复变函数与积分变换（修订版-复旦大学）课后的习题答案

复变函数与积分变换（修订版）课后答案（复旦大学出版社）

习题 七

1.证明：如果f(t)满足傅里叶变换的条件，当f(t)为奇函数时，则有

f(t)?解:

a(?)???2π2???0f(t)?cos?tdt?21?π?2π?2π?10t?cos?tdt?22π???10?cos?tdt???0b(?)?sin?td?

2π?π2π?210t?cos?tdt?2?1010td(sin?t)2其中b(?)?当f(t)?????t?sin?t10

?0f?t??sin?tdt

?sin?t?2tdtf(t)为偶函数时，则有

???2sin?4??2π???2sin?π?2sin???42?10t?d(cos?t)10???0a(w)?cos?td?

2?t?cos?t????4cos???10cos?tdt???其中a(?)?证明： 因为f(t)?????0f(t)?cos?tdt

????2?4sin???3?sint,t?6π3.计算函数f(t)??的傅里叶变换. ?2π?1????G(?)ei?td?其中G(?)为f(t)

??0,t?6π解:

F的傅里叶变换

G(?)???????????f(t)e?i?tdt???????f?(?)??????f(t)?e?i?tdt??6π?6πsint?e?i?tdtf(t)?(cos?t?isin?t)dt ??????6π?6πsint?(cos?t?isin?t)dt6π0?f(t)?cos?tdt?i?f(t)?sin?tdt

sint?sin?tdt??2i??当f(t)为奇函数时，f(t)?cos?t为奇函数，从而

isin6π?π(1??)2?????f(t)?cos?tdt?0

4.求下列函数的傅里叶变换 (1)f(t)?e解：

F?f?(?)???t

f(t)?sin?t为偶函数，从而

?????f(t)?sin?tdt?2???0??0f(t)?sin?tdt.

?????f(t)e?i?tdt??????e?|t|?e?i?tdt??????e?(|t|?i?t)dt

故G(?)??2i?f(t)?sin?tdt. 有

?0??et(1?i?)dt????0e?t(1?i?)dt?21??2G(??)??G(?)为奇数。

f(t)?12?(2)f(t)?t?eG(?)?(cos?t?isin?t)d??t2

?????G(?)?ei?td??12???? 解：因为

F[e?t2??=

12π?????G(?)?isin?td??iπ???0G(?)?sin?td?

]?π?e??24.而(e?t2)?e/?t2?(?2t)??2t?e?t2.

所以根据傅里叶变换的微分性质可得G(?)?Ft(?esinπt1?t2所以，当f(t)为奇函数时，有

f(t)????0b(?)?sin?td?.其中b(?)=2π???0?t2f(t)?sin?tdt.?)π?2i??2?4e

同理，当f(t)为偶函数时，有 f(t)?a(?)?(3)f(t)?解：

???0a(?)?cos?td?.其中

??0?π2f(t)?cos?tdt

2??t,2.在上一题中，设f(t)????0,t?1t?1.计算a(?)的

值.

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

G(?)?F(f)(?)???????sinπt1?t2?e?i?tdtL/2GL(?)??????sinπt1?t????2??L/2F(t)e?i?tdt.

?(cos?t?isin?t)dt?12[cos(π??)t?cos(π??)t]1?t22

dt证明当L??时，有

p.v.12π???i??i?sinπt?sin?t1?t1?t22dt??2i???0??0??0cos(π+?)tdt?i?cos(π-?)t1?tdt(利用留数定理)?G??L(?)ei?td??F(t)

?i??sin?,当|?|?π??2?0,当|?|?π.? 对所有的实数t成立. (书上有推理过程) 6.求符号函数 sgnt?变换.

??1,??|t|?1,tt?0t?0(4)f(t)?解:

G(?)??2???011?t11?t444

的傅里叶

?????e?i?tdt??????cos?t1?t4dt?i?????sin?t1?t4解:

dtcos?t1?tdt??????cos?t1?t4因s为Fu(tdt1?(?)π??)(?i?)把.函数

令R(z)=2211?z4gt与nu(t)(作比)较.

，则R(z)在上半平面有两个一级极

不难看出 sgn(t)?u(t)?u(?t).

22(?1?i).

