A television signal video and audio has a bandwidth of 45 MH

A television signal (video and audio) has a bandwidth of 4.5 MHz. This signal is sampled, quantized, and binary coded to obtain a PCM signal. Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate. If the samples are quantized into 1024 levels, determine the number of binary pulses required to encode each sample. Determine the binary pulse rate (bits per second) of the binary-coded signal, and the minimum t>and width required to transmit this signal.

Solution

f_sig =4.5MHz

So Nyquist rate fn = 2*f_sig =9 MHz

And sampling rate = 120% of fn

= 1.2*9 = 10.8MHz

B)

PCM levels = 1024 = 2¹⁰

So the total numbers of bits required = 10

C)

So the tranmission rate = 10 * sampling frequency

= 108 Mbits/sec