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).
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.
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?
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.