Delta modulator numerical (3 kHz, 10 kHz sampling)

Given: $f_m = 3$ kHz, $f_s = 10$ kHz, $A_{max} = 1$. Assume output LPF bandwidth $f_M = f_m = 3$ kHz.

i) Minimum step size to avoid slope overload

The staircase slope $\Delta f_s$ must be at least the maximum signal slope $2\pi f_mA$:

$$\begin{aligned} \Delta_{min} &= \frac{2\pi f_mA_{max}}{f_s} \\ &= \frac{2\pi \times 3000 \times 1}{10\,000} = 1.885\ \text{V} \end{aligned}$$

ii) SQNR for a sinusoidal signal

Signal power at maximum amplitude $S = A^2/2 = \dfrac{\Delta^2f_s^2}{8\pi^2f_m^2}$; granular noise after the LPF $N_q = \dfrac{\Delta^2}{3}\dfrac{f_M}{f_s}$:

$$\begin{aligned} \text{SQNR} &= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2f_M} = \frac{3}{8\pi^2}\left(\frac{10}{3}\right)^3 \\ &= 0.0380 \times 37.04 = 1.407 \\ \text{SQNR}_{dB} &= 10\log_{10}1.407 = 1.48\ \text{dB} \end{aligned}$$

This very low value shows that $f_s = 10$ kHz is far too low for DM; DM needs $f_s$ many times the Nyquist rate.

iii) Minimum transmission bandwidth

One bit per sample: $R_b = f_s = 10$ kbps.

$$B_{T,min} = \frac{R_b}{2} = 5\ \text{kHz}$$

Answer: $\Delta_{min} = 1.885$ V; SQNR = 1.407 (1.48 dB); $B_T = 5$ kHz (10 kHz if $B_T = R_b$ is taken).