ANSWERS - IES MasterAfter passing through frequency doubler, NBFM becomes, = c c m m f A cos 2 2 f t...
Transcript of ANSWERS - IES MasterAfter passing through frequency doubler, NBFM becomes, = c c m m f A cos 2 2 f t...
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)
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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.