The Laplace Transform
CHAPTER 4
Copyright © Jones and Bartlett;滄海書局 Ch4_2
Chapter Contents
4.1 Definition of the Laplace Transform4.2 The Inverse Transform and Transforms of Derivat
ives4.3 Translation Theorems4.4 Additional Operational Properties4.5 The Dirac Delta Function4.6 Systems of Linear Differential Equations
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4.1 Definition of Laplace Transform
Basic DefinitionIf f(t) is defined for t 0, then
(1)
b
bdttftsKdttftsK
00)(),(lim)(),(
If f(t) is defined for t 0, then(2)
is said to be the Laplace Transform of f.
Definition 4.1.1 Laplace Transform
0)()}({ dttfetf stL
Copyright © Jones and Bartlett;滄海書局 Ch4_4
Evaluate L {1}
Solution: Here we keep that the bounds of integral are 0 and in mind.From the definition
Since e-st 0 as t , for s > 0.
Example 1 Using Definition 4.1.1
sse
se
dtedte
sb
b
bst
b
b st
b
st
11limlim
lim)1()1(
0
00
L
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Evaluate L {t}
Solution:
Example 2 Using Definition 4.1.1
2
00
111}1{
1
1}{
ssss
dtess
tet st
st
L
L
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Evaluate L {e-3t}
Solution:
Example 3 Using Definition 4.1.1
3,3
1
3
}{
0
)3(
0
)3(
0
33
ss
se
dtetdeee
ts
tststtL
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Evaluate L {sin 2t}
Solution:
Example 4 Using Definition 4.1.1
00
0
2cos22sin
2sin}{sin2
dttess
te
dttet
stst
stL
00,2cos
2sdtte
sst
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Example 4 (2)
}2{sin4222 t
ssL
0,4
2}2{sin 2
s
stL
00
2sin22cos2
dttess
tes
stst
0,02coslim ste st
t Laplace transform of sin 2t↓
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L is a Linear Tramsform
We can easily verify that
(3) )()(
)}({)}({
)}()({
sGsF
tgtf
tgtf
LL
L
Copyright © Jones and Bartlett;滄海書局 Ch4_10
(a)
(b) (c)
(d) (e)
(f) (g)
Theorem 4.1.1 Transform of Some Basic Functions
s1
}1{ L
,3,2,1,!
}{ 1 nsn
t nnL
ase ta
1
}{L
22}{sinks
ktk
L 22}{cos
kss
tk
L
22){sinks
ktk
L
22}{coshks
stk
L
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Fig 4.1.1 Piecewise-continuous function
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A function f(t) is said to be of exponential order,
if there exists constants c, M > 0, and T > 0, such that
|f(t)| Mect for all t > T. See Fig 4.1, 4.2.
Definition 4.1.2 Exponential Order
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Fig 4.1.2 Function f is of exponential order
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Fig 4.1.3 Functions with blue graphs are of exponential order
tet ||tet 2cos2 tt ee ||
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If f(t) is piecewise continuous on [0, ) and of
exponential order c, then L {f(t)} exists for s > c.
