F?f?t??cos?0t?????以及
1F????0??F????0?????21?F????0??F????0???. 2?
?1?f?t??e?atsin?0t?u?t?
F?f?t??sin?0t?????证明：
解：G????F?f??????e?at?sin?0t?u?t??e?i?tdt??????e?at?sin?0t?e?i?tdt0???ei?0t+e?i?0t?F?f?t??cos?0t??F?f?t????2?1???ei?0t?e?i?0t????F?f?t????F?f?t????2??2??2??1??F????0??F????0???2?同理：
?ei?0t?e?i?0t?F?f?t??sin?0t??F?f?t???2i??1??F??F?f?t??ei?0t?f?t??e?i?0t??????2i1??F????0??F????0???2i?11.设
ei?0t?e?i?0t?i?t??e??edt02i1?????a?i????0???t1?????a?i????0???t??edt??edt2i02i0?0? ?a?i??2??02???at习题八
1.求下列函数的拉普拉斯变换.
(1)f(t)?sint?cost，(3)f(t)?sin2t
(4)f(t)?t，(5)f(t)?sinhbt
2(2)f(t)?e?4t，
π?t?0sint,?0,0?t??f?t????tg?t???2
0，?e,t?0?其他?计算f*g?t?. 解：f*g?t???????f(y)g?t?y?dy
当t?y?o时，若t?0,则f?y??0,故
1sin2t 2 1121L(f(t))?L(sin2t)??2?222s?4s?4
11?4t(2)L(f(t))?L(e)?
2s?4
1?cos2t2(3)f(t)?sint?2
解: (1)f(t)?sint?cost?f*g?t?=0.
若0?t?
1?cos2t1111122L(f(t))?L()?L(1)?(cos2t)????2?22222s2s?4s(s?4)
?2t0,0?y?t,则
tf*g?t???f(y)g?t?y?dy??e?y?sin?t?y?dy
02(4)L(t)?3s
(5)
2若t??2,0?t?y??2.?t??2?y?t.
ebt?e?bt1bt1?bt1111bL(f(t))?L()?L(e)?L(e)?????222222s?b2s?bs?b
2.求下列函数的拉普拉斯变换.
则f*g?t???tt??2e?y?sin?t?y?dy
t?0?0,?1??t,?0?t???故f*g?t?? sint?cost?e?22?1???e?t.t?1?e2?22?2,0?t?1?（1）f(t)??1,1?t?2?0,t?2 ?(2)f(t)??解: (1)
?????cost,0?t?π
?0,t?π?st1?st212.设u?t?为单位阶跃函数，求下列函数的傅里叶变换.
1L(f(t))??f(t)?edt??2?edt??e?stdt?(2?e?s?e?2s)001s
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(2)
L(f(t))????011?e?πs?πsf(t)?edt??cost?edt?(1?e)?20ss?1
?stπ?st(2)F(s)?L(f(t))?L(e?2t?sin5t)?5(s?2)2?25
3.设函数f(t)?cost??(其中函数t)?sint?ut(,)1(3)F(s)?L(f(t))?L(1?t?et)?L(1)?L(t?et)??L(?t?et)su(t)为阶跃函数, 求f(t)的拉普拉斯变换.
解:
L(f(t))????f(t)?e?stdt????cost??(t)?e?stdt????sint?u(t)?e?st000dt????cost??(t)?e?stdt????0sint?e?st??dt?cost?e?st11s2t?0?s2?1?1?s2?1?s2?14.求图8.5所表示的周期函数的拉普拉斯变换
解：
L(ft)?e?stdtT(t))??T0fT(1?e?as?1?ass2?as(1?e?as)
5. 求下列函数的拉普拉斯变换.
