Interacting Tank-in-Series System

President University Erwin Sitompul SMI 4/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 4System Modeling and Identification

http://zitompul.wordpress.com


Chapter 2 Linearization

Interacting Tank-in-Series System

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.


Interacting Tank-in-Series SystemChapter 2 Linearization

i 11 1 2

1 1

2 ( )q ah g h hA A

1 22 1 2 2

2 2

2 ( ) 2a ah 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 m5 10 m s

hhq

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:



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

1,02,0i,0

1 1

1 1 1,0 2,0

1 22 ( )h

hq

f a gh A h h

31 2 10 2 9.82 0.25 (0.638 0.319)

0.03135

1,02,0i,0

1 1

2 1 1,0 2,0

1 22 ( )h

hq

f a gh A h h

31 2 10 2 9.82 0.25 (0.638 0.319)

0.03135

1,02,0i,0

1

i 1

1hhq

fq A

10.25 4



1,02,0i,0

2 1

1 2 1,0 2,0

1 22 ( )h

hq

f a gh A h h

31 2 10 2 9.82 0.1 (0.638 0.319)

0.07838

1,02,0i,0

2 1 2

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

1 2 1 22 ( ) 2h

hq

f a ag gh A h h A h

3 31 2 10 2 9.8 1 2 10 2 9.82 0.1 (0.638 0.319) 2 0.1 0.319

0.15677

1,02,0i,0

2

i

0hhq

fq


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


1 1 1

1 1 2 11i

22 2 22

1 2 1

( ) ( )( )

( )( )

f f fh t h h qh t

q th tf f fh t

h h q

1 1i

22

( ) ( )0.03135 0.03135 4( )

( )0.07838 0.15677 0( )

h t h tq t

h th t



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 gh h qh t h t

q th t h tg g g

h h q

1 1i

2 2

( ) ( )1 0 0( )

( ) ( )0 1 0h t h t

q th t h t



: 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



i i,0

5.5 liters sq q





i i,0

7.5 liters sq q


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 t t tdt

x f x u

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

( )( )( )

ttt

xuy

: vector of state space variables: vector of input variables: vector of output variables


Solution of State Space EquationsChapter 3 State Space Process Models

0( ) ( ) ( ), (0)d t t tdt

x Ax Bu x x

0( ) ( ) ( )s s s s X x AX BU1 1

0( ) ( ) ( ) ( )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:


Chapter 3

After the inverse Laplace transformation,Solution of State Space Equations

State 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


1 10 2

A

11 1 1

0 2t s

es

A L

1

2 111 1 0 1

det0 2

ss s

s

L

Chapter 3

Solution of State Space EquationsState Space Process Models

Consider a matrix .Calculate .teA

1

1 11 ( 1)( 2)

102

s s s

s

L2

20

t t tt

t

e e ee

e

A =


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 ATis a diagonal matrix of the form

1

2

0 00

00 0 n

Λ

1 1( )te s Λ I ΛL

1

2

0 00

00 0 n

t

t

t

ee

e


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 1t t u t

x x

( ) 1 0 0 ( )y t t x

det( ) 0 A I

1 4 1 4det 0 3 0

0 0 2

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

• The eigenvalues of A



Canonical Transformation11( ) 0A I e

11

21

31

0 4 1 40 2 00 0 1

eee

0 1

100

e

22( ) 0A I e12

22

32

2 4 1 40 0 00 0 1

eee

0 2

210

e

33( ) 0A I e13

23

33

1 4 1 40 1 00 0 0

eee

0 3

104

e

1

2

3

0 00 00 0

Λ

1 0 00 3 00 0 2

1 2 3T e e e

1 2 10 1 00 0 4

The eigenvectors of A•



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

1

1 2 0.250 1 00 0 0.25

T

1

1 0 00 3 00 0 2

A T AT Λ

1

1,1

0.25

B T B 1 2 1 C CT

1 2 10 1 00 0 4

T

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

11 0 0( ) ( ) 1 ( )0 3 0

0.250 0 2t t u t

x x

( ) 1 2 1 ( )y t t x


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


Homework 4: Canonical Transformation

0 19 30 1( ) 1 0 0 ( ) 0 ( )

0 1 0 0t t u t

x x

( ) 0 2 1 ( )y t t x


Perform the canonical transformation for the following state space equation.


Homework 4: Canonical Transformation

0 1 0 0( ) 0 0 1 ( ) 0 ( )

6 11 6 1t t u t

x x

( ) 20 9 1 ( )y t t x

NEW

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

Interacting Tank-in-Series System

Documents

Transcript of Interacting Tank-in-Series System