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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]

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Preliminaries of MCT ctd….. The central concept on which the modern control theory is based,

is STATE.

State: The state of dynamic system is the smallest set of variables(referred to as state variables) such that the knowledge of thesevariables at together with the knowledge of input for0t t= 0t t>

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

completely determines the behavior of the process; that is,

The concept of state is not limited to physical/engineeringsystems but also biological systems, economic systems, socialsystems etc.

Deter mines

0 t t 0 0t t t tx(t) , and u(t) x(t) →

>= >

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State variables: The smallest set of variables that determinesthe behavior of the system

State vector:Vector of state variables. For ‘n’ state variables, astate vector of n-components to be constructed. Let ‘x’ be a

Preliminaries of MCT ctd…..


State space: It is an ‘n’ dimensional space whose coordinatesare state variables.

[ ]1 2 3 4x x x x . . . x=

1 2 3 nx ,x , x , , xK

Any state can be represented by a point in state space

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State space equation: It is a first order differential equation(or set of first order differential equations).

It generally involves three variables of interest.

State variables



Output variables

Let us consider

x x u,y x

= +=

&

State variables

Output equation

Input variablesState equation

Outputvariables

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There exists many kinds of state space representation.However the no. of states (state variables) are the same.

A higher order differential equation can be represented inthe form of set of first order differential equations. Hence



Let us consider a 2nd order dynamic system

y by cy u+ + =&& &

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The preceding example shows


1

1 2

x yx y x

== =& &

& && &


Note that the states and are the outputs of integrator.

Remember! The state space model for a system is a set of1st order differential equations.

2 2 1x y u y cy u x cx= = − − = − −

1x

2x

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Let us consider


1 1 1 2 n 1 2 p

2 2 1 2 n 1 2 p

3 3 1 2 n 1 2 p

x f (x , x , , x ,u , u , , u , t)

x f (x , x , , x , u ,u , , u , t)

x f (x , x , , x ,u , u , , u , t)

=

= =

& L L

& L L

& L L


n n 1 2 n 1 2 px f (x , x , , x , u ,u , ,u , t)

=

MM

& L L

a e equa ons

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1 1 1 2 n 1 2 p

2 2 1 2 n 1 2 p

3 3 1 2 n 1 2 p

y h (x , x , , x , u , u , , u , t)

y h (x , x , , x ,u ,u , , u , t)

y h (x , x , , x , u ,u , , u , t)

=

= =

L L

L L

L L



m m 1 2 n 1 2 p

y h (x , x , , x ,u ,u , , u , t)

=

M

M

L L

u pu equa on

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Combining the two


11 1

22 2

ux yux y

x(t) , y(t) , u(t)

= = = MM M

State Vector Output Vector Input Vector


pn m

1 1 2 n 1 2 p

2 1 2 n 1 2 p

n 1 2

ux yf (x ,x , ,x ,u ,u , ,u ,t)

f (x ,x , ,x ,u ,u , ,u ,t)

f(x)

f (x ,x

=

L L

L L

MM

1 1 2 n 1 2 p

2 1 2 n 1 2 p

n 1 2 p m 1 2 n 1 2 p

h (x ,x , ,x ,u ,u , ,u ,t)

h (x ,x , ,x ,u ,u , ,u ,t)

, h(x) =

, ,x ,u ,u , ,u ,t h (x ,x , ,x ,u ,u , ,u ,t)

L L

L L

MM

L L L L

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The resultant state space model is


x f (x , u , t)

y h (x , u , t)==

&


It may represents Linear System

Nonlinear System

Linear Time varying system

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The LTV system is represented by


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

x t A t x t B t u t

y t C t x t D t u t

= +

= +

&


The LTI system is represented by

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

x t Ax t Bu t

y t Cx t Du t

= +

= +

&This is the focusof this course

11

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Transfer function V/S State Space Two important Questions

Given a Transfer function G(s) = Y(s)/U(s), how to obtain thestate space representation?

Given the state space representation G: (A,B,C,D), how toobtain Transfer function G(s) = Y(s)/U(s)?


