Question
Download Solution PDFA voice signal band limited to 3.4 kHz is sampled at 8 kHz and pulse code modulated using 64 quantization levels. Ten such signals are time division multiplexed using on 5-bit synchronising word. The minimum channel band width will be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In time division multiplexing bandwidth (Rb)is given as:
Rb = (nN+a)Fs ;
where n is no of bits,
N is no of signals multiplexed,
a is synchronized bit per frame
Fs is sampled frequency
Calculation:
Quantization level 64 is given so,
n =
N = 10; a = 5; Fs = 8 kHz
Rb = (nN+a)Fs
= (6 × 10 + 5) × 8000
= 520 kHz
correct option is 4
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