Lecture 22: Partial Fraction Expansion of the Lapalace Transform
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Transcript Lecture 22: Partial Fraction Expansion of the Lapalace Transform
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
t0
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
t0
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
t0
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
t0
1
sa
1
s a
2
0 t< 0
f ( t ) n at
t0
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
t0
te
L ( f ( t ))
Laplace Transforms of
s a
2
2
Important Inverse Transforms
1
at
1
L
e
sa
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
n2
... 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
n2
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
n2
... 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
s0
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 2s 4
s s 1 s 5
G (s)
a0
s
a1
s 1
a2
s5
s 2s 4
a0 s
s s 1 s 5
s0
s 2s 4
a1 s 1
s s 1 s 5
s 2s 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
s5
0.15 e
5t
24
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 2s 4
s s 1 s 5
G (s)
a0
s
a1
s 1
s 2s 4
a2
s5
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
s5
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
1G 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
s0
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
an2
d
( s pi ) n G s
ds
a n3
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 12 s 3
b1
s 1
s3
0.6667
2
a1
c
s0
d 2
(s 2)
s
2
ds s 2 s 1 s 3
(s 2)
2
b2 s 1
s 2 s 12 s 3
2
1
3 s 10 s 7
s 12 s 3 s 14 s 3 2
s0
G (s)
0.6667
s
2
0.4444
s
s 1
2
1
3s 3s
2
s s 3 s 4 s 32
-0.0278
s 3
s 1
t
s0
s 1
1
g ( t ) 0.4444 0.6667 t te
0.4444
1
d
(s 2)
2
s 1
b1
s 2 s 12 s 3
ds
(s 2)
c s 3
s 2 s 12 s 3
2
.5 e
t
.5
s 1
0.0278
s3
0.0278 e
3t
0.5
s 1
Heaviside Expansion
1
H eaviside E xpansion Form ula: L
n
A bi b t
As
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