Communication Systems Engineering 通信系统工程 答案 299页

'SOLUTIONSMANUALCommunicationSystemsEngineeringSecondEditionJohnG.ProakisMasoudSalehiPreparedbyEvangelosZervas课后答案网www.hackshp.cnUpperSaddleRiver,NewJersey07458 Publisher:TomRobbinsEditorialAssistant:JodyMcDonnellExecutiveManagingEditor:VinceO’BrienManagingEditor:DavidA.GeorgeProductionEditor:BarbaraA.TillComposition:PreTEX,Inc.SupplementCoverManager:PaulGourhanSupplementCoverDesign:PMWorkshopInc.ManufacturingBuyer:IleneKahn
c2002PrenticeHallbyPrentice-Hall,Inc.UpperSaddleRiver,NewJersey07458Allrightsreserved.Nopartofthisbookmaybereproducedinanyformorbyanymeans,withoutpermissioninwritingfromthepublisher.课后答案网Theauthorandpublisherofthisbookhaveusedtheirbesteffortsinpreparingthisbook.Theseeffortsincludethedevelopment,research,andtestingofthetheoriesandprogramstodeterminetheireffectiveness.Theauthorandpublishermakenowarrantyofanykind,expressedorimplied,withregardtotheseprogramsorthedocumentationcontainedinthisbook.Theauthorandpublishershallnotbeliableinanyeventforincidentalorconsequentialdamagesinconnectionwith,orarisingoutof,thefurnishing,performance,oruseoftheseprograms.www.hackshp.cnPrintedintheUnitedStatesofAmerica10987654321ISBN0-13-061974-6PearsonEducationLtd.,LondonPearsonEducationAustraliaPty.Ltd.,SydneyPearsonEducationSingapore,Pte.Ltd.PearsonEducationNorthAsiaLtd.,HongKongPearsonEducationCanada,Inc.,TorontoPearsonEducac`ıondeMexico,S.A.deC.V.PearsonEducation—Japan,TokyoPearsonEducationMalaysia,Pte.Ltd.PearsonEducation,UpperSaddleRiver,NewJersey ContentsChapter2.........................................1Chapter3.......................................42Chapter4.......................................71Chapter5......................................114Chapter6......................................128Chapter7......................................161Chapter8......................................213Chapter9......................................250Chapter10.....................................283课后答案网www.hackshp.cniii Chapter2Problem2.11)
N2∞2
=x(t)−αiφi(t)dt−∞i=1
∞NN=x(t)−αφ(t)x∗(t)−α∗φ∗(t)dtiijj−∞i=1j=1
∞N
∞N
∞=|x(t)|2dt−αφ(t)x∗(t)dt−α∗φ∗(t)x(t)dtiijj−∞−∞−∞i=1j=1NN
∞+αα∗φ(t)φ∗dtijij−∞i=1j=1
∞NN
∞N
∞=|x(t)|2dt+|α|2−αφ(t)x∗(t)dt−α∗φ∗(t)x(t)dtiiijj−∞−∞−∞i=1i=1j=1Completingthesquareintermsofαiweobtain
∞N
∞2N
∞2
2=|x(t)|2dt−φ∗(t)x(t)dt+−φ∗(t)x(t)dtiαii−∞−∞−∞i=1i=1Thefirsttwotermsareindependentofα’sandthelasttermisalwayspositive.Thereforetheminimumisachievedfor
∞α=φ∗(t)x(t)dtii−∞whichcausesthelasttermtovanish.课后答案网2)Withthischoiceofαi’s
∞N
∞2
2=|x(t)|2dt−φ∗(t)x(t)dti−∞−∞i=1www.hackshp.cn
∞N=|x(t)|2dt−|α|2i−∞i=1Problem2.21)Thesignalx1(t)isperiodicwithperiodT0=2.Thus
1
11−j2πnt1−jπntx1,n=Λ(t)e2dt=Λ(t)edt2−12−1
0
11−jπnt1−jπnt=(t+1)edt+(−t+1)edt2−12000=1jte−jπnt+1e−jπnt+je−jπnt2πnπ2n2−12πn−111−1jte−jπnt+1e−jπnt+je−jπnt2πnπ2n202πn011jπn−jπn1−(e+e)=(1−cos(πn))π2n22π2n2π2n21 Whenn=0then
111x1,0=Λ(t)dt=2−12Thus∞11x1(t)=+2(1−cos(πn))cos(πnt)2π2n2n=12)x2(t)=1.Itfollowsthenthatx2,0=1andx2,n=0,∀n=0.3)ThesignalisperiodicwithperiodT0=1.Thus1
T0
1x=ete−j2πntdt=e(−j2πn+1)tdt3,nT00011e(−j2πn+1)−1=e(−j2πn+1)t=−j2πn+10−j2πn+1e−1e−1==√(1+j2πn)1−j2πn1+4π2n24)Thesignalcos(t)isperiodicwithperiodT1=2πwhereascos(2.5t)isperiodicwithperiodT2=0.8π.Itfollowsthenthatcos(t)+cos(2.5t)isperiodicwithperiodT=4π.ThetrigonometricFourierseriesoftheevensignalcos(t)+cos(2.5t)is∞ncos(t)+cos(2.5t)=αncos(2πt)T0n=1∞n=αncos(t)2n=1Byequatingthecoefficientsofcos(nt)ofbothsidesweobservethata=0forallnunlessn=2,52ninwhichcasea=a=1.Hencex=x=1andx=0forallothervaluesofn.25课后答案网4,24,524,n5)Thesignalx5(t)isperiodicwithperiodT0=1.Forn=0
11x=(−t+1)dt=(−1t2+t)=15,00202Forn=0www.hackshp.cn
1x=(−t+1)e−j2πntdt5,n011=−jte−j2πnt+1e−j2πnt+je−j2πnt2πn4π2n202πn0j=−2πnThus,∞11x5(t)=+sin2πnt2πnn=16)Thesignalx6(t)isperiodicwithperiodT0=2T.Wecanwritex6(t)as∞∞x6(t)=δ(t−n2T)−δ(t−T−n2T)n=−∞n=−∞2 ∞∞1jπnt1jπn(t−T)=eT−eT2T2Tn=−∞n=−∞∞1−jπnj2πnt=(1−e)e2T2Tn=−∞However,thisistheFourierseriesexpansionofx6(t)andweidentifyx6,nas1−jπn1n0nevenx6,n=(1−e)=(1−(−1))=12T2TTnodd7)ThesignalisperiodicwithperiodT.Thus,
T12−j2πntx7,n=δ(t)eTdtT−T21d−j2πntj2πn=(−1)eT=Tdtt=0T28)Thesignalx(t)isrealevenandperiodicwithperiodT=1.Hence,x=a/2or802f08,n8,n
14f0x8,n=2f0cos(2πf0t)cos(2πn2f0t)dt−14f0
1
14f04f0=f0cos(2πf0(1+2n)t)dt+f0cos(2πf0(1−2n)t)dt−1−14f04f01114f014f0=sin(2πf0(1+2n)t)|1+sin(2πf0(1−2n)t)|12π(1+2n)4f02π(1−2n)4f0n(−1)11=+π(1+2课后答案网n)(1−2n)9)Thesignalx9(t)=cos(2πf0t)+|cos(2πf0t)|isevenandperiodicwithperiodT0=1/f0.Itisequalto2cos(2πft)intheinterval[−1,1]andzerointheinterval[1,3].Thus04f04f04f04f0
14f0x9,n=2f0cos(2www.hackshp.cnπf0t)cos(2πnf0t)dt−14f0
1
14f04f0=f0cos(2πf0(1+n)t)dt+f0cos(2πf0(1−n)t)dt−1−14f04f01114f014f0=sin(2πf0(1+n)t)|1+sin(2πf0(1−n)t)|12π(1+n)4f02π(1−n)4f01π1π=sin((1+n))+sin((1−n))π(1+n)2π(1−n)2Thusxiszeroforoddvaluesofnunlessn=±1inwhichcasex=1.Whenniseven9,n9,±12(n=2l)thenl(−1)11x9,2l=+π1+2l1−2l3 Problem2.3Itfollowsdirectlyfromtheuniquenessofthedecompositionofarealsignalinanevenandoddpart.Neverthelessforarealperiodicsignal∞a0nnx(t)=+ancos(2πt)+bnsin(2πt)2T0T0n=1Theevenpartofx(t)isx(t)+x(−t)xe(t)=2∞1nn=a0+an(cos(2πt)+cos(−2πt))2T0T0n=1nn+bn(sin(2πt)+sin(−2πt))T0T0∞a0n=+ancos(2πt)2T0n=1Thelastistruesincecos(θ)isevensothatcos(θ)+cos(−θ)=2cosθwhereastheoddnessofsin(θ)providessin(θ)+sin(−θ)=sin(θ)−sin(θ)=0.Theoddpartofx(t)isx(t)−x(−t)xo(t)=2∞n−bnsin(2πt)T0n=1Problem2.4a)ThesignalisperiodicwithperiodT.Thus
T
T1−t−j2πnt1−(j2πn+1)txn=eeTdt=eTdtT0T0T1课后答案网−(j2πn+1)t1−(j2πn+T)=−eT=−e−1Tj2πn+1j2πn+TT01−TT−j2πn−T=[1−e]=[1−e]j2πn+TT2+4π2n2an−jbnIfwewritexn=2weobtainthetrigonometricFourierseriesexpansioncoefficientsaswww.hackshp.cn2T−T4πn−Tan=T2+4π2n2[1−e],bn=T2+4π2n2[1−e]b)Thesignalisperiodicwithperiod2T.Sincethesignalisoddweobtainx0=0.Forn=0
T
T1−j2πnt1t−j2πntxn=x(t)e2Tdt=e2Tdt2T−T2T−TT
T1−jπnt=teTdt2T2−T1jTnT2nT=te−jπTt+e−jπTt2T2πnπ2n2−T1jT2−jπnT2−jπnjT2jπnT2jπn=e+e+e−e2T2πnπ2n2πnπ2n2jn=(−1)πn4 ThetrigonometricFourierseriesexpansioncoefficientsare:n+12an=0,bn=(−1)πnc)ThesignalisperiodicwithperiodT.Forn=0
T123x0=x(t)dt=T−T22Ifn=0then
T12−j2πntxn=x(t)eTdtT−T2
T
T12−j2πnt14−j2πnt=eTdt+eTdtT−TT−T24TTj−j2πnt2j−j2πnt4=eT+eT2πn−T2πn−T24j−jπnjπn−jπn−jπn=e−e+e2−e22πn1n1n=sin(π)=sinc()πn222Notethatx=0fornevenandx=1(−1)l.ThetrigonometricFourierseriesexpansionn2l+1π(2l+1)coefficientsare:2la0=3,,a2l=0,,a2l+1=(−1),,bn=0,∀nπ(2l+1)d)ThesignalisperiodicwithperiodT.Forn=0
T课后答案网12x0=x(t)dt=T03Ifn=0then
T
T1−j2πnt133−j2πntxn=x(t)eTdt=teTdtTwww.hackshp.cn0T0T
2T
T13−j2πnt13−j2πnt+eTdt+(−t+3)eTdtTTT2TT332T3jT−j2πntT−j2πnt3=teT+eTT22πn4π2n203jTnT2nT−te−j2πTt+e−j2πTtT22πn4π2n22T32TTj−j2πnt33jT−j2πnt+eT+eT2πnTT2πn2T3332πn=[cos()−1]2π2n23ThetrigonometricFourierseriesexpansioncoefficientsare:432πna0=3,an=π2n2[cos(3)−1],bn=0,∀n5 e)ThesignalisperiodicwithperiodT.Sincethesignalisoddx0=a0=0.Forn=0
T
T1214−j2πntxn=x(t)dt=−eTdtT−TT−T22
T
T144−j2πnt12−j2πnt+teTdt+eTdtT−TTTT442T4jT−j2πntT−j2πnt4=teT+eTT22πn4π2n2−T4−TT1jT−j2πnt41jT−j2πnt2−eT+eTT2πn−TT2πnT24πnjn2sin(2)jnn=(−1)−=(−1)−sinc()πnπnπn2Forneven,sinc(n)=0andx=j.ThetrigonometricFourierseriesexpansioncoefficientsare:2nπn−1n=2lπlan=0,∀n,bn=22(−1)l[1+]n=2l+1π(2l+1)π(2l+1)f)ThesignalisperiodicwithperiodT.Forn=0
T13x0=x(t)dt=1T−T3Forn=0
0
T13−j2πnt133−j2πntxn=(t+2)eTdt+(−t+2)eTdtT−TTT0T33jTnT2n0=te−j2πTt+e−j2πTtT22πn4π2n2−T32T3课后答案网jT−j2πntT−j2πnt3−teT+eTT22πn4π2n200T2jT−j2πnt2jT−j2πnt3+eT+eTT2πn−TT2πn03312πn12πn=www.hackshp.cn−cos()+sin()π2n223πn3ThetrigonometricFourierseriesexpansioncoefficientsare:312πn12πna0=2,an=2π2n22−cos(3)+πnsin(3),bn=0,∀nProblem2.51)Thesignaly(t)=x(t−t0)isperiodicwithperiodT=T0.1
α+T0n−j2πtyn=x(t−t0)eT0dtT0α1
α−t0+T0n−j2π=x(v)eT0(v+t0)dvT0α−t0n
α−t0+T0n−j2πt01−j2πv=eT0x(v)eT0dvT0α−t0−j2πnt0=xneT06 whereweusedthechangeofvariablesv=t−t02)Fory(t)tobeperiodictheremustexistTsuchthaty(t+mT)=y(t).Buty(t+T)=x(t+T)ej2πf0tej2πf0Tsothaty(t)isperiodicifT=T0(theperiodofx(t))andf0T=kforsomekinZ.Inthiscase1
α+T0ny=x(t)e−j2πT0tej2πf0tdtnT0α1
α+T0(n−k)−j2πt=x(t)eT0dt=xn−kT0α3)Thesignaly(t)isperiodicwithperiodT=T0/α.
β+T
β+T01−j2πntαα−j2πnαtyn=y(t)eTdt=x(αt)eT0dtTβT0β1
βα+T0n−j2πv=x(v)eT0dv=xnT0βαwhereweusedthechangeofvariablesv=αt.4)1
α+T0ny=x(t)e−j2πT0tdtnT0α1nα+T01
α+T0nn−j2πt−j2πt=x(t)eT0−(−j2π)eT0dtT0αT0αT0n1
α+T0nn−j2πt=j2πx(t)eT0dt=j2πxnT0T0αT0Problem2.61
α+T01
α+T0∞j2πn∞j2πmx(t)y∗(t课后答案网)dt=xeT0ty∗e−T0tdtnmT0αT0αn=−∞m=−∞∞∞1
α+T0j2π(n−m)=xy∗eT0tdtnmn=−∞m=−∞T0α∞∞∞www.hackshp.cn=xy∗δ=xy∗nmmnnnn=−∞m=−∞n=−∞Problem2.7UsingtheresultsofProblem2.6weobtain1
α+T0∞x(t)x∗(t)dt=|x|2nT0αn=−∞Sincethesignalhasfinitepower1
α+T0|x(t)|2dt=K<∞T0α∞2Thus,n=−∞|xn|=K<∞.Thelastimpliesthat|xn|→0asn→∞.Toseethiswrite∞−MM∞|x|2=|x|2+|x|2+|x|2nnnnn=−∞n=−∞n=−Mn=M7 EachoftheprevioustermsispositiveandboundedbyK.Assumethat|xn|2doesnotconvergetozeroasngoestoinfinityandchoose
=1.Thenthereexistsasubsequenceofxn,xnk,suchthat|xnk|>
=1,fornk>N≥MThen∞∞|x|2≥|x|2≥|x|2=∞nnnkn=Mn=Nnk∞2Thiscontradictsourassumptionthatn=M|xn|isfinite.Thus|xn|,andconsequentlyxn,shouldconvergetozeroasn→∞.Problem2.8Thepowercontentofx(t)is
T
T012212Px=lim|x(t)|dt=|x(t)|dtT→∞T−TT002But|x(t)|2isperiodicwithperiodT0/2=1sothat2
T0/22T0/21P=|x(t)|2dt=t3=xT003T003FromParseval’stheorem1
α+T0∞a21∞P=|x(t)|2dt=|x|2=0+(a2+b2)xnnnT0αn=−∞42n=1Forthesignalunderconsideration−4nodd−2nodda=π2n2b=πnnn0neven0nevenThus,课后答案网∞∞11212=a+b322n=1n=18∞12∞1=+π4(2l+1)4π2(2l+1)2www.hackshp.cnl=0l=0But,∞1π2=(2l+1)28l=0andbysubstitutingthisinthepreviousformulaweobtain∞1π4=(2l+1)496l=0Problem2.91)Since(a−b)2≥0wehavethata2b2ab≤+22withequalityifa=b.Let11n2n2A=α2,B=β2iii=1i=18 Thensubstitutingαi/Aforaandβi/Bforbinthepreviousinequalityweobtainαiβi1αi21βi2≤+AB2A22B2withequalityifαi=A=korα=kβforalli.Summingbothsidesfromi=1tonweobtainβiBiinαβ1nα21nβ2iiii≤+AB2A22B2i=1i=1i=1nn12121212=2A2αi+2B2βi=2A2A+2B2B=1i=1i=1Thus,111nnn2n2αβ≤1⇒αβ≤α2β2iiiiiiABi=1i=1i=1i=1Equalityholdsifαi=kβi,fori=1,...,n.2)Thesecondequationistrivialsince|xiyi∗|=|xi||yi∗|.Toseethiswritexiandyiinpolarjθxjθy∗j(θx−θy)coordinatesasxi=ρxieiandyi=ρyiei.Then,|xiyi|=|ρxiρyieii|=ρxiρyi=|xi||yi|=|xi||yi∗|.Weturnnowtoprovethefirstinequality.Letzibeanycomplexwithrealandimaginarycomponentszi,Randzi,Irespectively.Then,n2nn2n2n2zi=zi,R+jzi,I=zi,R+zi,Ii=1i=1i=1i=1i=1nn=(zi,Rzm,R+zi,Izm,I)i=1m=1Since(zi,Rzm,I−zm,Rzi,I)2≥0weobtain(zz+zz)2≤(z2+z2)(z2+z2)i,Rm,R课后答案网i,Im,Ii,Ri,Im,Rm,IUsingthisinequalityinthepreviousequationwegetn2nnzi=(zi,Rzm,R+zi,Izm,I)i=1i=1m=1nnwww.hackshp.cn221221≤(zi,R+zi,I)2(zm,R+zm,I)2i=1m=12nnn221221221=(zi,R+zi,I)2(zm,R+zm,I)2=(zi,R+zi,I)2i=1m=1i=1Thus22nnnn221zi≤(zi,R+zi,I)2orzi≤|zi|i=1i=1i=1i=1=xy∗.Equalityisobtainedifzi,R=zm,R=korTheinequalitynowfollowsifwesubstituteziiizi,Izm,I1zi=zm=θ.3)From2)weobtain2nn∗xiyi≤|xi||yi|i=1i=19 But|xi|,|yi|arerealpositivenumberssofrom1)11nn2n2|x||y|≤|x|2|y|2iiiii=1i=1i=1Combiningthetwoinequalitiesweget211nn2n2∗22xiyi≤|xi||yi|i=1i=1i=1Frompart1)equalityholdsifαi=kβior|xi|=k|yi|andfrompart2)xiyi∗=|xiyi∗|ejθ.Therefore,thetwoconditionsare|xi|=k|yi|xi−yi=θwhichimplythatforalli,xi=KyiforsomecomplexconstantK.3)ThesameprocedurecanbeusedtoprovetheCauchy-Schwartzinequalityforintegrals.Aneasierapproachisobtainedifoneconsiderstheinequality|x(t)+αy(t)|≥0,forallαThen
∞
∞0≤|x(t)+αy(t)|2dt=(x(t)+αy(t))(x∗(t)+α∗y∗(t))dt−∞−∞
∞
∞
∞
∞=|x(t)|2dt+αx∗(t)y(t)dt+α∗x(t)y∗(t)dt+|a|2|y(t)|2dt−∞−∞−∞−∞∞∗∞∗Theinequalityistrueforx(t)y(t)dt=0.Supposethatx(t)y(t)dt=0andset−∞−∞∞2|x(t)|dt−∞α=−∞∗x(t)y(t)dt−∞Then,
课后答案网∞∞22∞2[|x(t)|dt]|y(t)|dt0≤−|x(t)|2dt+−∞−∞|∞x(t)y∗(t)dt|2−∞−∞and

1
1∞∞2∞2x(t)y∗(t)dt|x(t)|2dt|y(t)|2dt≤−∞www.hackshp.cn−∞−∞Equalityholdsifx(t)=−αy(t)a.e.forsomecomplexα.Problem2.101)UsingtheFouriertransformpair−α|t|F2α2α1e−→=α2+(2πf)24π2α2+f24π2andthedualitypropertyoftheFouriertransform:X(f)=F[x(t)]⇒x(−f)=F[X(t)]weobtain2α1−α|f|F=e4π2α224π2+tWithα=2πwegetthedesiredresult1−2π|f|F=πe1+t210 2)F[x(t)]=F[Π(t−3)+Π(t+3)]=sinc(f)e−j2πf3+sinc(f)ej2πf3=2sinc(f)cos(2π3f)3)F[x(t)]=F[Λ(2t+3)+Λ(3t−2)]32=F[Λ(2(t+))+Λ(3(t−)]2312fjπf312f−j2πf2=sinc()e+sinc()e322334)T(f)=F[sinc3(t)]=F[sinc2(t)sinc(t)]=Λ(f)Π(f).But
∞
1
f+122Π(f)Λ(f)=Π(θ)Λ(f−θ)dθ=Λ(f−θ)dθ=Λ(v)dv−∞−1f−1223Forf≤−=⇒T(f)=02
f+1f+13121221239For−0anda<0separately.Notethatintheaboveexpressionifa>1,thenx(at)isacontractedformofx(t)whereasifa<1,x(at)isanexpandedversionofx(t).Thismeansthatifweexpandasignalinthetimedomainitsfrequencydomainrepresentation(Fouriertransform)contractsandifwecontractasignalinthetimedomainitsfrequencydomainrepresentationexpands.Thisisexactlywhatoneexpectssincecontractingasignalinthetimedomainmakesthechangesinthesignalmoreabrupt,thus,increasingitsfrequencycontent.13 Problem2.14Wehave

F[x(t)y(t)]=∞∞x(τ)y(t−τ)dτe−j2πftdt−∞−∞

=∞x(τ)∞y(t−τ)e−j2πf(t−τ)dte−j2πfτdτ−∞−∞Nowwiththechangeofvariableu=t−τ,wehave

∞y(t−τ)e−j2πf(t−τ)dt=∞fy(u)e−j2πfudu−∞−∞=F[y(t)]=Y(f)and,therefore,
F[x(t)y(t)]=∞x(τ)Y(f)e−j2πfτdτ−∞=X(f)·Y(f)Problem2.15WestartwiththeFouriertransformofx(t−t0),
F[x(t−t)]=∞x(t−t)e−j2πftdt00−∞Withachangeofvariableofu=t−t0,weobtain
F[x(t−t)]=∞x(u)e−j2πft0e−j2πfudu0−∞
=e−j2πft0∞x(u)e−j2πfudu课后答案网−∞=e−j2πft0F[x(t)]Problem2.16

www.hackshp.cn

∗∞x(t)y∗(t)dt=∞∞X(f)ej2πftdf∞Y(f)ej2πf
tdfdt−∞−∞−∞−∞


=∞∞X(f)ej2πftdf∞Y∗(f)e−j2πf
tdfdt−∞−∞−∞


=∞X(f)∞Y∗(f)∞ej2πt(f−f
)dtdfdf−∞−∞−∞Nowusingpropertiesoftheimpulsefunction.
∞ej2πt(f−f
)dt=δ(f−f)−∞andtherefore


∞x(t)y∗(t)dt=∞X(f)∞Y∗(f)δ(f−f)dfdf−∞−∞−∞
=∞X(f)Y∗(f)df−∞14 wherewehaveemployedthesiftingpropertyoftheimpulsesignalinthelaststep.Problem2.17(Convolutiontheorem:)F[x(t)y(t)]=F[x(t)]F[y(t)]=X(f)Y(f)Thussinc(t)sinc(t)=F−1[F[sinc(t)sinc(t)]]=F−1[F[sinc(t)]·F[sinc(t)]]=F−1[Π(f)Π(f)]=F−1[Π(f)]=sinc(t)Problem2.18
∞F[x(t)y(t)]=x(t)y(t)e−j2πftdt−∞

