This document discusses inverse Laplace transformations and their applications. It covers three cases of inverse Laplace transformations: 1) simple poles that are real and unequal, 2) simple complex conjugate poles, and 3) repeated roots. The document also discusses using the partial fractions expansion method and method of residues to determine inverse Laplace transforms. It provides examples of using the "finger method" to find inverse transforms of functions with simple poles. Additionally, it discusses the initial value theorem and final value theorem, and provides an example of using Laplace transforms to solve a differential equation with zero initial conditions.

1. Admissions in India 2015Admissions in India 2015
By:
http://admission.edhole.com

2. Inverse Laplace TransformationsInverse Laplace Transformations
Dr. Holbert
February 27, 2008
Lect11 EEE 202 2
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3. Inverse LaplaceTransformInverse LaplaceTransform
Consider that F(s) is a ratio of polynomial
expressions
The n roots of the denominator, D(s) are called
the poles
◦ Poles really determine the response and
stability of the system
The m roots of the numerator, N(s), are called
the zeros
Lect11 EEE 202 3
)(
)(
)(
s
s
s
D
N
F =
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4. Inverse LaplaceTransformInverse LaplaceTransform
We will use partial fractions expansion
with the method of residues to determine
the inverse Laplace transform
Three possible cases (need proper
rational, i.e., n>m)
1. simple poles (real and unequal)
2. simple complex roots (conjugate pair)
3. repeated roots of same value
Lect11 EEE 202 4
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5. 1. Simple Poles1. Simple Poles
Simple poles are placed in a partial fractions
expansion
The constants, Ki, can be found from (use
method of residues)
Finally, tabulated Laplace transform pairs are
used to invert expression, but this is a nice form
since the solution is
Lect11 EEE 202 5
( ) ( )
( )( ) ( ) n
n
n
m
ps
K
ps
K
ps
K
pspsps
zszsK
s
+
++
+
+
+
=
+++
++
= 


2
2
1
1
21
10
)(F
ipsii spsK −=
+= )()( F
tp
n
tptp n
eKeKeKtf −−−
+++= 21
21)(admission.edhole.com

6. 2. Complex Conjugate Poles2. Complex Conjugate Poles
 Complex poles result in a Laplace transform of the
form
 The K1 can be found using the same method as for
simple poles
WARNING: the "positive" pole of the form –α+jβ
MUST be the one that is used
 The corresponding time domain function is
Lect11 EEE 202 6
+
++
−∠
+
−+
∠
=+
++
+
−+
=
)()()()(
)(
11
*
11
βα
θ
βα
θ
βαβα js
K
js
K
js
K
js
K
s F
βα
βα js
sjsK +−=
−+= )()(1 F
( ) ++= −
θβα
teKtf t
cos2)( 1admission.edhole.com

