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Test: Communication Systems - 1 - Electronics and Communication Engineering (ECE) MCQ


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25 Questions MCQ Test - Test: Communication Systems - 1

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

A cordless telephone using separate frequencies for transmission in base and portable units is known as

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

Separate frequencies for transmission from base and portable units allows two way transmission and is called duplex arrangement.

Test: Communication Systems - 1 - Question 2

For attenuation of high frequencies we should use

Detailed Solution for Test: Communication Systems - 1 - Question 2

Since Xc for high frequencies is low, the high frequencies are shunted to ground and are not transmitted.

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Test: Communication Systems - 1 - Question 3

A modem is classified as low speed if data rate handled is

Detailed Solution for Test: Communication Systems - 1 - Question 3

When data rate in bits per second is upto 600, modem is low speed.

Test: Communication Systems - 1 - Question 4

VSB modulation is preferred in TV because

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

VSB (vestigial side band) transmission transmits one side band fully and the other side band partially thus, reducing the bandwidth requirement.

Test: Communication Systems - 1 - Question 5

A woofer should be fed from the input through a

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

Woofer is a low frequency loud speaker covering the range 16 Hz to 500 Hz.

Test: Communication Systems - 1 - Question 6

The colour subcarrier and sidebands produced by its modulation with the chrominance signals are accommodated in the standard channel width by the process of

Detailed Solution for Test: Communication Systems - 1 - Question 6

In frequency interleaving the space between bundles is used for colour signal.

Test: Communication Systems - 1 - Question 7

In FM signal with a modulation index mf is passed through a frequency tripler. The wave in the output of the tripler will have a modulation index of

Detailed Solution for Test: Communication Systems - 1 - Question 7

Frequency multiplier increase the deviation,

.

Test: Communication Systems - 1 - Question 8

In colour TV receiver, varactor diode is used for

Detailed Solution for Test: Communication Systems - 1 - Question 8

In varactor diode the applied reverse bias controls the width and therefore capacitance of depletion layer.

This capacitance is used for tuning.

Test: Communication Systems - 1 - Question 9

The number of noise sources in a BJT are

Detailed Solution for Test: Communication Systems - 1 - Question 9

Shot noise, partition noise and thermal noise.

Test: Communication Systems - 1 - Question 10

Energy content of atmospheric noise

Detailed Solution for Test: Communication Systems - 1 - Question 10

Atmospheric noise decreases as frequency is increased.

Test: Communication Systems - 1 - Question 11

A 400 W carrier is amplitude modulated with m = 0.75. The total power in AM is

Detailed Solution for Test: Communication Systems - 1 - Question 11

.

Test: Communication Systems - 1 - Question 12

c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p(1008 x 103t)] in the FM signal (in terms of the Bessel Coefficients) is

Detailed Solution for Test: Communication Systems - 1 - Question 12

As, we known FM, s(t)

Frequency deviation = 3 . (AM Bandwidth) = 6 fm

and β= 6 is given in (d) only.

ωc + n ωm = (1008 x 103)2p

2p x 106 + n. 4p x 103 = (1008 x 103).2p

n = 4

So, β = 6, n = 4 hence answer is 5j4(6).

Test: Communication Systems - 1 - Question 13

Non-coherently detection is not possible for

Test: Communication Systems - 1 - Question 14

Tracking of extra terrestrial objects requires

Detailed Solution for Test: Communication Systems - 1 - Question 14

All the three are required.

Test: Communication Systems - 1 - Question 15

Assertion (A): Free space does not interfere with normal radiation and propagation of radio waves

Reason (R): Free space has no magnetic or gravitational fields.

Detailed Solution for Test: Communication Systems - 1 - Question 15

Since free space does not have magnetic and other fields, waves can propagate without any interference.

Test: Communication Systems - 1 - Question 16

In radar systems PRF stands for

Detailed Solution for Test: Communication Systems - 1 - Question 16

PRF in radar systems means pulse repetition frequency.

Test: Communication Systems - 1 - Question 17

Which of the following cannot be the Fourier series expansion of a periodic signal?

Detailed Solution for Test: Communication Systems - 1 - Question 17

x(t) = cos t + 0.5

not satisfies the Dirichlet condition.

The integration of constant term is ∞.

Test: Communication Systems - 1 - Question 18

Which of the following is the indirect way of FM generation?

Detailed Solution for Test: Communication Systems - 1 - Question 18

It generates FM through phase modulation.

Test: Communication Systems - 1 - Question 19

A telephone exchange has 9000 subscribers. If the number of calls originating at peak time is 10, 000 in one hour, the calling rate is

Detailed Solution for Test: Communication Systems - 1 - Question 19

Calling rate is the number of calls per subscriber.

Test: Communication Systems - 1 - Question 20

If C is capacity of noisy channel, (bits/s), δf is bandwidth Hz and S/N is signal to noise ratio, then

Detailed Solution for Test: Communication Systems - 1 - Question 20

This is Shannon-Hartley theorem.

Test: Communication Systems - 1 - Question 21

As the frequency increases, the absorption of ground wave by earth's surface

Detailed Solution for Test: Communication Systems - 1 - Question 21

Because of this reason ground waves can be used for frequencies upto about 1600 kHz only.

Test: Communication Systems - 1 - Question 22

The velocity of sound waves in air

Detailed Solution for Test: Communication Systems - 1 - Question 22

vT.

Test: Communication Systems - 1 - Question 23

The range of a cordless telephone is about

Detailed Solution for Test: Communication Systems - 1 - Question 23

The range depends on quality of instrument and is about 100 m.

Test: Communication Systems - 1 - Question 24

An earth mat for a communication tower consists of

Detailed Solution for Test: Communication Systems - 1 - Question 24

The number of radials is large and depth is about 30 cm.

Test: Communication Systems - 1 - Question 25

Degaussing in a picture tube means

Detailed Solution for Test: Communication Systems - 1 - Question 25

Since gauss is a unit of magnetic flux, degaussing means removal of magnetism.

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