《复变函数与积分变换(刘建亚)》作业答案
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《复变函数与积分变换》作业参考答案
习题1:
4、计算下列各式
(1) 3i(3?i)(1+3i); (3)
3; 2(3?i)6(5) z?1?3i234,求z,z,z; (7) 2?1。
解:(1) 3i(3?i)(1+3i)=3i(3+3i?i+3)=3i(2i+23)=?6+63i;
(3)
333(2?23i)3(2?23i)333?????i; 24?1288(3?i)2?23i(2?23i)(2?23i)(5) z2?1?3i?3i?3?2?23i13????i,
4422?1?3i1?3i?1?3????1, 224z3?z2?z?13z4?z3?z???i.
22(7) 因为?1?cos??isin?,所以
6?1?cos??2k?6?isin??2k?6,
即
k?0时,w0?cosk?1时,w1?cos?6?isin?6?31?i; 223?3??isin?i; 66k?2时,w2?cos5?5?31?isin???i; 66227?7?31?isin???i; 66229?9??isin??i; 66k?3时,w3?cosk?4时,w4?cosk?5时,w5?cos
11?11?31?isin??i. 6622习题2:
3、下列函数在何处可导?何处解析?在可导点求出其导数.
(2) f(z)?x2?iy; (4) f(z)?sinxchy?icosxshy (6) f(z)?az?b。 cz?d解:(2) 因为u(x,y)?x2,v(x,y)??y,
???u?x?2x,uy?0,vx?0,vy??1.
这四个一阶偏导数都连续,故u(x,y)和v(x,y)处处可微,但柯西-黎曼方程仅在x??成立,所以f(z)只在直线x??处不解析.
(4) 因为u(x,y)?sinxchy,v(x,y)?cosxshy,
1上21上可导,此时f?(z)x??1?2xx??1??1,但复平面上处222???u?x?cosxchy,uy?sinxshy,vx??sinxshy,vy?cosxchy.
这四个一阶偏导数都连续,故u(x,y)和v(x,y)处处可微,且满足柯西-黎曼方程,所以f(z)在复平面内解析,并且
ey?e?yey?e?y?f?(z)?u??isinx?x?ivx?cosxchy?isinxshy?cosx?22eye?yey?ixe?yix??cosx?isinx??. ?cosx?isinx???e??e2222ey?ix?e?y?ixe?iz?eiz???cosz22 (6)
?z?0limf(z??z)?f(z)1?a(z??z)?baz?b??lim??z?0?z?c(z??z)?d?zcz?d????limad?bcad?bc??z?0(cz?c?z?d)(cz?d)(cz?d)2ad?bcd外处处解析,且f?(z)?. 2c(cz?d)
所以,f(z)在除z??
4、指出下列函数的奇点. (1)
z?1z?2; (2) . 2222z(z?4)(z?1)(z?1)解:(1)
z2(z2?4)?(z?1)(4z3?8z)?3z4?4z3?4z2?8zf?(z)??422z(z?4)z4(z2?4)2??3z?4z?4z?8z3(z2?4)232
所以,f(z)的奇点为0,?2i.
(z?1)2(z2?1)?2(z?2)(z?1)(2z2?z?1)3z3?9z2?5z?3(2) f?(z)? ??422322(z?1)(z?1)(z?1)(z?1)所以,f(z)的奇点为?1,?i.
10、如果f(z)?u?iv在区域D内解析,并且满足下列条件之一,试证f(z)在D内是一常数.
(2) f(z)在D内解析;
?证明:由f(z)?u?iv在区域D内解析,知u(x,y)、v(x,y)在区域D内可微,且u?x?vy,????D内解析,知u?u?y??vx.同理,由f(z)在x??vy,uy?vx.
???D内是一从而我们得到u?x?vy?uy?vx?0,所以u(x,y)、v(x,y)皆为常数,故f(z)在
常数.
