Transcript No Slide Title

THE LAPLACE TRANSFORM
LEARNING GOALS
Definition
The transform maps a function of time into a function of a complex
variable
Two important singularity functions
The unit step and the unit impulse
Transform pairs
Basic table with commonly used transforms
Properties of the transform
Theorem describing properties. Many of them are useful as
computational tools
Performing the inverse transformation
By restricting attention to rational functions one can simplify
the inversion process
Convolution integral
Basic results in system analysis
Initial and Final value theorems
Useful result relating time and s-domain behavior
ONE-SIDED LAPLACE TRANSFORM
s    j
s  the integral is well defined
s  RoC 
It will be necessary to consider t  0 as the lower limit
To insure uniqueness of the transform one assumes f (t )  0 for t  0
A SUFFICIENT CONDITION FOR EXISTENCE OF LAPLACE TRANSFORM
Transform exists for
Re{s}    0
THE INVERSE TRANSFORM
Contour integral
in the complex plane
Evaluating the integrals can be quite time-consuming. For this reason we
develop better procedures that apply only to certain useful classes of function
TWO SINGULARITY FUNCTIONS
Unit step
(Important “test” function
in system analysis)
This function has derivative that is zero
everywhere except at the origin.
We will “define” a derivative for it
For positive time functions
f ( t )  f ( t ) u( t )
Using square pulses to approximate
an arbitrary function
The narrower the pulse the better
the approximation
Using the unit step to build functions
Computing the transform of the unit step

U ( s)  0 1 e
 sx
An example of Region of Convergence
(RoC)
T
dx  lim T   e  sx dx
0
T
 1

U ( s )  lim T    e  sx 
 s
0
1
e  sT
U ( s)   lim T 
s
s
Im(s )
( s    j )
Re{s}  0
1
e T e jT
U ( s )   lim T 
s
  j
1
U ( s)  ; s  Re{s}  0
s
Re(s )
RoC
To simplify question of RoC:
A special class of functions
RoC  s :Re{s}   
In this case the RoC is at least half a
plane. And any linear combination
of such signals will also have a RoC
that is a half plane
Complex Plane
(Good model for impact, lightning, and
other well known phenomena)
THE IMPULSE FUNCTION
These two conditions are not
feasible for “normal” functions
Approximations
to the impulse
Height is proportional
to area
Representation of the impulse
Sifting or sampling
property of the
impulse
For t0  t2 or t0  t1 the integral is NOT defined
for  (t ) the
 e  st0 In order to have a valid transform

lower limit is assumed t  0
Laplace transform
LEARNING BY DOING
0  t0    2
 cos
t0  10  2
0
We will develop properties that will
permit the determination of a large
number of transforms from a small
table of transform pairs
Linearity
Time shifting

Time truncation
Multiplication by exponential
Multiplication by time
Some properties will
be proved and used as
efficient tools in the
computation of Laplace
transforms
LEARNING EXAMPLE
LINEARITY PROPERTY
Find the transform for f (t )  e at

F ( s)   e
0
at  st
Homogeneity

e dt   e ( s a ) t dt
0

1  ( s  a )t
1
F ( s)  
e

sa
sa
0
Additivity
Follow immediately from the linearity
properties of the integral
APPLICATION
e  j t  e j t  st

e dt
0
2
1
1
 L[e  j t ]  L[e j t ]
2
2
a   j

Basic Table of Laplace Transforms
a  j
1 1
1 1

2 s  j 2 s  j
1 ( s  j )  ( s  j )

2 ( s  j )( s  j )
s
 2
s 2
F ( s) 
We develop properties that expand the
table and allow computation of transforms
without using the definition
With a similar use of linearity one shows