∞∞=X(θ)ej2πθtdθy(t)e−j2πftdt−∞−∞

∞∞=X(θ)y(t)e−j2π(f−θ)tdtdθ−∞−∞
∞=X(θ)Y(f−θ)dθ=X(f)Y(f)−∞Problem2.191)Clearly∞∞x1(t+kT0)=x(t+kT0−nT0)=x(t−(n−k)T0)课后答案网n=−∞n=−∞∞=x(t−mT0)=x1(t)m=−∞whereweusedthechangeofvariablewww.hackshp.cnm=n−k.2)∞x1(t)=x(t)δ(t−nT0)n=−∞Thisisbecause
∞∞∞
∞∞x(τ)δ(t−τ−nT0)dτ=x(τ)δ(t−τ−nT0)dτ=x(t−nT0)−∞n=−∞n=−∞−∞n=−∞3)∞∞F[x1(t)]=F[x(t)δ(t−nT0)]=F[x(t)]F[δ(t−nT0)]n=−∞n=−∞∞∞1n1nn=X(f)δ(f−)=X()δ(f−)T0n=−∞T0T0n=−∞T0T015 Problem2.201)ByParseval’stheorem
∞
∞
∞sinc5(t)dt=sinc3(t)sinc2(t)dt=Λ(f)T(f)df−∞−∞−∞whereT(f)=F[sinc3(t)]=F[sinc2(t)sinc(t)]=Π(f)Λ(f)But
∞
1
f+122Π(f)Λ(f)=Π(θ)Λ(f−θ)dθ=Λ(f−θ)dθ=Λ(v)dv−∞−1f−1223Forf≤−=⇒T(f)=02
f+1f+13121221239For−0Inthecaseofα=β
t−αt2ατ1αtIft<0⇒y(t)=eedτ=e−∞2α
0
tIft<0⇒y(t)=e−αte2ατdτ+e−αtdτ−∞0−αt0=ee2ατ+te−αt2α−∞1−αt=[+t]e2α20 5)Usingtheconvolutiontheoremweobtain01<|f|2Y(f)=Π(f)Λ(f)=f+1−10.5.Thisshowsthatlimt→∞8f(t)≥limt→∞16=∞.Thisshowsthatthesignalisnotenergy-type.Tocheckifthesignalispowertype,weobviouslyhavelim课后答案网1Te−2tcos2tdt=0.ThereforeT→∞T0
T12t2P=limecos(t)dtT→∞T021/4e2T(cos(T))2+1/4e2Tcos(T)sin(T)+1/8eT−3/8=limwww.hackshp.cnT→∞T=∞Thereforex2(t)isneitherpower-norenergy-type.3)
∞
∞E=(sgn(t))2dt=1dtx3−∞−∞=∞andhencethesignalisnotenergy-type.Tofindthepower
T12Px3=lim(sgn(t)))dtT→∞2T−T
T12=lim1dtT→∞2T−T1=lim2T=1T→∞2T22 4)Sincex4(t)isperiodic(oralmostperiodicwhenf1/f2isnotrational)thesignalisnotenergytype.Toseewhetheritispowertype,wehave
T12Px4=lim(Acos2πf1t+Bcos2πf2t)dtT→∞2T−T1
T=limA2cos22πft+B2cos22πft+2ABcos2πftcos2πftdt1212T→∞2T−TA2+B2=2Problem2.301)
T21j(2πf0t+θ)P=limAedtT→∞2T−T
T12=limAdtT→∞2T−T=A22)
T12P=lim1dtT→∞2T01=23)
T√E=limK2/tdtT→∞0√T=lim2K2tT→∞0课后答案网√=lim2K2TT→∞=∞therefore,itisnotenergy-type.Tofindthepowerwww.hackshp.cn1
T√P=limK2/tdtT→∞2T−T1√=lim2K2TT→∞2T=0andhenceitisnotpower-typeeither.Problem2.311)x(t)=e−αtu(t).ThespectrumofthesignalisX(f)=1andtheenergyspectraldensity−1α+j2πf21GX(f)=|X(f)|=α2+4π2f2Thus,−11−α|τ|RX(τ)=F[GX(f)]=e2α23 Theenergycontentofthesignalis1EX=RX(0)=2α2)x(t)=sinc(t).ClearlyX(f)=Π(f)sothatGX(f)=|X(f)|2=Π2(f)=Π(f).Theenergycontentofthesignalis
∞
12EX=Π(f)df=Π(f)df=1−∞−12∞3)x(t)=n=−∞Λ(t−2n).Thesignalisperiodicandthusitisnotoftheenergytype.Thepowercontentofthesignalis1
11
0
1P=|x(t)|2dt=(t+1)2dt+(−t+1)2dtx2−12−1001=11t3+t2+t+11t3−t2+t23−12301=3Thesameresultisobtainifwelet∞2nSX(f)=|xn|δ(f−)2n=−∞withx=1,x=0andx=2(seeProblem2.2).Then022l2l+1π(2l+1)∞P=|x|2Xnn=−∞18∞118π21=+=+=4π2(2l+1)44π2963课后答案网l=04)
T
T222TEX=lim|u−1(t)|dt=limdt=lim=∞T→∞−TT→∞0T→∞22Thus,thesignalisnotoftheenergytype.www.hackshp.cn
T1221T1PX=lim|u−1(t)|dt=lim=T→∞T−TT→∞T222Hence,thesignalisofthepowertypeanditspowercontentis1.Tofindthepowerspectraldensity2wefindfirsttheautocorrelationRX(τ).
T12RX(τ)=limu−1(t)u−1(t−τ)dtT→∞T−T2
T12=limdtT→∞Tτ1T1=lim(−τ)=T→∞T22Thus,S(f)=F[R(τ)]=1δ(f).XX224 T5)Clearly|X(f)|2=π2sgn2(f)=π2andEX=limT→∞2Tπ2dt=∞.Thesignalisnotofthe−2energytypefortheenergycontentisnotbounded.Considernowthesignal1txT(t)=Π()tTThen,XT(f)=−jπsgn(f)Tsinc(fT)and|X(f)|2
f
∞2T2SX(f)=lim=limπTsinc(vT)dv−sinc(vT)dvT→∞TT→∞−∞fHowever,thesquaredtermontherightsideisboundedawayfromzerosothatSX(f)is∞.Thesignalisnotofthepowertypeeither.Problem2.321)a)Ifα=γ,|Y(f)|2=|X(f)|2|H(f)|21=(α2+4π2f2)(β2+4π2f2)111=−β2−α2α2+4π2f2β2+4π2f2Fromthis,R(τ)=11e−α|τ|−1e−β|τ|andE=R(0)=1.Yβ2−α22α2βyy2αβ(α+β)Ifα=γthen2221GY(f)=|Y(f)|=|X(f)||H(f)|=(α2+4π2f2)2Theenergycontentofthesignalis
∞课后答案网1EY=−∞(α2+4π2f2)2
∞12α2α=df4α2−∞α2+4π2f2α2+4π2f2
∞
∞1−2α|t|1−2αt=edt=2edtwww.hackshp.cn4α2−∞4α20∞11−2αt1=2α2−2αe=4α30b)H(f)=1=⇒|H(f)|2=1.Theenergyspectraldensityoftheoutputisγ+j2πfγ2+4π2f221GY(f)=GX(f)|H(f)|=γ2+4π2f2Π(f)Theenergycontentofthesignalis
11211fγ2EY=df=arctan−1γ2+4π2f22πγ2π−1221fγ=arctanπγ4π25 c)Thepowerspectraldensityoftheoutputis∞2n2nSY(f)=|xn||H()|δ(f−)22n=−∞1∞|x|22l+12l+1=δ(f)+2δ(f−)4γ2γ2+π2(2l+1)22l=0∞1812l+1=δ(f)+δ(f−)4γ2π2(2l+1)4(γ2+π2(2l+1)2)2l=0Thepowercontentoftheoutputsignalis∞2n2PY=|xn||H()|2n=−∞18∞1π4π2=++−4γ2π2γ2(2l+1)4γ4(γ2+π2(2l+1)2)γ4(2l+1)2l=018π2π4π2∞1=+−+4γ2π2γ2968γ4γ4γ22l=0π2+(2l+1)1π22π2γ=−+tanh()3γ2γ4γ52wherewehaveusedthefactπx4x∞1ex−e−xtanh()=,tanh(x)=2πx2+(2l+1)2ex+e−xl=0d)Thepowerspectraldensityoftheoutputsignalis2111SY(f)=SX(f)|H(f)|=2γ2+4π2f2δ(f)=2γ2δ(f)Thepowercontentofthesignalis课后答案网
∞1PY=SY(f)df=2−∞2γe)X(f)=−jπsgn(f)sothat|X(f)|2=π2forallfexceptf=0forwhich|X(f)|2=0.Thus,theenergyspectraldensityoftheoutputiswww.hackshp.cnπ2G(f)=|X(f)|2|H(f)|2=Yγ2+4π2f2andtheenergycontentofthesignal
∞11f2π∞π2E=π2df=π2arctan()=Yγ2+4π2f22πγγ2γ−∞−∞2)a)h(t)=sinc(6t)=⇒H(f)=1Π(f)TheenergyspectraldensityoftheoutputsignalisG(f)=66YG(f)|H(f)|2andwithG(f)=1weobtainXXα2+4π2f2112f1fGY(f)=Π()=Π()α2+4π2f236636(α2+4π2f2)626 Theenergycontentofthesignalis
∞
311EY=GY(f)df=36α2+4π2f2df−∞−3312π=arctan(f)36(2απ)α−316π=arctan()36απαb)TheenergyspectraldensityisG(f)=1Π(f)Π(f)=1Π(f)andtheenergycontentoftheY36636output
1121EY(f)=df=36−1362c)∞221nnSY(f)=SX(f)|H(f)|=|xn|Π()δ(f−)36122n=−∞SinceΠ(n)isnonzeroonlyfornsuchthatn≤1andx=1,x=0andx=2(see12122022l2l+1π(2l+1)2Problem2.2),weobtain21122l+1SY(f)=δ(f)+|x2l+1|δ(f−)4·36362l=−321112l+1=δ(f)+δ(f−)1449π2(2l+1)42l=−3Thepowercontentofthesignalis12111.2253PY=+(1++)==+1449π281625144π2d)S(f)=1δ(f),|H(f)|2=1Π(f).Hence,S(f)=1Π(f)δ(f)=1δ(f).ThepowercontentX2366Y72672ofthesignalisP=∞1δ(f)df=1.Y−∞72课后答案网72e)y(t)=sinc(6t)1=πsinc(6t)1.However,convolutionwith1istheHilberttransformwhichtπtπtisknowntoconservetheenergyofthesignalprovidedthattherearenoimpulsesattheorigininthefrequencydomain(f=0).Thisisthecaseofπsinc(6t),sothat
∞
∞1fπ2
3π2E=π2sinc2(6t)dt=π2Π2()df=df=Ywww.hackshp.cn−∞−∞363636−36Theenergyspectraldensityis12f22GY(f)=Π()πsgn(f)3663)1istheimpulseresponseoftheHilberttransformfilter,whichisknowntopreservetheenergyπtoftheinputsignal.|H(f)|2=sgn2(f)a)Theenergyspectraldensityoftheoutputsignalis2GX(f)f=0GY(f)=GX(f)sgn(f)=0f=0SinceGX(f)doesnotcontainanyimpulsesattheorigin1EY=EX=2α27 b)Arguingasinthepreviousquestion2Π(f)f=0GY(f)=GX(f)sgn(f)=0f=0SinceΠ(f)doesnotcontainanyimpulsesattheoriginEY=EX=1c)∞22nSY(f)=SX(f)sgn(f)=|xn|δ(f−),n=02n=−∞But,x=0,x=1sothat2l2l+1π(2l+1)∞∞2n81nSY(f)=2|x2l+1|δ(f−2)=π2(2l+1)4δ(f−2)l=0l=0Thepowercontentoftheoutputsignalis8∞18π21PY===π2(2l+1)4π29612l=0d)S(f)=1δ(f)and|H(f)|2=sgn2(f).ThusS(f)=S(f)|H(f)|2=0,andthepowerX2YXcontentofthesignaliszero.e)Thesignal1hasinfiniteenergyandpowercontent,andsinceG(f)=G(f)sgn2(f),S(f)=tYXYS(f)sgn2(f)thesamewillbetruefory(t)=11.XtπtProblem2.33Notethat
∞
T122Px=Sx(f)df=lim|x(t)|dt−∞T→∞T−T课后答案网2Butintheinterval[−T,T],|x(t)|2=|x(t)|2sothat22T
T122Px=lim|xT(t)|dtT→∞T−Twww.hackshp.cn2UsingRayleigh’stheorem
T
∞12212Px=lim|xT(t)|dt=lim|XT(f)|dfT→∞T−TT→∞T−∞2
∞
∞11=limGxT(f)df=limGxT(f)dfT→∞T−∞−∞T→∞T∞ComparingthelastwithPx=−∞Sx(f)dfweseethat1Sx(f)=limGxT(f)T→∞T28 Problem2.34∞Lety(t)betheoutputsignal,whichistheconvolutionofx(t),andh(t),y(t)=h(τ)x(t−τ)dτ.−∞UsingCauchy-Schwartzinequalityweobtain
∞|y(t)|=h(τ)x(t−τ)dτ−∞
1
1∞2∞2≤|h(τ)|2dτ|x(t−τ)|2dτ−∞−∞
11∞2≤E2|x(t−τ)|2dτh−∞Squaringthepreviousinequalityandintegratingfrom−∞to∞weobtain
∞
∞
∞|y(t)|2dt≤E|x(t−τ)|2dτdth−∞−∞−∞∞∞2Butbyassumption−∞−∞|x(t−τ)|dτdt,Eharefinite,sothattheenergyoftheoutputsignalisfinite.∞ConsidertheLTIsystemwithimpulseresponseh(t)=n=−∞Π(t−2n).ThesignalisperiodicwithperiodT=2,andthepowercontentofthesignalisP=1.IftheinputtothissystemisH2theenergytypesignalx(t)=Π(t),then∞y(t)=Λ(t−2n)n=−∞whichisapowertypesignalwithpowercontentP=1.Y2Problem2.35FornoaliasingtooccurwemustsampleattheNyquistratefs=2课后答案网·6000samples/sec=12000samples/secWithaguardbandof2000fs−2W=2000=⇒fs=14000Thereconstructionfiltershouldnotpick-upfrequenciesoftheimagesofthespectrumX(f).Thenearestimagespectrumiscenteredatwww.hackshp.cnfsandoccupiesthefrequencyband[fs−W,fs+W].Thusthehighestfrequencyofthereconstructionfilter(=10000)shouldsatisfy10000≤fs−W=⇒fs≥16000Forthevaluefs=16000,KshouldbesuchthatK·f=1=⇒K=(16000)−1sProblem2.36Afx(t)=Asinc(1000πt)=⇒X(f)=Π()10001000ThusthebandwidthWofx(t)is1000/2=500.Sincewesampleatfs=2000thereisagapbetweentheimagespectraequalto2000−500−W=100029 ThereconstructionfiltershouldhaveabandwidthWsuchthat5001500Problem2.371)∞xp(t)=x(nTs)p(t−nTs)n=−∞∞=p(t)x(nTs)δ(t−nTs)n=−∞∞=p(t)x(t)δ(t−nTs)n=−∞Thus∞Xp(f)=P(f)·Fx(t)δ(t−nTs)n=−∞∞=P(f)X(f)Fδ(t−nTs)n=−∞∞1n课后答案网=P(f)X(f)δ(f−)Tsn=−∞Ts∞1n=P(f)X(f−)Tsn=−∞Ts1www.hackshp.cn2)Inordertoavoidaliasing>2W.FurthermorethespectrumP(f)shouldbeinvertibleforTs|f|0Hθ(f)=0f=0=cosθ−jsgn(f)sinθcosθ+jsinθf<0Thus,−11hθ(t)=F[Hθ(f)]=cosθδ(t)+sinθπt39 2)1xθ(t)=x(t)hθ(t)=x(t)(cosθδ(t)+sinθ)πt1=cosθx(t)δ(t)+sinθx(t)πt=cosθx(t)+sinθxˆ(t)3)
∞
∞|x(t)|2dt=|cosθx(t)+sinθxˆ(t)|2dtθ−∞−∞
∞
∞=cos2θ|x(t)|2dt+sin2θ|xˆ(t)|2dt−∞−∞
∞
∞+cosθsinθx(t)ˆx∗(t)dt+cosθsinθx∗(t)ˆx(t)dt−∞−∞∞2∞2∞∗But−∞|x(t)|dt=−∞|xˆ(t)|dt=Exand−∞x(t)ˆx(t)dt=0sincex(t)andˆx(t)areorthogonal.Thus,E=E(cos2θ+sin2θ)=ExθxxProblem2.591)z(t)=x(t)+jxˆ(t)=m(t)cos(2πf0t)−mˆ(t)sin(2πf0t)+j[m(t)cos(2#πf0t)−mˆ(t)sin(2#πf0t)=m(t)cos(2πf0t)−mˆ(t)sin(2πf0t)+jm(t)sin(2πf0t)+jmˆ(t)cos(2πf0t)=(m(t)+jmˆ(t))ej2πf0tThelowpassequivalentsignalisgivenby课后答案网x(t)=z(t)e−j2πf0t=m(t)+jmˆ(t)l2)TheFouriertransformofm(t)isΛ(www.hackshp.cnf).ThusΛ(f+f0)+Λ(f−f0)X(f)=−(−jsgn(f)Λ(f))211−δ(f+f0)+δ(f−f0)2j2j11=Λ(f+f0)[1−sgn(f+f0)]+Λ(f−f0)[1+sgn(f−f0)]22.1.............−f0−1−f0f0f0+1Thebandwidthofx(t)isW=1.40 3)z(t)=x(t)+jxˆ(t)=m(t)cos(2πf0t)+ˆm(t)sin(2πf0t)+j[m(t)cos(2#πf0t)+ˆm(t)sin(2#πf0t)=m(t)cos(2πf0t)+ˆm(t)sin(2πf0t)+jm(t)sin(2πf0t)−jmˆ(t)cos(2πf0t)=(m(t)−jmˆ(t))ej2πf0tThelowpassequivalentsignalisgivenbyx(t)=z(t)e−j2πf0t=m(t)−jmˆ(t)lTheFouriertransformofx(t)isΛ(f+f0)+Λ(f−f0)X(f)=−(jsgn(f)Λ(f))211−δ(f+f0)+δ(f−f0)2j2j11=Λ(f+f0)[1+sgn(f+f0)]+Λ(f−f0)[1−sgn(f−f0)]22..1................................−f0−f0+1f0−1f0课后答案网www.hackshp.cn41 Chapter3Problem3.1Themodulatedsignalisu(t)=m(t)c(t)=Am(t)cos(2π4×103t)200250π3=A2cos(2πt)+4sin(2πt+)cos(2π4×10t)ππ332003200=Acos(2π(4×10+)t)+Acos(2π(4×10−)t)ππ3250π3250π+2Asin(2π(4×10+)t+)−2Asin(2π(4×10−)t−)π3π3TakingtheFouriertransformofthepreviousrelation,weobtain2002002jπ2502−jπ250U(f)=Aδ(f−)+δ(f+)+e3δ(f−)−e3δ(f+)ππjπjπ133[δ(f−4×10)+δ(f+4×10)]2A32003200=δ(f−4×10−)+δ(f−4×10+)2ππ−jπ3250jπ3250+2e6δ(f−4×10−)+2e6δ(f−4×10+)ππ32003200+δ(f+4×10−)+δ(f+4×10+)ππ−jπ3250jπ3250+2e6δ(f+4×10−)+2e6δ(f+4×10+)ππThenextfiguredepictsthemagnitudeandthephaseofthespectrumU(f).|U(f)|A......................课后答案网.......................A/2−fc−250−fc−200−fc+200−fc+250fc−250fc−200fc+200fc+250ππππππππU(f)www.hackshp.cnπ
.............................6
−π
.............................6
Tofindthepowercontentofthemodulatedsignalwewriteu2(t)as2223200223200u(t)=Acos(2π(4×10+)t)+Acos(2π(4×10−)t)ππ223250π223250π+4Asin(2π(4×10+)t+)+4Asin(2π(4×10−)t−)π3π3+termsofcosineandsinefunctionsinthefirstpowerHence,
T222222AA4A4A2P=limu(t)dt=+++=5AT→∞−T2222242 Problem3.2u(t)=m(t)c(t)=A(sinc(t)+sinc2(t))cos(2πft)cTakingtheFouriertransformofbothsides,weobtainAU(f)=[Π(f)+Λ(f)](δ(f−fc)+δ(f+fc))2A=[Π(f−fc)+Λ(f−fc)+Π(f+fc)+Λ(f+fc)]2Π(f−f)=0for|f−f|<1,whereasΛ(f−f)=0for|f−f|<1.Hence,thebandwidthofcc2ccthebandpassfilteris2.Problem3.3ThefollowingfigureshowsthemodulatedsignalsforA=1andf0=10.Asitisobservedbothsignalshavethesameenvelopebutthereisaphasereversalatt=1forthesecondsignalAm2(t)cos(2πf0t)(rightplot).ThisdiscontinuityisshownclearlyinthenextfigurewhereweplottedAm2(t)cos(2πf0t)withf0=3.110.80.80.60.60.40.40.20.200-0.2-0.2-0.4-0.4-0.6-0.6-0.8-0.8-1-100.20.40.60.811.2课后答案网1.41.61.8200.20.40.60.811.21.41.61.820.80.60.4www.hackshp.cn0.20-0.2-0.4-0.6-0.8-100.20.40.60.811.21.41.61.82Problem3.412y(t)=x(t)+x(t)243 1=m(t)+cos(2πft)+m2(t)+cos2(2πft)+2m(t)cos(2πft)ccc21211=m(t)+cos(2πfct)+m(t)++cos(2π2fct)+m(t)cos(2πfct)244TakingtheFouriertransformoftheprevious,weobtain11Y(f)=M(f)+M(f)M(f)+(M(f−fc)+M(f+fc))22111+δ(f)+(δ(f−fc)+δ(f+fc))+(δ(f−2fc)+δ(f+2fc))428ThenextfiguredepictsthespectrumY(f)1/21/41/8-2fc-fc-2W2Wfc2fcProblem3.5u(t)=m(t)·c(t)=100(2cos(2π2000t)+5cos(2π3000t))cos(2πfct)Thus,1005U(f)=δ(f−课后答案网2000)+δ(f+2000)+(δ(f−3000)+δ(f+3000))22[δ(f−50000)+δ(f+50000)]55=50δ(f−52000)+δ(f−48000)+δ(f−53000)+δ(f−47000)2255+δ(f+52000)+www.hackshp.cnδ(f+48000)+δ(f+53000)+δ(f+47000)22Aplotofthespectrumofthemodulatedsignalisgiveninthenextfigure........................125.........................................50-53-52-48-47047485253KHzProblem3.6Themixedsignaly(t)isgivenbyy(t)=u(t)·xL(t)=Am(t)cos(2πfct)cos(2πfct+θ)A=m(t)[cos(2π2fct+θ)+cos(θ)]244 Thelowpassfilterwillcut-offthefrequenciesaboveW,whereWisthebandwidthofthemessagesignalm(t).Thus,theoutputofthelowpassfilterisAz(t)=m(t)cos(θ)2A22Ifthepowerofm(t)isPM,thenthepoweroftheoutputsignalz(t)isPout=PM4cos(θ).TheA2powerofthemodulatedsignalu(t)=Am(t)cos(2πfct)isPU=2PM.Hence,Pout12=cos(θ)PU2AplotofPoutfor0≤θ≤πisgiveninthenextfigure.PU0.50.450.40.350.30.250.20.150.10.05000.511.522.533.5Theta(rad)Problem3.71)Thespectrumofu(t)is20U(f)=[δ(f−fc)+δ(f+fc)]22课后答案网+[δ(f−fc−1500)+δ(f−fc+1500)4+δ(f+fc−1500)+δ(f+fc+1500)]10+[δ(f−fc−3000)+δ(f−fc+3000)4www.hackshp.cn+δ(f+fc−3000)+δ(f+fc+3000)]Thenextfiguredepictsthespectrumofu(t).......................................101/2...................................................5/2-1030-1015-1000-985-9700970985100010151030X100Hz2)Thesquareofthemodulatedsignalisu2(t)=400cos2(2πft)+cos2(2π(f−1500)t)+cos2(2π(f+1500)t)ccc+25cos2(2π(f−3000)t)+25cos2(2π(f+3000)t)cc+termsthataremultiplesofcosines45 Ifweintegrateu2(t)from−TtoT,normalizetheintegralby1andtakethelimitasT→∞,22Tthenallthetermsinvolvingcosinestendtozero,whereasthesquaresofthecosinesgiveavalueof1.Hence,thepowercontentatthefrequencyf=105HzisP=400=200,thepowercontent2cfc2atthefrequencyPfc+1500isthesameasthepowercontentatthefrequencyPfc−1500andequalto1,whereasP=P=25.2fc+3000fc−300023)u(t)=(20+2cos(2π1500t)+10cos(2π3000t))cos(2πfct)11=20(1+cos(2π1500t)+cos(2π3000t))cos(2πfct)102ThisistheformofaconventionalAMsignalwithmessagesignal11m(t)=cos(2π1500t)+cos(2π3000t)102211=cos(2π1500t)+cos(2π1500t)−102Theminimumofg(z)=z2+1z−1isachievedforz=−1anditismin(g(z))=−201.Since10220400z=−1isintherangeofcos(2π1500t),weconcludethattheminimumvalueofm(t)is−201.20400Hence,themodulationindexis201α=−4004)u(t)=20cos(2πfct)+cos(2π(fc−1500)t)+cos(2π(fc−1500)t)=5cos(2π(fc−3000)t)+5cos(2π(fc+3000)t)Thepowerinthesidebandsis112525Psidebands=+++=26课后答案网2222ThetotalpowerisPtotal=Pcarrier+Psidebands=200+26=226.TheratioofthesidebandspowertothetotalpowerisPsidebands26=www.hackshp.cnPtotal226Problem3.81)u(t)=m(t)c(t)=100(cos(2π1000t)+2cos(2π2000t))cos(2πfct)=100cos(2π1000t)cos(2πfct)+200cos(2π2000t)cos(2πfct)100=[cos(2π(fc+1000)t)+cos(2π(fc−1000)t)]2200[cos(2π(fc+2000)t)+cos(2π(fc−2000)t)]2Thus,theuppersideband(USB)signalisuu(t)=50cos(2π(fc+1000)t)+100cos(2π(fc+2000)t)46 2)TakingtheFouriertransformofbothsides,weobtainUu(f)=25(δ(f−(fc+1000))+δ(f+(fc+1000)))+50(δ(f−(fc+2000))+δ(f+(fc+2000)))AplotofUu(f)isgiveninthenextfigure...............................................50.................................25-1002-1001010011002KHzProblem3.9IfweletTpTpt+t−x(t)=−Π4+Π4TpTp22thenusingtheresultsofProblem2.23,weobtain∞v(t)=m(t)s(t)=m(t)x(t−nTp)n=−∞∞1nj2πnt=m(t)X()eTpTpn=−∞TpwhereTpTpnt+t−X()=F−Π4+Π4TpTpTpf=n22TpTpTp−j2πfTpj2πfTp=sinc(f)e4−e422f=n课后答案网TpTpnπ=sinc()(−2j)sin(n)222Hence,theFouriertransformofv(t)is∞www.hackshp.cn1nπnV(f)=sinc()(−2j)sin(n)M(f−)222Tpn=−∞Thebandpassfilterwillcut-offallthefrequenciesexcepttheonescenteredat1,thatisforn=±1.TpThus,theoutputspectrumis1111U(f)=sinc()(−j)M(f−)+sinc()jM(f+)2Tp2Tp2121=−jM(f−)+jM(f+)πTpπTp41111=M(f)δ(f−)−δ(f+)π2jTp2jTpTakingtheinverseFouriertransformofthepreviousexpression,weobtain41u(t)=m(t)sin(2πt)πTp47 whichhastheformofaDSB-SCAMsignal,withc(t)=4sin(2π1t)beingthecarriersignal.πTpProblem3.10Assumethats(t)isaperiodicsignalwithperiodTp,i.e.s(t)=nx(t−nTp).Then∞v(t)=m(t)s(t)=m(t)x(t−nTp)n=−∞∞1nj2πnt=m(t)X()eTpTpn=−∞Tp∞1nj2πnt=X()m(t)eTpTpn=−∞TpnwhereX()=F[x(t)]|f=n.TheFouriertransformofv(t)isTpTp∞1nj2πntV(f)=FX()m(t)eTpTpn=−∞Tp∞1nn=X()M(f−)Tpn=−∞TpTpThebandpassfilterwillcut-offallthefrequencycomponentsexcepttheonescenteredatf=±1.cTpHence,thespectrumattheoutputoftheBPFis111111U(f)=X()M(f−)+X(−)M(f+)TpTpTpTpTpTpInthetimedomaintheoutputoftheBPFisgivenby11j2π1t1∗1−j2π1tu(t)=X()m(t)eTp+X()m(t)eTpTpTpTpTp11j2π1t∗1−j2π1t=课后答案网m(t)X()eTp+X()eTpTpTpTp111=2Re(X())m(t)cos(2πt)TpTpTpAsitisobservedu(t)hastheformamodulatedDSB-SCsignal.TheamplitudeofthemodulatingsignalisA=12Re(X(1))andthecarrierfrequencyf=1.cTpTpwww.hackshp.cncTpProblem3.111)ThespectrumofthemodulatedsignalAm(t)cos(2πfct)isAV(f)=[M(f−fc)+M(f+fc)]2ThespectrumofthesignalattheoutputofthehighpassfilterisAU(f)=[M(f+fc)u−1(−f−fc)+M(f−fc)u−1(f−fc)]2MultiplyingtheoutputoftheHPFwithAcos(2π(fc+W)t)resultsinthesignalz(t)withspectrumAZ(f)=[M(f+fc)u−1(−f−fc)+M(f−fc)u−1(f−fc)]2A[δ(f−(fc+W))+δ(f+fc+W)]248 A2=(M(f+fc−fc−W)u−1(−f+fc+W−fc)4+M(f+fc−fc+W)u−1(f+fc+W−fc)+M(f−2fc−W)u−1(f−2fc−W)+M(f+2fc+W)u−1(−f−2fc−W))A2=(M(f−W)u−1(−f+W)+M(f+W)u−1(f+W)4+M(f−2fc−W)u−1(f−2fc−W)+M(f+2fc+W)u−1(−f−2fc−W))TheLPFwillcut-offthedoublefrequencycomponents,leavingthespectrumA2Y(f)=[M(f−W)u−1(−f+W)+M(f+W)u−1(f+W)]4ThenextfiguredepictsY(f)forM(f)asshowninFig.P-5.12.Y(f)-WW2)AsitisobservedfromthespectrumY(f),thesystemshiftsthepositivefrequencycomponentstothenegativefrequencyaxisandthenegativefrequencycomponentstothepositivefrequencyaxis.Ifwetransmitthesignaly(t)throughthesystem,thenwewillgetascaledversionoftheoriginalspectrumM(f).Problem3.12Themodulatedsignalcanbewrittenas课后答案网u(t)=m(t)cos(2πfct+φ)=m(t)cos(2πfct)cos(φ)−m(t)sin(2πfct)sin(φ)=uc(t)cos(2πfct)−us(t)sin(2πfct)whereweidentifyuc(t)=m(t)cos(www.hackshp.cnφ)asthein-phasecomponentandus(t)=m(t)sin(φ)asthequadraturecomponent.Theenvelopeofthebandpasssignalis%%V(t)=u2(t)+u2(t)=m2(t)cos2(φ)+m2(t)sin2(φ)ucs%=m2(t)=|m(t)|Hence,theenvelopeisproportionaltotheabsolutevalueofthemessagesignal.Problem3.131)Themodulatedsignalisu(t)=100[1+m(t)]cos(2π8×105t)=100cos(2π8×105t)+100sin(2π103t)cos(2π8×105t)+500cos(2π2×103t)cos(2π8×105t)=100cos(2π8×105t)+50[sin(2π(103+8×105)t)−sin(2π(8×105−103)t)]+250[cos(2π(2×103+8×105)t)+cos(2π(8×105−2×103)t)]49 TakingtheFouriertransformofthepreviousexpression,weobtainU(f)=50[δ(f−8×105)+δ(f+8×105)]153153+25δ(f−8×10−10)−δ(f+8×10+10)jj153153−25δ(f−8×10+10)−δ(f+8×10−10)jj+125δ(f−8×105−2×103)+δ(f+8×105+2×103)+125δ(f−8×105−2×103)+δ(f+8×105+2×103)=50[δ(f−8×105)+δ(f+8×105)]53−jπ53jπ+25δ(f−8×10−10)e2+δ(f+8×10+10)e253jπ53−jπ+25δ(f−8×10+10)e2+δ(f+8×10−10)e2+125δ(f−8×105−2×103)+δ(f+8×105+2×103)+125δ(f−8×105−2×103)+δ(f+8×105+2×103)|U(f)|......................125......................................50...........................25fc−2×103−fcfc+2×103fc−2×103fcfc+2×103U(f)π..................................2fc−103fc+103−π..................................2课后答案网2)TheaveragepowerinthecarrierisA21002P=c==5000carrier22Thepowerinthesidebandsiswww.hackshp.cn50250225022502Psidebands=+++=6500022223)Themessagesignalcanbewrittenasm(t)=sin(2π103t)+5cos(2π2×103t)=−10sin(2π103t)+sin(2π103t)+5Asitisseentheminimumvalueofm(t)is−6andisachievedforsin(2π103t)=−1ort=3+1k,withk∈Z.Hence,themodulationindexisα=6.4×1031034)Thepowerdeliveredtotheloadis|u(t)|21002(1+m(t))2cos2(2πfct)Pload==505050 Themaximumabsolutevalueof1+m(t)is6.025andisachievedforsin(2π103t)=1ort=20arcsin(1)20+k.Since2×103fthepeakpowerdeliveredtotheloadisapproximatelyequalto2π103103c(100×6.025)2max(Pload)==72.601250Problem3.141)u(t)=5cos(1800πt)+20cos(2000πt)+5cos(2200πt)1=20(1+cos(200πt))cos(2000πt)2Themodulatingsignalism(t)=cos(2π100t)whereasthecarriersignalisc(t)=20cos(2π1000t).2)Since−1≤cos(2π100t)≤1,weimmediatelyhavethatthemodulationindexisα=1.23)ThepowerofthecarriercomponentisP=400=200,whereasthepowerinthesidebandscarrier2400α2isPsidebands=2=50.Hence,Psidebands501==Pcarrier2004Problem3.151)Themodulatedsignaliswrittenasu(t)=100(2cos(2π103t)+cos(2π3×103t))cos(2πft)c=200cos(2π103t)cos(2πft)+100cos(2π3×103t)cos(2πft)cc=100cos(2π(f+103)t)+cos(2π(f−103)t)cc+50cos(2π(f+3×103)t)+cos(2π(f−3×103)t)ccTakingtheFouriertransformofthepreviousexpression,weobtain课后答案网U(f)=50δ(f−(f+103))+δ(f+f+103)cc+δ(f−(f−103))+δ(f+f−103)cc+25δ(f−(f+3×103))+δ(f+f+3×103)cc+δ(www.hackshp.cnf−(f−3×103))+δ(f+f−3×103)ccThespectrumofthesignalisdepictedinthenextfigure..............................50....................25−1003−1001−999−99799799910011003KHz2)Theaveragepowerinthefrequenciesfc+1000andfc−1000is1002Pfc+1000=Pfc−1000==50002Theaveragepowerinthefrequenciesfc+3000andfc−3000is502Pfc+3000=Pfc−3000==1250251 Problem3.161)TheHilberttransformofcos(2π1000t)issin(2π1000t),whereastheHilberttransformofsin(2π1000t)is−cos(2π1000t).Thusmˆ(t)=sin(2π1000t)−2cos(2π1000t)2)TheexpressionfortheLSSBAMsignalisul(t)=Acm(t)cos(2πfct)+Acmˆ(t)sin(2πfct)SubstitutingAc=100,m(t)=cos(2π1000t)+2sin(2π1000t)andˆm(t)=sin(2π1000t)−2cos(2π1000t)intheprevious,weobtainul(t)=100[cos(2π1000t)+2sin(2π1000t)]cos(2πfct)+100[sin(2π1000t)−2cos(2π1000t)]sin(2πfct)=100[cos(2π1000t)cos(2πfct)+sin(2π1000t)sin(2πfct)]+200[cos(2πfct)sin(2π1000t)−sin(2πfct)cos(2π1000t)]=100cos(2π(fc−1000)t)−200sin(2π(fc−1000)t)3)TakingtheFouriertransformofthepreviousexpressionweobtainUl(f)=50(δ(f−fc+1000)+δ(f+fc−1000))+100j(δ(f−fc+1000)−δ(f+fc−1000))=(50+100j)δ(f−fc+1000)+(50−100j)δ(f+fc−1000)Hence,themagnitudespectrumisgivenby&|Ul(f)|=502+1002(δ(f−fc+1000)+δ(f+fc−1000))√=10125(δ(f−fc+1000)+δ(f+fc−1000))Problem3.17课后答案网TheinputtotheupperLPFisuu(t)=cos(2πfmt)cos(2πf1t)1=[cos(2π(f1−fm)t)+cos(2π(f1+fm)t)]www.hackshp.cn2whereastheinputtothelowerLPFisul(t)=cos(2πfmt)sin(2πf1t)1=[sin(2π(f1−fm)t)+sin(2π(f1+fm)t)]2Ifweselectf1suchthat|f1−fm|W,thenthetwolowpassfilterswillcut-offthefrequencycomponentsoutsidetheinterval[−W,W],sothattheoutputoftheupperandlowerLPFisyu(t)=cos(2π(f1−fm)t)yl(t)=sin(2π(f1−fm)t)TheoutputoftheWeaver’smodulatorisu(t)=cos(2π(f1−fm)t)cos(2πf2t)−sin(2π(f1−fm)t)sin(2πf2t)52 whichhastheformofaSSBsignalsincesin(2π(f1−fm)t)istheHilberttransformofcos(2π(f1−fm)t).Ifwewriteu(t)asu(t)=cos(2π(f1+f2−fm)t)thenwithf1+f2−fm=fc+fmweobtainanUSSBsignalcenteredatfc,whereaswithf1+f2−fm=fc−fmweobtaintheLSSBsignal.Inbothcasesthechoiceoffcandf1uniquelydeterminef2.Problem3.18Thesignalx(t)ism(t)+cos(2πft).ThespectrumofthissignalisX(f)=M(f)+1(δ(f−f)+020δ(f+f0))anditsbandwidthequalstoWx=f0.Thesignaly1(t)aftertheSquareLawDeviceisy(t)=x2(t)=(m(t)+cos(2πft))210=m2(t)+cos2(2πft)+2m(t)cos(2πft)00211=m(t)++cos(2π2f0t)+2m(t)cos(2πf0t)22Thespectrumofthissignalisgivenby11Y1(f)=M(f)M(f)+δ(f)+(δ(f−2f0)+δ(f+2f0))+M(f−f0)+M(f+f0)24anditsbandwidthisW1=2f0.Thebandpassfilterwillcut-offthelow-frequencycomponentsM(f)M(f)+1δ(f)andthetermswiththedoublefrequencycomponents1(δ(f−2f)+δ(f+2f)).2400ThusthespectrumY2(f)isgivenbyY2(f)=M(f−f0)+M(f+f0)andthebandwidthofy2(t)isW2=2W.Thesignaly3(t)isy(t)=2m(t)cos2(2πft)=m(t)+m(t)cos(2πft)300withspectrum1Y3(t)=M(f)+(M(f−f0)+M(f+f0))课后答案网2andbandwidthW=f+W.Thelowpassfilterwilleliminatethespectralcomponents1(M(f−302f0)+M(f+f0)),sothaty4(t)=m(t)withspectrumY4=M(f)andbandwidthW4=W.Thenextfiguredepictsthespectraofthesignalsx(t),y1(t),y2(t),y3(t)andy4(t).www.hackshp.cn53 X(f).............12−f0−WWf0Y1(f)1......4−2f0−f0−W−f0+W−2W2Wf0−Wf0+W2f0Y2(f)−f0−Wf−f0+W0−Wf0+WY3(f)−f0−W−f0+W−WWf0−Wf0+WY4(f)−WWProblem3.191)课后答案网y(t)=ax(t)+bx2(t)=a(m(t)+cos(2πft))+b(m(t)+cos(2πft))200=am(t)+bm2(t)+acos(2πft)0+bcos2(2πft)+2bm(t)cos(2πft)www.hackshp.cn002)Thefiltershouldrejectthelowfrequencycomponents,thetermsofdoublefrequencyandpassonlythesignalwithspectrumcenteredatf0.ThusthefiltershouldbeaBPFwithcenterfrequencyfandbandwidthWsuchthatf−W>f−W>2WwhereWisthebandwidthofthe00M02MMmessagesignalm(t).3)TheAMoutputsignalcanbewrittenas2bu(t)=a(1+m(t))cos(2πf0t)aSinceAm=max[|m(t)|]weconcludethatthemodulationindexis2bAmα=a54 Problem3.201)WhenUSSBisemployedthebandwidthofthemodulatedsignalisthesamewiththebandwidthofthemessagesignal.Hence,W=W=104HzUSSB2)WhenDSBisused,thenthebandwidthofthetransmittedsignalistwicethebandwidthofthemessagesignal.Thus,W=2W=2×104HzDSB3)IfconventionalAMisemployed,thenW=2W=2×104HzAM4)UsingCarson’srule,theeffectivebandwidthoftheFMmodulatedsignaliskfmax[|m(t)|]Bc=(2β+1)W=2+1W=2(kf+W)=140000HzWProblem3.211)Thelowpassequivalenttransferfunctionofthesystemis1f+1|f|≤WHl(f)=2u−1(f+fc)H(f+fc)=2W2W213CDFwww.hackshp.cn1...........................(1−p)3-1012344)33k3−k23P(X>1)=p(1−p)=3p(1−p)+(1−p)kk=2Problem4.71)TherandomvariablesXandYfollowthebinomialdistributionwithn=4andp=1/4and1/2respectively.Thus04441334411p(X=0)==p(Y=0)==04428022472 134413334414p(X=1)==p(Y=1)==144281224224413332416p(X=2)==p(Y=2)==2442822243144133·4414p(X=3)==p(Y=3)==3442832244044131411p(X=4)==p(Y=4)==444284224SinceXandYareindependentwehave332681p(X=Y=2)=p(X=2)p(Y=2)==282410242)p(X=Y)=p(X=0)p(Y=0)+p(X=1)p(Y=1)+p(X=2)p(Y=2)+p(X=3)p(Y=3)+p(X=4)p(Y=4)3433·4234·223·421886=++++=21221221221221240963)p(X>Y)=p(Y=0)[p(X=1)+p(X=2)+p(X=3)+p(X=4)]+p(Y=1)[p(X=2)+p(X=3)+p(X=4)]+p(Y=2)[p(X=3)+p(X=4)]+p(Y=3)[p(X=4)]535=4096课后答案网5l4)Ingeneralp(X+Y≤5)=l=0m=0p(X=l−m)p(Y=m).Howeveritiseasiertofindp(X+Y≤5)throughp(X+Y≤5)=1−p(X+Y>5)becausefewertermsareinvolvedinthecalculationoftheprobabilityp(X+Y>5).Notealsothatp(X+Y>5|X=0)=p(X+Y>5|X=1)=0.www.hackshp.cnp(X+Y>5)=p(X=2)p(Y=4)+p(X=3)[p(Y=3)+p(Y=4)]+p(X=4)[p(Y=2)+p(Y=3)+p(Y=4)]125=4096Hence,p(X+Y≤5)=1−p(X+Y>5)=1−1254096Problem4.81)Sincelimx→∞FX(x)=1andFX(x)=1forallx≥1weobtainK=1.2)Therandomvariableisofthemixed-typesincethereisadiscontinuityatx=1.lim→0FX(1−
)=1/2whereaslim→0FX(1+
)=13)1113P(2)=1−P(X≤2)=1−FX(2)=1−1=0Problem4.91)x<−1⇒FX(x)=0
xx−1≤x≤0⇒F(x)=(v+1)dv=(1v2+v)=1x2+x+1X−12−122
0
x1210≤x≤1⇒FX(x)=(v+1)dv+(−v+1)dv=−x+x+−10221≤x⇒FX(x)=12)1171p(X>)=1−FX()=1−=22883)p(X>0,X<1)F(1)−F(0)12X2X3p(X>0|X<)===2p(X<1)1−p(X>1)7224)WefindfirsttheCDFp(X≤x,X>1)11F(x|X>)=p(X≤x|X>)=2X122p(X>)课后答案网2Ifx≤1thenp(X≤x|X>1)=0sincetheeventsE={X≤1}andE={X>1}aredisjoint.221212Ifx>1thenp(X≤x|X>1)=F(x)−F(1)sothat22XX2F(x)−F(1)1XX2FX(x|X>)=1www.hackshp.cn21−FX(2)DifferentiatingthisequationwithrespecttoxweobtainfX(x)x>1f(x|X>1)=1−FX(1)2X220x≤125)
∞E[X|X>1/2]=xfX(x|X>1/2)dx−∞
∞1=xfX(x)dx1−FX(1/2)12
∞1=8x(−x+1)dx=8(−1x3+1x2)1321222=374 Problem4.101)TherandomvariableXisGaussianwithzeromeanandvarianceσ2=10−8.Thusp(X>x)=Q(x)andσ10−4p(X>10−4)=Q=Q(1)=.15910−44×10−4p(X>4×10−4)=Q=Q(4)=3.17×10−510−4p(−2×10−410−4,X>0)p(X>10−4).159p(X>10|X>0)====.318p(X>0)p(X>0).53)y=g(x)=xu(x).ClearlyfY(y)=0andFY(y)=0fory<0.Ify>0,thentheequationy=xu(x)hasauniquesolutionx1=y.Hence,FY(y)=FX(y)andfY(y)=fX(y)fory>0.FY(y)isdiscontinuousaty=0andthejumpofthediscontinuityequalsFX(0).+−1FY(0)−FY(0)=FX(0)=2InsummarythePDFfY(y)equals1fY(y)=fX(y)u(y)+δ(y)2ThegeneralexpressionforfindingfY(y)cannotbeusedbecauseg(x)isconstantforsomeintervalsothatthereisanuncountablenumberofsolutionsforxinthisinterval.4)课后答案网
∞E[Y]=yfY(y)dy−∞
∞1=yfX(y)u(y)+δ(y)dy−∞2
∞2www.hackshp.cn1−yσ=√ye2σ2dy=√2πσ202π5)y=g(x)=|x|.Foragiveny>0therearetwosolutionstotheequationy=g(x)=|x|,thatisx1,2=±y.Hencefory>0fX(x1)fX(x2)fY(y)=+=fX(y)+fX(−y)|sgn(x1)||sgn(x2)|22−y=√e2σ22πσ2Fory<0therearenosolutionstotheequationy=|x|andfY(y)=0.
∞22−y2σE[Y]=√ye2σ2dy=√2πσ202π75 Problem4.111)y=g(x)=ax2.Assumewithoutlossofgeneralitythata>0.Then,ify<0theequationy=ax2hasnorealsolutionsandfY(y)=0.Ify>0therearetwosolutionstothesystem,namely&x1,2=y/a.Hence,fX(x1)fX(x2)fY(y)=+|g(x1)||g(x2)|&&fX(y/a)fX(−y/a)=&+&2ay/a2ay/a1−y=√√e2aσ2ay2πσ22)Theequationy=g(x)hasnosolutionsify<−b.ThusFY(y)andfY(y)arezerofory<−b.If−b≤y≤b,thenforafixedy,g(x)btheng(x)≤b0)=1−FX(0)=2ThusFY(y)isastaircasefunctionandwww.hackshp.cnfY(y)=FX(0)δ(y−b)+(1−FX(0))δ(y−a)4)Therandomvariabley=g(x)takesthevaluesyn=xnwithprobabilityp(Y=yn)=p(an≤X≤an+1)=FX(an+1)−FX(an)Thus,FY(y)isastaircasefunctionwithFY(y)=0ifyxN.ThePDFisasequenceofimpulsefunctions,thatisNfY(y)=[FX(ai+1)−FX(ai)]δ(y−xi)i=1Naiai+1=Q−Qδ(y−xi)σσi=176 Problem4.12Theequationx=tanφhasauniquesolutionin[−π,π],thatis22φ1=arctanxFurthermoresinφ1sin2φx(φ)===1+=1+x2cosφcos2φcos2φThus,fΦ(φ1)1fX(x)==|x(φ1)|π(1+x2)WeobservethatfX(x)istheCauchydensity.SincefX(x)isevenweimmediatelygetE[X]=0.However,thevarianceisσ2=E[X2]−(E[X])2X
∞21x=dx=∞π−∞1+x2Problem4.131)
∞
∞E[Y]=yfY(y)dy≥yfY(y)dy0
α∞≥αyfY(y)dy=αp(Y≥α)αThusp(Y≥α)≤E[Y]/α.2)Clearlyp(|X−E[X]|>
)=p((X−E[X])2>
2).Thususingtheresultsofthepreviousquestionweobtain22E[(X−E[X])2]σ2p(|X−E[X]|>
)=p((X−E[X])>
)≤=课后答案网
2
2Problem4.14Thecharacteristicfunctionofthebinomialdistributionisnjvknkn−kψX(v)=ep(1−p)kk=0nwww.hackshp.cnnjvkn−kjvn=(pe)(1−p)=(pe+(1−p))kk=0Thus(1)1djvn1jvn−1jvE[X]=mX=(pe+(1−p))=n(pe+(1−p))pjejdvv=0jv=0=n(p+1−p)n−1p=np(2)d2E[X2]=m=(−1)(pejv+(1−p))nXdv2v=0d=(−1)n(pejv+(1−p)n−1pjejvdvv=0=n(n−1)(pejv+(1−p))n−2p2e2jv+n(pejv+(1−p))n−1pejvv=0=n(n−1)(p+1−p)p2+n(p+1−p)p=n(n−1)p2+np77 Hencethevarianceofthebinomialdistributionisσ2=E[X2]−(E[X])2=n(n−1)p2+np−n2p2=np(1−p)Problem4.15ThecharacteristicfunctionofthePoissondistributionis∞λk∞(ejv−1λ)kψ(v)=ejvke−k=Xk!k!k=0k=0But∞ak=easothatψ(v)=eλ(ejv−1).Hencek=0k!X(1)1d1λ(ejv−1)jvE[X]=mX=ψX(v)=ejλe=λjdvv=0jv=0(2)d2djv−1E[X2]=m=(−1)ψ(v)=(−1)λeλ(e)ejvjXdv2Xdvv=0v=0=λ2eλ(ejv−1)ejv+λeλ(ejv−1)ejv=λ2+λv=0HencethevarianceofthePoissondistributionisσ2=E[X2]−(E[X])2=λ2+λ−λ2=λProblem4.16Fornodd,xnisoddandsincethezero-meanGaussianPDFiseventheirproductisodd.Sincetheintegralofanoddfunctionovertheinterval[−∞,∞]iszero,weobtainE[Xn]=0forneven.∞n22LetIn=−∞xexp(−x/2σ)dxwithneven.Then,
d∞x21x2n−1−n+1−In=nxe2σ2−xe2σ2dx=0dx−∞σ2d2
∞x22n+1x21x2课后答案网n−2−n−n+2−In=n(n−1)xe2σ2−xe2σ2+xe2σ2dxdx2−∞σ2σ42n+11=n(n−1)In−2−σ2In+σ4In+2=0Thus,Iwww.hackshp.cn=σ2(2n+1)I−σ4n(n−1)In+2nn−2√√withinitialconditionsI0=2πσ2,I2=σ22πσ2.Weprovenowthat√I=1×3×5×···×(n−1)σn2πσ2n√Theproofisbyinductiononn.Forn=2itiscertainlytruesinceI2=σ22πσ2.WeassumethattherelationholdsfornandwewillshowthatitistrueforIn+2.Usingthepreviousrecursionwehave√I=1×3×5×···×(n−1)σn+2(2n+1)2πσ2n+2√−1×3×5×···×(n−3)(n−1)nσn−2σ42πσ2√=1×3×5×···×(n−1)(n+1)σn+22πσ2ClearlyE[Xn]=√1Iand2πσ2nE[Xn]=1×3×5×···×(n−1)σn78 Problem4.171)fX,Y(x,y)isaPDFsothatitsintegraloverthesupportregionofx,yshouldbeone.
1
1
1
1fX,Y(x,y)dxdy=K(x+y)dxdy0000



1111=Kxdxdy+ydxdy00001111=Kx2y|1+y2x|1202000=KThusK=1.2)p(X+Y>1)=1−P(X+Y≤1)
1
1−x=1−(x+y)dxdy00
1
1−x
1
1−x=1−xdydx−dxydy0000
1
112=1−x(1−x)dx−(1−x)dx0022=33)ByexploitingthesymmetryoffX,Yandthefactthatithastointegrateto1,oneimmediatelyseesthattheanswertothisquestionis1/2.The“mechanical”solutionis:
1
1p(X>Y)=(x+y)dxdy课后答案网0y
1
1
1
1=xdxdy+ydxdy0y0y
111
11=x2dy+yxdy20y0y
1
1www.hackshp.cn12=(1−y)dy+y(1−y)dy0201=24)p(X>Y|X+2Y>1)=p(X>Y,X+2Y>1)/p(X+2Y>1)Theregionoverwhichweintegrateinordertofindp(X>Y,X+2Y>1)ismarkedwithanAinthefollowingfigure.y(1,1).A...x1/3x+2y=179 Thus
1
xp(X>Y,X+2Y>1)=(x+y)dxdy11−x32
11−x121−x2=x(x−)+(x−())dx12223
11511=x2−x−dx1848349=108
1
1p(X+2Y>1)=(x+y)dxdy01−x2
11−x11−x2=x(1−)+(1−())dx0222
1333=x2+x+dx0848111=3×1x3+3×1x2+3x830420807=8Hence,p(X>Y|X+2Y>1)=(49/108)/(7/8)=14/275)WhenX=YthevolumeunderintegrationhasmeasurezeroandthusP(X=Y)=06)ConditionedonthefactthatX=Y,thenewp.d.fofXisfX,Y(x,x)fX|X=Y(x)=1=2x.课后答案网0fX,Y(x,x)dxInwords,were-normalizefX,Y(x,y)sothatitintegratesto1ontheregioncharacterizedbyX=Y.11Theresultdependsonlyonx.Thenp(X>2|X=Y)=1/2fX|X=Y(x)dx=3/4.7)
1
1www.hackshp.cn1fX(x)=(x+y)dy=x+ydy=x+002
1
11fY(y)=(x+y)dx=y+xdx=y+0028)FX(x|X+2Y>1)=p(X≤x,X+2Y>1)/p(X+2Y>1)
x
1p(X≤x,X+2Y>1)=(v+y)dvdy01−v2
x333=v2+v+dv084813323=x+x+x888Hence,3x2+6x+3363f(x|X+2Y>1)=888=x2+x+Xp(X+2Y>1)77780
1E[X|X+2Y>1]=xfX(x|X+2Y>1)dx0
1363=x3+x2+x0777111=3×1x4+6×1x3+3×1x2=1774073072028Problem4.181)FY(y)=p(Y≤y)=p(X1≤y∪X2≤y∪···∪Xn≤y)Sincethepreviouseventsarenotnecessarilydisjoint,itiseasiertoworkwiththefunction1−[FY(y)]=1−p(Y≤y)inordertotakeadvantageoftheindependenceofXi’s.Clearly1−p(Y≤y)=p(Y>y)=p(X1>y∩X2>y∩···∩Xn>y)=(1−FX1(y))(1−FX2(y))···(1−FXn(y))Differentiatingthepreviouswithrespecttoyweobtain"n"n"nfY(y)=fX1(y)(1−FXi(y))+fX2(y)(1−FXi(y))+···+fXn(y)(1−FXi(y))i=1i=2i=n2)FZ(z)=P(Z≤z)=p(X1≤z,X2≤z,···,Xn≤z)=p(X1≤z)p(X2≤z)···p(Xn≤z)Differentiatingthepreviouswithrespecttozweobtain"n"n"nfZ(z)=fX1(z)FXi(z)+fX2(z)FXi(z)+···+fXn(z)FXi(z)i=1i=2i=nProblem4.19课后答案网

∞xx21∞x2−2−E[X]=xe2σ2dx=xe2σ2dx0σ2σ20HoweverfortheGaussianrandomvariableofzeromeanandvarianceσ2
1∞x22−2www.hackshp.cn√xe2σ2dx=σ2πσ2−∞Sincethequantityunderintegrationiseven,weobtainthat
1∞x212−2√xe2σ2dx=σ2πσ202Thus,(1√1πE[X]=2πσ2σ2=σσ222InordertofindVAR(X)wefirstcalculateE[X2].

