System Modeling and Identification
Lecture 4
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
SMI 4/1
Chapter 2
Linearization
Interacting Tank-in-Series System
Linearize the the interacting tank-in-series system for the
operating point resulted by the parameter values as
given in Homework 2.
 For qi, use the last digit of your Student ID.
For example: Kartika  qi= 8 liters/s.
 Submit the mdl-file and the screenshots of the MatlabSimulink file + scope.
qi
h1
v1
President University
q1
a1
Erwin Sitompul
h2
v2
a2
SMI 4/2
Chapter 2
Linearization
Interacting Tank-in-Series System
 The model of the system is:
h1 
qi
A1
A1
h2 
a1
2 g ( h1  h 2 ) 
A2

a1
 f1 ( h1 , h2 , q i )
2 g ( h1  h 2 )
a2
A2
2 gh 2
 f 2 ( h1 , h2 , q i )
y1  h1
 g 1 ( h1 , h2 , q i )
y 2  h2
 g 2 ( h1 , h2 , q i )
 As can be seen from the result of Homework 2, the
steady state parameter values, which are taken to be
the operating point, are:
h1,0  0.638 m
h 2 ,0  0.319 m
q i ,0  5  10
3
m
3
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s
Erwin Sitompul
SMI 4/3
Chapter 2
Linearization
Interacting Tank-in-Series System
 The linearization around the operating point
(h1,0, h2,0, qi,0) is performed as follows:
 f1
h
h1 ,0
1 h
2 ,0
q i,0
 f1
h
h1 ,0
2 h
2 ,0
q i,0
 f1
q
h1 ,0
i h
2 ,0
q i,0



1 a1
2g
2 A1
( h1,0  h 2 ,0 )
1 a1
2g
2 A1
( h1,0  h 2 ,0 )
1
A1
1

0.25
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
1 2  10
2

0.25
1 2  10
2
3
3
0.25
2  9.8
(0.638  0.319)
2  9.8
(0.638  0.319)
  0 .0 3 1 3 5
 0 .0 3 1 3 5
4
Erwin Sitompul
SMI 4/4
Chapter 2
Linearization
Interacting Tank-in-Series System
f 2
h
h1 ,0
1 h
2 ,0
q i,0
f 2
h
h1 ,0
2 h
2 ,0
q i,0

1 a1
2g
2 A2
( h1,0  h 2 ,0 )



2
1 a1
2g
2 A2
( h1,0  h2 ,0 )
1 2  10
2
0.1
3
1 2  10

3
(0.638  0.319)
0.1
1 a2
2g
2 A2
h2 ,0
2  9.8
(0.638  0.319)

2  9.8
1 2  10
2
0.1
3
 0 .0 7 8 3 8
2  9.8
0.319
  0 .1 5 6 7 7
f 2
q
h1 ,0
i h
2 ,0
q i,0
0
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Erwin Sitompul
SMI 4/5
Chapter 2
Linearization
Interacting Tank-in-Series System
  h (t ) 
 A   h (t )  B   qi (t )
  f1
   h (t )     h
1
1

 
   h 2 ( t )     f 2

  h1
   h (t ) 
1

   h 2 ( t ) 
 f1 
  f1 
 h 2    h1 ( t )    q1 


   q (t )
i


 f 2    h2 ( t )    f 2 



 h2 

q
 1
   0.03135
 
  0.07838
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0.03135    h1 ( t )   4 
      qi (t )

 0.15677    h 2 ( t )   0 
Erwin Sitompul
SMI 4/6
Chapter 2
Linearization
Interacting Tank-in-Series System
 y (t )  C   h (t )  D   qi (t )
 g1
 h
  h1 ( t ) 
 1




h
(
t
)
 2   g 2

  h1
  h1 ( t )   1

 
  h2 ( t )   0
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g1 
 g1
 h 2    h1 ( t )    q1




 g 2    h2 ( t )    g 2


 h2 
  q1


   q (t )
i



0    h1 ( t )   0 
      qi (t )

1    h2 ( t )   0 
Erwin Sitompul
SMI 4/7
Chapter 2
Linearization
Interacting Tank-in-Series System
President University
Erwin Sitompul
SMI 4/8
Chapter 2
Linearization
Interacting Tank-in-Series System
q i  q i,0
h1  h1,0
h 2  h 2,0
:
:
:
:
h1 ,
h2 ,
h1 ,
h2 ,
President University
original model
original model
linearized model
linearized model
Erwin Sitompul
SMI 4/9
Chapter 2
Linearization
Interacting Tank-in-Series System
q i  q i,0
 5 .5 liters s
:
:
:
:
h1 ,
h2 ,
h1 ,
h2 ,
President University
original model
original model
linearized model
linearized model
Erwin Sitompul
SMI 4/10
Chapter 2
Linearization
Interacting Tank-in-Series System
q i  q i,0
 7 .5 liters s
:
:
:
:
h1 ,
h2 ,
h1 ,
h2 ,
President University
original model
original model
linearized model
linearized model
Erwin Sitompul
SMI 4/11
Chapter 3
Analysis of Process Models
State Space Process Models
 Consider a continuous-time MIMO system with m input
variables and r output variables. The relation between
input and output variables can be expressed as:
d x (t )
 f
 x ( t ), u ( t ) 
dt
y ( t )  g  x ( t ), u ( t ) 
x ( t ) : vector of state space variables
u ( t ) : vector of input variables
y ( t ) : vector of output variables
President University
Erwin Sitompul
SMI 4/12
Chapter 3
State Space Process Models
Solution of State Space Equations
 Consider the state space equations:
d x (t )
 A x ( t )  B u ( t ),
dt
y (t )  C x (t )
x (0)  x 0
 Taking the Laplace Transform yields:
s X ( s )  x 0  A X ( s )  BU ( s )
X (s)  (s I  A)
1
Y (s)  C  (s I  A)
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1
x 0  (s I  A) BU (s)
1
x (0)  ( s I  A ) B U ( s ) 
1
Erwin Sitompul
SMI 4/13
Chapter 3
State Space Process Models
Solution of State Space Equations
 After the inverse Laplace transformation,
t
x (t )  e
At
x (0) 
e
A ( t  )
B u ( ) d 
0
t
y (t )  C e
At
x (0)  C  e
A ( t  )
B u ( ) d 
0
e
At
L
1
 ( s I  A ) 1 


