HindustanAeronauticsLtd. Electronics-Communication-Engineering-ECE Contributed by Dhisha updated on Aug 2020
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HAL Electronics and Communication Engineering Communication Systems practice questions with answers,HAL technical questions for practice HAL model questions with answers

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 tthe 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

Solution : 2.1

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

6. Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth?

(a) VSB

(b) DSB-SC

(c) SSB  (ans)

(d) AM

Solution :     VSB  → fm + fc

DBS - SC  → 2 fm

SSB  → fm

AM → 2 fm

Thus SSB has minimum bandwidth and it required minimum power i.e. 17% as compared to AM.

7.  A device with input x(t) and output y(t) is characteristic by : y (t) = x2(t). An FM signal with frequency deviation of 90 KHz and modulation signal bandwidth of 5 KHz is applied to this device. The bandwidth of the output signal is

(a) 370 KHz  (ans)

(b) 190 KHz

(c) 380 KHz

(d) 95 KHz

Solution : In present case

?f = 90; fm = 5

β = [?f / fm] = [90/5] = 18

FM equation

A cos [wct  + β = sin wmt]

= A cos [wct  + 18 sin wmt]

y(t) = x2 (t) = A2 cos2 [wct  + 18 Sin wmt]

Note : Cos2 q = [1 + Cos2q] / 2

If there is change in frequency the modulation index also changes in same ratio

y(t) = A2 [(1/2) + (1/2) Cos {2wct  + 36Sin wmt}]

y(t) = [(A2/2) + (A2/2) Cos {2wct  + 36Sin wmt}]

After the device,

β(new) = 36 = [?f(new) / fm]

?f(new) = 36 x 5 = 180

By carson's rule

Bandwidth = 2(?f +  fm)

= 2 (180 + 5)

Bandwidth = 370 kHz

8. 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?

Image - Qno - 28 (1)

(a) Image - Qno - 2.8 (a)

(b) Image - Qno - 2.8 (b)

(c) Image - Qno - 2.8 (c)  (ans)

(d) Image - Qno - 28 (d)

Solution : x(t) = u(t) - 2 u(t - 1) + u(t - 2) for x(t)

Image - Answer - 28 (I)

The impulse of a matched filter is delayed version of the input.

Now delay not given, so assumed to be zero.

∴ y (t) = x (t) * x(t) [n(t) = x(t)] where * represent convolution.

y(t) = r(t) - 4r (t - 1) + 6r (t - 2) - 4r (t-3) + r (t - 4)

Hence u (t) * u (t) = r(t) Ramp function.

Image - Answer - 28 (II)

9. Noise with uniform power spectral density of N0 W/Hz is passed through a filter H (w) = 2 exp (-jwtd) following by an ideal pass filter of bandwidth B Hz. The output noise power in Watts is

(a) 2 N0B   (ans)

(b) 4 N0B

(c) 8 N0B

(d) 16 N0B

Solution : Image - Answer - 29

PSD of input Gx(w) = N0 W/Hz. H1(w) = 2 e-jwtd

Gy1(w) = |H1(w) 2| Gx(w) = 4 N0 W/Hz

Gy2(w)| = H2(w)| Gy1 x (w)

= 4 4 N0 W/Hz, - B≤w≤B

B
Noise power = ∫ Gy2 (w) dW = 2 x 4 N0 B
-B

10. A carrier is phase modulated (PM) with frequency deviation of 10 KHz by a single tone frequency of 1 KHz. If the single tone frequency is increased to 2 KHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is

(a) 21 kHz

(b) 22 kHz

(c) 42 kHz

(d) 44 kHz   (ans)

Solution : ?f = 10 KHz     fm(new) = 2 KHz

fm = 1 KHz

By carson's Rule

BW = 2 (?f + fm) = 2 (10 + 1) = 22 KHz

? f(new) = 2 x 10 = 20

BW(new) = 2 (20 + 2) = 44 kHz