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The Maximum theoretical limit of channel is governed by Nyquist theorem, which says Bth = 2 x freq in hertz x log2L ( where L = number of level) Thus we have in this case Bth= 2x6000 = 12000 bits/sec = 12 KbpsFrom ans a we found that maximum channel capacity = 39949.2 bps Thus we have to achieve 50% i.e. = 20000 bps ( roughly)And we have from the earlier expression of ans b20000 = 2x 6000 x log2LOr log2L = 1.667
Now the number of levels have to be +ve integer. So L should be taken to the nearest integer that is 4 Thus we should have 4 discreet levels.
(NB cross check achieved bit rate with 4 levels = 2 x 6000x log2(4) = 2x6000x2 = 24000 bps i.e > 50%)
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