Communication miscellaneous Easy Questions and Answers | Page - 39

Communication miscellaneous

Armstrong FM transmitter performs frequency multiplication in stages—

1. to increase the overall S/N ratio
1. to reduce bandwidth
1. to find the desired value of carrier frequency as well as frequency deviation
1. for convenience

View Hint View Answer Discuss in Forum

Since the desired multiplication factors or carrier frequency and frequency deviation are not the same. So if the multiplication is done in one shot, the carrier frequency as well as the frequency deviation is multiplied by the same factor which is not desired. Hence, multiplication is done in stages.

Correct Option: C

Since the desired multiplication factors or carrier frequency and frequency deviation are not the same. So if the multiplication is done in one shot, the carrier frequency as well as the frequency deviation is multiplied by the same factor which is not desired. Hence, multiplication is done in stages.

The input to a linear delta modulator having a step-size δ = 0.62V is a sine wave with frequency f m and peak amplitude Em. If sampling frequency is f s = 40 kHz the combination of the sine wave frequency and the peak amplitude, where slope overload will take place is— Em fm

1. 0.3 V 8 kHz
1. 1.5 V 4 kHz
1. 1.5 V 2 kHz
1. 3.0 V 1 kHz

View Hint View Answer Discuss in Forum

For slope overload to take place
E_m ≥ δf_s
2πf_m

Hence we will check the condition for each case given in the form of option (A), (B), (C) and (D).
After solving we conclude that this equality is satisfied with
E_m = 1·5V and f_m = 4kHz
Hence alternative (B) is the correct choice.

Correct Option: B

For slope overload to take place
E_m ≥ δf_s
2πf_m

Hence we will check the condition for each case given in the form of option (A), (B), (C) and (D).
After solving we conclude that this equality is satisfied with
E_m = 1·5V and f_m = 4kHz
Hence alternative (B) is the correct choice.

A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δ = 0·62 is 100 mV. The modulator is tested with a 1 kHz sinusoidal signal. The maximum amplitude of this test signal required to avoid slope overload is—

1. 2.04 V
1. 1.08 V
1. 4.08 V
1. 2.16 V

View Hint View Answer Discuss in Forum

Condition to avoid slope-overload noise
δ ≥ 2πf_m.A_m
T_s

|A_m|_max = δf_s
2πf_m

= 100 × 10⁻³(10 × 3.4 × 2 × 10³)
2π3.4 × 10³

= 1.08V

Correct Option: B

Condition to avoid slope-overload noise
δ ≥ 2πf_m.A_m
T_s

|A_m|_max = δf_s
2πf_m

= 100 × 10⁻³(10 × 3.4 × 2 × 10³)
2π3.4 × 10³

= 1.08V

A PCM system uses a uniform quantizer followed by a 8- bit encoder. The bit rate of the system is equal to 10⁶bits/s. The maximum message bandwidth for which the system operates satisfactory is—

1. 25 MHz
1. 6.25MHz
1. 12.5 MHz
1. 50 MHz

View Hint View Answer Discuss in Forum

Let the message bandwidth ω Nyquist rate = 2_ω
bandwidth = 2_ω × (no. of bits/sec taken for quantization)
= 2_ω × 8 = 16_ω bits/sec
given that bit rate = 10⁸ bits/sec
Now, 10⁸ = 16_ω
ω = 100 ×100 = 10⁶ = 6.25 MHz
16

Correct Option: B

Let the message bandwidth ω Nyquist rate = 2_ω
bandwidth = 2_ω × (no. of bits/sec taken for quantization)
= 2_ω × 8 = 16_ω bits/sec
given that bit rate = 10⁸ bits/sec
Now, 10⁸ = 16_ω
ω = 100 ×100 = 10⁶ = 6.25 MHz
16

A binary channel with capacity 30 k bits/sec is available for PCM voice transmission. If signal is band limited to 3 kHz then the appropriate values of quantizing level L and the sampling frequency will be—

1. 64, 6 kHz
1. 32, 6 kHz
1. 128, 6 kHz
1. 32, 12 kHz

View Hint View Answer Discuss in Forum

f_s ≥ 2f_m
or
f_s = 2 × 3 MHz = 6 MHz
given the channel capacity
BW = 30 × 10³ bits/sec
∴ 30 × 10³ = nf_s
or
n ≤ 30 × 10³ ≤ 5 = 5
6 × 10³

Now, quantization level L = 2ⁿ = 2⁵ = 32

Correct Option: B

f_s ≥ 2f_m
or
f_s = 2 × 3 MHz = 6 MHz
given the channel capacity
BW = 30 × 10³ bits/sec
∴ 30 × 10³ = nf_s
or
n ≤ 30 × 10³ ≤ 5 = 5
6 × 10³

Now, quantization level L = 2ⁿ = 2⁵ = 32