0 votes 0 votes For a channel of $3 \mathrm{KHz}$ bandwidth and signal to noise ratio of $30 \mathrm{~dB}$, the maximum data rate is: $3000 \mathrm{bps}$ $6000 \mathrm{bps}$ $15000 \mathrm{bps}$ $30000 \mathrm{bps}$ Others ugcnetcse-june2008-paper2 + – admin asked Jan 6 • edited Jan 7 by makhdoom ghaya admin 328 views answer comment Share Follow See 1 comment See all 1 1 comment reply shwetah commented Sep 29, 2017 reply Follow Share From my point of view shannon's theorem is c=B*log2(1+SNR) so it's 3*10^3*log2 (1+30) = so it's nearest to 3*10^3 * 5 = 15000 bps.. so answer is "C". Correct if i'm wrong 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes Applying Shannon's Theorem we get answer as D amaity answered Sep 29, 2017 • edited Sep 29, 2017 by amaity amaity comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes C) 151000 bps Neeraj Chandrakar answered Oct 16, 2017 Neeraj Chandrakar comment Share Follow See all 0 reply Please log in or register to add a comment.