Question

The capacity of a Gaussian Channel is given by

1) C= 2 B (1+S/N) bits/s
2) C= B (1+S/N)^2 bits/s
3) C = B^2 (1+S/N) bits/s
4) C = B ( 1 + S/N) bits/s

Answer 1

Shannon-Hartley theorem says that:

Channel Capacity is  C = B Log_2 (1 + S/N)  bits/sec
   This is the theoretical limit on the transmission rate on the analog Gaussian channel.

B = Bandwidth of the channel
S = Average Signal power 
N = Average Additive Gaussian Noise power in the channel

Answer 2

Answer:

The capacity of a Gaussian Channel is given by $C=B log ({ 1+\frac{S}{N} ) bits/sec$

Explanation:

• The information capacity of Gaussian channel is

                            $C=\frac{1}{2} log ({ 1+\frac{S}{N} ) bits/sample$

       Where S is the signal power and N is the noise variance

• If the noise variance is zero, the capacity of the channel is infinite.
• For an analog signal sampled at the Nyquist rate, then the sampling frequency is $f= 2 B$. Then

                         $C=\frac{1}{2} log ({ 1+\frac{S}{N} )*2B =B log ({ 1+\frac{S}{N} ) bits/sec$

      where B is the bandwidth of the signal