Question

A signal having a bandwidth of 5 MHz is transmitted using the Pulse Code Modulation (PCM) scheme as follows. The signal is sampled at a rate of 50% above the Nyquist rate and quantized into 256 levels. The binary pulse rate of the PCM signal in Mbits per second is \(\underline{\hspace{1cm}}\).

Hint:

The binary pulse rate of a PCM system is determined by the sampling rate and the number of bits per sample.

Answer 1

Correct Answer: 120

The Nyquist rate is twice the bandwidth: \[ f_{\text{Nyquist}} = 2 \times 5 \, \text{MHz} = 10 \, \text{MHz} \] The sampling rate is 50% above the Nyquist rate: \[ f_{\text{sampling}} = 1.5 \times f_{\text{Nyquist}} = 1.5 \times 10 \, \text{MHz} = 15 \, \text{MHz} \] The quantization levels are 256, so the number of bits per sample is: \[ \text{Bits per sample} = \log_2(256) = 8 \, \text{bits} \] Thus, the binary pulse rate is: \[ \text{Pulse rate} = f_{\text{sampling}} \times \text{Bits per sample} = 15 \times 10^6 \times 8 = 120 \, \text{Mbits/s} \] Thus, the binary pulse rate of the PCM signal is \( 120 \, \text{Mbits/s} \).