5.5 Unit Step Function In applications, systems are often subjected to discontinuous forcing functions. (show real power of Laplace transform) t a a t a t u t t t u 1 0 0 ) ( 0 1 0 0 ) ( 為 typical “ engineering function” 如 electrical or mechanical driving force off/on. ● 把 f(t) 乘上 u(t-a) 可 produce all sorts of effects.
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5.5 Unit Step Function
In applications, systems are often subjected to discontinuous forcing functions.
(show real power of Laplace transform)
taat
atu
tt
tu
100
)(
0100
)(
為 typical “engineering function”
如 electrical or mechanical driving force off/on.
● 把 f(t) 乘上 u(t-a) 可 produce all sorts of effects.
如
而
ttf sin5)(
之部分關掉在會把 2)()2()( ttftutf
2byrightshift)()2()2( tftutf 則把t2
(t-shifting)
)()( btuatu
)()(
00)(
atuat
taatat
tf
)3(2)2(32)( tututf
)()()()(
0)(
00)(
)()()()()()(
)(0)(
)(
btuatutgtf
btbtatgat
tf
atuthatutgtgtf
atthattg
tf
而
1) 若要解之微分方程 右邊為
)()(
100
)(
atutf
taat
tf
即
s
etdetdeatuatuL
atuL
tftybyay
sa
a
tsts
tfL
0)()(
?)(
)()(
?)(
即
2) 若要解之微分方程 右邊為
)()()(
sin)()sin(
sinsin
sinsin
)sin(
?)()sin(
)()(
0)(
?)(
tfLeatuatfL
tLeatuatL
tLeorvLe
vdevevdev
atvtdeat
atuatL
tftybyay
sa
sa
sasa
vssaa
sav
ats
tfL
又
即
令
即
ta
)()sin()(
)sin(00
)(
atuattf
taatat
tf
即
)()()(
)()(
sFeatuatf
sFtfsaL
L
則
若即
)()()(
)()()()()(
)()()(
)()(
00)(
0
sFevdevfevdevf
tdeatftdeatuatfatuatfL
sFeatuatf
sFtf
savssasav
atsts
sa
證：
atv 令
)()()(1
1)()sin(
11
sin
1)(
11
)()(
22
sFeatuatfs
eatuats
t
seatu
s
sFtf
sa
sa
sa
得證即
：：令
：
也可反過來證
)()()(
)()(
)()(
)()()(proof
)()()(
1
1
0)(
0
atuatfsFe
tdeatuatf
tdeatfsFe
tddataat
defdefesFe
atuatfsFe
Lsa
ats
atssa
asssasa
Lsa
)(0a-0integrand
0
atu
即用
間在只要得證
把下限換為
Kreyszig P. 282 例 4.
)()()()()(
)(
21
11
231
)(
)()(23
)()(
)()(23
mtransforLaplace)(
0)0()0()()(23
1
2
2
2
2
tRtqsRsQLty
eetq
sssssQ
sRsQsssR
sY
sRsYss
tyyy
trtyyy
tt
用
求
otherwise0
211)(
ttr
)1(21
)1(21)1(210
1)(2
1
1)(2
1
1)(2)(
1
0
10
0||
21
21
)(
21
21
21
21
)()(
1)()()(
)()()(21if2.Case
)()()(1if1.Case
tt
tttt
tttt
t tt t
t tt
t
t
t
eety
eeeee
ee
tdetde
dee
rdtqr
dtqrtyt
dtqrtyt
t
此時
之值在哪個範圍看所要
)1(2)2(2)1()2(
2
1
)(2)(
1)(2)(
1
021
21
)(
21
)()(
0)(
2if3.Case
tttt
tt
t tt
t
t
eeeety
ee
dee
dtqty
r
t
其他區則
tL
tL
L
s
s
s
s
es
sssssse
s
s
tftusFeLsss
L
ssse
sY
se
sYss
se
tuL
tutryyy
2
11
2
1
1
1
21
22/1
112/1
)2)(1(1
11
fractionpartial11
