复变函数与积分变换（修订版）课后答案（复旦大学出版社）
令R(z)=2211?z4，则R(z)在上半平面有两个一级极
22sgt与nu(t)(作比)较.
点
????(1?i),(?1?i).
2222不难看出 sgn(t)?u(t)?u(?t). 故：
?R(t)?edt?2πi?Res[R(z)?e,i?ti?z(1?i)]?2πi?Res[R(z)?e,i?z(?1?i)]
F[sgn(t)]?F(u(t))?F(u(?t))??2i??π??(?)??(??)??2i?1iπ?π??(?)?[1i(??)?π??(??)]
故.
cos?t1?t4?????dt?Re[?????ei?t41?tdt]?122e?|?|/2(cos|?|2?sin|?|2)
(5) f(t)?解: G(?)??t1?t47.已知函数
f(t)的傅里叶变换
F(?)=π??(???0)??(???0)?,求f(t)
?t????t1?t4解:
?e?i?tdtf(t)?F(F(?))=????-1?????1?t????4?cos?tdt?i?4t?sin?t1?t4?2π????????1????π???(???0)??(???0)?ed?i?t?i?tdt
而F(cos?0t)=??cos?0t?eei?0tdt?e?i?0t??i?t?sin?t1?tdtz1?z4??e2?i?0tdt
?π[?(???0)??(???0)]同(4).利用留数在积分中的应用,令R(z)=则
?i???????
所以f(t)?cos?0t8.设函数f(t)的傅里叶变换F(?)，a为一常数. 证
t?sin?t1?t?e?|?|/4dt?(?i)Im(?????t?ei?t41?tdt)明
.
?[f(at)](?)???F?a?a1??. ?i22?sin?25.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
L/2解:F[f(at)](?)??????f(at)?e?i?tdt?1?a????f(at)?e?i?td(at)
GL(?)???L/2F(t)e?i?tdt. 当a>0时,令u=at.则 F[f(at)](?)?1a证明当L??时，有
p.v.12π??????f(u)?eu?i?adu????F?? a?a?11aF(???GL(?)ei?td??F(t)
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).
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. 解: 因
为
Fu(t1?(?)π??)(?i?)把.??1,??|t|?1,tt?0t?0的傅里叶
F?????F?f??t?????.
证明：
函数
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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?f??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??sint?cost?e?t?,故f*g???t?2?1??e?t.1?e2?2t?00?t?t??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????0??0ee?at?sin?0t?e?ei?0t?i?t?at?e2i?i?0ti?t1??e0??F?f?t??2??2?i?t??e0??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??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??F????0??F????0???(2)f(t)?e?4t，
11.设
?0,f?t????t?e,t?0t?0??sint,g?t????0，?0?t?其他π2
解: (1)f(t)?sint?cost?
L(f(t))?12L(sin2t)?1?1222sin2t ?12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
f*g?t?=0.
2s?4s?4
11?4t(2)L(f(t))?L(e)?
2s?4
1?cos2t2(3)f(t)?sint?2
L(f(t))?L(1?cos2t21111122)?L(1)?(cos2t)????2?2222s2s?4s(s?4)
若0?t?f*g?t???2t,0?y?t,则 f(y)g?t?y?dy??0?t0e?y?sin?t?y?dy
(4)L(t)?3s
(5)
L(f(t))?L(e?e2bt?bt22若t??2,0?t?y??2.?t??2?y?t.
1bt1?bt1111b)?L(e)?L(e)?????22222s?b2s?bs?b
则f*g?t???tt??2e?y?sin?t?y?dy
2.求下列函数的拉普拉斯变换.
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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）
?2,0?t?1?（1）f(t)??1,1?t?2?0,t?2 ?1(7)f(t)?t2?e?t (8)f(t)?t2?3t?2
解:(1)
f(t)? t2l?sinlt??12lt2l1[(?t)?sinlt]?sinlt)????2ls222?cost,0?t?π(2)f(t)??
0,t?π?F(s)?L(f(t))?L(???s?2s12l?L[(?t)?sinlt]s(s?l)222解: (1)
L(f(t))?1? ??0f(t)?edt??st?102?edt??edt?(2?e?e1s?st2?st1)2ls?l(l2)???22l(s?l)
?πs
(2)F(s)?L(f(t))?L(e?2t?sin5t)?5(s?2)?25
1s?L(?t?e)t(2)
L(f(t))????0f(t)?edt??cost?edt?(1?e)?20ss?1
1s1s?11s1(s?1)2?stπ?st1?πs1?e2(3)F(s)?L(f(t))?L(1?t?e)?L(1)?L(t?e)?tt3.设函数f(t)?cost??(t)?sint?u(t),其中函数
??()???
u(t)为阶跃函数, 求f(t)的拉普拉斯变换.
解:
L(f(t))??(4)
???0f(t)?edt??st???st???0cost??(t)?edt??sint?u(t)?edt0?st?st???stF(s)?L(f(t))?L(e?4t?cos4t)?s?4(s?4)?162
?????cost??(t)?edt??sint?edt0?stt?0?cost?e?1s?12?1?1s?12?s22s?1
(5)u(2t?4)??
