# 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 wi

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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 _____.

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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 fm = 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:

Rb = nfs = 2 × 100 = 200 bits/sec