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)????0b(?)?sin?td?
其中b(?)?2??π?0f?t??sin?tdt 当
f(t)
为
偶
函
数
时
，
则
有
f(t)????0a(w)?cos?td?
其中a(?)?2????0f(t)?cos?tdt
证明：
因为f(t)?1??G(?)ei?t2π???d?其中G(?)为f(t)的傅里叶变换
G(?)??????f(t)e?i?tdt??????f(t)?(cos?t?isin?t)dt ????????f(t)?cos?tdt?i???f(t)?sin?tdt 当f(t)为奇函数时，f(t)?cos?t为奇函数，从而
?????f(t)?cos?tdt?0
f(t)?sin?t为偶函数，从而
?????f(t)?sin?tdt?2???0f(t)?sin?tdt.
故G(?)??2i???0f(t)?sin?tdt. 有
G(??)??G(?)为奇数。
f(t)?12??????G(?)?ei?td??12??????G(?)?(cos?t?isin?t)d?=1??π???G(?)?isin?td??i??π?0G(?)?sin?td?
2所以，当f(t)为奇函数时，有
f(t)????b(?)?sin?td?.其中b(?2???0)=π0f(t)?sin?tdt.同理，当f(t)为偶函数时，有
f(t)????0a(?)?cos?td?.其中
a(?)?2??π?0f(t)?cos?tdt 22.在上一题中，设f(t)????t,t?1??0,t?1.计算a(?)的
值.
解:
a(?)?2??π?0f(t)?cos?tdt?2π?10t2?cos?tdt?2π???10?cos?tdt?2?1t2?cos?tdt?2?1?1t2π0π?0d(sin?t)?2π??t2?sin?t10?2
1π??0sin?t?2tdt?2sin?4π??????120t?d(cos?t)?2sin?π??4?????t?cos?t1?120?cos?tdt?0???2sin?4co???s?4sin???2???33.计算函数f(t)????sint,t?6π的傅里叶变换. ??0,t?6π解:
F?f?(?)??????f(t)?e?i?tdt??6π?6πsint?e?i?tdt??6π?6πsint?(cos?t?isin?t)dt
??2i?6π0sint?sin?tdt?isin6π?π(1??2)4.求下列函数的傅里叶变换 (1)f(t)?e?t
解： F?f?(?)??????f(t)e?i?tdt????e?|t|?e?i?tdt????e?(|t|?i?t)????dt
??0i?)??et(1?i?)dt????0e?t(1?dt?21??2(2)
f(t)?t?e?t2
解：因为
2F[e?t2]?π?e??4.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
2G(?)?F(t?e?t2)?π???42i?e
(3)f(t)?sinπt1?t2 解：
G(?)?F(f)(?)????????????sinπt?i?t?edt1?t2L/2sinπt?(cos?t?isin?t)dt1?t2GL(?)?
L???L/2?F(t)e?i?tdt.
?1[cos(π??)t?cos(π??)t]??i???sinπt?sin?t??1?t2dt??2i???201?t2dt?i???cos(π+?)t??cos(π-?)t01?t2dt?i?01?t2dt(利用留数定理)?????i2sin?,当|?|?π??0,当|?|?π.(4)f(t)?11?t4 解:
G(?)????1??1?t4e?i?tdt????cos?t??sin?t??1?t4dt?i???1?t4dt?2???cos?t??cos?t01?t4dt????1?t4dt令R(z)=11?z4，则R(z)在上半平面有两个一级极点
22(1?i),22(?1?i). ?????R(t)?ei?tdt?2πi?Res[R(z)?ei?z,22(1?i)]?2πi?Res[R(z)?ei?z,22(?1?i)] 故
.
???cos?t1?t4dt?Re[???ei?t1?|?|/2|?||?|
????1?t4dt]?22e(cos2?sin2)(5) f(t)?t1?t4 解:
G(?)????t?e?i?t??1?t4dt????t??1?t4?cos?tdt?i???t?sin?t??1?t4dt ??i???t?sin?t??1?t4dt同(4).利用留数在积分中的应用,令R(z)=z1?z4 则
?i???t?sin?t??1?t4dt?(?i)Im(???t?ei?t??1?t4dt).
