The Inverse Laplace Transform The University of Tennessee Electrical and Computer Engineering Department Knoxville, Tennessee wlg
The Inverse Laplace Transform The University of Tennessee Electrical and Computer Engineering Department Knoxville, Tennessee wlg.
Mar 31, 2015
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The Inverse Laplace Transform
The University of TennesseeElectrical and Computer Engineering Department
Knoxville, Tennessee
wlg
Inverse Laplace Transforms
Background:
To find the inverse Laplace transform we use transform pairsalong with partial fraction expansion:
F(s) can be written as;
)(
)()(
sQ
sPsF
Where P(s) & Q(s) are polynomials in the Laplace variable, s.We assume the order of Q(s) P(s), in order to be in properform. If F(s) is not in proper form we use long division and divide Q(s) into P(s) until we get a remaining ratio of polynomialsthat are in proper form.
Inverse Laplace Transforms
Background:There are three cases to consider in doing the partial fraction expansion of F(s).
Case 1: F(s) has all non repeated simple roots.
n
n
ps
k
ps
k
ps
ksF
...)(
2
2
1
1
Case 2: F(s) has complex poles:
...)))()((
)()(
*11
1
1
js
k
js
k
jsjssQ
sPsF
Case 3: F(s) has repeated poles.
)(
)(...
)(...
)())((
)()(
1
1
1
12
1
12
1
11
11
1
sQ
sP
ps
k
ps
k
ps
k
pssQ
sPsF
rr
r
(expanded)
(expanded)
Inverse Laplace Transforms
Case 1: Illustration:
Given:
)10()4()1()10)(4)(1(
)2(4)( 321
s
A
s
A
s
A
sss
ssF
274)10)(4)(1(
)2(4)1(| 11
ssss
ssA 94
)10)(4)(1(
)2(4)4(| 42
ssss
ssA
2716)10)(4)(1(
)2(4)10(| 103
ssss
ssA
)()2716()94()274()( 104 tueeetf ttt
Find A1, A2, A3 from Heavyside
Inverse Laplace Transforms
Case 3: Repeated roots.
When we have repeated roots we find the coefficients of the terms as follows:
|111
)()(1 psr
sFpsds
dk r
|121
)()(!2 12
2
psrsFps
ds
dk r
|11
)()()!( 1 psj
sFpsdsjr
dk r
jr
jr
Inverse Laplace Transforms
Case 3: Repeated roots. Example
2
1
1
2211
2 )3()3()3(
)1()(
K
K
A
s
K
s
K
s
A
ss
ssF
)(____________________)( 33 tuteetf tt ? ? ?
Inverse Laplace Transforms
Case 2: Complex Roots:
...)))()((
)()(
*11
1
1
js
K
js
K
jsjssQ
sPsF
F(s) is of the form;
K1 is given by,
jeKKK
jsjssQ
sPjsK js
||||
))(()(
)()(
111
1
11
|
Inverse Laplace Transforms
Case 2: Complex Roots:
js
eK
js
eK
js
K
js
K jj
11
*11
|||
tj
etej
etj
etej
eKjs
eK
js
eKL
jj
1||
||||111
2
)()(|
1|2
1||
tje
tjeateK
tjete
je
tjete
jeK
Inverse Laplace Transforms
)cos(||2|||
1111
teK
js
eK
js
eKL t
jj
Case 2: Complex Roots:
Therefore:
You should put this in your memory:
Inverse Laplace Transforms
Complex Roots: An Example.
For the given F(s) find f(t)
o
jj
j
jss
sK
ss
sA
js
K
js
K
s
AsF
jsjss
s
sss
ssF
js
s
10832.0)2)(2(
12
)2(
)1(
5
1
)54(
)1(
22)(
)2)(2(
)1(
)54(
)1()(
|
|
2|1
0|
11
2
2
*
Inverse Laplace Transforms
Complex Roots: An Example. (continued)
We then have;
jsjsssF
oo
2
10832.0
2
10832.02.0)(
Recalling the form of the inverse for complex roots;
)(108cos(64.02.0)( 2 tutetf ot
Inverse Laplace Transforms
Convolution Integral:
Consider that we have the following situation.
h(t)x(t) y(t)
x(t) is the input to the system.h(t) is the impulse response of the system.y(t) is the output of the system.
System
We will look at how the above is related in the time domainand in the Laplace transform.
Inverse Laplace Transforms
Convolution Integral:
In the time domain we can write the following:
ttdxthdhtxthtxty
00)()()()()()()(
In this case x(t) and h(t) are said to be convolved and theintegral on the right is called the convolution integral.
It can be shown that,
sHsXsYthtxL )()()()(
This is very important
* note
Inverse Laplace TransformsConvolution Integral:
Through an example let us see how the convolution integral and the Laplace transform are related.
We now think of the following situation:
x(t) y(t)
X(s) Y(s)
te 4
h(t)
H(s)
)4(
1
s
Inverse Laplace TransformsConvolution Integral:
From the previous diagram we note the following:
)()(;)()(;)()( thLsHtyLsYtxLsX
h(t) is called the system impulse response for the followingreason.
)()()( sHsXsY
If the input x(t) is a unit impulse, (t), the L(x(t)) = X(s) = 1.Since x(t) is an impulse, we say that y(t) is the impulseresponse. From Eq A, if X(s) = 1, then Y(s) = H(s). Since,
Eq A
.)(,
)()()()( 11
responseimpulsesystemthSo
thsHLresponseimpulsetysYL
Inverse Laplace TransformsConvolution Integral:
A really important thing here is that anytime you are givena system diagram as follows,
H(s)X(s) Y(s)
the inverse Laplace transform of H(s) is the system’simpulse response.
This is important !!
Inverse Laplace TransformsConvolution Integral:
Example using the convolution integral.
e-4t
x(t) y(t) = ?
t
tt
tt deededuety0
44
0
)(4)(4 )()(
)(4
1
4
1
4
1)( 444
0
44 | 0 tueeedeety ttt
t t
Inverse Laplace TransformsConvolution Integral:
Same example but using Laplace.
x(t) = u(t) s
sX1
)(
h(t) = e-4tu(t) 4
1)(
ssH
)(14
1)(
4
4141
4)4(
1)(
4 tuety
sss
B
s
A
sssY
t
Inverse Laplace TransformsConvolution Integral:
Practice problems:
?)(,)2(
3)(
2)()( thiswhat
ssYand
ssXIfa
).(),()()()()( 6 thfindtutetyandtutxIfb t
).(,)4(
2)()()()(
2tyfind
ssHandttutxIfc
Answers given on note page
)(2)(5.1)( 2 tuetth t
Inverse Laplace Transforms
Circuit theory problem:
You are given the circuit shown below.
+_
t = 0 6 k
3 k 1 0 0 F
+
_v(t)1 2 V
Use Laplace transforms to find v(t) for t > 0.
Circuit theory problem:
Inverse Laplace Transforms
We see from the circuit,
+_
t = 0 6 k
3 k 1 0 0 F
+
_v(t)1 2 V
voltsxv 49
312)0(
Circuit theory problem:
Inverse Laplace Transforms
+
_v c(t) i(t)
3 k
1 0 0 F
6 k
05)(
0)(
0)()(
tvdt
tdv
RC
tv
dt
tdv
tvdt
tdvRC
c
c
cc
c
c
Take the Laplace transformof this equations includingthe initial conditions on vc(t)
Circuit theory problem:
Inverse Laplace Transforms
)(4)(
5
4)(
0)(54)(
0)(5)(
5 tuetv
ssV
sVssV
tvdt
tdv
tc
c
cc
cc
Stop
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