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How you can Find the maximum data rate of a noiseless 4-kHz channel-using Analog encoding eg QPSK with 2 bits per sample?
Wiki User
∙ 16y ago
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.
Wiki User
∙ 16y ago
The maximum data rate of a noiseless channel can be calculated using the Nyquist formula: Data rate = 2 x bandwidth x log2(L) where L is the number of levels per sample. For QPSK with 2 bits per sample, L = 4 (since each sample can represent 2^2 = 4 levels), so the data rate would be 2 x 4 kHz x log2(4) = 16 kbps.
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