Question

In a PCM system, the signal m(t) = {sin (100 πt) + cos (100 πt)} V is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is _____.

Answer 1

Concept:

The Nyquist sampling rate is double of the maximum signal frequency.

For uniform quantizer, the step size is given by:

\({\rm{\Delta }} = \frac{{{V_{p - p}}}}{L}\);

L = 2ⁿ = number of encoding levels and n is the number of bits required to represent the given encoding levels.

The minimum data rate is given by:

\({R_b} = n{f_s}\) 

Calculation:

\(m\left( t \right) = sin\left( {100\pi t} \right) + cos\left( {100\pi t} \right)\;V\) 

Maximum signal frequency f_m = 50 Hz

Nyquist rate or Nyquist frequency will be:

\({f_N} = 2 \times {f_m} = 2 \times 50 = 100\;Hz\) 

Given that sampling frequency = N.R. = 100 Hz

Maximum signal value:

\({V_{max}} = \sqrt {{1^2} + {1^2}} = \sqrt 2 \) 

Peak to peak voltage:

\({V_{p - p}} = {V_{max}} - {V_{min}}\) 

\( = \sqrt 2 - \left( { - \sqrt 2 } \right) = 2\sqrt 2 \) 

Also, we can write:

\({2^n} = \frac{{{V_{p - p}}}}{{\rm{\Delta }}} = \frac{{2\sqrt 2 }}{{0.75}} = 3.77\) 

n = 2 bits 

∴ The minimum data rate will be:

R_b = nf_s = 2 × 100 = 200 bits/sec