EE 422G Notes: Chapter 5
Instructor: Zhang
Chapter 5 The Laplace Transform
5-1 Introduction
x(t)
y(t) output
input
Dynamic
System
(1) System analysis
A processor which
processes the input signal
to produce the output
static system: y(t) = ax(t) => easy (simple processing)
dynamic system:
dy ( n ) (t )
dy ( n 1) (t )
dx ( m ) (t )
 a1
 ...  a n y (t )  b0
 ...  bm x(t )
dt n
dt n 1
dt m
Can we determine y(t) for given u(t) easily?
Easier solution method
X(f)
Y(f )
Dynamic
System
H(f )
Y( f )  H( f )X ( f )
Algebraic equation, not
differential equation
y (t )  F 1 (Y ( f ))
(1) we have systematic way to obtain H(f) based on the differential equation
(2) we can obtain X(f)
 Fourier transform: an easier way
(2) Problem
Fourier transform of the input signal:
Page 5-1
EE 422G Notes: Chapter 5
Instructor: Zhang
X( f ) 

 x (t )e
 j 2ft
dt

| e  j 2ft | 1
if x(t) does not go to zero when t   and t  
 X(f) typically does not exist! (the existence of X(f) if not guaranteed)
A very strong condition, can not be satisfied by many signals!
(3) Solution:
Why should we care about t<0 for system analysis? We do not care!
x(t) does not go to zero (when t   ), but x(t )e t may!
(Will be much easier. )
 use “single-sided” Fourier transform of x(t )e t , instead of “doublesided” Fourier transform of x(t).


 ( x (t )e
t
)e
 jt

dt   x (t )e  (  j ) t dt
0
0
very useful, let’s use a new name for it: Laplace transform.
(4) Laplace Transform
Definition: Laplace transform of x(t)

L[ x(t )]  X ( s)   x(t )e  st dt
( s    j )
0
What is a Laplace transform of x(t)?
A time function? No, t has been eliminated by the integral with respect to t!
A function of s ( s is complex variable)
(5) System analysis using Laplace transform
Page 5-2
EE 422G Notes: Chapter 5
Instructor: Zhang
X(s)
Y(s )
Dynamic
System
G(s )
Inverse Laplace transform
Y ( s)  G( s) X ( s)
y (t )  L1 (Y ( s))
(6) How is the inverse transform defined?
  j
1
x (t ) 
X ( s )e st ds

2j   j
(7) Will we often use the definition of the inverse transform to find time
function?
No!
What will we do?
Express X(s) as sum of terms for which we know the inverse transforms!
5-2 Examples of Evaluating Laplace Transforms using the definition
(1) x(t)=1 and step function x(t)=u(t)


t 
1
L[ x(t )  u (t )]   x(t )e dt   e dt    e  st d ( st )
s t 0
0
0

e  st
s
t 
t 0
 st

e t  jt
e
s
 st
t 
t 0

e   j e  0  j 0
e

e
s
s
(| e  j | 1
e   0 , (if   0)
e    , (if
1
1
 (cos 0  j sin 0) 
s
s
1
 L(1)  L[u (t )] 
(Re(s))  0
s
  0)
(2) x(t )  et u(t )
L[e
t

u(t )]   e
0
t  st

e dt   e  (  s ) t dt
0
~
Define a new complex variable s  s  
Page 5-3
EE 422G Notes: Chapter 5


e
Instructor: Zhang
~
s t
dt
0

we know  e  st dt 
0

 e
~
s t
dt 
1
s
Re( s )  0
~
1
Re( s )  0
~
s
0

  e  (  s ) t dt 
0
 L[e t u(t )] 
 L[e t ] 
1
s 
Re( s   )  0
1
s 
Re( s   )  0 or
Re( s )   Re( )
1
s 
(3) x (t )   (t )

L[ (t )]    (t )e  st dt
0
 e  st
t 0
 e t e  jt
t 0
 e t (cost  j sin t )
t 0
1
No constraint on s.
5-2B Discussion: Convergence of the Laplace Transform

