This document discusses inverse Laplace transformations and contains an example lecture on the topic. It introduces three cases for inverse Laplace transformations based on the properties of the poles of the transformed function: simple real poles, complex conjugate poles, and repeated poles. Methods like partial fractions expansion and the method of residues are presented for handling each case. Other topics covered include the finger method, initial and final value theorems, and using Laplace transforms to solve differential equations.

1. Video Lectures for MBA
BY:
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2. Inverse Laplace
Transformations
2
DR. HOLBERT
FEBRUARY 27, 2008
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EEE 202 Lect11

3. Inverse Laplace Transform
3
Consider that F(s) is a ratio of polynomial
expressions
s s
( ) ( )
s
( )
N
D
F =
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
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4. Inverse Laplace Transform
4
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
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5. 1. Simple Poles
5
Simple poles are placed in a partial fractions
expansion
( ) ( )
n

= m

K
K
K
K s z s z
+ +
+
=
+ +
2
1
F ( ) 0 1
( s p )( s p ) ( s p
) s p
s p
s p
n
n
s
+
+
+
+ + 
+
2
1
1 2
The constants, Ki, can be found from (use method
of residues)
i i s pi K s p s =- = ( + ) F ( )
Finally, tabulated Laplace transform pairs are used
to invert expression, but this is a nice form since
the solution is
p t
f t = K e- p1 t + K e- p2 t ++ K e- n
n
1 2 ( )
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6. 2. Complex Conjugate Poles
6
Complex poles result in a Laplace transform of the form
Ð-
*
q
K
K
F s K K

( ) 1 1
s j
a b a b s + +
j
+
Ð
s + -
j
+ =
1 1
s + +
j
+
+ -
=
q
a b
a b
( ) ( ) ( ) ( )
The K1 can be found using the same method as for simple
poles
a b a b s j K s j s =- + = ( + - ) ( ) 1 F
WARNING: the "positive" pole of the form –a+jb MUST be
the one that is used
The corresponding time domain function is
+
f (t) = 2 K e-a t cos(b t +q ) + 1
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7. 3. Repeated Poles
7
When F(s) has a pole of multiplicity r, then F(s) is
written as
K
K
K
s s
( )
P
= r
11
12
1
( ) ( ) +  +
( ) +
r s p
2
1
s +
p
+
s +
p
=
1
( )
s s +
p
1
1 1
( )
Q
F
Where the time domain function is then
+
-
r
1
f t K e 1 K t e 1 K t 1
- - p t
p t p t e
= + + + -
 +r
( 
r
-
1 ) !
( )
11 12 1
That is, we obtain the usual exponential but
multiplied by t's
r
1
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8. 3. Repeated Poles (cont’d.)
8
The K1j terms are evaluated from
( ) [( ) ]
s p
1
d
-
j s p s
( )
r j
= +
F
1 1
!
1
r
r j
ds
r j
K
=-
-
-
This actually simplifies nicely until you reach s³
terms, that is for a double root (s+p1)²
( ) [( ) ]
K s p s K d
( ) 2 ( )
= + F = + F
s p
1
s p s
11 1
2
12 1
ds
=-
Thus K12 is found just like for simple roots
Note this reverse order of solving for the K values
s p
1
=-
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9. The “Finger” Method
Let’s suppose we want to find the inverse Laplace
transform of
F s s
= +
s s s
( ) 5( 1)
( 2)( 3)
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
+ +
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Lect11 EEE 202 9

10. 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
5
F s s
= +
s s s s
= +
( ) 5( 1) 5(0 1)
= 5 (1)
=
This result gives us the constant coefficient for
the inverse transform of that pole; here: e–0·t
Lect11 EEE 202 10
6
(2)(3)
( )(0 2)(0 3)
( 2)( 3)
+ +
+ +
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11. The Finger Method (cont’d)
Let’s ‘finger’ the 2nd and 3rd poles (s2 & s3)
= - +
5( 2 1)
s s
s s s s
= - +
5( 3 1)
= +
( ) 5( 1)
s s
= +
( ) 5( 1)
They have inverses of e–2·t and e–3·t
The final answer is then
( ) = 5 + 5
- - 10
-
f t e 2t e 3t
3
2
6
= -
5 ( 1)
= -
5 ( 2)
Lect11 EEE 202 11
5
10
3
( 3)( 1)
( 3)( 3 2)( 3)
( 2)( 3)
2
( 2)(1)
( 2)( 2)( 2 3)
( 2)( 3)
= -
- -
- - + +
+ +
=
-
- + - +
+ +
s s s s
F
F
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12. Initial Value Theorem
12
The initial value theorem states
F
f t =
s s
lim ( ) lim ( )
t s
® ®¥
0
Oftentimes we must use L'Hopital's Rule:
 If g(x)/h(x) has the indeterminate form 0/0 or ¥/ ¥ at x=c,
then
g x
'( )
g x
=
lim h x
'( )
( )
( ) lim
h x
x®c x®c
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13. Final Value Theorem
13
The final value theorem states
F
lim f (t) =
lim s (s)
t ®¥ s ®
0
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
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14. Initial and Final Value Theorems
The initial and final value theorems also provide
quick ways to somewhat check our answers
Example: the ‘finger’ method solution gave
( ) = 5 + 5
- - 10
-
f t e 2t e 3t
3
2
6
Substituting t=0 and t=∞ yields
= = + - = 5 + 15 - 20
=
5
6
( 0) 5 5
0 10
0
f t e e
10 15
f t e-¥ e-¥
2
( ) 5
6
0
6
3
2
6
= ¥ = + - =
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Lect11 EEE 202 14

15. Initial and Final Value Theorems
What would initial and final value theorems
find?
First, try the initial value theorem f = s s = 5 ( s
+
1)
(L'Hopital's
too)
s s
+ +
( 2)( 3)
5
F
s s
®¥ ®¥
+
= ¥
¥
f s
2 =
+
d
ds
s s s
2 5
(0) lim ( ) lim
5 ( 1)
=
5 6 lim
=
(0) lim
=
+ +
d
ds s
®¥ ®¥
s
Next, employ final value theorem
5 0
¥
5
6
5(1)
= =
(2)(3)
f s s s
¥ = = +
5 ( 1)
F
® ® s s
+ +
( 2)( 3)
( ) lim ( ) lim
s s
0 0
This gives us confidence with our earlier answer
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Lect11 EEE 202 15

16. Solving Differential Equations
16
Laplace transform approach automatically includes
initial conditions in the solution
d x t
( ) ( ) (0)
dt
ù
- = úû
d y t
s s x
ù
X
( ) 2
( ) (0) '(0)
2
2
s s s y y
dt
- - = úû
é
L
êë
é
L
êë
Y
Example: For zero initial conditions, solve
d y t + d y t
+ =
( ) 11 ( ) 30 ( ) 4 ( )
2
2
y t u t
dt
dt
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17. Class Examples
Find inverse Laplace transforms of
s s
+
s
( 1)
2
s s
s s
+ +
4 8
( )
Y
( )
2
=
=
Z
Drill Problems P5-3, P5-5 (if time permits)
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Lect11 EEE 202 17