Modern Control - German University in Cairoems.guc.edu.eg/Download.ashx?id=409&file=tut8_409.pdf · Modern Control MCTR 702 Dr. Ayman Ali El-Badawy Kalman Decomposition (controllability):

Modern Control

GUC Faculty of Engineering and Material Science

Department of Mechatronics

Tutorial #8

Controllability and Observability II

and Minimal Realization

Modern Control MCTR 702

Dr. Ayman Ali El-Badawy





Kalman Decomposition (controllability):

When the system is NOT completely controllable, we may want to decompose it to know

which modes are uncontrollable. And then see if the Eigen values of these modes are

already stable, then the system is stabilizable. This is by constructing a similarity transformation matrix from any rank(P) linearly

independent columns (basis of P) appended to them n-rank(P) other independent columns

such that the resulting matrix is non singular.





Kalman Decomposition (observability):

When the system is NOT completely observable, we may want to decompose it to know

which modes are unobservable. And then see if the Eigen values of these modes are already

stable, then the system is Detectable.





For the system Given in Problem two, consider the second input only and find

which modes are uncontrollable. Is the system stabilizable?

Solution:

0

3660

000

601

0

0

1

,

606

010

100

2

PBAABBP

BA

Problem 1:

|P| = 0 because the 3rd column is linearly dependent on the 1st and 2nd columns

(-6x1st -6x2nd). If it’s not noticeable, we perform row operations to find the basis

vectors of the P matrix. The decomposition is by constructing a transformation

matrix constituting of these basis vectors along with added independent vectors.


Department of Mechatronics Modern Control MCTR 702


u

w

w

w

w

w

w

BVBAVVA

V

0

0

1

100

061

060

0

0

1

ˆ,

100

061

060

ˆ

060

100

001

3

2

1

3

2

1

11

I have only one uncontrollable mode (W3). Its Eigen value is 1 (unstable). Then

this system is NOT stabilizable.





State space realization

The transfer function associated with this realization is given by;

The question that arises now is “can we reduce this realization ?”.i.e. is there

any state that could be cancelled?

Minimal Realization





Problem 1:

Compute a minimal realization for each of the following state equations.

Solution :

)1)(2(

1 + s

23s

1 + s

)(

)()(

1

0

32

10

0

0]11[)(

2

1

ss

ssU

sXsH

s

ssH

Due to pole-zero cancellation,

it is clear that the new H(s) is

given as





]1[],1[],2[

2

2

2

)2(

1

)(

)()(

111

1

CBA

uxx

uxx

uxx

L

ssU

sXsH

One state is cancelled, and the

irreducible realization (minimal

realization) is





Important Note:

If we apply controllability and observability Kalman decomposition to a

system (in either order)(the second being applied to separate SUB-

systems only), we get

uB

B

X

X

X

X

AA

A

AAAA

AA

X

X

X

X

oc

co

oc

oc

oc

co

oc

oc

oc

co

oc

oc

oc

co

0

0

00

000

00

43

242321

13

Du

X

X

X

X

CCy

oc

oc

oc

co

occo

]00[





The most systematic way, to get the minimal (irreducible) realization, is to perform

controllability and observability Kalman decomposition (in either order), the second being

applied to separate SUB-systems only, to get Aco , Bco, and Cco

Solution :

By first performing (controllability) Kalman decomposition, we find the controllability matrix

as

3271

710

100

][ 2BAABBP

Which is full row rank. This means that the

system is completely controllable. It is clear

that the system is already in the controllable

as it is given in CCF





Now, we apply observability Kalman decomposition

41715

130

013

2CA

CA

C

Q

rank = 2

The 3rd row depends on the first 2 rows

100

0.33-0.330

0.110.11-0.33

100

130

0131

U

u

z

z

z

z

z

z

1

0

0

100

0.33-0.330

0.110.11-0.33

100

0.33-0.330

0.110.11-0.33

71715

100

010

100

0.33-0.330

0.110.11-0.331

3

2

1

1

3

2

1





001ˆ C

u

z

z

z

z

z

z

1

1

0

3-45

045

010

3

2

1

3

2

1

coA coB

coC

01coC

Then, the minimal realization is given by the following matrices

1

0coB

45

10coA





And the transfer function is given by

54s

1)()(

2

1

sBAsICsH cococo

In this problem, it is clear that one state is cancelled due to pole-zero cancellation

Documents

Modern Control - German University in Cairoems.guc.edu.eg/Download.ashx?id=409&file=tut8_409.pdf · Modern Control MCTR 702 Dr. Ayman Ali El-Badawy Kalman Decomposition (controllability):