Consideranoiselesschannelwithabandwidthof3000
Hztransmittingasignalwithtwosignallevels(foreach
level,wesend1bits).Calculatethemaximumbitrate.
Example 3.34
We know for noiseless channel , Channel capacity or bit rate should be calculated
using the formula below:
Weneedtosend265kbpsoveranoiselesschannelwith
abandwidthof20kHz.Howmanysignallevelsdowe
need?Commentonyouranswer.
Solution
WecanusetheNyquistformulaasshown:
Example 3.36
L=2^(C/2B)
This is not a valid number. So possible values of n= 6 or 7
If n=6 then L=64, and C= 240 kbps
and if n=7 then L=128, C= 280kbps
Shannon’s Theorem
Shannon’s theorem gives the capacity of a
system in the presence of noise.
C = B log
2(1 + SNR)
Consideranextremelynoisychannelinwhichthevalue
ofthesignal-to-noiseratioisalmostzero.Inother
words,thenoiseissostrongthatthesignalisfaint.For
thischannelthecalculatethecapacity.
Example 3.37
Thismeansthatthecapacityofthischanneliszero
regardlessofthebandwidth.Inotherwords,wecannot
receiveanydatathroughthischannel.
Wecancalculatethetheoreticalhighestbitrateofa
regulartelephoneline.Atelephonelinenormallyhasa
bandwidthof3000.Thesignal-to-noiseratioisusually
3162.Forthischannelcalculatethechannelcapacity.
Example 3.38
Thismeansthatthehighestbitrateforatelephoneline
is34.860kbps.Ifwewanttosenddatafasterthanthis,
wecaneitherincreasethebandwidthofthelineor
improvethesignal-to-noiseratio.