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We measure the performance of a telephone line (4 KHz of
bandwidth). When the signal is 20 V, the noise is 6 mV.
What is the maximum data rate supported by this telephone
line?
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Answer / zeeshan ali hirani
Capacity=B*log2(1+SNR)=
4000 * (log(1 + (20 / 0.006)) / log(2)) = 46 812.7305 bps
| Is This Answer Correct ? | 76 Yes | 18 No |
Answer / shikamaru
Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps
| Is This Answer Correct ? | 22 Yes | 8 No |
Answer / vikas chauhan
snr = sig power/noise power
snr = 20Vsq / 6mVsq
and use the formula
| Is This Answer Correct ? | 4 Yes | 0 No |
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