15、求解下列方程:
z(2) e?1?0
z解:e??1,于是
z?Ln(?1)?ln1?iarg(?1)?2k?i=(2k?1)?i,k?Z
18、求Ln(?i),Ln(?3?4i)的值及主值. 解:Ln(?i)?ln?i?iarg(?i)?2k?i???2i?2k?i,所以其主值为??2i;
4Ln(?3?4i)?ln?3?4i?iarg(?3?4i)?2k?i?ln5?i(??arctan)?2k?i,所以其主
3值为ln5?i(??arctan 19、求e解:e
1?i1?i
4). 3?2
,e
1?i?4
,3i,(1?i)i的值.
?2?e?ei(?)2??????e?cos(?)?isin(?)???ie;
22??e1?i?4?e?e14i?4?2???2?24??e?cos?isin??4e??i?e?1?i?; ???44222????143i?eiLn3?ei(ln3?2k?i)?e?2k?+iln3?e?2k??cosln3?isinln3?; (1?i)i?eiln(1?i)?e
20、求1,(?2)解:122???i?ln2?i?2k?i?4???e1????2k????iln24???e1????2k???4??ln2ln2??cos?isin??.
22??2,1,i,(3?4i)1?i的值.
2k?i?ii?e2Ln1?e2?e?cos(22k?)?isin(22k?);
?22(?2)2?e2Ln(?2)2ln2?(2k?1)2?i?cos??(2k?1)??2????isin?(2k?1)2??;
?1?i?e?iLn1?e?i(2k?i)?e2k?;
i?eiiLni?e???i?i?2k?i?2???e1????2k???2??;
(3?4i)1?i?e(1?i)Ln(3?4i)?e4????(1?i)?ln5???arctan?i?2k?i?3?????e44??ln5?arctan?2k??i?ln5?arctan?2k??33???5e??2k?4cosln5???isinln5??,??arctan,k?Z????????3
22、解方程: (1) chz?0;
解:z?Arch0?Ln(0?0?1)?Lni??2k?
??1???i,k?Z. 2?习题3:
1、沿下列路径计算积分
?2?i0z2dz:
(1) 从原点至2?i的直线段;
(2) 从原点沿实轴至2,再由2铅直向上至2?i;
(3) 从原点沿虚轴至i,再由i沿水平方向向右至2?i. 解:(1) 从原点至2?i的直线段的复参数方程为z?x?ix1,dz?(1?i)dx,参数22x:0?2,所以
?2?i01111zdz??(1?i)3x2dx?(1?i)3x3?(2?i)3
023230222(2) 从原点沿实轴至2的直线段的复参数方程为z?x,参数x:0?2,由2铅直向上至2?i的直线段的复参数方程为z?2?iy,参数y:0?1,所以
?2?i0z2dz??z2dz??z2dz??x2dx??(2?iy)2idyC1C200221118i2111?x3??(?iy2?4y?4i)dy???2?4i=?i?(2?i)330033333
(3) 从原点沿虚轴至i的直线段的复参数方程为z?iy,参数y:0?1,由i沿水平方向向右至2?i的复参数方程为z?x?i,参数x:0?2,所以
?
2?i01z2dz??z2dz??z2dz??(iy)2idy??(x?i)2dxC1C200212i1i1???iy2dy??(x?i)2dx???(2?i)3??(2?i)300333322、分别沿y?x与y?x算出积分
?1?i0(x2?iy)dz的值.
解:y?x的复参数方程为z?(1?i)x,dz?(1?i)dx,参数x:0?1所以
??
