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