2222点

????(1?i),故：

F[sgn(t)]?F(u(t))?F(u(?t))?1iπ?π??(?)?[1i(??)?π??(??)]?R(t)?edt?2πi?Res[R(z)?e,i?ti?z(1?i)]?2πi?Res[R(z)?e,i?z(?1?i)]

?|?|2

2i?故.

cos?t1?t4?π??(?)??(??)??2i??????dt?Re[?????ei?t41?tdt]?122e?|?|/2(cos|?|2?sin)

7.已知函数f(t)的傅里叶变换

(5) f(t)?解: G(?)??t1?t4

F(?)=π??(???0)??(???0)?,求f(t)

解:

?t????t1?t4?e?i?tdt????f(t)?F(F(?))=-1?2π????????1????π???(???0)??(???0)?ed?i?t?i?t?????1?t????4?cos?tdt?i?4t?sin?t1?t4dt

而F(cos?0t)=??cos?0t?eei?0tdt?e?i?0t??i?t?sin?t1?tdtz1?z4??e2?i?0tdt

?π[?(???0)??(???0)]所以同(4).利用留数在积分中的应用,令R(z)=则

?i???????

f(t)?cos?0t8.设函数f(t)的傅里叶变换F(?)，a为一常数. 证明

t?sin?t1?t?e?|?|/4dt?(?i)Im(?????t?ei?t41?tdt)i22?sin?2. ?[f(at)](?)???F?a?a1??. ?5.设函数F(t)是解析函数，而且在带形区域Im(t)??内有界.定义函数GL(?)为

解:F[f(at)](?)??????f(at)?e?i?tdt?1?a????f(at)?e?i?td(at)

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

当a>0时,令u=at.则 F[f(at)](?)?1a当t?y?o时，若t?0,则f?y??0,故

u?i?a?????f(u)?edu????F?? a?a?11aF(f*g?t?=0.

当a<0时,令u=at,则F[f(at)](?)??故原命题成立.

9.设F????F?f????;证明

?a). 若0?t?f*g?t???2t,0?y?t,则 f(y)g?t?y?dy??0?t0e?y?sin?t?y?dy

若t?F?????F?2,0?t?y??2.?t??2?y?t.

?f??t?????.

证明：

F?f??t????????则f*g?t???tt??2e?y?sin?t?y?dy

t?00?t?t????????????????f??t??ef?u??ef?t??e?i?tdt???du?????f?u??edu??i?u

du??i????u?????f?u??e??i(??)?u???i?????t?dt?F????.?0,?1故f*g?????sint?cost?e?t?,t?2?1??e?t.1?e2?2?2

???210.设F????F?f????，证明：

F12.设u?t?为单位阶跃函数，求下列函数的傅里叶变换.

?1?f?t??e?at?f?t??cos?0t?????1212??F????0??F????0???以及

Fsin?0t?u?t?

?f?t??sin?0t???????F????0??F????0???.

解：G????F?f??????????????e?at?sin?0t?u?t??edt?i?t?i?tdt证明：

?eF?f?t??cos?0t??F?f?t???i?0t+e2?i?0t

????i?0t????0??0ee?at?sin?0t?e?ei?0t?i?t?at?e2i?i?0ti?t1??e0??F?f?t??2??2??e??F?f?t????2???????edt?dt??012i?12??F????0??F????0??????0e???a?i????0???t12i?e???a?i????0???tdt同理：

F?0?a?i????022

?i?t?i?t?e0?e0?f?t??sin?0t??F?f?t???2i??

???F??f?t??e2i12i1i?0t?i?0t????F??f?t??e????F????0??F????0???11.设

?0,f?t????t?e,t?0t?0??sint,g?t????0，?0?t?其他π2

计算f*g?t?. 解：f*g?t???????f(y)g?t?y?dy

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

习题八

1.求下列函数的拉普拉斯变换.

(1)f(t)?sint?cost，

(3)f(t)?sint 2(4)f(t)?t(5)f(t)?sinhbt， 解: (1)f(t)?sint?cost?sin2t

2

L(f(t))?12L(sin2t)?1?22L(f(t))?????0f(t)?edt??st???st???0cost??(t)?edt??sint?u(t)?edt0?st?st???st?????cost??(t)?edt??sint?edt0?stt?0(2)f(t)?e?4t，

?cost?e?1s?12?1?1s?12?s22s?12

4.求图8.5所表示的周期函数的拉普拉斯变换

12s?4s?4

11(2)L(f(t))?L(e?4t)?

