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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  Test: Communication Systems - 4 - Electronics and Communication Engineering (ECE) MCQ

Test: Communication Systems - 4 - Electronics and Communication Engineering (ECE) MCQ


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25 Questions MCQ Test - Test: Communication Systems - 4

Test: Communication Systems - 4 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Communication Systems - 4 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Communication Systems - 4 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Communication Systems - 4 below.
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Test: Communication Systems - 4 - Question 1
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24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM is

Detailed Solution for Test: Communication Systems - 4 - Question 1

Bandwidth ≈ 24 (ln2 128) 10 kHz = 1.68 MHz.

Test: Communication Systems - 4 - Question 2
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Resonant circuits are used in

Detailed Solution for Test: Communication Systems - 4 - Question 2

RF amplifiers use resonant loads (L and C in parallel).

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Test: Communication Systems - 4 - Question 3
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The bandwidth of DSB suppressed carrier modulation system when the modulating frequency varies between 500 Hz and 5 kHz is

Detailed Solution for Test: Communication Systems - 4 - Question 3

fm = 5 kHz - 500 Hz 4.5 kHz

Bandwidth = 2 fm 2 x 4.5 kHz = 9 kHz.

Test: Communication Systems - 4 - Question 4
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Linear diode detector utilizes
Detailed Solution for Test: Communication Systems - 4 - Question 4

Detection (also called demodulation) is due to rectification property of diode.

Test: Communication Systems - 4 - Question 5
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In the equivalent circuit of a moving coil cone type loudspeaker, the mass of the moving system is represented by
Detailed Solution for Test: Communication Systems - 4 - Question 5

Mass has inertia and is analogous to inductance.

Test: Communication Systems - 4 - Question 6
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Assertion (A): In India we use CCIR-B standard

Reason (R): Video signals are frequency modulated.
Detailed Solution for Test: Communication Systems - 4 - Question 6

Video signals are amplitude modulated.

Test: Communication Systems - 4 - Question 7
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CATV stands for
Detailed Solution for Test: Communication Systems - 4 - Question 7

In USA the first cable service was called community antenna television.

Test: Communication Systems - 4 - Question 8
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A geostationary satellite
Detailed Solution for Test: Communication Systems - 4 - Question 8

It orbits the earth in a 24 hour period. Thus it appears to be stationed over one spot of the globe.

Test: Communication Systems - 4 - Question 9
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Household tape recorders use
Detailed Solution for Test: Communication Systems - 4 - Question 9

For reasons of economy only one motor is used.

Test: Communication Systems - 4 - Question 10
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A 400 W carrier is modulated to a depth of 75 percent. The total power in modulated wave will be
Detailed Solution for Test: Communication Systems - 4 - Question 10

Sideband frequencies will be fc fm.

Test: Communication Systems - 4 - Question 11
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A line has L = 0.8 µH/m and C = 32 pF/m. Z0 =
Detailed Solution for Test: Communication Systems - 4 - Question 11

.

Test: Communication Systems - 4 - Question 12
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A line has standing waves with a voltage maximum of 35 mV and minimum of 17.5 mV VSWR is
Detailed Solution for Test: Communication Systems - 4 - Question 12

.

Test: Communication Systems - 4 - Question 13
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For a bit rate of 8 kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are
Detailed Solution for Test: Communication Systems - 4 - Question 13

where i is integer, and nc is fixed integer.

fi = Rb(nc + i) (3 + 1) x 8 32 kHz

(4 + 1)0.8 40 kHz.

Test: Communication Systems - 4 - Question 14
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If luminance signal Y = 0.3 R + 0.59 G + 0.11 B, the hue of R - Y colour is
Detailed Solution for Test: Communication Systems - 4 - Question 14

R - Y = R - 0.3R - 0.59C - 0.11B = 0.7R - 0.59G - 0.11B.

Test: Communication Systems - 4 - Question 15
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Trap circuits used in video IF amplifier of colour TV are
Detailed Solution for Test: Communication Systems - 4 - Question 15

High Q LC resonant circuit provides high selectivity.

Test: Communication Systems - 4 - Question 16
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A 1000 kHz carrier is simultaneously modulated with 300 Hz, 800 Hz and 2 kHz audio sine waves. Which of the following frequency is least likely to be present in the output?
Detailed Solution for Test: Communication Systems - 4 - Question 16

Frequency present in the sidebands is equal to = fc fm, fc 2fm, fc 3fm .

Test: Communication Systems - 4 - Question 17
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A 300 Hz sound wave of intensity 20 dB is transmitted by a 1500 kHz radio wave by frequency modulation. It is seen that instantaneous frequency varies between 1500.2 kHz and 1499.8 kHz, 300 times a second. If the intensity of sound wave is 100 dB.
Detailed Solution for Test: Communication Systems - 4 - Question 17

Since, the intensity of audio wave has been increased five times, the instantaneous frequency variation is also increased five times.

Test: Communication Systems - 4 - Question 18
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An analog signal is band limited to B Hz sampled at Nyquist rate and samples are quantized into 4 levels each with probability . The information rate is
Detailed Solution for Test: Communication Systems - 4 - Question 18

.

Signal should be sampled at frequency 2B Hz.

Hence, information rate = 2 (2B) = 4B bits/s.

Test: Communication Systems - 4 - Question 19
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If the average power of radar transmitters is 2 kW and the peak power of the transmitter is 1000 kW, what will be the duty cycle?
Detailed Solution for Test: Communication Systems - 4 - Question 19

.

Test: Communication Systems - 4 - Question 20
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The HV supply to a TV receiver is 25 kV and the total beam current for the three electron guns is 2000 μA. The power handled by IHVT is
Detailed Solution for Test: Communication Systems - 4 - Question 20

25 x 103 x 2000 x 10-6 = 50 W.

Test: Communication Systems - 4 - Question 21
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Some advantages of optical fibre cables are

  1. small diameter
  2. Immunity to cross talk and EM interference
  3. Laser and LED modulation methods lend themselves ideally to digital operation
Which of the above are correct?
Detailed Solution for Test: Communication Systems - 4 - Question 21

Fibre optic cables have all these advantages.

Test: Communication Systems - 4 - Question 22
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The contrast control in colour TV receiver is done by
Detailed Solution for Test: Communication Systems - 4 - Question 22

Luminance amplifier changes the luminance levels and provides contrast control.

Test: Communication Systems - 4 - Question 23
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The received signal frequency at any time of a super-heterodyne receiver having IF = 456 kHz is 1 MHz. The corresponding image signal is
Detailed Solution for Test: Communication Systems - 4 - Question 23

Image signal frequency = Received signal frequency - 2 x Intermediate frequency.

= 1 MHz - 2 x 456kHz

1 MHz - 912 kHz 88 kHz

which lies in medium bond.

Test: Communication Systems - 4 - Question 24
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The requirements of carrier frequency are
Detailed Solution for Test: Communication Systems - 4 - Question 24

Both should have a precise value and low drift.

Test: Communication Systems - 4 - Question 25
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If N bits are lumped together to get an N-bit symbol, the possible number of symbols is
Detailed Solution for Test: Communication Systems - 4 - Question 25

Each bit can have two levels. Hence 2N.

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