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Calculate the minimum bit rate for a channel having b.w. 3100 Hz and S/N ratio 10 dB
1 Answer
written 6.1 years ago by |
Given: - B = 3100 Hz
$\big(\frac{S}{N}\big)dB=10$
$∴10=10 log_{10}big(\frac{S}{N}\big) \\ ∴\frac{S}{N}=10$
∴ Maximum bit rate
$= R_{\max} = B \log_2 \big[1+\frac{S}{N}\big] \\ = 3100 \log_2(1+10) \\ =\frac{3100 \log_{10}11}{\log_{10}2} \\ = 10,724 bits/sec$