SQNR of delta modulation

In delta modulation the quantization noise has two parts: slope-overload noise and granular noise. SQNR is derived assuming the step $\Delta$ is just large enough to avoid slope overload, so only granular noise remains.

1. Maximum signal amplitude (no slope overload) For $m(t) = A\cos\omega_mt$ the maximum slope is $2\pi f_mA$. The staircase can change by $\Delta$ per $T_s$, so we need

$$2\pi f_mA \le \Delta f_s \;\Rightarrow\; A_{max} = \frac{\Delta f_s}{2\pi f_m}$$

2. Signal power

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

3. Granular noise power Without slope overload the error $e(t) = m(t) - m_q(t)$ lies in $(-\Delta, +\Delta)$ and is assumed uniform with pdf $1/(2\Delta)$:

$$\overline{e^2} = \int_{-\Delta}^{\Delta}e^2\frac{1}{2\Delta}\,de = \frac{\Delta^2}{3}$$

4. Effect of the output LPF The error waveform changes at the sampling rate, so its power spectral density is taken as roughly flat from 0 to $f_s$. The receiver LPF passes only $0$ to $f_M$ (message bandwidth), so

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

5. SQNR

$$\begin{aligned} \text{SQNR} &= \frac{S}{N_q} = \frac{\Delta^2f_s^2}{8\pi^2f_m^2}\cdot\frac{3f_s}{\Delta^2f_M} \\ &= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2f_M} \end{aligned}$$

With $f_m = f_M = W$:

$$\text{SQNR} = \frac{3}{8\pi^2}\left(\frac{f_s}{W}\right)^3 \approx 0.038\left(\frac{f_s}{W}\right)^3$$

Observations

• SQNR is independent of $\Delta$ (when $\Delta$ is set at the overload limit).
• SQNR $\propto f_s^3$: doubling $f_s$ improves SQNR by $10\log_{10}8 = 9$ dB.
• Compared with PCM (6 dB per bit, i.e. per unit increase of bandwidth factor), DM needs a much higher sampling rate for good quality. Example: $f_s/W = 32$ gives $0.038 \times 32\,768 \approx 1245$, i.e. about 31 dB.