LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
in General by (117k points)
closed by
The channel capacity under the Gaussian noise environment for a discrete memoryless channel with a bandwidth of 4 MHz and SNR of 31 is.
1. 20 Mbps
2. 4 Mbps
3. 8 kbps
4. 4 kbps

1 Answer

0 votes
by (59.8k points)
selected by
Best answer
Correct Answer - Option 1 : 20 Mbps


The capacity of a band-limited AWGN channel is given by the formula:

\(C = B{\log _2}\left( {1 + \frac{S}{N}} \right)\)

C = Maximum achievable data rate (in bits/sec)

B = channel bandwidth

\(\frac{S}{N}\) = Signal to Noise power

Note: In the expression of channel capacity, S/N is not in dB.


Given B.W. = 4 MHz, S/N = 31

The channel capacity will be:

\(C = (4\times 10^6)~{\log _2}\left( {1 + 31} \right)\)

C = 20 Mbps

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.