SQNR of uniformly quantized PCM

SQNR of uniformly quantized PCM

Let the message $m(t)$ have peak amplitude $m_p$ (range $-m_p$ to $+m_p$) and be quantized uniformly into $L$ levels.

Step size:

$$\Delta = \frac{2m_p}{L}$$

Quantization noise. The error $q$ is uniformly distributed over $(-\Delta/2, \Delta/2)$ with pdf $1/\Delta$:

$$N_q = E[q^2] = \int_{-\Delta/2}^{\Delta/2} q^2\frac{1}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}$$

Signal-to-quantization noise ratio. With signal power $P = \overline{m^2(t)}$:

$$\text{SQNR} = \frac{P}{N_q} = 3L^2\frac{P}{m_p^2}$$

In terms of bits. For $n$-bit binary coding, $L = 2^n$:

$$\text{SQNR} = \frac{3P}{m_p^2}\,2^{2n}, \qquad \text{SQNR}_{dB} = 10\log_{10}\frac{3P}{m_p^2} + 6.02\,n$$

Sinusoidal message with full-scale amplitude, $P = m_p^2/2$:

$$\text{SQNR} = 1.5\times 2^{2n}, \qquad \text{SQNR}_{dB} = 1.76 + 6.02\,n\ \text{dB}$$

(For a uniformly distributed signal, $P = m_p^2/3$ and SQNR $= 2^{2n}$, i.e. $6.02n$ dB.)

Relation between SQNR and bits

• SQNR grows exponentially with $n$ ($\propto 4^n$); in dB it grows linearly: each added bit improves SQNR by about 6 dB.
• Example: 7 bits → 43.9 dB; 8 bits → 49.9 dB (for a full-scale sine).
• But bandwidth $B_T = nf_s/2$ grows linearly with $n$. So SQNR improves exponentially with bandwidth: a key trade-off in PCM.