uses same carrier frequency for two

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different messages signals to occupy thesame transmission bandwidth.

65. (b)

Since x and y are statistically independent,

fXY (x, y) = fX(x) . fY(y)

66. (b)

Step size used in delta modulation is fixedwhere as variable step-size is used inadaptive delta modulation.

67. (b)

Let message bandwidth = B

then, Nyquist rate will be = 2B

Message BW of PCM system will be= 2B×8 = 16 B

Bit rate = Rb = 108

Rb = 16B = 108

B = 6.25 MHz

68. (d)

Given, pdf is

xf x 0.2 x 1 0.2 x 2

0.4 x 3 0.15 x 4 0.15 x 5

x54321

0.150.2

0.4

f (x)X

Mean, Ex[x] = 5

xx 0xf x

= 1 × 0.2 + 2 × 0.2 + 3 × 0.4 +4 × 0.15 + 5 × 0.15

= 3.15

69. (c)

Random processes for the time averagesequals the ensemble averages, are knownas ergodic processes.

70. (c)

If x(t) and y(t) are respectively the inputand output processes for an LTI systemthen,

Mean of the output,

Y X h t dt

where, X is mean of input.

2t 2tY

0

2 2e u t dt 4 e dt

2t

04 e 22

71. (a)

If x(t) is periodic then it can be describedexactly by a finite number of samples –corresponding to those in one period of x(t).So, let us first check whether x(t) is periodic.

T1 = period of 10 cos 2 16 t6 3

T2 = period of 4 sin 2 18 t8 4

1

2

T 1 4 4T 3 1 3

Which is a rational number.

Hence, x(t) is periodic. Now, to determineits period T.

1 1T LCM , 13 4

T = 3T1 = 4T2

The maximum frequency present in x(t) is4 Hz, which is the frequency of the sin 8 t component.

the minimum sampling frequency required= 8 sample per sec.

the number of samples in one period of x(t)is equal to 8 since T = 1 second and thesampling frequency is 8 samples persecond.

72. (b)

Since the signal fully loads the quantizer, itmeans that the Q levels cover the full rangeof –Am to +Am of the signal.

Hence, step size m2AQ

quantization noise mean-squared value

= 2

qN12

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2 2m m2 2

4A A12Q 3Q

S = average signal power = 2mA

2

2 22m

2q m

AS 3Q 3 QN 2 2A

73. (b)

2x t 3cos 250 t

Message frequency, fm = 2×125 = 250Hz

x(t) is sampled at regular interval of Tsecond.

To recover this signal without any distortion,

m1 2fT

m

1T2f

1T2 250

Tmax = 2ms

74. (a)

By reducing the pulse width of the pulseused in flat-top sampling, apesture effectcan be reduced.

If is pluse width of pulse and is thebandwidth of the message signal then to

reduce aperture effect

1 .2

75. (b)

0 Tt

g (t)z

gz(t) = u(t) – u(t – T)

Gz(s) = Ts1 e

s

76. (d)

Quantization characteristic of mid-tread typequantizer.

1 2 30.5–0.5–3 –2 –1

–1

–2

–3

–4

4

3

2

1

Output

Input

77. (d)

In linear delta modulator fixed step size isused which ensures that

max

dx tTs dt

Thus slope overload is avoided. But considerthe case when the message signal isrelatevely flat.

x (t)q

x(t)

From figure the granular noise arising fromthe ‘hunting’ that takes place when the signalis not changing much, will increase as thestep size is similar to the quantizationnoise of PCM.

78. (c)

T/2 T

As(t)

Impulse response of matched filter will be,

h(t) = s(T – t) = s(t)

H(f) = 2 j fTAT fTsinc e2 2

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79. (a)

Roll of factor, 11 f w

if 11,f 0

Transmission bandwidth,

T 1B 2w f 2w

80. (b)

Since FSK and PSK signals have aconstant envelope, they are immune toamplitude non-linearities which arise inmicrowave and radio channels. Hence FSK,PSK signals are preffered to ASK inbandpass data transmission over none linearchannel.

81. (a)

Probability of error of QPSK

= bE1 erfc2

Probability of error of BPSK

= bE1 erfc2

Probability of error of coherent BPSK

= bE1 erfc2 2

Probability of error of noncoherent BPSK

= bE1 exp2 2

Probability of error of DPSK

= bE1 exp2

82. (b)Given, M = 16, fs = 10 kbps

n = log2 M = log2 16 = 4Bit rate, Rb = nfs = 4 × 10 kbps = 40 kbps

83. (a)

Since BFSK, BPSK and MSK signals haveconstant envelope, they are immune toamplitude non-linearities which arise in radiochannels. Hence QAM is not preffered whenchannel is non-linear.

