《数字通信》第四版课后答案

CHAPTER2Problem2.1:P(Ai)=3
j=1P(Ai,Bj),i=1,2,3,4Hence:P(A1)=3
j=1P(A1,Bj)=0.1+0.08+0.13=0.31P(A2)=3
j=1P(A2,Bj)=0.05+0.03+0.09=0.17P(A3)=3
j=1P(A3,Bj)=0.05+0.12+0.14=0.31P(A4)=3
j=1P(A4,Bj)=0.11+0.04+0.06=0.21Similarly:P(B1)=4
i=1P(Ai,B1)=0.10+0.05+0.05+0.11=0.31P(B2)=4
i=1P(Ai,B2)=0.08+0.03+0.12+0.04=0.27P(B3)=4
i=1P(Ai,B3)=0.13+0.09+0.14+0.06=0.42Problem2.2:Therelationshipholdsforn=2(2-1-34):p(x1,x2)=p(x2|x1)p(x1)Supposeitholdsforn=k,i.e:p(x1,x2,...,xk)=p(xk|xk−1,...,x1)p(xk−1|xk−2,...,x1)...p(x1)Thenforn=k+1:p(x1,x2,...,xk,xk+1)=p(xk+1|xk,xk−1,...,x1)p(xk,xk−1...,x1)=p(xk+1|xk,xk−1,...,x1)p(xk|xk−1,...,x1)p(xk−1|xk−2,...,x1)...p(x1)Hencetherelationshipholdsforn=k+1,andbyinductionitholdsforanyn.1课后答案网:pY(y)=1|a|pXy−baProblem2.4:Relationship(2-1-44)gives:pY(y)=13a[(y−b)/a]2/3pXy−ba1/3Xisagaussianr.v.withzeromeanandunitvariance:pX(x)=1√2πe−x2/2Hence:pY(y)=13a√2π[(y−b)/a]2/3e−12(y−ba)2/3−10−8−6−4−2024681000.050.10.150.20.250.30.350.40.450.5ypdfofYa=2b=3Problem2.5:(a)Since(Xr,Xi)arestatisticallyindependent:pX(xr,xi)=pX(xr)pX(xi)=12πσ2e−(x2r+x2i)/2σ22课后答案网=(Xr+Xi)ejφ⇒Xr+Xi=(Yr+jYi)e−jφ=Yrcosφ+Yisinφ+j(−Yrsinφ+Yicosφ)⇒Xr=Yrcosφ+YisinφXi=−Yrsinφ+Yicosφ TheJacobianoftheabovetransformationis:J=∂Xr∂Yr∂Xi∂Yr∂Xr∂Yi∂Xi∂Yi=cosφ−sinφsinφcosφ=1Hence,by(2-1-55):pY(yr,yi)=pX((Yrcosφ+Yisinφ),(−Yrsinφ+Yicosφ))=12πσ2e−(y2r+y2i)/2σ2(b)Y=AXandX=A−1YNow,pX(x)=1(2πσ2)n/2e−xx/2σ2(thecovariancematrixMoftherandomvariablesx1,...,xnisM=σ2I,sincetheyarei.i.d)andJ=1/|det(A)|.Hence:pY(y)=1(2πσ2)n/21|det(A)|e−y(A−1)A−1y/2σ2Forthepdf’sofXandYtobeidenticalwerequirethat:|det(A)|=1and(A−1)A−1=I=⇒A−1=AHence,Amustbeaunitary(orthogonal)matrix.Problem2.6:(a)ψY(jv)=EejvY=Eejvni=1xi=Eni=1ejvxi=ni=1EejvX=ψX(ejv)nBut,pX(x)=pδ(x−1)+(1−p)δ(x)⇒ψX(ejv)=1+p+pejv⇒ψY(jv)=1+p+pejvn3课后答案网(b)E(Y)=−jdψY(jv)dv|v=0=−jn(1−p+pejv)n−1jpejv|v=0=npandE(Y2)=−d2ψY(jv)d2v|v=0=−ddvjn(1−p+pejv)n−1pejvv=0=np+np(n−1)p⇒E(Y2)=n2p2+np(1−p)Problem2.7:ψ(jv1,jv2,jv3,jv4)=Eej(v1x1+v2x2+v3x3+v4x4)E(X1X2X3X4)=(−j)4∂4ψ(jv1,jv2,jv3,jv4)∂v1∂v2∂v3∂v4|v1=v2=v3=v4=0From(2-1-151)ofthetext,andthezero-meanpropertyofthegivenrv’s:ψ(jv)=e−12vMvwherev=[v1,v2,v3,v4],M=[µij].Weobtainthedesiredresultbybringingtheexponenttoascalarformandthenperformingquadrupledifferentiation.Wecansimplifytheprocedurebynotingthat:∂ψ(jv)∂vi=−µive−12vMvwhereµi=[µi1,µi2,µi3,µi4].Alsonotethat:∂µjv∂vi=µij=µjiHence:∂4ψ(jv1,jv2,jv3,jv4)∂