1∞x2∞x223−−E[X]=xe2σ2dx=−xd[e2σ2]σ2002∞
∞2−x−x2=−xe2σ2+2xe2σ2dx00
∞xx22−2=0+2σe2σ2dx=2σ0σ281 Thus,222π2π2VAR(X)=E[X]−(E[X])=2σ−σ=(2−)σ22Problem4.20LetZ=X+Y.Then,
∞
z−yFZ(z)=p(X+Y≤z)=fX,Y(x,y)dxdy−∞−∞Differentiatingwithrespecttozweobtain
∞d
z−yfZ(z)=fX,Y(x,y)dxdy−∞dz−∞
∞d=fX,Y(z−y,y)(z−y)dy−∞dz
∞=fX,Y(z−y,y)dy−∞
∞=fX(z−y)fY(y)dy−∞wherethelastlinefollowsfromtheindependenceofXandY.ThusfZ(z)istheconvolutionoffX(x)andfY(y).WithfX(x)=αe−αxu(x)andfY(y)=βe−βxu(x)weobtain
zf(z)=αe−αvβe−β(z−v)dvZ0Ifα=βthen
zf(z)=α2e−αzdv=α2ze−αzu(z)Z−10Ifα=βthen
zαβf(z)=αβe−βze(β−α)vdv=e−αz−e−βzu(z)Z−1课后答案网0β−αProblem4.211)fX,Y(x,y)isaPDF,henceitsintegraloverthesupportingregionofx,andyis1.
∞
∞
∞
∞f(x,ywww.hackshp.cn)dxdy=Ke−x−ydxdyX,Y0y0y
∞
∞=Ke−ye−xdxdy0y
∞∞=Ke−2ydy=K(−1)e−2y=K10220ThusKshouldbeequalto2.2)
xxf(x)=2e−x−ydy=2e−x(−e−y)=2e−x(1−e−x)X00
∞∞f(y)=2e−x−ydy=2e−y(−e−x)=2e−2yYyy82 3)f(x)f(y)=2e−x(1−e−x)2e−2y=2e−x−y2e−y(1−e−x)XY=2e−x−y=f(x,y)X,YThusXandYarenotindependent.4)Ifx3/2>1/2.Problem4.23√√1)ClearlyX>r,Y>rimpliesthatX2>r2,Y2>r2sothatX2+Y2>2r2orX2+Y2>2r.√√ThustheeventE1(r)={X>r,Y>r}isasubsetoftheeventE2(r)={X2+Y2>2r|X,Y>0}andp(E1(r))≤p(E2(r)).2)SinceXandYareindependentp(E(r))=p(X>r,Y>r)=p(X>r)p(Y>r)=Q2(r)1√3)UsingtherectangulartopolartransformationV=X2+Y2,Θ=arctanYitisproved(seeXtextEq.4.1.22)that2课后答案网v−vfV,Θ(v,θ)=2e2σ22πσHence,withσ2=1weobtain&√
∞
πv22v−p(X2+Y2>2r|X,Y>0)=√e2dvdθwww.hackshp.cn2r02π
∞22∞1−v1−v=√ve2dv=(−e2)√42r42r1−r2=e4Combiningtheresultsofpart1),2)and3)weobtain221−r21−rQ(r)≤eorQ(r)≤e242Problem4.24ThefollowingisaprogramwritteninFortrantocomputetheQfunctionREAL*8x,t,a,q,pi,p,b1,b2,b3,b4,b5PARAMETER(p=.2316419d+00,b1=.31981530d+00,84 +b2=-.356563782d+00,b3=1.781477937d+00,+b4=-1.821255978d+00,b5=1.330274429d+00)C-pi=4.*atan(1.)C-INPUTPRINT*,’Enter-x-’READ*,xC-t=1./(1.+p*x)a=b1*t+b2*t**2.+b3*t**3.+b4*t**4.+b5*t**5.q=(exp(-x**2./2.)/sqrt(2.*pi))*aC-OUTPUTPRINT*,qC-STOPENDTheresultsofthisapproximationalongwiththeactualvaluesofQ(x)(takenfromtextTable4.1)aretabulatedinthefollowingtable.Asitisobservedaverygoodapproximationisachieved.xQ(x)Approximation1.1.59×10−11.587×10−11.56.68×10−26.685×10−22.2.28×10−22.276×10−22.56.21×10−36.214×10−33.1.35×10−31.351×10−33.52.33×10−42.328×10−44.3.17×10−53.171×10−54.53.40×10−63.404×10−65.2.87×10−72.874×10−7Problem4.25课后答案网Then-dimensionaljointGaussiandistributionis1−(x−m)C−1(x−m)tfX(x)=&e(2π)ndet(C)TheJacobianofthelineartransformationwww.hackshp.cnY=AXt+bis1/det(A)andthesolutiontothisequationisx=(y−b)t(A−1)tWemaysubstituteforxinfX(x)toobtainfY(y).1f(y)=exp−[(y−b)t(A−1)t−m]C−1Y(2π)n/2(det(C))1/2|det(A)|[(y−b)t(A−1)t−m]t1=exp−[yt−bt−mAt](At)−1C−1A−1(2π)n/2(det(C))1/2|det(A)|[y−b−Amt]1=exp−[yt−bt−mAt](ACAt)−1(2π)n/2(det(C))1/2|det(A)|[yt−bt−mAt]t85 ThusfY(y)isan-dimensionaljointGaussiandistributionwithmeanandvariancegivenbym=b+Amt,C=ACAtYYProblem4.261)ThejointdistributionofXandYisgivenby)11σ20XfX,Y(x,y)=2exp−XY0σ2Y2πσ2ThelineartransformationsZ=X+YandW=2X−YarewritteninmatrixnotationasZ11XX==AW2−1YYThus,(seeProb.4.25))11−1ZfZ,W(z,w)=exp−ZWM2πdet(M)1/22Wwhereσ20t2σ2σ2σ2ρZ,WσZσWZM=A2A=22=20σσ5σρZ,WσZσWσW√Fromthelastequalityweidentifyσ2=2σ2,σ2=5σ2andρZ,W=1/10ZW2)XFR(r)=p(R≤r)=p(≤r)Y
∞
yr
0
∞=fX,Y(x,y)dxdy+fX,Y(x,y)dxdy0课后答案网−∞−∞yrDifferentiatingFR(r)withrespecttorweobtainthePDFfR(r).Notethat
adf(x)dx=f(a)dab
adwww.hackshp.cnf(x)dx=−f(b)dbbThus,
∞d
yr
0d
∞FR(r)=fX,Y(x,y)dxdy+fX,Y(x,y)dxdy0dr−∞−∞dryr
∞
0=yfX,Y(yr,y)dy−yfX,Y(yr,y)dy0−∞
∞=|y|fX,Y(yr,y)dy−∞Hence,
∞y2r2+y2
∞1+r21−1−y2()fR(r)=|y|e2σ2dy=2ye2σ2dy−∞2πσ202πσ212σ211=2=2πσ22(1+r2)π1+r286 fR(r)istheCauchydistribution;itsmeaniszeroandthevariance∞.Problem4.27Thebinormaljointdensityfunctionis*11fX,Y(x,y)=&exp−×2πσ1σ21−ρ22(1−ρ2))(x−m1)2(y−m2)22ρ(x−m1)(y−m2)+−σ2σ2σ1σ2121+,=&exp−(z−m)C−1(z−m)t(2π)ndet(C)wherez=[xy],m=[m1m2]andσ2ρσ1σ2C=1ρσ1σ2σ221)With4−4C=−49weobtainσ2=4,σ2=9andρσσ=−4.Thusρ=−2.121232)ThetransformationZ=2X+Y,W=X−2YiswritteninmatrixnotationasZ21XX==AW1−2YYTheditributionfZ,W(z,w)isbinormalwithmeanm=mAt,andcovariancematrixC=ACAt.Hence214−42192C==1课后答案网−2−491−2256Theoff-diagonalelementsofCareequaltoρσZσW=COV(Z,W).ThusCOV(Z,W)=2.3)ZwillbeGaussianwithvarianceσ2=9andmeanZwww.hackshp.cn2mZ=[m1m2]=41Problem4.28√fX,Y(x,y)2πσYfX|Y(x|y)==%exp[−A]fY(y)2πσXσY1−ρ2X,Ywhere(x−mX)2(y−mY)2(x−mX)(y−mY)(y−mY)2A=+−2ρ−2(1−ρ2)σ22(1−ρ2)σ22(1−ρ2)σXσY2σ2X,YXX,YYX,YY1(y−mY)2σ2ρ2(x−mX)(y−mY)σX2XX,Y=2(1−ρ2)σ2(x−mX)+σ2−2ρσX,YXYY21ρσX=22x−mX+(y−mY)σ2(1−ρX,Y)σXY87 Thus2)11ρσXfX|Y(x|y)=√%exp−22x−mX+(y−mY)2πσX1−ρ2X,Y2(1−ρX,Y)σXσYwhichisaGaussianPDFwithmeanmX+(y−mY)ρσX/σYandvariance(1−ρ2)σ2.Ifρ=0X,YXthenfX|Y(x|y)=fX(x)whichimpliesthatYdoesnotprovideanyinformationaboutXorX,Yareindependent.Ifρ=±1thenthevarianceoffX|Y(x|y)iszerowhichmeansthatX|Yisdeterministic.Thisistobeexpectedsinceρ=±1impliesalinearrelationX=AY+bsothatknowledgeofYprovidesalltheinformationaboutX.Problem4.291)TherandomvariablesZ,WarealinearcombinationofthejointlyGaussianrandomvariablesX,Y.ThustheyarejointlyGaussianwithmeanm=mAtandcovariancematrixC=ACAt,wherem,CisthemeanandcovariancematrixoftherandomvariablesXandYandAisthetransformationmatrix.Thebinormaljointdensityfunctionis1+,f(z,w)=&exp−([zw]−m)C−1([zw]−m)tZ,Wn(2π)det(C)|det(A)|Ifm=0,thenm=mAt=0.Withσ2ρσ2cosθsinθC=22A=ρσσ−sinθcosθweobtaindet(A)=cos2θ+sin2θ=1andcosθsinθσ2ρσ2cosθ−sinθC=22−sinθcosθρσσsinθcosθσ2(1+ρsin2θ)ρσ2(cos2θ−sin2θ)=2222ρσ(cosθ−sinθ)σ(1−ρsin2θ)课后答案网2)SinceZandWarejointlyGaussianwithzero-mean,theyareindependentiftheyareuncorre-lated.Thisimpliesthat22ππcosθ−sinθ=0=⇒θ=+k,k∈Zwww.hackshp.cn42NotealsothatifXandYareindependent,thenρ=0andanyrotationwillproduceindependentrandomvariablesagain.Problem4.301)fX,Y(x,y)isaPDFanditsintegraloverthesupportingregionofxandyshouldbeone.
∞
∞fX,Y(x,y)dxdy−∞−∞



0022∞∞22K−x+yK−x+y=e2dxdy+e2dxdy−∞−∞π00π



K0x20y2K∞x2∞y2−−−−=e2dxe2dx+e2dxe2dxπ−∞−∞π00K1√=2(2π)2=Kπ2ThusK=188 2)Ifx<0then

01x2+y21x20y2−−−fX(x)=e2dy=e2e2dy−∞ππ−∞1−x21√1−x2=e22π=√e2π22πIfx>0then

∞1x2+y21x2∞y2−−−fX(x)=e2dy=e2e2dy0ππ01−x21√1−x2=e22π=√e2π22π21−xThusforeveryx,fX(x)=√e2whichimpliesthatfX(x)isazero-meanGaussianrandom2πvariablewithvariance1.SincefX,Y(x,y)issymmetrictoitsargumentsandthesameistruefortheregionofintegrationweconcludethatfY(y)isazero-meanGaussianrandomvariableofvariance1.3)fX,Y(x,y)hasnotthesameformasabinormaldistribution.Forxy<0,fX,Y(x,y)=0butabinormaldistributionisstrictlypositiveforeveryx,y.4)TherandomvariablesXandYarenotindependentforifxy<0thenfX(x)fY(y)=0whereasfX,Y(x,y)=0.5)



0022∞∞221−x+y1−x+yE[XY]=XYe2dxdy+e2dxdyπ−∞−∞π00



10x20y21∞x2∞y2−−−−=Xe2dxYe2dy+Xe2dxYe2dyπ−∞−∞π00112=(−1)(−1)+=π课后答案网ππThustherandomvariablesXandYarecorrelatedsinceE[XY]=0andE[X]=E[Y]=0,sothatE[XY]−E[X]E[Y]=0.fX,Y(x,y)6)IngeneralfX|Y(x,y)=fY(y)www.hackshp.cn.Ify>0,then0x<0fX|Y(x,y)=%x22−e2x≥0πIfy≤0,then0x>0fX|Y(x,y)=%x22−e2x<0πThus(22−xfX|Y(x,y)=e2u(xy)πwhichisnotaGaussiandistribution.Problem4.31)1(x−m)2+y2fX,Y(x,y)=exp−2πσ22σ289 Withthetransformation&YV=X2+Y2,Θ=arctanXweobtainfV,Θ(v,θ)=vfX,Y(vcosθ,vsinθ))v(vcosθ−m)2+v2sinθ=exp−2πσ22σ2)vv2+m2−2mvcosθ=exp−2πσ22σ2Toobtainthemarginalprobabilitydensityfunctionforthemagnitude,weintegrateoverθsothat
2πvv2+m2mvcosθ−fV(v)=e2σ2eσ2dθ02πσ2
vv2+m212πmvcosθ−=e2σ2eσ2dθσ22π022v−v+mmv=e2σ2I0()σ2σ2where
mv12πmvcosθI0()=eσ2dθσ22π0Withm=0weobtain2−vv2e2σ2v>0fV(v)=σ0v≤0whichistheRayleighdistribution.Problem4.321)LetXbearandomvariabletakingthevalues1,0,withprobability1and3respectively.Then,i44m=1·1+3·0=1.TheweaklawoflargenumbersstatesthattherandomvariableY=1nXXi444ni=1ihasmeanwhichconvergestomXiwithprobabilityone.UsingChebychev’sinequality(seeProblemσ2200014.13)wehavep(|Y−m|≥
)≤课后答案网Yforevery
>0.Hence,withn=2000,Z=X,m=Xi2i=1iXi4weobtainσ2σ2p(|Z−500|≥2000
)≤Y⇒p(500−2000
≤Z≤500+2000
)≥1−Y
2
221n1223ThevarianceσYofY=ni=1Xwww.hackshp.cniisnσXi,whereσXi=p(1−p)=16(seeProblem4.13).Thus,with
=0.001weobtain3/16p(480≤Z≤520)≥1−=.0632×10−11n2)UsingtheC.L.T.theCDFoftherandomvariableY=ni=1XiconvergestotheCDFoftherandomvariableN(m,√σ).HenceXin480520480520n−mXin−mXiP=p≤Y≤=Q−QnnσσWithn=2000,m=1,σ2=p(1−p)weobtainXi4n480−500520−500P=Q&−Q&2000p(1−p)2000p(1−p)20=1−2Q√=.68237590 Problem4.33Considertherandomvariablevectorx=[ωω+ω...ω+ω+···+ω]t11212nwhereeachωiistheoutcomeofaGaussianrandomvariabledistributedaccordingtoN(0,1).Sincemx,i=E[ω1+ω2+···+ωi)]=E[ω1]+E[ω2]+···+E[ωi]=0weobtainmx=0ThecovariancematrixisC=E[(x−m)(x−m)t]=E[xxt]xxThei,jelement(Ci,j)ofthismatrixisCi,j=E[(ω1+ω2+···+ωi)(ω1+ω2+···+ωj)]=E[(ω1+ω2+···+ωmin(i,j))(ω1+ω2+···+ωmin(i,j))]+E[(ω1+ω2+···+ωmin(i,j))(ωmin(i,j)+1+···+ωmax(i,j))]Theexpectationinthelastlineofthepreviousequationiszero.Thisistruesincealltheran-domvariablesinsidethefirstparenthesisaredifferentfromtherandomvariablesinthesecondparenthesis,andforuncorrelatedrandomvariablesofzeromeanE[ωkωl]whenk=l.Hence,Ci,j=E[(ω1+ω2+···+ωmin(i,j))(ω1+ω2+···+ωmin(i,j))]min(i,j)min(i,j)min(i,j)=E[ωkωl]=E[ωkωk]+E[ωkωl]k=1l=1k=1k,l=1k=lmin(i,j)=1=min(i,j)k=1课后答案网Thus11···1122C=.........www.hackshp.cn12···nProblem4.34TherandomvariableX(t0)isuniformlydistributedover[−11].Hence,mX(t0)=E[X(t0)]=E[X]=0AsitisobservedthemeanmX(t0)isindependentofthetimeinstantt0.Problem4.35mX(t)=E[A+Bt]=E[A]+E[B]t=0wherethelastequalityfollowsfromthefactthatA,Bareuniformlydistributedover[−11]sothatE[A]=E[B]=0.RX(t1,t2)=E[X(t1)X(t2)]=E[(A+Bt1)(A+Bt2)]=E[A2]+E[AB]t+E[BA]t+E[B2]tt211291 TherandomvariablesA,BareindependentsothatE[AB]=E[A]E[B]=0.Furthermore
122211311E[A]=E[B]=xdx=x|=26−13−1Thus11RX(t1,t2)=+t1t233Problem4.36Sincethejointdensityfunctionof{X(ti}ni=1isajointlyGaussiandensityofzero-meantheauto-correlationmatrixoftherandomvectorprocessissimplyitscovariancematrix.Thei,jelementofthematrixisRX(ti,tj)=COV(X(ti)X(tj))+mX(ti)mX(tj)=COV(X(ti)X(tj))=σ2min(t,t)ijProblem4.37SinceX(t)=Xwiththerandomvariableuniformlydistributedover[−11]weobtainfX(t1),X(t2),···,X(tn)(x1,x2,...,xn)=fX,X,···,X(x1,x2,...,xn)forallt1,...,tnandn.Hence,thestatisticalpropertiesoftheprocessaretimeindependentandbydefinitionwehaveastationaryprocess.Problem4.38Theprocessisnotwidesensestationaryfortheautocorrelationfunctiondependsonthevaluesoft1,t2andnotontheirdifference.Toseethissupposethatt1=t2=t.Iftheprocesswaswidesensestationary,thenRX(t,t)=RX(0).However,RX(t,t)=σ2tanditdependsontasitisopposedtoRX(0)whichisindependentoft.Problem4.39课后答案网IfaprocessX(t)isMthorderstationary,thenforalln≤M,and∆fX(t1)X(t2)···X(tn)(x1,x2,···,xn)=fX(t1+∆)···X(tn+∆)(x1,···xn)Ifweletn=1,thenwww.hackshp.cn
∞
∞mX(0)=E[X(0)]=xfX(0)(x)dx=xfX(0+t)(x)dx=mX(t)−∞−∞forallt.Hence,mx(t)isconstant.Withn=2weobtainfX(t1)X(t2)(x1,x2)=fX(t1+∆)X(t2+∆)(x1,x2),∀∆Ifwelet∆=−t1,thenfX(t1)X(t2)(x1,x2)=fX(0)X(t2−t1)(x1,x2)whichmeansthat
∞
∞Rx(t1,t2)=E[X(t1)X(t2)]=x1x2fX(0)X(t2−t1)(x1,x2)dx1dx2−∞−∞dependsonlyonthedifferenceτ=t1−t2andnotontheindividualvaluesoft1,t2.ThustheMthorderstationaryprocess,hasaconstantmeanandanautocorrelationfunctiondependentonτ=t1−t2only.Hence,itisawidesensestationaryprocess.92 Problem4.401)f(τ)cannotbetheautocorrelationfunctionofarandomprocessforf(0)=0f(0)for|τ|>1.Thusf(τ)cannotbetheautocorrelationfunctionofarandomprocess.4)f(τ)isevenandthemaximumisachievedattheorigin(τ=0).Wecanwritef(τ)asf(τ)=1.2Λ(τ)−Λ(τ−1)−Λ(τ+1)TakingtheFouriertransformofbothsidesweobtainS(f)=1.2sinc2(f)−sinc2(f)e−j2πf+ej2πf=sinc2(f)(1.2−2cos(2πf))AsweobservethepowerspectrumS(f)cantakenegativevalues,i.e.forf=0.Thusf(τ)cannotbetheautocorrelationfunctionofarandomprocess.Problem4.41AswehaveseeninProblem4.38theprocessisnotstationaryandthusitisnotergodic.Thisinaccordancetoourdefinitionofergodicityasapropertyofstationaryandergodicprocesses.Problem4.42Therandomvariableωtakesthevalues{1,2,...,6}withprobability1.Thusi6
∞E=EX2(t)dtX−∞

