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Calculate the maximum bit rate for a channel having b.w. 3100 Hz and S/N ratio 20 dB
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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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