84. (a)

As there are 6 faces for each die, there are36 pairs possible altogether. Each of thesecan occur in two ways if we do not bother

about which die has shown up whichnumber.

probability of any given pair of numbers

=1 1 126 6 18

information obtained whenever any pair ofnumbers shows up

= 2 21log log 18

18

Average information = 2118 log 18 bits

18

= 4.1703 bits/pair of numbers

85. (a)

C = 2SBlog 1N

= 2SBlog 1B

=4

62 6

101.5 10 log 11.5 10

= 14.38 kilo bits/second

86. (d)

Maximum entropy is given by H(s) = log2 Mwhere, M = size of alphabet

= log2 128 = log2 27

= 7 bits

87. (c)

The basic two requirements for optimalcodes are

(a) Minimum average length of a code word fora given set of source alphabet {X} and thesource symbol property set {P(xi)}

(b) Unique dicepher ability of the encodedsequence.

88. (a)

Mutual information I(X,Y) of channelrepresents the average amount ofinformation transferred through the channel,in bits per symbol.

I(X,Y) = I(Y,X) = H(Y) – H(Y/X)= H(X) – H(X/Y)

or I(X,Y) = H(X) + H(Y) – H(X/Y)

89. (c)

90. (d)

Bandwidth efficiency of FSK is given as

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=b 2

T

R 2log MB M

If M is increased, Numerator increasesslowly where as Denominator increasesfastly. So bandwidth eff iciency

decreases.

Also, bE decreases and comes closer to

the shannon limit.

91. (c)

(n,k) (7,4)

i.e. no. of message bit = 4

no. of code length = 7

no. of parity bits = 7 – 4 = 3

Ratio of parity bits to message bits = 34

Code rate = no. of message bitno. of code length =

47

92. (d)

We know that,

P A BP A BP B

P A B 0

So, A and B are mutually exclusive.

93. (c)

SNR = 3k2 22n, n = number of bits persample

rms

max

EkE

k is the ratio between the rms and peakvalues of the signal voltage.

10 log10 (SNR) = 10 log10 3 + 20 log10 k + 20n log10 2

SNRdB = 4.77 + kdB = 6.02 n40 = 4.77 + kdB + 6.02 nkdB < –10 dB

40 – 4.77 – 6.02 n < –10

or 40 4.77 10 n

6.02

n > 7.5 n = 8

94. (b)

B = bR 1.5441 1 1 Mbps2 2

= 1.544 MHzB = bandwidth, = roll off factor,

Rb = bit rate

95. (c)

4

ii 1

1E E4

2 2 221 2a a 4a 2a4

2 2 22a 4a a a

2 21 44a 11a4

96. (b)

n = T G

C

B 2BB

=6 3

325 10 2 20 10 416

60 10

97. (b)

98. (b)

99. (a)

100. (b)

101. (b)

102. (b)

103. (b)

104. (c)

Control System

105. (a)

The bilinear transformation z 1z 1

maps

the inside of the unit circle in the z-planeto left half of the -plane.

106. (a)

For an all-pass analog filter

| H( ) | = constant

If H(s) =1 s1 s

H( ) =1 j t1 j t

| H( ) | =2 2

2 2

1 11

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107. (d)

From the given pole-zero plot of the digitalfilter, the system function

H(z) = 6

1 1(z 1)(z 1) z z (z 2)(z 2)2 2

z

=2 2 2

61 1(z 1) z (z 4)

4z

= 6 4 2 21z z 1.25z (z 4)4

= 6 6 4 4 2 21z z 4z 1.25z 6z z 14

= 6 6 4 2z z 5.25z 6.25z 1 = 1 – 5.25z–2 + 6.25z–4 – 1z–6

h(n) = [1, 0, –5.25, 0, 6.25, 0, –1]

As the impulse response, h(n) = InverseZ.T of H(z) has only finite duration = 7samples, the given digital filter is an FIRfilter.

108. (d)

For stability poles lie inside unit circle andzeros can be anywhere

For minimum phase all pole and zeros lieinside the unit circle.

Poles = 0.5Zeros = 2

Therefore, non minimum and stable

109. (a)

In the design of IIR digital filters by theMethod of Impulse Invariance, the impulseresponse hd(n) of the digital filter is obtainedby sampling the impulse response, ha(t) of

the prototype analog filter. hd(n) = s

a t nTh (t) .

As no practical analog filter is band limited,problem of aliasing (overlapping of theadjacent analog spectra is bound to occur)occurs. Hence frequency response of digitalfilter is not a scaled version of the frequencyresponse of the analog filter. In the designof IIR digital filters by the method of Bilinear

Transformation, z 1sz 1

al iasing is

avoided, because the mapping from s-planeto z-plane is one to one and the entireimaginary axis of the s-plane is mappedinto the unit circle in the z-plane.

