# Inverse Laplace Transforms

Electronics - Devices

Nov 15, 2013 (4 years and 8 months ago)

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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

:

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
-

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)

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

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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
 

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)

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