Discussion
Twenty-four channels each band-limited to \( 4 \ \mathrm{kHz} \) are time division multiplexed. The minimum pulse rate is
Answer: C
Each channel needs its Nyquist rate and the slots share one train:
\[ 24 \times 8000 = 192 \ \mathrm{kHz} \]
giving 192 kHz, before any allowance for synchronisation. 96 kHz uses the signal bandwidth rather than twice it. 384 kHz doubles unnecessarily. And 48 kHz is far short. Multiplying the per-channel rate by the channel count is the first line of every multiplex calculation, and overhead is then added to it.
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