HW8-solution.pdf

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1、EE238B1-Winter2012SolutiontoHW#8SolutiontoHomeworkAssignment8SolutiontoProblem1:FromthedefinitionsofFTandinverseFT,Z∞Z∞1jωtjωtf2(t)=F2(ω)edωF1(−ω)=f1(t)edt.2π−∞−∞Thus,ZZZ∞∞∞1jωtf1(t)f2(t)dt=f1(t)F2(ω)edωdt−∞2π−∞−∞ZZZ∞∞∞1jωt1=F2(ω)f1(t)edtdω=F1(−ω)F2(ω)dω.2π−∞−∞2π−∞Bychanging

2、thevariableωto−ωinthelastintegrationoftheaboveformula,wecanshowthatitalsoR∞equals1F(ω)F(−ω)dω.2π−∞12SolutiontoProblem2:ApplyingthedualitypropertytoPair3ofTable4.1,wehave2a−a

3、ω

4、f(t)=⇐⇒F(ω)=2πet2+a2ThesignalenergycanbecalculatedasZ∞Z∞12−2a

5、ω

6、−2aω4π2πEf=(2π)edω=4πedω==.2π−∞02aaThe

7、signalenergycontainedwithinbandwidthBHz(2πBrad/s)isZ2πB
12−2a

8、ω

9、2π−4aπBE[−2πB,2πB]=(2π)edω=1−e.2π−2πBaE[−2πB,2πB]−4aπB−4aπB−ln0.010.366=1−e=99%⇒e=0.01⇒B==Hz.Ef4aπaSolutiontoProblem3:Theunitimpulseresponseis−1−112th(t)=F{H(ω)}=F=eu(−t).2−jωsinceh(t)isanon-causalsignal,thesyst

10、emisnon-causal.(a)f(t)=e−tu(t)⇐⇒F(ω)=1.Thus,1+jω1111Y(ω)=F(ω)H(ω)==+.(1+jω)(2−jω)31+jω2−jωThezero-stateresponseofthesystemtothegiveninputis−11−t12ty(t)=F{Y(ω)}=eu(t)+eu(−t).331EE238B1-Winter2012SolutiontoHW#8(b)f(t)=etu(−t)⇐⇒F(ω)=1.Thus,1−jω1111Y(ω)=F(ω)H(ω)==−.(1−jω)(2−jω)1−

11、jω2−jωThezero-stateresponseofthesystemtothegiveninputis−1t2ty(t)=F{Y(ω)}=eu(−t)−eu(−t).SolutiontoProblem4:test(a)Z1−st−s−ste11−eL{f(t)}=edt=−

12、0=.0ssROC:C(entirecomplexplane).(b)Z∞Z∞−2t−st−(s+2)tL{f(t)}=ecos(5t+θ)edt=ecos(5t+θ)dtZ0Z0∞∞1−(s+2)tj(5t+θ)1−(s+2)t−j(5t+θ)=eedt+eedt202

13、0Z∞Z∞1jθ−(s+2−j5)t1−jθ−(s+2+j5)t=eedt+eedt2020ejθ1−lime−(s+2−j5)te−jθ1−lime−(s+2−j5)tt→∞t→∞=+2s+2−j52s+2+j5Ast→∞,e−(s+2±j5)tconverges(to0)ifandonlyifRes>−2.Thus,1ejθe−jθ(s+2)cosθ−5sinθL{f(t)}=+=2s+2−j5s+2+j5(s+2)2+25ROC:Res>−2.SolutiontoProblem5:test(a)Z1e−st−s−111−e−s−se−s−s

14、t1−sF(s)=tedt=(−st−1)

15、0=e+=.0s2s2s2s2(b)Zπe−ste−sπ+1−stπF(s)=(sint)edt=(−ssint−cost)

16、0=.0s2+1s2+12