Subject: Computer Engineering

Topic: Electronic Circuits and Communication Fundamentals

Difficulty: Medium / High

Given B = 3100 Hz

S/N = 20 dB

But 20 dB = 10 log (S/N)

∴ S/N = 100

The maximum bit rate: -

$R_{\max }=B \log_2 \big[1+\frac{S}{N}\big] \\ =3100 log_2[1+100 ] \\ =\frac{(3100 \log_{10}101 )}{log_{10}2 } =20,640 bits/sec$

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