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5.求下列函数的拉普拉斯变换.
t,2t
(1)
(2)ftt()esin5,,ftlt()sin,,2l
t,4t(3)(4)ftt()1e,,,ftt()ecos4,,
(5)(6ftut()(24),,fttt()5sin23cos2,,
12,t2(7)(8)fttt()32,,,ftt()e,,
(1)
t1ftlttlt()sin[()sin],,,,,,22ll
t1FsLftLltLtlt()(())(sin)[()sin],,,,,,,22ll
112llss,,,,,,,,()2222222222()()lsllslsl,,,
5,2tFsLftLt,,,,
(2)()(())(esin5)2s,,
(2)25
1ttt(3)()(())(1e)
(1)(e)(e)FsLftLtLLtLt,,,,,,,,,,,s1111,,,,,()2ssss,,1
(1)
s,4,4t(4)FsLftLt()(())(ecos4),,,,2(4)16s,,
1,2t,,(5)ut(24),,,0,其他,
st,FsLftLututdt()(())((24))=(24)e,,,,,,0,12sts,,=e=edt,2s
(6)
FsLftLttLtLt()(())(5sin23cos2)5(sin2)3(cos2),,,,,
2103ss,,,,,,53222sss,,,444
13,,,
(1)()1,t222(7)FsLftLt,,,,,()(())(e)33
22,,ss,,()()
1222(8)FsLftLttLtLtLss()(())(32)()3()2
(1)(232),,,,,,,,,,s
6.记,对常数,若,证明sRe()ss,,,LfsFs[]()(),000
st0LfsFss[e]()(),,,0
证明:
ststst,00Lfsftdt[e]()e()e,,,,,0
,()()sstsst,,,00,,,,,,ftdtftdtFss()e()e()0,,00
()nn7记,证明:
FsLfts()[(t)()](),,,LfsFs[]()(),
当n=1时,
,,st,Fsftdt()()e,,,0
,,st,,,Fsftdt()[()e],,,0st,,,,,,,[()e]ftst,,,,,,,,,dttftdtLtft()e(()),,00s,
()nn所以,当n=1时,显然成立。
FsLfts()[(t)()](),,,
(1)1kk,,假设,当n=k-1时,有FsLfts()[(t)()](),,,
现证当n=k时
,,kst,,1
(1)k,dtftdt()()e,,,dFs(),()k0Fs(),,dsds
kst,,1,,,,,,,[()()e]tftkst,,,,,,dttftdt()()e,,00,s
k,,,Lfts[(t)()]()
8.记,如果a为常数,证明:
LfsFs[]()(),
1sLfatsF[()]()(),aa
设,由定义LfsFs[]()(),
,,udu,stLfatfatdtatutdt[()]()e.(,,),,,,,令,0aa
ss,,uu,,,,du1aa,,,,fufudu()e()e,,00aa
s1,F()aa
9.记,证明:
,,,ft()ft()st,,即,,,LFsds[]()e()dtFsds,,,s0stt
,,,,,stst,,Fsdsftdtdsftdsdt()[()e]()[e],,,,,,,,0sss
,,,,1()()ftftstst,,,,,,,,,ftdtdtL()[e]e[]s,,00ttt
10.计算下列函数的卷积
(1)
(2)11,tt,
t(3)(4)sinsinatat,t,e
(5)(6,,()()tft,,sinsinatat,
t解:
(1)1111,,,,dt,,0
t13
(2)tttdt,,,,,,,,(),06
(3)
ttt,,,,,,tttttddd,,,,,,,,,,eeeeee,,,,,,,,000
t,,ttt,,,,,,,,e[e]ee1dt,,0,0
(4)
tt1atataatdataatd,,,,,,,,,,,,,sinsinsinsin()[coscos
(2)],,002
t1,,atatsincos2a22
(5)
tt,,,,,,,,,,()()()()()()()tfttftdtftdt,,,,,,,,,,,,,,00
0t,,0,t,,,,,,()()()()fdfd,,,,,,,,,,,,(),0,,ftt,,0t
tt1sincossincos()[sinsin
(2)]tttdttd,,,,,,,,,,,,,,002
ttt,,,sinin
(2)tstd,,,022
t1t,,,sincos
(2)tt,024
tt1,,,,,ttttsin[coscos()]sin242
11.设函数f,g,h均满足当t<
0时恒为零,证明fgtgft,,,()()()()()()fghtfhtght,,,,,,以及
t0令,,tu=fgtfgftugu,,,,,,,,,,()()d()d,,,,,,tu,,,,证明:
0t
tt,,,,,,,,,,,ftugugfgft()d()d(),,ut,,,,00
t,,,hth()d,,,fgfg,,,(),,,,,,t,,,,0
tt,,,,,,,,,,fhtgh()d()d,,,,t,,,,,00
,,,fhtgft()()
12.利用卷积定理证明
tFs()Lftdt[()],,0s
t,gtftg()(),(0)0,,且gtftdt()(),,0证明:
设,则,LgtsLgtgsLgt[()][()](0)[()],,,,则
Lgt[()],所以Lgt[()],s
tFs()Lftdt[()],,0ds
13.求下列函数的拉普拉斯逆变换.
