## Answer 5.8

#### The theoretical maximum channel capacity for a bandpass PSK is identical to that for an ASK channel and is given by:
Cbandpass = B log2M,
For a non brick-wall filter, this expression must be modified to allow for the extra bandwidth of a practical filter shape as follows:
Cbandpass = (B log2M) / (1 + a)
Now M = 2 for binary signalling, thus:
C = (8000 x 1) / (1 + 1) = 4000 bps.