Question

The number of bits per sample in a PCM system with sinusoidal input is increased from n to n+1. The improvement in signal to quantization noise ratio will be

• A. n dB
• B. 2n dB
• C. 3 dB
• D. 6 dB

Hint:

For PCM, SQNR (in dB) \(\approx 1.76 + 6.02n\), where \(n\) is the number of bits per sample.
Each additional bit used for quantization increases the SQNR by approximately 6 dB. This means doubling the number of quantization levels (by adding 1 bit) quadruples the SQNR power ratio (\(10 \log_{10} 4 \approx 6.02\)).

Answer 1

The Correct Option is D

In a Pulse Code Modulation (PCM) system, the signal-to-quantization noise ratio (SQNR) for a full-scale sinusoidal input, when \(n\) bits are used per sample, is approximately given by:

\(\text{SQNR (dB)} \approx 1.76 + 6.02n\) 

where \(n\) is the number of bits per sample. This formula shows that for each additional bit used in quantization, the SQNR improves by approximately 6.02 dB (often rounded to 6 dB).

Let SQNR₁ be for \(n\) bits:

SQNR₁ \(\approx 1.76 + 6.02n\)

Let SQNR₂ be for \(n+1\) bits:

SQNR₂ \(\approx 1.76 + 6.02(n+1) = 1.76 + 6.02n + 6.02\)

Improvement in SQNR:

The improvement in SQNR is:

Improvement = SQNR₂ - SQNR₁

Improvement \(\approx (1.76 + 6.02n + 6.02) - (1.76 + 6.02n)\)

Improvement \(\approx 6.02 \text{ dB}\)

Rounding to the nearest integer or common approximation, this is 6 dB.

This "6 dB per bit" rule is a well-known characteristic of PCM systems.

Final Answer:

6 dB