(1)f(t)?t2l?sinlt(2)f(t)?e?2t?sin5t
(3)f(t)?1?t?et(4)f(t)?e?4t?cos4t
(5f(t)?u(2t?4) (6f(t)?5sin2t?3cos2t
1(7) f(t)?t2?e?t (8) f(t)?t2?3t?2
解:(1)
f(t)? t12l?sinlt??2l[(?t)?sinlt]F(s)?L(f(t))?L(t2l?sinlt)??12lL[(?t)?sinlt]??12l(l1?2lsss2?l2)???2l?(s2?l2)2?(s2?l2)2
?1s?(1s?1)??1s?1 (s?1)2(4)
F(s)?L(f(t))?L(e?4t?cos4t)?s?4(s?4)2?16
(5)u(2t?4)???1,t?2
?0,其他F(s)?L(f(t))?L(u(2t?4))=??0u(2t?4)?e?stdt
=??e?stdt=1?2s2se(6)
F(s)?L(f(t))?L(5sin2t?3cos2t)?5L(sin2t)?3L(cos2t)?5?2s10?3s s2?4?3?s2?4?s2?4(7)
1?(1?13F(s)?L(f(t))?L(t2?e?t)?2)?(2)3?3 (s??)2(s??)2(8)
F(s)?L(f(t))?L(t2?3t?2)?L(t2)?3L(t)?2L(1)?1(2s2s?3s?2)
6.记L[f](s)?F(s)，对常数s0，若
Re(s?s0)??0，证明L[es0t?f](s)?F(s?s0)
证明:
L[es0t?f](s)???es0t?f(t)?e?st0dt???f(t)?e(s0?s)tdt???00f(t)?e?(s?s0)tdt?F(s?s0)7 记L[f](s)?F(s)，证明:F(n)(s)?L[(?t)n?f(t)](s)
证明：当n=1时,
F(s)?????st0f(t)?edt
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F?(s)?[?f(t)?e?stdt]?0??????0???[f(t)?e?st]dt???t?f(t)?e?stdt??L(t?f(t))0?s
解：(1)1?1? (2) t?t?(3)
tt0?1?1d??t
0t3??(t??)d??t ?16所以，当n=1时, F(n)(s)?L[(?t)n?f(t)](s)显然成立。
假设，当n=k-1时, 有
t?et????et??d??et????e??d???et????de??000ttF(k?1)(s)?L[(?t)k?1?f(t)](s)
现证当n=k时
k?1?st(k?1)d(?t)?f(t)?edtdF(s)?(k)0F(s)??dsdsk?1??[(?t)???f(t)?e?st]??dt??(?t)k?f(t)?e?stdt00?s?L[(?t)k?f(t)](s)
????e[?e]??e??d??et?t?1tt00??t
(4)
tt1sinat?sinat??sina??sina(t??)d????[cosat?cos(2a??at)]d?002 1t?sinat?cos2at2a2(5)
8. 记L[f](s)?F(s)，如果a为常数,证明:
?(t??)?f(t)???(t??)?f(t??)d?????(t??)?f(t??)d(t??)00t??????(?)?f(?)d????(?)?f(?)d???0,f(t??),0???tt00ttt
L[f(at)](s)?1sF() aa (6)
证明：设L[f](s)?F(s)，由定义
sint?cost??sin??cos(t??)d??0t1t[sint?sin(2??t)]d?2?0L[f(at)]?????
??0??0f(at)?e?stdt.(令at?u,t?s?udu1??a?f(u)?eduaa?0udu,dt?)aaf(u)?es?ua1sF()aa
ttt?sint??sin(2??t)d?220t1?sint?cos(2??t)t024t1t?sint?[cost?cos(?t)]?sint242
11.设函数f, g, h均满足当t<0时恒为零，证明
9. 记L[f](s)?F(s)，证明:
?f(t)L[]??F(s)ds,即???f(t)?e?stdt???F(s)ds
s0sttf?g(t)?g?f(t)以及
(f?g)?h(t)?f?h(t)?g?h(t)
证明：
证明：
??sF(s)ds??[f(t)?edt]ds??f(t)?[?eds]dts0s???????st????st令t??=uf?g(t)??f(?)g?t???d????????f(t?u)?g?u?du0tt01f(t)?stf(t)??f(t)?[?e?st?]dt??edt?L[]s?00ttt10.计算下列函数的卷积
(1)1?1(2)t?t
(3)t?e (4)sinat?sinat
(5)?(t??)?f(t) (6sinat?sinat
t
??f(t?u)?g?u?du??g(?)?f?t???d??g?f(t)00tt
?f?g??h(t)??0?f(?)?g?????h?t???d???f????h(t??)?d???g(?)?h?t???d?00ttt
?f?h(t)?g?f(t)12.利用卷积定理证明
L[?f(t)dt]?0tF(s)s
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证明：设
g(t)??f(t)dt0tg?(t)?f(t),且g(0)?0，则
，则
因为
L?1(2)?sin2t22s?2
L[g?(t)]?sL[g(t)]?g(0)?sL[g(t)]L[g?(t)]，所以 sF(s)f(t)dt]?ds
所以
1stL?1(F(s))?L?1(??2)?sin2t4(s?4)24
(5)
?s?111g(t)??(?)du??L()0s?1u?1u?1t
L[g(t)]?L[?t0F(s)?ln其中
13. 求下列函数的拉普拉斯逆变换.
