CDS-II (3)

8/18/2019 CDS-II (3)

1/48

Control System Design-II

.

Department of Electrical EngineeringPakistan Institute of Engineering and Applied Sciences

P.O. Nilore Islamabad Pakistan

url: www.pieas.edu.pk/aqayyumEmail: [email protected]

8/18/2019 CDS-II (3)

2/48

Solving TimeSolving TimeSolving TimeSolving Time----Invariant state equationInvariant state equationInvariant state equationInvariant state equation---- A reviewA reviewA reviewA review

( )

( ) 2 k 0 1 2 k

Consider

x Ax Bu, (1) x t ???

Let us consider a homogenous state equation

x ax (2)

Let

x t b b t b t b t (3)

= + =

=

= + + + + +

&

&

L L

Control System Design-II by Dr. A.Q. Khan2

1 2 k

k 1 2 k

1 2 k 0 1 2 k

1 0

2 2

2 1 0 2 0

3

3

x t t t

subsitituting (3) and (4) into (2)

b 2b t kb t a (b b t b t b t ) (5)

we have

b ab1

2b ab a b b a b , Similarly2

1b a

6

−

−

= + + +

+ + + = + + + + +

=

= = ⇒ =

=

& L

L L L

3 k

0 0 k 0

1 1b a b , b a b

3 k = =L L

8/18/2019 CDS-II (3)

3/48

( )

( )

( ) ( ) ( )

2

0

0

2 2 at

As

1

x t 1 at at b2

At t 0, equation (3) im plies

b x 0 , therefore

1x t 1 at a t x 0 e x 0

2

Now consider

x Ax (6)

= + + +

=

=

= + + + =

=

L

L

&


( ) 20 1 2Analogy with scalar, we havex t b b t b t= + + +

( )

( ) ( ) ( ) ( )

k

k

k 1

1 2 k

2 2 At

b t (7)

x t b 2b t kb t (8)

proceeding in similar fashion, we

1x t 1 At A t x 0 e x 0 = (t)x 0 (9)

2

where (t ) is state transition

−

+ +

= + + +

= + + + = φ

φ

L L

& L

L

matrix.

8/18/2019 CDS-II (3)

4/48

Referring to equation (9), the term is of particular

interest. It is referred to as matrix exponential and analogously, it can be

represented as

This matrix exponential for an converges absolutely forall finite time.

Ate

k k At

k 0

A te

k!

∞

=

= ∑

n n×


( )

( )

At At

A B t At Bt

A B t At Bt

At At

d e Ae

dx

e e e if AB BA

e e e if AB BA

e e = I

+

+

−

• =

• = =

• ≠ ≠

•

8/18/2019 CDS-II (3)

5/48

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1

11 At

C onsider equation (6) and taking Laplace T ransform

sX (s) - x (0) AX s X s sI A x (0)

x t sI A x 0 x t e x 0

or

−

−−

= ⇒ = −

= − ⇒ =

L

Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach


( )

where

t : n n m atrix and is the unique solution

: State transition m atr

φ ×

( )

( ) ( )-1 -At

ix. It has all the inform ation about

the free motion o f the sys tem define by x Ax

0 I

Also note t =e = t

=

φ =

φ φ −

&

8/18/2019 CDS-II (3)

6/48

Case-I: If be distinct eigenvalues of A, then

will contain exponentials Case-II: If the matrix A is diagonal with , then

1 2 n, ,λ λ λL ( )tφ

1 2 nt t t

e ,e , eλ λ λ

L

1 2 n, ,λ λ λL

( )

1

2

n

t

t

t

e 0 0

0 e 0t

0 0 e

λ

λ

λ

φ =

L

L

M O O M

L


Case-III: if there is multiplicity of eigenvalues, then will contain terms like in addition to

1 1 1 2 n, , , ,λ λ λ λ λL

( )tφ 1 1t t2te , t eλ λ

1 2 nt t te ,e , eλ λ λ

L

8/18/2019 CDS-II (3)

7/48

Properties of ( )tφ

( )( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

1 2 1 2

At

A(0)

1 1At At 1

A t t A t A t

1 2 1 2

n

As t e

0 e I

t e e t or t t

t t e e e t t

t nt

− −− −

+

φ =

• φ = =

• φ = = = φ − φ = φ −

• φ + = = = φ φ

• φ = φ


An Example

( ) ( ) ( )2 1 1 0 2 0 t t t t t t• φ − φ − = φ −

8/18/2019 CDS-II (3)

