solution to homework 4

President University Erwin Sitompul SMI 5/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 5

System Modeling and Identification

http://zitompul.wordpress.com


Solution to Homework 40 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



Solution to Homework 4

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


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”.


Order Reduction



Order Reduction Starting 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



Order Reduction The 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



Order Reduction Result:

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?



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 Bu

1 1 z T AT z T Bu

Λ*B

y C x

y CT z

*C[nn] [nm] [rn]



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



Order Reduction Examining 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



The Procedure of Order Reduction Step 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)



The Procedure of Order ReductionWe now define a Dominance Measure,

* *ik kj

ikj

k

c bD

The value of Dikj will be small for large λk (a λ with large value, away from imaginary axis, is not a dominant λ).

The value of Dikj will be zero for cik* = 0

(the kth proper motion of the ith output is not observable). The value of Dikj will be zero for bkj

* = 0 (the kth proper motion of the jth input is not controllable).



The Procedure of Order Reduction The 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



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, for normalized system with gain equals to 1,if Mk >20, then λk is dominantif Mk <2, then λk is not dominantif 2<Mk<20, consider Sk

Result of Step 1:Out of n eigenvalues, p dominant eigenvalues are chosen: λ1, λ2, …, λp



The Procedure of Order Reduction Step 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

z d

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.



The Procedure of Order Reduction

* z Λz B u*

d 1d d

*nn n 2

z Λ z Bu

z Λ z B

0

01

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

*?




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




*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]




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]



The Procedure of Order Reduction Result 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


Recollecting the last example, the original state space is given as:


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.




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




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



Comparison: Original vs. Reduced-Order Model

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

s s

1 4 3

0 3 1

x x u

s1 0y x

Original Model

Reduced-Order Model



Comparison: Original vs. Reduced-Order Model

: Original model: Reduced-order model

• Similar overall response• Steady-state gain of

the reduced-order model can be tuned to match the original model• How to find the tuning

factor? 2 Extra Points



Homework 5A 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.



Homework 5 (New)Redo b) and c) of the previous homework problem if, for this time, only the first state is considered to be the significant state. This is equivalent to replacing the output equation with:

NEW

( ) 1 0 0 ( )t ty x

3 1 0 1

( ) 1 3 0 ( ) 0 ( )

3 5 6 4,5

t t u t

x x

For even Student-ID number, take x1 and x2 as significant states.For odd Student-ID number, take x1 and x3 as significant states.

solution to homework 4

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