s(1)F(s)?(s?1)(s?2)g(t)?L?1(
所以
11?)?e?t?et s?1s?1s2?8(2)F(s)?2(s?4)2
e?t?etet?e?tF(s)??L()?L()
tte?t?etet?e?tshtf(t)?L(F(s))????2?
ttt?11(3)F(s)?s(s?1)(s?2) s(4)F(s)?2(s?4)2(5)F(s)?ln
s2?2s?1122(6)F(s)?????s(s?1)2ss?1(s?1)2
所以
s?1s?1
s2?2s?1(6F(s)?s(s?1)2122L?1(F(s))?L?1(?)?L?1()?L?1()2ss?1(s?1)??1?2et?2tet?2tet?2et?114.利用卷积定理证明
s21??解：(1)F(s)?(s?1)(s?2)s?2s?1
L?1[L?1(2111?)?2L?1()?L?1()?2e2t?ets?2s?1s?2s?1
st]??sinat(s2?a2)2a
证明：
(2)
L?1[s2?83?121?1s2?431F(s)?22?L(2)?L(22)?sin2t?tcos2t(s?4)4s?42(s?4)42ssa1?1]?L(??)2222222(s?a)s?as?aa
sa,L(sinat)?
s2?a2s2?a2又因为
L(cosat)?1111???(3F(s)?s(s?1)(s?2)2ss?12(s?2)
?1故L(F(s))?所以，根据卷积定理
1?t1?2t?e?e22
(4)
F(s)?s1?4s12??????()?222222(s?4)4(s?4)4s?2
sa11??)?cosat?sinats2?a2s2?a2aat11t1??cosa???sin(at?a?)d???[sinat?sin(2a??at)]d?0aa02t??sinat2a L?1(15.利用卷积定理证明
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12tt?y2L[]?e?edys(s?1)π0
?1证明：
L?1[111]?L?1[?]s(s?1)ss?1 1112L?1(F(s))?L?1(2)?L?1(2)3s?16s?22 11?sint?sin2t)36(3)
s?2s?211F(s)?2???()?2222L?1[1s(s?1)]?2πet?t0e?y2dy
因为
L?1(1s)?1πt?12,L?1(1s?1)?et
所以，根据卷积定理有
L?1[11?1t2t?1(t?y)1tt?1s(s?1)]?π?t2?e?π?0y2edy?πe?0y2e?ydy?2et?te?ydy???令y?u??2et?te?u2du2?2et?t0e?y2π0π0πdy
16. 求下列函数的拉普拉斯逆变换.
(1)F(s)?1(s2?4)2(2)F(s)?1
s4?5s2?4 (3)F(s)?s?2(s2?4s?5)2
(4)F(s)?2s2?3s?3(s?1)(s?3)2
解:(1)
112(s2?4)1s2F(s)??4(s2?4)2?16?(s2?4)2?8?(s2?4)2?1212
16?s?4s2?4?8?(s2?4)2故
?112L(F(s))?L?1(21?1s?411s2?4)?8L((s2?4)2)?16sin2t?8t?cos2t
16(2)：
F(s)?1s4?5s2?4?13(11s2?1?s2?4)?1112 3(s2?1?2s2?22) 35 / 37
(s?4s?5)[(s?2)?1]2(s?2)?1
故L?1(F(s))?12t?e?2t?sint (4)
F(s)?2s2?3s?3ABCD(s?1)(s?3)2?s?1?s?3?(s?3)2?(s?3)3?A?113 4,B??4,C?2,D?3故
1?13F(s)?43s?1?4s?3?2(s?3)2?(s?3)3
且
(1s?3)???1(s?3)2,(1s?3)???2?1(s?3)3
所以
L?1(F(s))?1e?t?1e?3t?3t?e?3t?3t2?e?3t442
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)]?s2Y(s)?sy(0)?y?(0)?s2Y(s)?1
方程两边取拉氏变换,得