Communication miscellaneous Easy Questions and Answers | Page - 1

Communication miscellaneous

The d.c. power to a modulated class C amplifier is 500 W. The power that modulator must supply for 100% modulation applied at the output electrode—

1. 500 W
1. 250 W
1. 1000 W
1. 2000 W

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P_modulated = μ² P_dc
2

Given μ = 1 P_dc = 500 W

P_modulated or P_dc = 1² .500 = 250 W.
2

Correct Option: B

P_modulated = μ² P_dc
2

Given μ = 1 P_dc = 500 W

P_modulated or P_dc = 1² .500 = 250 W.
2

The allocation of power in each sidebands if percentage of modulation is changed to 80%—

1. 320W
1. 160W
1. 640W
1. 180W

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Here given μ = 0·8 and P^c = 1000 W
so,
P_t = P_c 1 + μ²
2

P_t = P_c 1 + 0.8² = 1320 W
2

Now,
P_t = P_c + P_USB + P_LSB
1320 = 1000 + 2P_USB (∴ P_USB = P_LSB)
P_USB = 320 = 160 W
2

Correct Option: B

Here given μ = 0·8 and P^c = 1000 W
so,
P_t = P_c 1 + μ²
2

P_t = P_c 1 + 0.8² = 1320 W
2

Now,
P_t = P_c + P_USB + P_LSB
1320 = 1000 + 2P_USB (∴ P_USB = P_LSB)
P_USB = 320 = 160 W
2

What is the percentage modulation?

1. 80%
1. 40%
1. 100%
1. 89.4%

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We know that,
P_t = P_c 1 + μ²
2

where, P_t= P_c + P_USB = 1000 + 200 + 200 = 1400 W P_c = 1000 W
Now,
1400 = 1000 1 + μ²
2

or   μ² = 0.4
2

or   μ² = 0.8
or   μ = 0.894
or   μ = 89.4%

Correct Option: D

We know that,
P_t = P_c 1 + μ²
2

where, P_t= P_c + P_USB = 1000 + 200 + 200 = 1400 W P_c = 1000 W
Now,
1400 = 1000 1 + μ²
2

or   μ² = 0.4
2

or   μ² = 0.8
or   μ = 0.894
or   μ = 89.4%

The percentage increase in signal power of a carrier amplitude modulated 100% by a square wave—

1. 50%
1. 100%
1. 200%
1. None of these

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Let the given square wave looks like as shown below.
Since the average value of square wave i.e.m(t) = 1 and, on 100% modulation the amplitude of the carrier will be twice for half the time and zero for the other half time. Now, P_t = P_c [1+ m²(t)] = P_c [1 + 1] = 2P_c
Percentage increase in power
= 2 P_c − P_c ×100 = 100
P_c

Correct Option: B

Let the given square wave looks like as shown below.
Since the average value of square wave i.e.m(t) = 1 and, on 100% modulation the amplitude of the carrier will be twice for half the time and zero for the other half time. Now, P_t = P_c [1+ m²(t)] = P_c [1 + 1] = 2P_c
Percentage increase in power
= 2 P_c − P_c ×100 = 100
P_c

Given f(t) = cos ω₁t + sin ω₂t then f_h(t) =?

1. sin ω₁t – cos ω₂t
1. sin ω₁t + cos ω₂t
1. sin ω₂t – cos ω₁t
1. None of these

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We know that Hilbert transform [f_h (t)] can be obtained by replacing the signal (– 90°) i.e.
f_h (t) = cos (ω₁t – 90°) + sin (ω₂t – 90°)
or f_h (t) = sin ω₁t – cos ω₂t
Hence alternative (A) is the correct choice.

Correct Option: A

We know that Hilbert transform [f_h (t)] can be obtained by replacing the signal (– 90°) i.e.
f_h (t) = cos (ω₁t – 90°) + sin (ω₂t – 90°)
or f_h (t) = sin ω₁t – cos ω₂t
Hence alternative (A) is the correct choice.