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Test: Communication Systems - 5 - Electronics and Communication Engineering (ECE) MCQ


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25 Questions MCQ Test - Test: Communication Systems - 5

Test: Communication Systems - 5 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Communication Systems - 5 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Communication Systems - 5 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 - 5 below.
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Test: Communication Systems - 5 - Question 1

A total of 16 equally probable events exist. If it is required to select one out of them, the number of required bits is

Detailed Solution for Test: Communication Systems - 5 - Question 1

24 = 16.

Test: Communication Systems - 5 - Question 2

When modulation index of an AM wave is increased from 0.5 to 1, the transmitted power

Detailed Solution for Test: Communication Systems - 5 - Question 2

.

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Test: Communication Systems - 5 - Question 3

Quantizing error occurs in

Detailed Solution for Test: Communication Systems - 5 - Question 3

An analog number cannot be converted into an exact digital number.

This is called quantizing error.

Test: Communication Systems - 5 - Question 4

In audio cassettes the width of each recording track is

Detailed Solution for Test: Communication Systems - 5 - Question 4

Each track is 2.5 mm wide.

Test: Communication Systems - 5 - Question 5

The most commonly used transistor amplifier circuit is

Detailed Solution for Test: Communication Systems - 5 - Question 5

Common emitter connection has high power gain, and good current and voltage gains.

Test: Communication Systems - 5 - Question 6

When energy is fed into open wire transmission line the energy may get dissipated due to

Detailed Solution for Test: Communication Systems - 5 - Question 6

All the three cause dissipation of energy as heat.

Test: Communication Systems - 5 - Question 7

Fourier series expansion of a real periodic signal with fundamental frequency is given by It is given that c3 = 3 + j5. then c-3 is

Detailed Solution for Test: Communication Systems - 5 - Question 7

For real signal

ck = c*-k

3 - i5.

Test: Communication Systems - 5 - Question 8

Loudness control gives

Detailed Solution for Test: Communication Systems - 5 - Question 8

For depth in sound the low frequency notes, (i.e., bass) need more amplification.

Test: Communication Systems - 5 - Question 9

If transmitted power is 75 kW, the field strength at 30 km distance will be about

Detailed Solution for Test: Communication Systems - 5 - Question 9

Using the derivation,

.

Test: Communication Systems - 5 - Question 10

A mobile telephone has a range of about

Detailed Solution for Test: Communication Systems - 5 - Question 10

Mobile telephone has about 50 km range.

Test: Communication Systems - 5 - Question 11

In the television system in India, each frame is scanned

Detailed Solution for Test: Communication Systems - 5 - Question 11

Each frame is divided into odd and even fields. Each field is scanned 25 times so that each frame is scanned 50 times.

Test: Communication Systems - 5 - Question 12

Each kilometer of travel in electromagnetic radiation means a time delay of

Detailed Solution for Test: Communication Systems - 5 - Question 12

Since velocity of em waves is 300 m/ms, each kilometer means time delay of 3.3 μs.

Test: Communication Systems - 5 - Question 13

A video monitor is exactly similar to TV receiver.

Detailed Solution for Test: Communication Systems - 5 - Question 13

Trinitron is a colour picture tube.

Test: Communication Systems - 5 - Question 14

A RADAR can be used to

Detailed Solution for Test: Communication Systems - 5 - Question 14

Radar is used for all the three purposes.

Test: Communication Systems - 5 - Question 15

In single tone AM modulation, the transmission efficiency for m = 1 is

Detailed Solution for Test: Communication Systems - 5 - Question 15

Since signal power = 0.5 PC and total power is 1.5 PC efficiency is 33.3%.

Test: Communication Systems - 5 - Question 16

AM amplifier having noise figure of 20 dB and available power gain of 15 dB is followed by a mixer circuit having noise figure of 9 dB. The overall noise figure as referred to input in dB is

Detailed Solution for Test: Communication Systems - 5 - Question 16

.

Test: Communication Systems - 5 - Question 17

If the oscillator output is modulated by audio frequencies upto 10 kHz, the frequency range occupied by the side bands will be

Detailed Solution for Test: Communication Systems - 5 - Question 17

Sideband frequencies will be fc fm.

Test: Communication Systems - 5 - Question 18

Which one of the following is non-resonant antenna?

Detailed Solution for Test: Communication Systems - 5 - Question 18

Rhombic antenna is a non-resonant antenna capable of operating over 3 to 30 MHz range.

Test: Communication Systems - 5 - Question 19

The efficiency of using a binary system for selection of 1 out of 13 equally probable events is

Detailed Solution for Test: Communication Systems - 5 - Question 19

log2 13 = 3.7 therefore, number of bits = 4.

.

Test: Communication Systems - 5 - Question 20

The most suitable method for detecting a modulated signal g(t) = (3 to 6 cos 2pfmt) cos 2pfct is

Detailed Solution for Test: Communication Systems - 5 - Question 20

Because it is an AM envelope.

Test: Communication Systems - 5 - Question 21

The time required for horizontal blanking is 16% of that for each horizontal line. If horizontal time is 63.5 μs, the horizontal blanking time for each line is about

Detailed Solution for Test: Communication Systems - 5 - Question 21

.

Test: Communication Systems - 5 - Question 22

A message signal with bandwidth 10 kHz is lower sideband SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then passed though a narrow band frequency modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be

Detailed Solution for Test: Communication Systems - 5 - Question 22

Bandwidth = 2Δf + 2B

But for NBFM, B < 1

BW ≈ 2fm 2fc1 2 x 106 Hz.

Test: Communication Systems - 5 - Question 23

When the number of quantising levels is 16 in PCM, the number of pulses in a code group will be

Detailed Solution for Test: Communication Systems - 5 - Question 23

2n = L n = 4.

Test: Communication Systems - 5 - Question 24

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor.

Detailed Solution for Test: Communication Systems - 5 - Question 24

Given n1 = 6, n2 = 8,

then (SNR)Q = 22(n2 - n1) 16.

Test: Communication Systems - 5 - Question 25

For 10 bit PCM system, the signal to quantization noise ratio is 62 dB. If the number of bits is increased by 2, then the signal to quantization noise ratio will

Detailed Solution for Test: Communication Systems - 5 - Question 25

= (SNR)Q + 6.02 x 2 (SNE)Q + 12 dB.

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