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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:

SNR|dB=20 dB10log(S/N)=20 S/N=102=100C=Blog2(1+S/N)=19974bps20kbps

if SNR is increased to 25 dB. determine increased channel capacity.

SNRdB=25 dB10log(S/N)=25 S/N=102.5=316.2C=Blog2(1+S/N)=24925bps25kbps

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