(1) To assure
 x (t )e
 st

dt   x (t )e t e  jt dt converge, 
0
enough such that x(t )e
 Re(s ) must be psotive
0
t
goes to zero when t goes to positive infinite
(2) Region of absolute convergence and pole
Page 5-4
EE 422G Notes: Chapter 5
Instructor: Zhang
(3) How to obtain Fourier transform form Laplace transform:
s  j
L[ x (t )]  X ( s )  X ( j )  F ( x (t ))
Important: why introduce Laplace transform; definition of Laplace transform as a
modification of Fourier transform; find the Laplace transforms of the three basic
functions based on the (mathematical) definition of Laplace transform.
Chapter structure
Part one: Definition (5.1)
Part two: Direct evaluation of Laplace transforms of simple timedomain function.
Easy? Not at all!
 complex (not simple) functions: even harder
 tools for application of Laplace transform in system analysis
Part three: Rest of the chapter
What tools: tools for easier Laplace transform evaluation
tools for easy inverse transform
Page 5-5
EE 422G Notes: Chapter 5
Instructor: Zhang
5-3 Some Laplace Transform theorems
(Tools for evaluating Laplace transform based on the Laplace
transforms of the basic functions)
5.3.1 Linearity
Assume x(t )  a1x1 (t )  a2 x2 (t ) ( a1 and a 2 are time independent)
X1( s)  L[ x1(t )],
X 2 ( s)  L[ x2 (t )]
then X ( s)  L[ x(t )]  a1 X1 ( s)  a2 X 2 ( s)
HW#2-1: Assume x(t )  a1 (t ) x1 (t )  a2 (t ) x2 (t ) , X1( s)  L[ x1(t )] and X 2 ( s)  L[ x2 (t )] .
(a) Is X ( s )  L[ x (t )] equal to a1 (t ) X1 ( s)  a2 (t ) X 2 ( s) ?
(Answer: Yes or No)
(b) If A1 ( s)  L[a1 (t )] , is it true
L[ x(t )]  A1( s) X1( s)  A2 ( s) X 2 ( s) ?
(Answer: Yes or No)
Example 5.1
(1) Find L(cos0t )
Key to solution : express (cos0t ) as linear combination of  (t ) , u (t ) ,
and/or e t :
L[ (t )]  1
let   j0
L[u(t )] 
1
s
L[e t ] 
1
s 
L[e  j 0t ] 
let    j0
1
s  j0
L[e j 0t ]  L[e  (  j 0 )t ] 
1
s  j0
Page 5-6
EE 422G Notes: Chapter 5
Instructor: Zhang
Can we use e j0t and e j0t to express cos(0t ) ?
e  j 0 t  cos(  0t )  j sin(  0t )
 cos( 0t )  j sin( 0t )
e j 0 t  cos( 0t )  j sin( 0t )
e  j 0t  e j 0t  2 cos( 0 t )
e  j 0 t  e j  0 t
 cos( 0 t ) 
2
1
 L[cos( 0 t )]  [ L(e  j 0t )  L(e j 0t )]
2
1 1
1 
 


2  s  j  0 s  j 0 


(2) Find L[sin0t ]
1  ( s  j  0 )  ( s  j 0 ) 


2  ( s  j 0 )( s  j 0 ) 
s
s 2  0
2
5-3-2 Transforms of Derivatives
Assume
X ( s )  L[ x (t )]
Then
 dx(t ) 

L
  sX ( s)  x(0 )
 dt 
Proof:

(1) Definition

dx (t )  st
 dx (t ) 
L

e dt   e  st dx (t )

 d (t )  0 dt
0
(2) Integration by parts:
General equation:
Page 5-7
EE 422G Notes: Chapter 5
Instructor: Zhang
b
 u(t )dv(t )  u(t )v(t )
t b
t a
b
  v(t )du (t )
a
a
(3) Use the above equation
v(t )  x(t ) ,
Why? If we assume