L[sin t ]  2
s 2
Additional entries for the table
LEARNING EXAMPLE
x(t )  3   (t )  3e
4 t
Application of Linearity
1
1
X ( s)  3  1  3
s
s4
Notice that the unit step is not
shown explicitly. Hence
3 and 3u(t )
LEARNING EXAMPLE
Find the Laplace transform for
x (t )  cos(3t   / 3)
x (t )  cos / 3 cos 3t  sin  / 3sin 3t
X ( s)  cos / 3
s
3

sin

/
3
s2  9
s2  9
are equivalent
MULTIPLICATION BY EXPONENTIAL


L[e f ( t )]   e f ( t )e dt   f ( t )e
 at
 at
 st
0
0
F (s  a)
LEARNING EXAMPLE
LEARNING EXAMPLE
 ( sa ) t
dt x(t )  e 2t cos / 3cos 4t  sin  / 3sin 4t 
a  2, b  4
s2
4
X ( s )  cos / 3

sin

/
3
( s  2) 2  16
( s  2) 2  16
y(t )  e cos(10t )
3 t
f ( t )  cos10t  F ( s ) 
s
(From table)
s  100
2
y( t )  e f ( t )  Y ( s )  F ( s  3) 
3 t
s3
( s  3)  100
2
New entries for the table of transform pairs
MULTIPLICATION BY TIME
LEARNING EXAMPLE Let u(t ) be the unit step
Find the transform of the ramp function
r (t )  tu(t )
Differentiation under an integral
1
s
d 1 1
tu (t )      2
ds  s  s
d1 2
t 2 u(t )    2   3
ds  s  s
u (t )  U ( s) 
By succesive applicatio n
of the property one shows
t n (u(t )) 
n!
s n1
This result, plus linearity, allows
computation of the transform of any
polynomial
Remember that we consider the functions
LEARNING BY DOING
to be zero for t  0. Hence
x(t )  1  2t  6t 3
x ( t )  x ( t ) u( t )
1
1
3!
X ( s)   2 2  6 4
s
s
s
TIME SHIFTING PROPERTY
LEARNING EXAMPLE
f (t )u(t )  F ( s)  f (t  t0 )u(t  t0 )  e  st 0 F ( s)
 1 1 t  3
f (t )  
0 elsewhere
f (t )
1
u(t  1)
1
3
t
 u(t  3)
f (t )  u(t  1)  u(t  3)

1
1 1
F ( s)  e  s  e 3s  e  s  e 3s
s
s s

LEARNING EXTENSION
FIND THE TRANSFORM FOR
f (t )  te ( t 1) u(t  1)  e ( t 1) u(t  1)
One can apply the time shifting property
if the time variable always appears as it
appears in the argument of the step. In
this case as t-1
f (t )  (t  1  1)e ( t 1) u(t  1)  e ( t 1) u(t  1)
f (t )  (t  1)e ( t 1) u(t  1)  e ( t 1) u(t  1)
e
( t 1)
u(t  1)
 (t  1)e ( t 1) u(t  1)
tu(t ) 
f (t )  e (t  1)e t u(t  1) g (t )
And apply the time truncation
property
f (t )  g(t )u(t  1)  F ( s)  e  s L[ g(t  1)]
g (t  1)  ete ( t 1)  te  t
L[ g (t  1)] 
1
( s  1) 2
The two properties are only different
representations of the same result
1
s2
te t u(t ) 
One could also write
1
( s  1) 2
 (t  1)e
 ( t 1)
e s
u (t  1) 
( s  1)2
PERFORMING THE INVERSE TRANSFORM
Simple, complex conjugate poles
FACT: Most of the Laplace transforms
that we encounter are proper rational
functions of the form
Zeros = roots of numerator
Poles = roots of denominator
mn

C1 ( s   )
C2 

 ...
2
2
2
2
(s   )  
(s   )  
KNOWN: PARTIAL FRACTION EXPANSION
If Q ( s )  Q1 ( s )Q2 ( s ) is a COPRIME factorization
of the denominato r with
deg(Qi )  ni (  ni  n), then
F ( s)  K 0 
Pole with multiplicity r
P1 ( s ) P2 ( s )

; deg( Pi )  ni
Q1 ( s ) Q2 ( s )
If m<n and the poles are simple
THE INVERSE TRANSFORM OF EACH
PARTIAL FRACTION IS IMMEDIATE.
WE ONLY NEED TO COMPUTE THE
VARIOUS CONSTANTS
SIMPLE POLES
 /( s  pi )
LEARNING EXAMPLE
F ( s) 
12( s  1)( s  3)
s( s  2)( s  4)( s  5)
Write the partial fraction expansion
K
K
K
K
F ( s)  1  2  3  4
s s2 s4 s5
Determine the coefficients (residues)
12 1 3 9