Theorem 4.1.2 Sufficient Conditions for Existence
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Example 5
Find L {f(t)} for
Solution:
0 ,22
20)}({
3
3
3
3
0
s
se
se
dtedtetf
sst
ststL
3,2
30,0)(
t
ttf
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Fig 4.1.5 Piecewise-continuous function in Ex 5
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4.2 The Inverse Transform and Transform of Derivatives
(a)
(b) (c)
(d) (e)
(f) (g)
Theorem 4.2.1 Some Inverse Transform
s1
1 1L
,3,2,1,!1
1
nsn
t nn L
ase ta 11L
221sin
ksk
tk L
221cos
kss
tk L
221sinh
ksk
tk L
221cosh
kss
tk L
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Example 1 Applying Theorem 4.2.1
Find the inverse transform of
(a) (b)
Solution:(a)
(b)
51 1
sL
71
21
sL
45
15
1
241!4
!411
tss
LL
tss
7sin7
17
77
17
12
12
1
LL
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L -1 is also linear
We can easily verify that
(1))}({)}({
)}()({11
1
sGsF
sGsF
LL
L
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Find
Solution:
(2)
Example 2 Termwise Division and Linearity
462
21
ss
L
462
21
ss
L
42
26
42
46
42
21
21
221
sss
sss
LL
L
tt 2sin32cos2
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Example 3 Partial Fractions and Linearity
Find
Solution:Using partial fractions
Then
(3)
If we set s = 1, 2, −4, then
)4)(2)(1(962
1
sssss
L
)4)(2)(1(962
sssss
)2)(1()4)(1()4)(2(
962
ssCssBssA
ss
421
sC
sB
sA
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(4) Thus
(5)
30/1,6/25,5/16 cBA
Example 3 (2)
)4)(2)(1(962
1
sssss
L
ttt eee 42
301
625
516
41
301
21
625
11
516 111
sssLLL
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Transform of Derivatives
(6)
(7)(8)
)}({ tf L
000)()()( dttfestfedttfe ststst
)}({)0( tfsf L
)0()()}({ fssFtf L
)}({ tf L
000)()()( dttfestfedttfe ststst
)}({)0( tfsf L)0()]0()([ ffssFs
)0()0()()}({ 2 fsfsFstf L)0()0()0()()}({ 23 ffsfssFstf L
Copyright © Jones and Bartlett;滄海書局 Ch4_26
If are continuous on [0, ) and are of
exponential order and if f(n)(t) is piecewise-continuous
On [0, ), then
where
Theorem 4.2.2 Transform of a Derivative
)1(,,, nfff
.)}({)( tfsF L
)0()0()0()(
)}({)1(21
)(
nnnn
n
ffsfssFs
tf
L
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Solving Linear ODEs
Then
(9)
(10)
)(01
1
1 tgyadt
yda
dtyd
a n
n
nn
n
n
1)1(
10 )0(,)0(,)0( n
n yyyyyy
)}({}{01
1
1 tgyadt
yda
dtyd
a n
n
nn
n
n LLLL
)]0()0()([ )1(1 nnnn yyssYsa
)(
)(
)]0()0()([
0
)2(211
sG
sYa
yyssYsa nnnn
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We have
(11)
where
)()()()( sGsQsYsP
)()(
)()(
)(sPsG
sPsQ
sY
01
1)( asasasP nn
nn
Copyright © Jones and Bartlett;滄海書局 Ch4_29
Find unknown that satisfies a DE and Initial condition
)(tyTransformed DE becomes an algebraic equation In )(sY
Apply Laplace transform L
Solve transformed equation for )(sY
Solution of original IVP
)(ty Apply Inverse transform 1L
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Solve
Solution:
(12)
(13)
Example 4 Solving a First-Order IVP
6)0(,2sin133 ytydtdy
}2{sin13}{3 tydtdy
LLL
426
)(36)( 2
ssYssY
426
6)()3( 2
ssYs
)4)(3(506
)4)(3(26
36
)( 2
2
2
sss
ssssY
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We can find A = 8, B = −2, C = 6Thus
43)4)(3(506
22
2
sCBs
sA
sss
)3)(()4(506 22 sCBssAs
462
38
)4)(3(506
)( 22
2
ss
ssss
sY
42
34
23
18)( 2
12
11
sss
sty LLL