0
0
  v(t )du (t )   x(t )de
 st
u(t )  e  st

 s  x(t )e st dt
0
X (s)  Lx(t )
u (t )  e  st ,
v(t )  x(t )


 dx(t ) 
  u (t )dv(t )   e  st dx(t )  [ L
 from (1) ]
dt


0
0
t 
b
u (t )v(t ) t 0   v(t )du (t )
a
t 

 e x(t ) t 0  s  x(t )e st dt
 st
0
t 
 e st x(t ) t 0  sL[ x(t )]
lim e  st x(t ) must go to zero. Otherwise,
t 

L[ x (t )]   x (t )e  st dt does not exist !

0
t 
e  st x (t ) t  0 
use
 e  s  0 x ( 0 )   x ( 0)
0  as lower limit => x(0  )

t 
 u (t )v(t ) t 0   v(t )du (t )
0

  x(0 )  sX ( s)  sX ( s)  x(0  )
Page 5-8
EE 422G Notes: Chapter 5
Instructor: Zhang
HW#2-2: Assume X ( s )  L[ x (t )] . Prove
 d 2 x(t ) 
2

(1)

L
  s X ( s)  sx(0 )  x (0 )
 dt 
 d ( n ) x (t ) 
HW#2-3: Express L 
 using L[ x (t )]
n
 dt

Example 5-2
Find i(t) using Laplace transform method
for t>0
Solution:
(1) Before switched from 1 to 2 at t=0
i
4
 2 A  i (0 )  2 A
2
(2) System equation (t>0)
di (t )
 Ri (t )  0
dt
di (t )

 2i (t )  0
dt
L
KVL:
( L  1H ) ( R  2ohm)
(3) Solve system equation using Laplace transform
 di (t )

 di (t ) 
L
 2i (t )  L 
  2 L[i (t )]
 dt

 dt 
 sI ( s )  i (0  )  2 I ( s )
 ( s  2) I ( s )  2  0
2
 I (s) 
s2
 i (t )  2e  2t u (t ) A
Page 5-9
EE 422G Notes: Chapter 5
Instructor: Zhang
5-3-3 Laplace Transform of an integral
t
Assume
y (t ) 
 x (  ) d ,
X ( s )  L[ x(t )]

Then
t
 X ( s ) y (0 )
L   x( )d  

s
s


where y (0 ) 
0
 x (  ) d

Proof :
t
 
L   x (  )d   
 
 0


 udv 
0
t 
uv t  0 
 0

 x (  )d e  st dt
 




  vdu
0
(1)
t 
uv t 0
 t
  e  st
   x( )d 
 
 s
 st
e
t 
s
 lim
t
 x (  ) d 
e
y (0  )

s
t 0
1

 s 0 0
s

t 
 x( )d
y(0  )

0
(2)

  st
0
0
 vdu   
e

1
1
x(t )dt    x (t )e  st dt   X ( s )
s
s 0
s
(3)
t
 y (0 ) X ( s ) X ( s ) y (0 )
L   x(  )d  



Proved!
s
s
s
s


Page 5-10
EE 422G Notes: Chapter 5
Instructor: Zhang
Example 5.3:
Find I(s) = L(i(t))
Solution:
(1) Differential equation
KVL :
x(t )  vL (t )  vC (t )  vR (t )



di (t )

v
(
t
)

L
L

dt

t

1
t
t
vC (t )   i (  )d
1

dv
(

)

i
(

)
d


C 

 C

C



v R (t )  Ri (t )
t

1

vC (t )  vC (  )   i (  )d
C 


t
1
vC (  )  0  vC (t )   i (  )d 

C 

(2) Laplace transform
dv (t )
i (t )  C C
dt
1
dvC (t )  i (t )dt
C
X ( s )  VL ( s )  VC ( s )  VR ( s )
VL ( s )  L[ sI ( s )  i (0 )]  LsI ( s )
i (0 )  zero
0