5、计算积分
1?i0(x2?iy)dz??(x2?ix)(1?i)dx?0151?i; 66y?x2的复参数方程为z?x?ix2,dz?(1?2xi)dx,参数x:0?1所以
1?i0(x?iy)dz??(x2?ix2)(1?2xi)dx?02151?i 66??Czdz的值,其中C为正向圆周: z(1) z?3
解:设C1是C内以被积函数的奇点z?0为圆心的正向圆周,那么
??Czzz?zdz??dz?dz?z???CC11zzzz1??C1zdz?3?2?i=6?i
6、试用观察法得出下列积分的值,并说明观察时所依据的是什么?C是正向圆周z?1: (1) (4)
??Cdz; (2) z?2??Cdz; (3) 2z?2z?3dz??Ccosz;
dz??C1; (5) z?3??Czezdz; (6)
dz??C?i??5?. ?z???z??2??2??dz??Cz?2?0,根据柯西积分定理; dz(2) ??Cz2?2z?3?0,根据柯西积分定理;
dz(3) ??Ccosz?0,根据柯西积分定理;
dz(4) ??C1?2?i,根据复合闭路定理;
z?3解:(1) (5)
??zedz?0,根据柯西积分定理;
Cz(6)
dz4?i???C?i??5?5?i,根据柯西积分定理及复合闭路定理.
?z???z??2??2??
7、沿指定曲线的正向计算下列积分: (1)
????Cezdz,C:z?3?1; z?3dz,C:z?a?a; 22z?a(2)
C(3)
??Ceiz4C:z?2i?dz,; 23z?1zdz,C:z?2; z?3(4)
??C(5)
dz??C(z2?1)(z3?1),C:z?r?1;
(6)
??Cz3coszdz,C为包围z?0的闭曲线;
(7)
dz3C:z?,; ??C(z2?1)(z2?4)2(8) (9)
sinz??Czdz,C:z?3;
???Ccosz2???z??2??dz,C:z?3;
ez(10) ??Cz5dz,C:z?1. ezz解:(1) ??Cz?3dz?2?i?e(2)
z?3?2?e3i;
dz1?2?i???Cz2?a2z?a??z??a?ai;
eizeiz?(3) ?; dz?2?i???Cz2?1z?iz?ie(4)
??Czdz?0; z?3(5)
dz??C(z2?1)(z3?1)?0;
(6)
??zC3coszdz?0;
?dz?1?dzdz??2??C(z2?1)(z2?4)2i?????C(z?i)(z2?4)??C(z?i)(z?4)?(7) ;
??1?11??2?2??02i?z?4z??iz?4z?i?(8) (9)
sinz??Czdz?2?i?sinzz?0?0;
???Ccosz??z???2??dz?22?i???sinz???2?i;
?1!z?2ez2?iz(10) ?dz??e5?Cz(5?1)!
?z?0?i12.
21、证明:u?x?y和v?22y都是调和函数,但是u?iv不是解析函数. 22x?y?2u?u?u?2u?2x,2?2,??2y,2??2, 证明:因为?x?x?y?y?v?2xy?2v6x2y?2y3?vx2?y2?2v2y3?6x2y?2,2?,,2?, ?22222322223?x(x?y)?x(x?y)?y(x?y)?y(x?y)所以
?2u?2u?2v?2v????2?0,2?2?0,且u?x?vy,uy??vx. 2?x?y?x?y即u?x2?y2和v?
22、由下列各已知调和函数求解析函数f(z)?u?iv,并写出z的表达式: (1) u?(x?y)(x2?4xy?y2); (2) v?y都是调和函数,但是u?iv不是解析函数.
x2?y2y,f(2)?0; 22x?y(3) u?2(x?1)y,f(2)??i.
解:(1) 因为f(z)?u?iv是调和函数,所以
?v?u?v?u????3x2?6xy?3y2,??3x2?6xy?3y2. ?x?y?y?x于是
v??(3x2?6xy?3y2)dy?g(x)?3x2y?3xy2?y3.