2s?4

1?cos2t2(3)f(t)?sint?2

?12

解：

L(fT(t))?T0?st

L(f(t))?L(1?cos2t211122)?L(1)?(cos2t)????2?2222s2s?4s(s?4)

11?fT(t)?e1?et2ldt?1?ass2?as?as(1?e?as)

(4)L(t)?3s

(5)

L(f(t))?L(e?e2bt?bt225. 求下列函数的拉普拉斯变换.

(1)f(t)?11111b?sinlt(2)f(t)?e?2t?sin5t

)?L(e)?L(e)?????22222s?b2s?bs?b

1bt?bt(3)f(t)?1?t?et(4)f(t)?e?4t?cos4t (5f(t)?u(2t?4)

2.求下列函数的拉普拉斯变换.

?2,0?t?1?（1）f(t)??1,1?t?2?0,t?2 ?

(6f(t)?5sin2t?3cos2t

1(7) f(t)?t2?e?t (8) f(t)?t2?3t?2 解:(1)

f(t)??cost,0?t?π(2)f(t)??

0,t?π? t2l?sinlt??12lt2l1[(?t)?sinlt]?sinlt)????2ls222解: (1)

L(f(t))?? ??0f(t)?edt??st?2?edt??edt?(2?e?e01s1?st2?st1F(s)?L(f(t))?L(?s?2s12l?L[(?t)?sinlt]s(s?l)222)

?πs(2)

??12ls?l(l2)???22l(s?l)L(f(t))????0f(t)?edt??cost?edt?(1?e)?20ss?1

?stπ?st1?πs1?e

(2)F(s)?L(f(t))?L(e?2t?sin5t)?5(s?2)?25

1s?L(?t?e)t23.设函数f(t)?cost??(t)?sint?u(t),其中函数

(3)F(s)?L(f(t))?L(1?t?e)?L(1)?L(t?e)?ttu(t)为阶跃函数, 求f(t)的拉普拉斯变换.

解:

?1s?(1s?1)??1s?1(s?1)2

(4)

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

F(s)?L(f(t))?L(e?4t?cos4t)?s?4(s?4)?162

F?(k)(s)?dF(k?1)(s)dsk?1?d??st??0(?t)k?1?f(t)?e?stdtds]dt??1,t?2(5)u(2t?4)??

0,其他?

F(s)?L(f(t))?L(u(2t?4))=?u(2t?4)?e0??st??0?[(?t)k?f(t)?e?s???0(?t)?f(t)?ek?stdt?L[(?t)?f(t)](s)

dt8. 记L[f](s)?F(s)，如果a为常数,证明:

L[f(at)](s)?1sF() aa=?e2??stdt=1se?2s(6)

F(s)?L(f(t))?L(5sin2t?3cos2t)?5L(sin2t)?3L(cos2t)?5?2s?42证明：设L[f](s)?F(s)，由定义

L[f(at)]???3?ss?42?10?3ss?42???0?saf(at)?eu?stdt.(令at?u,t???ua,dt?dua)(7)

1?(1??t1F(s)?L(f(t))?L(t?e)?22?3)3?()23???0f(u)?edua?1a?0f(u)?e?saudu

?1sF()aa(s??)2(s??)2

9. 记L[f](s)?F(s)，证明:

L[f(t)t]?(8)

1222F(s)?L(f(t))?L(t?3t?2)?L(t)?3L(t)?2L(1)?(2s?3s?2)s

6.记L[f](s)?F(s)，对常数s0，若

Re(s?s0)??0，证明L[es0t??sF(s)ds,即

???0f(t)t?e?stdt????sF(s)ds

证明：

?f](s)?F(s?s0)

??sF(s)ds???0???s[f(t)?edt]ds??s?st???0f(t)?[?eds]dts?st?st证明:

L[e?s0t???f](s)?f(t)?e??01?stf(t)?[?et]dt????0f(t)t?edt?L[f(t)t]

es0t?f(t)?e?stdt?(s?s0)t??0(s0?s)tdt???0f(t)?e(n)dt?F(s?s0)n10.计算下列函数的卷积

(1)1?1(2)t?t

t(3)t?e (4)sinat?sinat

(5)?(t??)?f(t) (6sinat?sinat

解：(1)1?1?

