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A telephone channel has a BN of 3,000 Hz and SNR 20 d B. Determine the channel capacity.
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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 (\mathrm{S} / \mathrm{N})=25 \\\\ &\mathrm{~S} / \mathrm{N}=10^{2.5}=316.2 \\\\ &\mathrm{C}=\mathrm{Blog}_2(1+\mathrm{S} / \mathrm{N})=24925 \mathrm{bps} \\\\ &\cong 25 \mathrm{kbps}\\ \end{aligned}\\ $

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