那么
?v?g?(x)?6xy?3y2??3x2?6xy?3y2, ?x则
g(x)??x3?C,
所以
v??x3?3x2y?3xy2?y3?C,
f(z)?(x3?3x2y?3xy2?y3)?i(?x3?3x2y?3xy2?y3)?iC3223?(1?i)??x?3x(iy)?3x(iy)?(iy)???iC
?(1?i)z3?iC?v?2xy?vx2?y2?(2) ,. ??x(x2?y2)2?y(x2?y2)2因为f(z)?u?iv是调和函数,所以
x2?y2?2xy(x?iy)211?, f?(z)?v??iv??i???yx(x2?y2)2(x2?y2)2(x2?y2)2(x?iy)2z2从而f(z)??1z?C. 由f(2)?0知C?12,所以f(z)?112?z.
(3) 因为f(z)?u?iv是调和函数,所以
?v?u?v?x???y??2(x?1),?y??u?x?2y. 于是
v??2ydy?g(x)?y2.
那么
?v?x?g?(x)??2(x?1), 则
g(x)??x2?2x?C,
所以
v??x2?2x?y2?C,
f(z)?(2xy?2y)?i(?x2?2x?y2)?iC??i??(x?iy)2?2(x?iy)?1???iC
??i(z?1)2?iC由f(2)??i知C?0,所以f(z)??i(z?1)2.
习题4:
1、下列数列?zn?是否收敛?若收敛,求其极限.
?n(1) z1?ni?i?nin?1?ni; (2) zn???1?2??; (3) zn?(?1)?n?1;n?i(4) z?n?e2.
1?ni1?n2?2ni1?n22n1?n2???i??1,虚解:(1) zn?,当n??时,实部1?ni1?n21?n21?n21?n2部
2n?0,所以?zn?收敛于?1. 21?n?ni??(2) zn??1???2?敛于0.
?5??ni?5?,当时??en????2???2???0,那么zn?0,所以?zn?收
?????n?n(3) 当n??时,实部(?1)n是发散的,所以?zn?发散. (4) zn?e?n?i2?cosn?n??isin,实部和虚部都发散,所以?zn?发散. 22
2、判断下列级数的收敛性与绝对收敛性:
??1?3?(1) ???1???i2?; (3)
n?n?1???n???n?n?1?e?n?i22n.
n3?1??1?解:(1) 记zn??1???i2,则当n??时Re(zn)??1???e,那么zn不趋近于
n?n??n?0,所以级数发散.
n?i2n?i2n(3)
?n?1?e?n2??1e收敛,即级数绝对收敛,所以收敛. ?22nn?1n?1n???
7、将下列各函数展成z的幂级数,并指出它们的收敛半径. (1)
12cosz. ; (3) 31?z?11解:(1) ???(?z3)n?1?z3?z6??. 331?z1?(?z)n?0因为??lim(?1)n?1(?1)nn???1,所以收敛半径R?1.
2n?cos2z?11??n(2z)cosz????(?1)?1?22?n?0(2n)!?(3)
2n?12n23456?2z12z2z2z??(?1)n??1?????(2n)!22!4!6!n?02
(?1)因为??limn??n?122n?1(2n?2)!n22n?1(?1)(2n)!?lim4?0,所以收敛半径R??.
n??(2n?1)(2n?2)
8、将下列各函数在指定点z0处展成泰勒级数,并指出它们的收敛半径. (3)
11,; (4) ,z0?1?i; (6) arctanz,z0?0. z??102z4?3z?1?n?1,则2??(n?1)(z?1)n.
zn?0f(n)(z0)(?1)n(n?1)!z?n?2解:(3) cn??n!n!因为??limn??z?z0n?1n?1,所以收敛半径R?1.
3n,则 ?(1?3i)n?1f(n)(z0)3nn!(4?3z)?n?1(4) cn??n!n!z?z0?13nn. ??z?(1?i)??4?3zn?0(1?3i)n?13n?1因为??limn??(1?3i)n?2z3103n,所以收敛半径. R??n?13(1?3i)102n?1??zzdz?2n2nnz(6) arctanz??. ???n?0(?z)dz???(?z)dz??(?1)01?z2002n?1n?0n?0(?1)n?1因为??limn??2n?3(?1)n?1,所以收敛半径R?1.
