Question

In a PCM system, speech signal bandlimited to 4 kHz is sampled at 1.5 times Nyquist rate and quantized using 256 levels. The bit rate required to transmit the signal will be

• A. 64 kbps
• B. 96 kbps
• C. 128 kbps
• D. 160 kbps

Hint:

Remember the key parameters involved in calculating the bit rate for PCM. Key values are sampling rate based on Nyquist rate, and the number of bits per sample based on the levels.

Answer 1

The Correct Option is B

Step 1: The Nyquist rate is given as twice the maximum frequency. In our case: \[ f_{nyquist} = 2 \times 4 kHz = 8kHz \] Step 2: The actual sampling frequency \(f_s\) is 1.5 times Nyquist rate: \[ f_s = 1.5 \times f_{nyquist} = 1.5 \times 8 kHz = 12 kHz \] Step 3: The number of quantization levels = 256. Number of bits, n, required is: \[ 2^n = 256 \] \[ n = \log_2{256} = 8 bits \] Step 4: The bit rate \(R_b\) is the product of sampling frequency and the number of bits: \[ R_b = f_s \times n = 12 \times 10^3 \times 8 = 96000 bps = 96 kbps \] Therefore the bit rate is 96 kbps.