7 记L[f](s)?F(s)，证明:F(s)?L[(?t)?f(t)](s) 证明：当n=1时, F(s)????0??f(t)?e?st?stdt

F?(s)?[??0f(t)?edt]??st?1?1d?0t?t 16t

3???0?[f(t)?e?s]dt?????0

t?f(t)?edt??L(t?f(t))n?st(2)t?t? (3)

?t0??(t??)d??所以，当n=1时, F(n)(s)?L[(?t)?f(t)](s)显然

t?e?tt?t0??ed??e????ed???e????de00t0t??t0t??tt??tt??成立。

假设，当n=k-1时, 有

F(k?1)??e[?e]??ed??e?t?1(4)

??

(s)?L[(?t)k?1?f(t)](s)

现证当n=k时

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

1sinat?sinat??sina??sina(t??)d????[cosat?cos(2a??at)]d?002tt(1)F(s)?

(2)F(s)?s(s?1)(s?2)s?8(s?4)222?tsinat?cos2at2a21

(5)

?(t??)?f(t)???(t??)?f(t??)d?????(t??)?f(t??)d(t??)00tt(3)F(s)?

1s(s?1)(s?2)

0t0,t?????t?(?)?f(?)d???0?(?)?f(?)d???f(t??),0???t (6)

sint?cost??t0sin??cos(t??)d??1t2?0[sint?sin(2??t)]d??t2sint?tt2?0sin(2??t)d??t2sint?14cos(2??t)t0?t2sint?14[cost?cos(?t)]?t2sint11.设函数f, g, h均满足当t<0时恒为零，证明

f?g(t)?g?f(t)以及

(f?g)?h(t)?f?h(t)?g?h(t)

证明：

f?g(t)??t0f(?)g?t???d??令?t????=u???0tf(t?u)?g?u?du??t0f(t?u)?g?tu?du??0g(?)?f?t???d??g?f(t)?tf?g??h(t)??0?f(?)?g?????h?t???d???t0f????h(t??)?d???t0g(?)?h?t???d??f?h(t)?g?f(t)12.利用卷积定理证明

L[?tf(t)dt]?F(s)0s

g(t)??tf(t)dt证明：设

0g?(t)?f(t),且g(0)?0，则

L[g?(t)]?sL[g(t)]?g(0)?sL[g(t)]，则

L[g(t)]?L[g?(t)]s，所以

L[?tf(t)dt]?F(s)0ds

13. 求下列函数的拉普拉斯逆变换.

(4)F(s)?s(s2?4)2

(5)F(s)?lns?1s?1

2(6F(s)?s?2s?1s(s?1)2

解：(1)F(s)?s(s?1)(s?2)?21s?2?s?1

L?1(2L?1(1?12tts?2?1s?1)?2s?2)?L(1s?1)?2e?e

(2) F(s)?s2?8?32L?1(2)?1L?1(s?4(s2?4)24s2?42(s2?4)2)?34sin2t?12tcos2t

(3F(s)?1s(s?1)(s?2)?12s?11s?1?2(s?2)

故L?1(F(s))?1?2t2?e?t12e? (4)F(s)?s??4s(s2?4)2??14(s2?4)2??14?(2s2?22)?因为

L?1(2s2?22)?sin2t 所以

L?1(F(s))?L?1(?1s4?t(s2?4)2)?4sin2t

(5)

F(s)?lns?1(11g(t)s?1???0u?1?u?1)du??L(t)

其中

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

g(t)?L(?11s?1?1s?1)?e?t?e

tL(?11s)?1πt?12,L(?11s?1t)?e

所以 F(s)??L(e?t?ett)?L(e?ettt?t所以，根据卷积定理有

)

L[?11s(s?1)et]?1π?t?12?e?2t2?πett0yet?12(t?y)dy?21π2etet?t0y2edy?y2?1?yf(t)?L(F(s))???1e?t?e?e?et?t?2?sht

?2?tedy??????y令y?u?e?u2du??tedyttt2(6)F(s)?s?2s?1122s(s?1)2??s?s?1?(s?1)2

所以

L?1(F(s))?L?1(?1?12?12s)?L(s?1)?L((s?1)2)??1?2et?2tet?2tet?2et?1

14.利用卷积定理证明 L?1[s(s2?a2)]?t2a?sinat

证明： L?1[s(s2?a2)2]?L?1(ss2?a2?a1s2?a2?a)

又因为

L(cosat)?ss2?a2,L(sinat)?as2?a2

所以，根据卷积定理

L?1(sa11s2?a2?s2?a2?a)?cosat?asinat??tcosa??1?sin(at?a?)d1td0a??a?1[sinat?sin(2a02??at)]??t2a?sinat15.利用卷积定理证明

L?1[1?2t2s(s?1)]πet?0e?ydy

证明：

L?1[1?111s(s?1)]?L[s?s?1] L?1[1s(s?1)]?2ett2π?0e?ydy

因为

π0π0π0

16. 求下列函数的拉普拉斯逆变换.

