IES-ESE Electronics-Communication-Engineering-ECE Contributed by Not specified updated on Mar 2019

UPSC engineering  services Electronics and telecommunication engineering exam model questions with answers and detailed explanations. UPSC IES-ESE Examinations previously asked questions with answers for preparing your examinations

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1. Consider a binary digital communication system with equally likely 0's  and 1's. When binary 0 is transmitted the voltage at the detector input can lie between the levels-0.25 V and + 0.25 V with equal probability : when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1V with equal probability. If the detector has a threshold of 2.0 V (i.e,. if the received signal is greater than 0.2V, the bit is taken as 1), the average bit error probability is

(a) 0.15 (ans)

(b) 0.2

(c) 0.05

(d) 0.5

Shaded area shows the probability of error

Average BEP = P(0)  P(1/0) + P(1) P(0/1)

P(0/1) = Pe (0)  =  error
total

=    0.25 - 0.2     =  0.05
0.25 - (0.25)       0.50

And  P(0/1) = Pe (1)  = 0.2 - 0   = 0.2
1 - 0

P(0) = P(1) = 1/2

BEP = 1/2 (0.2 + 0.1) = 0.15

2. A random variable X with uniform density in the interval 0 to 1 is Quantized as follows:

If      0 ? X ? 0.3,     xq = 0

If      0.3 ? X ? 1,     xq = 0.7

Where Xq is the quatized value of x

The root mean square value of the quantization noise is

(a) 0.573

(b) 0.198    (ans)

(c) 2.205

(d) 0.266

Solution : Since it is uniform as

xq = 0 in the range 0?x?0.3

xq = 0.7 in the range 0.3?x?1

The square mean value is

¥
s2  = ? (x - xq)2  f (x) dx
-¥

1
= ? (x - xq)2  f (x) dx
0

0.3                               0.1
= ? (x - 0)2  f (x) dx + ? (x - 0.7)2  f (x) dx
0                                0.3

0.3                                        1
= [x3/3]    + [x3/3  + 0.49 x - 1.4]
0                                          0.3

or    s2  = 0.039

The root mean square quantization noise

RMS = ?s2

= ?0.039 = 0.198

3. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2.

Group 1 Group 2  1  : FM   P : Slope overload  2  : DM   Q : m Law  3  :  PSK   R : Envelope detector  4  : PCM   S : Capture effect     T : Hilbert transform     U : Matched filter

(a) 1 - T,     2 - P,        3 - U,         4 - S

(b) 1 - S,     2U,           3 - P,         4 - T

(c) 1 - S,     2 - P,          3U,          4 - Q   (ans)

(d) 1 - U,    2 - R,         3 - S,         4 - Q

Solution :     FM --- Capture effect --- Receives only strong signal

DM ---- Slop over load Noise

PSK --- Matched filter

PCM - m law - Non linear quantization by using Companding with a law

V = log (1 +  m |M|)
log (1 +m )

4. There analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is

(a) 115.2 kbps

(b) 28.8 kbps

(c) 57.6 kbps   (ans)

(d) 38.4 kbps

Solution : The three analog Signals having BW 1200 Hz, 600Hz and 600 Hz are sampled at their respective Nyquist rate i.e. at 2400, 1200, 1200 sample/sec respectively.

The total of (2400 + 1200 + 1200) = 4800 sample/sec

The Bit rate = n. fs = (4800 sample/sec) x 12 = 57.6 Kbps

Where n = number of bit in a symbol

5. Find the correct match between group 1 and group 2.

Ground I

P - [1 + km (t)] A sin (wct)

Q - km (t) A sin (wct)

R - A sin [ w'c + k]'-¥  m (t) dt

S - A sin [wct  + k '?-¥ m (t) dt]

Solution :

Group II

W - Phase modulation

X - Frequency modulation

Y - Amplitude modulation

Z - DSB-SC modulation

P        Q        R        S
(a)  Z        Y        X        W

(b)  W       X        Y        Z

(c)   X       W       Z        Y

(d)   Y       Z       W        X  (ans)

Solution : The correct match is given below

[1 + km (t)] A sin (wct)   Amplitude modulation km (t) A sin (wct)   DB-SC modulation  A sin [w'c + k]'-¥  m (t) dt   Phase modulation  A sin [wct + k '?-¥ m (t) dt]   Frequency modulation