1  I ( s) 1
VC ( s )  
  i (  )d 
C s
s 


0
1
11

I ( s) 
i (  )d
Cs
s c 
1
1
I ( s )  vc ( 0  )
Cs
s
VR ( s )  RI ( s )

Page 5-11
EE 422G Notes: Chapter 5
Instructor: Zhang
1
1
I ( s )  vC (0 )  RI ( s )
Cs
s
1
X ( s )  vC (0 )
s
 I ( s) 
1
Ls 
R
Cs
sX ( s )  vC (0 )

1
Ls 2   sR
C
sX ( s )  vC (0 )

1 

R
Ls2   s 
LC 
L

X ( s )  LsI ( s ) 
5-3-4. Complex Frequency shift (s-shift) Theorem
Assume
y(t )  x(t )e t
X (s)  L[ x(t )]
Y (s)  L[ y(t )]
Y (s)  X ( s   )
1
1
L[u (t )]  ,
L[u (t )e t ] 

s
s 
0
s
L[cos  0t ]  2
L
[sin

t
]


0
s  02
s 2  02
0
s 
=> L[cos  0t e t ] 
L[sin  0t e t ] 
2
2
(s   )   0
(s   )2   0 2
Then
Example 5-4
Solution:
x(t )  L1[ X ( s)]  L1
Find
s 8
s 2  6s  13
s 8
( s  3)  5
 2
s  6s  13 s  6s  9  4
s3
(5 / 2) 2


( s  3) 2  2 2 ( s  3) 2  2 2
X (s) 
2
Page 5-12
EE 422G Notes: Chapter 5
Instructor: Zhang
5
x(t )  L1 [ X ( s)]  e 3t cos 2t  e 3t sin 2t
2
(t  0)
5-3-4 Delay Theorem
 question: How to express delayed function?
 Assume L[ x(t )]  L[ x(t )u (t )]  X ( s )
Then
Proof :
L[ x(t  t0 )u(t  t0 )]  e st 0 X (s)
(If (t0  0) , it will not be a delay!)
(t0  0)
L[ x(t  t 0 )u (t  t0 )]

  x(t  t0 )u (t  t 0 )e st dt
0
t0

  x(t  t0 )u (t  t 0 )e dt   x(t  t 0 )u (t  t 0 )e st dt
 st
0
t0


  x(t  t0 )e dt   x(t  t 0 )e s ( t t0 )st0 dt
 st
t0
e
t0
 st0

 x(t  t )e
 s ( t t 0 )
0
d (t  t 0 )
t0
 t t 0
 e
 st0

 x( )e
0
 s
d  e
 st0

 x(t )e
 st
dt  e st0 X ( s )
0
Page 5-13
EE 422G Notes: Chapter 5
Instructor: Zhang
Question: will L[ x(t  t0 )u(t  t0 )]  e  st 0 X (s) be true if t0  0 ?
No! (it will not be a delay)
Example 5-5: Square wave beginning at t = 0
T
1
1  20 s
1
1 
L[ xsq (t )]   2 e
 2 e T0s  2 e
s
s
s
s
  e
3T0
s
2
1
 2 e 2T0s
s
1
 T0 s
2
1
1
1
1
1
 2   2 2  2 3  2 4  ...
s
s
s
s
s
1 2
  (  2  3  ...)
s s



1
2
1 1  1 1 e

 
 
T
 0s
s s (1   ) s 1   s
1 e 2
5-3-5 Convolution
Signal 1: x1 (t )
Signal 2 : x2 (t )

y (t )  x1 (t ) * x 2 (t ) 
if
T0
s
2
 x ( ) x
1
2
(t   )d

x1 (t )  0
t  0

 y (t )   x1 ( ) x2 (t   )d
0
Page 5-14
EE 422G Notes: Chapter 5
if
Instructor: Zhang
x2 (t )  0
t  0
( x2 (t   )  0   t )
t
 y (t )   x1 ( ) x2 (t   )d
0
x1 (t )  0,
Therefore, if
x2 (t )  0
t  0
t