8/48

( ) ( )

( ) ( ){ }

( )

-1

1At

1

0 1For x Ax, where A=

2 3

Obtain t and t

Solution : As e t sI A

s 1Note that sI A2 s 3

s 3 11sI A

−

−

= − −

φ φ

= φ = −

−

− = +

+ − = −

&

-1L


( )

( )( )

Using partial fraction expansion

1t

s 1 s 2

φ =

+ +

-1L

( )-t 2 t -t 2 t t 2 t t 2 t

-t 2 t -2t t t 2 t 2t t

2 1 1 1

s 3 1 s 1 s 2 s 1 s 2

2 s 2 2 2 1s 1 s 2 s 2 s 1

2e e e e 2e e e e, t

2e 2e 2e e 2e 2e 2e e

− −

− −

− − + + + + +

= − − − + + + +

− − − −= ⇒ φ − =

− − − −

-1L

8/18/2019 CDS-II (3)

9/48

Solution to Homogenous State equations

n 1

n 1

n n

n p

Consider

x

ux Ax Bu (1)

A

B

Now

x Ax Bu

×

×

×

×

∈ ℜ

∈ ℜ= + − − − − − − − − −

∈ ℜ

∈ ℜ

− =

&

&

Convolution integral


( )

( ) ( ) ( )

At At At

t t

A At A

0 0

e x Ax e Bu multiply by e both sideIntegrating both sides

de x Ax d e Bu d e x

d

− − −

− τ − − τ

− =

− τ = τ τ ⇒ ττ∫ ∫

&

& ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

t t

A

0 0

t tA tAt A At

0 0

t

0

d e Bu d

e x t x 0 e Bu d x t e x 0 e Bu d

x t t x 0 t Bu d (2)

− τ

− τ− − τ

τ = τ τ

⇒ − = τ τ ⇒ = + τ τ

⇒ = φ + φ − τ τ τ − − − − − − − −

∫ ∫

∫ ∫

∫

8/18/2019 CDS-II (3)

10/48

Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1

Consider

x Ax Bu

The Laplace transform of the above system is

sX s x 0 AX s BU s sI A X s x 0 BU s

X s sI A x 0 BU s−

= +

− = + ⇒ − = + ⇒

= − +

&


( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )i

i

tA tAt

0

0 i

tA t t A t

t

x t e x 0 e Bu d

For initial time t t 0, then

x t e x 0 e Bu d

− τ

− −τ

⇒ = + τ τ

= ≠

= + τ τ

∫

∫

8/18/2019 CDS-II (3)

11/48

An ExampleAn ExampleAn ExampleAn Example

( ) ( )( )

Consider

x Ax Bu,

where

0 1 0 0 0 for t

8/18/2019 CDS-II (3)

12/48

Laplace Transform method:

Method of Diagonalization of A:

Computation of=?:

Some Further resultsAt

e

( ) ( ){ }1At

t e sI A −

φ = = −-1LDiscussed in

Previous slides

1if D P A P then−=


Caley-Hamilton Theorem and minimal polynomial theorem

Minimal polynomials of A involves distinct roots

Minimal polynomials of A involves repeated roots

A t D t 1

e P e P −=

8/18/2019 CDS-II (3)

13/48

Cayley-Hamilton Theorem Very useful theorem in linear algebra

after the names of two mathematicians: Arthur Cayley andWilliam Hamilton

It states, “Every square matrix over commutative ring (Real or”


For example if A is an square matrix withcharacteristics polynomial

---------- (1)

Then

-------- (2)

n n×

( )

n n 1

1 n 1 n

a a a 0−

−

φ λ = λ + λ + + λ + =L

( ) n n 11 n 1 nA A a A a A a I 0−

−φ = + + + + =L

8/18/2019 CDS-II (3)

14/48

While the CH-Theorem shows that , there may be a

polynomial having lower degree than , for which. The monic polynomial of lowest degree for which

is referred to as minimal ol nomial of A.

( ) 0φ λ =

( )ϕ λ ( )φ λ

( )A 0ϕ =

A 0=


The minimal polynomial plays important role in thecomputation of polynomials of n×n matrix.