2n?1
10、求下列各函数在指定圆环域的洛朗级数展开式: (2)
1,0?z?1,1?z?1???; 2z(1?z)1,在以i为中心的圆环域内;
z2(z?i)1,z?3.
(z?2)(z?3)(5)
(7)
?11?1??n??解:(2) 在0?z?1内,由于??z,且?,所以 2(1?z)1?zn?0?1?z??1??(n?1)zn, 2(1?z)n?0?1从而??(n?2)zn. 2z(1?z)n??1在1?z?1???内,由于
1?1,所以 z?1n11111??1??????????, z1?(z?1)z?11?1z?1n?0?z?1?z?1?1(?1)n从而. ??2n?3z(1?z)n?0(z?1)1?1??(5) 当0?z?i?1时,由于????2,且
z?z?n?11111??z?i?n(z?i), ??????????(?1)n?1z?izi?(z?i)i1?in?0?i?n?0iinn?1n?2??11nn(z?i)n?1n(z?i)所以2???(?1),从而2. ??(?1)n?1n?1ziz(z?i)in?1n?1当1?z?i??时,由于
i?1,所以 z?1n?11111??i?inn, ???????????(?1)zi?(z?i)z?i1?iz?in?0?z?i?n?0(z?i)n?1z?in?11?1??n(n?1)i且????2,从而2??(?1),所以 n?2zzz(z?i)??n?1n?1n(n?1)i. ??(?1)z2(z?i)n?1(z?i)n?3(7) 由于
23?1且?1,所以 zz1111?11????????(z?2)(z?3)z?3z?2z?1?3z1?2z? nnnnnn????1??3?3?2?2??13?2??????????????nn?1z?zzzzz????n?0n?0n?0n?0???
习题5:
1、求下列函数的孤立奇点并确定它们的类别,若是极点,指出它们的级. (1)
11sinz1ln(z?1)sin; (3) ; (4) ; (7) ; (11) .
z31?zzz2(ez?1)z(z2?1)211lim??,的孤立奇点.由于2222z?0z(z?1)z(z?1)解:(1) 易见z?0,z??i是f(z)?1??,所以z?0,z??i是极点.
z??iz(z2?1)2limz?0,一级极点,z??i,二级极点.
sinz(3) lim3??,所以z?0是极点.z?0,二级极点.
z?0zln(z?1)ln(z?1)?1,所以z?0是可去奇点;(4) 易见z?0是f(z)?的孤立奇点,且lim
z?0zz(7) z?0,三级极点,z?2k?i(k??1,?2,?),一级极点; (11) z?1,本性奇点.
5、求下列各函数在有限奇点处的留数.
z2112zsin(2) ; (3) ; (6) . 222zz?1?z??z?1?解:(2) 记f(z)?规则Ⅰ,
1,则易见0,?1是f(z)的孤立奇点,且他们都是一级极点.由2z?1?z?1?1,
z?01?z2Res[f(z),0]?lim?z?0?f(z)?limz?0Res[f(z),1]?lim?z?1?f(z)?limz?1z?1?11??,
z(1?z)2Res[f(z),?1]?lim?z?1?f(z)?limz??111??.
z??1z(1?z)2(3) 记f(z)?z2?z2?1?2,则f(z)有二级极点?i.由规则Ⅱ,
Res[f(z),i]?1d2izi, lim?z?if(z)?lim??????z?i(z?i)3(2?1)!z?idz?41d?2izi. lim?z?if(z)?lim????3??z??iz??i(2?1)!dz(z?i)411,则f(z)有本性奇点z0?0.因为sin在z0?0的去心邻域
zzRes[f(z),?i]?2(6) 记f(z)?zsin0?z??内的洛朗级数为
1?(?1)nz?2n?1 sin??zn?0(2n?1)!于是有
1?(?1)nz?2n?1zsin??zn?0(2n?1)!2?0?z???
