拉氏变换习题课

1!)(1)(1))((1))((1)1(mmatsmtasetstusLLLLL)(2)()(2)(1))((1)())(()(2)1()(witawetiwwtuwmmmiatFFFFF))()(()(sin))()(()(cosawawiatawawatFF2222)(cos)(sinassatasaatLLP51EX5.求下列函数的付利叶变换:））（）（（）（）（））（）（）（解：00220002|}1{21|}1{21})(()({21)(2)(sin000000FFFF)(sin)()10tutwtf)(sin)()20tutwetft202020200)(0000)(||)(sinsin)(sin)(sinwiLF）（解：)(cos)()30tutwetft2022020)(0000)(||)(coscos)(cos)(coswiwiwwsiwtwdtetwdtetutwetutweiwsiwstiwiwtttLF）（解：))()(1(|)1(|)()(00)(00000000）（）（）（解：FF)()()500ttuetftiw))()(1(|)1(|)()(020000wwi）（）（）（解：FF)()()60ttuetftiw||LLL-4ts→s+4s→s+4221.8ftecos4tcos4tss+4s+16s+4+16LLL22p921.6ft=5sin2t-3cos2t2s=5×-3×s+4s+4LL-t-t1.11ft=u1-e解：u1-e=ut1f(t)=u(t)=s2.为质LLLL1sint1因=arctan，所以由相似性，有tssinat11=arctan,sataa1sinat1a即=arctan,atassinata所以=arctants))()((batubatfL))((1|))((1|))()((1))()(()())((asFeasFeabtubtfabatubatfsFtfabsassbsassLLL设解：为质质LLL-ats→s+a2.4因ft=Fs,由相似性，有tf=aFasa在利用位移性，tef=aFas|a=aFa(s+a3.21因为（由位移性质）数质-3t2-3t222esin2t=s+3+4所以利用像函的微分性，有4s+3d2tesin2t=-=dss+3+4s+3+4LL21213tsdds积质所以t-3t0-3t2t-3t022222由分性，esin2tdt112=esin2tsss+3+4esin2tdt=23s121sss+3+4s+3+4LLL,-1-1-1-tt13ft=-F's,t1ds+1所以ft=-lntdss-11111=--=-e-ets+1s-1tLLL2t积质LLt-3t0-3t24由分性，esin2tdt4s+311=tesin2tsss+3+4数质LL∞s∞∞s22s1利用象函的微分性，有sinkt=sinktds=tksπssds=arctan|=-arctan=arccots+kk2kk2LL-3t∞-3ts∞22sesin2t=esin2tdst2s+3=ds=arccot(s+3)+4204tdtLL-3t-3tesin2t1esin2t=tst1s+3arccots2-1-1-1-13sfttdttttLLLL∞22s2t-tFss11ds2s-1s-1t=e-e41111-4s-14s+1111数变换atbtp1002.求下列函的Laplace逆：s2Fs=s-as-b1ab解：A部分分式法:Fs=-a-bs-as-b1abae-beFs=-=a-bs-as-ba-bLLL1B留数法：k122sts=sk=1ststatbt1212Fs=ResFseseseab=+e+es-bs-aa-ba-bL112222211222110312141111coscos2313233sssssssssttssLLLL11003.2p-1-2ts2=1-=δt-2es+2s+2LL222111003.6ln211211dpsdsssssss22-1-1-12-1-tts1sln=-t12s-ts-111=-=-e+e-2ttLLLL1003.(8)p计算LLLLLL-122-122-1-122-1-122-t-t-t1s+2s+21s+2s+211=*s+2s+2s+2s+211=*s+1+1s+1+1=esint*esint1=esint-tcost21111221113sssut由延迟性质：LL=LLLL-22-2s-2s22-sτt→t-τ1+es1e1e++sseFs=ft-τut-τFs|11121212|22ssstuttutts所以LLLL-2-222t→t-21+e1e+ss111053.p2222222t0t0ss+a1as1=sinat*cosataas+as+a1=sinaτcosat-τdτa1=[sinat+cos2aτ-at]dτ2at=sinat2aLL54对两边进变换t-t-∞∞0∞-t-t-iωtt-iωt-t-iωt-∞-∞0∞∞-1-iωt-1+iωt20022222p651x't-4xtdt=e解：e=eedt=eedt+eedt112=edt+edt=+=1-iω1+iω1+ω原方程行付氏得：12iωXsXs=iω1+ω-2iω1所以Xs=1+ω4+ω12iω2iω=-34+ω1+ωF11转质-βtβt2t-2t-tt2tt-t-2t11=ute,=ute翻性β+iωβ-iω1所以xt=u-te-ute+ute-u-te31e-et03=0t=01e-et03FF112212iω2iωxt=-34+ω1+ω11111=-+-32-iω2+iω1+iω1-iωFF54对两边进变换t-t-∞∞0∞-t-t-iωtt-iωt-t-iωt-∞-∞0∞∞-1-iωt-1+iωt20022222p651x't-4xtdt=e解：e=eedt=eedt+eedt112=edt+edt=+=1-iω1+iω1+ω原方程行付氏得：12iωXsXs=iω1+ω-2iω1所以Xs=1+ω4+ω12iω2iω=-34+ω1+ωF11转质-βtβt2t-2t-tt2tt-t-2t11=ute,=ute翻性β+iωβ-iω1所以xt=u-te-ute+ute-u-te31e-et03=0t=01e-et03FF112212iω2iωxt=-34+ω1+ω11111=-+-32-iω2+iω1+iω1-iωFF