canonical transformation
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Chapter 3. State Space Process Models. Canonical Transformation. Homework 4. Chapter 3. State Space Process Models. Canonical Transformation. As the result,. Chapter 3. State Space Process Models. Order Reduction. - PowerPoint PPT PresentationTRANSCRIPT
President University Erwin Sitompul SMI 5/1
Dr.-Ing. Erwin SitompulPresident University
Lecture 5
System Modeling and Identification
http://zitompul.wordpress.com
President University Erwin Sitompul SMI 5/2
Canonical TransformationHomework 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
1 5 2 3 3 2
Chapter 3 State Space Process Models
0 19 30
1 0 0
0 1 0
A
1
25
,5
1
e
( ) ii 0A I e
19 30
1 0
0 1
i
ii
i
0e 2
9
,3
1
e 3
4
.2
1
e
President University Erwin Sitompul SMI 5/3
Chapter 3 State Space Process Models
Canonical Transformation
1 2 3T e e e
25 9 4
5 3 2
1 1 1
1
0.0179 0.0893 0.1071
0.1250 0.3750 1.25
0.1429 0.2857 2.1429
T
1
5 0 0
0 3 0
0 0 2
A T AT
1
0.0179
0.1250
0.1429
B T B
9 7 5 C CT
As the result,
5 0 0 0.0179
( ) ( ) ( )0 3 0 0.1250
0 0 2 0.1429
t t u t
x x
( ) 9 7 5 ( )y t t x
President University Erwin Sitompul SMI 5/4
A mathematical model which is constructed using physical or chemical laws may have a high order due to the number of interacting components inside the model.
In many cases, a model with lower order is wanted because it is easier to handle.
Thus, it is wished that the order of a model can be made low. But, it is also wished that the model still can be interpreted physically.
The method in doing so is called “Order Reduction”.
Chapter 3 State Space Process Models
Order Reduction
President University Erwin Sitompul SMI 5/5
Chapter 3 State Space Process Models
Order ReductionStarting point:
A state space of order n x Ax Buy C x
We define:xs : part of x that contains states considered to be
significant, such as measured variables or other critical variables.
xr : the remaining part of x which are not the member of xs.
After the definition process, the states are to be reconstructed as:
s
r
xx
x
President University Erwin Sitompul SMI 5/6
Chapter 3 State Space Process Models
Order ReductionThe reconstruction of state vector yields an implication
on matrices A, B, and C. If the ith state is switched with the jth state,
xixj
ith column of C jth column of C
ith row of B jth row of B
ith column & ith row of A
jth column & jth
row A
President University Erwin Sitompul SMI 5/7
Chapter 3 State Space Process Models
Order ReductionResult:
A new state space of order p, as an approximation of the original state space of order n
ss x Ax Bu
sy C x
Question:How to find the ideal order p of the new state space?
•What is p equal to?
President University Erwin Sitompul SMI 5/8
Chapter 3 State Space Process Models
Order Reduction
x T z
The order reduction is performed using the Canonical Transformation:
•Modal Coordinate•Matrix of Eigenvectors of A
x Ax Bu
T z AT z Bu1 1 z T AT z T Bu
Λ *B
y C x
y CT z
*C[nn] [nm] [rn]
President University Erwin Sitompul SMI 5/9
Chapter 3 State Space Process Models
Order Reduction
* z Λz B u*y C z
As the results of the Canonical Transformation, in time and frequency domain we can write:
*1( ) ( ) ( )s s s Z I Λ B U*( ) ( )s sY C Z
* *1( ) ( ) ( )s s s Y C I Λ B U
1
1
1
1 2
1
0 0
0 0( )
0 0
s
s
s n
s
I Λ
1
2
0 0
0 0
0 0 n
Λ
President University Erwin Sitompul SMI 5/10
Chapter 3 State Space Process Models
Order ReductionExamining the output equations in frequency domain,
the relationship between the jth input Uj(s) and the ith output Yi(s) can be formulated as:
* *1( ) ( ) ( )s s s Y C I Λ B U
1
1
1
2
1
0 0
0 0
0 0
s
s
s n
* *
1
1( ) ( )
n
ik kji jk k
c bY s U ss
*1
1* *1
*
( ) ( ) ( ) j
i ji in
nj
b
Y s s U sc c
b
I Λ
President University Erwin Sitompul SMI 5/11
Chapter 3 State Space Process Models
The Procedure of Order ReductionStep 1:
The finding of dominant eigenvalues
The step response of the state space model is given as:
* *
1
1 1( )
n
ik kjik k
c bY ss s
* *
1
1( ) 1k
nt
i ik kjk k
y t c b e
•approaches –1 for stable λ,
can be normed
•measures the influence of λk in the connection between uj(t) and yi(t)
President University Erwin Sitompul SMI 5/12
Chapter 3 State Space Process Models
The Procedure of Order ReductionWe now define a Dominance Measure,
* *ik kj
ikj
k
c bD
The value of Dijk will be small for large λk (a λ with large value, away from imaginary axis, is not a dominant λ)
The value of Dijk will be zero for cik* = 0
(the kth proper motion of the ith output is not observable)
The value of Dijk will be zero for bkj* = 0
(the kth proper motion of the jth input is not controllable)
President University Erwin Sitompul SMI 5/13
Chapter 3 State Space Process Models
The Procedure of Order ReductionThe Dominance Measures of each eigenvalue are
further analyzed to yield two measures, the maximum and the sum:
1 1100 max max
r m
k ikji j
M D
1 1
100r m
k ikji j
S D
Maximum of Dominance Measure
Sum of Dominance Measure
President University Erwin Sitompul SMI 5/14
Chapter 3 State Space Process Models
The Procedure of Order Reduction The dominant eigenvalues are chosen according to the
following criteria:1. All unstable eigenvalues (λ>0) are dominant.2. The stable eigenvalues with the largest Mk are
dominant.Generally, if Mk>20, then λk is dominant
if Mk <2, then λk is not dominant
if 2<Mk<20, consider SkResult of Step 1:
Out of n eigenvalues, p dominant eigenvalues are chosen: λ1, λ2, …, λp
President University Erwin Sitompul SMI 5/15
Chapter 3 State Space Process Models
The Procedure of Order ReductionStep 2:
The calculation of A, B, and C~ ~ ~
The state vector z is now reorganized as follows:
1
1
p
p
n
z
zz
z
zd
n
z
z
Where:zd : states of z with dominant eigenvalues.zn : the remaining states of z, which are the states
with not dominant eigenvalues.