其中n?1的项的系数c?1??13!,所以
Res[f(z),0]??
6、利用留数定理计算下列积分. (1)
1 6dz22C,为圆周x?y?2(x?y) 22??(z?1)(z?1)C解:被积函数f(z)在圆周C的内部有一级极点z0?i和二级极点z1?1,由留数的计算规则Ⅰ、Ⅱ得
Res[f(z),i]?lim?z?i?f(z)?limz?iz?i11?, 2(z?1)(z?i)4Res[f(z),1]?1d?2z12lim??z?1?f(z)??lim2??.
?z?1(z?i)2(2?1)!z?1dz?2于是由留数定理得积分值
dz?i?2?iRes[f(z),i]?Res[f(z),1]?? ??22??(z?1)(z?1)2Ce2zdz (2) ?2?(z?1)z?2解:被积函数f(z)在z?2内有一个二级极点z0?1,由留数的计算规则Ⅱ得
Res[f(z),1]?1d2lim??z?1?f(z)??lim2e2z?2e2
?z?1(2?1)!z?1dz?于是由留数定理得积分值
e2z2dz?2?iRes[f(z),1]?4?ei 2??(z?1)z?2(4)
z?sinz??3zdz
2解:被积函数f(z)在z?知
3内有可去奇点z0?0,则Res[f(z),0]?0,所以由留数定理2z?sinz??3zdz?0
2(6)
??z?12esinzdz
z2(z2?1)1内有一个二级极点z0?0,由留数的计算规则Ⅱ得 2解:被积函数f(z)在z?1d2esinz(z2?1)cosz?2zesinzRes[f(z),0]?lim?zf(z)??1 ??limz?0(2?1)!z?0dz?(z2?1)2于是由留数定理得积分值
??z?12esinzdz?2?iRes[f(z),0]?2?i 22z(z?1)d?
5?3cos?i? 9、(1)
?2?0z2?1dz解:令z?e,则d??,cos??.于是
iz2zI??被积函数f(z)?2?0d?2dz ??25?3cos?iz?3z?10z?3?111z??在内有一个一级极点,其留数 z?123z?10z?33
111?1?Res[f(z),?]?lim?z??f(z)?lim?
13z??1?3?8z??3(z?3)33所以
21?I?2?i???
i82(5)
???0x2dx
(x2?1)(x2?4)x2解:R(x)?2是偶函数,而R(z)在上半平面内有一级极点z0?i和z1?2i,2(x?1)(x?4)且
z2i, Res[R(z),i]?lim?z?i?R(z)?lim?z?iz?i(z?i)(z2?4)6z2iRes[R(z),2i]?lim?z?2i?R(z)?lim2??,
z?2iz?2i(z?1)(z?2i)3所以
???0x21?ii??dx??2?i?????
(x2?1)(x2?4)2?63?6(6)
?????cosxdx
(x2?1)(x2?9)1,m?4,n?0,m?n?1,且R(z)在实轴上无孤立奇点,故
x4?10x2?9??解:R(x)?积分
?存在,所求积分I是它的实部.
??eixdx 22(x?1)(x?9)函数R(z)在上半平面有两个一级极点z0?i和z1?3i,而且
eizi, Res[R(z)e,i]?lim?z?i?R(z)e?lim??z?iz?i(z?i)(z2?9)16eizizeizi, Res[R(z)e,3i]?lim?z?3i?R(z)e?lim2?z?3iz?3i(z?1)(z?3i)48e3iziz从而
?????eixi???i2dx?2?i???3e?1? ??223?3(x?1)(x?9)?16e48e?24e所以
?
????cosx?2dx?3e?1? ?223(x?1)(x?9)24e习题8:
4、试求f(t)?e?t的傅氏变换.