CH-Theorem together with minimal polynomials helps in the

computation of function of matrix, e.g. exp(At), Cos(At) etc.

8/18/2019 CDS-II (3)

15/48

Computation of

Case-1: Minimal polynomials of A involves only

distinct roots

Let us assume that the minimal polynomial of A is m. Then

can be obtained by solving the following determinant equation

( ) AtA eφ =

( )Aφ

1

2

t2 m 1

1 1 1

t2 m 1

1 eλ−

λ−

λ λ λL

L


------------ (3)

Expanding this determinant equation about the last column, weobtain

----- (4)

m

2 2 2

tm 1 m 1 m 1

m m m

2 m 1 At

0

1 e

1 A A A e

λ− − −

−

=

λ λ λ

M M M M M M

M

L

( ) ( ) ( ) ( )At 2 m 10 1 2 m 1e t I t A t A t A −

−= α + α + α + + αL

Sylvester’s interpolation

formula

8/18/2019 CDS-II (3)

16/48

and

------ (5)

can be solved using the equations (5). Then can beobtained by substituting the computed

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1

2

m

t 2 m 10 1 1 2 1 m 1 1

t 2 m 1

0 1 2 2 2 m 1 2

t 2 m 1

0 1 m 2 m m 1 m

e t t t t

e t t t t

e t t t t

λ −−

λ −

−

λ −

−

= α + α λ + α λ + + α λ

= α + α λ + α λ + + α λ

= α + α λ + α λ + + α λ

L

L

M

L

'sα A te'sα


• If A is an n×n matrix and has distinct eigenvalues, then thenumber of the numbers of to be determined is m=n

• If A has multiple eigenvalues, but its minimal has only simpleroots then m < n.

( )k tα

8/18/2019 CDS-II (3)

17/48

An Example

1

2

1

2

Find

:

The eigenvalues are , .From the theory just discussed, we have

1 e

e

I A e

At

t

t

At

e

A

Solution

λ

λ

λ λ

λ

λ

= −

= = −

=

−

1 2

0 1

0 2

0 2

1 0


,

1

e

I A e

Expanding

t

At

−

−

− =

1 2

2

0 1

1 2 0

( )

( )

the determinant, we obtain

At t At t

t

At

t

e A I Ae e A I Ae

e e

e

− −

−

−

− + + − = ⇒ = + − −

=

2 2

2

2

12 2 0 22

11 1

2

0

8/18/2019 CDS-II (3)

18/48

( ) ( )

( ) ( )( ) ( )

1 2

:

,

where

Since , ; Then

At

t

t

Alternative

As

e t I t A

t t e t t e

λ

λ

α α

α α λα α λ

λ λ

= +

+ =+ =

= = −

1

2

0 1

0 1 1

0 1 2

0 2


( )

( ) ( ) ( ) ( )

( ) ( )

t t

t At t

t

t

t t e t e

Therefore

e e I e A

e

α

α α α− −

−−

−

=− = ⇒ = −

−

= + − =

0

2 2

0 1 1

22

2

11

2 12

11 11

1 22

0

8/18/2019 CDS-II (3)

19/48

Computation of

Case-2: Minimal polynomials of A involves multiple

roots

Let us assume that the minimal polynomial of A involves three

equal roots and the remaining aredistinct. Then using Sylvester's criterion, can be obtainedas

( ) AtA eφ =

( )1 2 3λ = λ = λ ( )4 5 m, ,λ λ λL

( )Aφ


(6)

( )( )

( )

1

1

1

m

2 tm 3

1 1

t2 m 2

1 1 1

t2 3 m 1

1 1 1 1

t2 3 m 1

m m m m

2 3 m 1 At

m 1 m 2 t0 0 1 3 e2 2

0 1 2 3 m 1 te

1 e 0

1 e

1 A A A A e

λ−

λ−

λ−

λ−

−

− −λ λ

λ λ − λ

λ λ λ λ =

λ λ λ λ

L

L

L

M M M M L M M

L

L

8/18/2019 CDS-II (3)

20/48

Equation (6) can be expanded in similar fashion as before

(7)and

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

1

1

1

2t m 3

2 3 1 m 1 1

t 2 m 1

1 2 1 3 1 m 1 1

t 2 m 1

m 1 m 2te t t t2 2

te t 2 t 3 t t

λ −

−

λ −

−

λ −

− −= α + α λ + + α λ

= α + α λ + α λ + + α λ

L

L

( ) ( ) ( ) ( )

