1
Inverse Laplace Transforms
There is no integral definition for finding an inverse Laplace transform.
Inverse Laplace transforms are found as follows:
1)
For simple functions: Use tables of Laplace transform pairs.
2)
For complex functions: Decompose the complex function into two or more
simple functions using Partial Fraction Expansion (PFE) and then find the inverse
transform of each function from a table of Laplace transform pairs.
Example
:
Find f(t) for F(s) = 16/(s+8)
Lecture #3 EGR 261
–
Signals & Systems
Read
:
Ch. 12, Sect. 1

9 in
Electric Circuits, 9
th
Edition
by Nilsson
Ch. 4, Sect. 1

3 and Sect. B.5 in
Linear Signals & Systems, 2
nd
Ed.
by Lathi
Handout on Bb and webpage
: Partial Fraction Expansion (using various calculators)
2
Partial Fraction Expansion
(or Partial Fraction Decomposition)
Partial Fraction Expansion (PFE) is used for functions whose inverse Laplace
transforms are not available in tables of Laplace transform pairs.
PFE involves decomposing a given F(s) into
F(s) = A
1
F
1
(s) + A
2
F
2
(s) + … + A
N
F
N
(s)
Where F
1
(s), F
2
(s), … , F
N
(s) are the Laplace transforms of known functions.
Then by applying the linearity and superposition properties:
f(t) = A
1
f
1
(t) + A
2
f
2
(t) + … + A
N
f
N
(t)
Example
:
Find f(t) for F(s) = 16s/(s
2
+ 4s + 29)
Lecture #3 EGR 261
–
Signals & Systems
3
In most engineering applications,
Finding roots of the polynomials yields:
where
z
i
= zeros of F(s)
and
p
i
are the poles of F(s)
Note that:
N(s) numerator in the form of a polynomia
l in s
F(s)
D(s) denominator in the form of a polynom
ial in s
1 2 M
1 2 N
K(s  z )(s  z ) (s  z )
F(s)
(s  p )(s  p ) (s  p )
i
i
s = z
s = p
F(s) 0
F(s)
Lecture #3 EGR 261
–
Signals & Systems
4
Poles and zeros in F(s)
Poles and zeros are sometimes plotted on the s

plane. This is referred to as a
pole

zero diagram
and is used heavily in later courses such as Control Theory for
investigating system stability and performance. Poles and zeros are represented on
the pole

zero diagram as follows:
x

represents a pole
o

represents a zero
Example
Sketch the pole

zero diagram for
the following function:
jw
s

plane
100(s + 2)(s + 5)
F(s)
(s + 4)(s + 1  j2)(s + 1 + j2)
Lecture #3 EGR 261
–
Signals & Systems
5
Surface plots used to illustrate F(s)
The names “poles” and “zeros” come from
the idea of using a
surface plot
to graph
the magnitude of F(s). If the surface,
which represents F(s), is something like a
circus tent, then the zeros of F(s) are like
“tent stakes” where the height of the tent
is zero and the poles of F(s) are like “tent
poles” with infinite height.
Example
A surface plot is
shown to the right.
Note:
Pole

zero diagrams and surface plots for F(s) are not key topics for this
course and will not be covered on tests. They are mentioned here as a brief
introduction to future topics in electrical engineering.
Lecture #3 EGR 261
–
Signals & Systems
6
An important requirement for using Partial Fractions Expansion
Show that expressing F(s) as
leads to an important requirement for performing Partial Fractions Expansion:
If F(s) does not satisfy the condition above, use long division to place it (the
remainder) in the proper form (to be demonstrated later).
order of N(s) < order of D(s)
N
1 2
1 2 N
A
N(s) A A
F(s) =
D(s) (s  p ) (s  p ) (s  p )
Lecture #3 EGR 261
–
Signals & Systems
7
Methods of performing Partial Fractions Expansion
:
1)
common denominator method
2)
residue method
3)
calculators/software
Example
: (
Simple roots)
Use PFE to decompose F(s) below and then find f(t). Perform PFE using:
1)
common denominator method
4s
F(s)
(s 1)(s 2)(s 3)
Lecture #3 EGR 261
–
Signals & Systems
8
Example
: (continued)
2)
residue method
4s
F(s)
(s 1)(s 2)(s 3)
3)
calculators (demonstrate with TI

