LINEAR CONTROL
SYSTEMS
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
lecture6
Lecture 6
Topics to be covered include:

State space representation.







State transition matrix.
Calculation of state transition matrix for time varying system.
Significance of state transition matrix and its property.
State transition equation.
State transition equation determined by state diagram.
Eigenvalues of the matrix A in state space form and
poles of transfer function
Similarity transformation.
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Dr. Ali Karimpour Feb 2013
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State Space Models
‫مدل فضای حالت‬
General form of state space model
LTI systems
If initial condition and input are defined, then x(t), y(t) ?
‫ چند است؟‬x(t) ‫ و‬y(t) ‫اگر شرائط اولیه و ورودی مشخص باشد مقدار‬
Start with homogeneous equation
x  Ax
.‫با معادله همگن شروع کنید‬
x(t0 )  x0
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State transition matrix ‫ماتریس انتقال حالت‬
x  Ax
x(t0  0)  x(0)
x(t )   (t ) x(0)
 (t ) : is state transition matrix ‫ماتریس انتقال حالت‬
If t0 is not zero
x(t )   (t  t0 ) x(t0 )
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Determination of state transition matrix
‫تعیین ماتریس انتقال حالت‬
Method I x  Ax x(t0 )  x0
sx( s)  x0  Ax( s)
1
‫روش اول‬
1
x( s)  ( sI  A) x0

x(t )  L ( sI  A)
1
x
 (t )  L (sI  A)
1
x(t )   (t ) x0
0
1

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Dr. Ali Karimpour Feb 2013
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Determination of state transition matrix
‫تعیین ماتریس انتقال حالت‬
Method II
x  Ax
x(t )  e x0
At
 (t )  e
x(t0 )  x0
‫روش دوم‬
x(t )   (t ) x0
At
1 2 2
1
3 3
e  I  At 
At 
A t  ...
1.2
1.2.3
At
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Dr. Ali Karimpour Feb 2013
lecture6
Determination of state transition matrix
‫تعیین ماتریس انتقال حالت‬
 x1 (t )  11 (t )
 x (t )   (t )
 2   21
12 (t )   x1 (0) 



22 (t )  x2 (0)
 x1 (t )  11 (t )
 x (t )   (t )
 2   21
12 (t )  1  11 (t ) 





22 (t ) 0 21 (t )
 x1 (t )  11 (t )
 x (t )   (t )
 2   21
12 (t )  0 12 (t ) 





22 (t ) 1  22 (t )
Method III
‫روش سوم‬
So ith column of Φ is the response of system to
0 
0 
 
. 
 
. 
1 
 
. 
. 
 
0 
ith
position
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Dr. Ali Karimpour Feb 2013
lecture6
Example 1 State transition matrix and x(t)
.‫ را بیابید‬x(t) ‫ ماتریس انتقال حالت و‬:1 ‫مثال‬
 2 1 
x  
x

 0  3
 2
x(0)   
1 


x(t )  L1 ( sI  A) 1 x0
 (t )  L1 ( sI  A) 1 
e 2t
 (t )  
 0
e
2 t

1   s  2
L 
 0

e
e  3t
3t



 1  
s  3  

2 t
 x1 (t )  e
 x (t )  
 2   0
1
 1


1  s  2
 L 
 0
 

1

( s  2)( s  3) 

1 
s  3 







e 2t  e 3t  2 3e 2t  e 3t 
   

 3t
 3t
1
e
e
   

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Dr. Ali Karimpour Feb 2013
lecture6
Property of state transition matrix for LTI systems
LTI ‫خواص ماتریس انتقال حالت در سیستمهای‬
1-
Φ(0) = I
2-
Φ-1(t)= Φ(-t)
3-
Φ(t2-t1)Φ(t1-t0) = Φ(t2-t0)
4-
Φk(t) = Φ(kt)
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Dr. Ali Karimpour Feb 2013
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State transition matrix for time varying systems
‫ماتریس انتقال حالت برای سیستمهای متغیر با زمان‬
0 0
x  
x

t 0 
x1 (t )  0
x2 (t )  tx1 (t )
 x1 (0)  1 
x(0)  
 

 x2 (0) 0
1

x(t )  
2
0.5t 
 x1 (0)  0
x(0)  
 

