PCM SQNR increases 6 dB per bit

In PCM with $n$ bits per sample, the quantizer has $L = 2^n$ levels. We find SQNR in terms of $n$.

1. Step size for a message in the range $\pm m_p$:

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

2. Quantization noise: the error $q$ is uniformly distributed in $(-\Delta/2, \Delta/2)$:

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

3. Signal power: take a full-load sinusoid $m(t) = m_p\cos\omega_mt$, so $S = m_p^2/2$.
4. SQNR:

$$\begin{aligned} \text{SQNR} &= \frac{S}{N_q} = \frac{m_p^2/2}{m_p^2/(3L^2)} = \frac{3}{2}L^2 = \frac{3}{2}\,2^{2n} \\ \text{SQNR}_{dB} &= 10\log_{10}\frac{3}{2} + 10\log_{10}2^{2n} \\ &= 1.76 + 20n\log_{10}2 \\ &= 1.76 + 6.02n\ \text{dB} \end{aligned}$$

5. Effect of one extra bit: going from $n$ to $n + 1$ bits,

$$\text{SQNR}_{dB}(n + 1) - \text{SQNR}_{dB}(n) = 6.02(n + 1) - 6.02n = 6.02 \approx 6\ \text{dB}$$

In ratio terms, $L$ doubles, $\Delta$ halves, and $N_q = \Delta^2/12$ falls by a factor of 4, i.e. $10\log_{10}4 = 6.02$ dB.

For a general signal, $\text{SQNR} = 3L^2\,\overline{m^2}/m_p^2$, i.e. $\text{SQNR}_{dB} = \alpha + 6.02n$, where $\alpha$ depends only on the signal's power-to-peak ratio. The 6 dB/bit rule holds for any signal.

Price paid: each extra bit increases bit rate $R_b = nf_s$ and transmission bandwidth $B_T \approx nW$ in proportion, so PCM trades bandwidth for SNR exponentially: SQNR $\propto 2^{2B_T/W}$.

Example: 7 bits give $1.76 + 42.1 = 43.9$ dB; 8 bits give $49.9$ dB, about 6 dB more.