Communication miscellaneous Easy Questions and Answers | Page - 20

Communication miscellaneous

For the data in Q. 35 and if capacitor used is 100 Pf then R—

1. 0·17 × 10¹³ Ω
1. 0·212 × 10¹¹ Ω
1. 17 × 10¹³ Ω
1. None of these

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We know that
Time constant τ = RC
or
R = τ = 0.212 × 10⁻³ = 0.212 × 10⁷ Ω
c 100 × 10 ⁻¹²

Correct Option: D

We know that
Time constant τ = RC
or
R = τ = 0.212 × 10⁻³ = 0.212 × 10⁷ Ω
c 100 × 10 ⁻¹²

Which one of the following statement is correct? In a ratio detector:

1. Linearity is worse than that of a phase discriminator
1. Stabilization is provided, against signal strength variations.
1. The output is twice of that obtainable from a similar phase discriminator.
1. The circuit is same as that in a discriminator, except that the diode connections are reversed.

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The ratio detector has one major advantage over the slope detector and Foster-Seeley discriminator for FM demodulation is that it is relatively immune to amplitude variations in its input signal.
The schematic diagram for the ratio detector is shown. As with the Foster-Seeley discriminator, a ratio detector has a single tuned circuit in the transformer secondary. Therefore, the operation of a ratio detector is similar to that of the Foster-Seeley discrimination. In fact, the diodes D₁ and D₂ are identical. However, with the ratio detector, one diode is reversed (D₂ ) and current l_d can flow around the outermost loop of the circuit. Hence alternative (D) is the correct choice.

Correct Option: D

The ratio detector has one major advantage over the slope detector and Foster-Seeley discriminator for FM demodulation is that it is relatively immune to amplitude variations in its input signal.
The schematic diagram for the ratio detector is shown. As with the Foster-Seeley discriminator, a ratio detector has a single tuned circuit in the transformer secondary. Therefore, the operation of a ratio detector is similar to that of the Foster-Seeley discrimination. In fact, the diodes D₁ and D₂ are identical. However, with the ratio detector, one diode is reversed (D₂ ) and current l_d can flow around the outermost loop of the circuit. Hence alternative (D) is the correct choice.

Which one of the following is represented by
V(t) = 5[cos (10⁶ πt) – sin (10³ πt) × sin (10⁶ πt)]

1. SSB upper sideband signal.
1. DSB suppressed carrier signal.
1. AM signal.
1. Narrow band FM signal.

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NA

Correct Option: D

NA

In a broadcast transmitter, the RF output is represented as
e(t) = 50 [1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 10⁶t) volt
What are the sidebands of the signals in radians?

1. 5 × 10³ and 9 × 10³
1. 5·91 × 10⁶ , 5·995 × 10⁶ , 6·005 × 10⁶ and 6·009 × 10⁶
1. 4 × 10³ , 1·4 × 10⁴
1. 1 × 10⁶ , 1·1 × 107, 3 × 10⁶ , 1·5 × 10⁷

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The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 10⁶t) volt
Let 5000 = ω_m1 or ω_m1 = 0·005 MHz
9000 = ω_m2 or ω_m2 = 0·009 MHz
6 × 106 = ω_c1 or ω_c1 = 6 MHz
e(t) = 50[cos ω_c1 t + 0·89 cos ω_m1t. cos ω_c1t + 0·30 sin ω_m2t]
Sidebands of the signals are at
ω_c1 ± ω_m1 and ω_c1 ± ω_m2, frequency
i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
6·005 × 10⁶, 5·995 × 10⁶, 6·009 × 10⁶, 5·991 × 10⁶
Hence alternative (B) is the correct choice.

Correct Option: B

The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 10⁶t) volt
Let 5000 = ω_m1 or ω_m1 = 0·005 MHz
9000 = ω_m2 or ω_m2 = 0·009 MHz
6 × 106 = ω_c1 or ω_c1 = 6 MHz
e(t) = 50[cos ω_c1 t + 0·89 cos ω_m1t. cos ω_c1t + 0·30 sin ω_m2t]
Sidebands of the signals are at
ω_c1 ± ω_m1 and ω_c1 ± ω_m2, frequency
i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
6·005 × 10⁶, 5·995 × 10⁶, 6·009 × 10⁶, 5·991 × 10⁶
Hence alternative (B) is the correct choice.

The sum of two signals e₁ = 3 sin (4π × 10³t) and e₁= 5 sin(4π × 256t) is sampled at 1024 Hz. The sampled signal is passed through a low pass filter with cut off at 2048 Hz. The output of the filter will contain components at—

1. 256 Hz and 10³ Hz
1. 256 Hz and 1024 Hz
1. 256 Hz Only
1. 1024 Hz Only

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Given that
e₁= 3 sin(4π × 10³t), f_m1 = 2 × 10³ Hz
e₂ = 5 sin(4 π × 256t), f_m2 = 256 Hz
sampling frequency, f_s = 1024 Hz
cut-off frequency of LPF, ω_c = 2048 Hz
Since for proper reconstruction of signal f_s ≥ 2f_m
For f_m1, fs less than 2f_m1 so it will pass through the LPF,
while for f_m2, f_s > 2f_m2 [i.e. 1024 > (2 × 256)]
So the output of the filter will contain 256 Hz component only.

Correct Option: C

Given that
e₁= 3 sin(4π × 10³t), f_m1 = 2 × 10³ Hz
e₂ = 5 sin(4 π × 256t), f_m2 = 256 Hz
sampling frequency, f_s = 1024 Hz
cut-off frequency of LPF, ω_c = 2048 Hz
Since for proper reconstruction of signal f_s ≥ 2f_m
For f_m1, fs less than 2f_m1 so it will pass through the LPF,
while for f_m2, f_s > 2f_m2 [i.e. 1024 > (2 × 256)]
So the output of the filter will contain 256 Hz component only.