Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 2 (Q.No. 1)
1.
For an ideal 3000 Hz channel, S/N ratio is
Answer: Option
Explanation:
= 26000/3000 - 1 = 3.
Discussion:
15 comments Page 1 of 2.
Saranya said:
9 years ago
Accoding to shannon's theorm, C = B * log_2 (1+s/n).
From that formula we can derive s/n as,
s/n = 2^(C/B) - 1.
In this case C = 2 * fm.
So, s/n = 2^(2fm/B) - 1.
From that formula we can derive s/n as,
s/n = 2^(C/B) - 1.
In this case C = 2 * fm.
So, s/n = 2^(2fm/B) - 1.
(7)
KCT said:
6 years ago
In sampling, the Nyquist rate is used to avoid alias or fold-over distortion. Fs = 2Fmax.
Nyquist rate is used along with hartley and Shannon's hartley theorem.
Nyquist rate is used along with hartley and Shannon's hartley theorem.
Rhea said:
6 months ago
This is only applicable to 1000 Hz what if the question is for let's say 2 Hz , then?
Anyone, please explain it to me?
Anyone, please explain it to me?
Hitesh said:
1 decade ago
If S/N = 2^B.W/F-1. Then it will always be 3. Can anyone tell me what is reason for this?
Anil kumar said:
9 years ago
Please explain the question in an easy method. I can't understand the question.
Asha said:
10 years ago
This solution is difficult to understand. Can anyone explain this?
Sankar said:
9 years ago
@Saranya.
You gave a best explanation, thank you.
You gave a best explanation, thank you.
Hitesh said:
1 decade ago
Is this s/n ratio is only for ideal channel?
Shiva said:
1 decade ago
His is valid for ideal case or any case?
Santosh said:
1 decade ago
What is the 6000 used in the solution?
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