∞∞=Eω2e−2tu2(t)dt=Eω2e−2tdti−1i−∞0
∞
∞16=课后答案网E[ω2]e−2tdt=i2e−2tdti006i=1
∞∞=91e−2tdt=91(−1e−2t)6062091=12www.hackshp.cnThustheprocessisanenergy-typeprocess.However,thisprocessisnotstationaryfor−t21−tmX(t)=E[X(t)=E[ωi]eu−1(t)=eu−1(t)6isnotconstant.Problem4.431)Wefindfirsttheprobabilityofanevennumberoftransitionsintheinterval(0,τ].pN(n=even)=pN(0)+pN(2)+pN(4)+···∞21ατ=1+ατ1+ατl=011=1+ατ(ατ)21−(1+ατ)21+ατ=1+2ατ93 Theprobabilityp(n=odd)issimply1−p(n=even)=ατ.TherandomprocessZ(t)takesNN1+2ατthevalueof1(attimeinstantt)ifanevennumberoftransitionsoccurredgiventhatZ(0)=1,orifanoddnumberoftransitionsoccurredgiventhatZ(0)=0.Thus,mZ(t)=E[Z(t)]=1·p(Z(t)=1)+0·p(Z(t)=0)=p(Z(t)=1|Z(0)=1)p(Z(0)=1)+p(Z(t)=1|Z(0)=0)p(Z(0)=0)11=pN(n=even)+pN(n=odd)221=22)TodetermineRZ(t1,t2)notethatZ(t+τ)=1ifZ(t)=1andanevennumberoftransitionsoccurredintheinterval(t,t+τ],orifZ(t)=0andanoddnumberoftransitionshavetakenplacein(t,t+τ].Hence,RZ(t+τ,t)=E[Z(t+τ)Z(t)]=1·p(Z(t+τ)=1,Z(t)=1)+0·p(Z(t+τ)=1,Z(t)=0)+0·p(Z(t+τ)=0,Z(t)=1)+0·p(Z(t+τ)=0,Z(t)=0)=p(Z(t+τ)=1,Z(t)=1)=p(Z(t+τ)=1|Z(t)=1)p(Z(t)=1)11+ατ=21+2ατAsitisobservedRZ(t+τ,t)dependsonlyonτandthustheprocessisstationary.Theprocessisnotcyclostationary.3)Sincetheprocessisstationary1PZ=RZ(0)=2Problem4.441)课后答案网mX(t)=E[X(t)]=E[Xcos(2πf0t)]+E[Ysin(2πf0t)]=E[X]cos(2πf0t)+E[Y]sin(2πf0t)=0wherethelastequalityfollowsfromthefactthatwww.hackshp.cnE[X]=E[Y]=0.2)RX(t+τ,t)=E[(Xcos(2πf0(t+τ))+Ysin(2πf0(t+τ)))(Xcos(2πf0t)+Ysin(2πf0t))]=E[X2cos(2πf(t+τ))cos(2πft)]+00E[XYcos(2πf0(t+τ))sin(2πf0t)]+E[YXsin(2πf0(t+τ))cos(2πf0t)]+E[Y2sin(2πf(t+τ))sin(2πft)]00σ2=[cos(2πf0(2t+τ))+cos(2πf0τ)]+2σ2[cos(2πf0τ)−cos(2πf0(2t+τ))]2=σ2cos(2πfτ)094 wherewehaveusedthefactthatE[XY]=0.ThustheprocessisstationaryforRX(t+τ,t)dependsonlyonτ.3)SincetheprocessisstationaryPX=RX(0)=σ2.4)IfσX2=σY2,thenmX(t)=E[X]cos(2πf0t)+E[Y]sin(2πf0t)=0andR(t+τ,t)=E[X2]cos(2πf(t+τ))cos(2πft)+X00E[Y2]sin(2πf(t+τ))sin(2πft)00σ2=X[cos(2πf(2t+τ))−cos(2πfτ)]+002σ2Y[cos(2πfτ)−cos(2πf(2t+τ))]002σ2−σ2=XYcos(2πf(2t+τ)+02σ2+σ2XYcos(2πfτ)02TheprocessisnotstationaryforRX(t+τ,t)doesnotdependonlyonτbutontaswell.HowevertheprocessiscyclostationarywithperiodT=1.NotethatifXorYisnotofzeromeanthen02f0theperiodofthecyclostationaryprocessisT=1.ThepowerspectraldensityofX(t)is0f0
T222212σX−σYσX+σYPX=limcos(2πf02t)+dt=∞T→∞T−T222Problem4.451)课后答案网∞mX(t)=E[X(t)]=EAkp(t−kT)k=−∞∞=E[Ak]p(t−kT)k=−∞www.hackshp.cn∞=mp(t−kT)k=−∞2)RX(t+τ,t)=E[X(t+τ)X(t)]∞∞=EAkAlp(t+τ−kT)p(t−lT)k=−∞l=−∞∞∞=E[AkAl]p(t+τ−kT)p(t−lT)k=−∞l=−∞∞∞=RA(k−l)p(t+τ−kT)p(t−lT)k=−∞l=−∞95 3)∞∞RX(t+T+τ,t+T)=RA(k−l)p(t+T+τ−kT)p(t+T−lT)k=−∞l=−∞∞∞=R(k+1−(l+1))p(t+τ−kT)p(t−lT)Ak
=−∞l
=−∞∞∞=R(k−l)p(t+τ−kT)p(t−lT)Ak
=−∞l
=−∞=RX(t+τ,t)wherewehaveusedthechangeofvariablesk=k−1,l=l−1.SincemX(t)andRX(t+τ,t)areperiodic,theprocessiscyclostationary.4)
T1R¯X(τ)=RX(t+τ,t)dtT0
T∞∞1=RA(k−l)p(t+τ−kT)p(t−lT)dtT0k=−∞l=−∞∞∞
T1=RA(n)p(t+τ−lT−nT)p(t−lT)dtT0n=−∞l=−∞1∞∞
T−lT=R(n)p(t+τ−nT)p(t)dtAT−lTn=−∞l=−∞∞
∞1=RA(n)p(t+τ−nT)p(t)dtT−∞n=−∞∞1=RA(n)Rp(τ−nT)Tn=−∞∞whereRp(τ−nT)=−∞p(t+τ−nT)p(t)dt=p(t)p(−t)|t=τ−nT5)课后答案网∞1SX(f)=F[R¯X(τ)]=FRA(n)Rp(τ−nT)Tn=−∞∞
∞1−j2πfτ=RA(n)Rp(τ−nT)edτT−∞www.hackshp.cnn=−∞∞
∞1−j2πf(τ
+nT)=RA(n)Rp(τ)edτT−∞n=−∞∞
∞1−j2πfnT−j2πfτ
=RA(n)eRp(τ)edτT−∞n=−∞But,Rp(τ)=p(τ)p(−τ)sothat
∞
∞
∞R(τ)e−j2πfτ
dτ=p(τ)e−j2πfτ
dτp(−τ)e−j2πfτ
dτp−∞−∞−∞=P(f)P∗(f)=|P(f)|2wherewehaveusedthefactthatforrealsignalsP(−f)=P∗(f).SubstitutingtherelationabovetotheexpressionforSX(f)weobtain|P(f)|2∞S(f)=R(n)e−j2πfnTXATn=−∞96 |P(f)|2∞=RA(0)+2RA(n)cos(2πfnT)Tn=1wherewehaveusedtheassumptionRA(n)=RA(−n)andthefactej2πfnT+e−j2πfnT=2cos(2πfnT)Problem4.461)TheautocorrelationfunctionofAn’sisRA(k−l)=E[AkAl]=δklwhereδklistheKronecker’sdelta.Furthermoret−T−j2πfT2P(f)=FΠ()=Tsinc(Tf)e2THence,usingtheresultsofProblem4.45weobtainS(f)=Tsinc2(Tf)X2)InthiscaseE[A]=1andR(k−l)=E[AA].Ifk=l,thenR(0)=E[A2]=1.Ifk=l,n2AklAk2thenR(k−l)=E[AA]=E[A]E[A]=1.ThepowerspectraldensityoftheprocessisAklkl4∞211SX(f)=Tsinc(Tf)+cos(2πkfT)22k=1t−3T/23)Ifp(t)=Π(3T)andAn=±1withequalprobability,then|P(f)|21−j2πf3T2SX(f)=RA(0)=3Tsinc(3Tf)e2TT=9Tsinc2(3Tf)Forthesecondpartthepowerspectraldensityis∞211SX(f)=9Tsinc(3Tf)+cos(2πkfT)课后答案网22k=1Problem4.471)E[Bn]=E[An]+E[An−1]=0.TofindtheautocorrelationsequenceofBn’swewriteRB(k−l)=Ewww.hackshp.cn[BkBl]=E[(Ak+Ak−1)(Al+Al−1)]=E[AkAl]+E[AkAl−1]+E[Ak−1Al]+E[Ak−1Al−1]Ifk=l,thenRB(0)=E[A2k]+E[A2k−1]=2.Ifk=l−1,thenRB(1)=E[AkAl−1]]=1.Similarly,ifk=l+1,RB(−1)=E[Ak−1Al]]=1.Thus,2k−l=0RB(k−l)=1k−l=±10otherwiseUsingtheresultsofProblem4.45weobtain|P(f)|2∞SX(f)=RB(0)+2RB(k)cos(2πkfT)Tk=1|P(f)|2=(2+2cos(2πfT))T97 2)ConsiderthesamplesequenceofAn’s{···,−1,1,1,−1,−1,−1,1,−1,1,−1,···}.Thenthecor-respondingsequenceofBn’sis{···,0,2,0,−2,−2,0,0,0,0,···}.ThefollowingfiguredepictsthecorrespondingsamplefunctionX(t).......Ifp(t)=Π(t−T/2),then|P(f)|2=T2sinc2(Tf)andthepowerspectraldensityisTS(f)=Tsinc2(Tf)(2+2cos(2πfT))XInthenextfigureweplotthepowerspectraldensityforT=1.43.532.521.510.50-5-4-3-2-10123453)IfBn=An+αAn−1,then1+α2k−l=0RB(k−l)=αk−l=±1课后答案网0otherwiseThepowerspectraldensityinthiscaseisgivenby|P(f)|22SX(f)=(1+α+2αcos(2πfT))TProblem4.48www.hackshp.cnIngeneralthemeanofafunctionoftworandomvariables,g(X,Y),canbefoundasE[g(X,Y)]=E[E[g(X,Y)|X]]wheretheouterexpectationiswithrespecttotherandomvariableX.1)mY(t)=E[X(t+Θ)]=E[E[X(t+Θ)|Θ]]where
E[X(t+Θ)|Θ]=X(t+θ)fX(t)|Θ(x|θ)dx
=X(t+θ)fX(t)(x)dx=mX(t+θ)wherewehaveusedtheindependenceofX(t)andΘ.Thus
T1mY(t)=E[mX(t+θ)]=mX(t+θ)dθ=mYT098 wherethelastequalityfollowsfromtheperiodicityofmX(t+θ).SimilarlyfortheautocorrelationfunctionRY(t+τ,t)=E[E[X(t+τ+Θ)X(t+Θ)|Θ]]=E[RX(t+τ+θ,t+θ)]
T1=RX(t+τ+θ,t+θ)dθT0
T1=RX(t+τ,t)dtT0wherewehaveusedthechangeofvariablest=t+θandtheperiodicityofRX(t+τ,t)2)|YT(f)|2|YT(f)|2SY(f)=Elim=EElimΘT→∞TT→∞T|XT(f)ej2πfθ|2|XT(f)|2=EElimΘ=EElimT→∞TT→∞T=E[SX(f)]=SX(f)1T3)SinceSY(f)=F[T0RX(t+τ,t)dt]andSY(f)=SX(f)weconcludethat
1TSX(f)=FRX(t+τ,t)dtT0Problem4.49UsingParseval’srelationweobtain
∞
∞f2S(f)df=F−1[f2]F−1[S(f)]dτXX−∞−∞
∞1(2)=−δ(τ)RX(τ)dτ课后答案网−∞4π21d2=−(−1)2R(τ)|4π2dτ2Xτ=01d2=−4π2dτ2RX(τ)|τ=0Also,www.hackshp.cn
∞SX(f)df=RX(0)−∞Combiningthetworelationsweobtain∞22−∞fSX(f)df1dWRMS=∞=−22RX(τ)|τ=0−∞SX(f)df4πRX(0)dτProblem4.50RXY(t1,t2)=E[X(t1)Y(t2)]=E[Y(t2)X(t1)]=RYX(t2,t1)Ifweletτ=t1−t2,thenusingthepreviousresultandthefactthatX(t),Y(t)arejointlystationary,sothatRXY(t1,t2)dependsonlyonτ,weobtainRXY(t1,t2)=RXY(t1−t2)=RYX(t2−t1)=RYX(−τ)99 TakingtheFouriertransformofbothsidesofthepreviousrelationweobtainSXY(f)=F[RXY(τ)]=F[RYX(−τ)]
∞=R(−τ)e−j2πfτdτYX−∞
∞∗=R(τ)e−j2πfτ
dτ=S∗(f)YXYX−∞Problem4.511)S(f)=N0,R(τ)=N0δ(τ).TheautocorrelationfunctionandthepowerspectraldensityofX2X2theoutputaregivenbyR(t)=R(τ)h(τ)h(−τ),S(f)=S(f)|H(f)|2YXYXWithH(f)=Π(f)wehave|H(f)|2=Π2(f)=Π(f)sothat2B2B2BN0fSY(f)=Π()22BTakingtheinverseFouriertransformofthepreviousweobtaintheautocorrelationfunctionoftheoutputN0RY(τ)=2Bsinc(2Bτ)=BN0sinc(2Bτ)22)TheoutputrandomprocessY(t)isazeromeanGaussianprocesswithvarianceσ2=E[Y2(t)]=E[Y2(t+τ)]=R(0)=BNY(t)Y0ThecorrelationcoefficientofthejointlyGaussianprocessesY(t+τ),Y(t)isCOV(Y(t+τ)Y(t))E[Y(t+τ)Y(t)]RY(τ)ρY(t+τ)Y(t)===σY(t+τ)σY(t)BN0BN0Withτ=1,wehaveR(1)=sinc(1)=0sothatρ=0.Hencethejointprobability2BY2BY(t+τ)Y(t)densityfunctionofY(t)andY(课后答案网t+τ)is221−Y(t+τ)+Y(t)fY(t+τ)Y(t)=e2BN02πBN0SincetheprocessesareGaussiananduncorrelatedtheyarealsoindependent.www.hackshp.cnProblem4.52Theimpulseresponseofadelaylinethatintroducesadelayequalto∆ish(t)=δ(t−∆).TheoutputautocorrelationfunctionisRY(τ)=RX(τ)h(τ)h(−τ)But,
∞h(τ)h(−τ)=δ(−(t−∆))δ(τ−(t−∆))dt−∞
∞=δ(t−∆)δ(τ−(t−∆))dt−∞
∞=δ(t)δ(τ−t)dt=δ(τ)−∞Hence,RY(τ)=RX(τ)δ(τ)=RX(τ)100 Thisistobeexpectedsinceadelaylinedoesnotalterthespectralcharacteristicsoftheinputprocess.Problem4.53Theconverseofthetheoremisnottrue.ConsiderforexampletherandomprocessX(t)=cos(2πf0t)+XwhereXisarandomvariable.ClearlymX(t)=cos(2πf0t)+mXisafunctionoftime.However,passingthisprocessthroughtheLTIsystemwithtransferfunctionfΠ(2W)withWK)=0fork≤K.Ifk>K,thenp(X=k,X>K)p(1−p)k−1p(X=k|X>K)==p(X>K)p(X>K)But,∞∞Kp(X>K)=p(1−p)k−1=p(1−p)k−1−(1−p)k−1k=K+1k=1k=111−(1−p)KK=p−=(1−p)1−(1−p)1−(1−p)sothatp(1−p)k−1p(X=k|X>K)=(1−p)KIfweletk=K+lwithl=1,2,...,thenp(1−p)K(1−p)l−1l−1p(X=k|X>K)==p(1−p)(1−p)Kthatisp(X=k|X>K)isthegeometricallydistributed.Hence,usingtheresultsofthefirstpartweobtain∞H(X|X>K)=−p(1−p)l−1log(p(1−p)l−1)2l=11−p=−log2(p)−log2(1−p)pProblem6.5H(X,Y课后答案网)=H(X,g(X))=H(X)+H(g(X)|X)=H(g(X))+H(X|g(X))But,H(g(X)|X)=0,sinceg(·)isdeterministic.Therefore,www.hackshp.cnH(X)=H(g(X))+H(X|g(X))Sinceeachterminthepreviousequationisnon-negativeweobtainH(X)≥H(g(X))EqualityholdswhenH(X|g(X))=0.Thismeansthatthevaluesg(X)uniquelydetermineX,orthatg(·)isaonetoonemapping.Problem6.6Theentropyofthesourceis6H(X)=−pilog2pi=2.4087bits/symboli=1Thesamplingrateisfs=2000+2·6000=14000Hz129 Thismeansthat14000samplesaretakenpereachsecond.Hence,theentropyofthesourceinbitspersecondisgivenbyH(X)=2.4087×14000(bits/symbol)×(symbols/sec)=33721.8bits/secondProblem6.7Considerthefunctionf(x)=x−1−lnx.Forx>1,df(x)1=1−>0dxxThus,thefunctionismonotonicallyincreasing.Since,f(1)=0,thelatterimpliesthatifx>1then,f(x)>f(1)=0orlnxf(1)=0orlnx0,lnx≤x−11/Nwithequalityifx=0.Applyingtheinequalitywithx=,weobtainpi11/Nln−lnpi≤−1NpiMultiplyingthepreviousbypiandadding,weobtainNNN11piln−pilnpi≤−pi=0NNi=1i=1i=1Hence,课后答案网N1NH(X)≤−piln=lnNpi=lnNNi=1i=1But,lnNistheentropy(innats/symbol)ofthesourcewhenitisuniformlydistributed(seeProblem6.2).Hence,forequiprobablesymbolstheentropyofthesourceachievesitsmaximum.www.hackshp.cnProblem6.8Supposethatqiisadistributionover1,2,3,...andthat∞iqi=mi=1i−1Letv=11−1andapplytheinequalitylnx≤x−1tov.Then,iqimmii−1i−11111ln1−−lnqi≤1−−1mmqimmMultiplyingthepreviousbyqiandadding,weobtain∞∞∞∞11i−111qln1−−qlnq≤(1−)i−1−q=0iiiimmmmi=1i=1i=1i=1130 But,∞i−1∞1111qiln1−=qiln()+(i−1)ln(1−)mmmmi=1i=1∞11=ln()+ln(1−)(i−1)qimmi=1∞∞11=ln()+ln(1−)iqi−qimmi=1i=111=ln()+ln(1−)(m−1)=−H(p)mmwhereH(p)istheentropyofthegeometricdistribution(seeProblem6.4).Hence,∞−H(p)−qilnqi≤0=⇒H(q)≤H(p)i=1Problem6.9Themarginalprobabilitiesaregivenby2p(X=0)=p(X=0,Y=k)=p(X=0,Y=0)+p(X=0,Y=1)=3k1p(X=1)=p(X=1,Y=k)=p(X=1,Y=1)=3k1p(Y=0)=p(X=k,Y=0)=p(X=0,Y=0)=3k2p(Y=1)=p(X=k,Y=1)=p(X=0,Y=1)+p(X=1,Y=1)=3kHence,11111H(X)=课后答案网−pilog2pi=−(log2+log2)=.91833333i=011111H(X)=−pilog2pi=−(log2+log2)=.91833333i=0211H(X,Y)=−www.hackshp.cnlog2=1.585033i=0H(X|Y)=H(X,Y)−H(Y)=1.5850−0.9183=0.6667H(Y|X)=H(X,Y)−H(X)=1.5850−0.9183=0.6667Problem6.10H(Y|X)=−p(x,y)logp(y|x)x,yBut,p(y|x)=p(g(x)|x)=1.Hence,logp(g(x)|x)=0andH(Y|X)=0.Problem6.111)H(X)=−(.05log2.05+.1log2.1+.1log2.1+.15log2.15+.05log2.05+.25log2.25+.3log2.3)=2.5282131 2)Afterquantization,thenewalphabetisB={−4,0,4}andthecorrespondingsymbolprobabil-itiesaregivenbyp(−4)=p(−5)+p(−3)=.05+.1=.15p(0)=p(−1)+p(0)+p(1)=.1+.15+.05=.3p(4)=p(3)+p(5)=.25+.3=.55Hence,H(Q(X))=1.4060.Asitisobservedquantizationdecreasestheentropyofthesource.Problem6.12Usingthefirstdefinitionoftheentropyrate,wehaveH=limH(Xn|X1,...Xn−1)n→∞=lim(H(X1,X2,...,Xn)−H(X1,X2,...,Xn−1))n→∞However,X1,X2,...Xnareindependent,sothatnn−1H=limH(Xi)−H(Xi)=limH(Xn)=H(X)n→∞n→∞i=1i=1wherethelastequalityfollowsfromthefactthatX1,...,Xnareidenticallydistributed.Usingtheseconddefinitionoftheentropyrate,weobtain1H=limH(X1,X2,...,Xn)n→∞nn1=limH(Xi)n→∞ni=11=limnH(X)=H(X)n→∞nThesecondlineofthepreviousrelationfollowsfromtheindependenceofX1,X2,...Xn,whereasthethirdlinefromthefactthatforaDMStherandomvariables课后答案网X1,...Xnareidenticallydistributedindependentofn.Problem6.13H=nlim→∞Hwww.hackshp.cn(Xn|X1,...,Xn−1)=lim−p(x1,...,xn)log2p(xn|x1,...,xn−1)n→∞x1,...,xn=lim−p(x1,...,xn)log2p(xn|xn−1)n→∞x1,...,xn=lim−p(xn,xn−1)log2p(xn|xn−1)n→∞xn,xn−1=limH(Xn|Xn−1)n→∞However,forastationaryprocessp(xn,xn−1)andp(xn|xn−1)areindependentofn,sothatH=limH(Xn|Xn−1)=H(Xn|Xn−1)n→∞132 Problem6.14H(X|Y)=−p(x,y)logp(x|y)=−p(x|y)p(y)logp(x|y)x,yx,y=p(y)−p(x|y)logp(x|y)=p(y)H(X|Y=y)yxyProblem6.151)Themarginaldistributionp(x)isgivenbyp(x)=yp(x,y).Hence,H(X)=−p(x)logp(x)=−p(x,y)logp(x)xxy=−p(x,y)logp(x)x,ySimilarlyitisprovedthatH(Y)=−x,yp(x,y)logp(y).p(x)p(y)2)Usingtheinequalitylnw≤w−1withw=,weobtainp(x,y)p(x)p(y)p(x)p(y)ln≤−1p(x,y)p(x,y)Multiplyingthepreviousbyp(x,y)andaddingoverx,y,weobtainp(x,y)lnp(x)p(y)−p(x,y)lnp(x,y)≤p(x)p(y)−p(x,y)=0x,yx,yx,yx,yHence,H(X,Y)≤−p(x,y)lnp(x)p(y)=−p(x,y)(lnp(x)+lnp(y))x,yx,y=−课后答案网p(x,y)lnp(x)−p(x,y)lnp(y)=H(X)+H(Y)x,yx,yEqualityholdswhenp(x)p(y)=1,i.ewhenX,Yareindependent.p(x,y)Problem6.16www.hackshp.cnH(X,Y)=H(X)+H(Y|X)=H(Y)+H(X|Y)Also,fromProblem6.15,H(X,Y)≤H(X)+H(Y).Combiningthetworelations,weobtainH(Y)+H(X|Y)≤H(X)+H(Y)=⇒H(X|Y)≤H(X)Supposenowthatthepreviousrelationholdswithequality.Then,p(x)−p(x)logp(x|y)=−p(x)logp(x)⇒p(x)log()=0p(x|y)xxxHowever,p(x)isalwaysgreaterorequaltop(x|y),sothatlog(p(x)/p(x|y))isnon-negative.Sincep(x)>0,theaboveequalityholdsifandonlyiflog(p(x)/p(x|y))=0orequivalentlyifandonlyifp(x)/p(x|y)=1.Thisimpliesthatp(x|y)=p(x)meaningthatXandYareindependent.133 Problem6.17Toshowthatq=λp1+λ¯p2isalegitimateprobabilityvectorwehavetoprovethat0≤qi≤1andiqi=1.Clearly0≤p1,i≤1and0≤p2,i≤1sothat0≤λp1,i≤λ,0≤λp¯2,i≤λ¯Ifweaddthesetwoinequalities,weobtain0≤qi≤λ+λ¯=⇒0≤qi≤1Also,qi=(λp1,i+λp¯2,i)=λp1,i+λ¯p2,i=λ+λ¯=1iiiiBeforeweprovethatH(X)isaconcavefunctionoftheprobabilitydistributiononXweshowthatlnx≥1−1.Sincelny≤y−1,wesetx=1sothat−lnx≤1−1⇒lnx≥1−1.Equalityxyxxholdswheny=1=1orelseifx=1.xH(λp1+λ¯p2)−λH(p1)−λH¯(p2)p1,ip2,i=λp1,ilog+λ¯p2,ilogiλp1,i+λp¯2,iiλp1,i+λp¯2,iλp1,i+λp¯2,iλp1,i+λp¯2,i≥λp1,i1−+λ¯p2,i1−ip1,iip2,i=λ(1−1)+λ¯(1−1)=0Hence,λH(p1)+λH¯(p2)≤H(λp1+λ¯p2)Problem6.18课后答案网Letpi(xi)bethemarginaldistributionoftherandomvariableXi.Then,nnH(Xi)=−pi(xi)logpi(xi)i=1i=1xiwww.hackshp.cn"n=−···p(x1,x2,···,xn)logpi(xi)x1x2xni=1Therefore,nH(Xi)−H(X1,X2,···Xn)i=1p(x1,x2,···,xn)=···p(x1,x2,···,xn)log2nx1x2xni=1pi(xi)2ni=1pi(xi)≥···p(x1,x2,···,xn)1−x1x2xnp(x1,x2,···,xn)=···p(x1,x2,···,xn)−···p1(x1)p2(x2)···pn(xn)x1x2xnx1x2xn=1−1=0134 wherewehaveusedtheinequalitylnx≥1−1(seeProblem6.17.)Hence,xnH(X1,X2,···Xn)≤H(Xi)i=12nwithequalityifi=1pi(xi)=p(x1,···,xn),i.e.amemorylesssource.Problem6.191)Theprobabilityofanallzerosequenceisn1p(X1=0,X2=0,···,Xn=0)=p(X1=0)p(X2=0)···p(Xn=0)=22)Similarlywiththepreviouscasen1p(X1=1,X2=1,···,Xn=1)=p(X1=1)p(X2=1)···p(Xn=1)=23)p(X1=1,···,Xk=1,Xk+1=0,···Xn=0)=p(X1=1)···p(Xk=1)p(Xk+1=0)···p(Xn=0)kn−kn111==2224)Thenumberofzerosoronesfollowsthebinomialdistribution.Hencekn−knn11n1p(kones)==k22k2课后答案网5)Incasethatp(Xi=1)=p,theanswersofthepreviousquestionschangeasfollowsp(X=0,X=0,···,X=0)=(1−p)n12np(X=1,X=1,···,X=1)=pn12np(firstkones,nextn−kzeros)=pk(1−p)n−kwww.hackshp.cnnkn−kp(kones)=p(1−p)kProblem6.20FromthediscussioninthebeginningofSection6.2itfollowsthatthetotalnumberofsequencesoflengthnofabinaryDMSsourceproducingthesymbols0and1withprobabilitypand1−prespectivelyis2nH(p).Thusifp=0.3,wewillobservesequenceshavingnp=3000zerosandn(1−p)=7000ones.Therefore,#sequenceswith3000zeros≈28813AnotherapproachtotheproblemisviatheStirling’sapproximation.Ingeneralthenumberofbinarysequencesoflengthnwithkzerosandn−konesisthebinomialcoefficientnn!=kk!(n−k)!135 TogetanestimatewhennandkarelargenumberswecanuseStirling’sapproximation√nnn!≈2πneHence,10000!110000#sequenceswith3000zeros=≈√103000!7000!212π30·70Problem6.211)Thetotalnumberoftypicalsequencesisapproximately2nH(X)wheren=1000andH(X)=−pilog2pi=1.4855iHence,#typicalsequences≈21485.52)ThenumberofallsequencesoflengthnisNn,whereNisthesizeofthesourcealphabet.Hence,#typicalsequences2nH(X)≈≈1.14510−30#non-typicalsequencesNn−2nH(X)3)Thetypicalsequencesarealmostequiprobable.Thus,p(X=x,xtypical)≈2−nH(X)=2−1485.54)Sincethenumberofthetotalsequencesis2nH(X)thenumberofbitsrequiredtorepresentthesesequencesisnH(X)≈1486.5)Themostprobablesequenceistheonewithalla3’sthatis{a3,a3,...,a3}.Theprobabilityofthissequenceis课后答案网n100011p({a3,a3,...,a3})==226)Themostprobablesequenceofthepreviousquestionisnotatypicalsequence.Ingeneralinatypicalsequence,symbola1isrepeated1000www.hackshp.cnp(a1)=200times,symbola2isrepeatedapproximately1000p(a2)=300timesandsymbola3isrepeatedalmost1000p(a3)=500times.Problem6.221)Theentropyofthesourceis4H(X)=−p(ai)log2p(ai)=1.8464bits/outputi=12)Theaveragecodewordlengthislowerboundedbytheentropyofthesourceforerrorfreereconstruction.Hence,theminimumpossibleaveragecodewordlengthisH(X)=1.8464.3)ThefollowingfiguredepictstheHuffmancodingschemeofthesource.TheaveragecodewordlengthisR¯(X)=3×(.2+.1)+2×.3+.4=1.9136 0.4010.30.61110.20.31111.114)ForthesecondextensionofthesourcethealphabetofthesourcebecomesA2={(a1,a1),(a1,a2),...(a4,a4)}andtheprobabilityofeachpairistheproductoftheprobabilitiesofeachcomponent,i.e.p((a1,a2))=.2.AHuffmancodeforthissourceisdepictedinthenextfigure.TheaveragecodewordlengthinbitsperpairofsourceoutputisR¯2(X)=3×.49+4×.32+5×.16+6×.03=3.7300TheaveragecodewordlengthinbitspereachsourceoutputisR¯1(X)=R¯2(X)/2=1.865.5)Huffmancodingoftheoriginalsourcerequires1.9bitspersourceoutputletterwhereasHuffmancodingofthesecondextensionofthesourcerequires1.865bitspersourceoutputletterandthusitismoreefficient.000(a4,a4).160010(a4,a3).1200100(a3,a4).120110(a3,a3).09000010(a4,a2).08010011(a2,a4).08110110(a3,a2).060课后答案网01010(a2,a3).06011110(a4,a1).0410101110(a2,a2).04www.hackshp.cn101111(a1,a4).04110110(a3,a1).03110111(a1,a3).03111110(a2,a1).0201111110(a1,a2).0201111111(a1,a1).011Problem6.23ThefollowingfigureshowsthedesignoftheHuffmancode.Notethatateachstepofthealgorithmthebrancheswiththelowestprobabilities(thatmergetogether)arethoseatthebottomofthetree.137 0120101401..1....011...1012n−2011111...102n−1011111...112n−11Theentropyofthesourceisn−11i1n−1H(X)=2ilog22+2n−1log22i=1n−111=2iilog22+2n−1(n−1)log22i=1n−1in−1=+2i2n−1i=1Inthewaythatthecodeisconstructed,thefirstcodeword(0)haslengthone,thesecondcodeword(10)haslengthtwoandsoonuntilthelasttwocodewords(111...10,111...11)whichhavelengthn−1.Thus,theaveragecodewordlengthisn−1in−1R¯=p(x)l(x)=+2i2n−1x∈Xi=1=21−(1/2)n−1=H(X)课后答案网Problem6.24Thefollowingfigureshowsthepositionofthecodewords(blackfilledcircles)inabinarytree.Althoughtheprefixconditionisnotviolatedthecodeisnotoptimuminthesensethatitusesmorebitsthatisnecessary.Forexampletheuppertwocodewordsinthetree(0001,0011)canbesubstitutedbythecodewords(000,001)(un-filledcircles)reducinginthiswaytheaveragewww.hackshp.cncodewordlength.Similarlycodewords1111and1110canbesubstitutedbycodewords111and110.01138 Problem6.25ThefollowingfiguredepictsthedesignofaternaryHuffmancode.0.22010.1801.50111.1712.15220.1301.2821.1222.052TheaveragecodewordlengthisR¯(X)=p(x)l(x)=.22+2(.18+.17+.15+.13+.10+.05)x=1.78(ternarysymbols/output)Forafaircomparisonoftheaveragecodewordlengthwiththeentropyofthesource,wecomputethelatterwithlogarithmsinbase3.Hence,H(X)=−p(x)log3p(x)=1.7047xAsitisexpectedH(X)≤R¯(X).Problem6.26IfDisthesizeofthecodealphabet,thentheHuffmancodingschemetakesDsourceoutputsanditmergesthemto1symbol.Hence,wehaveadecreaseofoutputsymbolsbyD−1.InKstepsofthealgorithmthedecreaseofthesourceoutputsis课后答案网K(D−1).IfthenumberofthesourceoutputsisK(D−1)+D,forsomeK,thenweareinagoodpositionsincewewillbeleftwithDsymbolsforwhichweassignthesymbols0,1,...,D−1.Tomeettheaboveconditionwithaternarycodethenumberofthesourceoutputsshouldbe2K+3.Inourcasethatthenumberofsourceoutputsissixwecanaddadummysymbolwithzeroprobabilitysothat7=2·2+3.ThefollowingfigureshowsthedesignoftheternaryHuffmancode.www.hackshp.cn0.401.17120.15021.1312220.10221.0512220.02Problem6.27ParsingthesequencebytherulesoftheLempel-Zivcodingschemeweobtainthephrases0,00,1,001,000,0001,10,00010,0000,0010,00000,101,00001,000000,11,01,0000000,110,0,...Thenumberofthephrasesis19.Foreachphraseweneed5bitsplusanextrabittorepresentthenewsourceoutput.139 DictionaryDictionaryCodewordLocationContents1000010000000200010000000103000111000001400100001000101500101000000100600110000100101170011110000110801000000100011009010010000001010100101000100010001101011000000100101201100101001111130110100001010011140111000000001011015011111100011116100000100001117100010000000011100181001011001111019000000Problem6.28I(X;Y)=H(X)−H(X|Y)=−p(x)logp(x)+p(x,y)logp(x|y)xx,y=−p(x,y)logp(x)+p(x,y)logp(x|y)x,yx,yp(x|y)p(x,y)=p(x,y)log=p(x,y)logp(x)p(x)p(y)课后答案网x,yx,yUsingtheinequalitylny≤y−1withy=1,weobtainlnx≥1−1.Applyingthisinequalitywithxxp(x,y)x=p(x)p(y)weobtainp(x,y)I(X;Y)=p(x,y)logx,ywww.hackshp.cnp(x)p(y)p(x)p(y)≥p(x,y)1−=p(x,y)−p(x)p(y)=0p(x,y)x,yx,yx,ylnx≥1−1holdswithequalityifx=1.ThismeansthatI(X;Y)=0ifp(x,y)=p(x)p(y)orinxotherwordsifXandYareindependent.Problem6.291)I(X;Y)=H(X)−H(X|Y).Sinceingeneral,H(X|Y)≥0,wehaveI(X;Y)≤H(X).Also(seeProblem6.30),I(X;Y)=H(Y)−H(Y|X)fromwhichweobtainI(X;Y)≤H(Y).Combiningthetwoinequalities,weobtainI(X;Y)≤min{H(X),H(Y)}2)Itcanbeshown(seeProblem6.7),thatifXandZaretworandomvariablesoverthesamesetXandZisuniformlydistributed,thenH(X)≤H(Z).FurthermoreH(Z)=log|X|,where|X|is140 thesizeofthesetX(seeProblem6.2).Hence,H(X)≤log|X|andsimilarlywecanprovethatH(Y)≤log|Y|.Usingtheresultofthefirstpartoftheproblem,weobtainI(X;Y)≤min{H(X),H(Y)}≤min{log|X|,log|Y|}Problem6.30BydefinitionI(X;Y)=H(X)−H(X|Y)andH(X,Y)=H(X)+H(Y|X)=H(Y)+H(X|Y).CombiningthetwoequationsweobtainI(X;Y)=H(X)−H(X|Y)=H(X)−(H(X,Y)−H(Y))=H(X)+H(Y)−H(X,Y)=H(Y)−(H(X,Y)−H(X))=H(Y)−H(Y|X)=I(Y;X)Problem6.311)Thejointprobabilitydensityisgivenbyp(Y=1,X=0)=p(Y=1|X=0)p(X=0)=
pp(Y=0,X=1)=p(Y=0|X=1)p(X=1)=
(1−p)p(Y=1,X=1)=(1−
)(1−p)p(Y=0,X=0)=(1−
)pThemarginaldistributionofYisp(Y=1)=
p+(1−
)(1−p)=1+2
p−
−pp(Y=0)=
(1−p)+(1−
)p=
+p−2
pHence,H(X)=−plog2p−(1−p)log2(1−p)H(Y)=−(1+2课后答案网
p−
−p)log2(1+2
p−
−p)−(
+p−2
p)log2(
+p−2
p)H(Y|X)=−p(x,y)log2(p(y|x))=−
plog2
−
(1−p)log2
x,y−(1−
)(1−p)log2(1−
)−(1−
)plog2(1−
)=−
logwww.hackshp.cn2
−(1−
)log2(1−
)H(X,Y)=H(X)+H(Y|X)=−plog2p−(1−p)log2(1−p)−
log2
−(1−
)log2(1−
)H(X|Y)=H(X,Y)−H(Y)=−plog2p−(1−p)log2(1−p)−
log2
−(1−
)log2(1−
)(1+2
p−
−p)log2(1+2
p−
−p)+(
+p−2
p)log2(
+p−2
p)I(X;Y)=H(X)−H(X|Y)=H(Y)−H(Y|X)=
log2
+(1−
)log2(1−
)−(1+2
p−
−p)log2(1+2
p−
−p)−(
+p−2
p)log2(
+p−2
p)2)ThemutualinformationisI(X;Y)=H(Y)−H(Y|X).AsitwasshowninthefirstquestionH(Y|X)=−
log2
−(1−
)log2(1−
)andthusitdoesnotdependonp.Hence,I(X;Y)141 ismaximizedwhenH(Y)ismaximized.However,H(Y)isthebinaryentropyfunctionwithprobabilityq=1+2
p−
−p,thatisH(Y)=Hb(q)=Hb(1+2
p−
−p)H(q)achievesitsmaximumvalue,whichisone,forq=1.Thus,b2111+2
p−
−p==⇒p=223)SinceI(X;Y)≥0,theminimumvalueofI(X;Y)iszeroanditisobtainedforindependentXandY.Inthiscasep(Y=1,X=0)=p(Y=1)p(X=0)=⇒
p=(1+2
p−
−p)por
=1.Thisvalueofepsilonalsosatisfies2p(Y=0,X=0)=p(Y=0)p(X=0)p(Y=1,X=1)=p(Y=1)p(X=1)p(Y=0,X=1)=p(Y=0)p(X=1)resultinginindependentXandY.Problem6.32I(X;YZW)=I(YZW;X)=H(YZW)−H(YZW|X)=H(Y)+H(Z|Y)+H(W|YZ)−[H(Y|X)+H(Z|XY)+H(W|XYZ)]=[H(Y)−H(Y|X)]+[H(Z|Y)−H(Z|YX)]+[H(W|YZ)−H(W|XYZ)]=I(X;Y)+I(Z|Y;X)+I(W|ZY;X)课后答案网=I(X;Y)+I(X;Z|Y)+I(X;W|ZY)Thisresultcanbeinterpretedasfollows:Theinformationthatthetripletofrandomvariables(Y,Z,W)givesabouttherandomvariableXisequaltotheinformationthatYgivesaboutXplustheinformationthatZgivesaboutX,whenYisalreadyknown,plustheinformationthatWprovidesaboutXwhenZ,Ywww.hackshp.cnarealreadyknown.Problem6.331)UsingBayesrule,weobtainp(x,y,z)=p(z)p(x|z)p(y|x,z).Comparingthisformwiththeonegiveninthefirstpartoftheproblemweconcludethatp(y|x,z)=p(y|x).ThisimpliesthatYandZareindependentgivenXsothat,I(Y;Z|X)=0.Hence,I(Y;ZX)=I(Y;Z)+I(Y;X|Z)=I(Y;X)+I(Y;Z|X)=I(Y;X)SinceI(Y;Z)≥0,wehaveI(Y;X|Z)≤I(Y;X)2)Comparingp(x,y,z)=p(x)p(y|x)p(z|x,y)withthegivenformofp(x,y,z)weobservethatp(y|x)=p(y)or,inotherwords,randomvariablesXandYareindependent.Hence,I(Y;ZX)=I(Y;Z)+I(Y;X|Z)=I(Y;X)+I(Y;Z|X)=I(Y;Z|X)142 SinceingeneralI(Y;X|Z)≥0,wehaveI(Y;Z)≤I(Y;Z|X)3)ForthefirstcaseconsiderthreerandomvariablesX,YandZ,takingthevalues0,1withequalprobabilityandsuchthatX=Y=Z.Then,I(Y;X|Z)=H(Y|Z)−H(Y|ZX)=0−0=0,whereasI(Y;X)=H(Y)−H(Y|X)=1−0=1.Hence,I(Y;X|Z)0.5,thenp(X˜)=0.If|X˜|≤0.5,thentherearefoursolutionstotheequationX˜=X−Q(X),whicharedenotedbyx1,x2,x3andx4.Thesolutionx1correspondstothecase−2≤X≤−1,x2isthesolutionfor−1≤X≤0andsoon.Hence,x1+2(˜x−1.5)+2−x3+2−(˜x+0.5)+2fX(x1)==fX(x3)==4444x2+2(˜x−0.5)+2−x4+2−(˜x+1.5)+2fX(x2)==fX(x4)==4444Theabsolutevalueof(X−Q(X))isoneforX=x1,...,x4.Thus,for|X˜|≤0.54fX(xi)f˜(˜x)=X|(xi−Q(xi))|i=1(˜x−1.5)+2(˜x−0.5)+2−(˜x+0.5)+2−(˜x+1.5)+2=+++4444=1156 Problem6.551)RX(t+τ,t)=E[X(t+τ)X(t)]=E[Y2cos(2πf(t+τ)+Θ)cos(2πft+Θ)]0012=E[Y]E[cos(2πf0τ)+cos(2πf0(2t+τ)+2Θ)]2andsince
12πE[cos(2πf0(2t+τ)+2Θ)]=cos(2πf0(2t+τ)+2θ)dθ=02π0weconcludethat123RX(t+τ,t)=E[Y]cos(2πf0τ)=cos(2πf0τ)222)ν3×4×RX(0)10log10SQNR=10log102=40xmaxThus,4νlog10=4orν=82ThebandwidthoftheprocessisW=f0,sothattheminimumbandwidthrequirementofthePCMsystemisBW=8f0.3)IfSQNR=64db,thenν=log(2·106.4)=124Thus,ν−ν=4morebitsareneededtoincreaseSQNRby24db.ThenewminimumbandwidthrequirementisBW=12f.0Problem6.56Supposethatthetransmittedsequenceisx.Ifanerroroccursattheithbitofthesequence,thenthereceivedsequencexis课后答案网x=x+[0...010...0]whereadditionismodulo2.Thustheerrorsequenceisei=[0...010...0],whichinnaturalbinarycodinghasthevalue2i−1.Ifthespacingbetweenlevelsis∆,thentheerrorintroducedbythechannelis2i−1∆.www.hackshp.cn2)νD=p(errorinibit)·(2i−1∆)2channeli=1ν1−4ν=p∆24i−1=p∆2bb1−4i=14ν−1=p∆2b33)Thetotaldistortionis24ν−1x2maxDtotal=Dchannel+Dquantiz.=pb∆+233·N4·x2max4ν−1x2max=pb+N233·N2157 orsinceN=2νx2x2D=max(1+4p(4ν−1))=max(1+4p(N2−1))total3·4νb3N2b4)E[X2]E[X2]3N2SNR==Dtotalx2max(1+4pb(N2−1))XE[X2]22IfweletX˘=xmax,thenx2=E[X˘]=X˘.Hence,max3N2X˘23·4νX˘2SNR==1+4pb(N2−1)1+4pb(4ν−1)Problem6.571)|x|log(1+µ)xmaxg(x)=sgn(x)log(1+µ)Differentiatingtheprevioususingnaturallogarithms,weobtain1µ/xmax2g(x)=sgn(x)ln(1+µ)(1+µ|x|)xmaxSince,fortheµ-lawcompanderymax=g(xmax)=1,weobtain2
∞ymaxfX(x)D≈dx3×4ν−∞[g(x)]2
x2max[ln(1+µ)]2∞2|x|2|x|=1+µ+2µfX(x)dx3×4νµ2−∞x2maxxmaxx2max[ln(1+µ)]222=1+µE[X˘]+2µE[|X˘|]3×课后答案网4νµ2x2max[ln(1+µ)]222=1+µE[X˘]+2µE[|X˘|]3×N2µ2whereN2=4νandX˘=X/xmax.2)www.hackshp.cnE[X2]SQNR=DE[X2]µ23·N2=x2[ln(1+µ)]2(µ2E[X˘2]+2µE[|X˘|]+1)max3µ2N2E[X˘2]=[ln(1+µ)]2(µ2E[X˘2]+2µE[|X˘|]+1)3)SinceSQNRunif=3·N2E[X˘2],wehaveµ2SQNRµlaw=SQNRunif222[ln(1+µ)](µE[X˘]+2µE[|X˘|]+1)=SQNRunifG(µ,X˘)158 whereweidentifyµ2G(µ,X˘)=[ln(1+µ)]2(µ2E[X˘2]+2µE[|X˘|]+1)3)ThetruncatedGaussiandistributionhasaPDFgivenby2K−x2σ2fY(y)=√ex2πσxwheretheconstantKissuchthat
24σx1−x12σ2K√exdx=1=⇒K==1.0001−4σx2πσx1−2Q(4)Hence,
2K4σx|x|−x2σ2E[|X˘|]=√exdx2πσx−4σx4σx
22K4σx−x2σ2=√xexdx42πσx2024σxK−x22σ2=√−σxex22πσx20K−2=√(1−e)=0.172522πInthenextfigureweplot10log10SQNRunifand10log10SQNRmu−lawvs.10log10E[X˘2]whenthelattervariesfrom−100to100db.Asitisobservedtheµ-lawcompressorisinsensitivetothedynamicrangeoftheinputsignalforE[X˘2]>1.200150课后答案网uniform100SQNR(db)50mu-law0www.hackshp.cn-50-100-80-60-40-20020406080100E[X^2]dbProblem6.58Theoptimalcompressorhastheformx12−∞[fX(v)]3dvg(x)=ymax1−∞−∞[fX(v)]3dvwhereymax=g(xmax)=g(1).
∞
1
0
11111[fX(v)]3dv=[fX(v)]3dv=(v+1)3dv+(−v+1)3dv−∞−1−10
113=2x3dx=02159 Ifx≤0,then
x
x
x+1x+111134[fX(v)]3dv=(v+1)3dv=z3dz=z3−∞−104034=(x+1)34Ifx>0,then
x
0
x
111131[fX(v)]3dv=(v+1)3dv+(−v+1)3dv=+z3dz−∞−1041−x334=+1−(1−x)344Hence,4g(1)(x+1)3−1−1≤x<0g(x)=4g(1)1−(1−x)30≤x≤1Thenextfiguredepictsg(x)forg(1)=1.Sincetheresultingdistortionis(seeEquation6.6.17)10.80.60.40.20g(x)-0.2-0.4-0.6-0.8-1-1-0.8-0.6-0.4-0.200.20.40.60.81课后答案网x
∞331113D=[fX(x)]3dx=12×4ν−infty12×4ν2wehavewww.hackshp.cnE[X2]32ν232ν116νSQNR==×4E[X]=×4·=4D99627Problem6.59Thesamplingrateisfs=44100meaningthatwetake44100samplespersecond.Eachsampleisquantizedusing16bitssothetotalnumberofbitspersecondis44100×16.Foramusicpieceofduration50min=3000sectheresultingnumberofbitsperchannel(leftandright)is44100×16×3000=2.1168×109andtheoverallnumberofbitsis2.1168×109×2=4.2336×109160 Chapter7Problem7.1TheamplitudesAmtakethevaluesdAm=(2m−1−M),m=1,...M2Hence,theaverageenergyis1Md2ME=s2=E(2m−1−M)2avmgM4Mm=1m=1d2M=E[4m2+(M+1)2−4m(M+1)]g4Mm=1d2MM=E4m2+M(M+1)2−4(M+1)mg4Mm=1m=12dM(M+1)(2M+1)2M(M+1)=Eg4+M(M+1)−4(M+1)4M62M2−1d2=Eg34Problem7.2Thecorrelationcoefficientbetweenthemthandthenthsignalpointsissm·snγmn=课后答案网|sm||sn|%wheres=(s,s,...,s)ands=±Es.Twoadjacentsignalpointsdifferinonlyonemm1m2mNmjNcoordinate,forwhichsmkandsnkhaveoppositesigns.Hence,www.hackshp.cnNsm·sn=smjsnj=smjsnj+smksnkj=1j=kEsEsN−2=(N−1)−=EsNNN1Furthermore,|sm|=|sn|=(Es)2sothatN−2γmn=NTheEuclideandistancebetweenthetwoadjacentsignalpointsis111%%2d=|s−s|2=E/N=4Es=2Esmn±2sNN161 Problem7.3a)Toshowthatthewaveformsψn(t),n=1,...,3areorthogonalwehavetoprovethat
∞ψm(t)ψn(t)dt=0,m=n−∞Clearly,
∞
4c12=ψ1(t)ψ2(t)dt=ψ1(t)ψ2(t)dt−∞0
2
4=ψ1(t)ψ2(t)dt+ψ1(t)ψ2(t)dt02
2
41111=dt−dt=×2−×(4−2)404244=0Similarly,
∞
4c13=ψ1(t)ψ3(t)dt=ψ1(t)ψ3(t)dt−∞01
11
21
31
4=dt−dt−dt+dt40414243=0and
∞
4c23=ψ2(t)ψ3(t)dt=ψ2(t)ψ3(t)dt−∞01
11
21
31
4=dt−dt+dt−dt40414243=0Thus,thesignalsψn(t)areorthogonal.课后答案网b)Wefirstdeterminetheweightingcoefficients
∞xn=x(t)ψn(t)dt,n=1,2,3www.hackshp.cn−∞
41
11
21
31
4x1=x(t)ψ1(t)dt=−dt+dt−dt+dt=0020212223
4
41x2=x(t)ψ2(t)dt=x(t)dt=0020
41
11
21
31
4x3=x(t)ψ3(t)dt=−dt−dt+dt+dt=0020212223Asitisobserved,x(t)isorthogonaltothesignalwaveformsψn(t),n=1,2,3andthusitcannotrepresentedasalinearcombinationofthesefunctions.Problem7.4a)Theexpansioncoefficients{cn},thatminimizethemeansquareerror,satisfy
∞
4πtcn=x(t)ψn(t)dt=sinψn(t)dt−∞04162 Hence,
4πt1
2πt1
4πtc1=sinψ1(t)dt=sindt−sindt04204224242πt2πt=−cos+cosπ40π4222=−(0−1)+(−1−0)=0ππSimilarly,
4
4πt1πtc2=sinψ2(t)dt=sindt0420442πt24=−cos=−(−1−1)=π40ππand
4πtc3=sinψ3(t)dt041
1πt1
2πt1
3πt1
4πt=sindt−sindt+sindt−sindt204214224234=0Notethatc,ccanbefoundbyinspectionsincesinπtisevenwithrespecttothex=2axisand124ψ1(t),ψ3(t)areoddwithrespecttothesameaxis.b)TheresidualmeansquareerrorEmincanbefoundfrom
∞3E=|x(t)|2dt−|c|2mini−∞i=1Thus,课后答案网
4πt2421
4πt16Emin=sindt−=1−cosdt−04π202π241πt1616=2−sin−=2−πwww.hackshp.cn20π2π2Problem7.5a)Asanorthonormalsetofbasisfunctionsweconsidertheset10≤t<111≤t<2ψ1(t)=ψ2(t)=0o.w0o.w12≤t<313≤t<4ψ3(t)=ψ4(t)=0o.w0o.wInmatrixnotation,thefourwaveformscanberepresentedass1(t)2−1−1−1ψ1(t)s2(t)−2110ψ2(t)=s3(t)1−11−1ψ3(t)s4(t)1−2−22ψ4(t)163 Notethattherankofthetransformationmatrixis4andtherefore,thedimensionalityofthewaveformsis4b)Therepresentationvectorsares1=2−1−1−1s2=−2110s3=1−11−1s4=1−2−22c)Thedistancebetweenthefirstandthesecondvectoris(%2√2d1,2=|s1−s2|=4−2−2−1=25Similarlywefindthat(%2√2d1,3=|s1−s3|=10−20=5(%2√2d1,4=|s1−s4|=111−3=12(%2√2d2,3=|s2−s3|=−3201=14(%2√2d2,4=|s2−s4|=−333−2=31(%2√2d3,4=|s3−s4|=013−3=19√Thus,theminimumdistancebetweenanypairofvectorsis课后答案网dmin=5.Problem7.6Asasetoforthonormalfunctionsweconsiderthewaveforms10≤t<111≤t<212≤t<3ψ1(t)=www.hackshp.cnψ2(t)=ψ3(t)=0o.w0o.w0o.wThevectorrepresentationofthesignalsiss1=222s2=200s3=0−2−2s4=220Notethats3(t)=s2(t)−s1(t)andthatthedimensionalityofthewaveformsis3.164 Problem7.7Theenergyofthesignalwaveformsm(t)is

M2∞2∞1E=s(t)dt=s(t)−s(t)dtmmk−∞−∞Mk=1
∞1MM
∞=s2(t)dt+s(t)s(t)dtmM2kl−∞−∞k=1l=11M
∞1M
∞−sm(t)sk(t)dt−sm(t)sl(t)dtM−∞M−∞k=1l=1MM12=E+Eδkl−EM2Mk=1l=112M−1=E+E−E=EMMMThecorrelationcoefficientisgivenby∞−∞sm(t)sn(t)dtγmn=11∞|s(t)|2dt2∞|s(t)|2dt2−∞m−∞n
MM1∞11=sm(t)−sk(t)sn(t)−sl(t)dtE−∞MMk=1l=1
∞MM
∞11=Esm(t)sn(t)dt+M2sk(t)sl(t)dt−∞−∞k=1l=1M
M
11∞1∞−sn(t)sk(t)dt+sm(t)sl(t)dtEM−∞M−∞k=1l=11ME−1E−1EM2MM1=课后答案网M−1=−M−1EMProblem7.8AssumingthatE[n2(t)]=σn2,weobtainwww.hackshp.cn

TTE[n1n2]=Es1(t)n(t)dts2(v)n(v)dv00
T
T=s1(t)s2(v)E[n(t)n(v)]dtdv00
T=σ2s(t)s(t)dtn120=0wherethelastequalityfollowsfromtheorthogonalityofthesignalwaveformss1(t)ands2(t).Problem7.9a)Thereceivedsignalmaybeexpressedasn(t)ifs0(t)wastransmittedr(t)=A+n(t)ifs1(t)wastransmitted165 Assumingthats(t)hasunitenergy,thenthesampledoutputsofthecrosscorrelatorsarer=sm+n,m=0,1√wheres0=0,s1=ATandthenoisetermnisazero-meanGaussianrandomvariablewithvariance