At 2 m 1

0 1 2 m 1e t I t A t A t A

−

−= α + α + α + + αL


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

m

1 2 1 3 1 m 1 1

t 2 m 1

1 2 2 3 2 m 1 2

t 2 m 1

0 1 m 2 m m 1 m

e t t t t

e t t t t

−

λ −

−

λ −

−

= α + α λ + α λ + + α λ

= α + α λ + α λ + + α λ

L

M

L

Extend the above formulation to other cases; e.g; two or

more sets of multiple roots. Try at least two examples

8/18/2019 CDS-II (3)

21/48

( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )

Example: Find

:

The eigenvalues are , .From the theory just discussed, we have

and

At

At

t t

t

e

A

Solution

e t I t A t Ate t t te t t

e t t t e

λ

λ

λ λ λ

α α αα α λ α α

α α λ α λ

=

= = =

= + += + ⇒ = +

= + + ⇒

1

2

1 2 3

2

0 1 2

2

1 2 1 1 2

2

0 1 2 2 2

2 1 4

0 2 00 3 1

2 1

2 4

( ) ( ) ( )t

t t

t t t

λ

α α α= + +3

2

0 1 2

2

2 4


( ) ( )

( )

( ) ( ) ( )

Solving the above set of equations, we have

,

Finally

t t t t t t

t t t

t t t t

At

t e e te t e e te

t e e te

e e e te

e t I t A t A

α α

α

α α α

= − + = − + −

= − +

− + −= + + =

0 1 3 2 3 0 1 2

2 2 2 2

0 1

2 2

2

2 2 2

2

0 1 2

4 3 2 4 4 3

3

12 12 13 t t

t

t t t

e e

e

e e e

+ − +

2

2

2

4 4

0 0

0 3 3

8/18/2019 CDS-II (3)

22/48

Controllability and ObservabilityControllability and ObservabilityControllability and ObservabilityControllability and Observability

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( ) ( )

0 0

n n p

0 0

m

Consider a dynamical system described byx t A t x t B t u t , x t x

y t C t x t

where

x t : State, x t : State at initial time t , u t : Control input,

y t : output/measurement vector. The solution x t is g

= + =

=

∈ ℜ ∈ ℜ ∈ ℜ∈ ℜ

ɺ

( ) ( ) ( ) ( ) ( ) ( )t

0 0

iven as

x t = t,t x t + t, B u dφ φ τ τ τ τ


( )

( ) ( ) ( ) ( )

( ) ( ) ( )

0t

0

0 0 0 0 n

where t,t is the transition matrix and it is the solution of

t,t =A t t,t , t ,t =I

In this course, we shall study LTI systems of the following form

x t Ax t Bu t ,

φ

φ φ φ

= +

ɺ

ɺ

( )( ) ( )

( ) ( )0

0 0

A t-t

0

x t xy t Cx t

The transition m atrix is

t, t =e

==

φ

8/18/2019 CDS-II (3)

23/48

A continuous time LTI system is said to be

reachable at time if it is possible by means ofany unconstrained control input vector to transferthe system from initial state to any other state in afinite interval of time.

A continuous time LTI system is said to becontrollable (to the zero state) at time if it is

0

t

0t


input vector to transfer the system from initialstate to origin.

State Controllability

Output Controllability

( )0x t

A system is said to controllable at time if it is possible by means ofan unconstrained control vector to transfer from any initial state

to any other state in a finite interval of time

0t

( )0x t

8/18/2019 CDS-II (3)

24/48

( )( )

( )

T h e o r e m : A n e c e s s a ry a n d s u ff ic ie n t c o n d i t io n s

fo r s y s t e m re a c h a b ility [c o n tro lla b ility ] a t t im e is

th a t W ,t is p o s it iv e d e f in ite fo r s o m e t

w h e re W .,. is a n n × n G ra m m ia n m a tr ix d e fin e d b y

W , t

τ

τ τ

τ φ τ

≥

= ( ) ( ) ( ) ( )

t

T T , B B t, d τ

λ λ λ φ λ λ∫


Uniform Reachability/Uniform ControllabilityStrong notion

It refers to the fact that the property is independent of initial

time.