2  4  5 10
12(1)(1)
K 2  ( s  2) F ( s) s 2 
1
(2)(2)(3)
12(3)(1) 36
K 3  ( s  4) F ( s) s 4 

(4)(2)(1) 8
12(4)(2)
32
K 4  ( s  5) F ( s) s 5 

(5)(3)(1)
5
K1  sF ( s) s 0 
Get the inverse of each term and write
the final answer
36
32
9

f (t )    e 2 t  e 4t  e 5t u(t )
8
5
 10

The step function is necessary to make
the function zero for t<0
“FORM” of the inverse transform


f (t )  K1  K 2e 2t  K 3e 4t  K 4e 5t u(t )
COMPLEX CONJUGATE POLES
| K1 | 
Euler' s Identity
e j  e  j 
cos 
2
f (t )  2 | K1 | e t cos(  t   )  ...
USING QUADRATIC FACTORS

f (t )  C1e t cos  t  C2e t sin  t  ...
The two forms are equivalent !
P1 ( s )
C1 ( s   )
C2 


 ...
2
2
2
2
2
2
Q1 ( s ) ( s   )  
(s   )  
(s   )  


Avoids using complex algebra.
Must determine the coefficients in different way
10( s  2)
s ( s 2  4 s  5)
s 2  4 s  5  ( s  2  j1)( s  2  j1)
LEARNING EXAMPLE Y ( s ) 
f (t )  2 | K1 | e t cos(  t   )  ...
 ( s  2) 2  1
10( s  2)
K0
K1
K1*
Y ( s) 



s( s  2  j1)( s  2  j1)
s s  2  j1 s  2  j1
MUST use radians in
10(2)
20
K 0  sY ( s) s 0 

4
exponent
(2  j1)(2  j1) 5
10( j1)
5
K1  ( s  2  j1)Y ( s) s 2 j1 

 j 2.678
(2  j1)( j 2)
5153.43  2.236  153.43  2.236e
y (t )   4  2  2.236e 2t cos(t  2.678)  u (t )
Using quadratic factors
2
10( s  2)
C0 C1 ( s  2)
C2
C
((
s

2
)
 1)  C1 ( s  2) s  C 2 s
0
Y ( s) 




2
2
2
s( s  4 s  5) s ( s  2)  1 ( s  2)  1
s ( s 2  4 s  5)
10( s  2)  C0 (( s  2)2  1)  C1 ( s  2) s  C2 s
s 2 : 0  C 0  C1
 C1  C0  4
s :10  4C 0  2C1  C 2
s 0 : 20  5C 0
 C2  2
 C0  4
y (t )  (C0  C1e2t cos t  C2e2t sin t )u (t )
Alternative way to determine coefficients
For s  0 : 20  5C0
For s  2 : 0  C0  2C2
For s  1: 10  2C0  C1  C2
MULTIPLE POLES
 1 
1
n1  pt
L-1 

t
e
n
 ( s  p)  (n  1)!
 /( s  p1 ) r
The method of identification of coefficients, or even the method of selecting
values of s, may provide a convenient alternative for the determination of the
residues
LEARNING EXAMPLE


1

f (t )   K11e t  K12 te t  K13  t 2e t   K 2e 2 t u(t ) Using identification of coefficients
2



10( s  3)
3
(
s

1
)
F
(
s
)

10(1)
( s  2)
K 2  ( s  2) F ( s) s 2 
 10
3
(1)
2
3
K
(
s

1
)
(
s

2
)

K
(
s

1
)(
s

2
)

K
(
s

2
)

K
(
s

1
)
11
12
13
2
10
(
2
)
K13  ( s  1)3 F ( s)

