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 nose ratio that can be obtained by the PCM system (rounded off to two decimal places) is _____

Answer 1

Concept:

For PCM, the noise power (N_p) is given by:

\({N_p} = \frac{{{{\rm{\Delta }}^2}}}{{12}}\)

Where \({\rm{\Delta }} = \frac{{{V_{peak - peak}}}}{{{2^n}}}\) 

For calculating n, the bit-rate and the bandwidth of message signal is given as:

Bitrate, R_b = nf_s

f_s = 2f_m (Nyquist criteria)

Signal power = (RMS value of signal)²

Calculation:

Given

RMS value of signal = 0.1 V

Signal power = (0.1)² = 0.01     ---(1)

Bit rate R_b = nf_s

R_b = n (2f_m)

50 = n (2 × 5)

n = 5

Step size will be:

\({\rm{\Delta }} = \frac{{{V_{peak - peak}}}}{{{2^n}}} = \frac{{2V}}{{{2^5}}} = \frac{1}{{{2^4}}}\)

Now, the Noise Power will be:

\(\frac{{{{\rm{\Delta }}^2}}}{{12}} = {\left( {\frac{1}{{{2^4}}}} \right)^2}\frac{1}{{12}} = \frac{1}{{{2^8} \times 12}}\)

∴ The required Signal to Noise Ratio (SNR) will be:

\(\frac{{Signal\;power}}{{Noise\;Power}} = \frac{{0.01}}{{\left( {\frac{1}{{{2^8} \times 12}}} \right)}}\)

= 30.72