(1)F(s)?1(2)(s2?4)2F(s)?142

s?5s?4

(3)F(s)?s?2(s2?4s?5)2 2(4)F(s)?2s?3s?3(s?1)(s?3)2

解:(1) F(s)?112(s2?4)1s2?4(s2?4)2?16?(s2?4)2?8?(s2?4)2

?12?1s2?416?s2?48?(s2?4)2故

2L?1(F(s))?112?1s?416L?(s2?4)?18L((s2?4)2)?116sin2t?18t?cos2t

(2)：

F(s)?1s4?5s2?4?13(1s2?1?1s2?4)?13(112

s2?1?2s2?22)L?1(F(s))?1L?1(113s2?1)?1L?(26s2?22)

?13sint?16sin2t)(3)

F(s)?s?2s?211(s2?4s?5)2?[(s?2)2?1]2??2((s?2)2?1)?

故L?1(F(s))?12t?e?2t?sint

(4)

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

F(s)??A?2s?3s?3(s?1)(s?3)14,B??14223?As?132?Bs?3?C(s?3)2?D(s?3)3y(t)?L[Y(s)]??Res[?1?Res[Y(s)?ek?1stst;sk](s?2)?est,C?,D?3

(s?2)?e(s?1)(s?1)(s?3)(s?2)?e3818st;?1]?Res[(s?1)(s?1)(s?3);1]

?Res[故

?344?2F(s)???23s?1s?3(s?3)(s?3)113??(s?1)(s?1)(s?3)?t;?3]14e?e?te?3t(2) 方程两边同时取拉氏变换,得

s?Y(s)?s?2?Y(s)?4?221s?12?5?ss?222

且

(1s?3)???1(s?3)2(s?1)Y(s)?4?,(1s?3)???2?1(s?3)31s?122?5?ss?25s22?(s?2)?s?2(s?1)s22Y(s)?4(s?1)(s?1)1?122?(s?1)(s?2)12222

?2(所以

L(F(s))??1s?1s?122?12)?s?(s?1s?2?12)?2s?1s?1?22

14e?t?14e?3t?32t?e?3t?3t?e2?3t??

s?1s?2?s22

17.求下列微分方程的解

(1)y???2y??3y?e?t,y(0)?0,y?(0)?1(2)y???y??4sint?5cos2t,y(0)??1,y?(0)??2

(3)y???2y??2y?2et?cos2t,y(0)?y?(0)?0

(4)y????y??e2t,y(0)?y?(0)?y??(0)?0 (5)y(4)?2y???y?0,y(0)?y?(0)?y???(0)?0,y??(0)?1

解: (1)设

L[y(t)]?Y(s),L[(y?(t)]?sY(s)?y(0)?sY(s),L[(y??(t)]?sY(s)?sy(0)?y?(0)?sY(s)?122y(t)?L[Y(s)]??2sint?cos2t

(3)方程两边取拉氏变换,得

s?Y(s)?2s?Y(s)?2Y(s)?2?2s?1(s?1)?12

(s?2s?2)Y(s)?Y(s)?2(s?1)[(s?1)?1]222(s?1)(s?1)?122??[1(s?1)?12]?

因为由拉氏变换的微分性质知，若L[f(t)]=F(s),则

L[(?t)?f(t)]?F?(s)方程两边取拉氏变换,得

s?Y(s)?1?2s?Y(s)?3Y(s)?(s?2s?3)Y(s)?22

即

1s?1s?2s?1

L[F?(s)]?(?t)?f(t)?(?t)?L[F(s)]?1?11s?1

因为L?1[所以

L{?1?1?

