digital communication

1) 16-level signalling is equal to 4 bits per sample, hence the sampling rate is:

5/4 = 1.25 MHz

So the theoretical min. system bandwidth, by Nyquist theorem is given by:

1.25/2 = 0.625 MHz

2) Excess bandwidth = 0.8 - 0.625 = 0.175 MHz

Now, the 2T (period of Nyquist rate) is 1/(1.25 x 106) s

So the roll-off factor is given by:

(0.175 x 106)/(1.25 x 106) = 0.14