0
0

  x1 ( ) x2 (t   )d   x1 ( ) x2 (t   )d 
t


0
0

 x1 ( ) x2 (t   )d
 L[  x1 ( ) x2 (t   )d ]  L[  x1 ( ) x2 (t   )d ]  L[  x1 ( ) x2 (t   )d ]
 


0
0
st
[
x
(

)
x
(
t


)
d

]
e
dt

x
(

)[
x
(
t


)
e
dt ]d
1
2
1
2



st
0 0

t
Look at


 s ( t   )  s
x
(
t


)
e
dt

x
(
t


)
e
e dt
2
2


 st
0
0
 t  
 e
 s
d  dt
 x
2
( )e  s d

t  0    
t     
x2 ( )  0 

   0 

e
 s

 s
 s
x
(

)
e
d


e
X 2 (s)
2

0
Then

Y ( s )   x1 ( )e  s X 2 ( s )d 
0

 X 2 ( s )  x1 ( )e  s d 
0
 X 1 (s) X 2 (s)
Page 5-15
EE 422G Notes: Chapter 5
Instructor: Zhang
5-3-7 Product
5-3-8 Initial Value Theorem
x(t )  L1[ X ( s )]
 x(0  )  lim sX ( s )
s 
Example: A demonstration where x(0) is obvious
x(t )  et cos 0tu(t )
It is evident: x(0)  e 0 cos 0  0  1
Using Laplace transform
X ( s )  L[ x(t )] 
s 
(s   )2  0 2
s(s   )
x(0)  lim sX ( s )  lim
s 
s  (s   )2   2
0
 lim
s 
 lim
s 
s 2  s
s 2  2s   2   0 2
d ( s 2  s ) / ds
d ( s 2  2s   2   0 2 ) / ds

 

2s  
d (2 s   ) / ds
2
 lim
 lim  1
s   2 s  2
s   d ( 2 s  2 ) / ds
s  2
 lim
5-3-9 Final Value Theorem: if x(t ) and dx(t ) / dt are Laplace transformable, then
lim x(t )  lim sX ( s)
t 
s0
(condition: sX (s ) has no poles on j  axis or in the right-half s-plan
or lim t  x(t ) exists)
5-3-10 Scaling
a>0: x(at)
 a times fast (if a>1)
or slow (if a<1) as x(t)

X ( s)  L[ x(t )]
What do we expect on L[ x(at )] ?
Page 5-16
EE 422G Notes: Chapter 5
Instructor: Zhang
s
L[ x(at )]  X ( ) ?
a

L[ x(at )]   x(at )e
 st
0
 ( at )
1
dt   x(at )e a d (at )
a0
a 0
s
s
  
1
1 s
a

x
(

)
e
d  X  

a 0
a0
a a
t 
  at
  
5-4
Inversion of Rational Functions
(1) Ways to find x(t ) from X (s )
1  
X ( s )e st dt
(1) x(t ) 

2j  
(Contour Integral)
(2) Transform pair
1
 u (t )
s
1
 e t u (t )
s 
Therefore
X ( s) 
1
 x(t )  e t u (t )
s 
(2) All kinds of Laplace Transform ? No!
We almost only see
b0 s m  b1s m 1  ...  bm 1s  bm s
e
n
n 1
s  a1s  ...an 1s  an
rational function X(s)
Delay
x(t )  L1[ X ( s)]
y (t )  L1[ X ( s)e s ]  x(t   )u (t   )
Page 5-17
EE 422G Notes: Chapter 5
Instructor: Zhang
 Consider Rational Functions only!
(3) Non proper Rational Function
proper Rational Function
Non proper
proper
(b0  0)
m>=n
m<n
Non proper => proper + Polynomial (using long division)
s 3  4s 2  6s  7
s5

s

1

s 2  3s  2
s 2  3s  2
s 1
s 2  3s  2 3
s  4s 2  6s  7
s 2  4s  7
s 2  3s  s
s5
How to find inverse Laplace transform for polynomials?
s n   ( n ) (t )
L1 ( s  1)  L1 ( s)  L1 (1)   (1) (t )   (t )
consider proper rational functions only!
(4) Proper Rational Functions: Partial Fraction Expansion
1

