Communication miscellaneous Easy Questions and Answers | Page - 48

Communication miscellaneous

Indicate which one of the following advantage of the phase cancellation method of obtaining SSB over the filter method is false?

1. Switching from one sideband to the other is simpler.
1. It is possible to generate SSB at any frequency.
1. SSB wilth lower audio frequency present can generated.
1. There are more balanced modulators, therefore, the carrier is suppressed better.

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The main advantage of phase cancellation or phase method is that it does not require any sharp cut-off filter. Here each modulator is carefully balanced in order to suppress the carrier, for this each modulator should have equal sensitivity to the baseband signal.
Hence alternative (D) is the most appropriate choice.

Correct Option: D

The main advantage of phase cancellation or phase method is that it does not require any sharp cut-off filter. Here each modulator is carefully balanced in order to suppress the carrier, for this each modulator should have equal sensitivity to the baseband signal.
Hence alternative (D) is the most appropriate choice.

When the modulation index of a wave is doubled, the antenna current is also doubled. The AM system being used is—

1. Single-sideband, full carrier.
1. Vestigial sideband.
1. Single-sideband, suppressed carrier
1. Double-sideband, full carrier

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NA

Correct Option: B

NA

Indicate the false statement regarding the advantages of SSB over double sideband, full-carrier AM—

1. More channel space is available.
1. Transmitter circuits must be stable, giving better reception.
1. The signal is more noise-resistant.
1. Much less power is required for the same signal strengths.

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NA

Correct Option: A

NA

The total transmitted power given by the AM signal
φ_AM(t) = 10 cos (2π × 10⁶t) + 5 cos(2π × 10⁶t)
cos (2π × 10³t) + 2 cos (2π × 10⁶t) + cos (4π × 10³t)

1. 57.25 W
1. 62.25 W
1. 56.25 W
1. None of these

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The given amplitude modulation equation is φ_AM (t) = 10 cos (2π × 10⁶t) + 5 cos (2π × 10⁶t) cos (2π × 10³t) + 2 cos (2π × 10⁶t) cos (4π × 10³t)
φ_AM (t) = 1+ 5 cos(2π × 10³t) + 2 cos(4π×10³t)
10 10

where,
μ₁ = 5 = 0.5
10

μ₂ = 2 = 0.2
10

A_c = 10
Now, total transmitted power
P_t = P_c 1 + μ_eff²
2

μ_eff² = μ²₁+μ²₂ = 0.5²+0.2²
or
μ_eff² = 0.25 + 0.4 = 0.29
Now
P_t = A² 1 + 0.29
2 2

= 10² ×1.145 = 57.25 W
2

Correct Option: A

The given amplitude modulation equation is φ_AM (t) = 10 cos (2π × 10⁶t) + 5 cos (2π × 10⁶t) cos (2π × 10³t) + 2 cos (2π × 10⁶t) cos (4π × 10³t)
φ_AM (t) = 1+ 5 cos(2π × 10³t) + 2 cos(4π×10³t)
10 10

where,
μ₁ = 5 = 0.5
10

μ₂ = 2 = 0.2
10

A_c = 10
Now, total transmitted power
P_t = P_c 1 + μ_eff²
2

μ_eff² = μ²₁+μ²₂ = 0.5²+0.2²
or
μ_eff² = 0.25 + 0.4 = 0.29
Now
P_t = A² 1 + 0.29
2 2

= 10² ×1.145 = 57.25 W
2

For AM modulated signal given by i = 10 + K₁ V_i + K₂ V_i²
where V_i = A_c cos ω _ct + m (t),
uses square law demodulator, find μ given that
A_c = 1V, A_m = 0·5 V, K₁ = 2 and K2 = 0·2

1. 0.5
1. 0.2
1. 1
1. 0.4

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We know that AM modulated wave represented by equation
x(t)=A_c (1 + μ cos ω_mt) cos ω_ct
s(t)=A_c cos ω_ct + μ·A_c cos ω_mt cos ω_ct …(A)
given that,
i _AM(t) = 10 + K₁V₁ + K₂V² _i
i_AM(t) = 10 + K₁A_c cos ω_ct + K₁A_m cos ω_mt + K₂A²_c cos²ω_ct + K²₂A²_m cos² ω_mt 2K₂A_cA_m cos ω_ct cos ω_mt …(B)
In order to calculate, modulation index μ.
Compare the coefficient of cos ωmt cos ωct in equation (A) and (B) we get
μA_c = 2K₂A_cA_m
or μ = 2K₂A_m = 2 × 0·2 × 0·5= 0·20 (˙.˙ K₂ = 0·2)
Hence alternative (B) is the correct choice.

Correct Option: B

We know that AM modulated wave represented by equation
x(t)=A_c (1 + μ cos ω_mt) cos ω_ct
s(t)=A_c cos ω_ct + μ·A_c cos ω_mt cos ω_ct …(A)
given that,
i _AM(t) = 10 + K₁V₁ + K₂V² _i
i_AM(t) = 10 + K₁A_c cos ω_ct + K₁A_m cos ω_mt + K₂A²_c cos²ω_ct + K²₂A²_m cos² ω_mt 2K₂A_cA_m cos ω_ct cos ω_mt …(B)
In order to calculate, modulation index μ.
Compare the coefficient of cos ωmt cos ωct in equation (A) and (B) we get
μA_c = 2K₂A_cA_m
or μ = 2K₂A_m = 2 × 0·2 × 0·5= 0·20 (˙.˙ K₂ = 0·2)
Hence alternative (B) is the correct choice.