SQNR of delta modulation for a single-tone signal

Setup. Delta modulation uses step size $\Delta$, sampling rate $f_s = 1/T_s$ and a reconstruction LPF of bandwidth $W$ at the receiver. The message is a single tone:

$$m(t) = A_m\cos 2\pi f_m t$$

Step 1: Maximum amplitude without slope overload

The staircase can change by at most $\Delta$ in $T_s$, so the maximum slope it can follow is $\Delta/T_s = \Delta f_s$. The maximum slope of the tone is

$$\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$, so

$$A_{m,max} = \frac{\Delta f_s}{2\pi f_m}$$

Step 2: Signal power

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

Step 3: Granular (quantization) noise power

Without slope overload, the error $q(t) = m(t) - m_q(t)$ stays within $\pm\Delta$. Assuming it is uniformly distributed in $(-\Delta, \Delta)$:

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

This noise power is spread approximately uniformly over $0$ to $f_s$. The LPF passes only $0$ to $W$:

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

Step 4: SQNR

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

Step 5: In dB

Since $10\log_{10}\frac{3}{8\pi^2} = -14.2$ dB,

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

Observations:

• SQNR $\propto f_s^3$: doubling the sampling rate improves SQNR by 9 dB (compared with 6 dB per extra bit in PCM).
• SQNR does not depend on $\Delta$ (at the maximum amplitude), because the allowed amplitude and noise both scale with $\Delta$.

Example: $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz:

$$\text{SQNR} = \frac{3(64000)^3}{8\pi^2(1000)^2(4000)} = 2490 \approx 33.96\ \text{dB}$$