Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Using Partial Fraction Expansion
Outline of Today’s Lecture
 Review
 Laplace Transform
 Inverse Laplace Transform
 Properties of the Laplace Transform
 Final Value Theorem
Laplace Transform
 Traditionally, Feedback Control Theory was initiated by using
the Laplace Transform of the differential equations to develop
the Transfer Function
 The was one caveat: the initial conditions were assumed to be
zero.
 For most systems a simple coordinate change could effect this
 If not, then a more complicated form using the derivative
property of Laplace transforms had to be used which could lead
to intractable forms
 While we derived the transfer function, G(s), using the
convolution equation and the state space relationships, the
transfer function so derived is a Laplace Transform under
zero initial conditions
Laplace Transform
 CAUTION: Some Mathematics is necessary!
 The Laplace transform is defined as
For an analytic function f ( t )
(i.e., integrable everyw here and e veryw here less than e
F (s) 


e
0
 st
s0 t
for finite s 0 )
f ( t ) dt  L  f ( t ) 
F  s  is the Laplace transform of f ( t )
s is a com plex num ber
Fortunately, we rarely have to use these integrals as there are other methods
Some Common Laplace Transforms
 The Laplace Transform of the
Impulse Function
t0
 0

1
 (t )  
0  t< 


 0 t  0
L  (t )   1
 The Laplace Transform of the
Step Function
 0 t< 0
1( t )  
1 t  0
L 1  t   
1
Unit Ramp:
 0 t< 0
f (t )  
t t  0
1
s
2nd power of t:
 0

f (t )   t 2

2
t< 0
t0
1
L ( f ( t )) 
s
3
 The Place Transform of the
nth power of t:
 0

f (t )   t n

 n!
s
 The Laplace Transform of a
L ( f ( t )) 
 The Laplace Transform of the
2
t< 0
t0
1
L ( f ( t )) 
s
n 1
Some Common Laplace Transforms
 Laplace Trans Form of the
exponentials:
 0
f ( t )    at
e
L ( f ( t )) 
trigonometric functions:
t< 0
t0
1
sa
1
s  a
2
 0 t< 0
f ( t )   n  at
t0
t e
L ( f ( t )) 
n 1

s 
2
2
0 t< 0

f (t )  
 cos  t t  0
L ( f ( t )) 
s
s 
2
2
0 t< 0

f ( t )    at
 e sin  t t  0
L ( f ( t )) 
n!
s  a
 0 t< 0
f (t )  
 sin  t t  0
L ( f ( t )) 
 0 t< 0
f ( t )    at
t0
 te
L ( f ( t )) 
 Laplace Transforms of

s  a
2

2
Important Inverse Transforms
1 
 at
1 
L 
e
sa
 1

1
 at
 1  e 
L 

 s s  a  a


1


1
1
 bt
 at
e

e

L 


  s  a   s  b   b  a


1
2


n

L  2
2 
 s  2  n   n 
1
n
1
e
2
 i n t

sin  n t 1  
2

0<  < 1
Properties of the
Laplace Transform
 Laplace Transforms have
several very import
properties which are useful
in Controls
 d

L
f ( t )   sF ( s )  f (0 )
 dt

 d2

d
2
L
f ( t )   s F ( s )  sf (0 ) 
f (0 )
dt
dt


L

t
0

f ( t ) dt 
F (s)
s
Now, you should see the advantage
of having zero initial conditions
Final Value Theorem
 If f(t) and its derivative satisfy the conditions for Laplace
Transforms, then
lim f ( t )  lim sF ( s )
t 
s 0
 This theorem is very useful in determining the steady state gain
of a stable system transfer function
 Do not apply this to an unstable system as the wrong
conclusions will be reached!
The Transfer Function
s
n
 a1 s
n 1
 a2s
n2
 ...  a n  2 s  a n 1 2  a n  y 0 e
2
b s  b s
s  a s
m
y ( t )  y (0) e
st

