Capacity=B*log2(1+SNR)= 4000 * (log(1 + (20 / 0.006)) / log(2)) = 46 812.

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?

Question Posted / zeeshan ali hirani

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Answer Posted / zeeshan ali hirani

Capacity=B*log2(1+SNR)=
4000 * (log(1 + (20 / 0.006)) / log(2)) = 46 812.7305 bps

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