Need for non-uniform quantization; one algorithm

Why non-uniform quantization is needed

• In uniform quantization the noise power $\Delta^2/12$ is the same for every level, so SQNR is proportional to signal power. Weak signals get poor SQNR.
• Speech has a wide dynamic range (about 40 dB) and small amplitudes are far more probable than large ones.
• To give soft talkers acceptable SQNR with uniform steps would need about 12–13 bits per sample.
• Non-uniform quantization uses small steps for small amplitudes and large steps for large amplitudes, making SQNR nearly constant over the range. 8 bits then give toll-quality speech (64 kbps).

Algorithm: companding with the µ-law

Non-uniform quantization is realised by companding: compressor → uniform quantizer → (channel) → expander.

m(t)->[Compressor]->[Uniform quantizer+encoder]->channel
                                                    |
out<-[Expander]<-[Decoder]<-------------------------+

1. Normalise the input: $x = m/m_p$, so $|x| \le 1$.
2. Compress using the µ-law:

$$y = \text{sgn}(x)\,\frac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}, \qquad \mu = 255$$

For small $x$, $y \approx \mu x/\ln(1 + \mu)$ (large gain); for large $x$ the curve is logarithmic (small gain). 3. Uniformly quantize $y$ into $L = 2^n$ levels and encode into $n$ bits (8 bits in telephony). 4. At the receiver decode and expand with the inverse law:

$$|x| = \frac{(1 + \mu)^{|y|} - 1}{\mu}$$

5. The combined effect: equal steps in $y$ correspond to small steps in $x$ near zero and large steps near $\pm 1$.

For $\mu \gg 1$, SQNR $\approx \dfrac{3L^2}{[\ln(1 + \mu)]^2}$, almost independent of signal level; with $\mu = 255$, $n = 8$ it is about 38 dB over a 40 dB input range.

In practice the curve is approximated digitally by 15 linear segments (µ-law, T1 systems) or 13 segments for the A-law ($A = 87.6$, E1 systems).