2ss,8
(1)Fs(),Fs(),
(2)22(4)s,
(1)
(2)ss,,
s1(3)(4)Fs(),Fs(),22(4)s,sss
(1)
(2),,
2s,1ss,,21(5)(6Fs()ln,Fs(),2s,1ss
(1),
s21解:
(1)Fs(),,,
(1)
(2)21ssss,,,,
2111,,,1112ttLLL,,,,,()2()()2eessss,,,,2121
22ss,,8321431,,11FsLLttt()()()sin2cos2,,,,,22222
(2)(4)442(4)42sss,,,
1111(3Fs(),,,,ssssss
(1)
(2)212
(2),,,,
11,,,12tt故LFs,,,(())ee22
ss1412,,(4)Fs()(),,,,,,,222222(4)4(4)42sss,,,
因为
2,1,Lt()sin222,s2
所以
1st,,11,,,,LFsLt(())()sin222,4(4)4s
sgt,111()(5)FsduL()ln()(),,,,,,0suut,,,111其中
11,,1ttgtL,,,,()()eess,,11
,tttteeee,,FsLL()()(),,,tt
,tttteeee,,sht1,ftLFs()(())2,,,,,,ttt
2ss,,21122(6)Fs(),,,,,22sssss
(1)1
(1),,,所以
122,,,,1111LFsLLL,,,,(())()()()2sss,,1
(1)
tttt,,,,,,,tt12e2e2e2e1
14.利用卷积定理证明
st,1Lat[]sin,,22()2saa,
ssa1,,11,,,LL[]()2222222,,,()sasasaa又因为
saLatLat(cos),(sin),,2222sasa,,
所以,根据卷积定理
sa11,1Latat()cossin,,,,2222sasaaa,,
tt111,,,,,,,cossin()[sinsin
(2)]aatadataatd,,,,,,,00aa2
t,,sinat2a
15.利用卷积定理证明
t212,,ty1Ldy,[]ee,0ss,
(1)π证明:
111,,11,,LL[][],s1,sss
(1)
t212,,ty1Ldy,[]ee,0ss,
(1)π因为
1,111t,,112LtL,,(),()es,1sπ
所以,根据卷积定理有
111,,,tt1121,,,1()ttyty222[]eyeeyeLtdydy,,,,,,,00ss
(1),πππ
ttt22222令yu,tytuty,,,2eeeeee,,,,,,,dydudy,,,000πππ
16.求下列函数的拉普拉斯逆变换.
11Fs(),
(1)
(2)Fs(),2242(4)s,ss,,54
2233ss,,s,2(3)Fs(),(4)Fs(),222(45)ss,,
(1)(3)ss,,
(1)
22112(4)14ss,,Fs(),,,,,222222(4)16(4)8(4)sss,,,21214s,,,,,2221648(4)ss,,
2121411s,,,,111故LFsLLttt(())()()sin2cos2,,,,,2221648(4)168ss,,
1111Fs()(),,,4222ssss,,,,54314
(2):
1112,,()2223122ss,,
1112,,,111,,LFsLL(())()()222,,3162ss
11,,sinsin2)tt36
ss,,2211,(3)Fs()(),,,,22222(45)[
(2)1]2
(2)1ssss,,,,,,
1,,12t故LFstt,,,(())esin2
2233ssABCD,,Fs(),,,,,223
(1)(3)13(3)(3)ssssss,,,,,,
113,,,,,,ABCD,,,3442
故
113,3442Fs(),,,,23ssss,,,,13(3)(3)且
1111,,,(),()2,,,,23ssss,,,,3(3)3(3)
113,,,,,13323tttt所以LFstt,,,,,,(())eee3e442
17.求下列微分方程的解
t,,,,
(1)yyyyy,,,,,23e,(0)0,(0)1
,,,
(2)yyttyy,,,,,,,4sin5cos2,(0)1,(0)2
t,,,,(3)yyytyy,,,,,,222ecos2,(0)(0)0
2t,,,,,,,(4)yyyyy,,,,,e,(0)(0)(0)0
(4),,,,,,,,(5)yyyyyyy,,,,,,,20,(0)(0)(0)0,(0)1
(1)设
LytYsLytsYsysYs[()](),[(()]()(0)(),,,,,22,,,LytsYssyysYs[(()]()(0)(0)()1,,,,,
方程两边取拉氏变换,得
12sYssYsYs,,,,,,()12()3()s,1
12s,2(23)()1ssYs,,,,,ss,,11
ss,,22Ys(),,2
(1)(23)
(1)
(1)(3)ssssss,,,,,,
为Y(s)的三个一级极点,则sss,,,,,1,1,3123
3,1stytLYssYss()[()]Re[()e;
],,,,kk,1
stst
(2)e
(2)ess,,,,,,,Re[;
1]Re[;
1]ss
(1)
(1)(3)
(1)
(1)(3)ssssss,,,,,,st
(2)es,,,,Re[;
3]s
(1)
(1)(3)sss,,,
131,,ttt3,,,,eee488
