Modern Control Tutorial 7 Presentation 390 (1) 408

Modern Control

GUC Faculty of Engineering and Material Science

Department of Mechatronics

Tutorial #7

Controllability and Observability

Modern Control MCTR 702

Dr. Ayman Ali El-Badawy

Tutorial #7 Controllability and Observability





Controllability

In other words, the controllability check answers the question “can the input

affect the state variables, consequently control them?” If the answer is yes for

all state variables, then the system is completely controllable. But, If there is

one of the state equations such that the state derivative depends on neither the

input nor another state variable that is dependent on the input, then the system

is NOT completely controllable.





Controllability Checks: 1- Modal Check

a. If the system is in the diagonal form, matrix A is diagonal, and the B matrix has no

zero rows, then the system is completely controllable.

b. If the system has repeated Eigen values, its A matrix can not be fully diagonalized

(it has Jordan blocks).

u

x

x

x

x

x

x

x

x

0

1

1

0

2000

1200

0010

0011

4

3

2

1

4

3

2

1

Example: Consider the following system, in the first Jordan block, the non-

zero entry in the matrix B corresponds to the bottom row controllable.

But, in the second Jordan block, the non-zero entry in the matrix B

corresponds to the upper row uncontrollable.









Controllability Checks: 2- Controllable Canonical Form Check





Controllability Checks: 3- General Check

If the controllability matrix P has full row rank (n), then the system is

completely controllable.

Remember: if P is a square matrix, we can check on the full rank by checking

the non zero determinant for P.





Problem 1:

Check the controllability of the following systems:

The system is in the diagonal form (the matrix A is diagonal), and the matrix B has

no zero rows. Thus the system is completely controllable.

The system is in the diagonal form (the matrix A is diagonal), and the matrix B has

a zero row. Thus the system is NOT completely controllable.

Performing the general check:

0

21

21

2

2,

P

PABABBP

the system is NOT completely controllable.





Problem 2:

Check the controllability of the following system:

If we look at the system with first input only, we will find that this input affects only

X2 which in turn does not affect any of the other state variables, i.e., the state

variables X1 and X3 are uncontrollable. Thus the system is NOT completely

controllable with respect to the first input only.

Considering the second input only, this input affects only X1 which in turn affects

X3 but not X2, i.e., the state variable X2 is uncontrollable. Thus the system is NOT

completely controllable with respect to the second input only.

Now each input fails individually to control the three state variables. And the

question is “ Can the two inputs together control the whole system?”

The controllability check can answer this question.

00

01

10

,

606

010

100

BA

Solution:


Department of Mechatronics Modern Control MCTR 702


360

01

60

60

01

00

00

01

10

360

01

60

)(,

60

01

002

2

P

ABABAAB

BAABBP

Notice the 1st, 2nd, and 4th columns represent the three standard basis vectors

(echelon form). Thus the P matrix has full rank and the system is completely

controllable using the two inputs together.

00

01

10

,

606

010

100

BA





Observability

In other words, the observability check answers the question “Do the state

variables affect the output? Consequently they can be extracted from the

output” If the answer is yes, for all state variables, then the system is

completely observable. But, If there is one of the state variables that affects

neither the output nor another state variable that affects the output, then the

system is NOT completely observable.





Observability Checks: 1- Modal Check

a. If the system is in the diagonal form, matrix A is diagonal, and the C matrix has no

zero columns, then the system is completely observable.

b. If the system has repeated Eigen values, its A matrix can not be fully diagonalized

(it has Jordan blocks).

A system in Jordan form is completely observable iff there is only one Jordan block for

Each Eigen value, and the column of the matrix corresponding to the

Variable of the first column of each Jordan block is non zero (the most left entry of the

block).

1001y

arbit.

2000

1200

0010

0011

4

3

2

1

4

3

2

1

u

x

x

x

x

x

x

x

x

Example: Consider the following system, in the first Jordan block, the non-

zero entry in the matrix C corresponds to the first column observable.

But in the second Jordan block, the non-zero entry in the matrix C corresponds

to the second column unobservable.

CMC ˆ









Observability Checks: 2- Observable Canonical Form Check





Observability Checks: 3- General Check

If the observability matrix Q has full row rank (n), then the system is

completely observable.

Remember: if Q is a square matrix, we can check the rank by checking the

non zero determinant for Q.





Problem 3:

Check the observability of the following systems:

The system is in the diagonal form (the matrix A is diagonal), and the matrix C has

no zero columns. Thus the system is completely observable.

The system is in the diagonal form (the matrix A is diagonal), and the matrix C has

a zero column. Thus the system is NOT completely observable.

Performing the general check:

11

1111, QCA

CA

CQ

the system is NOT completely observable.

11,11

02 c.

CA





Similarity Transformations and Canonical Forms

The transformation matrix T that transforms any system into the controllable

canonical form is given by: where: 1 CCFsysCCF PPT

2. Observable Canonical Form:

The transformation matrix T that transforms any system into the observable

canonical form is given by: where: 11 sysOCFOCF QQT

11 CCFOCF PQ

1. Controllable Canonical Form:

DDCTCBTBATTA ˆ,ˆ,ˆ,ˆ 11





3. Diagonal Canonical Form (Jordan):

a. The transformation matrix T that transforms any system into the diagonal

canonical form is the matrix constituted from the Eigen vectors

b. The transformation matrix that transforms the system, from its controllable

canonical form to its Jordan form, is the transposed Vandermonde matrix.





4. Duality Principle:

a. Consider the following two systems:

These two systems are called dual to each other;

i.e., if one system is controllable, the other is observable and vice versa.

b. The controllable canonical form and the observable canonical form are dual forms

to each other, and then:

CCFOCFTCCFOCF

TCCFOCF

TCCFOCF DDBCCBAA ,,,

DuCxy

BuAxx

DuzBy

uCzAz

T

TT

sys dualsyssys dualsys , PQQP





Problem 5:

For the following system, determine :

1. its controllable canonical form.

2. its observable canonical form.

3. its Jordan form, from the controllable from.

111,

0

1

1

,

606

010

100

CBA





Solution:

1. Controllable Canonical Form:

36

1

6

6

1

0

0

1

1

36

1

6

)(,

6

1

0

2

2

BAABBP

ABABAAB

Getting the Coefficients:

,6,0,5,1

65)66(65

606

010

10

0123

232

aaaa

AI





256,

1

0

0

,

5

1

0

0

0

1

6

0

0

0

1

1

6

6

5

6

6

6

0

0

1

0

1

5

1

5

0

36

1

6

6

1

0

0

1

1

0

0

1

0

1

1

11

2

2

1

CTCBTBATTA

a

a

a

PT

CCFCCFCCF

CCF

Then the transformation matrix is given by:

2. Observable Canonical Form:

We can use the principle of duality

CCFOCFTCCFOCF

TCCFOCF

TCCFOCF DDBCCBAA ,,,





100,

2

5

6

,

5

0

6

1

0

0

0

1

0

OCFOCFOCF CBA

3. Diagonal Canonical Form:

Solve the characteristic equation to get the Eigen values in order to construct the

Vandermonde matrix:

13923.226077.1

17321.42679.1

111

111

3923.227321.41

6077.12679.11

1,7321.4,2679.1

065 23

TVTV





131244.278756.2

0769.0

0504.0

1273.0

100

07321.40

002679.1

1

1

TCC

BTB

TATA

CCFJordan

CCFJordan

CCFJordan

Documents

Modern Control Tutorial 7 Presentation 390 (1) 408