Communication miscellaneous Easy Questions and Answers | Page - 8

Communication miscellaneous

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

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

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

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

A carrier is simultaneously modulated by two sine waves with modulation indices of 0.3 and 0.4, the total modulation index—

1. is 1
1. cannot be calculated unless the phase relations are known
1. is 0.5
1. is 0.7

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Given μ₁ = 0·3, μ₂ = 0·4
μ_eff = √μ₁² + μ₂²
= √(0.3)² + (0.4)²
= √0.09 + 0.16
= √0.25
= 0.5

Correct Option: C

Given μ₁ = 0·3, μ₂ = 0·4
μ_eff = √μ₁² + μ₂²
= √(0.3)² + (0.4)²
= √0.09 + 0.16
= √0.25
= 0.5