??i?e?|?|/2?2?sin25.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
证明当时，有
?p.v.12πG(?)ei?td??F(t)???L 对所有的实数t成立.
(书上有推理过程) 6.求符号函数 sgnt?t??1,t?|t|??0的傅里叶?1,t?0变换. 解: 因
为
F(u(t))?1i??π??(?).把函数
sgn(t)与u(t)作比较.
不难看出 sgn(t)?u(t)?u(?t). 故：
F[sgn(t)]?F(u(t))?F(u(?t))?1iπ?π??(?)?[1i(??)?π??(??)]
?2i??π??(?)??(??)??2i?7.已知函数f(t)的傅里叶变换
F(?)=π??(???0)??(???0)?,求f(t)
解:
f(t)?F-1(F(?))=1??2π???π???(???i?t0)??(???0)?ed?而F(cos????i?t0t)=???cos?0t?edt????ei?0t?e?i?0t?i?0t
??2?edt?π[?(???0)??(???0)]所以f(t)?cos?0t8.设函数f(t)的傅里叶变换F(?)，a为一常数. 证明
[f(at)](?)?1aF?????a??. 解:F[f(at)](?)?????i?t1??f(at)?edt?a????i???f(at)?etd(at) 当a>0时,令u=at.则
F[f(at)](?)?1a?????f(u)?e?iua?du?1???aF??a??
当a<0时,令u=at,则F[f(at)](?)??1aF(?a). 故原命题成立.
9.设F????F?f????;证明
F?????F?f??t?????.
证明：
F?f??t?????????f??t??e?i?tdt?????f?u??ei?u
????du????f?u??i????u?????i(??)?u?????edu????f?u??e?du??????f?t??e??i?????t?dt?F????.10.设F????F?f????，证明：
F?f?t??cos?0t?????12??F????0??F????0???以及
F?f?t??sin?10t?????2??F????0??F????0???.证明：
F?f?t??cos??ei?0t+e?i?0t?0t??F??f?t??2???1??ei?0t??e?i?0t??2??F??f?t??2???F??f?t??2?????12??F????0??F????0???同理：
F?f??ei? t??sin?0t?e?i?0t?0t??F??f?t??2i???12i?F??f?t??ei?0t???F??f?t??e?i?0t????12i??F????0??F????0???11.设
f?t????0,t?0??sint,0?t?π?e?t,t?0g?t?????0，2
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
f*g?t?=0.
若0?t??2,0?y?t,则
f*g?t???tf(y)g?t?y?dy??t0e?y?sin?t?y?dy
0若t???2,0?t?y??2.?t?2?y?t.
则f*g?t???tyt??e??sin?t?y?dy
2??0,t?0故1?f*g?t?????sint?cost?e?t?2?,0?t?2
?1?2e?t???1?e2?.t?212.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?atsin?0t?u?t?
解：G????F?f??????????e?at?sin?0t?u?t??e?i?tdt????e?at?sin??t00t?e?idt?????atei?0t?e?i?0t?0e?2i?ei?tdt?1?????a?i????0???t1?????a?i????0???t2i?0edt?2i?0edt??0?a?i??2??2 0
习题八
1.求下列函数的拉普拉斯变换.