解:f(t)的傅里叶变化为
F(?)????e??0????f(t)e?j?tdt??ete?j?tdt??e?te?j?tdt??00??(1?j?)tdt??e?(1?j?)tdt0??0??11(1?j?)t0?e?e?(1?j?)t??1?j??(1?j?)112???21?j?1?j???1
5、试求矩形脉冲f(t)??
?A,0?t??,的傅氏变换.
?0,其他解:f(t)的傅里叶变化为
F(?)??????f(t)e?j?tdt??Ae?j?tdt0?A?j?t?A(1?e?j??)?e?0?j?j?
6、求下列函数的傅氏积分:
?0,???t??1,??1,?1?t?0,?(1) f(t)??
?1,0?t?1,??0,1?t???.解:f(t)是(??.??)上的奇函数,则
a(?)?0,
b(?)?2????0f(?)sin??d??2??10sin??d???21?cos??,
?于是
f(t)??a(?)cos?td???b(?)sin?td?00??????
??0?21?cos?2??1?cos???sin?td???sin?td?
??0?22??1?t,t?1,7、求函数f(t)??的傅氏积分,并计算 2t?1??0,?a(?)?2????xcosx?sinxx?cosdx. 3x2解:f(t)是(??.??)上的偶函数,则
????0f(?)cos??d??2??10(1??2)cos??d??4(sin???cos?)??3,
b(?)?0,
于是
f(t)??a(?)cos?td???b(?)sin?td?00??????
??4(sin???cos?)0??3?cos?td??4????sin???cos?
0?3cos?td?10、求符号函数sgnt????1,t?0,的傅氏变换.(提示:sgnt?2u(t)?1.)
?1,t?0?1?2. ???(?)??2??(?)?j?j?????解:方法一:F[sgnt]?2F[u(t)]?2??(?)?2???0方法二:F(?)?
???sgnt?e?j?tdt???e???j?tdt??e?j?tdt?02. j?11、求函数f(t)?sin2tcost的傅氏变换. 解:f(t)?sin2tcost?sin(2t?t)?sin(2t?t)1??sin3t?sint?,则
22F[f(t)]?1?F[sin3t]?F[sint]?2j??[?(??3)??(??3)??(??1)??(??1)]2
15、利用位移性质计算下列函数的傅氏变换:
(1) u(t?C);(2)
1[?(t?a)??(t?a)] 2解:(1)
F[u(t?C)]?e?j?CF[u(t)]?e?j?C??1?1?j?C???(?)??e???(?); ?j??j?j?a?j?a??(t?a)??(t?a)?F[?(t?a)]?F[?(t?a)]e?e(2) F????cos?a. ?222??
23、求下列函数的傅氏变换: (2) f(t)?ej?0tu(t);(3) f(t)?ej?0tu(t?t0);(4) f(t)?ej?0ttu(t).
1???(?),由卷积j?j?0t解:(2) 记F(?)?F[e]?2??(???0),F2(?)?F[u(t)]?1定理有
11F[f(t)]?F1(?)?F2(?)?2?2??1?2??(???)???(???)0??d????j(???)??????1???(t)????(??t??0)?dt(令t????0) ???j(??t??0)????11???(??t??0)????(???0)j(??t??0)j(???)0t?01j?0t(3) 记F]?2??(???0),F2(?)?F[tu(t)]??1(?)?F[e?2?j???(?),由卷积
定理有
F[f(t)]???1?2??(???)??j??(???)d?0??2????(???)?????1????(t)???j??(??t??0)?dt(令t????0) 2??(??t??)0??11F1(?)?F2(?)?2?2?????11??j??(??t??)???j???(???0)022(??t??0)(???0)t?01?j?t0e???(?),由j?j?0t(4) 记F]?2??(???0),F2(?)?F[u(t?t0)]?1(?)?F[e卷积定理有
F[f(t)]???11F1(?)?F2(?)?2?2??1???(t)?e?j(??t??0)t0???j(??t??0)?