ttety t 2sin32cos28)( 3
Example 4 (2)
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Example 5 Solving a Second-Order IVP
Solve
Solution:
(14)
Thus
5)0(',1)0(,2'3" 4 yyeyyy t
}{}{23 42
2tey
dtdy
dtyd
LLLL
41
)(2)]0()([3)0()0()(2
ssYyssYysysYs
41
2)()23( 2
sssYss
)4)(2)(1(96
)4)(23(1
232
)(2
22
sssss
ssssss
sY
ttt eeesYty 421
301
625
516
)}({)( L
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4.3 Translation Theorems
Proof:L {eatf(t)} = e-steatf(t)dt = e-(s-a)tf(t)dt = F(s – a)
If L {f} = F(s) and a is any real number, then
L {eatf(t)} = F(s – a)
Theorem 4.3.1 First Translation Theorem
assat tftfe )}({)}({ LL
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Fig 4.3.1 Shift on s-axis
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Evaluate (a) (b)
Solution:
(a) (b)
Example 1 Using the First Translation Theorem
}{ 35 te tL }4cos{ 2 te tL
45
45335
)5(6!3
}{}{
sstte
ssss
t LL
16)2(2
16
}4{cos}4cos{
22
2
)2(2
ss
ss
tte
ss
sst LL
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Inverse Form of Theorem 4.3.1
(1)
where
)(})({)}({ 11 tfesFasF atass
LL
.)}({)( 1 sFtf L
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Evaluate (a) (b)
Solution:(a)
we have A = 2, B = 11
(2)
Example 2 Partial Fractions and Completing the square
21
)3(52
ss
L
643/52/
21
sss
L
22 )3(3)3(52
sB
sA
ss
BsAs )3(52
22 )3(11
32
)3(52
ssss
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Example 2 (2)
And
(3)
From (3), we have
(4)
211
21
)3(1
113
12
)3(52
ssss
LLL
teess tt 33
21 112
)3(52
L
tess
t
ss
3
32
12
1 1)3(
1
LL
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Example 2 (3)
(b) (5)
(6)
(7)
2)2(3/52/
643/52/
22
ss
sss
2)2(1
32
2)2(2
21
643/52/
21
21
21
sss
sss
LL
L
22
1
22
1
22
232
221
ssss sss
LL
tete tt 2sin32
2cos21 22
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Solve
Solution:
Example 3 IVP
17)0(',2)0(,9'6 32 yyetyyy t
)(9)]0()([6)0()0()(2 sYyssYysysYs 3)3(2s
)()96( 2 sYss3)3(
252
ss
)()3( 2 sYs3)3(
252
ss
)(sY52 )3(
2)3(52
sss
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Example 3 (2)
(8)
52 )3(2
)3(11
32
)(
sss
sY
51
211
)3(!4
!42
)3(1
113
12
)(
sss
ty
LLL
,1 3
32
1 t
ss
tes
L t
ss
ets
34
35
1 !4
L
ttt etteety 3433
121
112)(
Copyright © Jones and Bartlett;滄海書局 Ch4_42
Example 4 An IVP
Solve
Solution:
0)0(',0)0(,164 yyeyyy t
)(6)]0()([4)0()0()(2 sYyssYysysYs1
11
ss
)()64( 2 sYss)1(
12
sss
)(sY)64)(1(
122
sssss
643/52/
13/16/1
)( 2
sss
sssY
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Example 4 (2)
tetee ttt 2sin32
2cos21
31
61 22
2)2(2
232
2)2(2
21
11
311
61
)(
21
21
11
sss
sssY
LL
LL
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The Unit Step Function U (t – a) is
Definition 4.3.1 Unit Step Function
at
atat
,1
0,0)(U
See Fig 4.3.2.
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Copyright © Jones and Bartlett;滄海書局 Ch4_46
Copyright © Jones and Bartlett;滄海書局 Ch4_47
Also a function of the type
(9)is the same as
(10)Similarly, a function of the type
(11)
can be written as (12)
atth
attgtf
),(
0),()(
)()()()()()( atthattgtgtf UU
bt
btatg
at
tf
,0
),(
0,0
)(
)]()()[()( btattgtf UU
Copyright © Jones and Bartlett;滄海書局 Ch4_48
Express
in terms of U (t).Solution:From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0
f(t) = 20t – 20tU (t – 5)
Consider the function
(13) See Fig 4.3.5.