v1∂v2∂v3∂v4|V=0=µ12µ34+µ23µ14+µ24µ13Problem2.8:Forthecentralchi-squarewithndegressoffreedom:ψ(jv)=1(1−j2vσ2)n/24课后答案网ψ(jv)dv=jnσ2(1−j2vσ2)n/2+1⇒E(Y)=−jdψ(jv)dv|v=0=nσ2d2ψ(jv)dv2=−2nσ4(n/2+1)(1−j2vσ2)n/2+2⇒EY2=−d2ψ(jv)dv2|v=0=n(n+2)σ2Thevarianceisσ2Y=E(Y2)−[E(Y)]2=2nσ4Forthenon-centralchi-squarewithndegreesoffreedom:ψ(jv)=1(1−j2vσ2)n/2ejvs2/(1−j2vσ2)wherebydefinition:s2=ni=1m2i.dψ(jv)dv=jnσ2(1−j2vσ2)n/2+1+js2(1−j2vσ2)n/2+2ejvs2/(1−j2vσ2)Hence,E(Y)=−jdψ(jv)dv|v=0=nσ2+s2d2ψ(jv)dv2=−nσ4(n+2)(1−j2vσ2)n/2+2+−s2(n+4)σ2−ns2σ2(1−j2vσ2)n/2+3+−s4(1−j2vσ2)n/2+4ejvs2/(1−j2vσ2)Hence,EY2=−d2ψ(jv)dv2|v=0=2nσ4+4s2σ2+nσ2+s2andσ2Y=EY2−[E(Y)]2=2nσ4+4σ2s2Problem2.9:TheCauchyr.v.has:p(x)=a/πx2+a2,−∞x∞(a)E(X)=∞−∞xp(x)dx=0sincep(x)isanevenfunction.EX2=∞−∞x2p(x)dx=aπ∞−∞x2x2+a2dxNotethatforlargex,x2x2+a2→1(i.enon-zerovalue).Hence,EX2=∞,σ2=∞5课后答案网(b)ψ(jv)=EjvX=∞−∞a/πx2+a2ejvxdx=∞−∞a/π(x+ja)(x−ja)ejvxdxThisintegralcanbeevaluatedbyusingtheresiduetheoremincomplexvariabletheory.Then,forv≥0:ψ(jv)=2πja/πx+jaejvxx=ja=e−avForv0:ψ(jv)=−2πja/πx−jaejvxx=−ja=eavvTherefore:ψ(jv)=e−a|v|Note:analternativewaytofindthecharacteristicfunctionistousetheFouriertransformrelationshipbetweenp(x),ψ(jv)andtheFourierpair:e−b|t|↔1πcc2+f2,c=b/2π,f=2πvProblem2.10:(a)Y=1nni=1Xi,ψXi(jv)=e−a|v|ψY(jv)=Eejv1nni=1Xi=ni=1EejvnXi=ni=1ψXi(jv/n)=e−a|v|/nn=e−a|v|(b)SinceψY(jv)=ψXi(jv)⇒pY(y)=pXi(xi)⇒pY(y)=a/πy2+a2.(c)Asn→∞,pY(y)=a/πy2+a2,whichisnotGaussian;hence,thecentrallimittheoremdoesnothold.ThereasonisthattheCauchydistributiondoesnothaveafinitevariance.Problem2.11:Weassumethatx(t),y(t),z(t)arereal-valuedstochasticprocesses.Thetreatmentofcomplex-valuedprocessesissimilar.(a)φzz(τ)=E{[x(t+τ)+y(t+τ)][x(t)+y(t)]}=φxx(τ)+φxy(τ)+φyx(τ)+φyy(τ)6课后答案网(b)Whenx(t),y(t)areuncorrelated:φxy(τ)=E[x(t+τ)y(t)]=E[x(t+τ)]E[y(t)]=mxmySimilarly:φyx(τ)=mxmyHence:φzz(τ)=φxx(τ)+φyy(τ)+2mxmy(c)Whenx(t),y(t)areuncorrelatedandhavezeromeans:φzz(τ)=φxx(τ)+φyy(τ)Problem2.12:Thepowerspectraldensityoftherandomprocessx(t)is:Φxx(f)=∞−∞φxx(τ)e−j2πfτdτ=N0/2.Thepowerspectraldensityattheoutputofthefilterwillbe:Φyy(f)=Φxx(f)|H(f)|2=N02|H(f)|2Hence,thetotalpowerattheoutputofthefilterwillbe:φyy(τ=0)=∞−∞Φyy(f)df=N02∞−∞|H(f)|2df=N02(2B)=N0BProblem2.13:MX=E[(X−mx)(X−mx)],X=X1X2X3,mxisthecorrespondingvectorofmeanvalues.7课后答案网=E[(Y−my)(Y−my)]=E[A(X−mx)(A(X−mx))]=E[A(X−mx)(X−mx)A]=AE[(X−mx)(X−mx)]A=AMxAHence:MY=µ110µ11+µ1304µ220µ11+µ310µ11+µ13