Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)

Non-uniform quantization and companding

In non-uniform quantization the step size is small for low amplitudes and large for high amplitudes. Since speech has mostly small amplitudes and a wide dynamic range, this keeps SQNR nearly constant for loud and soft talkers.

It is realised by companding:

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

1. Compressor applies a logarithmic gain: weak signals amplified more than strong ones. µ-law: $y = \text{sgn}(x)\dfrac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}$, $\mu = 255$; or A-law with $A = 87.6$.
2. Uniform quantizer then acts on the compressed signal; equal steps in $y$ are small steps in $x$ near zero.
3. Expander at the receiver applies the inverse law, restoring the original amplitude relation. In practice the curve is a 13-segment (A-law) or 15-segment (µ-law) piecewise linear approximation.

Numerical

Given: $W = 4$ kHz, range $\pm 2.4$ V ($m_p = 2.4$ V), $R_b = 64$ kbps. Assume sampling at the Nyquist rate and a full-scale sinusoidal signal.

Bits per sample

$$\begin{aligned} f_s &= 2W = 8\ \text{kHz} \\ n &= \frac{R_b}{f_s} = \frac{64\,000}{8000} = 8\ \text{bits}, \quad L = 2^8 = 256 \end{aligned}$$

Quantization noise power

$$\begin{aligned} \Delta &= \frac{2m_p}{L} = \frac{4.8}{256} = 0.01875\ \text{V} \\ N_q &= \frac{\Delta^2}{12} = \frac{(0.01875)^2}{12} = 2.93 \times 10^{-5}\ \text{V}^2 \end{aligned}$$

SQNR

$$\begin{aligned} S &= \frac{m_p^2}{2} = \frac{2.4^2}{2} = 2.88\ \text{V}^2 \\ \text{SQNR} &= \frac{2.88}{2.93 \times 10^{-5}} = 98\,304 \\ \text{SQNR}_{dB} &= 10\log_{10}98\,304 = 49.93\ \text{dB} \end{aligned}$$

(Check: $1.76 + 6.02 \times 8 = 49.92$ dB.)

Minimum bandwidth for TDM of 10 signals

$$\begin{aligned} R_{b,total} &= 10 \times 64 = 640\ \text{kbps} \\ B_{T,min} &= \frac{R_{b,total}}{2} = 320\ \text{kHz} \end{aligned}$$

Answer: $n = 8$ bits; $N_q = 2.93 \times 10^{-5}$ V² (normalised to 1 Ω); SQNR = 49.93 dB; $B_T = 320$ kHz.