Example 5 A Piecewise-Defined Function
5,0
50,20)(
t
tttf
atatf
atatatf
),(
0,0)()( U
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Copyright © Jones and Bartlett;滄海書局 Ch4_50
Fig 4.3.6 Shift on t-axis
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Proof:
If F(s) = L {f}, and a > 0, then L {f(t – a)U (t – a)} = e-asF(s)
Theorem 4.3.2 Second Translation Theorem
dtatatfedtatatfea
sta st )()()()(
0UU
)}()({ atatf UL
0)( dtatfe st
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Theorem 4.3.2 proof
Let v = t – a, dv = dt, then
If f(t) = 1, then f(t – a) = 1, F(s) = L {1} = 1/s,
(14)eg: The L.T. of Fig 4.3.4 is
0
)( )( dvvfe avs )}({)(0
tfedvvfee assvas L
se
atas
)}({UL
se
se
s
tttfss 32
31
2
)}3({)}2({3}1{2)}({
ULULLL
)}()({ atatf UL
Copyright © Jones and Bartlett;滄海書局 Ch4_53
Inverse Form of Theorem 4.7
(15))()()}({1 atatfsFe as UL
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Example 6 Using Formula (15)
Evaluate (a) (b)
Solution:(a) then
(b) then
se
s21
41
L
2/
21
9se
ss L
tesFssFa 41 )}({),4/(1)(,2 L
)2(4
1 )2(421
tee
sts UL
tsFsssFa 3cos)}({),9/()(,2/ 12 L
223cos
92/
21 tte
ss s UL
Copyright © Jones and Bartlett;滄海書局 Ch4_55
Alternative Form of Theorem 4.3.2
Since , then
The above can be solved. However, we try another approach.Let u = t – a,
That is,
(16)
4)2(4)2( 22 ttt
)}2(4)2()2(4)2()2{(
)}2({2
2
ttttt
tt
UUUL
UL
0
)( )()()}()({ duaugedttgeattg aus
a
stUL
)}({)}()({ atgeattg as LUL
Copyright © Jones and Bartlett;滄海書局 Ch4_56
Find
Solution:With g(t) = cos t, a = , then
g(t + ) = cos(t + )= −cos tBy (16),
Example 7 Second Translation Theorem – Alternative Form
)}({cos ttUL
ss es
stett
1}{cos)}({cos 2LUL
Copyright © Jones and Bartlett;滄海書局 Ch4_57
Solve
Solution:We find f(t) = 3 cos tU (t −), then
(17)
Example 8 An IVP
tt
ttf
,sin3
0,0)(
5)0(,)(' ytfyy
)()0()( sYyssY ses
s
13 2
)()1( sYs ses
s
13
5 2
sss e
ss
es
ess
sY
111
11
23
15
)( 22
Copyright © Jones and Bartlett;滄海書局 Ch4_58
Example 8 (2)
It follows from (15) with a = , then
Thus
(18)
See Fig 4.3.7.
)()sin(1
1,)(
11
21)(1
tte
stee
ssts ULUL
)()cos(12
1
tte
ss s UL
)()cos(23
)()sin(23
)(23
5)( )( ttttteety tt UUU
)(]cossin[23
5 )( tttee tt U
tttee
tett
t
,cos2/3sin2/32/35
0,5)(
Copyright © Jones and Bartlett;滄海書局 Ch4_59
Fig 4.3.7 Graph of function (18) in Ex 8
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Remember that the DE of a beam is
(19)
Beams
)(4
4
xwdx
ydEI
Copyright © Jones and Bartlett;滄海書局 Ch4_61
A beam of length L is embedded at both ends as Fig 4.3.8. Find the deflection of the beam when the load is given by
Solution:We have the boundary conditions: y(0) = y(L) = 0, y’(0) = y’(L) = 0. By (10),
Example 9
LxL
LxxL
wxw
2/,0
2/0,2
1)( 0
22
12
1)( 00
Lxx
Lwx
Lwxw U
222
2 0 Lx
Lxx
LL
wU
Copyright © Jones and Bartlett;滄海書局 Ch4_62
Fig 4.3.8 Embedded beam with a variable load in Ex 9
Copyright © Jones and Bartlett;滄海書局 Ch4_63
Example 9 (2)
Transforming (19) into
where c1 = y”(0), c3 = y(3)(0)
2
220 1122 Lse
sssL
L
w
2665
043
31
222
0)3(4
1122)(
