Communication miscellaneous Easy Questions and Answers | Page - 31

Communication miscellaneous

A signal m(t) band-limited to 3 kHz is sampled at a rate
33 1 %
3
higher than the Nyquist rate. The max acceptable error in the sample amplitude (max. quantization error) is 0.5% of peak amplitude V_p. The quantized samples are binary coded. Find out the minimum bandwidth required to transmit the encoded binary signal. If 24 such signals are TDM, determine the minimum bandwidth required to transmit the multiplexed signal—

1. 1536 kHz
1. 768 kHz
1. 472 kHz
1. 1344 kHz

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We know that minimum bandwidth required to transmit 24 channel through TDM (Time Division Multiplexing)
= nN f_s
2

Where, N = no. of bits
n = no. of channel
f_s = sampling frequency
According to question,
f_s = 33% greater that the Nyquist rate
= 1 + 1 .2.f_m
3

or
f_s = 4 ×2×3kHz = 8kHz
3

For the quantization step size δ the maximum quantization error is ± δ/2 and given as δ/2 = 0·5% of V_p
or
1 2V_p = 0.5 .V_p
2 L 100

Where,
δ = 2V_p
L

V_p = maximum amplitude
L = no. of level = 2ⁿ
N > 7 i.e. for binary coding we must take N = 8
Now,
B_{T_min} = N.n. f_s
2

= 8 × 24 × 8 kHz = 768 kHz
2

Correct Option: B

We know that minimum bandwidth required to transmit 24 channel through TDM (Time Division Multiplexing)
= nN f_s
2

Where, N = no. of bits
n = no. of channel
f_s = sampling frequency
According to question,
f_s = 33% greater that the Nyquist rate
= 1 + 1 .2.f_m
3

or
f_s = 4 ×2×3kHz = 8kHz
3

For the quantization step size δ the maximum quantization error is ± δ/2 and given as δ/2 = 0·5% of V_p
or
1 2V_p = 0.5 .V_p
2 L 100

Where,
δ = 2V_p
L

V_p = maximum amplitude
L = no. of level = 2ⁿ
N > 7 i.e. for binary coding we must take N = 8
Now,
B_{T_min} = N.n. f_s
2

= 8 × 24 × 8 kHz = 768 kHz
2

The AM signal that occupies the greatest bandwidth is for the same amplitude—

1. 5 kHz Square wave
1. 5 kHz Sine wave
1. 5 kHz Triangular wave
1. 5 kHz Saw tooth wave

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The bandwidth of a signal is defined as the range of frequencies where most of the energy or power lies. So for the same amplitude, square wave passes highest energy or power among the given wave.
Hence alternative (A) is the correct choice.

Correct Option: A

The bandwidth of a signal is defined as the range of frequencies where most of the energy or power lies. So for the same amplitude, square wave passes highest energy or power among the given wave.
Hence alternative (A) is the correct choice.

In figure given below
m(t) = 2 sin 2πt
t

s(t) = cos 200 πt and n(t) = sin 199 πt . The output y(t)
t

1. sin 2πt
  2t
1. sin 2πt + sin πt cos 3πt
  2t t
1. sin 2πt + sin 0.5πt cos 1.5πt
  2t t
1. sin 2πt + sin πt cos 0.75πt
  2t t

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m(t) s(t) = y₁(t)
=
= 2 sin (2πt) cos(2πt)
t

or
y₁(t) = 2 sin (2πt) cos(2πt)
t
...(A)
y₂(t) = y₁(t) + n(t)
= sin 202πt - sin 198πt + sin 198πt
t t
...(B)
Now y(t) = y₂(t).s(t)
= [sin 202πt - sin 198πt + sin 199πt]cos 200πt
t

or
y(t) = 1 [sin (402πt) + sin(2πt) − {sin(389πt) − sin2πt} + sin(399πt) − sin(πt)]
2t
...(C)
After filtering equation (C) through LPF with cut-off frequency 1 Hz we get
or
y(t) = sin (2πt) + sin (2πt) − sin (πt)
2t

or
y(t) = sin (2πt) + 2sin (0.5t) cos(1.5πt)
2t

or
y(t) = sin (2πt) + sin 0.5t cos 1.5πt
2t t

Hence alternative (C) is the correct choice.

Correct Option: C

m(t) s(t) = y₁(t)
=
= 2 sin (2πt) cos(2πt)
t

or
y₁(t) = 2 sin (2πt) cos(2πt)
t
...(A)
y₂(t) = y₁(t) + n(t)
= sin 202πt - sin 198πt + sin 198πt
t t
...(B)
Now y(t) = y₂(t).s(t)
= [sin 202πt - sin 198πt + sin 199πt]cos 200πt
t

or
y(t) = 1 [sin (402πt) + sin(2πt) − {sin(389πt) − sin2πt} + sin(399πt) − sin(πt)]
2t
...(C)
After filtering equation (C) through LPF with cut-off frequency 1 Hz we get
or
y(t) = sin (2πt) + sin (2πt) − sin (πt)
2t

or
y(t) = sin (2πt) + 2sin (0.5t) cos(1.5πt)
2t

or
y(t) = sin (2πt) + sin 0.5t cos 1.5πt
2t t

Hence alternative (C) is the correct choice.

Consider a system shown below. Let X(f) and Y(f) denotes the Fourier transform of x(t) and y(t) respectively. The positive frequencies where Y (f) has spectral peak are—

1. 1 kHz and 24 kHz
1. 2 kHz and 24 kHz
1. 1 kHz and 14 kHz
1. 2 kHz and 14 kHz

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From given figure since x(f) has spectral peak at 1kHz, output of balanced modulator will give
(f_c ± f_m) i.e. (10 ± 1)
which passes through the HPF with cut-off frequency ω_c = 10 kHz. So output after HPF get only 11 kHz signal, again after mixing with balanced modulator the output frequencies will be (f_c ± f₁) kHz i.e. 24 kHz and 2 kHz. Hence alternative (B) is the correct choice.

Correct Option: B

From given figure since x(f) has spectral peak at 1kHz, output of balanced modulator will give
(f_c ± f_m) i.e. (10 ± 1)
which passes through the HPF with cut-off frequency ω_c = 10 kHz. So output after HPF get only 11 kHz signal, again after mixing with balanced modulator the output frequencies will be (f_c ± f₁) kHz i.e. 24 kHz and 2 kHz. Hence alternative (B) is the correct choice.

The resonant frequency of an RF amplifier is 1 MHz and its bandwidth is 10 kHz. The Q factor will be—

1. 10
1. 100
1. 0.01
1. 0.1

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Q = Resonant frequency = 1 × 10⁶ = 100
Bandwidth 10 × 10³

Correct Option: B

Q = Resonant frequency = 1 × 10⁶ = 100
Bandwidth 10 × 10³