Correct Answer - Option 1 : 20 Mbps
Concept:
The capacity of a band-limited AWGN channel is given by the formula:
\(C = B{\log _2}\left( {1 + \frac{S}{N}} \right)\)
C = Maximum achievable data rate (in bits/sec)
B = channel bandwidth
\(\frac{S}{N}\) = Signal to Noise power
Note: In the expression of channel capacity, S/N is not in dB.
Calculation:
Given B.W. = 4 MHz, S/N = 31
The channel capacity will be:
\(C = (4\times 10^6)~{\log _2}\left( {1 + 31} \right)\)
C = 20 Mbps