信号检测与估计-第三章-答案

SelectedProblemSolutionsfromChapter3Problem3.1(a)Sinceh(t)=ae−at,itfollowsthatH(f)=Z∞−∞h(t)e−2πftdt=aZ∞0e−t(a+2πf)dt=aa+2πf(1)WeknowthatSy(f)=|H(f)|2Sx(f),sowecanwriteSy(f)=a2a2+4π2f2Hx(f)(2)(b)Inthiscase,H(f)=aZT0e−t(a+2πf)dt=ae−aTa+2πfe−2πfT(3)andSy(f)=a2e−2aTa2+4π2f2Sx(f)(4)Problem3.2Usingthedefinitionsofbandpassnoiseprocesses,wecanwriteE{z(ti)z(tj)}=E{[nI(ti)+nQ(ti)][nI(tj)+nQ(tj)]}=E{nI(ti)nI(tj)}−E{nQ(ti)nQ(tj)}+E{nI(tj)nQ(ti)}+E{nI(ti)nQ(tj)}(5)Sinceitisassumedthatthisbandpassprocessisstationary,wecanrewritethiscorrelationasE{z(ti)z(tj)}=RI(τ)−RQ(τ)+[RQI(τ)+RIQ(τ)](6)whereτ=ti−tj.UsingthepropertiesofEq.3.61inthetextcompletestheproof.AsimilarproofcaneasilybewrittenforE{z∗(ti)z∗(tj)}.Problem3.3Using3.68inthetext,itfollowsthatRI(τ)=N0δ(τ),RQ(τ)=N0δ(τ),andRIQ(τ)=0.Inordertogettothedesiredanswer,wemakethefollowingreasonableassumptions:1.Theintegralofcos2πfctfrom0toTisapproximatelyzero.2.Theintegralofsin2πfctfrom0toTisalsoapproximatelyzero.3.Theintegratepartsofthecorrelatorssuppressany“double-frequency”terms.Withtheseassumptions,andusingsometrigonometricidentities,wecanwritex=ZT0y(t)cos2πfctdt=AT2cosθ+12ZT0nI(t)dt(7)Sincethenoiseprocessesarezero-mean,itfollowsthatE{x}=AT2cosθ(8)Similarly,zcanbewrittenasz=AT2sinθ−12ZT0nQ(t)dt(9)Again,thezero-meanaspectoftheprocessgivesthedesiredresult.FromEq.7,wecanwritethevarianceofxasV{x}=14ZT0ZT0E{nI(t1)nI(t2)}dt1dt2(10)Evaluatingtheinnerintegral,wegetV{x}=N04ZT0dt2=N0T4(11)Asimilarderivationcanbedoneforz.UsingEqs.7and9,wecanwritethecrosscorrelationasZT0ZT0E{nI(t1)nQ(t2)}dt1dt2(12)whichiszerobecauseRIQ(τ)=0forallvalues.Problem3.6First,letusdefinearandomvariabler=ρ2(t)=|v(t)|2=(xI(t)+Ar)2+(xQ(t)+Ai)2(13)StartingwithEq.3.110inthetext,wecanderivetheanswerorwecangodirectlytoEq.B.31toseethatp(r)=12σ2exp−r+γ22σ2I0√rγσ2(14)whereγ2=A2i+A2r.Sinceρ=√r,asimplechangeofvariableresultsinp(ρ)=ρσ2exp−ρ2+γ22σ2I0ργσ2(15)Problem3.7(a)FromthestatementofMercer’sTheorem,Rx(t1,t2)=∞Xk=1λkfk(t1)f∗k(t2)(16)fromwhichitfollowsthatRx(t1,t1)=∞Xk=1λkfk(t1)f∗k(t1)(17)Integratingfrom0toT,thisisZT0Rx(t1,t1)dt1=∞Xk=1λkZT0fk(t1)f∗k(t1)dt1=∞Xk=1λk(18)(b)Again,westartwithMercer’sTheoremtogetRx(t1,t2)Rx(t2,t3)=∞Xk=1∞Xl=1λkλlfk(t1)f∗k(t2)fl(t2)f∗l(t3)(19)Integratingovert2,wehaveZT0Rx(t1,t2)Rx(t2,t3)dt2=∞Xk=1λ2kfk(t1)f∗k(t3)(20)Replacingt3byt1,wehaveZT0Rx(t1,t2)Rx(t2,t1)dt2=∞Xk=1λ2kfk(t1)f∗k(t1)(21)Integratingovert1andusingthefactthatRx(t1,t2)=Rx(t2,t1)(22)(forrealfunctions)givesusZT0ZT0R2x(t1,t2)dt1dt2=∞Xk=1λ2k(23)