1T1Tσ2=E√n(t)dt√n(τ)dτnT0T0
T
T1=E[n(t)n(τ)]dtdτT00
T
TN0N0=δ(t−τ)dtdτ=2T002Theprobabilitydensityfunctionforthesampledoutputis21−rf(r|s0)=√eN0πN0√21−(r−AT)f(r|s1)=√eN0πN0Sincethesignalsareequallyprobable,theoptimaldetectordecidesinfavorofs0ifPM(r,s0)=f(r|s0)>f(r|s1)=PM(r,s1)otherwiseitdecidesinfavorofs1.Thedecisionrulemaybeexpressedas√√√s022PM(r,s0)(r−AT)−r−(2r−AT)AT>=eN0=eN0<1PM(r,s1)s1orequivalently课后答案网s1>1√rN01−pr<√ln=η4Ebps2TheaverageprobabilityoferrorisP(e)=P(e|s1课后答案网)P(s1)+P(e|s2)P(s2)=pP(e|s1)+(1−p)P(e|s2)
η
∞=pf(r|s1)dr+(1−p)f(r|s1)dr−∞η
√2
√2η1(r−Eb)∞1(r+Eb)−−=p√www.hackshp.cneN0dr+(1−p)√eN0dr−∞πN0ηπN0
η12
∞21−x1−x=p√e2dx+(1−p)√e2dx2π−∞2πη2where11112Eb22Eb2η1=−+ηη2=+ηN0N0N0N0Thus,11112Eb22Eb2P(e)=pQ−η+(1−p)Q+ηN0N0N0N0b)Ifp=0.3andEb=10,thenN0P(e)=0.3Q[4.3774]+0.7Q[4.5668]=0.3×6.01×10−6+0.7×2.48×10−6=3.539×10−6167 Ifthesymbolsareequiprobable,then12Eb√−6P(e)=Q[]=Q[20]=3.88×10N0Problem7.12a)TheoptimumthresholdisgivenbyN01−pN0η=√ln=√ln24Ebp4Ebb)Theaverageprobabilityoferroris(η=√N0ln2)4Eb
∞√P(e)=p(a=−1)√1e−(r+Eb)2/N0drmηπN0
η√+p(a=1)√1e−(r−Eb)2/N0drm−∞πN0√√2η+Eb1Eb−η=Q&+Q&3N0/23N0/2&11&22N0/Ebln22Eb12Eb2N0/Ebln2=Q++Q−34N03N04Problem7.13a)Themaximumlikelihoodcriterionselectsthemaximumoff(r|sm)overtheMpossibletrans-mittedsignals.WhenM=2,theMLcriteriontakestheforms1课后答案网f(r|s1)><1f(r|s2)s2or,sincewww.hackshp.cn1√2f(r|s)=√e−(r−Eb)/N01πN01√2f(r|s)=√e−(r+Eb)/N02πN0theoptimummaximum-likelihooddecisionruleiss1>r<0s2b)Theaverageprobabilityoferrorisgivenby
∞1√
01√P(e)=p√e−(r+Eb)2/N0dr+(1−p)√e−(r−Eb)2/N0dr0πN0−∞πN0168 √
∞1x2
−2Eb/N01x2−−=p√√e2dx+(1−p)√e2dx2Eb/N02π−∞2π112Eb2Eb=pQ+(1−p)QN0N012Eb=QN0Problem7.14a)Theimpulseresponseofthefiltermatchedtos(t)ish(t)=s(T−t)=s(3−t)=s(t)wherewehaveusedthefactthats(t)isevenwithrespecttothet=T=3axis.22b)Theoutputofthematchedfilteris
ty(t)=s(t)s(t)=s(τ)s(t−τ)dτ00t<0A2t0≤t<1A2(2−t)1≤t<222A(t−2)2≤t<3=22A(4−t)3≤t<4A2(t−4)4≤t<5A2(6−t)5≤t<606≤tAscetchofy(t)isdepictedinthenextfigure2A2................课后答案网

A2......



www.hackshp.cn135246c)Attheoutputofthematchedfilterandfort=T=3thenoiseis
TnT=n(τ)h(T−τ)dτ0
T
T=n(τ)s(T−(T−τ))dτ=n(τ)s(τ)dτ00Thevarianceofthenoiseis

TTσ2=En(τ)n(v)s(τ)s(v)dτdvnT00
T
T=s(τ)s(v)E[n(τ)n(v)]dτdv00169
T
TN0=s(τ)s(v)δ(τ−v)dτdv200
TN022=s(τ)dτ=N0A20d)Forantipodalequiprobablesignalstheprobabilityoferroris1SP(e)=QNowhereSistheoutputSNRfromthematchedfilter.SinceNo24Sy(T)4A==NoE[n2]N0A2Tweobtain14A2P(e)=QN0Problem7.15a)TakingtheinverseFouriertransformofH(f),weobtain1e−j2πfTh(t)=F−1[H(f)]=F−1−F−1j2πfj2πft−T=sgn(t)−sgn(t−T)=2Π2Tb)Thesignalwaveform,towhich课后答案网h(t)ismatched,isT−t−TT−ts(t)=h(T−t)=2Π2=2Π2=h(t)TTt−TwherewehaveusedthesymmetryofΠ2withrespecttothet=Taxis.www.hackshp.cnT2Problem7.16IfgT(t)=sinc(t),thenitsmatchedwaveformish(t)=sinc(−t)=sinc(t).Since,(seeProblem2.17)sinc(t)sinc(t)=sinc(t)theoutputofthematchedfilteristhesamesincpulse.If2TgT(t)=sinc((t−))T2thenthematchedwaveformis2Th(t)=gT(T−t)=sinc((−t))=gT(t)T2170 wherethelastequalityfollowsfromthefactthatg(t)isevenwithrespecttothet=Taxis.TheT2outputofthematchedfilterisy(t)=F−1[g(t)g(t)]TTT2T=F−1Π(f)e−j2πfT42T2TT=sinc((t−T))=gT(t−)2T22Thustheoutputofthematchedfilteristhesamesincfunction,scaledbyTandcenteredatt=T.2Problem7.171)Theoutputoftheintegratoris
t
ty(t)=r(τ)dτ=[si(τ)+n(τ)]dτ00
t
t=si(τ)dτ+n(τ)dτ00Attimet=Twehave1
T
TE
Tby(T)=si(τ)dτ+n(τ)dτ=±T+n(τ)dτ00T0Thesignalenergyattheoutputoftheintegratoratt=Tis12EbEs=±T=EbTTwhereasthenoisepower

课后答案网TTPn=En(τ)n(v)dτdv00
T
T=E[n(τ)n(v)]dτdv00
T
TN0N0www.hackshp.cn=δ(τ−v)dτdv=T2002Hence,theoutputSNRisEs2EbSNR==PnN02)ThetransferfunctionoftheRCfilteris1H(f)=1+j2πRCfThus,theimpulseresponseofthefilteris1−th(t)=eRCu−1(t)RC171 andtheoutputsignalisgivenby
t1−t−τy(t)=r(τ)eRCdτRC−∞
t1−t−τ=(si(τ)+n(τ))eRCdτRC−∞
t
t1−tτ1−tτ=eRCsi(τ)eRCdτ+eRCn(τ)eRCdτRC0RC−∞Attimet=Tweobtain
T
T1−Tτ1−Tτy(T)=eRCsi(τ)eRCdτ+eRCn(τ)eRCdτRC0RC−∞Thesignalenergyattheoutputofthefilteris
T
T1−2TτvEs=eRCsi(τ)si(v)eRCeRCdτdv(RC)200
2T1−2TEbτ=eRCeRCdτ(RC)2T02TEbT2−=eRCeRC−1TEbT2−=1−eRCTThenoisepowerattheoutputofthefilteris
T
T1−2TPn=eRCE[n(τ)n(v)]dτdv(RC)2−∞−∞
T
T1−2TN0τ+v=eRCδ(τ−v)eRCdτdv(RC)2−∞−∞2
T1−2TN02τ=eRCeRCdτ(RC)2−∞2课后答案网1−2TN02T1N0=eRCeRC=2RC22RC2Hence,Es4EbRC−T2SNR==1−eRCwww.hackshp.cnPnTN03)ThevalueofRCthatmaximizesSNR,canbefoundbysettingthepartialderivativeofSNRwithrespecttoRCequaltozero.Thus,ifa=RC,thenϑSNR−TT−T−TT=0=(1−ea)−ea=−ea1++1ϑaaaSolvingthistranscendentalequationnumericallyfora,weobtainTT=1.26=⇒RC=a=a1.26Problem7.181)Thematchedfilteris−1t+1,0≤t(r1,r2)·m2and(√r1,r2)·m1>(r1,r2)·m3.Since(r1,r2)·m1=Tr1,(r1,r2)·m2=Tr2and(r1,r2)·m3=−Tr2,thepreviousconditionsarewrittenas课后答案网r1>r2andr1>−r2SimilarlywefindthatR2isthesetofpoints(r1,r2)thatsatisfyr2>0,r2>r1andR3istheregionsuchthatr2<0andr2<−r1.TheregionsR1,R2andR3areshowninthenextfigure.www.hackshp.cnR2R10R35)Ifthesignalsareequiprobablethen,P(e|m)=P(|r−m|2>|r−m|2|m)+P(|r−m|2>|r−m|2|m)1121131√Whenm1istransmittedthenr=[T+n1,n2]andtherefore,P(e|m1)iswrittenas√√P(e|m1)=P(n2−n1>T)+P(n1+n2<−T)175 Since,n1,n2arezero-meanstatisticallyindependentGaussianrandomvariables,eachwithvarianceN0,therandomvariablesx=n−nandy=n+narezero-meanGaussianwithvarianceN.212120Hence,

√21∞x21−Ty−−P(e|m1)=√√e2N0dx+√e2N0dy2πN0T2πN0−∞111TTT=Q+Q=2QN0N0N0√Whenm2istransmittedthenr=[n1,n2+T]andtherefore,√√P(e|m2)=P(n1−n2>T)+P(n2<−T)11T2T=Q+QN0N0Similarlyfromthesymmetryoftheproblem,weobtain11T2TP(e|m2)=P(e|m3)=Q+QN0N0SinceQ[·]ismomononicallydecreasing,weobtain112TTQN0P(s2)N0p2r(t)s1(t)dtr(t)s1(t)dt=s1(t)(s1(t)−s2(t))dt+n<0−∞s2wherenisazeromeanGaussianrandomvariablewithvariance
∞
∞σ2=(s(τ)−s(τ))(s(v)−s(v))E[n(τ)n(v)]dτdvn1212−∞−∞179
T
TN0=(s1(τ)−s2(τ))(s1(v)−s2(v))δ(τ−v)dτdv002
TN02=(s1(τ)−s2(τ))dτ20N
T
T2Aτ20=−Adτ200TN0A2T=23Since

∞TAt2Ats1(t)(s1(t)−s2(t))dt=−Adt−∞0TTA2T=6theprobabilityoferrorP(e|s1)isgivenbyA2TP(e|s1)=P(+n<0)62
−AT216x=%exp−dxA2TNA2TN02π0−∞2661A2T=Q6N0Similarlywefindthat1A2TP(e|s2)=Q6N0andsincethetwosignalsareequiprobable,theaverageprobabilityoferrorisgivenby课后答案网11P(e)=P(e|s1)+P(e|s2)2211A2TEs=Q=Q6N02N0whereEsistheenergyofthetransmittedsignals.www.hackshp.cnProblem7.23a)ThePDFofthenoisenisλ−λ|n|f(n)=e2TheoptimalreceiverusesthecriterionAAf(r|A)−λ[|r−A|−|r+A|]>>=e<1=⇒r<0f(r|−A)−A−A180 Theaverageprobabilityoferroris11P(e)=P(e|A)+P(e|−A)22
0
∞11=f(r|A)dr+f(r|−A)dr2−∞20
0
∞1−λ|r−A|1−λ|r+A|=λ2edr+λ2edr2−∞20λ
−Aλ
∞=e−λ|x|dx+e−λ|x|dx4−∞4A−A∞=λ1eλx+λ−1e−λx4λ−∞4λA1−λA=e2b)Thevarianceofthenoiseis
∞2λ−λ|x|2σn=exdx2−∞
∞−λx22!2=λexdx=λ=0λ3λ2Hence,theSNRisA2A2λ2SNR==22λ2andtheprobabilityoferrorisgivenby√√1−λ2A21−2SNRP(e)=e=e22ForP(e)=10−5weobtain课后答案网√ln(2×10−5)=−2SNR=⇒SNR=58.534=17.6741dBIfthenoisewasGaussian,then12Eb√www.hackshp.cnP(e)=Q=QSNRN0whereSNRisthesignaltonoiseratioattheoutputofthematchedfilter.WithP(e)=10−5we√findSNR=4.26andthereforeSNR=18.1476=12.594dB.Thustherequiredsignaltonoiseratiois5dBlesswhentheadditivenoiseisGaussian.Problem7.24Theenergyofthetwosignalss1(t)ands2(t)isE=A2TbThedimensionalityofthesignalspaceisone,andbychoosingthebasisfunctionas√10≤t2AA0:−r1−r2T)=P(n1−n2>T+0.5)
21∞−x2σ2=&endx2πσn2T+0.5Theaverageprobabilityoferroris11P(e)=P(e|s1)+P(e|s2)22
2
21T−1−x1∞−x=&e2σn2dx+&e2σn2dx22πσn2−∞22πσn2T+0.5TofindthevalueofTthatminimizestheprobabilityoferror,wesetthederivativeofP(e)withrespecttoTequaltozero.UsingtheLeibnitzruleforthedifferentiationofdefiniteintegrals,weobtain22(T−1)(T+0.5)ϑP(e)1−2−2=&e2σn−e2σn=0ϑT22πσn2or(T−1)2=(T+0.5)2=⇒T=0.25Thus,theoptimaldecisionruleiss1>r1−r2<0.25s2课后答案网Problem7.31a)Theinnerproductofsi(t)andsj(t)is
∞
∞nnsi(t)sj(t)dtwww.hackshp.cn=cikp(t−kTc)cjlp(t−lTc)dt−∞−∞k=1l=1nn
∞=cikcjlp(t−kTc)p(t−lTc)dt−∞k=1l=1nn=cikcjlEpδklk=1l=1n=Epcikcjkk=1nThequantityk=1cikcjkistheinnerproductoftherowvectorsCiandCj.SincetherowsofthematrixHnareorthogonalbyconstruction,weobtain
∞ns(t)s(t)dt=Ec2δ=nEδijpikijpij−∞k=1Thus,thewaveformssi(t)andsj(t)areorthogonal.186 b)UsingtheresultsofProblem7.28,weobtainthatthefiltermatchedtothewaveformnsi(t)=cikp(t−kTc)k=1canberealizedasthecascadeofafiltermatchedtop(t)followedbyadiscrete-timefiltermatchedtothevectorCi=[ci1,...,cin].Sincethepulsep(t)iscommontoallthesignalwaveformssi(t),weconcludethatthenmatchedfilterscanberealizedbyafiltermatchedtop(t)followedbyndiscrete-timefiltersmatchedtothevectorsCi,i=1,...,n.Problem7.32a)TheoptimalMLdetectorselectsthesequenceCithatminimizesthequantityn&D(r,C)=(r−EC)2ikbikk=1Themetricsofthetwopossibletransmittedsequencesarew&n&D(r,C)=(r−E)2+(r−E)21kbkbk=1k=w+1andw&n&D(r,C)=(r−E)2+(r+E)22kbkbk=1k=w+1Sincethefirsttermoftherightsideiscommonforthetwoequations,weconcludethattheoptimalMLdetectorcanbaseitsdecisionsonlyonthelastn−wreceivedelementsofr.ThatisC2n&n&22>(rk−Eb)−(rk+Eb)<0k=w+1k=w+1课后答案网C1orequivalentlyC1n>www.hackshp.cnrk<0k=w+1C2√b)Sincerk=EbCik+nk,theprobabilityoferrorP(e|C1)is&nP(e|C1)=PEb(n−w)+nk<0k=w+1n&=Pnk<−(n−w)Ebk=w+1n22Therandomvariableu=k=w+1nkiszero-meanGaussianwithvarianceσu=(n−w)σ.Hence1
√1−Eb(n−w)x2Eb(n−w)P(e|C1)=&exp(−)dx=Q2π(n−w)σ2−∞2π(n−w)σ2σ2187 SimilarlywefindthatP(e|C2)=P(e|C1)andsincethetwosequencesareequiprobable1Eb(n−w)P(e)=Qσ2c)TheprobabilityoferrorP(e)isminimizedwhenEb(n−w)ismaximized,thatisforw=0.Thisσ2impliesthatC1=−C2andthusthedistancebetweenthetwosequencesisthemaximumpossible.Problem7.331)Thedimensionalityofthesignalspaceistwo.Anorthonormalbasissetforthesignalspaceisformedbythesignals%%2,0≤t)21www.hackshp.cn1
∞x2A2T−=√e2N0dx=Q2πN0A2T2N02wherewehaveusedthefactthen=n2−%n1isazero-meanGaussianrandomvariablewithvarianceA2TN0.SimilarlywefindthatP(e|s1)=Q2N0,sothat111A2TP(e)=P(e|s1)+P(e|s2)=Q222N04)Thesignalwaveformψ(T−t)matchedtoψ(t)isexactlythesamewiththesignalwaveform121ψ2(T−t)matchedtoψ2(t).Thatis,%T2,0≤tr1−r22−V)=Q2−√88N0N0Ifs2(t)istransmitted,then1Tr(t)=s2(t)+s2(t−)+n(t)课后答案网22Theoutputofthesampleratt=Tandt=Tisgivenby2r1=n1((2T3A2Tr2=A++n2T42T4www.hackshp.cn15A2T=+n228TheprobabilityoferrorP(e|s2)is115A2T5A2TVP(e|s2)=P(n1−n2>+V)=Q+√2828N0N0Thus,theaverageprobabilityoferrorisgivenby11P(e)=P(e|s1)+P(e|s2)22111A2TV15A2TV=Q2−√+Q+√28N0N0228N0N0189 ϑP(e)TheoptimalvalueofVcanbefoundbysettingequaltozero.UsingLeibnitzruletoϑVdifferentiatedefiniteintegrals,weobtain1212ϑP(e)A2TV5A2TV=0=2−√−+√ϑV8N0N028N0N0orbysolvingintermsofV11A2TV=−826)Letabefixedtosomevaluebetween0and1.Then,ifweargueasinpart5)weobtain1A2TP(e|s1,a)=P(n2−n1>2−V(a))81A2TP(e|s2,a)=P(n1−n2>(a+2)+V(a))8andtheprobabilityoferroris11P(e|a)=P(e|s1,a)+P(e|s2,a)22ϑP(e|a)Foragivena,theoptimalvalueofV(a)isfoundbysettingequaltozero.BydoingsoweϑV(a)findthat1aA2TV(a)=−42ThemeansquareestimationofV(a)is11
11A2T
11A2TV=V(a)f(a)da=−ada=−042082Problem7.34课后答案网Forbinaryphasemodulation,theerrorprobabilityis112EbA2TP2=Q=Qwww.hackshp.cnN0N0WithP2=10−6wefindfromtablesthat1A2T=4.74=⇒A2T=44.9352×10−10N0Ifthedatarateis10Kbps,thenthebitintervalisT=10−4andtherefore,thesignalamplitudeis&A=44.9352×10−10×104=6.7034×10−3Similarlywefindthatwhentherateis105bpsand106bps,therequiredamplitudeofthesignalisA=2.12×10−2andA=6.703×10−2respectively.Problem7.351)TheimpulseresponseofthematchedfilterisA(T−t)cos(2πf(T−t))0≤t≤Ts(t)=u(T−t)=Tc0otherwise190 2)Theoutputofthematchedfilteratt=Tis
Tg(T)=u(t)s(t)|=u(T−τ)s(τ)dτt=T02
TA22=(T−τ)cos(2πfc(T−τ))dτT202
Tv=T−τA22=vcos(2πfcv)dvT20A2v3v21vcos(4πfv)T=+−sin(4πfv)+cT264×2πf8×(2πf)3c4(2πf)2ccc0A2T3T21Tcos(4πfcT)=+−sin(4πfcT)+T264×2πfc8×(2πfc)34(2πfc)23)Theoutputofthecorrelatoratt=Tis
Tq(T)=u2(τ)dτ02
TA22=τcos(2πfcτ)dτT20However,thisisthesameexpressionwiththecaseoftheoutputofthematchedfiltersampledatt=T.Thus,thecorrelatorcansubstitutethematchedfilterinademodulationsystemandviseversa.Problem7.361)Thesignalr(t)canbewrittenas&&r(t)=±2Pscos(2πfct+φ)+2Pcsin(2πfct+φ)1%−1Ps=2(Pc+Ps)sin2πfct+φ+antanPc1&−1Pc=课后答案网2PTsin2πfct+φ+ancosPTwherean=±1aretheinformationsymbolsandPTisthetotaltransmittedpower.AsitisobservedthesignalhastheformofaPMsignalwhere1−1Pcwww.hackshp.cnθn=ancosPTAnymethodusedtoextractthecarrierphasefromthereceivedsignalcanbeemployedatthereceiver.Thefollowingfigureshowsthestructureofareceiverthatemploysadecision-feedbackPLL.TheoperationofthePLLisdescribedinthenextpart.t=Tbv(t)T×b(·)dtThreshold0DFPLLcos(2πfct+φ)2)Atthereceiverthesignalisdemodulatedbycrosscorrelatingthereceivedsignal1&−1Pcr(t)=2PTsin2πfct+φ+ancos+n(t)PT191 withcos(2πfct+φˆ)andsin(2πfct+φˆ).Thesampledvaluesattheoutputofthecorrelatorsare1&1r1=2PT−ns(t)sin(φ−φˆ+θn)+nc(t)cos(φ−φˆ+θn)221&1r2=2PT−ns(t)cos(φ−φˆ+θn)+nc(t)sin(φˆ−φ−θn)22wherenc(t),ns(t)arethein-phaseandquadraturecomponentsofthenoisen(t).Ifthedetectorhasmadethecorrectdecisiononthetransmittedpoint,thenbymultiplyingr1bycos(θn)andr2bysin(θn)andsubtractingtheresults,weobtain(afterignoringthenoise)1&rcos(θ)=2Psin(φ−φˆ)cos2(θ)+cos(φ−φˆ)sin(θ)cos(θ)1nTnnn21&rsin(θ)=2Pcos(φ−φˆ)cos(θ)sin(θ)−sin(φ−φˆ)sin2(θ)2nTnnn21&e(t)=r1cos(θn)−r2sin(θn)=2PTsin(φ−φˆ)2Theerrore(t)ispassedtotheloopfilteroftheDFPLLthatdrivestheVCO.Asitisseenonlythephaseθnisusedtoestimatethecarrierphase.3)Havingacorrectcarrierphaseestimate,theoutputofthelowpassfiltersampledatt=Tbis11&−1Pcr=±2PTsincos+n2PT11&Pc=±2PT1−+n2PT11Pc=±2PT1−+n2PTwherenisazero-meanGaussianrandomvariablewithvariance

