A telephone channel has a BN of 3,000 Hz and SNR 20 d B. Determine the channel capacity.

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written 2.4 years ago by

prajapatijaimin • 3.3k

Solution:

$\begin{aligned}\\ &\left.\mathrm{SNR}\right|_{\mathrm{dB}}=20 \mathrm{~dB} \\\\ &10 \log (\mathrm{S} / \mathrm{N})=20 \\\\ &\mathrm{~S} / \mathrm{N}=10^2=100 \\\\ &\mathrm{C}=\mathrm{B} \log _2(1+\mathrm{S} / \mathrm{N}) \\\\ &=19974 \mathrm{bps} \cong 20 \mathrm{kbps}\\ \end{aligned}\\$

$\rightarrow$ if $S N R$ is increased to $25 \mathrm{~dB}$ . determine increased channel capacity.

$ \begin{aligned}\\ &\mathrm{SNR}_{\mathrm{dB}}=25 \mathrm{~dB} \\\\ &10 \log …

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