What is the required bandwidth of the PCM system for 256 quantization levels when 48 telephone channels, each bandlimited to 4 kHz, are to be time-division multiplexed by this PCM?

What is the required bandwidth of the PCM system for 256 quantization levels when 48 telephone channels, each bandlimited to 4 kHz, are to be time-division multiplexed by this PCM? Correct Answer 3.072 MHz

Concept:

The bandwidth of a PCM system for an encoded signal sampled at a frequency of f_s is given by:

B.W. = n f_s

f_s = Sampling frequency

n = number of bits used for encoding.

n is related to the number of quantization levels (L) as:

L = 2ⁿ

or n = log₂L

Calculation:

Since the sampling frequency is not mentioned, we'll assume it to be sampled at the Nyquist rate, i.e.

f_s = 2f_m

f_m = Maximum frequency present at the modulating signal.

∴ For the given bandlimited signal with a frequency 4 kHz, the sampling frequency will be:

fs = 2 × 4 = 8 kHz

With L = 256, the number of bits will be:

n = log₂ 256 = log₂ 2⁸

n = 8 bits

Now for 48 such channels, the required bandwidth will be:

B.W. = 48 × n × f_s = 48 × 8 × 8000

B.W. = 3.072 MHz