(2)方程两边同时取拉氏变换,得
1s2sYssYs,,,,,,,,()2()45222ss,,12
1s2
(1)()45
(2)sYss,,,,,,,222ss,,12
452ss,Ys(),,,222222
(1)
(1)
(1)
(2)
(1)sssss,,,,,
11112s,,,,,,,2()()s2222222ssssss,,,,,,111211
2s,,,222ss,,12
1ytLYstt()[()]2sincos2,,,,
(3)方程两边取拉氏变换,得
s,12sYssYsYs,,,,,,()2()2()22
(1)1s,,
2
(1)s,2(22)()ssYs,,,2
(1)1s,,
2
(1)1s,,Ys()[],,,222[
(1)1]
(1)1ss,,,,
因为由拉氏变换的微分性质知,若L[f(t)]=F(s),则
LtftFs[()()](),,,
即
,11,LFstfttLFs[()]()()()[()],,,,,,
1,1tLt,,[]esin2s,,
(1)1
2
(1)1s,,,11,LL{}[()],,222[
(1)1]
(1)1ss,,,,
1t,1,,,,,,()[]esintLtt2
(1)1s,,
t故有,,ytt,,,esint
(4)方程两边取拉氏变换,设L[y(t)]=Y(s),得
132,,,sYssysyysYsy,,,,,,,,,,()(0)(0)(0)()(0)s,2
13sYssYs,,,,()()s,2
111Ys(),,,22ssssss,,,,2
(1)
(2)
(1)故
113,,,,,12323ttttytLYs,,,,,,,()[()]eete3te442
(5)设L[y(t)]=Y(s),则
LytsYsysYs[(()]()(0)(),,,,
22,,,LytsYssyysYs[(()]()(0)(0)(),,,,,
323,,,,,,LytsYssysyysYs[(()]()(0)(0)(0)()1,,,,,,,,(4)4324,,,,,,LytsYssysysyysYss[(()]()(0)(0)(0)(0)(),,,,,,,,,,,
方程两边取拉氏变换,,得42sYsssYsYs,,,,,,()2()()0
42(21)()ssYss,,,,
ss1211,Ys()(),,,,,,22222
(1)2
(1)21sss,,,
s111,,11,,,,,,,ytLLtt()[][()]sin222,,
(1)212ss
18.求下列微分方程组的解
t,,xxy,,,e,
(1)xy(0)(0)1,,,t,yxy,,,,322e,,
(2)
,xygt,,2(),,,xxyy(0)(0)(0)(0)0,,,,,,,,,xyy,,,0,
(1)设
LxtXsLytYs[(()](),[(()](),,
LxtsXsxsXs[(()]()(0)()1,,,,,,
LytsYsysYs[(()]()(0)()1,,,,,,,
微分方程组两式的两边同时取拉氏变换,得
1,sXsXsYs,,,,,()1()(),,s,1,2,sYsXsYs,,,,,()13()2(),s,1,
得
s,YssXs()
(1)()...
(1),,,,,s,1,21s,,3()
(2)()1...
(2)XssYs,,,,,,,ss,,11,
(2)代入
(1),得
ss,13()
(2)[
(1)()]XsssXs,,,,,,ss,,11
2sssss,,,,1
(2)12
(1)()ssXs,,,,,sss,,,111
1t故于是有Xsxt()()e...(3),,s,1
(3)代入
(1),得
11stYssyt()
(1)()e,,,,,,,sss,,,111
(2)设
LxtXsLytYsLgtGs[(()](),[(()](),[(()](),,,
,LxtsXsLytsYs[(()](),[(()](),,,,
22,,,,LxtsXsLytsYs[(()](),[(()](),,,,,
方程两边取拉氏变换,得sXssYsGs,,,,()2()()...
(1),
22,,sXssYsYs,,,,,()()()0...2,
(1)
(2),,,s得
sYsGs()()...(3),,,2s,1
t,1?
,,,,ytLYsgttgd()[()]()*coscos,,,,t,,,0(3)代入
(1):
ssXssGsGs,,,,,,()2[()]()2s,1
即:
2221ss,sXsGsGs,,,,,()
(1)()()22ss,,11
21,s12s,,XsGsGs()()(),,,,,,22s,,ss1,,,s,1
,,,,,,,xtLXstgtgtd()[()](12cos)()(12cos)(),,,,0
txtgtd()(12cos)(),,,,,,,,0
tytgtd()()cos(),,,,,,,,0
19.求下列方程的解
t,
(1)xtxtdt()()e23,,,,,,,,0t
(2)yttydt()()(),,,,,,,,0解:
(1)设L[x(t)]=X(s),方程两边取拉氏变换,得
123XsXs()(),,,,2sss,1
123,sXs()[1],,2ss,1
2(23)
(1)352352,,,,,ssssXs(),,,,,,3323sssss
2,,,,,xttt()35
(2)设L[y(t)]=Y(s),方程两边取拉氏变换,得
1YsLtyt()(()),,,2s
11YsYs()(),,,22ss1Ys(),2s,1
111,,,,,,ytLYsLsht()(())()2s1,