Asked by Learner

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

Answers

Answered by bobpursley
C=Bandwidth*log<sub>2</sub>(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.

Answered by Learner
@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?
Answered by bobpursley
I reduced it to 10 because of significant events rules.
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