8/18/2019 CDS-II (3)

25/48

Theorem: For LTI systems given in before, the pair (A,B) issaid to be completely state controllable if and only if

n 1rank B AB A B n− = ⋯

( ) ( ) ( ) ( ) ( ) ( )t

0

0

Proof : The solution is

x t = t x 0 + t, B u d , t 0

Apply the definition of state controllability,

φ φ τ τ τ τ =∫


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1

1

1

t

1 1

0

t

0

At 2

0 1 2 n 1

.e., x t 0, t ere ore

0= t x 0 + t B u d

x 0 B u d

using e a I a A a A a− −

=

φ φ − τ τ τ τ ⇒

= − φ −τ τ τ τ

φ −τ = = τ + τ + τ + + + τ

∫

∫ ⋯

( )

n 1

n 1k

kk 0

A

a A

−

−

=

= τ

∑

8/18/2019 CDS-II (3)

26/48

----- (1)

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( )

1 1

1

t tn 1 n 1k k

k kk 0 k 00 0

t

k k

0

1

n 12k n 1

k

Now

x 0 a A B u d A B a u d

Let a u d

x 0 A B B AB A B

− −

= =

−−

= − τ τ τ τ = − τ τ τ

β = τ τ τ

β β = − β = − = Μβ

∑ ∑∫ ∫

∫

∑ ⋯

Controllability matrix


For system to be completely state controllable, equationmust holds. This requires that matrix M must have n linearly

independent row/column vectors. Rank (M)= n For , the dim of M is n×np. The Rank (M) should be n

for CSS.

k 0

n

=

β

pu ∈ ℜ

8/18/2019 CDS-II (3)

27/48

[ ] ( )

1 1

2 2

x x1 1 1u

x 0 1 x 0

1 1B AB Rank 1 2

0 0

Hence the system is not completely state cont

M

rollable

M

= + −

= = ⇒ = ≠

ɺ

ɺ

Example-I

Example-II


[ ] ( )

1 1

2 2

x x1 1 0u

x 0 1 x 1

0 1B AB Rank 21 1

The system is completely state controlla

M

ble

M

= + −

= = ⇒ = −

ɺ

ɺ

8/18/2019 CDS-II (3)

28/48

Controllability conditions in s-plane

Necessary and sufficient conditions: No cancellationof poles and zeros

The system losses its controllability duringcancellation of poles and zeros

An Example


This system is not CSC.

Check it in state space??

( )( )

( )( )( )

X s s 2.5U s s 2.5 s 1

+= + −

8/18/2019 CDS-II (3)

29/48

1 1 2

2 1 2

Controllab

3

2 1.5ility ?

4

An Example

???

x x x u

x x x u

= − + +

= − + +

&

&


8/18/2019 CDS-II (3)

30/48

1 1 2

2 1 2

1

A n E x a m p le

Appa ran t lyC o n tro llab ili ty ???? Y E S

S o lu t ion

32 4

:

1 .5 x x x u x x x u

y x

= − + += − + +

=

⇒

&

&


[ ]

C h eck the con tro llab i liy m a trix . H e re

3 1 1,

2 1 .5 41 1

4 4

A B

B A M B

− = =

−

= =

( ), R a n k 1 M =

8/18/2019 CDS-II (3)

31/48

1 1 2

2 1 2

An Example

Apparantly

Let us proceed as follows

32 1.5 4

Controllability ???? YES

Solution:

x x x u x x x u

= − + += − +

⇒

+

&

&


[ ]

Check the controllabiliy matrix. Here3 1 1

,2 1.5 4

1 1,

44

A B

B A B M

− = = −

= =

( )Rank 1 M =

Interesting

8/18/2019 CDS-II (3)

32/48

( ) ( ) ( ) ( )

( )

( )

( )

( )

( )

( ) ( )

1 1 1

2

2

F r o m t h e s t a t e - s p a c e m o d e l , w e h a v e

1 . 5 2 . 5 2 . 5

1 . 5 2 . 5 2 . 5

2 . 5 2 . 5

2 . 5 11 . 5 2 . 5

x x x u u

s s Y s s U s

Y s s s

U s s ss s

+ − = + →+ − = +

+ +

= = + −+ −

&& & &

L

Remarks


The system can not be controllable if it is not reducible to controllablecanonical form B should not be in the eigen-space of A.Alternatively B should not be in the null space of A

ConclusionActuator position is very important

8/18/2019 CDS-II (3)

33/48

1 1 2 1 2

2 2

3 3 1

1

A n o th e r E x a m p

4 2

2

l

3

e

x x x u u

x x

x x u

y x

= − + + +

= −

= − +

=

&

&

&


C o n tro lla b ility ????