 x2 (0) 1 
0 
x(t )   
1 
 1
 (t )  
2
0
.
5
t

0

1
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Dr. Ali Karimpour Feb 2013
lecture6
State transition equation
x  Ax  Bu
‫معادله انتقال حالت‬
x(t0 )  x0
sx(s)  x0  Ax(s)  Bu (s)
1
1
x( s)  ( sI  A) x0  ( sI  A) Bu ( s)
t
x(t )   (t ) x0    (t   )Bu ( )d
0
State transition equation
Convolution
integral
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Dr. Ali Karimpour Feb 2013
lecture6
Example 2 Find state transition equation for unit step
.‫ معادله انتقال حالت را با فرض اعمال پله واحد بیابید‬:2 ‫مثال‬
 2 1  0
 x1 (0) 
x  
x   u x(0)  


 x2 (0)
 0  3 1 


x(t )  L1 ( sI  A) 1 x0
 (t )  L1 ( sI  A) 1 

1   s  2
L 
 0

e 2t
 (t )  
 0
 1  
s  3  

1
 1


1  s  2
 L 
 0
 

e 2t  e 3t 

 3t
e

1

( s  2)( s  3) 

1 
s  3 







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Dr. Ali Karimpour Feb 2013
lecture6
Example 2 Find state transition equation for unit step
.‫ معادله انتقال حالت را با فرض اعمال پله واحد بیابید‬:2 ‫مثال‬
 2 1  0
x  
x   u

 0  3 1 
 x1 (0) 
x(0)  

x
(
0
)
 2 
e 2t
 (t )  
 0
e 2t  e 3t 

 3t
e

t
x(t )   (t ) x0    (t   )bu( )d
0
2 ( t  )
t e
e 2t e 2t  e 3t   x1 (0) 
e 2(t  )  e 3(t  )  0
 
x(t )  
  u ( )d


3( t  )
 3t
0
e
e
0
 1 
0
  x2 (0)
2 ( t  )
e 2t e 2t  e 3t   x1 (0) 
t e
 e 3( t  ) 
 
x(t )  
d


3( t  )
 3t
0
e
e


0
  x2 (0)
.....
 
..... 13
Dr. Ali Karimpour Feb 2013
lecture6
Example 3 Find x1(s) and x2(s)
.‫را بیابید‬
x2(s)
‫و‬
x1(s)
‫ معادله‬:3 ‫مثال‬
 2 1  0
 x1 (0) 
x  
x   u x(0)  


 x2 (0)
 0  3 1 
x(s)  (sI  A) 1 x(0)  (sI  A) 1 Bu (s)
s  2
x( s )  
 0
 1
 x1 ( s )   s  2
 x ( s)  
 2   0

1
1

s  2
x(0)  

s3 
 0
1
1
 0
u ( s)



s  3  1 
1
1
1
1



 1

x1 (0) 
x 2 ( 0) 
u ( s)




( s  2)( s  3)  x1 (0)  ( s  2)( s  3)
s2
( s  2)( s  3)
( s  2)( s  3)

u ( s )  



1   x 2 ( 0)  
1
1
1



x 2 ( 0) 
u (s)





s3 
s3


s3
s3

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Dr. Ali Karimpour Feb 2013
lecture6
ِExample 4: Derive x1(s) by using state diagram.
 2 1  0
x  
x   u

 0  3 1 
x2 (0)
u(s)
1
x2 s -1
-3
x2
x1 (0)
s -1
1
x1 s -1 x1
x2
-2
s -1
x1
1
1
1
x1 ( s) 
x1 (0) 
x2 (0) 
u ( s)
s2
( s  2)( s  3)
( s  2)( s  3)
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Dr. Ali Karimpour Feb 2013
lecture6
Relation between SS model and TF model.
‫ارتباط بین معادالت فضای حالت و تابع انتقال‬
adj ( sI  A)
G( s)  C
BD
sI  A
Poles of G(s) ?
Eigenvalues of A (modes) ?
‫ ؟‬A ‫ارتباط بین قطبهای تابع انتقال و مقادیر ویژه‬
Poles of G(s)

Eigenvalues of A
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Dr. Ali Karimpour Feb 2013
lecture6
Example 5: Find the poles of transfer function
and the eigenvalues of A
1
0  0 
 0
x   0
0
1  x  0u
 6  11  6 1 
y  [1 1 0] x
s 1
SI  A  0
s
‫ مطلوبست قطبهای‬:5‫مثال‬
A ‫تابع انتقال و مقادیر ویژه‬
0
 1  s 3  6s 2  11s  6  ( s  1)( s  2)( s  3)
6 11 s  6
eig(A)
s 1
1
s 1


