拉氏变换习题课

1、1!)(1)(1)(1)(1)1( mmatsmtasetstusl ll ll ll ll l )(2)()(2)(1)(1)()()(2) 1 ()(witawetiwwtuwmmmiat f ff ff ff ff f)()()(sin)()()(cosawawiatawawat f ff f2222)(cos)(sinassatasaat l ll lp51 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)(sin000000wwwwiwwwwiwiwiwituetueituieetutw

2、wwwwwwtiwtiwtiwtiw f ff ff ff f)(sin)() 10tutwtf )(sin)() 20tutwetft 202020200)(0000)(| )(sinsin)(sin)(sinwiwwwswtwdte twdtetutwetutweiwsiwstiwiwttt l lf f）（解解：)(cos)() 30tutwetft 2022020)(0000)(| )(coscos)(cos)(coswiwiwwsiwtwdtetwdtetutwetutweiwsiwstiwiwttt l lf f）（解解：)()(1(| )1(|)()(00)(00000000ww

3、wwiewiwettuttuewwitwwwwitwwwtiw ）（）（）（解解：f ff f)()() 500ttuetftiw )()(1(|)1(|)()(020000wwiwwwiwittuttuewwwwwwtiw ）（）（）（解解：f ff f)()() 60ttuetftiw |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 l

4、l ll l2 22 2p p9 92 21 1. .6 6f 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 s 2.为质l ll ll ll l1 1s si i n nt t1 1因因= = a ar rc

5、 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 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 s )()(batubatfl l)(1|)(1|)()(1)()()()(asfeasfeabtubtfabatubatfsftfabsassbsass l ll ll l 设设解解

6、： 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在在利利用用位位移移性性，t te ef f= = a af f a as s | |a a= = a af f a a( (s s+ + a a3. 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+

7、+ 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，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

8、 ss s+ + 3 3+ + 4 4s s+ + 3 3+ + 4 4lll ,- - 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llt t- -3 3t t0 0- -3 3t t2 24 4 由由分分性性，e es s i i n n2

9、 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 4 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 kk k2 2k kk k 2ll- -3

10、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 2 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 2

11、s s2 2t t- - t 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 1 111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- -

12、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 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 l 1

13、12222211222110312141111coscos2313233sssssssssttssl ll ll ll l 1100 3. 2p- -1 1- -2 2t ts s2 2= =1 1- -= =t t - - 2 2e es s+ + 2 2s 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

14、- - 2 2t tt t l ll ll ll l1003.(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 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

15、nt t * * e es si in nt t1 1= =e es si in nt t - - t tc co os st t2 2 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 ll l- -2 2- -2 22 22

16、2t tt t- -2 21 1+ + e e1 1e e+ +s ss s 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 t

17、2 2a al ll l 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= =+ +=

18、=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+ +- -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+ +f 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

19、 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 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