拉氏变换习题课ppt课件

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1、11!)(1)(1)(1)(1)1(mmatsmtasetstusL LL LL LL LL L)(2)()(2)(1)(1)()()(2)1()(witawetiwwtuwmmmiat F FF FF FF FF F2)()()(sin)()()(cosawawiatawawat F FF F2222)(cos)(sinassatasaat L LL L3P51 EX5.求下列函数的付利叶变换求下列函数的付利叶变换:）（）（）（）（）（）（）（解解：00220002|121|121)()(21)(2)(sin000000wwwwiwwwwiwiwiwituetueituieetutwwwww

2、wwtiwtiwtiwtiw F FF FF FF F)(sin)()10tutwtf 4)(sin)()20tutwetft 202020200)(0000)(|)(sinsin)(sin)(sinwiwwwswtwdte twdtetutwetutweiwsiwstiwiwttt L LF F）（解解：5)(cos)()30tutwetft 2022020)(0000)(|)(coscos)(cos)(coswiwiwwsiwtwdtetwdtetutwetutweiwsiwstiwiwttt L LF F）（解解：6)()(1(|)1(|)()(00)(00000000wwwwiewiw

3、ettuttuewwitwwwwitwwwtiw ）（）（）（解解：F FF F)()()500ttuetftiw 7)()(1(|)1(|)()(020000wwiwwwiwittuttuewwwwwwtiw ）（）（）（解解：F FF F)()()60ttuetftiw 8|L LL LL L-4 4t ts ss s+4 4s ss s+4 42 22 21 1.8 8f f t te e c co os s 4 4t tc co os s 4 4t ts ss s+4 4s s+1 16 6s s+4 4+1 16 6 L LL LL L2 22 2p p9 92 21 1.6 6f

4、f t t=5 5s s i i n n2 2t t-3 3c co os s 2 2t t2 2s s=5 5-3 3s s+4 4s s+4 4L LL L-t t-t t1 1.1 1 1 1f ft t=u u1 1-e e解解：u u1 1-e e=u ut t1 1f f(t t)=u u(t t)=s s910 2.为质L LL LL LL L1 1s si i n nt t1 1因因=a ar rc ct ta an n，所所以以由由相相似似性性，有有t ts ss si i n na at t1 11 1=a ar rc ct ta an n,s sa at ta aa a1

5、 1s si i n na at t1 1a a即即=a ar rc ct ta an n,a at ta as ss si i n na at ta a所所以以=a ar rc ct ta an nt ts s11 )()(batubatfL L)(1|)(1|)()(1)()()()(asFeasFeabtubtfabatubatfsFtfabsassbsass L LL LL L 设设解解：12 为质质L LL LL L-a at ts ss s+a a2 2.4 4 因因f f t t=F F s s,由由相相似似性性，有有t tf f=a aF F a as sa a在在利利用用位位

6、移移性性，t te ef f=a aF F a as s|a a=a aF F a a(s s+a a133.14 21因为（由位移性质）数质-3 3t t2 2-3 3t t2 22 22 2e es s i i n n2 2t t=s s+3 3+4 4所所以以利利用用像像函函的的微微分分性性，有有4 4 s s+3 3d d2 2t t e es s i i n n2 2t t=-=d ds ss s+3 3+4 4s s+3 3+4 4LL 21213tsdds 积质所以t t-3 3t t0 0-3 3t t2 2t t-3 3t t0 02 22 22 22 22 2 由由分分性性

7、，e es si in n2 2t td dt t1 11 12 2=e es si in n2 2t ts ss ss s+3 3+4 4e es si in n2 2t td dt t=2 2 3 3s s1 12 21 1s ss ss s+3 3+4 4s s+3 3+4 4LLL15 ,-1 1-1 1-1 1-t tt t1 13 3f f t t=-F F s s,t t1 1d ds s+1 1所所以以f f t t=-l l n nt td ds ss s-1 11 11 11 11 1=-=-e e-e et ts s+1 1s s-1 1t tL LL LL L 2t积质

8、LLt t-3 3t t0 0-3 3t t2 24 4 由由分分性性，e es s i i n n2 2t t d dt t4 4 s s+3 31 11 1=t t e es s i i n n2 2t ts ss ss s+3 3+4 41617 数质LLs ss s2 22 2s s1 1 利利用用象象函函的的微微分分性性，有有s s i i n nk kt t=s s i i n nk kt td ds s=t tk ks ss ss sd ds s=a ar rc ct t a an n|=-a ar rc ct t a an n=a ar rc cc co ot ts s+k k

9、k k2 2k kk k 2LL-3 3t t-3 3t ts s2 22 2s se e s s i i n n2 2t t=e e s s i i n n2 2t t d ds st t2 2s s+3 3=d ds s=a ar rc cc co ot t(s s+3 3)+4 42 218 04tdtLL-3 3t t-3 3t te es s i i n n2 2t t1 1e es s i i n n2 2t t=t ts st t1 1s s+3 3a ar rc cc co ot ts s2 2 -1-1-1-13sf ttdttttLLLL2 22 2s s2 2t t-t