1122)0()0(")(
Ls
Ls
esss
LEIL
w
s
c
sc
sY
esss
LEIL
wysysYs
)0()0()0()0()( 234 yysysyssYsEI
Copyright © Jones and Bartlett;滄海書局 Ch4_64
Thus
412
311 !3
!3!2
!2
)(
sc
sc
xy
LL
2/
61
61
510 !5
!51!5
!51!4
!42/2 Lse
sssL
EILw
LLL
2225
6062
55403221 L
xL
xxxL
EIL
wx
cx
cU
Example 9 (3)
Copyright © Jones and Bartlett;滄海書局 Ch4_65
Example 9 (4)
Applying y(L) = y’(L) = 0, then
Thus
01920
49
62
40
3
2
2
1 EI
LwLc
Lc
0960
85
2
302
21 EI
LwLcLc
EILwcEILwc 40/9,960/23 022
01
2225
60
803
192023
)(
5540
3022
0
Lx
Lxxx
LEIL
w
xEILw
xEILw
xy
U
Copyright © Jones and Bartlett;滄海書局 Ch4_66
4.4 Additional Operational Properties
Multiplying a Function by tn
that is, Similarly,
)}({)()]([
)(
00
0
ttfdtttfedttfes
dttfedsd
dsdF
stst
st
L
)}({)}({ tfdsd
ttf LL
)}({)}({
)}({)}({)}({
2
2
2
tfdsd
tfdsd
dsd
ttfdsd
ttfttft
LL
LLL
Copyright © Jones and Bartlett;滄海書局 Ch4_67
If F(s) =L {f(t)} and n = 1, 2, 3, …, then
Theorem 4.4.1 Derivatives of Transform
)()1()}({ sFdsd
tft n
nnn L
Copyright © Jones and Bartlett;滄海書局 Ch4_68
Example 1 Using Theorem 4.4.1
Find L {t sin kt}
Solution:With f(t) = sin kt, F(s) = k/(s2 + k2), then
22222 )(2
}{sin}sin{
ksks
ksk
dsd
ktdsd
ktt
LL
Copyright © Jones and Bartlett;滄海書局 Ch4_69
Different approaches
Theorem 4.3.1:
Theorem 4.4.1:
23
233
)3(11
}{}{
sstte
ssss
t LL
2233
)3(1
)3(3
1}{}{
ss
sdsd
edsd
te tt LL
Copyright © Jones and Bartlett;滄海書局 Ch4_70
Solve
Solution:
or
From example 1,
Thus
Example 2 An IVP
1)0(,0)0(,4cos16 xxtxx
222
22
)16(161
)(
161)()16(
ss
ssX
ss
sXs
kttks
kssin
)(2
2221
L
ttst
ss
stx
4sin81
sin41
)16(8
81
164
41
)( 221
21
LL
Copyright © Jones and Bartlett;滄海書局 Ch4_71
Convolution
A special product of f * g is defined by
(2)and is called the convolution of f and g. The convolution is a function of t, eg:
(3)
Note: f * g = g * f
t
dtgfgf0
)()(*
)cossin(21
)sin(sin0
ttt ettdtete
Copyright © Jones and Bartlett;滄海書局 Ch4_72
Proof:
If f(t) and g(t) are piecewise continuous on [0, ) and
of exponential order, then
Theorem 4.4.2 Convolution Theorem
)()()}({)}({}{ sGsFtgtfgf LLL
0
)(
0
0 0
)(
00
)()(
})()(
)()()()(
dgedf
ddgfe
dgedfesGsF
s
ss
Copyright © Jones and Bartlett;滄海書局 Ch4_73
Theorem 4.4.2 proof
Holding fixed, let t = + , dt = d
The integrating area is the shaded region in Fig 4.4.1.Changing the order of integration:
dttgedfsGsF st )()()()(0
}{
)()(
)()()()(
00
00
gf
dtdtgfe
dtgfdtesGsF
tst
tst
L
Copyright © Jones and Bartlett;滄海書局 Ch4_74
Fig 4.4.1 Changing order of integration from t first to first
Copyright © Jones and Bartlett;滄海書局 Ch4_75
Example 3 Transform of a Convolution
Find
Solution:Original statement = L {et * sin t}
t
dte0
)sin( L
)1)(1(1
11
11
22
ssss
Copyright © Jones and Bartlett;滄海書局 Ch4_76
Inverse Transform of Theorem 4.9
L -1{F(s)G(s)} = f * g(4)
Look at the table in Appendix III,
(5)
222
3
)(2
}cos{sinks
kktktkt
L
Copyright © Jones and Bartlett;滄海書局 Ch4_77
Example 4 Inverse Transform as a Convolution
Find
Solution:Let
then
(6)
22
21
)(1ks
L
22
1)()(
kssGsF
ktkks
kk
tgtf sin11
)()( 221
L
t
dtkkkks 02222
1 )(sinsin1
