# 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 ___.

## Answer (Detailed Solution Below) 120

__Concept:__

The bit rate is defined as:

\(R_b = n × f_s = \frac{1}{t_s}\)

n: Number of bits

f_{s}: Sampling frequency

L: Number of levels

L = 2^{n}

__Calculation:__

Consider the given signal as shown:

The bandwidth is 5 Mhz

Nyquist rate = 2 × f_{m} = 2 × 5 Mhz

Nyquist rate = 10 MHz

The sampling rate is 50% above the Nyquist rate

\(\frac{1}{t_s} = 1.5 \times 10 MHz = 15 MHz\)

L = 256 = 2^{n}

**n = 8**

The bit rate is:

\(R_b = \frac{1}{t_s} = 15 MHz \times 8 = 120 Mbps\)