7. 3. Repeated Poles3. Repeated Poles
When F(s) has a pole of multiplicity r, then F(s)
is written as
Where the time domain function is then
That is, we obtain the usual exponential but
multiplied by t's
Lect11 EEE 202 7
( ) ( ) ( )
 +
+
++
+
+
+
=
+
= r
r
r
ps
K
ps
K
ps
K
pss
s
s
1
1
2
1
12
1
11
11
1
)(
)(
)(
Q
P
F
( )
 +
−
+++= −
−
−− tp
r
r
tptp
e
r
t
KetKeKtf 111
!1
)(
1
11211
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8. 3. Repeated Poles (cont’d.)3. Repeated Poles (cont’d.)
The K1j terms are evaluated from
This actually simplifies nicely until you reach s³
terms, that is for a double root (s+p1)²
Thus K12 is found just like for simple roots
 Note this reverse order of solving for the K values
Lect11 EEE 202 8
( )
( )[ ]
1
)(
!
1
11
ps
r
jr
jr
j sps
ds
d
jr
K
−=
−
−
+
−
= F
( ) ( )[ ]
1
1
)()(
2
111
2
112
ps
ps
sps
ds
d
KspsK
−=
−=
+=+= FF
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9. The “Finger” MethodThe “Finger” Method
Let’s suppose we want to find the inverse Laplace
transform of
We’ll use the “finger” method which is an easy way
of visualizing the method of residues for the case of
simple roots (non-repeated)
We note immediately that the poles are
s1 = 0 ; s2 = –2 ; s3 = –3
)3)(2(
)1(5
)(
++
+
=
sss
s
sF
Lect11 EEE 202 9
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10. The Finger Method (cont’d)The Finger Method (cont’d)
For each pole (root), we will write down the
function F(s) and put our finger over the term that
caused that particular root, and then substitute that
pole (root) value into every other occurrence of ‘s’
in F(s); let’s start with s1=0
This result gives us the constant coefficient for the
inverse transform of that pole; here: e–0·t
Lect11 EEE 202 10
6
5
)3)(2(
)1(5
)30)(20)((
)10(5
)3)(2(
)1(5
)( ==
++
+
=
++
+
=
ssss
s
sF
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11. The Finger Method (cont’d)The Finger Method (cont’d)
Let’s ‘finger’ the 2nd
and 3rd
poles (s2 & s3)
They have inverses of e–2·t
and e–3·t
The final answer is then
tt
eetf 32
3
10
2
5
6
5
)( −−
−+=
Lect11 EEE 202 11
3
10
)1)(3(
)2(5
)3)(23)(3(
)13(5
)3)(2(
)1(5
)(
2
5
)1)(2(
)1(5
)32)(2)(2(
)12(5
)3)(2(
)1(5
)(
−
=
−−
−
=
++−−
+−
=
++
+
=
=
−
−
=
+−+−
+−
=
++
+
=
ssss
s
s
ssss
s
s
F
F
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12. InitialValueTheoremInitialValueTheorem
The initial value theorem states
Oftentimes we must use L'Hopital's Rule:
◦ If g(x)/h(x) has the indeterminate form 0/0 or
∞/ ∞ at x=c, then
Lect11 EEE 202 12
)(lim)(lim
0
sstf
st
F
∞→→
=
)('
)('
lim
)(
)(
lim
xh
xg
xh
xg
cxcx →→
=
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13. FinalValueTheoremFinalValueTheorem
The final value theorem states
The initial and final value theorems are useful
for determining initial and steady-state
conditions, respectively, for transient circuit
solutions when we don’t need the entire time
domain answer and we don’t want to perform
the inverse Laplace transform
Lect11 EEE 202 13
)(lim)(lim sstf
0st
F
→∞→
=
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14. Initial and FinalValueTheoremsInitial and FinalValueTheorems
The initial and final value theorems also provide
quick ways to somewhat check our answers
Example: the ‘finger’ method solution gave
Substituting t=0 and t=∞ yields
tt
eetf 32
3
10
2
5
6
5
)( −−
−+=
6
5
2
15
10
6
5
)(
0
6
20155
3
10
2
5
6
5
)0( 00
=−+=∞=
=
−+
=−+==
∞−∞−
eetf
eetf
Lect11 EEE 202 14
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15. Initial and FinalValueTheoremsInitial and FinalValueTheorems
What would initial and final value theorems find?
First, try the initial value theorem (L'Hopital's too)
Next, employ final value theorem
This gives us confidence with our earlier answer
0
5
52
5
lim
65
)1(5
lim)0(
)3)(2(
)1(5
lim)(lim)0(
2
=
∞
=
+
=
++
+
=
∞
∞
=
++
+
==
∞→∞→
∞→∞→
sss
s
f
ss
s
ssf
s
ds
d
ds
d
s
ss
F
6
5
)3)(2(
)1(5
)3)(2(
)1(5
lim)(lim)(
00
==
++
+
==∞
→→ ss
s
ssf
ss
F
Lect11 EEE 202 15
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16. Solving Differential EquationsSolving Differential Equations
Laplace transform approach automatically
includes initial conditions in the solution
Example: For zero initial conditions, solve
Lect11 EEE 202 16
)0(')0()(
)(
)0()(
)(
2
2
2
yysss
dt
tyd
xss
dt
txd
−−=





−=





Y
X
L
L
)(4)(30
)(
11
)(
2
2
tuty
dt
tyd
dt
tyd
=++
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17. Class ExamplesClass Examples
Find inverse Laplace transforms of
Drill Problems P5-3, P5-5 (if time permits)
84
)(
)1(
)(
2
2
++
=
+
=
ss
s
s
s
s
s
Z
Y
Lect11 EEE 202 17
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