nconvolutio
)1()1()()2)(1(
1
)2)(1()(
)(23
)1(
)1()(23
也可用可用
然後用先找
例： 0)0()0(otherwise0
11)(
yyt
tr
)1(2)1(
22
2
022
)(20
0
1
21
21
)1()(
21
21
21
2
1)(
1
)()()()(
tt
tttt
t
tt
tt
tt
eetusY
eeee
e
deee
deevgf
edegf
vgfvgfsVsGsFL
)2(21
21
)1(21
21
)2()2()1()1()()()(
)1()1()(21
21
)(
22/1
112/1
)2)(1(1
)()(
)2)(1(1
1)23(
)(
)(2)(3)()(Sol
)1()()()2()1()()(
0)0()0()(23
)2(2)2()1(2)1(
2111
1
2
22
22
tueetuee
tutftutfesFLesFLsYL
tutfesFL
eetf
sss
ssssF
eesss
esss
esY
se
se
sYssYsYsA
ttrBtututrA
yytryyy
tttt
ss
s
tt
ssss
ss
令另法
：
例：
1L 1L 1L
tt
tttt
tt
eeeeee
eeee
ee
ty
2242
)2(2)1(2)2()1(
)1(2)2(
)(21
)(
21
21
21
210
)(
即
1t
21 t
t2
2 3 4 5t
)(tf
)3()2()()(sin)3(sin)2(sin)(sinsin)(
)(.1
tututututtuttuttutttf
tf寫出
)6()4()2()(
21
sin4)(
)(.2
tututututtf
tf
寫出
tt
tt
tf
2sin20
02)(
2 3 4
ss
ss
ess
es
tLes
es
ttuLtuLtuLtfL
ttttutututtutututf
tfsFtf
22
2
11
21
2
sin21
2
)2sin()2()(2)(2)(2.step
)2(sin)2()(2)(2)2sin()2()(2)(2)(
functionstepunitofin terms)(1.step)(mtransforLaplace)(
變為不一定要把
先寫出解：
之求
ttttt
tf
tuttutttuttututtsFL
atuatfsFe
ttttfs
ssss
sF
es
sL
se
Les
Ls
LsFL
es
se
se
sssF
tfsF
sa
ss
s
sss
cos20
202)(
)(cos)2()2(22)()cos()2(4)2()2(22)(now
)()()(
cos422)(1
422)(e
14
22)(
1422
)(
)()(
1
222
21
212
21
211
222
22
即
用
即項若沒有
例：
求有反之
Ex. 3 of P. 271 LC circuit
)()(1
)()(
)()(1)(
)(
2
2
tEtQCtd
QdR
tdQd
L
tdtQd
ti
tEtdtiCtd
tidLtiR
Now, E(t) 如圖：
ttt
tE50
521200
)(
2t
5
E(t)
0
)(,0)0(,)0(,0 0 tQQQQR 求本題
)open0tswitch,0)0(( 之前為在 i
)5(cos1)5()2(cos1)2(cos)(
)5(cos1)5()2(cos1)2(1
)(/1
)cos1(1
sin1
)(1
cos)(
/1/
/1)(
11)(
1
11)(
1)0()0()(
mtransforLaplace
)()(1
)5()2()(step1.
00
252
22
20221
00
522
02
0
5200
2
520
0
2
0
1
0
ttuttuCEtQtQ
ttuttuL
eess
L
tdss
L
EtQtQ
eeCLss
LECLs
sQsQ
es
es
EQsLsQC
sL
es
es
EsQC
QQssQsL
tEtQC
QLtutuEtE
C
Lss
t
ss
ss
ss
Q
？
用
微分為
暫不看
：Solve
CL12
d
dtgf
sGsFL
ssL
t
t
0
0
1
221
1sin
)()(
)()(
11
nconvolutioUse ：
比較麻煩
之方法需分成二段若用
可解及用
55220
3Chap.
)0(domainentireon)(TransformLaplace)( ttQatu
比較麻煩
之方法需分成二段若用
可解及用
55220
3Chap.
)0(domainentireon)(TransformLaplace)( ttQatu
11
sin)()sin(
11
sin)(sin)(sin
442
44)2()2(
2?)2(
)()()(
)(
)(
)(
)()()(
2
2
232
22222
32
2
0
0)(
setLetutL
setLetLetutL
ssse
ttLetLetutLs
tL
tutL
atfLeatutf
atfLe
tdatfee
vdavfe
atvlettdtfeatutfL
ss
sss
s
ss
L
sa
sa
tssa
avs
ats
或
例：
一樣
已知
回到例：
即
也可用
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