Need for non-uniform quantization; SQNR of DM

Need for non-uniform quantization

In uniform quantization the step size $\Delta$ is fixed, so the quantization noise power $\Delta^2/12$ is the same for all signal levels. Hence:

• Weak signals (low amplitude) get a poor SQNR, while loud signals get a high SQNR.
• Speech has a wide dynamic range (about 40 dB) and small amplitudes occur much more often than large ones.
• To keep weak signals acceptable with uniform steps, many more bits would be needed.

Non-uniform quantization uses small steps for small amplitudes and large steps for large amplitudes. This gives nearly constant SQNR over a wide range of input levels with fewer bits (8 bits instead of about 12 for telephony). It is implemented by companding ($\mu$-law, A-law).

SQNR of delta modulation (no slope overload)

Let the message be a single tone $m(t) = A_m\cos 2\pi f_m t$, step size $\Delta$, sampling rate $f_s = 1/T_s$, and the receiver LPF bandwidth $W$.

1. Condition for no slope overload. The staircase can rise by at most $\Delta$ per $T_s$:

$$\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m \le \frac{\Delta}{T_s} \;\Rightarrow\; A_{m,max} = \frac{\Delta f_s}{2\pi f_m}$$

2. Signal power (with maximum amplitude):

$$S = \frac{A_m^2}{2} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}$$

3. Granular noise power. With no slope overload, the error $q(t)$ lies within $\pm\Delta$ and is uniformly distributed:

$$E[q^2] = \frac{1}{2\Delta}\int_{-\Delta}^{\Delta} q^2\,dq = \frac{\Delta^2}{3}$$

This power is spread uniformly over $0$ to $f_s$. The output LPF passes only $W$, so

$$N_q = \frac{\Delta^2}{3}\cdot\frac{W}{f_s}$$

4. SQNR:

$$\text{SQNR} = \frac{S}{N_q} = \frac{\Delta^2 f_s^2/(8\pi^2 f_m^2)}{\Delta^2 W/(3f_s)} = \frac{3 f_s^3}{8\pi^2 f_m^2 W}$$

In dB:

$$\text{SQNR}_{dB} = 30\log_{10} f_s - 20\log_{10} f_m - 10\log_{10} W - 14.2$$

So SQNR improves by 9 dB each time $f_s$ is doubled. Example: $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz gives SQNR $= 2490 \approx 33.96$ dB.