0
m 1
1
n 1
n
1
st
  b0 s  b1 s
m
 b2 s
 a2s
m 2
n2
m 1
 b2 s
m 2
 ...  b n  2 s  b n 1 s  b n  e
 ...  b n  2 s  b n 1 s  b n 
2
2
 ...  a n  2 s  a n 1 2  a n 
2
e
st
T he transfer function form is then
b0 s  b1 s
m
y (t )  G ( s )u (t )  G ( s ) 
s  a1 s
n
m 1
n 1
 b2 s
 a2s
m 2
n2
 ...  b n  2 s  b n 1 s  b n
2
 ...  a n  2 s  a n 1 2  a n
2

b(s)
a(s)
W hen w e factor G (s) in the L aplace (or s ) dom ain, w e get the form
G (s) 
K
s
r

m
i 1
n
i 1
( s  zi )
( s  pi )
k
i 1
q
i 1
( s  2  i ni s   ni )
2
( s  2  i ni s   ni )
2
O ur problem is then, how do w e express this in the tim e dom ain?
W e w ill not have tables that w ould corre spond to a relatively com plex function.
T he answ er: W e m ust use P artial Fraction E xpansion
st
Partial Fraction Expansion
 When using Partial Fraction Expansion, our objective is to
turn the Transfer Function
G (s) 
K
s
r

m
i 1
n
i 1
( s  zi )
( s  pi )
k
i 1
q
i 1
( s  2  i ni s   ni )
2
( s  2  i ni s   ni )
2
into a sum of fractions where the denominators are the
factors of the denominator of the Transfer Function:
G (s) 
K
s
r

A1 ( s )
s  p1

A2 ( s )
s  p2
 ... 
An ( s )
s  pn

B1 ( s )
s  2  1 n 1 s   n 1
2
 ... 
Bq ( s )
s  2  q nq s   nq
Then we use the linear property of Laplace Transforms and
the relatively easy form to make the Inverse Transform.
2
Partial Fraction Expansion
 There are three cases that we need to consider in expanding
the transfer function:
 Case 1: All of the roots are real and distinct
K
G (s) 
m
i 1
( s  zi )

k
2
i 1
n
i 1
( s  2  i  ni s   ni )
( s  pi )
 Case 2: Complex Conjugate Roots
G (s) 
K
m
i 1
( s  zi )

q
i 1
k
( s  2  i  ni s   ni )
2
i 1
( s  2  i i s   i )
2
 Case 3: Repeated roots
K
G (s) 
s
r

n
i 1
( s  pi )
m
i 1
q
n 1
( s  zi )
( s  pi )
2

k
i 1
r
q 1
( s  2  i ni s   ni )
2
( s  p i ) ...
3
q
i 1
( s  2  i i s   i )
2
k
Case 1: Real and Distinct Roots
G (s) 
...
s
n
i 1
( s  pi )
P ut the transfer function in the form of
G (s) 
a0

s
a1
s  p1

a2
s  p2
 ... 
an
s  pn
w here the a i are called the residue at the pole p i
and determ ined by
a 0  sG  s 
a 3  ( s  p3 )G ( s ) s  p
s0
a 1  ( s  p1 ) G  s 
s   p1
a2   s  p2  G  s 
s   p2
3
...
an   s  pn  G  s 
s   pn
Case 1: Real and Distinct Roots
Example
G (s) 
s  2s  4
s  s  1  s  5 
G (s) 
a0

s
a1
s 1

a2
s5
 s  2s  4 
a0  s 
 s  s  1   s  5  



s0
 s  2s  4 
a1   s  1  
 s  s  1   s  5  


 s  2s  4 
a2   s  5 
 s  s  1   s  5  


G (s) 
1.6
s

 0.75
s 1
g ( t )  1.6  0.75 e
t

0.15
s5
 0.15 e
5t
24
1 5

s  1

s  5
 1.6
1 3
1  4
  0.75
3  1
5  4
 0.15
Case 1: Real and Distinct Roots
An Alternative Method (usually difficult)
G (s) 
K
m
i 1
( s  zi )
s
k
2
i 1
n
i 1
( s  2  i n i s   n i )
( s  pi )
P ut the transfer function in the form of
G (s) 
a0
s
K
m
i 1

a1
s  p1
( s  zi )

k
i 1
a2
s  p2
 ... 
an
s  pn
( s  2  i n i s   n i ) 
2
a 0  s  p 1   s  p 2   ...   s  p n 
 a 1 s  s  p 2   ...   s  p n 
 a 2 s  s  p 1   s  p 3   ...   s  p n 
 ...
 a n s  s  p 1   s  p 2   ... 
B oth sides are exam pled and the coeffic ients equated
to form n+ 1 linear e quations w hich are then solved for the u nknow n coefficients
T his m ethod can be used for the other ca ses as w ell w ith adjustm ents
Case 1: Real and Distinct Roots
Example
G (s) 
s  2s  4
s  s  1  s  5 
G (s) 
a0

s
a1
s 1
s  2s  4 

a2
s5
a 0  s  1   s  5   a1s  s  5   a 2 s  s  1 
s  6 s  8  a 0  s  6 s  5   a1  s  5 s   a 2  s  s 
2
2
2
2
 a 0  a1  a 2  1
 a 1  a 2   0 .6
 a 1   0 .7 5

