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Uniform vs non-uniform quantization; 16-level PCM numerical

Short answer

Uniform quantization uses a constant step size, giving poor SQNR for weak signals; non-uniform uses small steps near zero and larger steps at high amplitudes, providing nearly constant SQNR via companding. For 16-level PCM: bits per sample = log₂16 = 4 bits; minimum sampling rate = 8 kHz; bit rate = 4 × 8000 = 32 kbps.

Differentiate between uniform and non-uniform quantization. The information in an analog waveform with maximum frequency 4 kHz is to be transmitted over a 16-level PCM system. a) What would be the maximum number of bits per sample? b) What is the minimum sampling rate and bit rate?

Uniform vs non-uniform quantization

PointUniformNon-uniform
Step sizeConstant Δ\Delta over the whole rangeSmall near zero, larger at high amplitudes
Noise powerΔ2/12\Delta^2/12, same at all levelsVaries; small for weak signals
SQNRProportional to signal power; poor for weak signalsNearly constant over wide range
RealisationUniform ADCCompressor + uniform ADC + expander (companding)
LawsLinearµ-law (μ=255\mu = 255), A-law (A=87.6A = 87.6)
Bits for toll-quality speechAbout 12–138
Used inInstrumentation, video, audio with uniform pdfTelephony (speech)

Non-uniform quantization suits speech because small amplitudes are much more likely and the dynamic range is large.

Numerical

Given: fmax=W=4f_{max} = W = 4 kHz, L=16L = 16 levels.

a) Bits per sample

n=log⁡2L=log⁡216=4 bitsn = \log_2 L = \log_2 16 = 4\ \text{bits}

b) Minimum sampling rate and bit rate

fs,min=2W=2×4=8 kHzRb=nfs=4×8000=32 kbps\begin{aligned} f_{s,min} &= 2W = 2 \times 4 = 8\ \text{kHz} \\ R_b &= nf_s = 4 \times 8000 = 32\ \text{kbps} \end{aligned}

Answer: (a) 4 bits/sample; (b) fs=8f_s = 8 kHz (8000 samples/s), Rb=32R_b = 32 kbps.

✓ Answer checked by Claude Opus 5.5 (high effort)

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