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1. Maximum data rate of a channel for a noiseless 3-kHz binary channel is
So, data rate = 2*3000 log22 bps = 6000 bps.
Solution:
Maximum data rate = 2Hlog2V bps, where H is the bandwidth, V is the discrete levels. Here H is 3 kHz and V is 2.So, data rate = 2*3000 log22 bps = 6000 bps.