interacting tank-in-series system
DESCRIPTION
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 q i , use the last digit of your Student ID. For example: Kartika q i = 8 liters/s. - PowerPoint PPT PresentationTRANSCRIPT
President University Erwin Sitompul SMI 4/1
Dr.-Ing. Erwin SitompulPresident University
Lecture 4System Modeling and Identification
http://zitompul.wordpress.com
President University Erwin Sitompul SMI 4/2
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.
President University Erwin Sitompul SMI 4/3
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:
President University Erwin Sitompul SMI 4/4
Interacting Tank-in-Series SystemChapter 2 Linearization
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
President University Erwin Sitompul SMI 4/5
Interacting Tank-in-Series SystemChapter 2 Linearization
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
President University Erwin Sitompul SMI 4/6
i( ) ( ) ( )t t q t h A h B
Interacting Tank-in-Series SystemChapter 2 Linearization
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
President University Erwin Sitompul SMI 4/7
Interacting Tank-in-Series SystemChapter 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 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
President University Erwin Sitompul SMI 4/8
Interacting Tank-in-Series SystemChapter 2 Linearization
President University Erwin Sitompul SMI 4/9
Interacting Tank-in-Series SystemChapter 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
President University Erwin Sitompul SMI 4/10
Interacting Tank-in-Series SystemChapter 2 Linearization
i i,0
5.5 liters sq q
: h1, original model: h2, original model: h1, linearized model: h2, linearized model
President University Erwin Sitompul SMI 4/11
Interacting Tank-in-Series SystemChapter 2 Linearization
: h1, original model: h2, original model: h1, linearized model: h2, linearized model
i i,0
7.5 liters sq q
President University Erwin Sitompul SMI 4/12
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
President University Erwin Sitompul SMI 4/13
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:
President University Erwin Sitompul SMI 4/14
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
President University Erwin Sitompul SMI 4/15
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 =
President University Erwin Sitompul SMI 4/16
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
President University Erwin Sitompul SMI 4/17
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
President University Erwin Sitompul SMI 4/18
Chapter 3 State Space Process Models
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•
President University Erwin Sitompul SMI 4/19
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 0.250 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
President University Erwin Sitompul SMI 4/20
Make yourself familiar with the canonical transformation. Obtain the canonical form of the state space below.
Chapter 3 State Space Process Models
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
President University Erwin Sitompul SMI 4/21
Perform the canonical transformation for the following state space equation.
Chapter 3 State Space Process Models
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)