?1??j(???)t02??(???)e???(???)d?0??????j(???)?????(??t??0)?d?(令t????0)
???11e?j(??t??0)t0???(??t??0)?e?j(???0)t0???(???0)j(??t??0)j(???0)t?0习题9:
2、求下列函数的拉氏变换:
?1,0?t?1,?(1) f(t)???1,1?t?5,
?0,t?5?(3) f(t)?cost?(t)?sintu(t). 解:(1) L[f(t)]????01f(t)edt??edt??e?stdt?(1?2e?s?e?5s).
01s?st1?st5(3) L[f(t)]????0f(t)edt???st??01s2(1?sint)edt?1?2?.
s?1s2?1?st
3、求下列周期函数的拉氏变换:
?sint,0?t??,2?(1) f(t)以为周期且在一个周期内的表达式为f(t)??.
0,??t?2??T?11?stL[f(t)]?f(t)edt?sinte?stdt?sT?0?sT?01?e1?e解:
?11?(j?s)t1??edt??sinte?(j?s)tdt??2?s?01?e2j0(1?e??s)(s2?1)??
4、求下列函数的拉氏变换: (1) f(t)?(t?1)e; (2) f(t)?5sin2t?3cost; (3) f(t)?1?te;
(6) f(t)?ecoskt(k为实常数);
tt2t(9) f(t)?te?3tsin2t; (10) f(t)?t?e0t?3tsin2tdt;
e?3tsin2t(11) f(t)?.
tL[f(t)]?L[t2et?2tet?et]?L[t2et]?2L[tet]?L[et]解:(1)
211s2?4s?5??2???32(s?1)(s?1)s?1(s?1)3103s? s2?4s2?1
(2) L[f(t)]?5L[sin2t]?3L[cost]?(3) L[f(t)]?L[1]?L[te]?t11?; 2s(s?1)(6) F(s)?L[coskt]?s?1sL[f(t)]?F(s?1)?,则由位移性质有;
s2?k2(s?1)2?k224(s?3)?L[f(t)]??F(s)?,则; 222(s?3)?4[(s?3)?4]2?te?3tsin2tdt??1F(s),从而 L,则
????0?s(s?3)2?4(9) F(s)?L[e?3tsin2t]?(10) F(s)?L[e?3tsin2t]?2d?1?2(3s?12s?13); L[f(t)]???F(s)??222ds?ss[(s?3)?4]?(11) F(s)?L[e??3tsin2t]?2,则
(s?3)2?4s?3s?3?arccot. 22L[f(t)]??F(s)ds?s?2?arctan
(9) f(t)?te?3tsin2t; (10) f(t)?t?e0t?3tsin2tdt;
e?3tsin2t(11) f(t)?.
tL[f(t)]?L[t2et?2tet?et]?L[t2et]?2L[tet]?L[et]解:(1)
211s2?4s?5??2???32(s?1)(s?1)s?1(s?1)3103s? s2?4s2?1
(2) L[f(t)]?5L[sin2t]?3L[cost]?(3) L[f(t)]?L[1]?L[te]?t11?; 2s(s?1)(6) F(s)?L[coskt]?s?1sL[f(t)]?F(s?1)?,则由位移性质有;
s2?k2(s?1)2?k224(s?3)?L[f(t)]??F(s)?,则; 222(s?3)?4[(s?3)?4]2?te?3tsin2tdt??1F(s),从而 L,则
????0?s(s?3)2?4(9) F(s)?L[e?3tsin2t]?(10) F(s)?L[e?3tsin2t]?2d?1?2(3s?12s?13); L[f(t)]???F(s)??222ds?ss[(s?3)?4]?(11) F(s)?L[e??3tsin2t]?2,则
(s?3)2?4s?3s?3?arccot. 22L[f(t)]??F(s)ds?s?2?arctan
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