Correct Answer - Option 2 : 30 Kbps
Concept:
Shannon’s channel capacity is the maximum bits that can be transferred error-free. Mathematically, this is defined as:
\( C = B~{\log _2}\left( {1 + \frac{{\rm{S}}}{{\rm{N}}}} \right){\rm{bits}}\)
B = Bandwidth of the channel
\(\frac{S}{N}=\) Signal to noise ratio
Note: In the expression of channel capacity, S/N is not in dB.
Calculation:
Given B = 3 kHz and SNR = 1023
Channel capacity will be:
\({\rm{C}} = 3~{\log _2}\left( {1 + 1023} \right){\rm{kbits}}/{\rm{sec}}\)
C = 30 kbps