TbTbσ2=E课后答案网n(t)n(τ)cos(2πft+φ)cos(2πfτ+φ)dtdτncc00
TN0b2=cos(2πfct+φ)dt20N0=4www.hackshp.cnNotethatTbhasbeennormalizedto1sincetheproblemhasbeenstatedintermsofthepoweroftheinvolvedsignals.Theprobabilityoferrorisgivenby12PTPcP(error)=Q1−N0PTThelossduetotheallocationofpowertothepilotsignalisPcSNRloss=10log101−PTWhenPc/PT=0.1,thenSNRloss=10log10(0.9)=−0.4576dBThenegativesignindicatesthattheSNRisdecreasedby0.4576dB.192 Problem7.371)Ifthereceivedsignalisr(t)=±gT(t)cos(2πfct+φ)+n(t)thenbycrosscorrelatingwiththesignalattheoutputofthePLL12ψ(t)=gt(t)cos(2πfct+φˆ)Egweobtain
1
T2Tr(t)ψ(t)dt=±g2(t)cos(2πft+φ)cos(2πft+φˆ)dtTcc0Eg0
1T2+n(t)gt(t)cos(2πfct+φˆ)dt0Eg1
2Tg2(t)=±Tcos(2π2ft+φ+φˆ)+cos(φ−φˆ)dt+ncEg021Eg=±cos(φ−φˆ)+n2wherenisazero-meanGaussianrandomvariablewithvarianceN0.Ifweassumethatthesignal2s1(t)=gT(t)cos(2πfct+φ)wastransmitted,thentheprobabilityoferroris1EgP(error|s1(t))=Pcos(φ−φˆ)+n<0211Egcos2(φ−φˆ)2Escos2(φ−φˆ)=Q=QN0N0whereEs=Eg/2istheenergyofthetransmittedsignal.Asitisobservedthephaseerror课后答案网φ−φˆreducestheSNRbyafactorSNR=−10logcos2(φ−φˆ)loss102)Whenφ−φˆ=45o,thenthelossduetothephaseerroriswww.hackshp.cn2o1SNRloss=−10log10cos(45)=−10log10=3.01dB2Problem7.381)TheclosedlooptransferfunctionisG(s)/sG(s)1H(s)===√1+G(s)/ss+G(s)s2+2s+1Thepolesofthesystemaretherootsofthedenominator,thatis√√−2±2−411ρ1,2==−√±j√222Sincetherealpartoftherootsisnegative,thepoleslieinthelefthalfplaneandtherefore,thesystemisstable.193 2)WritingthedenominatorintheformD=s2+2ζωs+ω2nnweidentifythenaturalfrequencyoftheloopasω=1andthedampingfactorasζ=√1.n2Problem7.391)TheclosedlooptransferfunctionisKG(s)/sG(s)Kτ1H(s)====1+G(s)/ss+G(s)τ1s2+s+Ks2+1s+Kτ1τ1Thegainofthesystematf=0is|H(0)|=|H(s)|s=0=12)Thepolesofthesystemaretherootsofthedenominator,thatis√−1±1−4Kτ1ρ1,2==2τ1Inorderforthesystemtobestabletherealpartofthepolesmustbenegative.SinceKisgreaterthanzero,thelatterimpliesthatτ1ispositive.Ifinadditionwerequirethatthedampingfactorζ=√1islessthan1,thenthegainKshouldsatisfythecondition2τ1K1K>4τ1Problem7.40ThetransferfunctionoftheRCcircuitisR+12Cs1+R2Cs1+τ2sG(s)=课后答案网1==R1+R2+Cs1+(R1+R2)Cs1+τ1sFromthelastequalityweidentifythetimeconstantsasτ2www.hackshp.cn=R2C,τ1=(R1+R2)CProblem7.41Assumingthattheinputresistanceoftheoperationalamplifierishighsothatnocurrentflowsthroughit,thenthevoltage-currentequationsofthecircuitareV2=−AV11V1−V2=R1+iCsV1−V0=iRwhere,V1,V2istheinputandoutputvoltageoftheamplifierrespectively,andV0isthesignalattheinputofthefilter.EliminatingiandV1,weobtainR11+VCs2R=R1V111+Cs1+−AAR194 IfweletA→∞(idealamplifier),thenV21+R1Cs1+τ2s==V1RCsτ1sHence,theconstantsτ1,τ2oftheactivefilteraregivenbyτ1=RC,τ2=R1CProblem7.42UsingthePythagoreantheoremforthefour-phaseconstellation,wefind222dr1+r1=d=⇒r1=√2Theradiusofthe8-PSKconstellationisfoundusingthecosinerule.Thus,2222odd=r2+r2−2r2cos(45)=⇒r2=%√2−2Theaveragetransmittedpowerofthe4-PSKandthe8-PSKconstellationisgivenbyd2d2P4,av=,P8,av=√22−2Thus,theadditionaltransmittedpowerneededbythe8-PSKsignalis2d2P=10log10√=5.3329dB(2−2)d2Weobtainthesameresultsifweusetheprobabilityoferrorgivenby&πPM=2Q2ρssin课后答案网MwhereρsistheSNRpersymbol.Inthiscase,equalerrorprobabilityforthetwosignalingschemes,impliesthatsinπ2π2πρ8,s4ρ4,ssin=ρ8,ssin=⇒10log10=20log10π=5.3329dB4www.hackshp.cn8ρ4,ssin8Problem7.43TheconstellationofFig.P-7.43(a)hasfourpointsatadistance2Afromtheoriginandfourpoints√atadistance22A.Thus,theaveragetransmittedpoweroftheconstellationis1√P=4×(2A)2+4×(22A)2=6A2a8√Thesecondconstellationhasfourpointsatadistance7Afromtheorigin,twopointsatadis-√tance3AandtwopointsatadistanceA.Thus,theaveragetransmittedpowerofthesecondconstellationis1√√9P=4×(7A)2+2×(3A)2+2A2=A2b82SincePb0theexpressionfortheprobabilityoferrortakestheform1
VT1
∞P(error)=p(r|s1(t))dr+p(r|s0(t))dr202VT
2√
1VTrr+EbrE1∞rr2−b−=e2σ2I0dr+e2σ2dr20σ2σ22Vσ2T204 TheoptimumthresholdlevelisthevalueofVT√thatminimizestheprobabilityoferror.However,whenEb1theoptimumvalueisclosetoEbandwewillusethisthresholdtosimplifytheN02analysis.TheintegralinvolvingtheBesselfunctioncannotbeevaluatedinclosedform.InsteadofI0(x)wewillusetheapproximationexI0(x)≈√2πxwhichisvalidforlargex,thatisforhighSNR.Inthiscase√√E11
VTrr2+EbrE1
2br√−b−(r−Eb)2/2σ2e2σ2I0dr≈√edr20σ2σ2202πσ2EbThisintegralisfurthersimplifiedifweobservethatforhighSNR,theintegrandisdominantinthe√vicinityofEbandtherefore,thelowerlimitcanbesubstitutedby−∞.Also1(r1√≈2πσ2Eb2πσ2andtherefore,√1√
Eb
Eb(√√12r−(r−E)2/2σ2121−(r−E)2/2σ2√ebdr≈ebdr202πσ2Eb2−∞2πσ211Eb=Q22N0Finally1
1E1∞2rr2b−P(error)=Q+√eN0dr22N2EbN00211E1−Ebb≤Q+e4N0课后答案网22N02Problem7.54(a)FourphasePSKIfweuseapulseshapehavingaraisedcosinespectrumwitharolloffα,thesymbolrateisdeter-minedfromtherelationwww.hackshp.cn1(1+α)=500002THence,1105=T1+αwhereW=105Hzisthechannelbandwidth.Thebitrateis22×105=bpsT1+α(b)BinaryFSKwithnoncoherentdetectionInthiscaseweselectthetwofrequenciestohaveafrequencyseparationof1,where1istheTTsymbolrate.Hence1f1=fc−2T1f2=f+c+2T205 wherefcisthecarrierinthecenterofthechannelband.Thus,wehave1=500002Torequivalently15=10THence,thebitrateis105bps.(c)M=4FSKwithnoncoherentdetectionInthiscasewerequirefourfrequencieswithadjacentfrequenciesseparationof1.Hence,weselectT1.5111.5f1=fc−|,f2=fc−,f3=fc+,f4=fc+T2T2TTwherefisthecarrierfrequencyand1=25000,or,equivalently,c2T1=50000TSincethesymbolrateis50000symbolspersecondandeachsymbolconveys2bits,thebitrateis105bps.Problem7.55a)Fornrepeatersincascade,theprobabilityofioutofnrepeaterstoproduceanerrorisgivenbythebinomialdistributionnin−iPi=p(1−p)iHowever,thereisabiterrorattheoutputoftheterminalreceiveronlywhenanoddnumberofrepeatersproducesanerror.Hence,theoverallprobabilityoferrorisnP=P=pi(1−p)n−inoddii=oddLetPevenbetheprobabilitythatanevennumberofrepeatersproducesanerror.Then课后答案网nP=pi(1−p)n−ievenii=evenandtherefore,nwww.hackshp.cnnin−inPeven+Podd=p(1−p)=(p+1−p)=1ii=0OnemorerelationbetweenPevenandPoddcanbeprovidedifweconsiderthedifferencePeven−Podd.Clearly,nin−inin−iPeven−Podd=p(1−p)−p(1−p)iii=eveni=oddanin−inin−i=(−p)(1−p)+(−p)(1−p)iii=eveni=odd=(1−p−p)n=(1−2p)nwheretheequality(a)followsfromthefactthat(−1)iis1forievenand−1wheniisodd.SolvingthesystemPeven+Podd=1P−P=(1−2p)nevenodd206 weobtain1nPn=Podd=(1−(1−2p))2b)Expandingthequantity(1−2p)n,weobtainnn(n−1)2(1−2p)=1−n2p+(2p)+···2Since,p1wecanignoreallthepowersofpwhicharegreaterthanone.Hence,1−6−4Pn≈(1−1+n2p)=np=100×10=102Problem7.56Theoverallprobabilityoferrorisapproximatedby1EbP(e)=KQN0%Thus,withP(e)=10−6andK=100,weobtaintheprobabilityofeachrepeaterP=QEb=rN010−8.TheargumentofthefunctionQ[·]thatprovidesavalueof10−8isfoundfromtablestobe1Eb=5.61N0Hence,therequiredEbis5.612=31.47N0Problem7.57a)Theantennagainforaparabolicantennaofdiameter课后答案网Dis2πDGR=ηλIfweassumethattheefficiencyfactoris0.5,thenwithcwww.hackshp.cn3×108λ===0.3mD=3×0.3048mf109weobtainGR=GT=45.8458=16.61dBb)TheeffectiveradiatedpowerisEIRP=PTGT=GT=16.61dBc)ThereceivedpowerisPTGTGR−9PR=2=2.995×10=−85.23dB=−55.23dBm4πdλ207 NotethatactualpowerinWattsdBm=10log1010−3=30+10log10(powerinWatts)Problem7.58a)TheantennagainforaparabolicantennaofdiameterDis2πDGR=ηλIfweassumethattheefficiencyfactoris0.5,thenwithc3×108λ===0.3mandD=1mf109weobtainGR=GT=54.83=17.39dBb)TheeffectiveradiatedpowerisEIRP=PTGT=0.1×54.83=7.39dBc)ThereceivedpowerisPTGTGR−10PR=2=1.904×10=−97.20dB=−67.20dBm4πdλProblem7.59Thewavelengthofthetransmittedsignalis课后答案网3×108λ==0.03m10×109Thegainoftheparabolicantennais22πDπ105GR=ηwww.hackshp.cn=0.6=6.58×10=58.18dBλ0.03ThereceivedpowerattheoutputofthereceiverantennaisPTGTGR3×101.5×6.58×105−13PR=(4πd)2=4×1072=2.22×10=−126.53dBλ(4×3.14159×0.03)Problem7.60a)SinceT=3000K,itfollowsthatN=kT=1.38×10−23×300=4.14×10−21W/Hz0Ifweassumethatthereceivingantennahasanefficiencyη=0.5,thenitsgainisgivenby22πD3.14159×505GR=ηλ=0.53×108=5.483×10=57.39dB2×109208 Hence,thereceivedpowerlevelisPTGTGR10×10×5.483×105−13PR=d2=1082=7.8125×10=−121.07dB(4πλ)(4×3.14159×0.15)b)IfEb=10dB=10,thenN0−1−13PREb7.8125×10−17R==×10=1.8871×10=18.871Mbits/secN0N04.14×10−21Problem7.61TheoverallgainofthesystemisGtot=Ga1+Gos+GBPF+Ga2=10−5−1+25=29dBHence,thepowerofthesignalattheinputofthedemodulatorisP=(−113−30)+29=−114dBs,demThenoise-figureforthecascadeofthefirstamplifierandthemultiplierisFos−10.5100.5−1F1=Fa1+=10+=3.3785Ga101WeassumethatF1isthespotnoise-figureandtherefore,itmeasurestheratiooftheavailablePSDoutofthetwodevicestotheavailablePSDoutofanidealdevicewiththesameavailablegain.Thatis,Sn,o(f)F1=Sn,i(f)Ga1GoswhereSn,o(f)isthepowerspectraldensityofthenoiseattheinputofthebandpassfilterandSn,i(f)isthepowerspectraldensityattheinputoftheoverallsystem.Hence,课后答案网−175−30−0.5−20Sn,o(f)=1010×10×10×3.3785=3.3785×10Thenoise-figureofthecascadeofthebandpassfilterandthesecondamplifierisFa−1100.5−120.2F2=FBPFwww.hackshp.cn+G=10+10−0.1=4.307BPFHence,thepowerofthenoiseattheoutputofthesystemisP=2S(f)BGGaF=7.31×10−12=−111.36dBn,demn,oBPF22Thesignaltonoiseratioattheoutputofthesystem(inputtothedemodulator)isPs,demSNR==−114+111.36=−2.64dBPn,demProblem7.62Thewavelengthofthetransmissionisc3×108λ===0.75mf4×109209 If1MHzisthepassbandbandwidth,thentherateofbinarytransmissionisRb=W=106bps.Hence,withN0=4.1×10−21W/HzweobtainPREb6−211.5−13=Rb=⇒10×4.1×10×10=1.2965×10N0N0Thetransmittedpowerisrelatedtothereceivedpowerthroughtherelation2PTGTGRPRdPR=d2=⇒PT=4π(4π)GTGRλλSubstitutinginthisexpressionthevaluesGT=100.6,GR=105,d=36×106andλ=0.75weobtainPT=0.1185=−9.26dBWProblem7.63SinceT=2900+150=3050K,itfollowsthatN=kT=1.38×10−23×305=4.21×10−21W/Hz0Thetransmittingwavelengthλisc3×108λ===0.130mf2.3×109Hence,thegainofthereceivingantennais22πD3.14159×646GR=η=0.55=1.3156×10=61.19dBλ0.130andtherefore,thereceivedpowerlevelisPTGTGR17×102.7×1.3156×106−12PR=(4πd)2=课后答案网1.6×10112=4.686×10=−113.29dBλ(4×3.14159×0.130)IfEb/N0=6dB=100.6,then−1−12PREb4.686×10−0.69R==×10=4.4312×10=4.4312Gbits/secN0N04.www.hackshp.cn21×10−21Problem7.64Inthenondecision-directedtimingrecoverymethodwemaximizethefunctionΛ(τ)=y2(τ)2mmwithrespecttoτ.Thus,weobtaintheconditiondΛ2(τ)dym(τ)=2ym(τ)=0dτdτmSupposenowthatweapproximatethederivativeofthelog-likelihoodΛ2(τ)bythefinitedifferencedΛ2(τ)Λ2(τ+δ)−Λ2(τ−δ)≈dτ2δ210 Then,ifwesubstitutetheexpressionofΛ2(τ)inthepreviousapproximation,weobtaindΛ2(τ)dτmym2(τ+δ)−mym2(τ−δ)=2δ
2
21=r(t)u(t−mT−τ−δ)dt−r(t)u(t−mT−τ+δ)dt2δmwhereu(−t)=gR(t)istheimpulseresponseofthematchedfilterinthereceiver.However,thisistheexpressionoftheearly-lategatesynchronizer,wherethelowpassfilterhasbeensubstitutedbythesummationoperator.Thus,theearly-lategatesynchronizerisacloseapproximationtothetimingrecoverysystem.Problem7.65Anon-offkeyingsignalisrepresentedass1(t)=Acos(2πfct+θc),0≤t≤T(binary1)s2(t)=0,0≤t≤T(binary0)Letr(t)bethereceivedsignal,thatisr(t)=s(t;θc)+n(t)wheres(t;θ)iseithers(t)ors(t)andn(t)iswhiteGaussiannoisewithvarianceN0.Thec122likelihoodfunction,thatistobemaximizedwithrespecttoθcovertheinterval[0,T],isproportionalto
T22Λ(θc)=exp−[r(t)−s(t;θc)]dtN00MaximizationofΛ(θc)isequivalenttothemaximizationofthelog-likelihoodfunction课后答案网
T22ΛL(θc)=−[r(t)−s(t;θc)]dtN002
T4
T2
T=−r2(t)dt+r(t)s(t;θ)dt−s2(t;θ)dtccN00N00N00Sincethefirsttermdoesnotinvolvetheparameterofinterestwww.hackshp.cnθcandthelasttermissimplyaconstantequaltothesignalenergyofthesignalover[0,T]whichisindependentofthecarrierphase,wecancarrythemaximizationoverthefunction
TV(θc)=r(t)s(t;θc)dt0Notethats(t;θc)cantaketwodifferentvalues,s1(t)ands2(t),dependingonthetransmissionofabinary1or0.Thus,amoreappropriatefunctiontomaximizeistheaveragelog-likelihood
T
T11V¯(θc)=r(t)s1(t)dt+r(t)s2(t)dt2020Sinces2(t)=0,thefunctionV¯(θc)takestheform
T1V¯(θc)=r(t)Acos(2πfct+θc)dt20211 SettingthederivativeofV¯(θc)withrespecttoθcequaltozero,weobtain
TϑV¯(θc)1=0=r(t)Asin(2πfct+θc)dtϑθc20
T
T11=cosθcr(t)Asin(2πfct)dt+sinθcr(t)Acos(2πfct)dt2020Thus,themaximumlikelihoodestimateofthecarrierphaseisT0r(t)Asin(2πfct)dtθc,ML=−arctanT0r(t)Acos(2πfct)dt课后答案网www.hackshp.cn212 Chapter8Problem8.11)ThefollowingtableshowsthevaluesofEh(W)/TobtainedusinganadaptiverecursiveNewton-Cotesnumericalintegrationrule.WT0.51.01.52.02.53.0Eh(W)/T0.22530.34420.37300.37480.34790.3750AplotofEh(W)/TasafunctionofWTisgiveninthenextfigure0.40.350.30.250.2Energy/T0.150.10.0500.511.522.533.5WT2)ThevalueofEh(W)asW→∞is
∞
TlimE课后答案网(W)=g2(t)dt=g2(t)dthTTW→∞−∞0
T212πT=1+cost−dt40T2
T1T2πT=+cost−dtwww.hackshp.cn420T2
1T2πT+1+cos2t−dt80T2TT3T=+==0.3750T488Problem8.2Wehavea+n−1withProb.124y=a+n+1withProb.124a+nwithProb.12Bysymmetry,Pe=P(e|a=1)=P(e|a=−1),hence,11311Pe=P(e|a=−1)=P(n−1>0)+Pn−>0+Pn−>024242111311=Q+Q+Q2σn42σn42σn213 Problem8.3a)Ifthetransmittedsignalis∞r(t)=anh(t−nT)+n(t)n=−∞thentheoutputofthereceivingfilteris∞y(t)=anx(t−nT)+ν(t)n=−∞wherex(t)=h(t)h(t)andν(t)=n(t)h(t).Ifthesamplingtimeisoffby10%,thenthesamplesattheoutputofthecorrelatoraretakenatt=(m±1)T.Assumingthatt=(m−1)Twithout1010lossofgenerality,thenthesampledsequenceis∞11ym=anx((m−T−nT)+ν((m−)T)1010n=−∞∞1IfthesignalpulseisrectangularwithamplitudeAanddurationT,thenn=−∞anx((m−10T−nT)isnonzeroonlyforn=mandn=m−1andtherefore,thesampledsequenceisgivenby111ym=amx(−T)+am−1x(T−T)+ν((m−)T)10101092121=amAT+am−1AT+ν((m−)T)101010Thepowerspectraldensityofthenoiseattheoutputofthecorrelatoris2N0222Sν(f)=Sn(f)|H(f)|=ATsinc(fT)2Thus,thevarianceofthenoiseis
∞2课后答案网N0222N0221N02σnu=ATsinc(fT)df=AT=AT−∞22T2andtherefore,theSNRis222292(AT)812ATSNR==www.hackshp.cn10N0A2T100N0Asitisobserved,thereisalossof10log81=−0.9151dBduetothemistiming.10100b)Recallfromparta)thatthesampledsequenceis9212ym=amAT+am−1AT+νm1010A2TThetermam−110expressestheISIintroducedtothesystem.Ifam=1istransmitted,thentheprobabilityoferroris11P(e|am=1)=P(e|am=1,am−1=1)+P(e|am=1,am−1=−1)22
−A2T2
−8A2T2νν1−110−=√eN0A2Tdν+√eN0A2Tdν2πN0A2T−∞2πN0A2T−∞1112A2T1822A2T=Q+Q2N0210N0214 SincethesymbolsofthebinaryPAMsystemareequiprobablethepreviousderivedexpressionistheprobabilityoferrorwhenasymbolbysymboldetectorisemployed.ComparingthiswiththeprobabilityoferrorofasystemwithnoISI,weobservethatthereisanincreaseoftheprobabilityoferrorby111822A2T12A2TPdiff(e)=Q−Q210N02N0Problem8.41)ThepowerspectraldensityofX(t)isgivenby12Sx(f)=Sa(f)|GT(f)|TTheFouriertransformofg(t)issinπfT−jπfTGT(f)=F[g(t)]=ATeπfTHence,|G(f)|2=(AT)2sinc2(fT)Tandtherefore,S(f)=A2TS(f)sinc2(fT)=A2Tsinc2(fT)xa2)Ifg1(t)isusedinsteadofg(t)andthesymbolintervalisT,then12Sx(f)=Sa(f)|G2T(f)|T12222=(A2T)sinc(f2T)=4ATsinc(f2T)T3)Ifweprecodetheinputsequenceas课后答案网bn=an+αan−3,then1+α2m=0Rb(m)=αm=±3www.hackshp.cn0otherwiseandtherefore,thepowerspectraldensitySb(f)isS(f)=1+α2+2αcos(2πf3T)bToobtainanullatf=1,theparameterαshouldbesuchthat3T1+α2+2αcos(2πf3T)=0=⇒α=−1|1f=34)Theanswertothisquestionisno.ThisisbecauseSb(f)isananalyticfunctionandunlessitisidenticaltozeroitcanhaveatmostacountablenumberofzeros.Thispropertyoftheanalyticfunctionsisalsoreferredasthetheoremofisolatedzeros.215 Problem8.51)Thepowerspectraldensityofs(t)isσ21S(f)=a|G(f)|2=|G(f)|2sTTTTTheFouriertransformGT(f)ofthesignalg(t)ist−Tt−3TG(f)=FΠ4−Π4TTT22TT−j2πfTTT−j2πf3T=sinc(f)e4−sinc(f)e42222TT−j2πfTj2πfT−j2πfT=sinc(f)e2e4−e422TTT−j2πfT=sinc(f)sin(2πf)2je2224Hence,222T2T|GT(f)|=Tsinc(f)sin(2πf)24andtherefore,2T2TSs(f)=Tsinc(f)sin(2πf)242)Iftheprecodingschemebn=an+kan−1isused,then1+k2m=0Rb(m)=km=±10otherwiseThus,课后答案网S(f)=1+k2+2kcos(2πfT)bandthereforethespectrumofs(t)is22T2TSs(f)=(1+k+2kcos(2πfT))Tsinc(f)sin(2πf)www.hackshp.cn24Inordertoproduceafrequencynullatf=1wehavetochoosekinsuchawaythatT1+k2+2kcos(2πfT)=1+k2+2k=0|f=1/TTheappropriatevalueofkis−1.3)Iftheprecodingschemeofthepreviouspartisused,theninordertohavenullsatfrequenciesf=n,thevalueoftheparameterkshouldbesuchthat4T1+k2+2kcos(2πfT)=1+k2=0|f=1/4TAsitisobserveditisnotpossibletoachievethedesirednullswithrealvaluesofk.Insteadofthepre-codingschemeofthepreviouspartwesuggestpre-codingoftheformbn=an+kan−2216 Inthiscase1+k2m=0Rb(m)=km=±20otherwiseThus,S(f)=1+k2+2kcos(2π2fT)bandthereforeS(n)=0fork=1.b2TProblem8.6a)ThepowerspectraldensityoftheFSKsignalmaybeevaluatedbyusingequation(8.5.32)withk=2(binary)signalsandprobabilitiesp=p=1.Thus,whentheconditionthatthecarrier012phaseθ0andandθ1arefixed,weobtain∞1nn2n12S(f)=2|S0()+S1()|δ(f−)+|S0(f)−S1(f)|4Tbn=−∞TbTbTb4TbwhereS0(f)andS1(f)aretheFouriertransformsofs0(t)ands1(t).Inparticular,
TbS(f)=s(t)e−j2πftdt0001
T2Ebbj2πft∆f=cos(2πf0t+θ0)edt,f0=fc−Tb021=12EbsinπTb(f−f0)+sinπTb(f+f0)e−jπfTbejθ02Tbπ(f−f0)π(f+f0)Similarly,
TbS(f)=s(t)e−j2πftdt1101=12EbsinπTb(f−f1)+sinπTb(f+f1)e−jπfTbejθ12课后答案网Tbπ(f−f1)π(f+f1)∆fwheref1=fc+2.ByexpressingS(f)as∞1n2n2n∗nnS(f)=2|S0()|+|S1()|+2Re[S0()S1()]δ(f−)4Tbn=−∞www.hackshp.cnTbTbTbTbTb1+|S(f)|2+|S(f)|2−2Re[S(f)S∗(f)]01014Tbwenotethatthecarrierphasesθ0andθ1affectonlythetermsRe(S0S1∗).Ifweaverageovertherandomphases,thesetermsdropout.Hence,wehave∞1n2n2nS(f)=2|S0()|+|S1()|δ(f−)4Tbn=−∞TbTbTb1+|S(f)|2+|S(f)|2014Tbwhere2TbEbsinπTb(f−fk)sinπTb(f+fk)|Sk(f)|=+,k=0,12π(f−fk)π(f+fk)NotethatthefirstterminS(f)consistsofasequenceofsamplesandthesecondtermconstitutesthecontinuousspectrum.217 b)ItisapparentfromS(f)thattheterms|S(f)|2decayproportionallyas1.alsonotethatk(f−fk)2222TbEbsinπTb(f−fk)sinπTb(f+fk)|Sk(f)|=+2π(f−fk)π(f+fk)becausetheproductsinπTb(f−fk)sinπTb(f+fk)×≈0π(f−fk)π(f+fk)duetotherelationthatthecarrierfrequencyf1.cTbProblem8.71)Theautocorrelationfunctionoftheinformationsymbols{an}is∗12Ra(k)=E[ana+n+k]=×|an|δ(k)=δ(k)4Thus,thepowerspectraldensityofv(t)is1212SV(f)=Sa(f)|G(f)|=|G(f)|TTt−TwhereG(f)=F[g(t)].Ifg(t)=AΠ(2),weobtain|G(f)|2=A2T2sinc2(fT)andtherefore,TS(f)=A2Tsinc2(fT)VInthenextfigureweplotSV(f)forT=A=1.10.90.80.70.60.5Sv(f)0.4课后答案网0.30.20.10-5-4www.hackshp.cn-3-2-1012345frequencyf2)Ifg(t)=Asin(πt)Π(t−T/2),then2T1111−j2πfTG(f)=Aδ(f−)−δ(f+)Tsinc(fT)e22j42j4AT11−j(2πfT+π)=[δ(f−)−δ(f+)]sinc(fT)e22244AT−jπ[(f−1)T+1]11−jπT=e42sinc((f−)T)−sinc((f−)T)e2244Thus,222AT2121|G(f)|=sinc((f+)T)+sinc((f−)T)44411πT−2sinc((f+)T)sinc((f−)T)cos442218 andthepowerspectralofthetransmittedsignalis2AT2121SV(f)=sinc((f+)T)+sinc((f−)T)44411πT−2sinc((f+)T)sinc((f−)T)cos442InthenextfigureweplotSV(f)fortwospecialvaluesofthetimeintervalT.TheamplitudeofthesignalAwassetto1forbothcases.0.450.80.40.70.350.6T=10.30.50.25T=20.4Sv(f)0.2Sv(f)0.30.150.20.10.050.100-5-4-3-2-1012345-5-4-3-2-1012345frequencyffrequencyf3)Thefirstspectralnullofthepowerspectrumdensityinpart1)isatposition1Wnull=TThe3-dBbandwidthisspecifiedbysolvingtheequation:1SV(W3dB)=SV(0)2Sincesinc2(0)=1,weobtain211sinc(W3dB课后答案网T)=2=⇒sin(πW3dBT)=√πW3dBT2Solvingthelatterequationnumericallywefindthat1.39160.443W3dB==www.hackshp.cnπTTTofindthefirstspectralnullandthe3-dBbandwidthforthesignalwithpowerspectraldensityinpart2)weassumethatT=1.Inthiscase2A2121SV(f)=sinc((f+))+sinc((f−))444andasitisobservedthereisnovalueoffthatmakesSV(f)equaltozero.Thus,Wnull=∞.Tofindthe3-dBbandwidthnotethatA21A2SV(0)=2sinc()=1.6212444Solvingnumericallytheequation1A2SV(W3dB)=1.621224wefindthatW3dB=0.5412.Asitisobservedthe3-dBbandwidthismorerobustasameasureforthebandwidthofthesignal.219 Problem8.8ThetransitionprobabilitymatrixPis010110011P=211001010Hence,102124461211014264P2=andP4=4011216462412016442andtherefore,2446100−11426401−104Pγ=1646240−1106442−1001−400410−4401==−γ1604−404400−4Thus,P4γ=−1γandbypre-multiplyingbothsidesbyPk,weobtain4k+41kPγ=−Pγ4Problem8.9课后答案网a)TakingtheinverseFouriertransformofH(f),weobtain−1ααh(t)=F[H(f)]=δ(t)+δ(t−t0)+δ(t+t0)www.hackshp.cn22Hence,ααy(t)=s(t)h(t)=s(t)+s(t−t0)+s(t+t0)22b)Ifthesignals(t)isusedtomodulatethesequence{an},thenthetransmittedsignalis∞u(t)=ans(t−nT)n=−∞Thereceivedsignalistheconvolutionofu(t)withh(t).Hence,∞ααy(t)=u(t)h(t)=ans(t−nT)δ(t)+δ(t−t0)+δ(t+t0)22n=−∞∞∞∞αα=ans(t−nT)+ans(t−t0−nT)+ans(t+t0−nT)22n=−∞n=−∞n=−∞220 Thus,theoutputofthematchedfilters(−t)atthetimeinstantt1is∞
∞w(t1)=ans(τ−nT)s(τ−t1)dτn=−∞−∞∞
∞α+ans(τ−t0−nT)s(τ−t1)dτ2−∞n=−∞∞
∞α+ans(τ+t0−nT)s(τ−t1)dτ2−∞n=−∞Ifwedenotethesignals(t)s(t)byx(t),thentheoutputofthematchedfilteratt1=kTis∞w(kT)=anx(kT−nT)n=−∞∞∞αα+anx(kT−t0−nT)+anx(kT+t0−nT)22n=−∞n=−∞c)Witht0=Tandk=ninthepreviousequation,weobtainwk=akx0+anxk−nn=kαααα+akx−1+anxk−n−1+akx1+anxk−n+12222n=kn=kαααα=akx0+x−1+x1+anxk−n+xk−n−1+xk−n+12222n=kThetermsunderthesummationistheISIintroducedbythechannel.Problem8.10a)Eachsegmentofthewire-linecanbeconsideredasabandpassfilterwithbandwidth课后答案网W=1200Hz.Thus,thehighestbitratethatcanbetransmittedwithoutISIbymeansofbinaryPAMisR=2W=2400bpsb)TheprobabilityoferrorforbinaryPAMtransmissioniswww.hackshp.cn12EbP2=QN0Hence,usingmathematicaltablesforthefunctionQ[·],wefindthatP2=10−7isobtainedfor12EbEb=5.2=⇒=13.52=11.30dBN0N0c)ThereceivedpowerPRisrelatedtothedesiredSNRperbitthroughtherelationPREb=RN0N0Hence,withN0=4.1×10−21weobtainP=4.1×10−21×1200×13.52=6.6518×10−17=−161.77dBWR221 SincethepowerlossofeachsegmentisLs=50Km×1dB/Km=50dBthetransmittedpowerateachrepeatershouldbePT=PR+Ls=−161.77+50=−111.77dBWProblem8.11Thepulsex(t)havingtheraisedcosinespectrumiscos(παt/T)x(t)=sinc(t/T)1−4α2t2/T2Thefunctionsinc(t/T)is1whent=0and0whent=nT.Ontheotherhandcos(παt/T)1t=0g(t)==1−4α2t2/T2boundedt=0Thefunctiong(t)needstobecheckedonlyforthosevaluesoftsuchthat4α2t2/T2=1orαt=T.2However,cos(πx)cos(παt/T)lim=lim2αt→T1−4α2t2/T2x→11−x2andbyusingL’Hospital’srulecos(πx)πππlim2=limsin(x)=<∞x→11−xx→1222Hence,1n=0x(nT)=课后答案网0n=0meaningthatthepulsex(t)satisfiestheNyquistcriterion.Problem8.12SubstitutingtheexpressionofXrcwww.hackshp.cn(f)inthedesiredintegral,weobtain
∞
−1−α
1−α2TTπT1−α2TXrc(f)df=1+cos(−f−)df+Tdf−∞−1+α2α2T−1−α2T2T
1+α2TTπT1−α+1+cos(f−)df1−α2α2T2T
−1−α
1+α2TT1−α2TT=df+T+df−1+α2T1−α22T2T
−1−α
1+α2TπT1−α2TπT1−α+cos(f+)df+cos(f−)df−1+αα2T1−αα2T2T2T
0
απTTπT=1+cosxdx+cosxdx−αα0αT
αTπT=1+cosxdx=1+0=1−ααT222 Problem8.13LetX(f)besuchthatTΠ(fT)+U(f)|f|<1V(f)|f|<1Re[X(f)]=TIm[X(f)]=T0otherwise0otherwisewithU(f)evenwithrespectto0andoddwithrespecttof=1Sincex(t)isreal,V(f)isodd2Twithrespectto0andbyassumptionitisevenwithrespecttof=1.Then,2Tx(t)=F−1[X(f)]
1
1
12Tj2πft2Tj2πftTj2πft=X(f)edf+X(f)edf+X(f)edf−1−11T2T2T
1
12Tj2πftTj2πft=Tedf+[U(f)+jV(f)]edf−1−12TT
1Tj2πft=sinc(t/T)+[U(f)+jV(f)]edf−1T1ConsiderfirsttheintegralT1U(f)ej2πftdf.Clearly,−T
1
0
1Tj2πftj2πftTj2πftU(f)edf=U(f)edf+U(f)edf−1−10TTandbyusingthechangeofvariablesf=f+1andf=f−1forthetwointegralsontheright2T2Thandsiderespectively,weobtain
1Tj2πftU(f)edf−1T
1
1−jπt2T课后答案网1j2πf
tjπt2T1j2πf
t=eTU(f−)edf+eTU(f+)edf−12T−12T2T2T
1ajπt−jπt2T1j2πf
t=eT−eTU(f+)edf−12T2T
1π2T1j2πf
t=2jsin(t)www.hackshp.cnU(f+)edfT−12T2Twhereforstep(a)weusedtheoddsymmetryofU(f)withrespecttof=1,thatis2T11U(f−)=−U(f+)2T2T1FortheintegralT1V(f)ej2πftdfwehave−T
1Tj2πftV(f)edf−1T
0
1j2πftTj2πft=V(f)edf+V(f)edf−10T
1
1−jπt2T1j2πf
tjπt2T1j2πf
t=eTV(f−)edf+eTV(f+)edf−12T−12T2T2T223 However,V(f)isoddwithrespectto0andsinceV(f+1)andV(f−1)areeven,thetranslated2T2Tspectrasatisfy
1
12T1j2πf
t2T1j2πf
tV(f−)edf=−V(f+)edf−12T−12T2T2THence,
1π2T1j2πf
tx(t)=sinc(t/T)+2jsin(t)U(f+)edfT−12T2T
1π2T1j2πf
t−2sin(t)U(f+)edfT−12T2Tandtherefore,1n=0x(nT)=0n=0Thus,thesignalx(t)satisfiestheNyquistcriterion.Problem8.14ThebandwidthofthechannelisW=3000−300=2700HzSincetheminimumtransmissionbandwidthrequiredforbandpasssignalingisR,whereRistherateoftransmission,weconcludethatthemaximumvalueofthesymbolrateforthegivenchannelisRmax=2700.IfanM-aryPAMmodulationisusedfortransmission,theninordertoachieveabit-rateof9600bps,withmaximumrateofRmax,theminimumsizeoftheconstellationisM=2k=16.Inthiscase,thesymbolrateis9600R==2400symbols/seckandthesymbolintervalT=1课后答案网=1sec.Theroll-offfactorαoftheraisedcosinepulseusedforR2400transmissionisisdeterminedbynotingthat1200(1+α)=1350,andhence,α=0.125.Therefore,thesquaredrootraisedcosinepulsecanhavearoll-offofα=0.125.Problem8.15www.hackshp.cnSincethebandwidthoftheideallowpasschannelisW=2400Hz,therateoftransmissionisR=2×2400=4800symbols/secThenumberofbitspersymbolis14400k==34800Hence,thenumberoftransmittedsymbolsis23=8.Ifaduobinarypulseisusedfortransmission,thenthenumberofpossibletransmittedsymbolsis2M−1=15.Thesesymbolshavetheformbn=0,±2d,±4d,...,±12dwhere2distheminimumdistancebetweenthepointsofthe8-PAMconstellation.Theprobabilitymassfunctionofthereceivedsymbolsis8−|m|P(b=2md)=,m=0,±1,...,±764224 Anupperboundoftheprobabilityoferrorisgivenby(see(8.4.33))121π6kEb,avPM<21−QM24M2−1N0WithPM=10−6andM=8weobtainkEb,av3=1.3193×10=⇒Eb,av=0.088N0Problem8.16a)Thespectrumofthebasebandsignalis1212SV(f)=Sa(f)|Xrc(f)|=|Xrc(f)|TTwhereT=1and2400T0≤|f|≤14TX(f)=T(1+cos(2πT(|f|−1))1≤|f|≤3rc24T4T4T0otherwiseIfthecarriersignalhastheformc(t)=Acos(2πfct),thenthespectrumoftheDSB-SCmodulatedsignal,SU(f),isASU(f)=[SV(f−fc)+SV(f+fc)]2AsketchofSU(f)isshowninthenextfigure.2AT课后答案网2-fc-3/4T-fc-fc+3/4Twww.hackshp.cnfc-3/4Tfcfc+3/4Tb)Assumingbandpasscoherentdemodulationusingamatchedfilter,thereceivedsignalr(t)isfirstpassedthroughalinearfilterwithimpulseresponsegR(t)=Axrc(T−t)cos(2πfc(T−t))Theoutputofthematchedfilterissampledatt=Tandthesamplesarepassedtothedetector.Thedetectorisasimplethresholddevicethatdecidesifabinary1or0wastransmitteddependingonthesignoftheinputsamples.Thefollowingfigureshowsablockdiagramoftheoptimumbandpasscoherentdemodulator.t=TBandpassDetectorr(t)matchedfilter.(ThresholdgR(t)device)225 Problem8.17a)IfthepowerspectraldensityoftheadditivenoiseisSn(f),thenthePSDofthenoiseattheoutputoftheprewhiteningfilterisS(f)=S(f)|H(f)|2νnpInorderforSν(f)tobeflat(whitenoise),Hp(f)shouldbesuchthat1Hp(f)=&Sn(f)2)Lethp(t)betheimpulseresponseoftheprewhiteningfilterHp(f).Thatis,hp(t)=F−1[Hp(f)].Then,theinputtothematchedfilteristhesignal˜s(t)=s(t)hp(t).Thefrequencyresponseofthefiltermatchedto˜s(t)isS˜(f)=S˜∗(f)e−j2πft0==S∗(f)H∗(f)e−j2πft0mpwheret0issomenominaltime-delayatwhichwesamplethefilteroutput.3)Thefrequencyresponseoftheoverallsystem,prewhiteningfilterfollowedbythematchedfilter,is∗2−j2πftS∗(f)0−j2πft0G(f)=S˜m(f)Hp(f)=S(f)|Hp(f)|e=eSn(f)4)Thevarianceofthenoiseattheoutputofthegeneralizedmatchedfilteris
∞
∞|S(f)|2σ2=S(f)|G(f)|2df=dfn−∞−∞Sn(f)Atthesamplinginstantt=t0=课后答案网T,thesignalcomponentattheoutputofthematchedfilteris
∞
∞y(T)=Y(f)ej2πfTdf=s(τ)g(T−τ)dτ−∞−∞
∞S∗(f)
∞|S(f)|2=S(f)df=dfwww.hackshp.cn−∞Sn(f)−∞Sn(f)Hence,theoutputSNRisy2(T)
∞|S(f)|2SNR==dfσ2−∞Sn(f)Problem8.18ThebandwidthofthebandpasschannelisW=3300−300=3000HzInordertotransmit9600bpswithasymbolrateR=2400symbolspersecond,thenumberofinformationbitspersymbolshouldbe9600k==42400Hence,a24=16QAMsignalconstellationisneeded.Thecarrierfrequencyfcissetto1800Hz,whichisthemid-frequencyofthefrequencybandthatthebandpasschanneloccupies.Ifapulse226 withraisedcosinespectrumandroll-offfactorαisusedforspectralshaping,thenforthebandpasssignalwithbandwidthWR=1200(1+α)=1500andα=0.25Asketchofthespectrumofthetransmittedsignalpulseisshowninthenextfigure.1/2Tf-3300-1800-300300600180033003000Problem8.19ThechannelbandwidthisW=4000Hz.(a)BinaryPSKwithapulseshapethathasα=1.Hence21(1+α)=20002Tand1=2667,thebitrateis2667bps.T(b)Four-phasePSKwithapulseshapethathasα=1.From(a)thesymbolrateis1=2667and2Tthebitrateis5334bps.(c)M=8QAMwithapulseshapethathasα=1.From(a),thesymbolrateis1=2667and2Thencethebitrate3=8001bps.T(d)BinaryFSKwithnoncoherentdetection.Assumingthatthefrequencyseparationbetweenthetwofrequenciesis∆f=1,where1isthebitrate,thetwofrequenciesaref+1andf−1.TTc2Tc2TSinceW=4000Hz,wemayselect1=1000,or,equivalently,1=2000.Hence,thebitrateis2TT2000bps,andthetwoFSKsignalsareorthogonal.课后答案网(e)FourFSKwithnoncoherentdetection.Inthiscaseweneedfourfrequencieswithseparationof1betweenadjacentfrequencies.Weselectf=f−1.5,f=f−1,f=f+1,andT1cT2c2T3c2Tf=f+1.5,where1=500Hz.Hence,thesymbolrateis1=1000symbolspersecondand4cT2TTsinceeachsymbolcarriestwobitsofinformation,thebitrateis2000bps.(f)M=8FSKwithnoncoherentdetection.Inthiscasewerequireeightfrequencieswithfrequencywww.hackshp.cnseparationof1=500Hzfororthogonality.Sinceeachsymbolcarries3bitsofinformation,theTbitrateis1500bps.Problem8.201)ThebandwidthofthebandpasschannelisW=3000−600=2400HzSinceeachsymboloftheQPSKconstellationconveys2bitsofinformation,thesymbolrateoftransmissionis2400R==1200symbols/sec2Thus,forspectralshapingwecanuseasignalpulsewitharaisedcosinespectrumandroll-offfactorα=1,thatisT12π|f|Xrc(f)=[1+cos(πT|f|)]=cos224002400227 IfthedesiredspectralcharacteristicissplitevenlybetweenthetransmittingfilterGT(f)andthereceivingfilterGR(f),then(1π|f|1GT(f)=GR(f)=cos,|f|<=120012002400TAblockdiagramofthetransmitterisshowninthenextfigure.anG(f)×toChannelTQPSKcos(2πfct)2)Ifthebitrateis4800bps,thenthesymbolrateis4800R==2400symbols/sec2InordertosatisfytheNyquistcriterion,thethesignalpulseusedforspectralshaping,shouldhavethespectrumfX(f)=TΠW√fThus,thefrequencyresponseofthetransmittingfilterisGT(f)=TΠW.Problem8.21ThebandwidthofthebandpasschannelisW=4KHz.Hence,therateoftransmissionshouldbelessorequalto4000symbols/sec.Ifa8-QAMconstellationisemployed,thentherequiredsymbolrateisR=9600/3=3200.Ifasignalpulsewithraisedcosinespectrumisusedforshaping,themaximumallowableroll-offfactorisdeterminedby1600(1+α)=2000whichyieldsα=0.25.Sinceαislessthan50%,weconsideralargerconstellation.Witha16-QAM课后答案网constellationweobtain9600R==24004andwww.hackshp.cn1200(1+α)=20000rα=2/3,whichsatisfiestherequiredconditions.TheprobabilityoferrorforanM-QAMconstellationisgivenbyP=1−(1−P√)2MMwhere113EavP√=21−√QMM(M−1)N0WithPM=10−6weobtainP√=5×10−7andthereforeM113Eav=5×10−72×(1−)Q415×2×10−10UsingthelastequationandthetabulationoftheQ[·]function,wefindthattheaveragetransmittedenergyisE=24.70×10−9av228 NotethatifthedesiredspectralcharacteristicXrc(f)issplitevenlybetweenthetransmittingandreceivingfilter,thentheenergyofthetransmittingpulseis
∞
∞
∞g2(t)dt=|G(f)|2df=X(f)df=1TTrc−∞−∞−∞Hence,theenergyEav=PavTdependsonlyontheamplitudeofthetransmittedpointsandthesymbolintervalT.SinceT=1,theaveragetransmittedpoweris2400Eav−9−7Pav==24.70×10×2400=592.8×10TIfthepointsofthe16-QAMconstellationareevenlyspacedwithminimumdistancebetweenthemequaltod,thentherearefourpointswithcoordinates(±d,±d),fourpointswithcoordinates22(±3d,±3d),fourpointswithcoordinates(±3d,±d),andfourpointswithcoordinates(±d,±3d).222222Thus,theaveragetransmittedpoweris1161d29d210d2P=(A2+A2)=4×+4×+8×=20d2avmcms2×162224i=1SincePav=592.8×10−7,weobtain1Pavd==0.0017220Problem8.22Theroll-offfactorαisrelatedtothebandwidthbytheexpression1+α=2W,orequivalentlyTR(1+α)=2W.ThefollowingtableshowsthesymbolrateforthevariousvaluesoftheexcessbandwidthandforW=1500Hz.α.25.33.50.67.751.00R240022562000179617141500Problem8.23课后答案网Thefollowingtableshowstheprecodedsequence,thetransmittedamplitudelevels,thereceivedsignallevelsandthedecodedsequence,whenthedatasequence10010110010modulatesaduobinarytransmittingfilter.Dataseq.dn:www.hackshp.cn10010110010Precodedseq.pn:011100100011Transmittedseq.an:-1111-1-11-1-1-111Receivedseq.bn:0220-200-2-202Decodedseq.dn:10010110010Problem8.24Thefollowingtableshowstheprecodedsequence,thetransmittedamplitudelevels,thereceivedsignallevelsandthedecodedsequence,whenthedatasequence10010110010modulatesamodifiedduobinarytransmittingfilter.Dataseq.dn:10010110010Precodedseq.pn:0010111000010Transmittedseq.an:-1-11-1111-1-1-1-11-1Receivedseq.bn:20020-2-20020Decodedseq.dn:10010110010229 Problem8.25LetX(z)denotetheZ-transformofthesequencexn,thatisX(z)=xz−nnnThentheprecodingoperationcanbedescribedasD(z)P(z)=mod−MX(z)whereD(z)andP(z)aretheZ-transformsofthedataandprecodedsequencesrespectively.Forexample,ifM=2andX(z)=1+z−1(duobinarysignaling),thenD(z)−1P(z)==⇒P(z)=D(z)−zP(z)1+z−1whichinthetimedomainiswrittenaspn=dn−pn−1andthesubtractionismod-2.However,theinversefilter1existsonlyifx,thefirstcoefficientofX(z)isrelativelyprimeX(z)0withM.Ifthisisnotthecase,thentheprecodedsymbolspncannotbedetermineduniquelyfromthedatasequencedn.Problem8.26Inthecaseofduobinarysignaling,theoutputofthematchedfilterisx(t)=sinc(2Wt)+sinc(2Wt−1)andthesamplesxn−maregivenby课后答案网1n−m=0xn−m=x(nT−mT)=1n−m=10otherwiseTherefore,themetricµ(a)intheViterbialgorithmbecomeswww.hackshp.cnµ(a)=2anrn−anamxn−mnnm=2ar−a2−aannnnn−1nnn=an(2rn−an−an−1)nProblem8.27Theprecodingfortheduobinarysignalingisgivenbypm=dmpm−1Thecorrespondingtrellishastwostatesassociatedwiththebinaryvaluesofthehistorypm−1.Forthemodifiedduobinarysignalingtheprecodingispm=dm⊕pm−2230 Hence,thecorrespondingtrellishasfourstatesdependingonthevaluesofthepair(pm−2,pm−1).Thetwotrellisesaredepictedinthenextfigure.Thebrancheshavebeenlabelledasx/y,wherexisthebinaryinputdatadmandyistheactualtransmittedsymbol.Notethatthetrellisforthemodifiedduobinarysignalhasmorestates,buttheminimumfreedistancebetweenthepathsisdfree=3,whereastheminimumfreedistancebetweenpathsfortheduobinarysignalis2.(pm−2,pm−1)ModifiedDuobinary000/-1Duobinarypm−11/10/-10011/1

