QAM -Lec16 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Cummenication system

1. Dr. Suad El-Geder
EC 331
Computer Engineer Department
QAM
LECTURE 16

2. ▪M-aryModulation:TransmittingMoreBitsPerSymbol
▪M-arymodulation(pronounced"M-airy")isafundamentalconceptindigital
communicationthatreferstoanymodulationschemewhereasinglesymbolcanrepresent
morethanonebitofinformation.
▪The"M"inM-arystandsforthenumberofdistinctsymbolsthatthemodulationscheme
canuse.Sinceeachsymbolcarriesacertainnumberofbits,increasingMallowsfora
higherdatarateforagivenbandwidth.

3. ▪TounderstandM-ary,it'scrucialtodistinguishbetweenabitandasymbol:
oBit:Abinarydigit(0or1),thesmallestunitofdigitalinformation.
oSymbol:Adistinctwaveform,state,orsignalcharacteristic(e.g.,aspecificamplitude,phase,orfrequency)that
representsagroupofbits.
▪In M-arymodulation, each symbol represents kbits, where M=2
??????
. Therefore, k=log
2??????.
▪Examples:
oIf M=2 (e.g., BPSK, Binary PSK), then k=log
22​=1 bit per symbol. This is the simplest case, often not explicitly
called M-ary, but it technically is M=2.
oIf M=4 (e.g., QPSK, Quadrature PSK), then k=log
24=2 bits per symbol.
oIf M=8 (e.g., 8-PSK, 8-FSK, or some 8-QAM), then k=log
28​=3 bits per symbol.

4. ▪TheprimarymotivationsforusingM-arymodulationschemesare:
1.IncreasedSpectralEfficiency:Foragivenbandwidth,M-arymodulationallowsyoutotransmit
morebitspersecond.Sinceeachsymbolcarriesmorebits,thesymbolrate(symbolspersecond)can
belowerthanthebitrate,whichtranslatestoanarrowerbandwidthrequirementforagivenbitrate.
oAnalogy:Imagineatrain.Ifeachtraincar(symbol)carriesonlyonepassenger(bit),youneedmanycars.
Ifeachcar(symbol)cancarry8passengers(bits),youneedfewercarsforthesamenumberofpassengers,
andthusashortertrain(lessbandwidth).
2.HigherDataRates:Conversely,forafixedbandwidth,M-arymodulationdirectlytranslatestohigher
achievabledatarates(bitspersecond).Thisiscrucialformodernapplicationslikehigh-definition
videostreaming,5Gwirelesscommunication,andfiberopticnetworks.

5. ▪QuadratureAmplitudeModulationisamethodofamplitudemodulationthatallowstwodifferentsignalssent
simultaneouslyonthesamecarrierfrequency,effectivelydoublingthebandwidththatcanbetransmitted.
▪Bymodulatingtwocarriersatexactlythesamefrequency,butshiftedby90◦,boththeamplitude(ASK)and
phase(PSK)ofthecarrierismodulated:QAM.
▪NobinaryQAMmethods(M=2
1
=2),butlotsofhigher-orderQAM:
1)4QAM(M=2
2
=4signallevels;2bitspersymbol,likeQPSK)
2)16QAM(M=2
3
=8signallevels;3bitspersymbol)
3)16QAM(M=2
4
=16signallevels;4bitspersymbol)
4)64QAM(M=2
6
=64signallevels;6bitspersymbol)
5)256QAM(M=2
8
=256signallevels;8bitspersymbol)...etc.

7. ▪Baudreferstotherateofchangeofasignalonthetransmissionmediumafterencodingandmodulation
haveoccurred.
▪Hence,baudisaunitoftransmissionrate,modulationrate,orsymbolrateand,therefore,thetermssymbols
persecondandbaudareoftenusedinterchangeably.
▪Mathematically,baudisthereciprocalofthetimeofoneoutputsignalingelement,andasignalingelement
mayrepresentseveralinformationbits.Baudisexpressedas
baud =
1
�
??????
✓wherebaud=symbolrate(baudpersecond).??????
�=timeofonesignalingelement(seconds).

8. ▪InformationCapacity,Bits,andBitRate:
owhereI=informationcapacity(bitspersecond)
oB=bandwidth(hertz)
o=transmissiontime(seconds)
▪FromEquation1,itcanbeseenthatinformationcapacityisalinearfunctionofbandwidthand
transmissiontimeandisdirectlyproportionaltoboth.
▪Ifeitherthebandwidthorthetransmissiontimechanges,adirectlyproportionalchangeoccursinthe
informationcapacity.
▪Thehigherthesignal-to-noiseratio,thebettertheperformanceandthehighertheinformationcapacity.
Equation 1

9. ▪Mathematically stated, the Shannon limit_forinformation capacity is:
owhere I = information capacity (bps)
oB = bandwidth (hertz)
o
??????
??????
= signal-to-noise power ratio (unitless)
▪In addition, since baud is the encoded rate of change, it also equals the bit rate divided by the number of
bits encoded into one signaling element. Thus,
oWhere, ??????
??????=channelcapacity(bps)
oN is the number of bits encoded into each signaling element.

11. ▪Example:Inadigitaltelephoneexchangevoicesignalisdigitalizedusing8-bitsPCM.Calculatefinal
bitrateofsystem.Voice:300HZ–3.4KH.
▪fs≥2fmsamples/sec
Criticalrate/Nyquistrate:fs=2fm
⇒fs=2×3.4k=6800samples/sec
Bitrate:??????
�=n.fsbits/sec
=8bits/sample×6800samples/sec=54400bits/sec

12. 2
Note:
ForbinaryPCM(assumingaNyquistpulseshape,
whichistypicalfortheoreticalcalculationsunless
otherwisespecified),theminimumtransmission
bandwidthrequiredishalfofthebitrate.Thisis
alsoknownastheNyquistbandwidthorminimum
bandwidthforbasebandtransmission.

14. ▪Example 4: From the plot in the figure shown:
Solution:

16. Example :
Solution:

17. Homework

20. Question 3:

21. Question :
Solution:

22. Question :
Solution:

23. Question 6:
Solution: