Ch3 2 Data communication and networking

Data communication and networking by neha g. kurale

1. DATA RATE LIMITS
Averyimportantconsiderationindatacommunications
ishowfastwecansenddata,inbitspersecond,overa
channel.Dataratedependsonthreefactors:
1.Thebandwidthavailable
2.Thelevelofthesignalsweuse
3.Thequalityofthechannel(thelevelofnoise)
Noiseless Channel: Nyquist Bit Rate
Noisy Channel: Shannon Capacity
Using Both Limits
Topics discussed in this section:

2. Increasing the levels of a signal
increases the probability of an error
occurring, in other words it reduces the
reliability of the system. Why??
Note

3. Capacity of a System
The bit rate of a system increases with an
increase in the number of signal levels we
use to denote a symbol.
A symbol can consist of a single bit or “n”
bits.
The number of signal levels = 2
n
.
As the number of levels goes up, the spacing
between level decreases -> increasing the
probability of an error occurring in the
presence of transmission impairments.

4. Nyquist Theorem
Nyquist gives the upper bound for the bit rate
of a transmission system by calculating the
bit rate directly from the number of bits in a
symbol (or signal levels) and the bandwidth
of the system (assuming 2 symbols/per cycle
and first harmonic).
Nyquist theorem states that for a noiseless
channel:
C = 2 B log
22
n
C= capacity in bps
B = bandwidth in Hz

5. DoestheNyquisttheorembitrateagreewiththe
intuitivebitratedescribedinbasebandtransmission?
Solution
Theymatchwhenwehaveonlytwolevels.Wesaid,in
basebandtransmission,thebitrateis2timesthe
bandwidthifweuseonlythefirstharmonicintheworst
case.However,theNyquistformulaismoregeneralthan
whatwederivedintuitively;itcanbeappliedtobaseband
transmissionandmodulation.Also,itcanbeapplied
whenwehavetwoormorelevelsofsignals.
Example 3.33

6. Consideranoiselesschannelwithabandwidthof3000
Hztransmittingasignalwithtwosignallevels.The
maximumbitratecanbecalculatedas
Example 3.34

7. Considerthesamenoiselesschanneltransmittinga
signalwithfoursignallevels(foreachlevel,wesend2
bits).Themaximumbitratecanbecalculatedas
Example 3.35

8. Weneedtosend265kbpsoveranoiselesschannelwith
abandwidthof20kHz.Howmanysignallevelsdowe
need?
Solution
WecanusetheNyquistformulaasshown:
Example 3.36
Sincethisresultisnotapowerof2,weneedtoeither
increasethenumberoflevelsorreducethebitrate.Ifwe
have128levels,thebitrateis280kbps.Ifwehave64
levels,thebitrateis240kbps.

9. Shannon’s Theorem
Shannon’s theorem gives the capacity
of a system in the presence of noise.
C = B log
2(1 + SNR)

10. Consideranextremelynoisychannelinwhichthevalue
ofthesignal-to-noiseratioisalmostzero.Inother
words,thenoiseissostrongthatthesignalisfaint.For
thischannelthecapacityCiscalculatedas
Example 3.37
Thismeansthatthecapacityofthischanneliszero
regardlessofthebandwidth.Inotherwords,wecannot
receiveanydatathroughthischannel.

11. Wecancalculatethetheoreticalhighestbitrateofa
regulartelephoneline.Atelephonelinenormallyhasa
bandwidthof3000.Thesignal-to-noiseratioisusually
3162.Forthischannelthecapacityiscalculatedas
Example 3.38
Thismeansthatthehighestbitrateforatelephoneline
is34.860kbps.Ifwewanttosenddatafasterthanthis,
wecaneitherincreasethebandwidthofthelineor
improvethesignal-to-noiseratio.

12. Thesignal-to-noiseratioisoftengivenindecibels.
AssumethatSNR
dB=36andthechannelbandwidthis2
MHz.Thetheoreticalchannelcapacitycanbecalculated
as
Example 3.39

13. Forpracticalpurposes,whentheSNRisveryhigh,we
canassumethatSNR+1isalmostthesameasSNR.In
thesecases,thetheoreticalchannelcapacitycanbe
simplifiedto
Example 3.40
Forexample,wecancalculatethetheoreticalcapacityof
thepreviousexampleas

14. Wehaveachannelwitha1-MHzbandwidth.TheSNR
forthischannelis63.Whataretheappropriatebitrate
andsignallevel?
Solution
First,weusetheShannonformulatofindtheupper
limit.
Example 3.41

15. TheShannonformulagivesus6Mbps,theupperlimit.
Forbetterperformancewechoosesomethinglower,4
Mbps,forexample.ThenweusetheNyquistformulato
findthenumberofsignallevels.
Example 3.41 (continued)

16. The Shannon capacity gives us the
upper limit; the Nyquist formula tells us
how many signal levels we need.
Note