Chapter 14

1 Chapter14 MODULATIONINTRODUCTIONA swehaveseeninpreviousthreechapters, ,thesourceinformationisimpresseduponacar rier-wave(essentiallyasinusoidofacertain frequency) ( ,audio,voltagepulsetraincarryingdigitali nformation) forexample,therepresentationofsampledsig nalsbytheamplitude, (2,6,9)thatwecanwritethissinusoidalcarri ersignalas:c(t)=Acos(2 fct+ )( )Here,Aiscalledtheamplitudeand ,dataareusedtomodulateorchangeitsamplitu de,frequency, , (digitalmodulation) ,wealsodescribeaspecialcaseofdigitalmodu lationthatisveryimportantfortransmission ofinformationusingmodems quadratureamplitudemodulation(QAM) atechniquecalled675676 ThePhysicalLayerofCommunicationsSystemso rthogonalfrequencydivisionmultiplexing(O FDM). , , (smallerwavelengths) , , ,thistimetoanintermediatefrequency(IF) (multiplexing). , ,thecharacteristicsofthemodulatedsinusoi d(suchasamplitude,frequencyorphase) (AM)andfrequencymodulation(FM) , , , EncoderChannel EncoderModulatorChannelDe-modulatorChann el DecoderSource [1].

2 ! ! , , , , (1G)oranalogcellularsystemsintheUSA, ,webrieflyconsideredamodelofadigitalcomm unicationsystemwherewehaveasource,asourc eencoder,andachannelencoderonthetransmit tersideandthecorrespondingchanneldecoder , , , ,ifthealphabetisbinary,thetwopossiblesym bolsare0and1andinformationissimplyalongs equenceof0 sand1 binary digitalsignalrepresentsthe zerosymbol usingaspecificsignalthatlastsforaduratio nofTssecondsandthe onesymbol , , ,mostsystemsareconstructedsuchthatM= , ,ifm1,m2,m3,andm4arethesymbolsofa4-arysy stem,wecanassociatethe dibit 00tom1,01tom2, (expressedinunitsofbaud).Thebitrate(ifM= 2k) s,thenthesymbolrateis1 (k=4oritisaM=16-aryalphabet) ,themessageismappedtotheamplitude,freque ncy,phase(oracombinationofthese) , , , , , (QAM).Inanalogmodulationwewereinterested intheSNR,butrecallfromChapter10thatindig italmodulationweareinterstedinthebiterro rrate(BER)asafunctionoftheratiooftheener gyperbit(Eb)tothevalueofthenoisePSD(N0) ,likeanalogmodulationwewouldliketomaximi zetheefficiencywithwhichweusetheavailabl ebandwidthindigitalmodulationwequantifyt hespectralefficiencyfortheamountofbandwi dthWrequiredtotransmitatagivendatarateRq uantifiedas = , ,therearetradeoffsbetweentheBERforagiven EbN0and.

3 690 ThePhysicalLayerofCommunicationsSystemst imetimeAmplitudeAmplitude100% Depth50% Depth0timeAmplitudetimeAmplitude10010010 0100 BasebandBasebandTS2TS3 , signalconstellation. Withasignalconstellation, ,thealphabethastwovalues 0 and 1. InASK,a 0 ismappedtooneamplitudevalueanda 1 ,a 0 ismappedtoonefrequencyvalueanda 1 ,a 0 ismappedtoonephasevalueanda 1 , (BASK).InBASK, ,si(t)=Aicos(2 fct+ ),0 t Tsfori=1,2( )Thetransmitterwilltransmits1(t)whentheb itiszeroands2(t) ,werefertothemodulationschemeashaving100 % (wherethetwoamplitudevaluesareAand0) ,wesaythatthemodulationdepthis50%. , zero bitisgivenby:Ezero= Ts0s21(t)dt=A21 Ts0cos2(2 fct+ )dt=A212 Ts0[1+cos(2 (2 fc)t+2 )]dt A212Ts( )Theapproximationisanequalityiffc=kTsand isacloseapproximationiffc$ one bitissimilarlyequalto:Eone=A222Ts( )Theaverageenergyperbitisgivenby:Eb,av=T s4[A21+A22]( )Hereweassumethatthenumberof 0 sandthenumberof 1 ,theaverageenergyperbitcanbecalculatedto beEb=A2Ts4andEb= , ( ).