(1)f(t)?sint?cost，
L(f(t))??f(t)?e?stdt??cost??(t)?e?stdt??sint?u(t)?e?stdt000??????(2)f(t)?e?4t??cost??(t)?e?stdt??sint?e?stdt??0????，?cost?e?st 11s2?1?2?2t?0?2s?1s?1s?1(3)f(t)?sin2t (4)f(t)?t2，(5)f(t)?sinhbt 1 解: (1)f(t)?sint?cost?sin2t
2
L(f(t))?12L(sin2t)?1212?s2?4?s2?4
(2)L(f(t))?1L(e?4t)?1
2s?4
(3)f(t)?sin2t?1?cos2t
2 L(f(t))?L(1?cos2t11111222)?2L(1)?2(cos2t)?2?s?2?s2?4?s(s2?4)
(4)L(t2)?2
s3 (5)
L(f(t))?L(ebt?e?bt2)?12L(ebt)?12L(e?bt)?12?1s?b?11b2?s?b?s2?b2
2.求下列函数的拉普拉斯变换.
?2,0?t
（1）f(t)???1?1,1?t?2??0,t?2 (2)f(t)???cost,0?t?π?0,t?π
解: (1)
L(f(t))????f(t)?e?stdt??12?e?stdt??2e?stdt?1s(2?e?s?e?2s001)
(2)
L(f(t))????t)?e?stdt??πcost?e?st0dt?1s(1?e)?1?e?πs0f(?πss2?1
3.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:
4.求图8.5所表示的周期函数的拉普拉斯变换
解：
L(f?T0fT(t)?e?stdtT(t))?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)t)?t2?e?t f( (8) f(t)?t2?3t?2
解:(1)
f(t)? t?sinlt??12l2l[(?t)?sinlt]F(s)?L(f(t))?L(t12l?sinlt)??2lL[(?t)?sinlt]??12l(l1?2lsss2?l2)???2l?(s2?l2)2?(s2?l2)2
(2)F(s)?L(f(t))?L(e?2t?sin5t)?5(s?2)2?25
(3)F(s)?L(f(t))?L(1?t?et)?L(1)?L(t?et)?1s?L(?t?et)?1s?(1s?1)??11 s?(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=12se?2s(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?1)3F(s)?L(f(t))?L(t2?e?t)?2?()3?23 (s??)2(s??)2(8)
F(s)?L(f(t))?L(t2?3t?2)?L(t2)?3L(t)?2L(1)?1s(2s2?3s?2)
6.记L[f](s)?F(s)，对常数s0，若
Re(s?ss0t0)??0，证明L[e?f](s)?F(s?s0)
证明:
L[es0t?f](s)???0es0t?f(t)?e?stdt???f(t)?(s0?s)t0edt???f(t)?e?(s?s0)t0dt?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)????0f(t)?e?stdt
F?(s)?[???0f(t)?e?stdt]?
?????[f(t)?e?st]?sdt?????00t?f(t)?e?stdt??L(t?f(t))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)(s)?L[(?t)k?1?f(t)](s)
现证当n=k时
??1)F(k)(s)?dF(k?(s)ds?d?(?t)k?1?f(t)?e?st0dtds??[(?t)k?1?f(t)?e?st??]0?sdt????0(?t)k?f(t)?e?stdt?L[(?t)k?f(t)](s)
8. 记L[f](s)?F(s)，如果a为常数,证明:
L[f(at)](s)?1aF(sa) 证明：设L[f](s)?F(s)，由定义
L[f(at)]????0f(at)?e?stdt.(令at?u,t?udua,dt?a)????f(u)?e?saudu?1???f(u)?e?sau0aa0du?1saF(a)
9. 记L[f](s)?F(s)，证明:
L[f(t)t]???sF(s)ds,即???f(t)0t?e?stdt???sF(s)ds证明：
????stsF(s)ds??[f(t)?e?dt]ds????f(t)?[??s0se?stds]dt????f(t)?[?1e?st???f(t)?stf(t)0ts]dt??0t?edt?L[t]
10.计算下列函数的卷积
(1)1?1(2)t?t
(3)t?e t
(4)sinat?sinat
(5)?(t??)?f(t) (6sinat?sinat
解：(1)1?1?t
?01?1d??t
(2) t?t??t??(t??)d??1306t (3)
t?et??t??et??d??et??t??e??d???ett00??0??de????et[?e??]t0??te??d??et?t?
01(4)