Question

In a certain telemetry system, there are four analog signals, in which one signal is bandlimited to 3.6 kHz and the remaining three signals are bandlimited to 1.5 kHz. If these signals are transmitted through an 8-bit PCM system using time division multiplexing, then the minimum channel bandwidth required will be __________kHz.

• A. 64.8

Answer 1

The correct option is A 64.8
Nyquist rate
$3.6kHz\times 2=7.2kHz$

$1.5kHz\times 2=3kHz$

$7200=12\times 600$

$3000=5\times \times 600$

So, minimum sampling rate can be achieved by assingning 12 time slot for signal having bandwidth of 3.6 kHz. 5 time slots for each signals having bandwidth of 1.5 kHz and rotating the commutator with a speed of 600 rotations/second.

So, $f_s=(12+5+5+5)600=16,200$ samples/second

bit rate $R_b=n{f}_s=8\times 16.2kbps=129.6kbps$

$(BW{)}_{min}=\frac{R_b}{2}=64.8kHz$