86, TI

89, and MathCAD)
Lecture #3 EGR 261
–
Signals & Systems
9
Repeated roots
A term in the decomposition with a repeated root in the denominator could in
general be represented as:
(Note that in general the order of the numerator should be 1 less than the order of
the denominator).
F(s) above is inconvenient, however, since it is not the transform of any easily
recognizable function. An equivalent form for F(s) works better since each part is a
known transform:
2
As B
F(s)
(s  p)
1 2
2
A A
F(s)
(s  p) (s  p)
F(t)
F(s)
Ke
at
u(t)
K
s + a
Kte
at
u(t)
2
K
s + a
?
1 2
2
K s + K
s + a
Lecture #3 EGR 261
–
Signals & Systems
10
Example
: (
Repeated roots)
Find f(t) for F(s) shown below.
2
2s
F(s)
(s 2) (s 3)
Lecture #3 EGR 261
–
Signals & Systems
11
Example
: (
Repeated roots)
Find f(t) for F(s) shown below.
3
2
3
s
1
2s
F(s)
Lecture #3 EGR 261
–
Signals & Systems
12
Complex roots
Complex roots always yield sine and/or cosine terms in the time domain. Complex
roots may be handled in one of two ways:
F(t)
F(s)
Kcos(wt)u(t)
2
2
w
+
s
K(s)
Ksin(wt)u(t)
2
2
w
+
s
K(w)
Ke
at
cos(wt)u(t)
2
2
w
+
a)
(s
a)
K(s
Ke
at
sin(wt)u(t)
2
2
w
+
a)
(s
K(w)
Also note that cosine and sine terms can be
represented as a single cosine term with a
phase angle using the identity shown below:
1)
using quadratic factors
–
Leave the portion
of F(s) with complex roots as a 2
nd
order
term and manipulate this term into the form
of the transform for sine and cosine
functions (with or without exponential
damping). Keep the transform pairs shown
to the right in mind:
2 2 1
Useful trigonometric identity:
B
Acos(wt) + Bsin(wt) = A + B cos wt  tan
A
(or using phasors, convert (A,B) to pol
ar form)
Lecture #3 EGR 261
–
Signals & Systems
13
2)
using complex roots
–
a complex term can be represented using complex linear
roots as follows:
*
1 2
2 2
A s A
F(s)
(s 2 s w ) (s jw) (s + jw)
B B
where the two terms with complex roots will yield a single time

domain term that is
represented in phasor form as
2 2B
B
or in time

domain form as 2Be
t
cos(wt +
)
*
Note that may be found as usual using t
he residue method. It is not necessary
to fin
d .
B B
The two methods for handling complex roots are summarized in the table below.
F(t)
F(s)
Ke
at
cos(wt)u(t)
2
2
w
+
a)
(s
a)
K(s
Ke
at
sin(wt)u(t)
2
2
w
+
a)
(s
K(w)
2Be
at
cos(wt +
)u(t)
*
B B
+
s +  jw s + + jw
Quadratic factor method
Complex linear root method
Lecture #3 EGR 261
–
Signals & Systems
14
Example
: (
Complex roots)
Find f(t) for F(s) shown below. Use both methods
described above and show that the results are equivalent.
1)
Quadratic factor method
2
2s
F(s)
(s 1)(s 2s 10)
Lecture #3 EGR 261
–
Signals & Systems
15
Example
: (
continued)
2)
Complex linear root method
2
2s
F(s)
(s 1)(s 2s 10)
Lecture #3 EGR 261
–
Signals & Systems
16
Example
: (
Time

delayed function)
Find f(t) for
2s
2e
F(s)
(s 3)(s 4)
Example
: (
Order of numerator too large)
Find f(t) for
2
2s 3
F(s)
s(s 4)
Lecture #3 EGR 261
–
Signals & Systems
Comments 0
Log in to post a comment