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Uniform vs non-uniform quantization; companding laws

Short answer

Uniform quantization has equal step size, constant noise power, but SQNR falls for weak signals. Non-uniform uses small steps near zero, large steps at high amplitude, giving nearly constant SQNR. Speech has wide dynamic range and peaked PDF near zero, so non-uniform quantization (companding) maintains quality with fewer bits. µ-law (μ=255) and A-law (A=87.6) compress logarithmically for large amplitudes, expand at receiver. Both are implemented as piecewise linear segments.

Differentiate between uniform quantization and non-uniform quantization. Why is non-uniform quantization done for speech signal? Explain about companding laws.

Uniform vs non-uniform quantization

PointUniformNon-uniform
Step sizeEqual everywhereSmall near zero, large at high amplitude
Quantization noiseSame Δ2/12\Delta^2/12 for all levelsSmall for weak, larger for strong signals
SQNRFalls for weak signalsNearly constant over range
ImplementationSimple ADCCompressor + uniform ADC + expander
Suited toSignals with uniform pdfSpeech, wide dynamic range

Why non-uniform quantization for speech

  • Speech has a large dynamic range (about 40 dB between loud and soft talkers).
  • Its amplitude pdf is peaked near zero: small amplitudes occur most often.
  • With uniform steps, SQNR ∝\propto signal power, so quiet talkers and soft sounds get poor SQNR.
  • Non-uniform steps give small steps where samples are most likely, so SQNR becomes almost independent of level, and 8 bits give quality that would need about 12 bits uniformly.

Companding laws

Companding = compressing at the transmitter + expanding at the receiver. With x=m/mpx = m/m_p normalised input (∣x∣≤1|x| \le 1):

µ-law (North America, Japan; μ=255\mu = 255):

y=sgn(x) ln⁡(1+μ∣x∣)ln⁡(1+μ)y = \text{sgn}(x)\,\frac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}

A-law (Europe, ITU-T E1 systems; A=87.6A = 87.6):

y={sgn(x)A∣x∣1+ln⁡A,0≤∣x∣≤1Asgn(x)1+ln⁡(A∣x∣)1+ln⁡A,1A≤∣x∣≤1y = \begin{cases} \text{sgn}(x)\dfrac{A|x|}{1 + \ln A}, & 0 \le |x| \le \dfrac{1}{A} \\[2mm] \text{sgn}(x)\dfrac{1 + \ln(A|x|)}{1 + \ln A}, & \dfrac{1}{A} \le |x| \le 1 \end{cases}
  • μ=0\mu = 0 or A=1A = 1 gives uniform quantization.
  • Both laws are nearly linear for small inputs and logarithmic for large inputs.
  • In practice they are implemented as 13-segment (A-law) or 15-segment (µ-law) piecewise linear approximations.

✓ Answer checked by Claude Opus 5.5 (high effort)

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