TDM of 24 voice signals: PAM bandwidth and PCM sampling rate

Given: $N = 24$ voice channels, $f_m = 3.4$ kHz.

(i) PAM-TDM, Nyquist sampling

Nyquist rate for each channel:

$$f_s = 2 f_m = 2 \times 3.4 = 6.8\ \text{kHz}$$

Total pulse rate of the TDM signal:

$$r = N f_s = 24 \times 6.8\ \text{kHz} = 163.2\ \text{k pulses/s}$$

The minimum channel bandwidth needed for TDM-PAM is half the pulse rate. This is the Nyquist bandwidth, the same as $N f_m$:

$$\begin{aligned} B_{min} &= \frac{N f_s}{2} = \frac{24 \times 6.8}{2} \\ &= 81.6\ \text{kHz} \;(= 24 \times 3.4\ \text{kHz}) \end{aligned}$$

Answer (i): Minimum channel bandwidth = 81.6 kHz.

(ii) PCM with 8-bit encoder, $R_b = 1.5 \times 10^6$ b/s

Each frame has one 8-bit sample from each channel:

$$R_b = N \, n \, f_s$$ $$\begin{aligned} f_s &= \frac{R_b}{N n} = \frac{1.5 \times 10^6}{24 \times 8} \\ &= \frac{1.5 \times 10^6}{192} = 7812.5\ \text{Hz} \end{aligned}$$

Check: $7812.5 > 6800$ Hz (the Nyquist rate), so there is no aliasing, with a guard band of $7812.5 - 6800 = 1012.5$ Hz.

Answer (ii): Sampling rate $f_s = 7812.5$ samples/s ≈ 7.81 kHz.