)(1 L
Copyright © Jones and Bartlett;滄海書局 Ch4_78
Example 4 (2)
Now recall thatsin A sin B = (1/2) [cos (A – B) – cos (A+B)]
If we set A = k, B = k(t − ), then
3
02
022221
2cossin
cos)2(sin21
21
]cos)2([cos21
)(1
kktktkt
kttkkk
dkttkkks
t
t
L
Copyright © Jones and Bartlett;滄海書局 Ch4_79
Transform of an Integral
When g(t) = 1, G(s) = 1/s, then
(7)
(8)
ssF
dft )(
)(0
L
s
sFdf
t )()( 1
0L
Copyright © Jones and Bartlett;滄海書局 Ch4_80
Examples:
ttdss
ttdss
tdss
t
t
t
cos121
)sin()1(
1
sin)cos1()1(
1
cos1sin)1(
1
2
0231
0221
021
L
L
L
Copyright © Jones and Bartlett;滄海書局 Ch4_81
Volterra Integral Equation
t
dthftgtf0
)()()()( (9)
Copyright © Jones and Bartlett;滄海書局 Ch4_82
Example 5 An Integral Equation
Solve
Solution:First, h(t-) = e(t-), h(t) = et. From (9)
Solving for F(s) and using partial fractions
)(for )(3)(0
2 tfdefettft tt
11
)(1
123)( 3
s
sFss
sF
12166
)( 43
sssssF
tett
sssstf
213
11
21!3!2
3)(
32
114
13
1 LLLL
Copyright © Jones and Bartlett;滄海書局 Ch4_83
Series Circuits
From Fig 4.4.2, we have
(10)
which is called the integrodifferential equation.
)()(1
)(0
tEdiC
tRidtdi
Lt
Copyright © Jones and Bartlett;滄海書局 Ch4_84
Example 6 An Integrodifferential Equation
Determine i(t) in Fig 4.4.2, when L = 0.1 h, R = 2 , C = 0.1 f, i(0) = 0, and
E(t) = 120t – 120tU (t – 1)
Solution:Using the data, (10) becomes
And then
)1(120120)(10)(21.00
ttditidtdi t
U
ss e
se
ssssI
sIssI111
120)(
10)(2)(0.1 22
Copyright © Jones and Bartlett;滄海書局 Ch4_85
Example 6 (2)
sss
s
ss
es
es
es
essss
es
essss
sI
22
2
222
)10(1
)10(10/1
10100/1
100/1)10(
10/110
100/11/1001200
)10(1
)10(1
)10(1
1200)(
)1()1(1080120
)]1([12)]1(1[12)()1(1010
)1(1010
tette
teettitt
tt
U
UU
Copyright © Jones and Bartlett;滄海書局 Ch4_86
Example 6 (3)
Written as a piecewise- defined function
(11)
1 ,)1(10801201212
10 ,1201212)(
)1(1010)1(1010
1010
tetteee
tteeti
tttt
tt
Copyright © Jones and Bartlett;滄海書局 Ch4_87
Fig 4.4.3
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Post Script – Green’s Function Redux
By applying the Laplace transform to the initial-value problem
where a and b are constants, we find that the transform of y(t) is
where F(s) = L {f(t)}. By rewriting the foregoing transform as the product
,0)0( ,0)0( ),( yytfbyyay
basssF
sY
2
)()(
)(1
)( 2 sFbass
sY
Copyright © Jones and Bartlett;滄海書局 Ch4_89
We can use the inverse form of the convolution theorem (4) to write the solution of the IVP as
(12)
where and On the other hand, we know from (9) of Section 3.10
that the solution of the IVP is also given by
(13)where G(t, ) is the Green’s function for the differential equation.
t
dftgty0
)()()(
)(1
21 tg
bass
L ).()}({1 tfsF L
t
dftGty0
)() ,()(
Copyright © Jones and Bartlett;滄海書局 Ch4_90
By comparing (12) and (13) we see that the Green’s function for the differential equation is related to by
(14)For example, for the initial-value problem y(0) = 0, y’(0) = 0 we find
Thus from (14) we see that the Green’s function for the DE is G(t, ) = g(t - ) = 1/2 sin 2(t - ). See Example 4 in Section 3.10.
