Communication miscellaneous Easy Questions and Answers | Page - 33

Communication miscellaneous

If the carrier of a 100 per cent modulated AM wave is suppressed, the percentage power saving will be—

1. 50
1. 150
1. 100
1. 66.66

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Percentage power saving
= Power saved × 100
Total power

= P_c × 100
P_c + sideband power

= P_c = 100 3 × 100 = 66.67
P_c + P_c/2
2

∵ sideband power = μ² A_c² = P_c
4 2

Correct Option: D

Percentage power saving
= Power saved × 100
Total power

= P_c × 100
P_c + sideband power

= P_c = 100 3 × 100 = 66.67
P_c + P_c/2
2

∵ sideband power = μ² A_c² = P_c
4 2

The message signal contain three frequencies 5 kHz, 10 kHz, 20 kHz respectively. The bandwidth of the AM signal is—

1. 40kHz
1. 10 kHz
1. 20 kHz
1. 30 kHz

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BW = 2x maximum frequency of the modulating signal.
= 2 × 20 = 40 kHz

Correct Option: A

BW = 2x maximum frequency of the modulating signal.
= 2 × 20 = 40 kHz

A superheterodyne receiver is to operate in the frequency range of 550 kHz-1650 kHz with the intermediate frequency of 450 kHz. The receiver is tuned to 700. The capacitance ratio R = C_max/C_min of the local oscillator would be—

1. 4.41
1. 2.1
1. 3
1. 9

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Given that frequency range (550-1650) kHz, where 1650 kHz is maximum and 550 is minimum value
f_IF = 450 kHz
When the local oscillator frequency is kept higher, the maximum to minimum capacitance ratio required is
R = C_max
2100 ² = 4.41
C_min 1000

Correct Option: A

Given that frequency range (550-1650) kHz, where 1650 kHz is maximum and 550 is minimum value
f_IF = 450 kHz
When the local oscillator frequency is kept higher, the maximum to minimum capacitance ratio required is
R = C_max
2100 ² = 4.41
C_min 1000

An AM modulation has output x_c(t) = A cos 400πt + B cos 380πt + B cos 400πt
The carrier power is 100 W and the efficiency is 40%. The value of A and B are—

1. 14.14, 16.33
1. 50, 10
1. 22.36,13.46
1. None of these

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Given modulator output
x_c(t) = A cos 400 πt + B cos 380 πt + B cos 420 πt …(i)
Carrier power = 100 W
Efficiency = 40%
Equation (i) can be represented in standard equation as

x_ct = A 1 + B
cos 20 πt cos 400 πt
A

...(ii)
Given that

P_c = 100 = A_c²
2
(here A_c = A)

μ = B
A

A² = 200
or
A = √200 = 14.14

Efficiency = μ²
2 + μ²

0.4 = (B/A)²
2 +(B/A)²

or

0.4 = B²
2A² + B²

or
0.8A² + 0.4B² = B²
or

B² = 0.8A²
= 4A² = 4 × 200
= 800
0.6 3 3 3

or

B = √ 800 =
√266.67 = 16.33
3

Correct Option: A

Given modulator output
x_c(t) = A cos 400 πt + B cos 380 πt + B cos 420 πt …(i)
Carrier power = 100 W
Efficiency = 40%
Equation (i) can be represented in standard equation as

x_ct = A 1 + B
cos 20 πt cos 400 πt
A

...(ii)
Given that

P_c = 100 = A_c²
2
(here A_c = A)

μ = B
A

A² = 200
or
A = √200 = 14.14

Efficiency = μ²
2 + μ²

0.4 = (B/A)²
2 +(B/A)²

or

0.4 = B²
2A² + B²

or
0.8A² + 0.4B² = B²
or

B² = 0.8A²
= 4A² = 4 × 200
= 800
0.6 3 3 3

or

B = √ 800 =
√266.67 = 16.33
3

An AM modulator has output x_c (t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440 πt The AM modulation efficiency is—

1. 0.01
1. 0.02
1. 0.03
1. 0.04

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The given modulator output
x_c(t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440πt x_c(t) may be written as
x_c(t) = 40 1 + 1 cos40πt cos400πt
5
...(1)
On comparing equation (1) with standard equation
x_c(t )= A_c (1 + μ cos ω_mt) cos ω_ct
μ = 1 ,A_c = 40
5

Efficiency = μ²
= (1/5)² = 1/25
= 1
2 + μ²
2 + (1/5)² 50 + 1/25 51

= 0.02
Hence alternative (B) is correct choice.

Correct Option: B

The given modulator output
x_c(t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440πt x_c(t) may be written as
x_c(t) = 40 1 + 1 cos40πt cos400πt
5
...(1)
On comparing equation (1) with standard equation
x_c(t )= A_c (1 + μ cos ω_mt) cos ω_ct
μ = 1 ,A_c = 40
5

Efficiency = μ²
= (1/5)² = 1/25
= 1
2 + μ²
2 + (1/5)² 50 + 1/25 51

= 0.02
Hence alternative (B) is correct choice.