 The solution of state space equations depends on the
roots of the characteristic equation:
d et( s I  A )  0
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Erwin Sitompul
SMI 4/14
Chapter 3
State Space Process Models
Solution of State Space Equations
1
Consider a matrix A  
0
At
Calculate e .
e
At
 s 1
1  
 L 
  0
1 

s  2

1
1 
L 
s  1
 det  0


 1

1  s  1
 L 
 0
 
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1
1
.
2 



s  2
1   0

s  2
1  

s  1 


1


( s  1)( s  2)  

1

 
s2

Erwin Sitompul
 e
At
et
=
 0
e
2 t
t
e 

2 t
e

SMI 4/15
Chapter 3
State Space Process Models
Canonical Transformation
 Eigenvalues of A, λ1, …, λn are given as solutions of the
equation det(A–λI) = 0.
 If the eigenvalues of A are distinct, then a nonsingular
matrix T exists, such that:
ΛT
1
AT
is a diagonal matrix of the form
 1

0

Λ


0
0
2
0
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0 


0 

n 
 e
Λt
L
1
e 1t

0
 ( s I  Λ) 1   




 0
Erwin Sitompul
0
e
2t
0
SMI 4/16
0 


0 

n t
e 
Chapter 3
State Space Process Models
Canonical Transformation
 Example
Perform the canonical transformation to the state space
equations below
1

x (t )  0

 0
y ( t )  1
4
3
0
0
1 4

0
x (t ) 

 2 
13 4 


1
u (t )


 1 
0  x (t )
d et( A   I )  0
 1  

det 
0

 0

4
3  
0
1 4  

 (  1   )(  3   )(  2   )  0
0

 2     1   1
2  3
• The eigenvalues of A
3   2
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Erwin Sitompul
SMI 4/17
Chapter 3
State Space Process Models
Canonical Transformation
( A  1 I ) e 1  0
0

0

 0
4
2
0
 1 4   e11 
1 
 
 
0
e 21  0  e 1  0
 
 
 0 
 1   e 31 
( A  2 I )e 2  0
2

0

 0
4
0
0
 1 4   e12 
2
 
 
0
e 22  0  e 2  1
 
 
 0 
1   e 32 
4
1
0
 1 4   e13 
1 
 
 
0
e 23  0  e 3  0
 
 
 4 
0   e 33 
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Erwin Sitompul
0

0

 3 
0
2
0
1

 0

 0
0 

0

 2 
0
3
0
The eigenvectors of A
•
T  e1
( A  3 I ) e 3  0
1

0

 0
 1

Λ 0

 0
1

 0

 0
e2
2
1
0
e3 
1

0

4 
SMI 4/18
Chapter 3
State Space Process Models
Canonical Transformation
The equivalence transformation can now be done, with
x = T~
x. Then, the state space equations
1
1

A  T AT  Λ  0

 0
 1 
1


B T B 
,
1


 0 .2 5 
0
3
0
0 

0

 2 
C  C T  1
1

T  0

 0
2
1
T
1
1

 0

 0
2
1
0
1

0

4 
2
1
0
As the result, we obtain a state space
in canonical form,
1

x (t )  0

 0
y ( t )  1
0
3
0
2
0 

0 x (t ) 

 2 
 1 


1
u (t )


 0.25 
1 x ( t )
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Erwin Sitompul
SMI 4/19
 0 .2 5 

0

0 .2 5 
Chapter 3
State Space Process Models
Homework 4: Canonical Transformation
 Make yourself familiar with the canonical
transformation. Obtain the canonical form of the state
space below.
0

x (t )  1

 0
y (t )   0
19
0
1
2
 30 

0
x (t ) 

0 
1 
 
0 u (t )
 
 0 
1 x ( t )
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Chapter 3
State Space Process Models
Homework 4: Canonical Transformation
 Perform the canonical transformation for the following
state space equation.
 0

x (t )  0

  6
 11
y ( t )   20
9
1
0
0 

1 x (t ) 

 6 
0 
 
0 u (t )
 
 1 
1 x ( t )
 Hint: Learn the following functions in Matlab:
[V,D] = eig(X)
NEW
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