)(1
21 tg
bass
L
)() ,( tgtG),(4 tfyy
)(4 tfyy
)(2sin21
41
21 tgt
s
L
Copyright © Jones and Bartlett;滄海書局 Ch4_91
Periodic Function
f(t + T) = f(t)
If f(t) is piecewise continuous on [0, ), of exponential order, and periodic with period T, then
Theorem 4.4.3 Transform of a Periodic Function
T st
sT dttfee
tf0
)(1
1)}({L
Copyright © Jones and Bartlett;滄海書局 Ch4_92
Use the same transform method
T
stT st dttfedttfetf )()()}({0
L
Theorem 4.4.3 proof
T stsT
sTT st
sT
T
st
dttfee
tf
tfedttfetf
tfedttfe
0
0
)(1
1)}({
)}({)()}({
)}({)(
L
LL
L
Copyright © Jones and Bartlett;滄海書局 Ch4_93
Find the L. T. of the function in Fig 4.4.4.
Solution:We find T = 2 and
From Theorem 4.4.3,
(15)
Example 7
21 ,0
10 ,1)(
t
tTE
)1(
111
101
11
)}({ 2
1
02 s
s
sst
s esse
edte
etE
L
Copyright © Jones and Bartlett;滄海書局 Ch4_94
Fig 4.4.4
Copyright © Jones and Bartlett;滄海書局 Ch4_95
Example 8 A Periodic Impressed Voltage
The DE
(16)Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.4.4.
Solution:
or
(17)
Because and
)(tERidtdi
L
s
s
eLRssL
sI
essRIsLsI
11
)/(/1
)(
)1(1
)()(
sss-s eee
e321
11
LRsRL
sRL
LRss ///
)/(1
Copyright © Jones and Bartlett;滄海書局 Ch4_96
Then i(t) is described as follows and see Fig 4.4.5:
(15)
...1111
)( 2
ss eeLRssR
sI
43 ,
32 ,1
21 ,
10 ,1
)(
)3()2()1(
)2()1(
)1(
teeee
teee
tee
te
ti
tttt
ttt
tt
t
Example 8 (2)
Copyright © Jones and Bartlett;滄海書局 Ch4_97
Fig 4.4.5
Copyright © Jones and Bartlett;滄海書局 Ch4_98
4.5 The Dirac Delta Function
Unit ImpulseSee Fig 4.5.1(a). Its function is defined by
(1)
where a > 0, t0 > 0.
For a small value of a, a(t – t0) is a constant function of large magnitude. The behavior of a(t – t0) as a 0, is called unit impulse, since it has the property . See Fig 4.5.1(b).
1)(0 0 dttt
att
attata
att
tta
0
00
0
0
,0
,21
0 ,0
)(
Copyright © Jones and Bartlett;滄海書局 Ch4_99
Fig 4.5.1 Unit impulse
Copyright © Jones and Bartlett;滄海書局 Ch4_100
The Dirac Delta Function
This function is defined by (t – t0) = lima0 a(t – t0)(2)The two important properties:
(i)
(ii) , x > t0
The unit impulse (t – t0) is called the Dirac delta function.
1)(0 0 x
dttt
0
00 ,0
,)(
tt
tttt
Copyright © Jones and Bartlett;滄海書局 Ch4_101
Proof:
The Laplace Transform is
(4)
For t0 > 0, (3)
Theorem 4.5.1 Transform of the Dirac Delta Function
0)}({ 0stett L
))](()(([21
)( 000 attatta
tta UU
saee
e
se
se
att
sasast
atsats
a
2
21
)}({
0
00 )()(
0L
Copyright © Jones and Bartlett;滄海書局 Ch4_102
When a 0, (4) is 0/0. Use the L’Hopital’s rule, then (4) becomes 1 as a 0.Thus ,
Now when t0 = 0, we have
00
2lim)(lim)(
00
00
stsasa
a
sta
ae
saee
etttt
LL
1)( tL
Copyright © Jones and Bartlett;滄海書局 Ch4_103
Example 1 Two Initial-Value Problems
Solve subject to (a) y(0) = 1, y’(0) = 0(b) y(0) = 0, y’(0) = 0Solution:(a) s2Y – s + Y = 4e-2s
Thusy(t) = cos t + 4 sin(t – 2)U (t – 2)
Since sin(t – 2) = sin t, then
(5)See Fig 4.5.2.