t n e t
u (t )
n!
( s   ) n 1
s 
 e t cos  0 tu(t )
2
(s   ) 2   0
 sum of
0
 e t sin  0 tu(t )
2
2
(s   )   0
1   (t )
1
 u (t )
s
1
 e t
s 
Page 5-18
EE 422G Notes: Chapter 5
Instructor: Zhang
Let’s look at examples, and then summarize!
Techniques:
Common Denominator
Specific value of s
Heaviside’s Expansion
Matlab
Factorize first!
Expand second!
Find coefficients third!
Example 5-9: Simple Factors
Y ( s) 
10
s 2  10s  16
Solution:
(1) Factorize and expand Y ( s ) 
10
A
B


( s  8)(s  2) s  8 s  2
(2) Common Denominator Methods
10
A( s  2)  B( s  8)

( s  8)( s  2)
( s  8)( s  2)
 A( s  2)  B( s  8)  10
 A  B  0  A  B

2 A  8B  10  2 B  8B  10
 B  10 / 6  5 / 3
A  5 / 3
5 1
5 1
Y (s)   
 
3 s 8 3 s2
5
5
 y (t )  ( e 8t  e 2t )u (t )
3
3
specific values of s
10
A
B


( s  8)( s  2) s  8 s  2
s 0 10
A B

 
8 2 8 2
s 2 10
A B

 
10  4 10 4
Can you solve for A and B?
Page 5-19
EE 422G Notes: Chapter 5
Instructor: Zhang
Heaviside Expansion
and
10
A
B


( s  8)( s  2) s  8 s  2
10
B( s  8) s 8 10

 A

 A  5 / 3
s2
s2
8 2
s 2
10
A( s  2)
10


BB
 5/3
s 8
s 8
28
Example 5-10 Imaginary Roots
15s 2  25s  20
Y ( s)  2
( s  1)(s  2)(s  8)
Solution: what do we have:
Y (s) 
Same as real roots!
A
A1
A
A
 2  3  4
s  j s  j s 2 s 8
A
A1 ( s  j )  A2 ( s  j )
A
 3  4
2
s2 s8
s 1
A
( A  A2 ) s  ( A2  A1 ) j
A
 1
 3  4
2
s 2 s8
s 1
Y (s) 
A1+A2 must be real number
(-A1+A2)j must be real number
Y ( s) 
A3
c1s  c2
A4


s2  1 s  2 s  8
Heaviside Expansion => A3=1 and A4= - 2.
c sc
15s 2  25s  20
1
2
Y ( s)  2
 12 2 

( s  1)( s  2)( s  8)
s 1 s  2 s  8
s=1 =>
15  25  20 c1  c2 1  2

 
2 3 9
2
3 9
s=2 =>
15  4  25  2  20 2c1  c2 1  2

 
5  4  10
5
4 10
Page 5-20
EE 422G Notes: Chapter 5
Instructor: Zhang
Can we solve for c1 and c2?
c1=1 c2=1
s 1
1
2


s2 1 s  2 s  8
s
1
1
2
 2
 2


s 1 s 1 s  2 s  8
=> [cos t  sin t  e 2t  2e 8t ]u(t )
Y (s) 
Too complex: use MATLAB
Example 5-11 Repeated linear Factors
10s
( s  2) 2 ( s  8)
A3
A
A
 1  2 
s  8 s  2 ( s  2) 2
Y (s) 
Example 5-12
10 s
( s  2) 3 ( s  8)
A3
A
A
A4
 1  2 

s  8 s  2 ( s  2) 2 ( s  2) 3
Y (s) 
Example 5-13 Complex - Conjugate Factors
2s 2  6s  6
Y ( s) 
( s  2)( s 2  2s  2)
A3
A1
A2
2s 2  6s  6