g ( s )  d  c( sI  A) b  3
2
( s  1)( s  2)( s  3) ( s  2)( s  3)
s  6s  11s  6
1
[num,den]=ss2tf(A,b,c,0), g=tf(num,den) , g=minreal(g)
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Dr. Ali Karimpour Feb 2013
lecture6
Example5: Continue
1
0  0 
 0
x   0
0
1  x  0u
 6  11  6 1 
y  [1 1 0] x
1  3
2  2
3  1
Eigenvalues of A
A ‫مقادیر ویژه‬
‫ ادامه‬:5‫مثال‬
G( s) 
1
( s  2)( s  3)
p1  3
p2  2
Poles of system
‫قطبهای سیستم‬
What happened to -1 ??!!??
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Dr. Ali Karimpour Feb 2013
lecture6
Change of variables (Similarity transformation)
)‫تغییر متغیرها ( تبدیالت همانندی‬
x  Ax  bu
y  cx  du
w1  x1  x3
w2  x1  2 x2

w3  x2  x3
x  Ax  bu
y  cx  du

Let
A is 3  3
b is 3 1
c is 1 3
d is 11
1 0  1
w  1 2 0  x  Px
0 1 1 
1
P 1w  AP w  bu
y  cP 1w  du
w Aˆ w  bˆu
y  cˆw  dˆu
1
1 0  1
x  P 1w  1 2 0  w
0 1 1 


Â
w P A P 1w  Pbu
y  cP 1w  du
b̂
ĉ
d̂
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Dr. Ali Karimpour Feb 2013
lecture6
Example 6: A similarity transformation example
‫ یک مثال از تبدیل همانندی‬:6 ‫مثال‬
 3  6  4  6 
x   1
2
2  x   3u
 1  6  6   4 
y  [ 2 2  1]x
 2 - 1 2 
P 1   1 1 - 1
 - 1 - 1 2 
1  2
2  4
3  1
0
 2 0
Aˆ  PAP1   0  1 0
 0 0  4
Eigenvalues of A
A ‫مقادیر ویژه‬
 4.9
bˆ  Pb   0 
 3 
cˆ  cP 1  [  0.4 0.6 0]
dˆ  d  0
Eigenvector corresponding to -2
 2
w   0
 0
y  [  0 .4
0
  4 .9 
 1 0  w   0 u
 3 
0  4
0
1  2
2  4
3  1
Eigenvalues of Â
 ‫مقادیر ویژه‬
0.6 0]w
New system is very simpler than older
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Dr. Ali Karimpour Feb 2013
Similarity transformation and eigenvalues
lecture6
‫تغییر متغیرها و مقادیر ویژه‬
x Ax  bu
y  cx  du
z Aˆ z  bˆu
y  cˆz  dˆu
Eigenvalues of A
A ‫مقادیر ویژه‬
Eigenvalues of Â
 ‫مقادیر ویژه‬


sI  A  0
sI  Aˆ  0
1
1
sI  Aˆ  sI  PAP 1  PsP  PAP  P sI  A P 1 

sI  Aˆ  sI  A
sI  A
Similarity transformation
does not affect the
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eigenvalues of the systemDr. Ali Karimpour Feb 2013
lecture6
Exercise
1- Find the state diagram of the following system
 2 1 3 0
x   0  3 2 x  1 u
 1 4  1  0
2- Find the transfer function of the following system.
a) By use of the formula between the two representation.
 2 1 3
b) By use of state diagram


x   0
 1
0 
 3 2  x  1 u
4  1  0
y  [1 2 0]x
3- Find y(t) for initial condition [1 3 -1]T and the unit step as
 2 1 3 0
input.
x   0  3 2  x  1 u

  
 1 4  1  0
22
y  [1 2 Dr.0Ali
]x Karimpour Feb 2013
lecture6
Exercise (cont.)
4- The state transition matrix and state transition equation
of following system.
 1 3 
1 
X (t )  
 X (t )  2 r (t )
0

4


 
Answer is:
 (t )  L1 ( sI  A) 1 
e  t
X (t )  
0
3
 1

1
t
2




s

1

3


  L1  s  1 s  5s  4   e
 L1  
  0 s  4 
1  0
 0



s  4 
 ( t  )
t e
e t  e 4t 
X
(
0
)


0 0
e  4t 

e t  e  4 t 

e  4t 
e  (t  )  e 4(t  )  1 
   R( )dt
e 4 (t  )  2
5- Suppose x1(0)=1, x2(0)=3 and x3(0)=2 and r(t)=u(t) . Find x(t)
and c(t) for t≥0
 1 4 0 
1 
X (t )   2  2 1  X (t )  1  r (t )
 0 0  3
0
c(t )  [1 0
0] X (t )
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Dr. Ali Karimpour Feb 2013