Lecture 2: The Laplace Transform
• Laplace transform definition
• Relation between time and Laplace
domains
ME 431 Lecture 2
• Laplace transform properties
• Initial and Final Value Theorem
• Introduction to MATLAB
1
• The Laplace transform is a mathematical operation
that takes an equation from being a function of
time, t, to being a function of the Laplace variable,
s

L[ f (t )]   f (t )e dt  F (s)
 st
ME 431 Lecture 2
The Laplace Transform
0
• Some mathematical operations become much
simpler in the Laplace domain
• We will never solve this integral, will use tables
2
unit impulse
Item No.
f(t)
1.
δ(t)
unit step
2.
1(t)
unit ramp
3.
t
4.
tn
5.
e-at
6.
sin (ωt)
7.
cos (ωt)
F(s)
t
1
t
t
1
1
s
1
s2
n!
s n+1
1
s+a
ω
s 2 + ω2
s
s 2 + ω2
ME 431 Lecture 2
Table of Laplace pairs on
pages 18-19
3
Properties of the Laplace Transform
Table of Laplace
properties on page 20
L[af (t )  bg (t )] 
aL[ f (t )]  bL[ g (t )] 
aF ( s )  bG ( s )
- constants factor out and Laplace operation
distributes over addition and subtraction
- note: L[ f (t )  g (t )]  F ( s )  G ( s )
ME 431 Lecture 2
1. Linearity
4
Properties of the Laplace Transform
often zero
 f (t )dt 
F (s)  
 t 0


L  f (t )dt 



s
s
 f (t )dt dt 
 f (t )dt 
F (s)   
 t 0  
 t 0


L   f (t )dt dt  2 



s
s2
s
3. Differentiation
 df 
L    sF ( s )  f (0)
 dt 
d f
L 2
 dt
2

2

s
F ( s )  sf (0)  f (0)


These properties turn
differential equations
into algebraic
equations
ME 431 Lecture 2
2. Integration
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Properties of the Laplace Transform
4. Multiplication by e-at
f (t )]  F (s  a)
- important for damped response
Example:
L[e
 at
cos t ]
f(t)
Note: roots of
denominator (poles) in
Laplace domain = roots
of characteristic
equation in the time
domain
s
from Laplace pairs table, F ( s ) = L[cos t ]  2
s  2
sa
then from prop above, F ( s  a) 
( s  a)2   2
ME 431 Lecture 2
L[e
 at
6
Properties of the Laplace Transform
5. Time shift
F (s), a  0
- important for analyzing time delays
ME 431 Lecture 2
L[ f (t  a)1(t  a)]  e
 as
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Properties of Laplace
Transform
dF ( s )
L[tf (t )]  
ds
2
d F ( s)
2
L[t f (t )] 
2
ds
n
n
n d F (s)
L[t f (t )]  (1)
ds n
ME 431 Lecture 2
6. Multiplication by t
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Example
0 for t<0
• FindL[2te3t  5]
3t
= 2L[te ]  5L[1(t )] (by property 1)
1
 L[t ]  2
s
1
s
1
 L[te ] 
(by property 4)
2
( s  3)
3t
2
5
=

2
( s  3)
s
Example
Laplace/Time Domain Relationship
Initial Value Theorem

f (0 )  lim sF (s), if the lim exists
s 
ME 431 Lecture 2
• Previously, saw how poles of X(s) relate to x(t)
• Two further relationships between X(s) and x(t):
Final Value Theorem
f ()  lim f (t )  lim sF ( s),
t 
s 0
if [poles of sF ( s )]  0
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Example
• Find the initial value of f(t), where
s3
F ( s) 
s( s 2  6s  13)

1
s2
s( s  3)
( s  3)
f (0)  lim sF ( s )  lim 2
 lim 2
s 
s  s ( s  6 s  13)
s  ( s  6 s  13) 1
 2
s
1 3
 2
 lim s s
s 
6 13
1  2
s s
0
Example
• Find the final value of f(t), where
s3
F ( s) 
s( s 2  6s  13)
s( s  3)
f ()  lim f (t )  lim sF ( s)  lim 2
t 
s 0
s 0 s ( s  6 s  13)
( s  3)
3
 lim 2

s 0 ( s  6 s  13)
13
poles of sF ( s)  3  2 j,
since <0, limit exists
ME 431 Lecture 2
MATLAB Introduction
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