Question

A message signal having peak-to-peak value of 2 V, root mean square value of 0.1 V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is .

Hint:

The maximum signal to quantization noise ratio in a PCM system depends on the number of bits per sample and is given by \( \text{SNR}_{\text{max}} = \frac{3 \cdot (2^n - 1)^2}{2^n} \).

Answer 1

Correct Answer: 30

The maximum signal to quantization noise ratio \( \text{SNR}_{\text{max}} \) for a uniform quantizer is given by: \[ \text{SNR}_{\text{max}} = \frac{3 \cdot (2^{n} - 1)^2}{2^n} \] where \( n \) is the number of bits per sample. The number of bits per sample can be calculated from the given transmission rate and bandwidth: \[ n = \frac{\text{Transmission rate}}{\text{Sampling rate}} = \frac{50 \, \text{kbps}}{2 \times 5 \, \text{kHz}} = 5 \, \text{bits/sample} \] Substituting \( n = 5 \) into the SNR formula: \[ \text{SNR}_{\text{max}} = \frac{3 \cdot (2^5 - 1)^2}{2^5} = \frac{3 \cdot (31)^2}{32} \approx 30.48 \] Thus, the maximum signal to quantization noise ratio is \( \boxed{30.00} \).