Communication miscellaneous Easy Questions and Answers | Page - 4

Communication miscellaneous

The most suitable method for detecting a modulated signal (2·5 + 5 cos ω_mt) cos ω_mt is—

1. envelope circuit
1. synchronous detector
1. ratio detector
1. Both (A) and (B)

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Given modulated signal
s(t) = (2·5 + 5 cos ω_mt) cos ω_c t
or
s(t) = 2·5 (1 + 2 cos ω_mt) cos ω_ct
Since here modulation index, μ = 2 > 1. Hence signal can’t be detected by envelop detector and synchronous detector.
Hence alternative (B) is the correct choice.

Correct Option: B

Given modulated signal
s(t) = (2·5 + 5 cos ω_mt) cos ω_c t
or
s(t) = 2·5 (1 + 2 cos ω_mt) cos ω_ct
Since here modulation index, μ = 2 > 1. Hence signal can’t be detected by envelop detector and synchronous detector.
Hence alternative (B) is the correct choice.

The positive RF peaks of an AM voltage rise to a maximum value of 12 V and drop to a minimum value of 4V. The modulation index assuming single tone modulation is—

1. 3
1. 1/3
1. 1/4
1. 1/2

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Given,
E_max = 12V, E_min = 4V
μ =
E_max − E_min
E_max + E_min

μ = 12 − 4 = 8 = 1
12 + 4 16 2

Correct Option: D

Given,
E_max = 12V, E_min = 4V
μ =
E_max − E_min
E_max + E_min

μ = 12 − 4 = 8 = 1
12 + 4 16 2

Amplitude modulation is used for broadcasting because—

1. it is more noise immune than other modulation systems
1. compared with other systems it requires less transmitting power
1. its use avoids receiver complexity
1. no other modulation system can provide the necessary bandwidth for high fidelity

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NA

Correct Option: C

NA

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

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