President University Erwin Sitompul SMI 4/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 4

System Modeling and Identification

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Chapter 2 Linearization

Homework 3

v1

qi

h1 h2

v2

q1

a1 a2

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 Matlab-Simulink file

+ scope.

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Solution to Homework 3Chapter 2 Linearization

i 11 1 2

1 1

2 ( )q a

h g h hA A

1 22 1 2 2

2 2

2 ( ) 2a a

h g h h ghA A

The model of the system is:

1 1 2 i( , , )f h h q

2 1 2 i( , , )f h h q

1,0

2,03 3

i,0

0.638 m0.319 m

5 10 m s

hh

q

1 1y h

2 2y h1 1 2 i( , , )g h h q

2 1 2 i( , , )g h h q

As can be seen from the result of Homework 2, the steady state parameter values, which are taken to be the operating point, are:

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Solution to Homework 3Chapter 2 Linearization

The linearization around the operating point (h1,0, h2,0, qi,0) is performed as follows:

1,0

2,0

i,0

1 1

1 1 1,0 2,0

1 2

2 ( )hhq

f a g

h A h h

31 2 10 2 9.8

2 0.25 (0.638 0.319)

0.03135

1,0

2,0

i,0

1 1

2 1 1,0 2,0

1 2

2 ( )hhq

f a g

h A h h

31 2 10 2 9.8

2 0.25 (0.638 0.319)

0.03135

1,0

2,0

i,0

1

i 1

1hhq

f

q A

1

0.25 4

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Solution to Homework 3Chapter 2 Linearization

1,0

2,0

i,0

2 1

1 2 1,0 2,0

1 2

2 ( )hhq

f a g

h A h h

31 2 10 2 9.8

2 0.1 (0.638 0.319)

0.07838

1,0

2,0

i,0

2 1 2

2 2 1,0 2,0 2 2,0

1 2 1 2

2 ( ) 2hhq

f a ag g

h A h h A h

3 31 2 10 2 9.8 1 2 10 2 9.8

2 0.1 (0.638 0.319) 2 0.1 0.319

0.15677

1,0

2,0

i,0

2

i

0hhq

f

q

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i( ) ( ) ( )t t q t h A h B

Solution to Homework 3Chapter 2 Linearization

1 1 1

1 1 2 11i

22 2 22

1 2 1

( ) ( )( )

( )( )

f f f

h t h h qh tq t

h tf f fh th h q

1 1i

22

( ) ( )0.03135 0.03135 4( )

( )0.07838 0.15677 0( )

h t h tq t

h th t

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Solution to Homework 3Chapter 2 Linearization

i( ) ( ) ( )t t q t y C h D

1 1 1

1 2 11 1i

2 22 2 2

1 2 1

( ) ( )( )

( ) ( )

g g g

h h qh t h tq t

h t h tg g g

h h q

1 1i

2 2

( ) ( )1 0 0( )

( ) ( )0 1 0

h t h tq t

h t h t

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Solution to Homework 3Chapter 2 Linearization

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Solution to Homework 3Chapter 2 Linearization

: h1, original model: h2, original model

: h1, linearized model: h2, linearized model

i i,0

1 1,0

2 2,0

q qh hh h

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Solution to Homework 3Chapter 2 Linearization

i i,0

5.5 liters sq q

: h1, original model: h2, original model

: h1, linearized model: h2, linearized model

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Solution to Homework 3Chapter 2 Linearization

: h1, original model: h2, original model

: h1, linearized model: h2, linearized model

i i,0

7.5 liters sq q

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Chapter 3

Analysis of Process Models

System Modeling and Identification

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State Space Process ModelsChapter 3 Analysis of 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 tt t

dt

xf x u

( ) ( ), ( )t t ty g x u

( )

( )

( )

t

t

t

x

u

y

: vector of state space variables

: vector of input variables

: vector of output variables

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Solution of State Space EquationsChapter 3 State Space Process Models

0

( )( ) ( ), (0)

d tt t

dt

xAx Bu x x

0( ) ( ) ( )s s s s X x AX BU

1 10( ) ( ) ( ) ( )s s s s X I A x I A BU

Consider the state space equations:

1 1( ) ( ) (0) ( ) ( )s s s s Y C I A x I A BU

( ) ( )t ty C x

Taking the Laplace Transform yields:

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Chapter 3

After the inverse Laplace transformation,

Solution of State Space EquationsState Space Process Models

( )

0

( ) (0) ( )t

t tt e e d A Ax x Bu

( )

0

( ) (0) ( )t

t tt e e d A Ay C x C Bu

1 1( )te s A I AL

The solution of state space equations depends on the roots of the characteristic equation:

det( ) 0s I A

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1 1

0 2

A

1

1 1 1

0 2t s

es

A L

1

2 111 1 0 1

det0 2

s

s s

s

L

Chapter 3

Solution of State Space EquationsState Space Process Models

Consider a matrix .

Calculate .teA

1

1 1

1 ( 1)( 2)

10

2

s s s

s

L2

20

t t tt

t

e e ee

e

A =

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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:

Chapter 3

Canonical TransformationState Space Process Models

1Λ T AT

is a diagonal matrix of the form

1

2

0 0

0

0

0 0 n

Λ

1 1( )te s Λ I ΛL

1

2

0 0

0

0

0 0 n

t

t

t

e

e

e

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ExamplePerform the canonical transformation to the state space equations below

Chapter 3 State Space Process Models

Canonical Transformation

1 4 1 4 13 4

( ) 0 3 0 ( ) 1 ( )

0 0 2 1

t t u t

x x

( ) 1 0 0 ( )y t t x

det( ) 0 A I

1 4 1 4

det 0 3 0

0 0 2

( 1 )( 3 )( 2 ) 01 1 2 3 3 2

• The eigenvalues of A

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Chapter 3 State Space Process Models

Canonical Transformation11( ) 0A I e

11

21

31

0 4 1 4

0 2 0

0 0 1

e

e

e

0 1

1

0

0

e

22( ) 0A I e

12

22

32

2 4 1 4

0 0 0

0 0 1

e

e

e

0 2

2

1

0

e

33( ) 0A I e

13

23

33

1 4 1 4

0 1 0

0 0 0

e

e

e

0 3

1

0

4

e

1

2

3

0 0

0 0

0 0

Λ

1 0 0

0 3 0

0 0 2

1 2 3T e e e

1 2 1

0 1 0

0 0 4

The eigenvectors of A•

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Chapter 3 State Space Process Models

Canonical TransformationThe equivalence transformation can now be done, with x = T x. Then, the state space equations

~

1

1 2 0.25

0 1 0

0 0 0.25

T

1

1 0 0

0 3 0

0 0 2

A T AT Λ

1

1

,1

0.25

B T B 1 2 1 C CT

1 2 1

0 1 0

0 0 4

T

As the result, we obtain a state space in canonical form,

11 0 0

( ) ( ) 1 ( )0 3 0

0.250 0 2

t t u t

x x

( ) 1 2 1 ( )y t t x

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Make yourself familiar with the canonical transformation. Obtain the canonical form of the state space below.

Chapter 3 State Space Process Models

Homework 4

0 19 30 1

( ) 1 0 0 ( ) 0 ( )

0 1 0 0

t t u t

x x

( ) 0 2 1 ( )y t t x

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Perform the canonical transformation for the following state space equation.

Chapter 3 State Space Process Models

Homework 4 (New)

0 1 0 0

( ) 0 0 1 ( ) 0 ( )

6 11 6 1

t t u t

x x

( ) 20 9 1 ( )y t t x

NEW

Hint: Learn the following functions in Matlab: [V,D] = eig(X)