1/1
10
1
0/-1
11

Problem8.281)Theoutputofthematchedfilterdemodulatoris∞
∞y(t)=akgT(τ−kTb)gR(t−τ)dτ+ν(t)−∞k=−∞∞=akx(t−kTb)+ν(t)k=−∞where,sinπtcosπtx(t)=g(t)g(t)=TTTRπtt2T1−4T2Hence,课后答案网∞y(mTb)=akx(mTb−kTb)+v(mTb)k=−∞11=am+am−1+am+1+ν(mTb)ππTheterm1a+1arepresentstheISIintroducedbydoublingthesymbolrateoftransmission.πm−1πm+1www.hackshp.cn2)InthenextfigureweshowonetrellisstagefortheMLsequencedetector.SincethereispostcursorISI,wedelaythereceivedsignal,usedbytheMLdecodertoformthemetrics,byonesample.Thus,thestatesofthetrelliscorrespondtothesequence(am−1,am),andthetransitionlabelscorrespondtothesymbolam+1.Twobranchesoriginatefromeachstate.Theupperbranchisassociatedwiththetransmissionof−1,whereasthelowerbranchisassociatedwiththetransmissionof1.am+1(am−1,am)-1-1-1-11-11-111-11-1111231 Problem8.29a)TheoutputofthematchedfilteratthetimeinstantmTis1ym=amxk−m+νm=am+am−1+νm4kTheautocorrelationfunctionofthenoisesamplesνmisN0E[νkνj]=xk−j2Thus,thevarianceofthenoiseis2N0N0σν=x0=22√Ifasymbolbysymboldetectorisemployedandweassumethatthesymbols√am=am−1=Ebhavebeentransmitted,thentheprobabilityoferrorP(e|am=am−1=Eb)is&&P(e|am=am−1=Eb)=P(ym<0|am=am−1=Eb)√
525&1−4Eb−νm=P(νm<−Eb)=√eN0dνm4πN0−∞%1
−52Eb214N0−ν52Eb=√e2dν=Q2π−∞4N0√Ifhoweveram−1=−Eb,then1&&3&32EbP(e|am=Eb,am−1=−Eb)=P(Eb+νm<0)=Q44N0√√SincethetwosymbolsEb,−Ebareusedwithequalprobability,weconcludethat&&P(e)=P(e|am=Eb)=P(e|am=−Eb)课后答案网11152Eb132Eb=Q+Q24N024N0b)Inthenextfigureweplottheerrorprobabilityobtainedinpart(a)(logwww.hackshp.cn10(P(e)))vs.theSNRperbitandtheerrorprobabilityforthecaseofnoISI.Asitobservedfromthefigure,therelativedifferenceinSNRoftheerrorprobabilityof10−6is2dB.-2-2.5-3-3.5-4-4.5log(P(e)-5-5.5-6-6.5-767891011121314SNR/bit,dB232 Problem8.30Thepowerspectraldensityofthenoiseattheoutputofthematchedfilteris2N0N01πfSν(f)=Sn(f)|GR(f)|=|X(f)|=cos()22W2WHence,theautocorrelationfunctionoftheoutputnoiseis
∞−1N01πfj2πfτRν(τ)=F[Sν(f)]=cos()edf2−∞W2W
∞N01πf−jπfj2πf(τ+1)=cos()e2We4Wdf2−∞W2W
∞N0j2πf(τ+1)=X(f)e4Wdf2−∞N01=x(τ+)24Wandtherefore,N01N0112N0Rν(0)=x()=sinc()+sinc(−)=24W222π1N0312N0Rν(T)=Rν()=sinc()+sinc()=2W2223πSincethenoiseisofzeromean,thecovariancematrixofthenoiseisgivenbyRν(0)Rν(T)2N011C==13Rν(T)Rν(0)π31Problem8.31LetSirepresentthestatethatthedifferencebetweenthetotalnumberofaccumulatedzerosand课后答案网thetotalnumberofaccumulatedonesisi,withi=−2,...,2.Thestatetransitiondiagramofthecorrespondingcodeisdepictedinthenextfigure.1www.hackshp.cn111 S2 S1 S0 S−1 S−20000Thestatetransitionmatrixis0100010100D=010100010100010Settingdet(D−λI)=0,weobtainλ5−4λ3+3λ=0.Therootsofthecharacteristicequationare√λ=0,±1,±3Thus,√C=log2λmax=log23=.7925233 Problem8.32Thestatetransitionmatrixofthe(0,1)runlength-limitedcodeis11D=10TheeigenvaluesofDaretherootsofdet(D−λI)=−λ(1−λ)−1=λ2−λ−1Therootsofthecharacteristicequationare√1±5λ1,2=2Thus,thecapacityofthe(0,1)runlength-limitedcodeis√1±5C(0,1)=log2()=0.69422Thecapacityofa(1,∞)codeisfoundfromTable8.3tobe0.6942.Asitisobserved,thetwocodeshaveexactlythesamecapacity.Thisresultistobeexpectedsincethe(0,1)runlength-limitedcodeandthe(1,∞)codeproducethesamesetofcodesequencesoflengthn,N(n),witharenamingofthebitsfrom0to1andviseversa.Forexample,the(0,1)runlength-limitedcodewitharenamingofthebits,canbedescribedasthecodewithnominimumnumberof1’sbetween0’sinasequence,andatmostone1betweentwo0’s.Intermsof0’s,thisissimplythecodewithnorestrictionsonthenumberofadjacent0’sandnoconsecutive1’s,thatisthe(1,∞)code.Problem8.33LetS0representthestatethattherunningpolarityiszero,andS1thestatethatthereexistssomepolarity(dccomponent).ThefollowingfiguredepictsthetransitionstatediagramoftheAMIcode课后答案网1/s(t)0/0S0S10/0www.hackshp.cn1/−s(t)Thestatetransitionmatrixis11D=11TheeigenvaluesofthematrixDcanbefoundfromdet(D−λI)=0=⇒(1−λ)2−1=0orλ(2−λ)=0Thelargestrealeigenvalueisλmax=2,sothatC=log2λmax=1234 Problem8.34Let{bk}beabinarysequence,takingthevalues1,0dependingontheexistenceofpolarizationatthetransmittedsequenceuptothetimeinstantk.FortheAMIcode,bkisexpressedasbk=ak⊕bk−1=ak⊕ak−1⊕ak−2⊕...where⊕denotesmodulotwoaddition.Thus,theAMIcodecanbedescribedastheRDScode,withRDS(=bk)denotingthebinarydigitalsummodulo2oftheinputbits.Problem8.35Definingtheefficiencyaskefficiency=nlog23weobtainCodeEfficiency1B1T0.6333B2T0.9494B3T0.8446B4T0.949Problem8.36a)ThecharacteristicpolynomialofDis1−λ12det(D−λI)=det=λ−λ−11−λTheeigenvaluesofDaretherootsofthecharacteristicpolynomial,thatis课后答案网√1±5λ1,2=2√Thus,thelargesteigenvalueofDisλ=1+5andthereforemax2www.hackshp.cn√1+5C=log2=0.69422b)Thecharacteristicpolynomialisdet(D−λI)=(1−λ)2withrootsλ1,2=1.Hence,C=log21=0.Thestatediagramofthiscodeisdepictedinthenextfigure.10S0S11c)Asitisobservedthesecondcodehaszerocapacity.Thisresultistobeexpectedsincewiththesecondcodewecanhaveatmostn+1differentsequencesoflengthn,sothat11C=limlog2N(n)=limlog2(n+1)=0n→∞nn→∞n235 Then+1possiblesequencesare0...01...1(nsequences)34563456kn−kandthesequence11...1,whichoccursifwestartfromstateS1.Problem8.37a)Thetwosymbols,dotanddash,canberepresentedas10and1110respectively,where1denoteslineclosureand0anopenline.Hence,theconstraintsofthecodeare•A0isalwaysfollowedby1.•Onlysequenceshavingoneorthreerepetitionsof1,areallowed.Thenextfiguredepictsthestatediagramofthecode,wherethestateS0denotesthereceptionofadotoradash,andstateSidenotesthereceptionofiadjacent1’s.11S1S2S3001S0b)Thestatetransitionmatrixis01001010D=00011000c)ThecharacteristicequationofthematrixDisdet(课后答案网D−λI)=0=⇒λ4−λ2−1=0Therootsofthecharacteristicequationare√1√11+521−52λ1,2=±λ3,4=±www.hackshp.cn22Thus,thecapacityofthecodeis√11+52C=log2λmax=log2λ1=log2=0.34712Problem8.38ThestatediagramofFig.P-8-38describesarunlengthconstrainedcode,thatforbidsanysequencecontainingarunofmorethanthreeadjacentsymbolsofthesamekind.Thestatetransitionmatrixis000100100100010100D=001010001001001000236 ThecorrespondingtrellisisshowninthenextfigureHistory111!!!11!!!1!!!...0""!""!""!00"""000"""Problem8.39Thestatetransitionmatrixofthe(2,7)runlength-limitedcodeisthe8×8matrix01000000001000001001000010001000D=10000100100000101000000110000000Problem8.40ThefrequencyresponseoftheRCfilteris1j2πRCf1C(f)==R+11+j2πRCfj2πRCfTheamplitudeandthephasespectrumofthefilterare1课后答案网12|C(f)|=,Θc(f)=arctan(−2πRCf)1+4π2(RC)2f2Theenvelopedelayis1dΘc(f)1−2πRCRCTc(f)=−=−=2πdfwww.hackshp.cn2π1+4π2(RC)2f21+4π2(RC)2f2wherewehaveusedtheformulad1duarctanu=dx1+u2dxProblem8.411)TheenvelopedelayoftheRCfilteris(seeProblem8.40)RCTc(f)=1+4π2(RC)2f2AplotofT(f)withRC=10−6isshowninthenextfigure237 x10-7109.9999.9989.9979.9969.995Tc(f)9.9949.9939.9929.9919.990500100015002000250030003500400045005000Frequency(f)2)ThefollowingfigureisaplotoftheamplitudecharacteristicsoftheRCfilter,|C(f)|.Thevaluesoftheverticalaxisindicatethat|C(f)|canbeconsideredconstantforfrequenciesupto2000Hz.Sincethesameistruefortheenvelopedelay,weconcludethatalowpasssignalofbandwidth∆f=1KHzwillnotbedistortedifitpassestheRCfilter.1|C(f)|0.9990500100015002000250030003500400045005000课后答案网Frequency(f)Problem8.42LetGT(f)andGR(f)bethefrequencyresponseofthetransmittingandreceivingfilter.Then,theconditionforzeroISIimplieswww.hackshp.cnT0≤|f|≤14TG(f)C(f)G(f)=X(f)=T[1+cos(2πT(|f|−1)]1≤|f|≤3TRrc2T4T4T0|f|>34TSincetheadditivenoiseiswhite,theoptimumtransmittingandreceivingfiltercharacteristicsaregivenby(seeExample8.6.1)11|Xrc(f)|2|Xrc(f)|2|GT(f)|=1,|GR(f)|=1|C(f)|2|C(f)|2Thus,1T211+0.3cos2πfT0≤|f|≤4T1|G(f)|=|G(f)|=T(1+cos(2πT(|f|−1)2TRT1≤|f|≤32(1+0.3cos2πfT)4T4T0otherwise238 Problem8.43A4-PAMmodulationcanaccommodatek=2bitspertransmittedsymbol.Thus,thesymbolintervaldurationisk1T==sec96004800Since,thechannel’sbandwidthisW=2400=1,inordertoachievethemaximumrateof2Ttransmission,R=1,thespectrumofthesignalpulseshouldbemax2TfX(f)=TΠ2WThen,themagnitudefrequencyresponseoftheoptimumtransmittingandreceivingfilteris(seeSection8.6.1andExample8.6.1)1214f24ff1+,|f|<2400|GT(f)|=|GR(f)|=1+Π=240024002W0otherwiseProblem8.441)Theequivalentdiscrete-timeimpulseresponseofthechannelis1h(t)=hnδ(t−nT)=0.3δ(t+T)+0.9δ(t)+0.3δ(t−T)n=−1Ifby{cn}wedenotethecoefficientsoftheFIRequalizer,thentheequalizedsignalis1qm=cnhm−nn=−1whichinmatrixnotationiswrittenas课后答案网0.90.30.c−100.30.90.3c0=10.0.30.9c10Thecoefficientsofthezero-forceequalizercanbefoundbysolvingthepreviousmatrixequation.Thus,www.hackshp.cnc−1−0.4762c0=1.4286c1−0.47622)Thevaluesofqmform=±2,±3aregivenby1q2=cnh2−n=c1h1=−0.1429n=−11q−2=cnh−2−n=c−1h−1=−0.1429n=−11q3=cnh3−n=0n=−11q−3=cnh−3−n=0n=−1239 Problem8.451)Theoutputofthezero-forceequalizeris1qm=cnxmnn=−1Withq0=1andqm=0form=0,weobtainthesystem1.00.1−0.5c−10−0.21.00.1c0=10.05−0.21.0c10Solvingtheprevioussystemintermsoftheequalizer’scoefficients,weobtainc−10.000c0=0.980c10.1962)Theoutputoftheequalizeris0m≤−4c−1x−2=0m=−3c−1x−1+c0x−2=−0.49m=−20m=−1qm=1m=00m=1c0x2+x1c1=0.0098m=2c1x2=0.0098m=30m≥4Hence,theresidualISIsequenceis课后答案网residualISI={...,0,−0.49,0,0,0,0.0098,0.0098,0,...}anditsspanis6symbols.www.hackshp.cnProblem8.46TheMSEperformanceindexatthetimeinstantkisNJ(ck)=Eck,nyk−n−ak62n=−NIfwedefinethegradientvectorgkasϑJ(ck)gk=2ϑckthenitslthelementisNϑJ(ck)1gk,l==E2ck,nyk−n−akyk−l2ϑck,l2n=−N=E[−ekyk−l]=−E[ekyk−l]240 Thus,thevectorgkis−E[ekyk+N]g=..k.=−E[ekyk]−E[ekyk−N]whereykisthevectoryk=[yk+N···yk−N]T.Sincegˆk=−ekyk,itsexpectedvalueisE[gˆk]=E[−ekyk]=−E[ekyk]=gkProblem8.471)If{cn}denotethecoefficientsofthezero-forceequalizerand{qm}isthesequenceoftheequal-izer’soutputsamples,then1qm=cnxm−nn=−1where{xk}isthenoisefreeresponseofthematchedfilterdemodulatorsampledatt=kT.Withq−1=0,q0=q1=Eb,weobtainthesystemEb0.9Eb0.1Ebc−100.9EbEb0.9Ebc0=Eb0.1Eb0.9EbEbc1EbThesolutiontothesystemisc−1c0c1=0.2137−0.38461.32482)Thesetofnoisevariables{νk}attheoutputofthesamplerisaGaussiandistributedsequencewithzero-meanandautocorrelationfunctionN0x|k|≤2R(k)=2kν课后答案网0otherwiseThus,theautocorrelationfunctionofthenoiseattheoutputoftheequalizerisRn(k)=Rν(k)c(k)c(−k)wherec(k)denotesthediscretetimeimpulseresponseoftheequalizer.Therefore,theautocorrela-www.hackshp.cntionsequenceofthenoiseattheoutputoftheequalizeris0.9402k=01.3577k=±1N0Eb−0.0546k=±2Rn(k)=20.1956k=±30.0283k=±40otherwiseTofindanestimateoftheerrorprobabilityforthesequencedetector,weignoretheresidualinterferenceduetothefinitelengthoftheequalizer,andweonlyconsiderpathsoflengthtwo.Thus,ifwestartatstatea0=1andthetransmittedsymbolsare(a1,a2)=(1,1)anerrorismadebythesequencedetectorifthepath(−1,1)ismoreprobable,giventhereceivedvaluesofr1andr2.Themetricforthepath(a1,a2)=(1,1)is−1r1−2Ebµ2(1,1)=[r1−2Ebr2−2Eb]Cr2−2Eb241 whereN0Eb0.94021.3577C=21.35770.9402Similarly,themetricofthepath(a1,a2)=(−1,1)is−1r1µ2(−1,1)=[r1r2]Cr2Hence,theprobabilityoferrorisP2=P(µ2(−1,1)<µ2(1,1))anduponsubstitutionofr1=2Eb+n1,r2=2Eb+n2,weobtainP2=P(n1+n2<−2Eb)Sincen1andn2arezero-meanGaussianvariables,theirsumisalsozero-meanGaussianwithvarianceN0EbN0Ebσ2=(2×0.9402+2×1.3577)=4.595822andtherefore18EbP2=Q4.5958N0ThebiterrorprobabilityisP2.2Problem8.48Theoptimumtapcoefficientsofthezero-forceequalizercanbefoundbysolvingthesystem1.00.30.0c−100.21.00.3c0=1课后答案网0.00.21.0c10Hence,c−1−0.3409c0=1.1364www.hackshp.cnc1−0.2273b)Theoutputoftheequalizeris0m≤−3c−1x−1=−0.1023m=−20m=−1qm=1m=00m=1c1x1=−0.0455m=20m≥3Hence,theresidualISIsequenceisresidualISI={...,0,−0.1023,0,0,0,−0.0455,0,...}242 Problem8.491)IfweassumethatthesignalpulsehasdurationT,thentheoutputofthematchedfilteratthetimeinstantt=Tis
Ty(T)=r(τ)s(τ)dτ0
T=(s(τ)+αs(τ−T)+n(τ))s(τ)dτ0
T
T=s2(τ)dτ+n(τ)s(τ)dτ00=Es+nwhereEsistheenergyofthesignalpulseandnisazero-meanGaussianrandomvariablewithvarianceσ2=N0Es.Similarly,theoutputofthematchedfilteratt=2Tisn2
T
Ty(2T)=αs2(τ)dτ+n(τ)s(τ)dτ00=αEs+n2)Ifthetransmittedsequenceis∞x(t)=ans(t−nT)n=−∞withantakingthevalues1,−1withequalprobability,thentheoutputofthedemodulatoratthetimeinstantt=kTisyk=akEs+αak−1Es+nkThetermαak−1EsexpressestheISIduetothesignalreflection.IfasymbolbysymboldetectorisemployedandtheISIisignored,thentheprobabilityoferroris课后答案网11P(e)=P(error|an=1,an−1=1)+P(error|an=1,an−1=−1)2211=P((1+α)Es+nk<0)+P((1−α)Es+nk<0)221112(1+www.hackshp.cnα)2Es12(1−α)2Es=Q+Q2N02N03)TofindtheerrorrateperformanceoftheDFE,weassumethattheestimationoftheparameterαiscorrectandthattheprobabilityoferrorateachtimeinstantisthesame.Sincethetransmittedsymbolsareequiprobable,weobtainP(e)=P(erroratk|ak=1)=P(erroratk−1)P(erroratk|ak=1,erroratk−1)+P(noerroratk−1)P(erroratk|ak=1,noerroratk−1)=P(e)P(erroratk|ak=1,erroratk−1)+(1−P(e))P(erroratk|ak=1,noerroratk−1)=P(e)p+(1−P(e))q243 wherep=P(erroratk|ak=1,erroratk−1)1=P(erroratk|ak=1,ak−1=1,erroratk−1)21+P(erroratk|ak=1,ak−1=−1,erroratk−1)211=P((1+2α)Es+nk<0)+P((1−2α)Es+nk<0)221112(1+2α)2Es12(1−2α)2Es=Q+Q2N02N0andq=P(erroratk|ak=1,noerroratk−1)12Es=P(Es+nk<0)=QN0SolvingforP(e),weobtain%Q2EsqN0P(e)==%%%1−p+q12(1+2α)2Es12(1−2α)2Es2Es1−Q−Q+Q2N02N0N0Asketchofthedetectorstructureisshowninthenextfigure.InputrkThreshold++Outputˆakdevice−Estimate×Delayα课后答案网Problem8.50AdiscretetimetransversalfilterequivalenttothecascadeofthetransmittingfiltergT(t),thechannelc(t),thematchedfilteratthereceiverwww.hackshp.cngR(t)andthesampler,hastapgaincoefficients{ym},where0.9m=0ym=0.3m=±10otherwiseThenoiseνk,attheoutputofthesampler,isazero-meanGaussiansequencewithautocorrelationfunctionE[νν]=σ2y,|k−l|≤1klk−lIftheZ-transformofthesequence{ym},Y(z),assumesthefactorizationY(z)=F(z)F∗(z−1)thenthefilter1/F∗(z−1)canfollowthesamplertowhitethenoisesequenceνk.Inthiscasetheoutputofthewhiteningfilter,andinputtotheMSEequalizer,isthesequenceun=akfn−k+nkk244 wherenkiszeromeanGaussianwithvarianceσ2.TheoptimumcoefficientsoftheMSEequalizer,ck,satisfy(see(8.6.35))1cnRu(n−k)=Rua(k),k=0,±1n=−1where1R(n−k)=E[uu]=ff+σ2δul−kl−nmm+n−kn,km=0yn−k+σ2δn,k,|n−k|≤1=0otherwisef−k,−1≤k≤0Rua(k)=E[anun−k]=0otherwiseWithY(z)=0.3z+0.9+0.3z−1=(f+fz−1)(f+fz)0101weobtaintheparametersf0andf1as√√±0.7854±0.1146f0=√,f1=√±0.1146±0.7854Theparametersf0andf1shouldhavethesamesignsincef0f1=0.3.However,thesignitselfdoesnotplayanyroleifthedataaredifferentiallyencoded.Tohaveastableinversesystem1/F∗(z−1),weselectf0andf1insuchawaythatthezeroofthesystemF∗(z−1)=f0+f1zisinsidetheunit√√circle.Thus,wechoosef0=0.1146andf1=0.7854andtherefore,thedesiredsystemfortheequalizer’scoefficientsis√0.9+0.10.30.0c−1√0.78540.30.9+0.10.3c0=0.11460.00课后答案网.30.9+0.1c10Solvingthissystem,weobtainc−1=0.8596,c0=0.0886,c1=−0.0266www.hackshp.cnProblem8.511)Thespectrumofthebandlimitedequalizedpulseis1∞n−jπnfn=−∞x()eW|f|≤WX(f)=2W2W0otherwise12+2cosπf|f|≤W=2WW0otherwise11+1cosπf|f|≤W=WW0otherwisewhereW=12Tb245 2)Thefollowingtableliststhepossibletransmittedsequencesoflength3andthecorrespondingoutputofthedetector.-1-1-1-4-1-11-2-11-10-11121-1-1-21-11011-121114Asitisobservedthereare5possibleoutputlevelsb,withprobabilityp(b=0)=1,mm4p(b=±2)=1andp(b=±4)=1.m4m83)ThetransmittingfilterGT(f),thereceivingfilterGR(f)andtheequalizerGE(f)satisfytheconditionGT(f)GR(f)GE(f)=X(f)ThepowerspectraldensityofthenoiseattheoutputoftheequalizerisS(f)=S(f)|G(f)G(f)|2=σ2|G(f)G(f)|2νnREREWithπT50−πT50|f|GT(f)=GR(f)=P(f)=e2thevarianceoftheoutputnoiseis
∞
∞X(f)2σ2=σ2|G(f)G(f)|2df=σ2dfνRE−∞−∞GT(f)
W|1+cosπf|224W=σdf−Wπ2T2W2e−2πT50|f|50课后答案网8σ2
Wπf2=1+cose2πT50fdfπ2T502W20WThevalueofthepreviousintegralcanbefoundusingtheformula
eaxcosnbxdxwww.hackshp.cn
1an−12axn−2=(acosbx+nbsinbx)excosbx+n(n−1)becosbxdxa2+n2b2Thus,weobtain22πT+π128σ2πT50W150W2T50σ=×e−1+νπ2T2W22πT4π2T2+4π2505050W24πT50−e2πT50W+14π2T2+π250W2Tofindtheprobabilityoferrorusingasymbolbysymboldetector,wefollowthesameprocedureasinSection8.4.3.Theresultsarethesamewiththatobtainedfroma3-pointPAMconstellation(0,±2)usedwithaduobinarysignalwithoutputlevelshavingtheprobabilitymassfunctiongiveninpartb).Anupperboundofthesymbolprobabilityoferroris246 P(e)1|bm=0)P(bm=0)+2P(|ym−2|>1|bm=2)P(bm=2)+2P(ym+4>1|bm=−4)P(bm=−4)=P(|ym|>1|bm=0)[P(bm=0)+2P(bm=2)+P(bm=−4)]7=P(|ym|>1|bm=0)8But
2∞22P(|y|>1|b=0)=√e−x/2σνdxmm2πσν1Therefore,141P(e)=Q8σνProblem8.52Sincethepartialresponsesignalhasmemorylengthequalto2,thecorrespondingtrellishas4stateswhichwelabelas(an−1,an).Thefollowingfigureshowsthreeframesofthetrellis.Thelabelsofthebranchesindicatetheoutputofthepartialresponsesystem.Asitisobservedthefreedistancebetweenmergingpathsis3,whereastheEuclideandistanceisequaltod=22+42+22=24E(an−1,an)(-1,-1)-4-4-4-2(-1,1)-2 #$0(1,-1) # #$(1,1)Problem8.53课后答案网a)Thealternativeexpressionfors(t)canberewrittenass(t)=ReaQ(t−nT)nnj2πfwww.hackshp.cncnT=Reaneg(t−nT)[cos2πfc(t−nT)+jsin2πfc(t−nT)]n=Reang(t−nT)[cos2πfcnT+jsin2πfcnT][cos2πfc(t−nT)+jsin2πfc(t−nT)]n=Reang(t−nT)[cos2πfcnTcos2πfc(t−nT)−sin2πfcnTsin2πfc(t−nT)n+jsin2πfcnTcos2πfc(t−nT)+jcos2πfcnTsin2πfc(t−nT)]=Reang(t−nT)[cos2πfct+jsin2πfct]n=Reag(t−nT)ej2πfctnn=s(t)soindeedthealternativeexpressionfors(t)isavalidone.247 b)j2πfnTeq(t)q(t)e-j2πfnTanra"nr++aa"ni-+niq^(t)q^(t)ModulatorDemodulator(withphaserotator)(withphasederotator)Problem8.54a)Theimpulseresponseofthepulsehavingasquare-rootraisedcosinecharacteristic,isanevenfunction,i.e.,xSQ(t)=xSQ(−t),i.e.,thepulseg(t)isanevenfunction.Weknowthattheproductofanevenfunctiontimesanevenfunctionisanevenfunction,whiletheproductofanevenfunctiontimesanoddfunctionisanoddfunction.Henceq(t)isevenwhileˆq(t)isoddandtheirproductq(t)ˆq(t)hasoddsymmetry.Therefore,
∞
(1+β)/2Tq(t)ˆq(t)dt=q(t)ˆq(t)dt=0−∞−(1+β)/2Tb)Wenoticethatwhenfc=k/T,wherekisaninteger,thentherotator/derotatorofacarrierlessQAMsystem(describedinProblem8.53)givesatrivialrotationofanintegernumberoffullcircles(2πkn),andthecarrierlessQAM/PSKisequivalenttoCAP.Problem8.55课后答案网TheanalogsignalisN−11j2πkt/Tx(t)=√Xke,0≤tσ2SinceC>0,weobtain1σ2σ2log2≤1−h(
)=⇒≤D2D22(1−h())andtherefore,theminimumvalueofthedistortionattainableattheoutputofthechannelisσ2Dmin=22(1−h())2)ThecapacityoftheadditiveGaussianchannelis1PC=log21+22σnHence,1σ2σ2σ22log2D≤C=⇒22C≤D=⇒P≤D1+σ2nTheminimumattainabledistortionisσ2Dmin=1+Pσ2课后答案网n3)Herethesourcesamplesaredependentandthereforeonesampleprovidesinformationabouttheothersamples.Thismeansthatwecanachievebetterresultscomparedtothememorylesscaseatagivenrate.Inotherwordsthedistortionatagivenrateforasourcewithmemoryislessthanthedistortionforacomparablesourcewithmemory.DifferentialcodingmethodsdiscussedinChapter4aresuitableforsuchsources.www.hackshp.cnProblem9.14ThecapacityofthechannelofthechannelisgivenbyC=maxI(X;Y)=max[H(Y)−H(Y|X)]p(x)p(x)LettheprobabilityoftheinputsC,BandAbep,qand1−p−qrespectively.FromthesymmetryofthenodesB,Cweexpectthattheoptimumdistributionp(x)willsatisfyp(B)=p(C)=p.TheentropyH(Y|X)isgivenbyH(Y|X)=p(x)H(Y|X=x)=(1−2p)H(Y|X=A)+2pH(Y|X=B)=0+2ph(0.5)=2pTheprobabilitymassfunctionoftheoutputisp(Y=1)=p(x)p(Y=1|X=x)=(1−2p)+p=1−pp(Y=2)=p(x)p(Y=2|X=x)=0.5p+0.5p=p257 Therefore,C=max[H(Y)−H(Y|X)]=max(h(p)−2p)ppTofindtheoptimumvalueofpthatmaximizesI(X;Y),wesetthederivativeofCwithrespecttopequaltozero.Thus,ϑC1−1=0=−log2(p)−p+log2(1−p)−(1−p)−2ϑppln(2)(1−p)ln(2)=log2(1−p)−log2(p)−2andtherefore1−p1−p1log2=2=⇒=4=⇒p=pp5Thecapacityofthechannelis12C=h()−=0.7219−0.4=0.3219bits/transmission55Problem9.15Thecapacityofthe“product”channelisgivenbyC=maxI(X1X2;Y1Y2)p(x1,x2)However,I(X1X2;Y1Y2)=H(Y1Y2)−H(Y1Y2|X1X2)=H(Y1Y2)−H(Y1|X1)−H(Y2|X2)≤H(Y1)+H(Y2)−H(Y1|X1)−H(Y2|X2)=I(X1;Y1)+I(X2;Y2)andtherefore,C=maxI(课后答案网X1X2;Y1Y2)≤max[I(X1;Y1)+I(X2;Y2)]p(x1,x2)p(x1,x2)≤maxI(X1;Y1)+maxI(X2;Y2)p(x1)p(x2)=C1+C2Theupperboundisachievablebychoosingtheinputjointprobabilitydensitywww.hackshp.cnp(x1,x2),insuchawaythatp(x1,x2)=˜p(x1)˜p(x2)where˜p(x1),˜p(x2)aretheinputdistributionsthatachievethecapacityofthefirstandsecondchannelrespectively.Problem9.161)LetX=X1+X2,Y=Y1+Y2andp(y1|x1)ifx∈X1p(y|x)=p(y2|x2)ifx∈X2theconditionalprobabilitydensityfunctionofYandX.WedefineanewrandomvariableMtakingthevalues1,2dependingontheindexiofX.NotethatMisafunctionofXorY.This258 isbecauseX1∩X2=∅andtherefore,knowingXweknowthechannelusedfortransmission.ThecapacityofthesumchannelisC=maxI(X;Y)=max[H(Y)−H(Y|X)]=max[H(Y)−H(Y|X,M)]p(x)p(x)p(x)=max[H(Y)−p(M=1)H(Y|X,M=1)−p(M=2)H(Y|X,M=2)]p(x)=max[H(Y)−λH(Y1|X1)−(1−λ)H(Y2|X2)]p(x)whereλ=p(M=1).Also,H(Y)=H(Y,M)=H(M)+H(Y|M)=H(λ)+λH(Y1)+(1−λ)H(Y2)SubstitutingH(Y)inthepreviousexpressionforthechannelcapacity,weobtainC=maxI(X;Y)p(x)=max[H(λ)+λH(Y1)+(1−λ)H(Y2)−λH(Y1|X1)−(1−λ)H(Y2|X2)]p(x)=max[H(λ)+λI(X1;Y1)+(1−λ)I(X2;Y2)]p(x)Sincep(x)isfunctionofλ,p(x1)andp(x2),themaximizationoverp(x)canbesubstitutedbyajointmaximizationoverλ,p(x1)andp(x2).Furthermore,sinceλand1−λarenonnegative,weletp(x1)tomaximizeI(X1;Y1)andp(x2)tomaximizeI(X2;Y2).Thus,C=max[H(λ)+λC1+(1−λ)C2]λTofindthevalueofλthatmaximizesC,wesetthederivativeofCwithrespecttoλequaltozero.Hence,dC2C1dλ=0=−log2(λ)+log2(1−λ)+C1−C2=⇒λ=2C1+2C2Substitutingthisvalueofλintheexpressionfor课后答案网C,weobtain2C12C12C1C=H2C1+2C2+2C1+2C2C1+1−2C1+2C2C22C12C12C12C1=−2C1+2C2logwww.hackshp.cn22C1+2C2−1−2C1+2C2log22C1+2C22C12C1+2C1+2C2C1+1−2C1+2C2C22C12C2=log(2C1+2C2)+log(2C1+2C2)2C1+2C222C1+2C22=log(2C1+2C2)2HenceC=log(2C1+2C2)=⇒2C=2C1+2C222)2C=20+20=2=⇒C=1Thus,thecapacityofthesumchannelisnonzeroalthoughthecomponentchannelshavezerocapacity.Inthiscasetheinformationistransmittedthroughtheprocessofselectingachannel.259 3)Thechannelcanbeconsideredasthesumoftwochannels.ThefirstchannelhascapacityC1=log21=0andthesecondchannelisBSCwithcapacityC2=1−h(0.5)=0.ThusC=log(2C1+2C2)=log(2)=122Problem9.171)TheentropyofthesourceisH(X)=h(0.3)=0.8813andthecapacityofthechannelC=1−h(0.1)=1−0.469=0.531Ifthesourceisdirectlyconnectedtothechannel,thentheprobabilityoferroratthedestinationisP(error)=p(X=0)p(Y=1|X=0)+p(X=1)p(Y=0|X=1)=0.3×0.1+0.7×0.1=0.12)SinceH(X)>C,somedistortionattheoutputofthechannelisinevitable.TofindtheminimumdistortionwesetR(D)=C.ForaBernoullitypeofsourceh(p)−h(D)0≤D≤min(p,1−p)R(D)=0otherwiseandtherefore,R(D)=h(p)−h(D)=h(0.3)−h(D).IfweletR(D)=C=0.531,weobtainh(D)=0.3503=⇒D=min(0.07,0.93)=0.07TheprobabilityoferrorisP(error)≤D=0.073)Forreliabletransmissionwemusthave课后答案网H(X)=C=1−h(
).Hence,withH(X)=0.8813weobtain0.8813=1−h(
)=⇒
<0.016or
>0.984Problem9.181)Therate-distortionfunctionoftheGaussiansourceforwww.hackshp.cnD≤σ2is1σ2R(D)=log22DHence,withσ2=4andD=1,weobtain1R(D)=log24=1bits/sample=8000bits/sec2ThecapacityofthechannelisPC=Wlog21+N0WInordertoaccommodatetherateR=8000bps,thechannelcapacityshouldsatisfyR(D)≤C=⇒R(D)≤4000log2(1+SNR)Therefore,log2(1+SNR)≥2=⇒SNRmin=3260 2)Theerrorprobabilityforeachbitis12Ebpb=QN0andtherefore,thecapacityoftheBSCchannelis12EbC=1−h(pb)=1−hQbits/transmissionN012Eb=2×4000×1−hQbits/secN0Inthiscase,theconditionR(D)≤Cresultsin12EbEb1≤1−h(pb)=⇒Q=0orSNR=→∞N0N0Problem9.191)Themaximumdistortioninthecompressionofthesourceis
∞
10D=σ2=S(f)df=2df=40maxx−∞−102)Therate-distortionfunctionofthesourceis1logσ20≤D≤σ21log400≤D≤40R(D)=22D=22D0otherwise0otherwise3)WithD=10,weobtain课后答案网1401R=log2=log24=12102Thus,therequiredrateisR=1bitpersampleor,sincethesourcecanbesampledatarateof20samplespersecond,therateisR=20bitspersecond.4)Thecapacity-costfunctioniswww.hackshp.cn1PC(P)=log21+2Nwhere,
∞
4N=Sn(f)df=df=8−∞−4Hence,1PPC(P)=log2(1+)bits/transmission=4log2(1+)bits/sec288Therequiredpowersuchthatthesourcecanbetransmittedviathechannelwithadistortionnotexceeding10,isdeterminedbyR(10)≤C(P).Hence,P20≤4log2(1+)=⇒P=8×31=2488261 Problem9.20ThedifferentialentropyoftheLaplaciannoiseis(seeProblem6.36)h(Z)=1+lnλwhereλisthemeanoftheLaplaciandistribution,thatis
∞
∞1−zE[Z]=zp(z)dz=zeλdz=λ00λThevarianceofthenoiseis
∞22221−z2222N=E[(Z−λ)]=E[Z]−λ=zeλdz−λ=2λ−λ=λ0λInthenextfigureweplotthelowerandupperboundofthecapacityofthechannelasafunctionofλ2andforP=1.AsitisobservedtheboundsaretightforhighSNR,smallN,buttheybecomelooseasthepowerofthenoiseincreases.3.532.5UpperBound21.51LowerBound0.5课后答案网0-20-15-10-50510NdBProblem9.21www.hackshp.cnBothchannelscanbeviewedasbinarysymmetricchannelswithcrossoverprobabilitytheproba-bilityofdecodingabiterroneously.Since,%Q2Ebantipodalsignalingp=%N0bQEborthogonalsignalingN0thecapacityofthechannelis%1−hQ2EbantipodalsignalingC=%N01−hQEborthogonalsignalingN0InthenextfigureweplotthecapacityofthechannelasafunctionofEbforthetwosignalingN0schemes.262 10.90.8AntipodalSignalling0.70.60.5CapacityC0.40.3OrthogonalSignalling0.20.10-10-8-6-4-20246810SNRdBProblem9.22ThecodewordsofthelinearcodeofExample9.5.1arec1=[00000]c2=[10100]c3=[01111]c4=[11011]Sincethecodeislineartheminimumdistanceofthecodeisequaltotheminimumweightofthecodewords.Thus,dmin=wmin=2Thereisonlyonecodewordwithweightequalto2andthisisc2.Problem9.23TheparitycheckmatrixofthecodeinExample9.5.3is课后答案网11100H=0101001001Thecodewordsofthecodearewww.hackshp.cnc1=[00000]c2=[10100]c3=[01111]c4=[11011]AnyofthepreviouscodewordswhenpostmultipliedbyHtproducesanall-zerovectoroflength3.ForexamplecHt=[1⊕100]=[000]2cHt=[1⊕11⊕11⊕1]=[000]4263 Problem9.24Thefollowingtablelistsallthecodewordsofthe(7,4)Hammingcodealongwiththeirweight.SincetheHammingcodesarelineardmin=wmin.Asitisobservedfromthetabletheminimumweightis3andthereforedmin=3.No.CodewordsWeight1000000002100011033010001134001010135000111146110010147101001148100100139011011041001011003110011010312111000031311010104141011100415011100141611111117Problem9.25TheparitycheckmatrixHofthe(15,11)Hammingcodeconsistsofallbinarysequencesoflength4,excepttheallzerosequence.ThesystematicformofthematrixHis111000111011000t100110110110100H=[P|I4]=010101101110010课后答案网001011011110001Thecorrespondinggeneratormatrixis1110011010www.hackshp.cn1010011011010101G=[I11|P]=1001111110111010110111011111111Problem9.26LetCbean(n,k)linearblockcodewithparitycheckmatrixH.WecanexpresstheparitycheckmatrixintheformH=[h1h2···hn]wherehiisann−kdimensionalcolumnvector.Letc=[c1···cn]beacodewordofthecodeCwithlnonzeroelementswhichwedenoteasci1,ci2,...,cil.Clearlyci1=ci2=...=cil=1and264 sincecisacodewordcHt=0=ch+ch+···+ch1122nn=ci1hi1+ci2hi2+···+cilhil=hi1+hi2+···+hil=0ThisprovesthatlcolumnvectorsofthematrixHarelineardependent.Sinceforalinearcodetheminimumvalueofliswminandwmin=dmin,weconcludethatthereexistdminlineardependentcolumnvectorsofthematrixH.NowweassumethattheminimumnumberofcolumnvectorsofthematrixHthatarelineardependentisdminandwewillprovethattheminimumweightofthecodeisdmin.Lethi1,hi2,...,hdminbeasetoflineardependentcolumnvectors.Ifweformavectorcwithnon-zerocomponentsatpositionsi1,i2,...,idmin,thencHt=ch+···+c=0i1i1idminwhichimpliesthatcisacodewordwithweightdmin.Therefore,theminimumdistanceofacodeisequaltotheminimumnumberofcolumnsofitsparitycheckmatrixthatarelineardependent.ForaHammingcodethecolumnsofthematrixHarenon-zeroanddistinct.Thus,notwocolumnshi,hjaddtozeroandsinceHconsistsofallthen−ktuplesasitscolumns,thesumhi+hj=hmshouldalsobeacolumnofH.Then,hi+hj+hm=0andthereforetheminimumdistanceoftheHammingcodeis3.Problem9.27Thegeneratormatrixofthe(n,1)repetitioncodeisa1×nmatrix,consistedofthenon-zerocodeword.Thus,G=1|1···1Thisgeneratormatrixisalreadyinsystematicform,sothattheparitycheckmatrixisgivenby课后答案网110···01010H=............www.hackshp.cn100···1Problem9.281)TheparitycheckmatrixHeoftheextendedcodeisan(n+1−k)×(n+1)matrix.Thecodewordsoftheextendedcodehavetheformce,i=[ci|x]wherexis0iftheweightofciisevenand1iftheweightofciisodd.Sincece,iHte=[ci|x]Hte=0andciHt=0,thefirstn−kcolumnsofHtecanbeselectedasthecolumnsofHtwithazeroaddedinthelastrow.Inthiswaythechoiceofxisimmaterial.ThelastcolumnofHteisselectedinsuchawaythattheeven-parityconditionissatisfiedforeverycodewordce,i.Notethatifce,ihasevenweight,thenc+c+···+c=0=⇒c[11···1]t=0e,i1e,i2e,in+1e,i265 foreveryi.ThereforethelastcolumnofHteistheall-onevectorandtheparitycheckmatrixoftheextendedcodehastheformt1101101101111101000tH=Ht=1001=1010100ee010101100100011111111100012)Theoriginalcodehasminimumdistanceequalto3.Butforthosecodewordswithweightequaltotheminimumdistance,a1isappendedattheendofthecodewordstoproduceevenparity.Thus,theminimumweightoftheextendedcodeis4andsincetheextendedcodeislinear,theminimumdistanceisde,min=we,min=4.3)Thecodinggainoftheextendedcodeis3Gcoding=de,minRc=4×=1.71437Problem9.29Ifnocodingisemployed,wehave112EbPpb=Q=QN0RN0whereP10−6==5课后答案网RN0104×2×10−11Thus,√p=Q[5]=1.2682×10−2bandtherefore,theerrorprobabilityfor11bitsisP=1−(1−p)11≈0.1310errorin11bitswww.hackshp.cnbIfcodingisemployed,thensincetheminimumdistanceofthe(15,11)Hammingcodeis3,11dminEs3Espe≤(M−1)Q=10QN0N0whereEsEbP11=Rc=Rc=×5=3.6667N0N0RN015Thus√p≤10Q3×3.6667≈4.560×10−3eAsitisobservedtheprobabilityoferrordecreasesbyafactorof28.Ifharddecisionisemployed,thendminp≤(M−1)dminpi(1−p)dmin−ieibbdmin+1i=2266 %whereM=10,d=3andp=QRP=2.777×10−2.Hence,minbcRN0p=10×(3×p2(1−p)+p3)=0.0227ebbbInthiscasecodinghasdecreasedtheerrorprobabilitybyafactorof6.Problem9.30Thefollowingtableshowsthestandardarrayforthe(7,4)Hammingcode.e1e2e3e4e5e6e71000000010000000100000001000000010000000100000001c100000001000000010000000100000001000000010000000100000001c210001100000110110011010101101001110100001010001001000111c301000111100011000001101100110101011010011101000010100010c400101011010101011010100001010011101001000100101110010100c500011111001111010111100111110000111000101100011010001110c611001010100101100010111101011101101110000111001111100100c710100110010011111001110000111011011101011110100011010010c810010010001001110100110110011000001100110110010111001000c901101101110110001011001001100111110011001001101000110111c1001011001101100000110001111000100100010100001011100101101c1100110101011010011101000010100010010001111000110000011011c1211100000110000101000011000001111000111010011100101110001c1311010100101010100101011110101100010110111011010001101011c1410111000011100111110010011001010100101100010111101011101c1501110011111001001100101010010110001011110101110110111000c1611111110111111101111111011111110111111101111111011111110Asitisobservedthereceivedvectory=[1110100]isinthe7thcolumnofthetableundertheerrorvectore5.Thus,thereceivedvectorwillbedecodedasc=y+e5=[1110000]=c12Problem9.31Thegeneratorpolynomialofdegree课后答案网m=n−kshoulddividethepolynomialp6+1.Sincethepolynomialp6+1assumesthefactorizationp6+1=(p+1)3(p+1)3=(p+1)(p+1)(p2+p+1)(p2+p+1)weobservethatm=n−kcantakeanyvaluefrom1to5.Thus,k=n−mcanbeanynumberin[1,5].Thefollowingtableliststhepossiblevaluesofwww.hackshp.cnkandthecorrespondinggeneratorpolynomial(s).kg(p)1p5+p4+p3+p2+p+12p4+p2+1orp4+p3+p+13p3+14p2+1orp2+p+15p+1Problem9.32Togeneratea(7,3)cycliccodeweneedageneratorpolynomialofdegree7−3=4.Since(seeExample9.6.2))p7+1=(p+1)(p3+p2+1)(p3+p+1)=(p4+p2+p+1)(p3+p+1)=(p3+p2+1)(p4+p3+p2+1)267 eitheroneofthepolynomialsp4+p2+p+1,p4+p3+p2+1canbeusedasageneratorpolynomial.Withg(p)=p4+p2+p+1allthecodewordpolynomialsc(p)canbewrittenasc(p)=X(p)g(p)=X(p)(p4+p2+p+1)whereX(p)isthemessagepolynomial.ThefollowingtableshowstheinputbinarysequencesusedtorepresentX(p)andthecorrespondingcodewords.InputX(p)c(p)=X(p)g(p)Codeword0000000000000011p4+p2+p+10010111010pp5+p3+p2+p0101110100p2p6+p4+p3+p21011100011p+1p5+p4+p3+10111001101p2+1p6+p3+p+11001011110p2+pp6+p5+p4+p1110010111p2+p+1p6+p5+p2+11100101Sincethecycliccodeislinearandtheminimumweightiswmin=4,weconcludethattheminimumdistanceofthe(7,3)codeis4.Problem9.33UsingTable9.1wefindthatthecoefficientsofthegeneratorpolynomialofthe(15,11)codearegiveninoctalformas23.Since,thebinaryexpansionof23is010011,weconcludethatthegeneratorpolynomialisg(p)=p4+p+1Theencoderforthe(15,11)cycliccodeisdepictedinthenextfigure.c(p)X(p)+课后答案网+Problem9.34TheithrowofthematrixGhastheformgi=[0···010www.hackshp.cn···0pi,1pi,2···pi,n−k],1≤i≤kwherepi,1,pi,2,...,pi,n−karefoundbysolvingtheequationpn−i+ppn−k−1+ppn−k−2+···+p=pn−imodg(p)i,1i,2i,n−kThus,withg(p)=p4+p+1weobtainp14modp4+p+1=(p4)3p2modp4+p+1=(p+1)3p2modp4+p+1=(p3+p2+p+1)p2modp4+p+1=p5+p4+p3+p2modp4+p+1=(p+1)p+p+1+p3+p2modp4+p+1=p3+1p13modp4+p+1=(p3+p2+p+1)pmodp4+p+1=p4+p3+p2+pmodp4+p+1=p3+p2+1268 p12modp4+p+1=p3+p2+p+1p11modp4+p+1=(p4)2p3modp4+p+1=(p+1)2p3modp4+p+1=(p2+1)p3modp4+p+1=p5+p3modp4+p+1=(p+1)p+p3modp4+p+1=p3+p2+pp10modp4+p+1=(p2+1)p2modp4+p+1=p4+p2modp4+p+1=p2+p1p9modp4+p+1=(p2+1)pmodp4+p+1=p3+pp8modp4+p+1=p2+1modp4+p+1=p2+1p7modp4+p+1=(p+1)p3modp4+p+1=p3+p+1p6modp4+p+1=(p+1)p2modp4+p+1=p3+p2p5modp4+p+1=(p+1)pmodp4+p+1=p2+pp4modp4+p+1=p+1modp4+p+1=p+1Thegeneratorandtheparitycheckmatrixofthecodearegivenby11001111011011111111010111G=1101010101110110111001011010011111101011001000011110101100100H=00111101011001011101011001课后答案网0001Problem9.351)Letg(p)=p8+p6+p4+p2+1bethegeneratorpolynomialofan(n,k)cycliccode.Then,n−k=8andtherateofthecodeiswww.hackshp.cnk8R==1−nnTherateRisminimumwhen8ismaximumsubjecttotheconstraintthatRispositive.Thus,nthefirstchoiceofnisn=9.However,thegeneratorpolynomialg(p)doesnotdividep9+1andtherefore,itcannotgeneratea(9,1)cycliccode.Thenextcandidatevalueofnis10.Inthiscasep10+1=g(p)(p2+1)andtherefore,n=10isavalidchoice.TherateofthecodeisR=k=2=1.n1052)Inthenexttablewelistthefourcodewordsofthe(10,2)cycliccodegeneratedbyg(p).InputX(p)Codeword0000000000000011010101010110p101010101011p+11111111111269 Asitisobservedfromthetable,theminimumweightofthecodeis5andsincethecodeislineardmin=wmin=5.3)Thecodinggainofthe(10,2)cycliccodeinpart1)is2Gcoding=dminR=5×=110Problem9.361)Foreverynpn+1=(p+1)(pn−1+pn−2+···+p+1)whereadditionsaremodulo2.Sincep+1dividespn+1itcangeneratea(n,k)cycliccode,wherek=n−1.2)Theithrowofthegeneratormatrixhastheformgi=[0···010···0pi,1]wherepi,1,i=1,...,n−1,canbefoundbysolvingtheequationspn−i+p=pn−imodp+1,1≤i≤n−1i,1Sincepn−imodp+1=1foreveryi,thegeneratorandtheparitycheckmatrixaregivenby1···0|1........G=....,H=[11···1|1]0···1|13)Avectorc=[c1,c2,...,cn]isacodewordofthe(n,n−1)cycliccodeifitsatisfiestheconditioncHt=0.But,1课后答案网t1cH=0=c.=c1+c2+···cn..1Thus,thevectorcbelongstothecodeifithasanevenweight.Therefore,thecycliccodegeneratedbythepolynomialp+1isasimpleparitycheckcode.www.hackshp.cnProblem9.371)UsingtheresultsofProblem9.31,wefindthattheshortestpossiblegeneratorpolynomialofdegree4isg(p)=p4+p2+1TheithrowofthegeneratormatrixGhastheformgi=0···010···0pi,1···pi,4wherepi,1,...,pi,4areobtainedfromtherelationp6−i+pp3+pp2pp+p=p6−i(modp4+p2+1)i,1i,2i,3i,4Hence,p5modp4+p2+1=(p2+1)pmodp4+p2+1=p3+pp4modp4+p2+1=p2+1modp4+p2+1=p2+1270 andtherefore,101010G=010101Thecodewordsofthecodearec1=[000000]c2=[101010]c3=[010101]c4=[111111]2)Theminimumdistanceofthelinear(6,2)cycliccodeisdmin=wmin=3.Therefore,thecodecancorrectdmin−1ec==1error23)Anupperboundoftheblockerrorprobabilityisgivenby1dminEspe=(M−1)QN0WithM=2,dmin=3andEsEbP21N=RcN=RcRN=6×2×6×104×2×10−6=1.3889000weobtain√p=Q3×1.3889=2.063×10−2eProblem9.38Theblockgeneratedbytheinterleavingisa5课后答案网×23blockcontaining115binarysymbols.SincetheGolaycodecancorrectdmin−17−1ec===322bitspercodeword,theresultingblockcancorrectasingleburstoferrorsofdurationlessorequalto5×3=15bits.www.hackshp.cnProblem9.391-Cmaxisnotingeneralcyclic,becausethereisnoguaranteethatitislinear.Forexampleletn=3andletC1={000,111}andC2={000,011,101,110},thenCmax=C1∪C2={000,111,011,101,110},whichisobviouslynonlinear(forexample111⊕110=001∈Cmax)andthereforecannotbecyclic.2-Cminiscyclic,thereasonisthatC1andC2arebothlinearthereforeanytwoelementsofCminarebothinC1andC2andthereforetheirlinearcombinationisalsoinC1andC2andthereforeinCmin.TheintersectionsatisfiesthecyclicpropertybecauseifcbelongstoCminitbelongstoC1andC2andthereforeallcyclicshiftsofitbelongtoC1andC2andthereforetoCmin.AllcodewordpolynomialscorrespondingtotheelementsofCminaremultiplesofg1(p)andg2(p)andthereforemultipleofLCM{g1(p),g2(p)},whichinturndividespn+1.Foranyc∈Cmin,wehavew(c)≥d1andw(c)≥d2,thereforetheminimumdistanceofCminisgreaterthanorequaltomax{d1,d2}.271 Problem9.401)Sinceforeachtimeslot[mT,(m+1)T]wehaveφ1(t)=±φ2(t),thesignalsaredependentandthusonlyonedimensionisneededtorepresentthemintheinterval[mT,(m+1)T].Inthiscasethedimensionalityofthesignalspaceisupperboundedbythenumberofthedifferenttimeslotsusedtotransmitthemessagesignals.2)Ifφ1(t)=αφ2(t),thenthedimensionalityofthesignalspaceovereachtimeslotisatmost2.Sincetherearenslotsoverwhichwetransmitthemessagesignals,thedimensionalityofthesignalspaceisupperboundedby2n.3)Letthedecodingrulebethatthefirstcodewordisdecodedwhenrisreceivedifp(r|x1)>p(r|x2)Thesetofrthatdecodeintox1isR1={r:p(r|x1)>p(r|x2)}Thecharacteristicfunctionofthissetχ1(r)isbydefinitionequalto0ifr∈R1andequalto1ifr∈R1.Thecharacteristicfunctioncanbeboundedas1p(r|x2)21−χ1(r)≤p(r|x1)Thisinequalityistrueifχ(r)=1becausetherightsideisnonnegative.Itisalsotrueifχ(r)=0becauseinthiscasep(r|x2)>p(r|x1)andtherefore,1p(r|x2)p(r|x2)21≤=⇒1≤p(r|x1)p(r|x1)Giventhatthefirstcodewordissent,thentheprobabilityoferroris课后答案网

