SQNR for uniform quantization in terms of levels and bits

SQNR is the ratio of signal power to quantization noise power at the output of a quantizer.

Step 1: Step size

Let the message $m(t)$ lie in $[-m_p, m_p]$ and be quantized uniformly into $L$ levels. Then the step size is

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

Step 2: Quantization noise power

The quantization error $q = m - m_q$ lies in $[-\Delta/2, \Delta/2]$. For a fine quantizer it is uniformly distributed, with pdf $f(q) = 1/\Delta$:

$$\begin{aligned} N_q = E[q^2] &= \int_{-\Delta/2}^{\Delta/2} q^2\cdot\frac{1}{\Delta}\,dq = \frac{1}{\Delta}\left[\frac{q^3}{3}\right]_{-\Delta/2}^{\Delta/2} \\ &= \frac{\Delta^2}{12} = \frac{(2m_p/L)^2}{12} = \frac{m_p^2}{3L^2} \end{aligned}$$

Step 3: SQNR in terms of levels

With message power $P = \overline{m^2(t)}$:

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

Step 4: SQNR in terms of bits

With binary coding of $n$ bits per sample, $L = 2^n$:

$$\text{SQNR} = \frac{3P}{m_p^2}\,2^{2n} = \frac{3P}{m_p^2}\,4^n$$ $$\text{SQNR}_{dB} = 10\log_{10}\!\left(\frac{3P}{m_p^2}\right) + 6.02\,n\ \text{dB}$$

Special case: full-scale sinusoid

For $m(t) = m_p\cos\omega_m t$, $P = m_p^2/2$:

$$\text{SQNR} = \frac{3L^2}{2} = 1.5L^2 = 1.5\times 2^{2n}$$ $$\text{SQNR}_{dB} = 1.76 + 6.02\,n\ \text{dB}$$

Example: $n = 8$ bits gives SQNR $= 1.76 + 48.16 = 49.92$ dB.

Conclusions:

• SQNR increases as the square of the number of levels.
• SQNR increases exponentially with bits: each extra bit adds about 6 dB.
• But each extra bit increases the bit rate $n f_s$ and the bandwidth $n f_s/2$, so there is a trade-off between SQNR and bandwidth.