( s  2)[( s  1) 2  1] s  2 s  1  j s  1  j
Example 5-14 Repeated Quadratic Factors
Page 5-21
EE 422G Notes: Chapter 5
Instructor: Zhang
s 4  5s 3  12s 2  7 s  15
Y (s) 
( s  2)(s 2  1) 2
A
B s  c B s  c2
 1  12 1  22
s2
s  1 ( s  1) 2
* Summary of Partial–Fraction Expansion
(1) Expansion Structure:
Simple Roots (including complex conjugate)
=>
Aj
could be complex.
s  j
Repeated Roots: m multiplicity
=>
Bm
B1
B2


...

s   j (s   j )2
(s   j )m
real number or complex number
(2) Avoid complex number
For complex conjugates:  j  a  jb
Aj
s  j

Aj*
s   j*

Bk
Bk *

(s   j )k
s   j*

cs  D
(s  a) 2  b 2

k

cs  D
[(s  a ) 2  b 2 ]k
(3) Inverse Laplace transform
Aj
s  j
 jt
 Aje
u (t )
 t
t k 1e j
 Ak
u (t )
k
(
k

1
)!
(s   j )
Aj
k2
Page 5-22
EE 422G Notes: Chapter 5
Instructor: Zhang
t k 1
 jt
 j *t


[
A
e

B
e
]u (t )
j
j
(k  1)!
(s   j ) k (s   j * ) k
Aj
Bj
 j  a  jb
t k 1 at

e [ A j (cos bt  j sin bt )  B j (cos bt  j sin bt ]u (t )
 j *  a  jb ( k  1)!
Matlab use for Partial – Fraction Expansion.
Have to be memorized:
(1) Table 5-2 all except for No. 7
(2) Table 5-3 No. 1-No. 7
Page 5-23
EE 422G Notes: Chapter 5
Instructor: Zhang
Appendix: Partial-Faction Expansion with MatLab
1.
2.
3.
Command (MatLab Help)
An Introduction
An Example: Complex Conjugate
1. Command (MatLab Help)
» help residue
RESIDUE Partial-fraction expansion (residues).
[R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of
a partial fraction expansion of the ratio of two polynomials B(s)/A(s).
If there are no multiple roots,
B(s)
R(1)
R(2)
R(n)
---- = -------- + -------- + ... + -------- + K(s)
A(s) s - P(1) s - P(2)
s - P(n)
Vectors B and A specify the coefficients of the numerator and
denominator polynomials in descending powers of s. The residues
are returned in the column vector R, the pole locations in column
vector P, and the direct terms in row vector K. The number of
poles is n = length(A)-1 = length(R) = length(P). The direct term
coefficient vector is empty if length(B) < length(A), otherwise
length(K) = length(B)-length(A)+1.
If P(j) = ... = P(j+m-1) is a pole of multplicity m, then the
expansion includes terms of the form
R(j)
R(j+1)
R(j+m-1)
-------- + ------------ + ... + -----------s - P(j) (s - P(j))^2
(s - P(j))^m
[B,A] = RESIDUE(R,P,K), with 3 input arguments and 2 output arguments,
converts the partial fraction expansion back to the polynomials with
coefficients in B and A.
Warning: Numerically, the partial fraction expansion of a ratio of
polynomials represents an ill-posed problem. If the denominator
polynomial, A(s), is near a polynomial with multiple roots, then
small changes in the data, including roundoff errors, can make
arbitrarily large changes in the resulting poles and residues.
Problem formulations making use of state-space or zero-pole
representations are preferable.
See also POLY, ROOTS, DECONV.
»
2.
An Introduction
Page 5-24
EE 422G Notes: Chapter 5
Instructor: Zhang
Page 5-25
EE 422G Notes: Chapter 5
Instructor: Zhang
Page 5-26
EE 422G Notes: Chapter 5
Instructor: Zhang
Page 5-27
EE 422G Notes: Chapter 5
Instructor: Zhang
Page 5-28
EE 422G Notes: Chapter 5
Instructor: Zhang
3. An Example: Complex Conjugate
Page 5-29