P(error|x1)=···p(r|x1)drRN−R1

=···p(r|x1)[1−χ1(r)]drRN

1p(r|x2)2≤···p(r|x1)drwww.hackshp.cnRNp(r|x1)

%=···p(r|x1)p(r|x2)drRN4)Theresultfollowsimmediatelyifweusetheunionboundontheprobabilityoferror.Thus,assumingthatxmwastransmitted,thentakingthesignalsxm
,m=m,oneatatimeandignoringthepresenceoftherest,wecanwrite

%P(error|xm)≤···p(r|xm)p(r|xm
)dr
RN1≤m≤Mm
=m5)Letr=xm+nwithnanN-dimensionalzero-meanGaussianrandomvariablewithvarianceperdimensionequaltoσ2=N0.Then,2p(r|xm)=p(n)andp(r|xm
)=p(n+xm−xm
)272 andtherefore,

%···p(r|xm)p(r|xm
)drRN

|n|2|n+xm−x|21−1−m
=···e2N0e2N0dnNNRN(πN0)4(πN0)4|xm−x|2

2|n|2+|xm−x|2/2+2n·(xm−x)−m
1−m
m
=e4N0···e2N0dnNRN(πN)20xm−x|xm−x|2

|n+m
|2−m
1−2=e4N0···eN0dnNRN(πN)202|xm−x|−m
=e4N0Usingtheunionboundinpart4,weobtain2−|xm−xm
|P(error|xm(t)sent)≤e4N01≤m
≤Mm
=mProblem9.411)Theencoderforthe(3,1)convolutionalcodeisdepictedinthenextfigure.k=1+.n=3+2)Thestatetransitiondiagramforthiscodeisdepictedinthenextfigure.0/000课后答案网#..000/0111/1110/00101101/100www.hackshp.cn1/1100/01011.&.1/1013)Inthenextfigurewedrawtwoframesofthetrellisassociatedwiththecode.Solidlinesindicateaninputequalto0,whereasdottedlinescorrespondtoaninputequalto1..000.00...111...011........01100..............001............10110........010....11................101273 4)Thediagramusedtofindthetransferfunctionisshowninthenextfigure.XdD2NJDJ32DNJDJDJ"Xa
XcXbXa

DNJUsingtheflowgraphresults,weobtainthesystem3Xc=DNJXa
+DNJXbXb=DJXc+DJXdX=D2NJX+D2NJXdcd2Xa

=DJXbEliminatingXb,XcandXdresultsin63Xa

DNJT(D,N,J)==222Xa
1−DNJ−DNJTofindthefreedistanceofthecodewesetN=J=1inthetransferfunction,sothatD6T(D)=T(D,N,J)|==D6+2D8+4D10+···1N=J=121−2DHence,dfree=65)Sincethereisnoselfloopcorrespondingtoaninputequalto1suchthattheoutputistheallzerosequence,thecodeisnotcatastrophic.Problem9.42Thenumberofbranchesleavingeachstatecorrespondtothenumberpossibledifferentinputstotheencoder.Sincetheencoderateachstatetakes课后答案网kbinarysymbolsatitsinput,thenumberofbranchesleavingeachstateofthetrellisis2k.ThenumberofbranchesenteringeachstateisthenumberofpossiblekLcontentsoftheencodershiftregisterthathavetheirfirstk(L−1)bitscorrespondingtothatparticularstate(notethatthedestinationstateforabranchisdeterminedbythecontentsofthefirstk(L−1)bitsoftheshiftregister).Thismeansthatthenumberofbranchesisequaltothenumberofpossibledifferentcontentsofthelastkbitsoftheencoder,i.e.,2k.www.hackshp.cnProblem9.431)Thestatediagramofthecodeisdepictedinthenextfigure0/000#..000/0111/1110/10101101/1000/1101/01011.&.1/001274 2)ThediagramusedtofindthetransferfunctionofthecodeisdepictedinthenextfigureDNJDNJD2J322DNJDJDJ"Xa
XcXbXa

DNJUsingtheflowgraphrelationswewrite3Xc=DNJXa
+DNJXbX=D2JX+D2JXbcdXd=DNJXc+DNJXd2Xa

=DJXbEliminatingXb,XcandXd,weobtain73Xa

DNJT(D,N,J)==32Xa
1−DNJ−DNJThus,D7T(D)=T(D,N,J)|==D7+D8+D9+···1N=J=131−D−D3)Theminimumfreedistanceofthecodeisdfree=74)Thefollowingfigureshows7framesofthetrellisdiagramusedbytheViterbidecoder.Itisassumedthattheinputsequenceispaddedbytozeros,sothattheactuallengthoftheinformationsequenceis5.ThenumbersonthenodesindicatetheHammingdistanceofthesurvivorpaths.ThedeletedbrancheshavebeenmarkedwithanX.Inthecaseofatiewedeletedthelowerbranch.Thesurvivorpathattheendofthedecodingisdenotedbyathickline.课后答案网1101101101110101011002443467......X..X....X.......2..2..5.XX.66X.....................X......X........1www.hackshp.cn..3X....3....4....5......X...X.X............3....4....5....4........X........X........X.Theinformationsequenceis11000andthecorrespondingcodeword111010110011000...5)Anuppertothebiterrorprobabilityofthecodeisgivenby1ϑT2(D,N)p¯b≤√kϑNN=1,D=4p(1−p)ButϑT2(D,N)ϑD7ND7==ϑNϑN1−(D+D3)N(1−DN−D3N)2andsincek=1,p=10−5,weobtainD7p¯≤≈4.0993×10−16b(1−D−D3)2√D=4p(1−p)275 Problem9.441)Thestatediagramofthecodeisdepictedinthenextfigure0/000#..000/0111/1010/11101101/1100/1001/01011.&.1/0012)ThediagramusedtofindthetransferfunctionofthecodeisdepictedinthenextfigureDNJDNJDJ232DNJDJDJ"Xa
XcXbXa

D2NJUsingtheflowgraphrelationswewrite22Xc=DNJXa
+DNJXbX=DJX+D3JXbdcXd=DNJXd+DNJXc2课后答案网Xa

=DJXbEliminatingXb,XcandXd,weobtain62473824Xa

DNJ+DNJ−DNJT(D,N,J)==42352623Xa
1−DNJ−DNJ−DNJ+DNJThus,www.hackshp.cnD6+D7−D8678T1(D)=T(D,N,J)|N=J=1=1−D−D4−D5+D6=D+2D+D+···3)Theminimumfreedistanceofthecodeisdfree=6.Thepath,whichisatadistancedfreefromtheallzeropath,isdepictedwithadoublelineinthenextfigure..000..00......101......011.....01............110........................111...............10......................100010................11................................001276 4)Thefollowingfigureshows6framesofthetrellisdiagramusedbytheViterbialgorithmtodecodethesequence{111,111,111,111,111,111}.ThenumbersonthenodesindicatetheHammingdistanceofthesurvivorpathsfromthereceivedsequence.ThebranchesthataredroppedbytheViterbialgorithmhavebeenmarkedwithanX.Inthecaseofatieoftwomergingpaths,wedeletethelowerpath.Thedecodedsequenceis{101,111,011,101,111,011}whichcorrespondstotheinformationsequence{x1,x2,x3,x4}={1,0,0,1}followedbytwozeros.11111111111111111136253400......X..XX....X.101..101...1.0114.2301101......X.............%.%.(X...((111..1111..4%(....2(....3X10...%..X..X.....(..(.....3(X....5(....411(........(........X.Problem9.45ThecodeofProblem9.41isa(3,1)convolutionalcodewithL=3.Thelengthofthereceivedsequenceyis15.Thismeansthat5symbolshavebeentransmitted,andsinceweassumethattheinformationsequencehasbeenpaddedbytwo0’s,theactuallengthoftheinformationsequenceis3.Thefollowingfiguredepicts5framesofthetrellisusedbytheViterbidecoder.Thenumbersonthenodesdenotethemetric(Hammingdistance)ofthesurvivorpaths.Inthecaseofatieoftwomergingpathsatanode,wehavepurgedthelowerpath.1012001301111103111600......X.000.000.)000++000...+XX.111.1111.0115+701........+!!*.).X...%.%.((..1..4%X....4001..001%..X(10....(课后答案网110....4...6(11........X.(.101Thedecodedsequenceis{111,001,011,000,000}andcorrespondstotheinformationsequence{1,0,0}followedbytwozeros.www.hackshp.cnProblem9.461)Theencoderforthe(3,1)convolutionalcodeisdepictedinthenextfigure.+k=1+n=3+2)Thestatetransitiondiagramforthiscodeisshownbelow277 0/000#..000/0111/1110/10101101/1001/0100/11011.&.1/0013)Inthenextfigurewedrawtwoframesofthetrellisassociatedwiththecode.Solidlinesindicateaninputequalto0,whereasdottedlinescorrespondtoaninputequalto1..000.00...111...011........01100..............101............10010........110....11................0014)Thediagramusedtofindthetransferfunctionisshowninthenextfigure.XdDNJD2J322DNJDJDJ"Xa课后答案网
XcXbXa

DNJUsingtheflowgraphresults,weobtainthesystem3Xc=DNJXa
+DNJXbwww.hackshp.cnX=D2JX+D2JXbcdXd=DNJXc+DNJXd2Xa

=DJXbEliminatingXb,XcandXdresultsin73Xa

DNJT(D,N,J)==32Xa
1−DNJ−DNJTofindthefreedistanceofthecodewesetN=J=1inthetransferfunction,sothatD7T(D)=T(D,N,J)|==D7+D8+D9+···1N=J=131−D−DHence,dfree=75)Sincethereisnoselfloopcorrespondingtoaninputequalto1suchthattheoutputistheallzerosequence,thecodeisnotcatastrophic.278 Problem9.47UsingthediagramofFigure9.28,weseethatthereareonlytwowaystogofromstateXa
tostateXa

withatotalnumberofones(sumoftheexponentsofD)equalto6.Thecorrespondingtransitionsare:D2DDD2Path1:Xa
→Xc→Xd→Xb→Xa

D2DDD2Path2:Xa
→Xc→Xb→Xc→Xb→Xa

Thesetwopathscorrespondtothecodewordsc1=0,0,1,0,1,0,1,1,0,0,0,0,...c2=0,0,0,1,0,0,0,1,1,1,0,0,...Problem9.481)Thestatetransitiondiagramandtheflowdiagramusedtofindthetransferfunctionforthiscodearedepictedinthenextfigure.0/00#..000/101/010/01DNJ0110Xd1/11NJD2J0/111/0011DNJDJDJ.&.Xa
Xc"XbXa

1/10D2NJThus,课后答案网2Xc=DNJXa
+DNJXbX=DJX+D2JXbcdXd=NJXc+DNJXdwww.hackshp.cnXa

=DJXbandbyeliminatingXb,XcandXd,weobtain33Xa

DNJT(D,N,J)==32Xa
1−DNJ−DNJTofindthetransferfunctionofthecodeintheformT(D,N),wesetJ=1inT(D,N,J).Hence,D3NT(D,N)=1−DN−D3N2)TofindthefreedistanceofthecodewesetN=1inthetransferfunctionT(D,N),sothatD3T(D)=T(D,N)|==D3+D4+D5+2D6+···1N=131−D−DHence,dfree=3279 3)Anupperboundonthebiterrorprobability,whenharddecisiondecodingisused,isgivenby1ϑT(D,N)P¯b≤√kϑNN=1,D=4p(1−p)SinceϑT(D,N)ϑD3ND3==ϑNN=1ϑN1−(D+D3)NN=1(1−(D+D3))2withk=1,p=10−6weobtainD3P¯≤=8.0321×10−9b(1−(D+D3))2√D=4p(1−p)Problem9.491)Letthedecodingrulebethatthefirstcodewordisdecodedwhenyiisreceivedifp(yi|x1)>p(yi|x2)Thesetofyithatdecodeintox1isY1={yi:p(yi|x1)>p(yi|x2)}Thecharacteristicfunctionofthissetχ1(yi)isbydefinitionequalto0ifyi∈Y1andequalto1ifyi∈Y1.Thecharacteristicfunctioncanbeboundedas(seeProblem9.40)1p(yi|x2)21−χ1(yi)≤p(yi|x1)Giventhatthefirstcodewordissent,thentheprobabilityoferrorisP(error|x1)=p(yi|x1)=p(yi|x1)[1−χ1(yi)]yi∈Y−Y1yi∈Y1%课后答案网p(yi|x2)2≤p(yi|x1)=p(yi|x1)p(yi|x2)p(yi|x1)yi∈Yyi∈Y2n%=p(yi|x1)p(yi|x2)i=1www.hackshp.cnwhereYdenotesthesetofallpossiblesequencesyi.Since,eachelementofthevectoryicantaketwovalues,thecardinalityofthesetYis2n.2)Usingtheresultsofthepreviouspartwehave112n%2np(yi|x1)p(yi|x2)P(error)≤p(yi|x1)p(yi|x2)=p(yi)p(yi)p(yi)i=1i=1112n2n%p(x1|yi)p(x2|yi)=p(yi)=2p(yi)p(x1|yi)p(x2|yi)p(x1)p(x2)i=1i=1However,giventhevectoryi,theprobabilityoferrordependsonlyonthosevaluesthatx1andx2aredifferent.Inotherwords,ifx1,k=x2,k,thennomatterwhatvalueisthekthelementofyi,itwillnotproduceanerror.Thus,ifbydwedenotetheHammingdistancebetweenx1andx2,thenp(x|y)p(x|y)=pd(1−p)d1i2i280 andsincep(y)=1,weobtaini2ndddP(error)=P(d)=2p2(1−p)2=[4p(1−p)]2Problem9.501)
∞21−vQ(x)=√e2dv2πx√
∞v=2t1−t2=√edtπ√x2
∞12−t2=edt2π√x21x=erfc√222)Theaveragebiterrorprobabilitycanbeboundedas(see(9.7.16))11∞E1∞&P¯≤af(d)Q2Rdb=af(d)Q2RdγbdcdcbkN0kd=dfreed=dfree1∞&=adf(d)erfc(Rcdγb)2kd=dfree∞%1=ad+dfreef(d+dfree)erfc(Rc(d+dfree)γb)2kd=11&∞≤erfc(Rdγ)af(d+d)e−Rcdγbcfreebd+dfreefree2kd=1But,课后答案网∞∞T(D,N)=aDdNf(d)=aDd+dfreeNf(d+dfree)dd+dfreed=dfreed=1andtherefore,∞ϑT(D,N)www.hackshp.cnd+dfree=ad+dfreeDf(d+dfree)ϑNN=1d=1∞=DdfreeaDdf(d+d)d+dfreefreed=1SettingD=e−Rcγbinthepreviousandsubstitutingintheexpressionfortheaveragebiterrorprobability,weobtain1&RϑT(D,N)P¯≤erfc(Rdγ)ecdfreeγbbcfreeb2kϑNN=1,D=e−RcγbProblem9.51Thepartitionofthe8-PAMconstellationinfoursubsetsisdepictedinthefigurebelow.281 -7-5-3-1135701-7-315-5-1370-,101-,-,-71-35-17-532)Thenextfigureshowsoneframeofthetrellisusedtodecodethereceivedsequence.Eachbranchconsistsoftwotransitionswhichcorrespondtoelementsinthesamecosetinthefinalpartitionlevel.-7,1-5,3-5,3 #-7,1$-3,5 #-1,7$-1,7-3,5TheoperationoftheViterbialgorithmforthedecodingofthesequence{−.2,1.1,6,4,−3,−4.8,3.3}isshownschematicallyinthenextfigure.Ithasbeenassumedthatwestartattheallzerostateandthatasequenceofzerosterminatestheinputbitstreaminordertocleartheencoder.ThenumbersatthenodesindicatetheminimumEuclideandistance,andthebrancheshavebeenmarkedwiththedecodedtransmittedsymbol.ThepathsthathavebeenpurgedaremarkedwithanX.-.2课后答案网1.164-3-4.83.31.441.4526.457.0515.4511.0914.38111XXXX333X3-5-510.245.0510.45X15.05-511.05355X1X-325.45www.hackshp.cn6.05511.4511.05-17XX-114.29XX14.656.057.057.05X5-3Transmittedsequence:1353-5-33282 Chapter10Problem10.11)Thewavelengthλis3×1083λ=m=m10910Hence,theDopplerfrequencyshiftisu100Km/hr100×103×10fD=±=±3=±Hz=±92.5926Hzλm3×360010Theplussignholdswhenthevehicletravelstowardsthetransmitterwhereastheminussignholdswhenthevehiclemovesawayfromthetransmitter.2)ThemaximumdifferenceintheDopplerfrequencyshift,whenthevehicletravelsatspeed100km/hrandf=1GHz,is∆fDmax=2fD=185.1852HzThisshouldbethebandwidthoftheDopplerfrequencytrackingloop.3)ThemaximumDopplerfrequencyshiftisobtainwhenf=1GHz+1MHzandthevehiclemovestowardsthetransmitter.Inthiscase3×108λmin=m=0.2997m109+106andtherefore100×103fDmax==92.6853Hz0.2997×3600Thus,theDopplerfrequencyspreadis课后答案网Bd=2fDmax=185.3706Hz.Problem10.21)SinceTm=1second,thecoherencebandwidth1Bcb==0.5Hzwww.hackshp.cn2TmandwithBd=0.01Hz,thecoherencetimeis1Tct==100/2=50seconds2Bd(2)SincethechannelbandwidthWbcb,thechannelisfrequencyselective.(3)SincethesignaldurationTTct,thechannelisslowlyfading.283 (4)TheratioW/Bcb=10.Hence,inprincipleuptotenthorderdiversityisavailablebysubdividingthechannelbandwidthinto10subchannels,eachofwidth0.5Hz.IfweemploybinaryPSKwithsymboldurationT=10seconds,thenthechannelbandwidthcanbesubdividedinto25subchannels,eachofbandwidth2=0.2Hz.Wemaychoosetohave5thorderfrequencydiversityTandforeachtransmission,thus,have5paralleltransmissions.Thus,wewouldhaveadatarateof5bitspersignalinterval,i.e.,abitrateof1/2bps.Byreducingtheorderofdiversity,wemayincreasethedatarate,forexample,withnodiversity,thedataratebecomes2.5bps.(5)Toanswerthequestionwemayusetheapproximaterelationfortheerrorprobabilitygivenby(10.1.37),orwemayusetheresultsinthegraphshowninFigure10.1.10.Forexample,forbinaryPSKwithD=4,theSNRperbitrequiredtoachieveanerrorprobabilityof10−6is18dB.ThisthetotalSNRperbitforthefourchannels(withmaximalrationcombining).Hence,theSNRperbitperchannelisreducedto12dB(afactoroffoursmaller).Problem10.3TheRayleighdistributionisαe−α2/2σα2,α>0σ2p(α)=α0,otherwiseHence,theprobabilityoferrorforthebinaryFSKandDPSKwithnoncoherentdetectionaveragedoverallpossiblevaluesofαis
2∞1−cαEbα22P=eN0e−α/2σαdα2202σα1
∞−α2cEb+1N02σ2=αeαdα2σ20αBut,
∞2n+1−ax2n!xedx=,(a>0)02an+1sothatwithn=0weobtain课后答案网1
∞−α2cEb+111N02σ2P2=αeαdα=2σα202σα22cEb+1N02σα211==Eb2σα22[cρ¯+1]2c+1bwww.hackshp.cnN0Eb2σα21where¯ρb=N0.Withc=1(DPSK)andc=2(FSK)wehave1,DPSKP=2(1+¯ρb)21,FSK2+¯ρb284 Problem10.4(a)cos2πf1t
×

MatchedFilter1()2+
×

MatchedFilter1()2r(t)
sin2πf1t1cos2πf2t
×

MatchedFilter2()2++


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MatchedFilter2()2sin2πf2tsampleatt=kTcos2πf1tDetectoroutput

×

MatchedFilter1()2selectthelarger+
×

MatchedFilter1()2r(t)
sin2πf1t+

2cos2πf2t课后答案网
×

MatchedFilter2()2+
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MatchedFilter2www.hackshp.cn()2sin2πf2t(b)TheprobabilityoferrorforbinaryFSKwithsquare-lawcombiningforD=2isgiveninFigure10.1.10.TheprobabilityoferrorforD=1isalsogiveninFigure10.1.10.NotethatanincreaseinSNRbyafactorof10reducestheerrorprobabilitybyafactorof10whenD=1andbyafactorof100whenD=2.Problem10.5√(a)risaGaussianrandomvariable.IfEbisthetransmittedsignalpoint,then&E(r)=E(r1)+E(r2)=(1+k)Eb≡mrandthevarianceisσ2=σ2+k2σ2r12285 Theprobabilitydensityfunctionofris2(r−mr)1−2f(r)=√e2σr2πσrandtheprobabilityoferroris
0P2=f(r)dr−∞
−mr21σr−x=√e2dx2π−∞1m2=Qrσ2rwherem2r(1+k)2Eb=σr2σ12+k2σ22Thevalueofkthatmaximizesthisratioisobtainedbydifferentiatingthisexpressionandsolvingforthevalueofkthatforcesthederivativetozero.Thus,weobtainσ2k=1σ22Notethatifσ1>σ2,thenk>1andr2isgivengreaterweightthanr1.Ontheotherhand,ifσ2>σ1,thenk<1andr1isgivengreaterweightthanr2.Whenσ1=σ2,k=1.Inthiscasem22Ebr=σ2σ2r1(b)Whenσ2=3σ2,k=1,and213212mr(1+3)Eb4Eb==σr2σ2+1(3σ2)3σ2课后答案网1911Ontheotherhand,ifkissettounitywehavem2r4EbEb==www.hackshp.cnσr2σ12+3σ12σ12Therefore,theoptimumweightingprovidesagainof410log=1.25dB3Problem10.61)Theprobabilityoferrorforafixedvalueofais12a2EPe(a)=QN0sincethegivenatakestwopossiblevalues,namelya=0anda=2withprobabilities0.1and0.9,respectively,theaverageprobabilityoferroris110.18E8EPe=+Q=0.05+Q2N0N0286 (2)AsE→∞,P→0.05N0e(3)Theprobabilityoferrorforfixedvaluesofa1anda2is12(a21+a22)EPe(a1,a2)=QN0Inthiscasewehavefourpossiblevaluesforthepair(a1,a2),namely,(0,0),(0,2),(2,0),and(2,2),withcorrespondingprobabilities).01,0.09,0.09and0.81.Hence,theaverageprobabilityoferroris110.018E16EPe=+0.18Q+0.81Q2N0N0(4)AsE→∞,P→0.005,whichisafactorof10smallerthanin(2).N0eProblem10.7Weassumethattheinputbits0,1aremappedtothesymbols-1and1respectively.TheterminalphaseofanMSKsignalattimeinstantnisgivenbykπθ(n;a)=ak+θ02k=0whereθ0istheinitialphaseandakis±1dependingontheinputbitatthetimeinstantk.Thefollowingtableshowsθ(n;a)fortwodifferentvaluesofθ0(0,π),andthefourinputpairsofdata:{00,01,10,11}.θ0b0b1a0a1θ(n;a)000-1-1−π001-1100101-1001111ππ00-1-10课后答案网π01-11ππ101-1ππ11112πProblem10.81)Theenvelopeofthesignaliswww.hackshp.cn%|s(t)|=|sc(t)|2+|ss(t)|212Eb2πt2Eb2πt=cos+sinTb2TbTb2Tb12Eb=TbThus,thesignalhasconstantamplitude.287 2)Thesignals(t)hastheformofthefour-phasePSKsignalwithπtgT(t)=cos,0≤t≤2Tb2TbHence,itisanMSKsignal.Ablockdiagramofthemodulatorforsynthesizingthesignalisgiveninthenextfigure.a2n××SerialSerial/77s(t)Parallelcos(πt)cos(2πft)+2TbcdataanDemux−π−π22××a2n+13)Asketchofthedemodulatorisshowninthenextfigure.t=2Tb2Tb××(·)dtThreshold0r(t)77πtParalleltocos(2πfct))cos()Serial2Tb−π−π22t=2Tb××2Tb(·)dtThreshold0Problem10.9Sincep=2,misodd(m=1)and课后答案网M=2,thereareNs=2pM=8phasestates,whichwedenoteasSn=(θn,an−1).The2p=4phasestatescorrespondingtoθnare*8www.hackshp.cnπ3πΘs=0,,π,22andtherefore,the8statesSnare*8ππ3π3π(0,1),(0,−1),,1,,−1,(π,1),(π,−1),,1,,−12222Havingatourdisposalthestate(θn,an−1)andthetransmittedsymbolan,wecanfindthenewphasestateasanπ(θn,an−1)−→(θn+an−1,an)=(θn+1,an)2Thefollowingfigureshowsoneframeofthephase-trellisofthepartialresponseCPMsignal.288 (θn,an−1)(θn+1,an)(0,1)..(0,1)...(0,−1)......(0,−1).........π,1)..........(π,1)(2...........2.........π..........(π(2,−1)..........2,−1)................(π,1).........(π,1)...................(π,−1)..........(π,−1).......3π........3π(,1).......(,1)2.....2.......3π..3π(,−1)...(,−1)22ThefollowingisasketchofthestatediagramofthepartialresponseCPMsignal.(π,1)(π,1)111(3π,1)2
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/(3π,−1)(,−1)22课后答案网(π,−1)Problem10.101)ForafullresponseCPFSKsignal,Lisequalto1.Ifh=2,thensincemiseven,therearep3terminalphasestates.Ifh=3,thenumberofstatesisN=2p.4s2)WithL=3andh=2,thenumberofstatesiswww.hackshp.cnN=p22=12.WhenL=3andh=3,the3s4numberofstatesisNs=2p22=32.Problem10.11(a)ThecodinggainisH1Rcdmin=×10=5(7dB)2(b)TheprocessinggainisW/R,whereW=107HzandR=2000bps.Hence,W107==5×103(37dB)R2×103289 (c)Thejammingmargingivenby(10.3.43)isPJ=W+(CG)−EbPsdBRdBdBJ0dB=37+7−10=34dBProblem10.12TheprobabilityoferrorforDSspreadspectrumwithbinaryPSKmaybeexpressedas12W/RbP2=QPJ/PSwhereW/RistheprocessinggainandPJ/PSisthejammingmargin.Ifthejammerisabroadband,WGNjammer,thenPJ=WJ0PS=Eb/Tb=EbRbTherefore,12EbP2=QJ0whichisidenticaltotheperformanceobtainedwithanon-spreadsignal.Problem10.13Weassumethattheinterferenceischaracterizedasazero-meanAWGNprocesswithpowerspectraldensityJ0.Toachieveanerrorprobabilityof10−5,therequiredEb/J0=10.Then,byusingtherelationin(10.3.40)and(10.3.44),wehaveW/R=W/R=EbPN/PSNu−1J0W/R=Eb(N−1)J0u课后答案网W=REb(N−1)J0uwhereR=104bps,Nu=30andEb/J0=10.Therefore,www.hackshp.cnW=2.9×106HzTheminimumchiprateis1/Tc=W=2.9×106chips/sec.Problem10.14Toachieveanerrorprobabilityof10−6,werequireEb=10.5dBJ0dBThen,thenumberofusersoftheCDMAsystemisN=W/Rb+1uEb/J0=1000+1=89users11.3IftheprocessinggainisreducedtoW/Rb=500,then500Nu=+1=45users11.3290 Problem10.15(a)Wearegivenasystemwhere(PJ/PS)dB=20dB,R=1000bpsand(Eb/J0)dB=10dB.Hence,usingtherelationin(10.3.40)weobtainW=PJ+Eb=30dBRdBPSdBJ0dBW=1000RW=1000R=106Hz(b)Thedutycycleofapulsejammerforworst-casejammingis∗0.70.7α===0.07Eb/J010Thecorrespondingprobabilityoferrorforthisworst-casejammingis0.0820.082−3P2===8.2×10Eb/J010Problem10.16Theradiosignalpropagatesatthespeedoflight,c=3×108m/sec.Thedifferenceinpropagationdelayforadistanceof300metersis300Td==1µsec3×108TheminimumbandwidthofaDSspreadspectrumsignalrequiredtoresolvethepropagationpathsisW=1MHz.Hence,theminimumchiprateis10课后答案网6chipspersecond.Problem10.17(a)WehaveNu=15userstransmittingatarateof10,000bpseach,inabandwidthofW=1MHz.The
b/J0iswww.hackshp.cnEW/R106/104100===J0Nu−11414=7.14(8.54dB)(b)Theprocessinggainis100.(c)WithNu=30andEb/J0=7.14,theprocessinggainshouldbeincreasedtoW/R=(7.14)(29)=207Hence,thebandwidthmustbeincreasedtoW=2.07MHz.291 Problem10.18(a)Thelengthoftheshift-registersequenceisL=2m−1=215−1=32767bitsForbinaryFSKmodulation,theminimumfrequencyseparationis2/T,where1/Tisthesymbol(bit)rate.Thehoprateis100hops/sec.SincetheshiftregisterhasN=32767statesandeachstateutilizesabandwidthof2/T=200Hz,thenthetotalbandwidthfortheFHsignalis6.5534MHz.(b)TheprocessinggainisW/R.Wehave,W6.5534×1064==6.5534×10bpsR100(c)IfthenoiseisAWGwithpowerspectraldensityN0,theprobabilityoferrorexpressionis11EbW/RP2=Q=QN0PN/PSProblem10.19(a)Ifthehoppingrateis2hops/bitandthebitrateis100bits/sec,then,thehoprateis200hops/sec.Theminimumfrequencyseparationfororthogonality2/T=400Hz.SincethereareN=32767statesoftheshiftregisterandforeachstateweselectoneoftwofrequenciesseparatedby400Hz,thehoppingbandwidthis13.1068MHz.(b)TheprocessinggainisW/R,课后答案网whereW=13.1068MHzandR=100bps.HenceW=0.131068MHzR(c)TheprobabilityoferrorinthepresenceofAWGNisgivenby(10.3.61)withN=2chipsperhop.www.hackshp.cnProblem10.20a)ThetotalSNRforthreehopsis20∼13dB.ThereforetheSNRperhopis20/3.Theprobabilityofachiperrorwithnoncoherentdetectionis1−Ecp=e2N02whereEc/N0=20/3.TheprobabilityofabiterrorisP=1−(1−p)2b=1−(1−2p+p2)=2p−p2−Ec1−Ec=e2N0−eN02=0.0013292 b)Inthecaseofonehopperbit,theSNRperbitis20,Hence,1−EcPb=e2N021−10=e2=2.27×10−5Thereforethereisalossinperformanceofafactor57AWGNduetosplittingthetotalsignalenergyintothreechipsand,then,usingharddecisiondecoding.Problem10.21(a)Wearegivenahoppingbandwidthof2GHzandabitrateof10kbs.Hence,W2×1095==2×10(53dB)R104(b)Thebandwidthoftheworstpartial-bandjammerisα∗W,whereα∗=2/(E/J)=0.2b0Henceα∗W=0.4GHz(c)Theprobabilityoferrorwithworst-casepartial-bandjammingise−1e−1P2=(Eb/J0)=10=3.68×10−2Problem10.22课后答案网TheprocessinggainisgivenasW=500(27dB)RbThe(Eb/J0)requiredtoobtainanerrorprobabilityof10−5forbinaryPSKis9.5dB.Hence,thejammingmarginiswww.hackshp.cnPJ=W−EbPSdBRbdBJ0dB=27−9.5=17.5dBProblem10.23Ifthejammerisapulsejammerwithadutycycleα=0.01,theprobabilityoferrorforbinaryPSKisgivenas12W/RbP2=αQPJ/PSForP2=10−5,andα=0.01,wehave12W/Rb−3Q=10PJ/PS293 Then,W/Rb500==5PJ/PSPJ/PSandPJ=100(20dB)PSProblem10.24∞c(t)=cnp(t−nTc)n=−∞Thepowerspectraldensityofc(t)isgivenby12Sc(f)=Sc(f)|P(f)|Tcwhere|P(f)|2=(AT)2sinc2(fT),T=1µseccccandSc(f)isthepowerspectraldensityofthesequence{cn}.Sincetheautocorrelationofthesequence{cn}isperiodicwithperiodLandisgivenasL,m=0,±L,±2L,...Rc(m)=−1,otherwisethen,Rc(m)canberepresentedinadiscreteFourierseriesasL−11j2πmk/LRc(m)=rC(k)e,m=0,1,...,L−1Lk=0where{rc(k)}aretheFourierseriescoefficients,whicharegivenas课后答案网L−1r(k)=R(m)e−j2πkm/L,k=0,1,...,L−1ccm=0andrc(k+nL)=rc(k)forn=0,±1,±2,....Thelattercanbeevaluatedtoyieldwww.hackshp.cnL−1−j2πkm/Lrc(k)=L+1−m=0e1,k=0,±L,±2L,...=L+1,otherwiseThepowerspectraldensityofthesequence{cn}maybeexpressedintermsof{rc(k)}.Thesecoefficientsrepresentthepowerinthespectralcomponentsatthefrequenciesf=k/L.Therefore,wehave∞1kSc(f)=rc(k)δf−LLTck=−∞Finally,wehave∞21kkSc(f)=rc(k)Pδf−LTcLTcLTck=−∞294 Problem10.25Withoutlossofgenerality,letusassumethatL1