),2(4" tyy
14
1)( 2
2
2
se
ss
sYs
2 ,sin4cos
20,cos)(
ttt
ttty
Copyright © Jones and Bartlett;滄海書局 Ch4_104
Fig 4.5.2
Copyright © Jones and Bartlett;滄海書局 Ch4_105
Example 1 (2)
(b)
Thus y(t) = 4 sin(t – 2)U (t – 2)and
(6)
14
)( 2
2
se
sYs
2 ,sin4
20,0
)2()2sin(4)(
tt
t
ttty U
Copyright © Jones and Bartlett;滄海書局 Ch4_106
Fig 4.5.3
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4.6 Systems of Linear DEs
Coupled StringsIn example 1, we will deal with
(1))(
)(
12222
1221111
xxkxm
xxkxkxm
Copyright © Jones and Bartlett;滄海書局 Ch4_108
Example 1 Example 4 of Section 3.12 Revised
Use L.T. to solve
(2)where x1(0) = 0, x1’(0) = 1, x2(0) = 0, x2’(0) = −1.
Solution: s2X1(s)– sx1(0) – x1’(0) + 10X1(s) – 4X2(s) = 0 −4X1(s) + s2X2(s) – sx2(0) – x2’(0) + 4X2(s) = 0
Rearrange:(s2 + 10)X1(s) – 4 X2(s) = 1 −4X1(s) + (s2 + 4)X2(s) = −1 (3)
044
04 10
221
211
xxx
xxx
Copyright © Jones and Bartlett;滄海書局 Ch4_109
Example 1 (2)
Solving (3) for X1:
Use X1(s) to get X2(s)
tttx
sssss
sX
32sin53
2sin10
2)(
125/6
25/1
)12)(2()(
1
2222
2
1
tttx
sssss
sX
32sin10
32sin
52
)(
125/3
25/2
)12)(2(6
)(
2
2222
2
2
Copyright © Jones and Bartlett;滄海書局 Ch4_110
Example 1 (3)
Then
(4)tttx
tttx
32sin10
32sin
52
)(
32sin53
2sin10
2)(
2
1
Copyright © Jones and Bartlett;滄海書局 Ch4_111
Networks
From Fig 4.6.1, we have
(5) 0
)(
122
2
uidtdi
RC
tERidtdi
L
Copyright © Jones and Bartlett;滄海書局 Ch4_112
Solve (5) where E(t) = 60 V, L = 1 h, R = 50 ohm, C = 10-4 f, i1(0) = i2(0) = 0.
Solution:We have
Then sI1(s) + 50I2(s) = 60/s−200I1(s) + (s + 200)I2(s) = 0
Example 2 An Electric Network
0)10(50
6050
1224
21
iidtdi
idtdi
Copyright © Jones and Bartlett;滄海書局 Ch4_113
Example 2 (2)
Solving the above
Thus
222
221
)100(120
1005/65/6
)100(12000
)(
)100(60
1005/65/6
)100(1200060
)(
ssssssI
ssssss
sI
tt
tt
teeti
teeti
1001002
1001001
12056
56
)(
6056
56
)(
Copyright © Jones and Bartlett;滄海書局 Ch4_114
Double Pendulum
From Fig 4.6.2, we have
(6)
0)()( 1121221212
121 glmmllmlmm
022212122222 glmllmlm
Copyright © Jones and Bartlett;滄海書局 Ch4_115
Please verify that when
the solution of (6) is
(7)
See Fig 4.6.3.
Example 3 Double Pendulum
,1)0(,1)0(,16,1,3 212121 llmm
0)0(',0)0(' 21
ttt
ttt
2cos23
32
cos21
)(
2cos43
32
cos41
)(
2
1
Copyright © Jones and Bartlett;滄海書局 Ch4_116
Fig 4.6.3
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