10、tF F s ss s1 11 1d ds s2 2 s s-1 1s s-1 1t t=e e-e e4 41 11 11 11 1-4 4 s s-1 14 4 s s+1 119 111数变换a at tb bt tp p1 10 00 02 2.求求下下列列函函的的L La ap pl l a ac ce e逆逆：s s2 2F F s s=s s-a a s s-b b1 1a ab b解解：A A 部部分分分分式式法法:F F s s=-a a-b b s s-a as s-b b1 1a ab ba ae e-b be eF F s s=-=a a-b bs s-a as s-b

11、 ba a-b bL LL LL L 1B留数法：k k1 12 22 2s s t ts s=s sk k=1 1s st ts st ta at tb bt t1 12 21 12 2F F s s=R R e es s F F s se es se es se ea ab b=+e e+e es s-b bs s-a aa a-b ba a-b bL L20 112222211222110312141111coscos2313233sssssssssttssL LL LL LL L21 1100 3.2p-1 1-2 2t ts s2 2=1 1-=t t-2 2e es s+2 2s

12、s+2 2L LL L 22211100 3.6ln211211dpsdsssssss2 22 2-1 1-1 1-1 12 2-1 1-t tt ts s1 1s sl l n n=-t t1 12 2s s-t ts s-1 11 11 1=-=-e e+e e-2 2t tt t L LL LL LL L221003.(8)p计 算 L LL LL LL LL LL L-1 12 22 2-1 12 22 2-1 1-1 12 22 2-1 1-1 12 22 2-t t-t t-t t1 1s s+2 2s s+2 21 1s s+2 2s s+2 21 11 1=*s s+2 2s

13、s+2 2s s+2 2s s+2 21 11 1=*s s+1 1+1 1s s+1 1+1 1=e es si in nt t*e es si in nt t1 1=e es si in nt t-t tc co os st t2 223 1111221113sssu t由延迟性质：L LL L=L LL LL LL L-2 22 2-2 2s s-2 2s s2 22 2-s s t tt t-1 1+e es s1 1e e1 1e e+s ss se eF F s s=f f t t-u u t t-F F s s|11121212|22ssstu ttu tts 所以L LL LL

14、 LL L-2 2-2 22 22 2t tt t-2 21 1+e e1 1e e+s ss s2425 111053.p2 22 22 22 22 22 22 2t t0 0t t0 0s ss s+a a1 1a as s1 1=s si in na at t*c co os sa at ta aa as s+a as s+a a1 1=s si in na a c co os sa a t t-d da a1 1=s si in na at t+c co os s 2 2a a-a at t d d2 2a at t=s si in na at t2 2a aL LL L2627282

15、9 54对两边进变换t t-t t-0 0-t t-t t-i i t tt t-i i t t-t t-i i t t-0 0-1 1-i i t t-1 1+i i t t2 20 00 02 22 22 22 22 2p p6 65 51 1x x t t-4 4x x t t d dt t=e e解解：e e=e ee ed dt t=e e e ed dt t+e e e ed dt t1 11 12 2=e ed dt t+e ed dt t=+=1 1-i i 1 1+i i 1 1+原原方方程程行行付付氏氏得得：1 12 2i i X X s sX X s s=i i 1 1+

16、-2 2i i 1 1所所以以X X s s=1 1+4 4+1 12 2i i 2 2i i=-3 3 4 4+1 1+F30 11转 质-t tt t2 2 t t-2 2 t t-t tt t2 2t tt t-t t-2 2 t t1 11 1=u u t te e,=u ut te e翻翻 性性+i i-i i 1 1所所以以x x t t=u u-t te e-u u t te e+u u t te e-u u-t te e3 31 1e e-e et t 0 03 3FF 112 22 21 12 2i i 2 2i i x x t t=-3 34 4+1 1+1 11 11 1

17、1 11 1=-+-3 32 2-i i 2 2+i i 1 1+i i 1 1-i i FF31 54对两边进变换t t-t t-0 0-t t-t t-i i t tt t-i i t t-t t-i i t t-0 0-1 1-i i t t-1 1+i i t t2 20 00 02 22 22 22 22 2p p6 65 51 1x x t t-4 4x x t t d dt t=e e解解：e e=e ee ed dt t=e e e ed dt t+e e e ed dt t1 11 12 2=e ed dt t+e ed dt t=+=1 1-i i 1 1+i i 1 1+

18、原原方方程程行行付付氏氏得得：1 12 2i i X X s sX X s s=i i 1 1+-2 2i i 1 1所所以以X X s s=1 1+4 4+1 12 2i i 2 2i i=-3 3 4 4+1 1+F32 11转 质-t tt t2 2 t t-2 2 t t-t tt t2 2t tt t-t t-2 2 t t1 11 1=u u t te e,=u ut te e翻翻 性性+i i-i i 1 1所所以以x x t t=u u-t te e-u u t te e+u u t te e-u u-t te e3 31 1e e-e et t 0 03 3FF 112 22 21 12 2i i 2 2i i x x t t=-3 34 4+1 1+1 11 11 11 11 1=-+-3 32 2-i i 2 2+i i 1 1+i i 1 1-i i FF