4 Thebasebandsignalcanberecoveredattherece iverusingthesametechniquesasAM(enveloped etectionorcoherentdetection). , , someexamplesaretelevisionremotes, (BFSK).InBFSK, ,si(t)=Acos(2 fit),0 t Tsfori=1,2( ) (t)whenthebitiszeroands2(t) (t)ands2(t)whichinvolvesmultiplicationof thetwosignalsandintegrationoveronesymbol periodgivenby: Ts0s1(t)s2(t)dt=A2 Ts0cos(2 f1t)cos(2 f2t)dt=A22 Ts0[cos(2 (f1 f2)t)+cos(2 (f1+f2)t)]dt( )Theterm cos(2 (f1+f2)t) cos(2 (f1 f2)t) ,orthogonalFSKensuresthatthereisnocorrel ationbetweens1(t)ands2(t),thatis: Ts0s1(t)s2(t)dt=0( )InthecaseoforthogonalFSK,f1 f2=1 Tssothattheintegrationin cos(2 (f1 f2)t) (OFDM)lateroninthischapter( ).AswesawinthecaseofBASK,theenergyperbit inBFSK(forbotha 0 anda 1 ) , , ,anFSKsignal,likeFM, (Chapter10).RecallthataManchesterpulseco nsistsoftwo half (liketheAdvancedMobilePhoneSystem AMPS).Recently,FSKhasfoundapplicationsin lowpowerwirelessnetworkslikeBluetooth, (BPSK).

5 InBPSK, ,si(t)=Acos(2 fct+ i),0 t Tsfori=1,2( )Thetransmitterwilltransmits1(t)whentheb itiszeroands2(t) radians,themaximumpossiblephasedifferenc ebetweenthetwobitsis .Itiscommontoassumethat 1=0and 2= inwhichcase,thetwosignalswillbe:s1(t)=Ac os(2 fct),0 t Tss2(t)=cos(2 fct+ )= Acos(2 fct),0 t Ts( ) ,wecanseethatthereisareversalofphasewhen thebitchangesandso, (t)= s2(t)andwecanviewBPSKasBASK whereA1=AandA2= , ,BPSK isequivalenttoantipodalorbipolarsignalin gwithnon-return-to-zero(NRZ) , ,BPSK signals,likephasemodulation, , , ,althoughdifferentmodulationschemesareus eddependingonthetransmissionrate, dualBPSK fortransmissionsfromthecellphonetowertom obiledevices(downlink).Here,therearetwoB PSK signals, ,thesourceproducesoneofMsymbolsmifori=1, 2,3, , (t) , :si(t)=Aicos(2 fct+ ),0 t Tsfori=1,2,3, ,M( ) (2i 1 M) , , (inthiscased) 3, 1, :s1(t)=cos(2 fct+ ),0 t Tss2(t)= cos(2 fct+ ),0 t Tss3(t)=3cos(2 fct+ ),0 t Tss4(t)= 3cos(2 fct+ ),0 t TsNotethatonceagain, , :si(t)=Acos[2 fct+2 (i M2)!]

