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Test: Analog Communication - 2

20 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Analog Communication - 2

Description

Attempt Test: Analog Communication - 2 | 20 questions in 60 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions

*Answer can only contain numeric values

QUESTION: 1

An AM signal is represented in the domain as S (-c) = 4 cos 1800T + 10 cos 20002 + 4cos2200T then the bandwidth of the AM signal is _____ Hz

Solution:

QUESTION: 2

A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.

A.
100
B.
50
C.
70
D.
90

Solution:

Carrier frequency f_c = 100MHz
Modulating frequency f_m = 10 KHz
Frequency deviation Δf = 500 KHz
Modulation index of FM signal is given by
m_f = Δf/f_m
= 500 * 10³/ 10 * 10³
= 50

QUESTION: 3

Consider a system shown in the figure-1. Let X(f) and Y(f) represent the Fourier transforms of x(t) and y(t) respectively. The ideal HPF has cutoff frequency of 10 KHZ. X(f) is shown in the figure-2. The positive frequency where Y(f) has spectral peaks are

A.
1 KHz and 20 KHz
B.
2 KHz and 24 KHz
C.
1 KHz and 14 KHz
D.
2 KHz and 14 KHz

Solution:

2 KHz S. 24 KHz

QUESTION: 4

Three message signals each band limited to 10 kHz, 10 kHz, and 5 KHz are transmitted through a channel using frequency division multiplexing. The modulation techniques used are any, D513-SC, SSB respectively. The guard band is 2 KHz. Band width of the channel is

A.
49 kHz
B.
45 kHz
C.
53 kHz
D.
57 kHz

Solution:

BW after AM modulation = 2 x 10 = 20 KHz

BW after DSS-SC modulation = 2 x10 = 20 KHz BW after SSB modulation = 5 KHz

Guard band width = 2 x 2k=4 k

∴ Total BW after frequency multiplexing = 49 KHz

QUESTION: 5

A signal '4 t is band limited to the frequency range 0 to f_n. The signal frequency is translated by multiplying it by the signal V, t —cos2πf_ct . The value of f_c such that the bandwidth of the translated signal is 1% of the carrier frequency f_c will be

A.
100 f_rn
B.
200 fm
C.
2 fm
D.
50 f_m

Solution:

BW after multiplexing =

*Answer can only contain numeric values

QUESTION: 6

The peak amplitude of an AM signal for tone modulation with carrier having maximum amplitude as 5 V, varies from 2 to 8 V then the sideband power is _________________________ (W)

Solution:

QUESTION: 7

Consider a frequency multiplier shown in the figure below.ss

A NBFM signalX Mal t with f, = 200KHz and f, ranges 50 Hz to 150 KHz is given as an input to frequency multiplier. The Maximum Frequency deviation of the NBFM is 25 Hz. The maximum frequency deviation at the output is 75 KHz.

Q. The required frequency multiplication n is

A.
1000
B.
2000
C.
3000
D.
4000

Solution:

QUESTION: 8

Consider a frequency multiplier shown in the figure below.ss

Q. The carrier frequency at the output is

A.
100 MHz
B.
150 MHz
C.
600 iviHz
D.
225 MHz

Solution:

QUESTION: 9

SSB modulation may be viewed as a hybrid form of amplitude modulation and frequency modulation. Assuming a message signal m(t) = sin t and a carrier frequency 'f_c', the instantaneous frequency for the lower side band is

Solution:

For SSB modulation, the modulated wave is

QUESTION: 10

Given

In a certain experimental conducted with f_m = 1 kHz and increasing A_m(starting from 0 volts), it is found that the carrier component of the FM signal is reduced to zero for the first time when A_m = 2 volts.

The frequency sensitivity of the modulator is

A.
2.71 kHz/ V
B.
1.95 kHz/ V
C.
1.22 kHz / V
D.
0.71 kHz/ V

Solution:

When A_m = 2V, carrier component = 0

ie. J₀ (03) = 0 for the 1^st time

From the curve, corresponding (3=2.44

*Answer can only contain numeric values

QUESTION: 11

If E_x and E_y represent the energies of m(t) and y(t) respectively, then _______________

Solution:

M_h (ω) = -jM (6)) sgn(ω), (The given system is a Hilbert transformer) The π of a Hilbert transformer is

H (ω) = -jsgn (ω)

QUESTION: 12

Determine the spectrum of y(t) in the system shown below.

A.
B.
C.
Both
D.
None

Solution:

Output of AM Modulator

QUESTION: 13

Consider the scheme shown

If x (t) = m (t) cos (co_ct) then y (t) is

A.
m(t)
B.
C.
D.

Solution:

QUESTION: 14

The above schematic generates FM signal as a wideband frequency modulator

using the indirect method. The output FM signal is characterized by f_c = 100 MHz, Δf = 75kHz (min). If the phase modulator has a modulation index of β= 0.3 radians, then n₁, n₂ are respectively

A.
25, 50
B.
75, 50
C.
50, 25
D.
50, 75

Solution:

The lowest modulation frequency i.e.,

Produces a frequency deviation

The highest modulation frequency ie.

Produces a frequency deviation of

Let us take the case of 100Hz as the worst case

Solving (1) & (2) we get,

n₁=75 , n₂ =50

QUESTION: 15

What is the standard IF frequency for AM receivers?

A.
100 KHz
B.
456 KHz
C.
1.07 MHz
D.
10.7 MHz

Solution:

In AM receivers, the standard intermediate frequency is approximately 455 KHz.

QUESTION: 16

Consider a message signal m(t) = 2sin(10007t)

Phase sensitivity constant Kp=5rad/V

Frequency sensitivity constant K_f=75KHz/V

Q. If amplitude of message signal is doubled, then bandwidth for FM and PM will be

A.
601KHz, 21KHz
B.
300.5KHz, 21KHz
C.
601KHz, 10.5KHz
D.
300.5KHz, 10.5KHz

Solution:

Δf = 75x 4 = 300KHz ,

B_FM= 2(300 + 0.5) = 601KHz, Δ f= 2 x K_p x 2 x 500 = 10KHz,B_pm= 2(10 + 0.5) = 21KHz

QUESTION: 17

A 20 KW and 1 MHz carrier is frequency modulated by a sinusoidal signal so that the maximum frequency deviation is 50 KHz and the message signal is band limited to 500 KHz. The generated FM signal is amplitude modulated using carrier of 1 GHz.

Q. The bandwidth of the generated signal is