6
a

5
a

a

6





0
1
2
5
a

a


3
.6
 1
 a 2  0 .1 5
2
 5 a  8  a  1 .6
0
 0
G (s) 
1 .6
s

0 .7 5
s 1
g ( t )  1 .6  0 .7 5 e

t
0 .1 5
s5
 0 .1 5 e
5t
Case 2: Complex Conjugate Roots
G (s) 
...
...
q
( s  2  i i s   i )
2
i 1
2
W e can either solve this using the m etho d of m atching coefficients
w hich is usually m ore difficult or by a m ethod sim ilar to that
previously used as follow s:

s  2  1 1 s   1  s   1 1   1  1  1
2
then the term
2
A( s )
s  2  i i s   i
2
2

 s   
1
a1
s   1 1   1  1  1
2
proceeding as before

 s   

 1G s
a 1  s   1 1   1  1  1 G  s 
a2
1
2
1
 1  1
2
 1  1  1
2
1
s    1 1   1  1  1
2
s    1 1   1  1  1
2


a2
s   1 1   1  1  1
2
Case 2: Complex Conjugate Roots
Example
G (s) 
15( s  2 )
s ( s  2 s  25)
2
a0


a1
s  1  5 0.2  1

2
s
a1
s  1  5 0.04  1
5 0.04  1  i 4.899


15( s  2 )
a0  s 

2
 s ( s  2 s  25) 

15  2
 1.2
25
s0


15( s  2 )
a 1   s  1  i 4.899  

2
 s ( s  2 s  25) 
a1 
s   1  i 4.899
15    1  i 4.899   2 
  1  i 4.899  (  1  i 4.899  1  i 4.899 )


15( s  2 )
a 2   s  1  i 4.899  

2
 s ( s  2 s  25) 
G (s) 
1.2

-0.6000 + 1.4085i
s  1  i 4.899
s
G (s) 
1.2

s
g ( t )  1.2 



15( s  2 )


 s ( s  1  i 4.899 ) 
s   1  i 4.899
 -0.6000 + 1.4085i
 -0.6000 - 1.4085i
s   1  i 4.899
-0.6000 - 1.4085i
s  1  i 4.899
12.6  1.12 s
s  2 s  25
2
12.6
25
5
1  0.04
e
 i5t


sin 5 t 1  0.04 
1.12
1  0.04
e
 i5t

1
sin  5 t 1  0.04  tan

1  0.04 

0.2

Case 3: Repeated Roots
...
G (s) 
...( s  p i ) ...
n
F orm the equation w ith the repeated term s expanded as
G ( s )  ... 
an
( s  pi )
a n  ( s  pi ) G ( s )
n

a n 1
( s  pi )
n
s   pi
d
a n 1 
 ( s  pi ) n G  s  

ds 
an2 
d
 ( s  pi ) n G  s 

ds 
a n3 
d
s p
2
2
s p
3
n

(
s

p
)
G  s  
i
3 
ds
s p
...
a1 
d
n 1
ds
n 1
 ( s  pi ) n G  s  


s p
n 1
 ... 
a1
s  pi
...
Case 3: Repeated Roots
Example
(s  2)
G (s) 
s
G (s) 
2
a2
s
2
 s  1  s  3 
2

a1

s
b2
 s  1
2



(s  2)
a2  s 

 s 2  s  12  s  3  


b1
s 1
s3
 0.6667
2
a1 
c

s0

d  2
(s  2)
s 

2
ds   s 2  s  1   s  3   

 


