How you can Find the maximum data rate of a noiseless 4-kHz channel-using Analog encoding eg QPSK with 2 bits per sample?

Answer:

Answer

for 4KHz then for noisy channel using Shannon theorem, sampling rate will be 8K samples/sec. So with 2 bit encoding, 2 bits are sent per sample. So the data rates is 8000 samples / sec * 2 bits = 16000bits / sec = 16kbps.

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