6 Ft],0 t Tsfori=1,2,3, ,M( )TheMfrequencieswillbefc+(i M2)!ffori=1,2,3, , ! , :si(t)=Acos(2 fct+ i),0 t Tsfori=1,2,3, ,M( )Thephase iistypicallygivenby i=2 M(i 1)+ (CDMA anddigitalTDMAinNorthAmerica). , 2, ,3 4,thefourphaseswillbe 4,3 4,5 4and7 4-QPSK,avariationofQPSK,thesymbolsarepic kedalternativelyfromthesetwoschemes(cons tant=0andconstant= 4) (multiplepositiveandnegative)amplitudesa reusedwiththetwophase-shiftedcarriers,th emodulationschemeiscalledquadratureampli tudemodulation(QAM). :si(t)=Ai,Icos(2 fct)+Ai,Qsin(2 fct),0 t Tsfori=1,2,3, ,M( ) :si(t)=Aicos(2 fct+ i),0 t Tsfori=1,2,3, ,M( )696 ThePhysicalLayerofCommunicationsSystemsw hereAi= A2i,I+A2i,Qand i= tan 1(Ai,QAi,I).Soitispossibleforustothinkof QAMasamixofbothamplitudeandphaseshiftkey ingsincethemessagemiismappedtoacarrierwi thamplitudeAiandphase ,itiscommoninM-QAMtopickthein-phaseandqu adrature-phaseamplitudessuchthattheyareo ftheform(2i 1 M) , ,recently, , , [1],[2].

7 Inmostcases,weassumethatthereceivedsigna lisonlycorruptedbyadditivewhiteGaussiann oise(AWGN) (t)forsomei 1,2,3, ,M,thereceivedsignalwillbe:r(t)=si(t)+n( t),0 t Ts( )Thegoalofthereceiveristodeterminewhatsi (t)wastransmittedgiventhatr(t) (t)was, ,r(t)iscorruptedbynoiseanditispossibleth atthereceiverwillsometimesdeterminethatt hetransmittedsignalwassj(t)wherej(=iwhen si(t) (Eb) (100% modulation ) 0 andabit 1. ,eventhevisualdifferencebetweenthe 0 bitandthe 1 ,whoafteralldonotsensevoltagesthatwell:- ).Asthereceivedsignalgetsnoisier, Tx signal with two bitsBASK Rx signalBASK Rx signalBASK Rx signalBASK Rx signalBASK Tx signal with two Rx signalBASK Rx signalEb/N0=10dBEb/N0=7dBEb/N0= :r(t)=Aicos(2 fct)+n(t),0 t Ts( )TorecoverthenumberAi, (a).Thatis,thereceivercomputes:Z= Ts0r(t)cos(2 fct)dt= Ts0 Aicos2(2 fct)dt+ Ts0n(t)cos(2 fct)dt=Ai2+ Ts0n(t)cos(2 fct)dt=Ai2+n( )wheren= Ts0n(t)cos(2 fct)dtisaGaussianrandomvariable(withzero meanandavariancethatisafunctionofN02) with threshold(a) BASK/BPSK receiverCompare with thresholds(a) MPSK/QAM receiverr(t)cos(2 fct)sin(2 fct)ZZ1Z2Ts0 TsTs00r(t)cos(2 fct) , ,undernoise-freeconditions, , 0 whena 1 ,fromacommonsenseperspective,ifZisaboveA i4,thereceiverdecidesthata 0 wastransmittedanda 1 , ,thereceiverneedstodetectwhetherthephase is0 or180.

8 Asimplewayofdeterminingthiswouldbetoperf ormthefollowingcomputation:Z= Ts0r(t)cos(2 fct)dt( ) ,thecomputednumberZwillbeasfollows:Z={A Ts0cos2(2 fct)dt+ Ts0n(t)cos(2 fct)dtifthephaseis0 A Ts0cos2(2 fct)dt+ Ts0n(t)cos(2 fct)dtifthephaseis180 ( )MODULATION699 Notethatinthenoiselesscase(n(t)=0),Zwill havealargepositivevalue(A2)whenthephasei s0 andalargenegativevalue( A2)whenthephaseis180 . ,thereceiverdecidesthatthe 0 bitwastransmittedifZispositiveandthe 1 ( ). ,thevalueofZmaybemovedtowardszeroandinso mecases, , :r(t)=Acos(2 fct+ i)+n(t)( ) ,thereceiverwillmultiplythereceivedsigna lbybothasineandacosinecarrier(thatareloc allygenerated) (b). :Z1=A Ts0cos(2 fct+ i)cos(2 fct)dt+ Ts0n(t)cos(2 fct)dt=A2cos( i)+n1( ) :Z2=A Ts0cos(2 fct+ i)sin(2 fct)dt+ Ts0n(t)sin(2 fct)dt=A2sin( i)+n2( )Inthenoiselesscase,thereceivermakesuseo ftheorderedpair(Z1,Z2)=(A2cos i,A2sin i) ,letussupposethat i=90 = (Z1,Z2)as(A2cos( 2),A2sin( 2))=(0,A2).}

