Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 2 (Q.No. 9)
9.
If transmission bandwidth is doubled in FM, SNR is
Answer: Option
Explanation:
SNR changes in the ratio of square of change in bandwidth.
Discussion:
18 comments Page 1 of 2.
Sandeep kumar said:
3 years ago
B is correct and because SNR is directly proportional to (BW)^2.
Dhshsh said:
4 years ago
B is the correct answer.
Pranjal ghosh said:
4 years ago
If you see it practical situation, when we are increasing the BW the noise insertion window has now increased letting a large spectral of noise power, so SNR should decrease in that case.
Saurabh kumar said:
5 years ago
According to Shannon Hartley;
C= B log_2(1+S/N)
S/N=2^(C/B) -1.
when B is double.
i.e SNR decreases by 4 times.
C= B log_2(1+S/N)
S/N=2^(C/B) -1.
when B is double.
i.e SNR decreases by 4 times.
Manoj Mondal said:
7 years ago
According to Shannon Hartley criteria, the channel capacity is constant then it should be answer B.
Prasad said:
7 years ago
Formula is directly proportional to β square as formula =3/2(β)2.
And beta is proportional to bandwidth. Snr(output)/snr(input) is proportional to bandwidth square. Here asking in transmission side we have to find snr(input). So, it is option C.
And beta is proportional to bandwidth. Snr(output)/snr(input) is proportional to bandwidth square. Here asking in transmission side we have to find snr(input). So, it is option C.
Shreya singh said:
8 years ago
SNR is directly proportional to (B.W)^2.
Shreya singh said:
8 years ago
Option B is correct. I too agree.
Kshitij khandelwal said:
8 years ago
Yes, answer should be B. Since the S/N is directly proportional to B.W.
Ayush kumar said:
8 years ago
Answer will be B because SNR is directly proportional to square of the bandwidth.
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