8/18/2019 CDS-II (3)

34/48

Output Controllability

This can also be defined in the manner similar to statecontrollability.

The system is said to be completely output controllable if it ispossible to construct an unconstrained control vector u(t)that will transfer the output y(t) to origin from any initialcondition.

Let us consider


The output controllability matrix

should have a rank m

( ) ( ) ( ) ( )( ) ( )

0 0x t Ax t Bu t , x t xy t Cx t

= + ==

ɺ

n 1

OM CB CAB CA B−

= ⋯

8/18/2019 CDS-II (3)

35/48

Stabilizability For partially controllable systems, if the controllable modes

are stable and the unstable modes are controllable, thesystem is said to be stabilizable.

Remember: Controllability for continuous-time LTI system

implies Reachability. However for discrete-time LTI system this


does not hold.

8/18/2019 CDS-II (3)

36/48

Observability and Reconstructability Definition: The system is said to be observable, if every state x(t0 ) can

be determined from the observation y(t) over finite interval of timet0≤t ≤t1.

In completely observable system, every transition in stateaffects output (or every element of the output vector in caseof multiple output systems)


e cu ty, n pract ce, ar se w t t e use o state ee accontrol when all the system states are not available forfeedback.

The observability concept is useful in solving the problem of

reconstructing the un-measurable states of the system fromthe measurable states in shortest possible length of time.

8/18/2019 CDS-II (3)

37/48

Consider

Definition: Let y(t;t0,x,u) be the output of the above system to the

initial state x0. The present state x0 is unobservable if the (future)output y(t;t0,x,0)=0 for all t ≥ t0

Definition: The present state x0 is unconstructible if the (past) output

( ) ( ) ( )

( ) ( )0 0x t Ax t , x t x

y t Cx t

= =

=

ɺ


( )y ; ,x,σ τ σ τ = ∀ ≤0 0

The LTI system shown above or the pair (A,C) is

completely observable if and only if it is completely state

reconstructable

8/18/2019 CDS-II (3)

38/48

Complete Observability and ReconstructabilityConsider

( ) ( ) ( )

( ) ( )0 0x t Ax t , x t x

y t Cx t

= =

=

ɺ


The above system (the pair (C,A)) is said to be observable ifand only if the following observability matrix is full rank

( )n

OM C A C A C −∗ ∗ ∗ ∗ ∗ =

1

⋮ ⋮ ⋯ ⋮

Consider the system describe before, the output is

8/18/2019 CDS-II (3)

39/48

( ) ( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

At

n

At k k

k

n k n

k n

k

y , p

Recall that , therefore

y t Ce x

e t A

y t C t A x t Cx t CAx t CA x

C

CA

t t t

α

α α α α

α α α

−

=

−−

−=

−

=

=

= = + + +

=

∑

∑

1

0

11

0 1 1

0

0

0

0 0 0 ⋯

⋯

⋯ ⋯ OM α

=


n CA −

1

⋮

⋯

( )

nm n

n

For Complete observability, the observability matrix s

Observability matrix

Sometimes in literature

OM

OM C A

is used as obser

hould be Full Rank

vability matri

C A C

.

×

−∗ ∗ ∗ ∗ ∗

∈ ℜ

=

=

1

⋮ ⋮ ⋯ ⋮

x

An Example:

8/18/2019 CDS-II (3)

40/48

[ ]

( )

Here

OM C A C

x Ax Bu,

An

R

Example:

where

A ,

y Cx

an

B ,C

k OM ∗ ∗ ∗

= = = − −

= = ⇒ =

= + =

1 1 01 0

2

1

1

1 1

1

20

ɺ


Hence the system is completely state observable

[ ]

Now Find the obserbabiliy for the following ???

x x u, y x

= + = − − −

0 1 0 0

0 0 1 4 5 10

6 11 6 1

ɺ

8/18/2019 CDS-II (3)

41/48

Analogous to Controllability, the necessary and sufficient

condition for complete observability is that no cancellationoccur in the transfer function or transfer matrix.