 20 
( s  2)
s  1
(1)
d
d  10( s  3) 
s 3 : 0  K11  K 2
K12 
( s  1)3 F ( s)
 

s  1
ds
ds  s  2  s 1
2
s
: 0  4 K11  K12  3K 2
10( s  2)  10( s  3)
10


 10
2
( s  2) 2
(
s

2)
s1: 10  5 K  3K  K  3K
s 1
s 1

K11 
K11 

2

1 d
( s  1)3 F ( s)
2
2! ds


s  1
1 d  10
2! ds ( s  2)2
1 102( s  2) 
10

 10
2 ( s  2) 4 s 1 ( s  2)3 s 1
11
12
13
2
s 0 : 30  2 K11  2 K12  2 K13  K 2
CONVOLUTION INTEGRAL
CLAIM: Given an ODE
dn y
d n 1 y
dmx
 an1 n 1  ...a0 y  bm m  ...  b0 x
dt n
dt
dt
there exists a function, h( t ), t  0, such that
Shifting
t
y (t )   h(t   ) x ( )d  h(t )  x (t )
0
is a particular solution of the equation for t  0
(Actually, the zero state response)
RESULT : If f1 , f 2 , are positive time functions
F ( s)  F1 ( s)F2 ( s)
EXAMPLE
FIND Y ( s )
t
y (t )   e ( t  x ) y ( x )dx  t ; t  0
PROOF
0
y (t )  e t  y (t )  t  Y ( s ) 
1 
1

1 
Y ( s)  2
s
 s  2
1
1
Y ( s)  2
s2
s
Y ( s) 
s2
s 2 ( s  3)
Using convolution to determine a network response
LEARNING EXAMPLE
Network function
H ( s) 
V0 ( s)
10

VS ( s ) s  5
V0 ( s) 
Input
VS ( s ) 
1
s
VS (s)
10 1

s5 s
v0 (t )  e 5t u(t )  u(t )
RESULT : If f1 , f 2 , are positive time functions
F ( s)  F1 ( s)F2 ( s)
For t  0
v0 (t )  10 e
5 ( t   )
V0 ( s)
V0 ( s)  H ( s)VS ( s)
10
 10e 5t u(t )
s5
1
 u( t )
s
t
H (s )
d  10e
5 t
t
e
5
d
0

t
 10e
5 t
 1 5 
 5 e 
0

v0 (t )  2e 5 t e 5 t  1  2 1  e 5t ; t  0
In general convolution is not an efficient approach to determine the
output of a system. But it can be a very useful tool in special cases
INITIAL AND FINAL VALUE THEOREMS
These results relate behavior of a function in the time domain with the behavior
of the Laplace transform in the s-domain
INITIAL VALUE THEOREM
Assume that both f (t ), df
dt
, have Laplace
transform. Then
lim t 0 f (t )  lim s  sF ( s )
df
]  sF ( s )  f (0)
dt
And if the derivative is transformable then
L[
lim s  L[
df
]0
dt
FINAL VALUE THEOREM
Assume that both f (t ), df
, have Laplace
dt
transform and that lim t  f (t ) exists. Then
lim t  f (t )  lim s 0 sF ( s )
NOTE : lim t  f (t ) will exist if F ( s ) has poles
with negative real part and at most a single
pole at s  0

df
 st
 dt (t )e dt  sF ( s)  f (0)
0
Taking limits as s  0

df
 dt (t )dt  lim s0 sF ( s)  f (0)
0
LEARNING EXAMPLE
LEARNING EXTENSION
10( s  1)
( s  1) 2
Given F ( s ) 
.
Given F ( s ) 
.
2
s ( s  2 s  2)
s ( s  2)( s 2  2 s  2)
Determine the initial and final values for f (t ) Determine the initial and final values for f (t )
Clearly, f(t) has Laplace transform. And
sF(s) -f(0) is also defined.
f (0)  lim s sF ( s)
10( s  1)
0
f (0)  lim s 2
s  2s  2
( s  1) 2
f (0)  lim s 
0
( s  2)( s 2  2 s  2)
F(s) has one pole at s=0 and the others
have negative real part. The final value
theorem can be applied.
lim t  f (t )  lim s0 sF ( s)
lim t  f (t )  lim s0
10( s  1)
5
2
s  2s  2
lim t 
( s  1) 2
1
f (t )  lim s 0

( s  2)( s 2  2 s  2) 4
NOTE : Computing the inverse one gets
f (t )  5  5 2e  t cos(t 
3
)
4
Laplace