Given,
Bandwidth=2.3Mhz=2.3×106hz
(A) When SNRdb=12
As we know
SNRdb=10log10(SNR)
SNR=101.2
=15.84
So Maximum rate=Bandwidth×log2(1+SNR)
=2.3×log2(1+15.84)
= 2.3 log2(16.84)
=4.073Mbps
(b) when SNRdb=18
SNRdb=10×log10(SNR)
SNR=101.8
=63.09
So Maximum rate=Bandwidth×log2(1+SNR)
=2.3×log2(1+63.09)
=2.3×log2(64.09)
=6mbps
(c) When SNRdb=25
SNRdb=10×log10(SNR)
SNR=102.5
=316.22
So Maximum rate=Bandwidth×log2(1+SNR)
=2.3×log2(1+316.22)
=2.3×log2(317.22)
=8.30mbps
Tradeoff factor:-For digital communication between two geographical apart stations the quality of received signals at the receiver end is an important factor. To achieve the desired quality of received
signal at receiver end We can have a trade-off between major perimeters .i.e. Bandwidth, bit width and hardware complexity.
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Thanks for your help
Thank you so much, I really appreciate.
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