Signals and Systems
Chapter 4 the Laplace Transform
4.4 The Inverse Laplace Transform
Review
desXsX tj)(
2
1)(1-F
Multiplying both sides by te
desXtx ts)(
2
1)(
s =σ+ jω , ds =jdω
j
j
ts dsesXj
tx
)(
2
1)(
tetx )(
ttjt etxFdteetxsX
)()()(
The Inverse Laplace Transform:
This equation states that x(t) can be represented as a weighted integral of complex exponentials.The contour of integration is the straight line in the s-plane corresponding to all points s satisfying Re(s)=
j
j
ts dsesXj
tx
)(
2
1)(
The formal evaluation of the integral for a general X(s) requires the use of contour integration (围线积分 ) in the complex plane,a topic that we will not consider here.The inverse Laplace transform can be determined by partial-fraction expansion.
For example: )2)(1(
1
sssX determine x(t).
Firstly, perform a partial-fraction expansion to obtain:
.21)2)(1(
1
s
B
s
A
sssX
Secondly, evaluate the coefficients:
.1
,1
B
A
thus ,the partial-fraction expansion for X(s) is :
.2
1
1
1
sssX
2Re2
1)(2
s
stue Lt
1Re1
1)(
s
stue Lt
So , )()()( 2 tuetuetx tt
Partial-fraction ExpansionThe procedure consists of expanding the rational algebraic expression into a linear combination of lower order terms.
011
1
011
1
)(
)()(
bsbsbsb
asasasa
sB
sAsX n
nn
n
mm
mm
rationalproper is , if sXnm
)())((
)())((
)(
)()(
21
21
nn
mm
pspspsb
zszszsa
sB
sAsX
sXpppp n ofpoles,,, 321
sXzzzz m ofzeros,,, 321
m
i i
i
as
AsX
1
)(
Assuming no multiple-order poles and that the order of the denominator polynomial is greater than the order of the numerator polynomial,we can expand X(s) in the form:
ROC σ> -ai (right sided signal) σ< -ai (left sided signal)
)()( tueAtx taii
i
)()( tueAtx taii
i
Adding the inverse transforms of the individual terms ,then yields the inverse transform of X(s).
Discussing 1 : no multiple-order poles,the first order
,.....,, 321 npppp
n
n
ps
A
ps
A
ps
AsX
2
2
1
1)(
)()()()( 112
211 ps
ps
Aps
ps
AAsXps
n
n
1)()( 11 pssXpsA
iii pssXpsA )()(
different poles,real numbers or complex numbers
tpn
tptptp neAeAeAeAtx ....)( 321321
321
321
s
A
s
A
s
AsX
11 )()1(
s
sXsA 1)3)(2)(1(
332)1(
1
2
ssss
sss
22 )()2(
s
sXsA 5)3)(2)(1(
332)2(
2
2
ssss
sss
33 )()3(
s
sXsA 6)3)(2)(1(
332)3(
3
2
ssss
sss
For example: )3)(2)(1(
332 2
sss
sssX
3
6
2
5
1
1)(
sss
sX
)2j1)(2j1)(2(
32
sss
ssX
2j12j1221
s
B
s
B
s
A
5
7)2(
2
ssXsA
5
2j1
)2j1)(2j1)(2(
3)2j1(
2j1
2
1
s
sss
ssB
)52)(2(
3)( 2
2
sss
ssX 1
)( 2sin5
22cos
5
1e2 )(e
5
7 2 tutttutx tt
For example:
Homework
Page 250 #4-4(3)(6)(9)(15)