President University Erwin Sitompul SMI 5/16
Chapter 3 State Space Process Models
The Procedure of Order Reduction
* z Λz B u*
d 1d d
*nn n 2
z Λ z Bu
z Λ z B
0
0 1
d
p
Λ 0
0
1
n
p
n
Λ 0
0
The canonical form of the state space, including the dominance consideration, is:
d* *1 2
n
zy C C z
*y C z
•What is B1*, B2
* ?
•What is C1*, C2
*?
President University Erwin Sitompul SMI 5/17
Chapter 3 State Space Process Models
The Procedure of Order Reduction
x T z
The transformation matrix T is also reorganized, considering dominant states (eigenvalues):
11 12s d
21 22r n
x T T z
x T T z
•Chosen freely according to a certain technical criteria defined by model maker
•Chosen according to a mathematical criteria the Dominance Measures
The matrix multiplication yields:
11 12s d n x T z T z 11s d x T z
•Omitted, because not significant/dominant
111 sd
z T x
President University Erwin Sitompul SMI 5/18
Chapter 3 State Space Process Models
The Procedure of Order Reduction
*d 1d d
*nn n 2
z Λ z Bu
z Λ z B
0
0
We go back now to the state equations:
*d 1d d z Λ z B u
1 *11 d 11 11 1s s
x T Λ T x T B u
A B
111 sd
z T x
[pp] [pm]
President University Erwin Sitompul SMI 5/19
Chapter 3 State Space Process Models
The Procedure of Order Reduction
C
* 1111 sy C T x
Also, we go back to the output equations:
d* *1 2
n
zy C C z
* *1 2d n y C z C z
•Omitted, because not significant/dominant
111 sd
z T x
[rp]
President University Erwin Sitompul SMI 5/20
Chapter 3 State Space Process Models
The Procedure of Order ReductionResult of Step 2:
Out of a state space of order n, we will have a state space of order p:
s s x Ax Bu
sy C x
President University Erwin Sitompul SMI 5/21
Recollecting the last example, the original state space is given as:
Chapter 3 State Space Process Models
Example: Order Reduction
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
After canonical transformation:11 0 0
( ) ( ) 1 ( )0 3 0
0.250 0 2
t t u t
x x
( ) 1 2 1 ( )y t t x
Simplify the 3rd order state space into a 2nd order state space using the Order Reduction, if the significant states are xs =[x1 x2]T.
President University Erwin Sitompul SMI 5/22
Chapter 3 State Space Process Models
Example: Order Reduction
1
1 2 0.25
0 1 0
0 0 0.25
T
1 2 1
0 1 0
0 0 4
T
1
1 0 0
0 3 0
0 0 2
Λ T AT
* 1
1
1
0.25
B T B
* 1 2 1 C CT
n=3 k=[1…n]m=1 j=[1…m]
r=1 i=[1…r]
* *11 11
1111
c bD
* *12 21
1212
c bD
(1)(1)
( 1)
1
(2)(1)
( 3)
0.667
* *13 31
1313
c bD
(1)(0.25)
( 2)
0.125
1 1100 max max
r m
k ikji j
M D
1 111100M D 100
2 121100M D 66.7
3 131100M D 12.5
•Dominant
•Dominant
President University Erwin Sitompul SMI 5/23
Chapter 3 State Space Process Models
Example: Order Reduction
1 1
2 2
3 3
1 2 1
0 1 0
0 0 4
x z
x z
x z
11T 111 d 11
A T Λ T
1 2 1 0 1 2
0 1 0 3 0 1
1 4
0 3
*11 1B T B
1 2 1
0 1 1
3
1
* 1111C C T
1 21 2
0 1
1 0
s s
1 4 3
0 3 1
x x u
s1 0y x
1 0 0
0 3 0
0 0 2
Λ
*
1
1
0.25
B
* 1 2 1C
dΛ
*1B
*1C
President University Erwin Sitompul SMI 5/24
Chapter 3 State Space Process Models
Order Reduction Homework 5
A linear time-invariant system is given as below:3 1 0 1
( ) 1 3 0 ( ) 0 ( )
3 5 6 4.5
t t u t
x x
0 1 0( ) ( )
0 0 1t t
y x
a) Calculate the eigenvalues and the eigenvectors of the system.b) A second order model is now wished to approximate the system.
The second and the third state are chosen to be the significant states. Perform the Order Reduction based on the chosen significant states. Regarding the Dominance Measure, which eigenvalues of the original model should be considered in the new reduced-order model?
c) Write the complete reduced-order model in state space form. Hint: This model must be a second order model.