Question

A TV signal (Audio and Video) has bandwidth of 4.5 MHz. The signal is sampled,quantized and binary-coded to obtain PCM signal.
(a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate.
(b) If the sample are quantized into 1024 levels, determine the number of binary pulse required to encode each sample.
(c) Determine the binary pulse rate (bit per second) of the binary-coded signal and minimum bandwidth required to transmit this signal

Answer 1

a)Since the sampling rate should be 2 times of highest frequency:
We have Nyquist rate=4.5x2=9Mhz
20% of Nyquist rate=1.8Mhz
b) since 2⁹=1024, to quantized into 1024 levels we need 9 no of binary pulses.

c) pulse rate is bit lenght/time=9/1s=9s

Minimum bandwidth is 2xnyquist frequency i.e 9Mhz

Answer 2

Nyquist rate = 2×4.5×10^6=9MHz 20% of Nyquist rate = 1.8MHz Sampling rate is of two time of Nyquist rate=9+1.8= 10.8 MHz B)2^10= 10 number of binary pulses C) Binary pulse rate = bit length × sampling rate = 10.8M×10 = 108Mbits