0,其他?
F(s)?L(f(t))?L(u(2t?4))=?u(2t?4)?e0??st?1,t?24.求图8.5所表示的周期函数的拉普拉斯变换
dt=?e2??stdt=1s
e?2s(6)
F(s)?L(f(t))?L(5sin2t?3cos2t)?5L(sin2t)?3L(cos2t)?5?2s?42?3?ss?42?10?3ss?42
解：
L(fT(t))?T0?st(7)
1?(1??t1?fT(t)?e1?et2ldt?1?ass2?as?as(1?e?asF(s)?L(f(t))?L(t2?e)?2?3)3?()23
)
(8)
(s??)2(s??)25. 求下列函数的拉普拉斯变换.
(1)f(t)??sinlt(2)f(t)?e
?4t?2t?sin5t
?cos4t(3)f(t)?1?t?e(4)f(t)?e(5f(t)?u(2t?4)
t1222F(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
?f](s)?F(s?s0)
(6f(t)?5sin2t?3cos2t
证明:
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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）
L[e?s0t?f](s)?f(t)?e??0es0t?f(t)?e?stdt?(s?s0)t?dt?F(s?s0)n?sF(s)ds???0???s[f(t)?edt]ds??st?s?st???0f(t)?[?eds]dts?st??st??0(s0?s)tdt???0f(t)?e(n)??f(t)?[?et1]dt????0f(t)t?edt?L[f(t)t]
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]??st10.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? (2) t?t?(3)
t0???0?[f(t)?e?s]dt?????0t?f(t)?edt??L(t?f(t))?st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
F(k?1)?1?1d?0t?t 16t
3?t0??(t??)d??(s)?L[(?t)k?1?f(t)](s)
t?e?tt???ed??e????ed???e????de00t0t??t0t??tt??tt??现证当n=k时
F?(k)(s)?dF(k?1)(s)dsk?1?d??st??0(?t)k?1?f(t)?e?stdt??e[?e]??ed??e?t?1(4)
dt??
ds]dt???0?[(?t)k?f(t)?e?s???0(?t)?f(t)?ek?sttt1sinat?sinat??sina??sina(t??)d????[cosat?cos(2a??at)]d?002?L[(?t)?f(t)](s) ?12asinat?cos2at2t
8. 记L[f](s)?F(s)，如果a为常数,证明:
L[f(at)](s)?1sF() aa(5)
?(t??)?f(t)???(t??)?f(t??)d?????(t??)?f(t??)d(t??)00tt证明：设L[f](s)?F(s)，由定义
L[f(at)]???????(?)?f(?)d????(?)?f(?)d???f(t??),0???t0,t??t00t
???0?saf(at)?eu?stdt.(令at?u,t???ua,dt?dua (6)
)???0f(u)?esdua?1a?0f(u)?e?sausint?cost??t2t2t2sint?sint?sint?t21414du??0t0sin??cos(t??)d??12?[sint?sin(2??t)]d?0tt1asin(2??t)d?t0F()a
9. 记L[f](s)?F(s)，证明:
L[f(t)t]???cos(2??t)[cost?cos(?t)]?t2sint
??sF(s)ds,即
???0f(t)t?e?stdt???sF(s)ds
11.设函数f, g, h均满足当t<0时恒为零，证明
f?g(t)?g?f(t)证明：
以及
(f?g)?h(t)?f?h(t)?g?h(t)证明：
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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?g(t)????t0t0f(?)g?t???d????????f(t?u)?g?u?dut令t??=u0F(s)?f(t?u)?g?u?du?t?t031?L()?L()?sin2t?tcos2t22222(s?4)4s?42(s?4)42?1?1s?82321s?42g(?)?f?t???d??g?f(t)
?f?g??h(t)???f(?)?g?????h?t???d?0
(3F(s)?1?1?1?1??tt0f????h(t??)?d???0g(?)?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.
(1)F(s)?s(s?1)(s?2) s2(2)F(s)??8(s2?4)2
(3)F(s)?1s(s?1)(s?2)
(4)F(s)?s(s2?4)2
(5)F(s)?lns?1s?1
s2(6F(s)??2s?1s(s?1)2
解：(1)F(s)?s(s?1)(s?2)?2s?2?1s?1
L?1(2?1(1?12tts?2?1s?1)?2Ls?2)?L(1s?1)?2e?e
(2)
s(s?1)(s?2)2ss?12(s?2)
故L?1(F(s))?1?t?12?e2e?2t (4)F(s)?s?1?4s(s2?4)2?4?(s2?4)2??14?(2s2?22)?因为
L?1(2s2?22)?sin2t 所以
L?1(F(s))?L?1(?1?s4(s2?4))?t24sin2t
(5)
F(s)?lns?1s?1???t)0(1u?1?1u?1)du??L(g(t)
其中
g(t)?L?1(11ts?1?s?1)?e?t?e
所以
F(s)??L(e?t?ett?tt)?L(e?et)
tttf(t)?L?1(F(s))??e??ee?tt?e?t?2?shtt
2(6)F(s)?s?2s?112s(s?1)2??s?s?1?2(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[st(s2?a2)]?2a?sinat
证明：
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