A noisy channel which has a bandwidth of 400KHz and theoretical channel capacity of 4.01Mbps.

Question

According to Shannon Capacity formula:
a. Calculate the Signal Power received at the receiver end if the Noise Power is .001 Watt.
b. Calculate the signal to noise ratio of the channel in dB (SNRdB).

Answer 1

C=Bandwidth*log₂(1+S/N)

4.01e6=400e3log2(1+S/N)
10=log2(1+S/N)
1024=1+S/N
S/N=1024
S=1023*.001 watts= about 1 watt

db gain= 10 log (S/N)=30 db

checkthis, it has been 30 years since I used Shannon.

Answer 2

@bobpursley You did:
4.01e6=400e3log2(1+S/N)
10=log2(1+S/N)
Now question is that after solving C and B i.e., C/B we get following result:
=4010000/400000
=401/40
=10.025
But you wrote only 10 here. Question is why?

Answer 3

I reduced it to 10 because of significant events rules.