1(s?1)?12]?e?sintt

Y(s)?s?2(s?1)(s?2s?3)2?s?2(s?1)(s?1)(s?3)2(s?1)[(s?1)?1]?12}??L[(21?11(s?1)?1t2)?]s1??1,s2?1,s3??3为Y(s)的三个一级极点,

??(?t)L[则

(s?1)?1t2]?t?e?sint

故有y?t??t?e?sint

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

(4)方程两边取拉氏变换,设L[y(t)]=Y(s),得

s?Y(s)?s?y(0)?s?y?(0)?y??(0)?s?Y(s)?y(0)?s?Y(s)?s?Y(s)?Y(s)?1?12332得

s?Y(s)?(s?1)X(s)?...(1)??s?1? ?3X(s)?(s?2)?Y(s)?2?1?s?1...(2)?s?1s?1?1s?21s?2?1s(s?2)(s?1)2

(2)代入(1)，得

3X(s)?(s?2)?[(s?1)X(s)?(s?s?1)X(s)?2s?2s(s?1)ss?1?]?2s?1s?1s?1故

y(t)?L[Y(s)]??11e?t?1e?2t?3s?1s?1t?e?3t?3t?e2?3t

?s(s?2)s?1s?s?1

442(5)设L[y(t)]=Y(s),则

L[(y?(t)]?sY(s)?y(0)?sY(s),L[(y??(t)]?s2?Y(s)?sy(0)?y?(0)?s2Y(s)L[(y???(t)]?s3?Y(s)?s2?y(0)?sy?(0)?y??(0)?s3Y(s)?1L[(y(4)(t)]?s4?Y(s)?s3?y(0)?s2?y?(0)?sy??(0)?y???(0)?s4?Y(s)?s

方程两边取拉氏变换,,得

s4?Y(s)?s?2s2?Y(s)?Y(s)?0(s4?2s2?1)?Y(s)?sY(s)?s(s2?1)2?12?2s(s2?1)2??12?(1s2?1)?故

y(t)?L?1[s]?L?1[?1?(11(s2?1)22s2?1)?]?2t?sint18.求下列微分方程组的解

(1)??x??x?y?et ?x(0)?y(0)?1??y??3x?2y?2?et

?x??2y??g(t)(2) ?x???y???y?x(0)?x?(0)?y(0)?y?(0)?0?0

解:(1) 设

L[(x(t)]?X(s),L[(y(t)]?Y(s)L[(x?(t)]?s?X(s)?x(0)?s?X(s)?1L[(y?(t)]?s?Y(s)?y(0)?s?Y(s)?1, 微分方程组两式的两边同时取拉氏变换,得 ???s?X(s)?1?X(s)?Y(s)?1?s?1? ?s?Y(s)?1?3X(s)?2Y(s)?2?s?1故X(s)?1于是有x(t)?ets?1...(3)(3)代入(1)，得

Y(s)?(s?1)?1s?1?ss?1?1s?1?y(t)?et

(2)设

L[(x(t)]?X(s),L[(y(t)]?Y(s),L[(g(t)]?G(s)L[(x?(t)]?s?X(s),L[(y?(t)]?s?Y(s)L[(x??(t)]?s2?X(s),L[(y??(t)]?s2?Y(s),方程两边取拉氏变换,得

?s?X(s)?2s?Y(s)

?G(s)...(1)??s2?X(s)?s2?Y(s)?Y(s)?0...?2?(1)?s?(2),得

Y(s)??ss2?1?G(s)...(3)

?y(t)?L?1[Y(s)]??g(t)*cost???tg0?cos?t???d?

(3)代入(1)：

s?X(s)?2s?[?ss2?1?G(s)]?G(s)即：2s?X(s)?(1?2ss2?1)G(s)?1?s2s2?1?G(s)

2X(s)?1?s?s?G(s)??s2?1??1?s?2s1?s2???G(s)所以

?x(t)?L?1[X(s)]?(1?2cost)?g(t)??t0(1?2cos?)?g(t??)d?故

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复变函数与积分变换（修订版）课后答案（复旦大学出版社）

x(t)??t0(2)设L[y(t)]=Y(s), 方程两边取拉氏变换,得

(1?2cos?)?g(t??)d?ty(t)???g(?)?cos(t??)d?0Y(s)?L(t?y(t))?1s22

Y(s)?Y(s)?19.求下列方程的解

(1)x(t)??x(t??)?e?d??2t?30t1s2?Y(s)?11s

解：(1)设L[x(t)]=X(s), 方程两边取拉氏变换,(2)y(t)??(t??)?y(?)d??t0ts?1?1?12?y(t)?L(Y(s))?L(1s?12)?sht得

X(s)?X(s)?1s?1?2s2?3sX(s)[1?12?3ss?1]?s2X(s)?(2?3s)(s?1)??3s2?5s?2s3s3??3s?5s2?2s3?x(t)??3?5t?t2

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