110. (c)

Remember the property of linear phase FIRfil ters with a real impulse responseregarding the location of its zeros:

If j1z re is a complex zero, then 3 more

complex zeros occur as given below:

j2

1

1 1z ez r

* j3 1z z re

and * j4 2

1z z er

are also its zeros

For a zero at j4

11z e ,2

the largest set

of remaining zeros that can be obtainedfrom this information are

j j j4 4 41 e , 2e , 2e

2

111. (d)

DF and equivalent lattice implementationare given as shown in figure.

z–1 z–1

0.4 0.64

Output y[n]

Input x[n]

Figure a

Output y[n]

z–1 z–1 –k2

–k2–k1

–k1

Figure b

Input x[n]

–k2

Write the output y(n) in terms of x(n) byseeing the forward paths along the arrowsfrom x(n) to y(n):

For the DF,

y(n) = x(n) + 0.4x(n – 1) + 0.64x(n – 2)...(1)

For the lattice,

y(n) = x(n) – k1x(n – 1) + k1k2x(n – 1)– k2x(n – 2)

= x(n) + k1(–1 + k2)x(n – 1) –k2x(n – 2) ...(2)

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Comparing (1) and (2),

k2 = –0.64, k1(–1 + k2) = 0.4

k1 =0.41.64 = –0.244

112. (a)a0 OutputInput

a1

a2

z–1

z–1

X(z) Y(z)

W(z)

W(z) = X(z) + (a1z–1 + a2z

–2)Y(z) ...(1)Y(z) = a0W(z) ...(2)

= a0X(z) + a0(a1z–1 + a2z

–2)Y(z)...(3)

H(z) = 01 2

0 1 2

aY(z)X(z) 1 a (a z a z )

= 1 2

1 20

11 a z a z

a

...(4)

Comparing (4) with the given T.F

H(z) = 1 21

1 0.7z 0.13z

0

1a = 1 or a0 = 1, a1 = 0.7, a2 = –0.13

113. (d)

Normalized digital angular frequencies:

p22 0.4

10

s32 0.6

10

s

s

pp

tan tan(0.3 )2 1.894tan(0.2 )

tan2

114. (c)

Given:

H(s) = 2 2a

s ah(t) = sin(at)u(t)

h(nT) = sin(anT)u(nT)

=jaTn jaTne eu(nT) u(nT)2j 2j

H(z) = jaT jaT1 z 1 z2j 2jz e z e

The poles of H(z) are ejaT and e–jaT

115. (c)

Low pass-to-Low pass

Transformation:

1 ˆ1 zzz

... (1)

orˆj

jˆj

1 eee

Where is the frequency of the prototypeis, is the frequency after transformation.

It can be shown that

ˆsin2

ˆsin2

...(2)

Given: Notch frequency of prototype digitalfilter,

fN = 60 Hz with fS = 400 Hz,

N602 0.3 rad

400

Notch frequency after transformation,

nf 120 Hz

N2 120ˆ 0.6 rad

400

N N N Nˆ ˆ0.15 , 0.45

2 2

Using the formula in equation (2),

sin(0.15 ) 0.46sin(0.45 )

116. (a)

Magnitude response of the Butterworth filteris given by

22N

c

1H(j )1

p

s

2 1000 r/s,2 1500 r/s

2Np

c

1 0.81

1

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2Ns

c

1 0.04

1

2Np

c

110.81

2Ns

c

110.04

2Np

c

0.190.81

2Ns

c

0.960.04

2Ns

p

0.96 0.81 102.30.04 0.19

s

p2Nlog log102.3 2.01

s

p

2.01 2.01 2.012N 11.4log(1.5) 0.1761log

N = 5.71

N should be taken as 6.

117. (d)

The system function, H(z) is calculated fromthe figure shown

Output

Input

1–k2

x(n)

y(n)

–k k

z–1

W(z)

W(z) = X(z) + kz–1 W(z)

W(z) = 1X(z)

1 kzY(z) = –kX(z) + (1 – k2)z–1W(z)

Y(z) = –kX(z) + (1 – k2)z–11

X(z)1 kz

2 1

1Y(z) 1 k zkX(z) 1 kz

1

1Y(z) k zX(z) 1 kz

118. (b)

From the given feedback diagram

Y(z) = X(z) + az–1Y(z)

T.F = H(z) = 1Y(z) 1 zX(z) z a1 az

H(z) has pole at z = a

For stability, the poles should lie inside theunit circle in the z-plane.

i.e., |z| < 1 or |a| < 1 or –1 < a < +1

So system will transit from stable tounstable state as |a| become greater than1.

