Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps

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 Posted / shikamaru

Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps

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