If there is some cancellation of Poles/Zeros, the

observability will be lost.

0 1 0 0

An Example


[ ]

( ) ( )

x x u

y x

Rank OM C A C A C ∗ ∗ ∗ ∗

= + − − − =

= = = 2

0 0 1 06 11 6 1

4 5 1

2

ɺ

Rank of the observability matrix

8/18/2019 CDS-II (3)

42/48

Let us see what happens to Poles and zeros cancellation.

( ) ( ) ( ) ( )

( ) ( )( )

( )

x x x

Y s Y s s s X s s s

X s X s

U s

= + + ⇒

= + + = = + +

1

1 1 1

2

1

1

4 5

4 5 1 4

ɺ ɺɺConsider the output equation

Also Computing from stat

y

e equations, we have


( )( ) ( )( )( )

( )

( )

( )

( )

( )

( )

( )( )

( )( )( )( )

( )( )

X s

U s s s s s s s

Y s Y s X s s s

U s X s U s s s s

s

s s

= =+ + + + + +

+ +

= = + + ++

=+ +

1

3 2

1

1

1 1

6 11 6 1 2 3

1 4

1 2 3

4

2 3 Poles and Zero Cancellation

occurs

h h l d ll

8/18/2019 CDS-II (3)

43/48

Let us see what happens to Poles and zeros cancellation.

( ) ( ) ( ) ( )

( ) ( )( )

( )

x x x

Y s Y s s s X s s s

X s X s

U s

= + + ⇒

= + + = = + +

1

1 1 1

2

1

1

4 5

4 5 1 4

ɺ ɺɺConsider the output equation

Also Computing from stat

y

e equations, we have


( )( ) ( )( )( )

( )

( )

( )

( )

( )

( )

( )( )

( )( )( )( )

( )( )

X s

U s s s s s s s

Y s Y s X s s s

U s X s U s s s s

s

s s

= =+ + + + + +

+ +

= = + + ++

=+ +

1

3 2

1

1

1 1

6 11 6 1 2 3

1 4

1 2 3

4

2 3 Poles and Zero Cancellation

occurs

A CSC l b CSO d

8/18/2019 CDS-II (3)

44/48

A CSC system may not necessarily be CSO and vice versa

[ ]

Illusrativde Example0 1 0

0.4 1.3 1

0.8 1

0 1

x x u

y x

= + − −

=

&


( )

1 1.30.8

Show if there is Poles Zero Cancellation or not??????

0.4; Rank 1

1 0.5OM OM NOT CSO

− −

= = ⇒ −

8/18/2019 CDS-II (3)

45/48

T r y th i s o n e

2 0 0

0 2 00 3 1

x x

=

&


8/18/2019 CDS-II (3)

46/48

Duality properties

The Linear time-invariant system

x Ax Bu

y Cx

= +=

ɺ

Symmetry properties existsObservability (Reconstructability) Controllability (Reachability)


* *

T T T

1

s t e ua o t e sy

z A z Bv

y C x

stem an v ce versa

where A A ,B C ,C B =

= +

=

= =1

1

1

1

1ɺ

The principle of duality is useful in that it allows the statereconstruction problem can be casted as dual control problem and

vice versa.

8/18/2019 CDS-II (3)

47/48

Controllability and Observability indicesControllability and Observability indicesControllability and Observability indicesControllability and Observability indices

−

− = n 1

1

Consi

der the controlability matrix

If there exists a least integer q such that

M B AB A B⋯

Controllability index


= µ

Then this least integer refferred to as is calle controlability indexd

⋯

8/18/2019 CDS-II (3)

48/48

( ) − = *

n 1* * * * *

Consider the observability matrix

If there exists a least integer v, such that

OM C C A C A⋯

Observability index

Controllability and Observability indicesControllability and Observability indicesControllability and Observability indicesControllability and Observability indices


( ) −

= ≤ ≤µ

n 1* * * * *

Then this least integer refferred to

C C

as

Rank( ) n

is call

ed

1 v n

obs

A C A

erva

⋯

bility index

•Very powerful concept• Quite often used in minimum order observer design

Documents

CDS-II (3)