A message signal of 20 cos (2π × 10 4 t), sampled at the Nyquist
| A message signal of 20 cos (2π × 104t), sampled at the Nyquist rate is given to a 10-bit PCM system. The resulting binary sequence is represented with rectangular pulses and is transmitted through free space by using ASK binary signaling scheme. The transmission bandwidth is:
A. 200 kHz
B. 400 kHz
C. 600 kHz
D. 100 kHz
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
The transmission bandwidth for ASK scheme is given by:
Bandwidth = 2Rb (for Rectangular pulses)
Rb = Bit rate given by:
Rb = nfs
n = number of bits used in PCM encoding.
fs = Sampling frequency
Calculation:
Given n = 10
fm = Message signal frequency \(= \frac{{2\pi \times {{10}^4}}}{{2\pi }} = 10kHz\)
fs = Nyquist Rate = 2 fm = 20 kHz.
Rb = nfs = 10 × 20 kbps
= 200 kbps.
So the Transmission Bandwidth = 200 k × 2
= 400 kHz.