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Q1:

An 8 level encoding scheme is used in a PCM system of 10 kHz channel BW. The channel capacity is

A 80 kbps

B 60 kbps

C 30 kbps

D 18 kbps

ANS:B - 60 kbps

C = 2B log2 M; M = 8 = 23, log2 M = 3 bits = 2 x 10 x 103 x 3 = 60 kbps.



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