Question

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor

• A. 6
• B. 8
• C. 12
• D. 16

Hint:

PCM SQNR. Signal-to-quantization noise ratio (SQNR) improves by approximately 6 dB for each additional bit used in quantization. A 6 dB increase corresponds to a factor of 4 in power ratio. An increase of 2 bits (\(\Delta n = 2\)) leads to an improvement of \(2 \times 6 = 12\) dB, which is a power ratio factor of \(4^{\Delta n = 4^2 = 16\). SQNR \(\propto 2^{2n\).

Answer 1

The Correct Option is D

In a Pulse Code Modulation (PCM) system, the signal-to-quantization noise ratio (SQNR) is related to the number of bits (\(n\)) used per sample (code word length)
Assuming a uniform quantizer and full-scale sinusoidal input, the SQNR in decibels (dB) is approximately: $$ \text{SQNR (dB)} \approx 6
02n + (1)76 $$ This shows that for each additional bit used in quantization, the SQNR improves by approximately 6 dB
Alternatively, the quantization noise power is inversely proportional to the square of the number of quantization levels (\(L\)), and \(L = 2^n\)
Signal power depends on the signal itself
SQNR (as a power ratio, not dB) is proportional to \(L^2\)
$$ \text{SQNR} \propto L^2 = (2^n)^2 = 2^{2n} $$ Let SQNR\(_1\) be the ratio for \(n_1 = 6\) bits, and SQNR\(_2\) be the ratio for \(n_2 = 8\) bits
$$ \frac{\text{SQNR}_2}{\text{SQNR}_1} = \frac{2^{2n_2}}{2^{2n_1}} = 2^{2(n_2 - n_1)} $$ The increase in the number of bits is \(\Delta n = n_2 - n_1 = 8 - 6 = 2\)
The improvement factor is: $$ 2^{2(\Delta n)} = 2^{2(2)} = 2^4 = 16 $$ The signal-to-quantization noise ratio improves by a factor of 16
In dB, the improvement is \(6
02 \times \Delta n = 6
02 \times 2 \approx 12\) dB
A 12 dB increase corresponds to a power ratio factor of \(10^{12/10} = 10^{(1)2} \approx 15
85 \approx 16\)