9 Thusthereceiverdecidesthatthecarrierphas eis (A2,0),( A2,0)and(0, A2).Thesefourpossibilitiescorrespondtoth efourphases 2,0, and3 , , , ,thereceiverwillmultiplythereceivedsigna lwithlocallygeneratedcarriersofallMfrequ encies, ,thereceivermultipliesthereceivedsignalr (t)byalocallygeneratedcosine, ( ) ,thecontinuoustimesignalr(t) (t)fori=1,2,3, , ( ).WewillseethatinthecaseofASK,N=1,inthec aseofPSKandQAM,N=2,andinthecaseofFSK,N= ,PSKandQAM,thesignalconstellationisthesa measthephasordiagramofthesignals(ignorin gthe2 fct). , (t),wecanthenwriteinvectornotation:Z=si+ n( ) , (t)wastransmitted, m= , (t),i=1,2,3, ,MthatlastforadurationTssecondseach,itis possibletoestablishthattheycanberepresen tedaselementsofafinitevectorspace(seeCha pter6)spannedbyasetoforthonormalbasisfun ctions 1(t), 2(t), , N(t)whereN , ,weassumethatthe706 ThePhysicalLayerofCommunicationsSystems 2(t) 1(t) 3(t) T3s1s2s3s4 T3 ,wecaneasilyexpressthesignalsasvectorsas follows:s1=( T3,0,0)s2=( T3, T3,0)( )s3=(0, T3, T3)s4=( T3, T3, T3)( )Thefoursignalsareplottedinthevectorsubs pacespannedby 1(t), 2(t)and 3(t) (t)eachofdurationTsseconds,itispossiblet odecomposethesesignalsintoanN( M)dimensionalvectorspacespannedbyNbasisf unctions j(t) :si(t)=N j=1sij j(t)ORsi=(si1,si2, ,siN)( )Wecancomputethecomponentsofsi(t),namely sijusingthefollowingexpression.

10 Sij= Ts0si(t) j(t)dt( )MODULATION707 Decision Boundary k(t) l(t)S2dij/2nsisjZ i(t)S1 j(t) (t)with j(t) (n1,n2, ,nN).Thecomponentsofnareindependentandid enticallydistributed( (t)wastransmitted,thereceivedsignalr(t)= si(t)+n(t).Wecanwriter(t)asavectorZ=(Z1, Z2, ,ZN)where:Zj= Ts0r(t) j(t)dt= Ts0si(t) j(t)dt+ Ts0n(t) j(t)dt=sij+nj( )Invectornotation,Z=si+n( ) decisionboundary , , , , ,ifthenoisevectorhasacomponentthatislarg erthandij2inthecorrectdirectionalongthel inejoiningsiandsj, , , ,itispossibletoshowthattheprobabilitytha tthereceiverpickss1giventhats2wastransmi ttedorviceversaisgivenby:Pe=12erfc d24N0 ( ) ( )and( ),wecanseethatthesignalsdependonlyoncos( 2 fct).Itcanbeshownthat: 1(t)= 2 Tscos(2 fct),( ) 1(t) (t) ,Eb=A22Ts A= 2 EbTs( )WecanrewritethesignalsforBPSKas:si(t)= Acos((2 fct),0 t Tsorsi(t)= 2 EbTscos(2 fct),0 t Ts( )Clearly,s1(t)= Eb 1(t)ands2(t)= Eb 1(t).Invectornotation,s1= Ebands2= (a).))