(s  2)
2
b2   s  1  

 s 2  s  12  s  3  


2

1
3 s  10 s  7


  s  12  s  3   s  14  s  3 2

s0
G (s) 
0.6667
s
2

0.4444
s

s  1
2

1
3s  3s
 2

 s  s  3 s 4  s  32

 -0.0278
s  3
 s  1
t
s0
s  1
1
g ( t )  0.4444  0.6667 t  te
 0.4444
1


d 
(s  2)
2
  s  1 
b1 

 s 2  s  12  s  3   
ds 





(s  2)
c   s  3 

 s 2  s  12  s  3  






2
 .5 e

t
.5
s 1

0.0278
s3
 0.0278 e
3t




 0.5
s  1
Heaviside Expansion

1
H eaviside E xpansion Form ula: L 




n
 A  bi   b t
As 
i

  
e
B  s   i 1  d
B  bi  
 ds

w here bi are the n distinct roots of B ( s )
E xam ple:
G (s) 
15( s  2)
s ( s  2 s  25)
2
 1  i 4.899 , and 1  i 4.899
R oots of the denom inator are 0,
d
ds
L
1
B  s   ( s  2 s  25)  2 s  2 s  5 s  4 s  25
2
G  s  
g (t ) 
30
25
e
0t
2
2
 15( s  2) 
st
  5 s 2  4 s  25  e i
 s  si
i 1 
3

15. 73.485i
 94.00  29.3 94i
e
g ( t )  1.2    0.3680  0.6667i  e
  1  i 4.899  t
  1  i 4.89 9  t

15. 73.485i
 94.00  29.394i
e
  1  i 4.899  t
   0.3680  0.6667i  e
  1  i 4.899  t
Laplace Space Formulation
 Example: The Nose Wheel
k  250, 000
kt  5, 000, 000
b  125, 000
m  250, 000
m u  50
 m z  bz  kz  bz u  kz u

 m u z u  bz u  ( k  k t ) z u  k t z r  bz  kz
  m s 2  bs  k  Z ( s )   bs  k  Z u  s 


2
  m u s  bs  k  k t  Z u  s    bs  k  Z  s   k t Z r  s 
ms
2
 bs  k  Z (s)  bs  k 
 bs  k  Z  s   ktZ r  s 
m
s  bs  k  kt 
2
u
Example
Z (s) 
 bs  k 
ms
2
Z  s    bs  k  k t Z r  s 
 bs  k   m u s  bs  k  k t 
2
2
2


 bs  k 
 bs  k  k t Z r  s 
1 

Z
s

 
2
2
2
2

m s  bs  k   m u s  bs  k  k t  
m s  bs  k   m u s  bs  k  k t 




  m s 2  bs  k   m s 2  bs  k  k    bs  k  2
u
t

2
2

m s  bs  k   m u s  bs  k  k t 


Z (s) 
Z (s) 

 bs  k  k t Z r  s 
 Z (s) 
2
2

m s  bs  k   m u s  bs  k  k t 


 bs  k  k t Z r  s 
ms
2
 bs  k   m u s  bs  k  k t    bs  k 
2
2
 bs  k  k t Z r  s 
4
3
2
m m u s  b  m  m u  s   k  m  m u   ktm  s  b(4k
 k t ) s  kk t
Now, we have a good view of the system structure
so that we can choose and adjust values, if needed
Example
 bs  k  k t
mmu
Z (s) 
s 
4
b m  mu 
s 
3
k m  m   k m 
u
mmu
Z (s) 
Z (s) 
Zr s
t
s 
2
b(4k  kt )
mmu
s
mmu
kk t
mmu
 50000. s  100000.  Z r  s 
s  2500.5 s  105001. s  60000. s  100000.
4
3
2
 50000. s  100000.  Z r  s 
 s  2458   s  42.15   s  0.2779  i 0.9423   s  0.2779  i 0.9423 
0.01 

For a 1 cm step input  Z r ( s ) 

s 

Z (s) 
a0
s

a
 s  2458 

b
 s  42.15 

a   s  2458  Z ( s ) s   2458  7.827 * 10
c
 s  0.2779  i 0.9423 

d
 s  0.2779  i 0.9423 
5
b   s  42.15  Z ( s ) s   42.15   0.004737
c   s  0.2779  i 0.9 423  Z ( s ) s   0.2779  i 0.9 423  0.00 23 26  i 0.004493
d   s  0.2779  i 0.9423  Z ( s ) s   0.2779  i 0.9423  0.002326  i 0.004493
Summary
Next Class: Nyquist Stability Criteria