0.5 < a < 1.5

119. (d)

Given H(z) = 2

2z 1

z 0.81

Frequency response,

j2

jj2

1 eH e0.81 e

Take / 2 rad ...(1)

If 2 , 3 , 5 , ...etc

or 3, ...etc

2 2

The frequency to be rejected by the filter,

f = 50 Hz

or 100 r/s

Normalizing it to the sampling frequency, fs

s

100 radf

...(2)

ss

100 , f 200 Hzf 2

120. (c)

Given fs = 9 kHz

Filter T.f = H(z) = 1 – z–6

Frequency response,

j j6H(e ) 1 e 0

If 6 0, 2 , 4 ,...etc.,

Taking 6 2 ,

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rad3

...(1)

If f1(Hz) is the analog frequency for which

j1 1H e 0 the 2 f r/sec

Normalizing to the sampling frequency,fs(Hz)

11

s

f2 rad

f ...(2)

From (1) and (2),

1

s

2 ff 3

s1

f 9f kHz 1.5 kHz6 6

121. (a)

An FIR filter of length ‘N’ with impulseresponse h(n) satisfies the condition,

h(n) = ±h(N – 1 – n)i.e., h(0) = h(N – 1)

h(1) = h(N – 2), etc.

For a 4th order FIR filter, N = 5h(n) = [h(0) h(1) h(2) h(3) h(4)]

h(N – 1 – n) = h(4 – n)= [h(4) h(3) h(2) h(1) h(0)]

Given H(z) = (1 + 2z–1 + z–2)G(z)

By observing the symmetry

G(z) should be (3 + 2z–1 + z–2)

which can be verified as shown below:H(z) = (1 + 2z–1 + 3z–2) ×

(3 + 2z–1 + z–2)= 3 + 8z–1 + 14z–2 + 8z–3 + 3z–4

Where h(0) = h(4), h(1) = h(3), h(2) = h(2)

122. (b)

Given H(z) =1 3

0 1 33

3

p p z p z1 d z

The T.F corresponds to a 3rd order digitalfilter

DF-I realization requires 2N delays DF-IIrealization or canonic realization withminimum number of delay elementsrequires N delays

Where N is the order of the filter

DF-I requires 6 delays and DF-II requires3 delays

123. (c)

H(z) = 01 2

2

b,

1 z a z a2 is real.

H(z) =2

02

2

b zz z a

z2 – z + a2 = 0

z = 21 1 4a2

The transfer function BIBO stable meansall poles lies inside the unit circle.

21 1 4a1

2

Substitute a2 = 1/2 Then

11 1 4 1 j 1 j22 2 2 2

1 j 1 1 1 0.707 1,2 2 4 4 2 stable

Substitute a2 = –3/2

31 1 41 7 1 72

2 2 2

1 2.645 1.822 12

unstable

Substitute a2 = –3/4

31 1 41 4 1 24

2 2 2

1 2 3 1.5 1,2 2

unstable

Substitute a2 = 3/2

31 1 41 5 1 5j2

2 2 2

1 5j 1 25 262 4 4 4

= 2.549 > 1

Unstable.

From the given option “a2 = 1/2” satisfyingthe conditions

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124. (d)

x[n] =sin( n / 6)

( n) and h(n) = csin( n)

( n)

jX(e )

1

–/6 /6

jH(e )

1

–c c

j j jY e X(e ).H e

y(n) = x(n) cGiven / 6

jY(e )

1

–/6 /6

y(n) =sin( n / 6)

( n)

147. (d)

For commercial PCM generally 8-bit codeis used.

148. (a)

When an alternative sequence of 1s and 0sis occurring, this indicates that thepossibility of granular noise is high.Consequently, the DAC will automaticallyrevert to its minimum step size and thusreduce the magnitude of the noise error.So, quantization noise is reduced anddistortion will be less.

Thus, Both the assertion and reason arecorrect and reason is correct explanationfor assertion.

149. (a)

SSB signal

= c cc c

A A ˆm t cos t m t sin t2 2

where, Ac = amplitude of carrier

c = frequency of carrier

m(t) = message signal = m mA cos t

m t = H.T. of m(t)

SSB signal = c mc m

A Acos t

2

Thus, envelope of SSB = c mA A2

is

constant.

To recover the SSB signal synchronousdetector is used.

150. (a)B.W 6.28 rad/sec.

2 fm 6.28 rad/sec.fm 1 rad/sec.

Now, according to sampling theorem, signalis fully preserved when it is sampled atNyquist rate or greater than Nyquist ratethus,

s mf 2f

s sm

1 1T T2f 2

Ts 0.5 Hz

So, assertion and reason both are correctand reason is correct explanation of theassertion.

Documents

ANSWERS - IES MasterAfter passing through frequency doubler, NBFM becomes, = c c m m f A cos 2 2 f t sin2 f t f = c c m m 2 f A cos 2 2f t sin2 f t f Thus, new carrier frequency =