Let us consider a system G with the following SS-form

( ) ( ) ( )

( ) ( ) ( )

: x t Ax t Bu t

G

y t Cx t Du t

= +

= +

&

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Laplace Transform of the state equation is

( ) ( )

( ) ( ) ( )1

( ) (0)

(0)

0

sX s x AX s BU s

sI A X s x BU s

X s sI A x BU s−

− = + ⇒

− = + ⇒

= − +

State Space Transfer function


Laplace Transform of the output equation is

Substituting X(s) in the above equation

Note that for T/F, x(0) = 0, then

( ) ( ) ( )Y s CX s DU s= +

( ) ( ) ( )( )1 (0) ( )Y s C sI A x BU s DU s−= − + +

( )( )

1

( )

Y sC sI A B D

U s

−= − +

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Note that A,B,C,D are constant matrices

As

characteristics polynomial of G(s)

Correlation between transfer function

and state space model Ctd……

( )1 ( )

( ) adj sI A

G s C sI A B D C B DsI A

− −= − + = +

−


Poles of G(s) are the eigenvalues of A. Hence G(s) can be written as

where Q(s) is a polynomial in s. If A,B,C,D are time varying or uncertain, then it is difficult to derive

equivalent transfer function. For a nonlinear system, transfer function representation is also

difficult to obtain.

Q(s)G(s)

sI A=

−

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Some important facts: Recall that the characteristic equation is obtained by equating

the denominator of the transfer function to ZERO.

In SS-representation, (sI-A) is very important. Thecharacteristic equation in this case is |sI-A| = 0.


- .some properties

If the coefficients of A are real, its eigenvalues are either real orin complex conjugate pairs

The eig(A) = eig(transpose(A))

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An Example Example 2: Obtain the transfer function from the state-

space model derived in Example 1.

Solution: Given that

1 1 0

A , B = , C 1 0 , D 0k b 1

= = =


[ ]

[ ]

1

1

22

m m mG (s ) C (s I A ) B D

s 1 0

1 0 0k b 1s

m m m

b0s 1

1 1m = 1 0 1

b k k m s b s k s s s m

m m m

−

−

− −

= − +

− = + +

+

= + + + + −

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State-representation for MIMO system Consider a multiple input multiple output (MIMO) system

with ‘p’ inputs and ‘m’ outputs; i.e.;

11

22

uy

uy u

= =


The transfer function matrix is given aspm

uy

( ) ( )

1 m pY(s)

G s C sI A B DU(s)

− ×

= = − + ∈ ℜ

Note that the Eigenvalues of A does not necessarily imply

the closed loop poles

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Transfer function State SpaceTransfer function State SpaceTransfer function State SpaceTransfer function State Space

Transform T/F into n-th order differential equation

Get the SS- representation then.

Consider an n-th order differential equation


Let1

n n

n y a y a y u

−

+ + + =LLLLLL

1

2 1

3 2

1n

n

x y

x y x

x y x

x y

−

=

= =

= =

=

& &

&& &

M

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1 2

2 3

1 1 2 1

1 1

2 2

0 1 0 0 0

0 0 1 0 0

n n n n

x x

x x

x u a x a x a x

x x

x xu

−

=

=

= − − −

= +

&

&M

& L

& L

& L


Let the output of this system be1 1 1n nn na a a x x−

− − − & L

[ ]

1

2

1 1 0 0

n

x

x y x

x

= =

LM

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In compact form

where

0 1 0 0 0

x Ax Bu

y Cx Du

= +

= +

&

L

L


[ ]

1 1

,

1

1 0 0 , 0

n n

A B

a a a

C D

−

= =

− − −

= =

M O O M M

L

L

20

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Case: When derivatives are present in

ControlConsider an n-th order differential equation with in n

derivatives in control channel1 1 .

1 0 1 1

Let

n n n n

n n n y a y a y b u b u b u b u

− −

−+ + + = + + + +L L


1 0

2 0 1 1 1

3 0 1 2 2 2

1 1 2

0 1 2 1 1 1

n n n

n n n n n

x y ß u

x y ß u ß u x ß u

x y ß u ß u ß u x ß u

x y ß u ß u ß u ß u x ß u− − −

− − − −

= −= − − = −

= − − − = −

= − − − − − = −

& & &

&& && & &

M

& &L21

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0 1

0 0

1 1 1 0

2 2 1 1 2 0

1 1 1 1 0

w h e r e , , a r e d e t e r m i n e d f r o m

n

n n n n n

ß ß ß

ß b

ß b a ß

ß b a ß a ß

ß b a ß a ß a ß− −

=

= −

= − −

= − − − −

L

M

L L


W i t h t h i s c h o i c e o f s t a t e v a r i a b le s , t h ee x i s t a n c e a n d u n i q u n e s s o f s ta t e e q u a t i o n

i s g u a r a n t e e d . L e t u s c

1 2 1

2 3 2

o n s t r u c t

s t a t e s p a c e e q u a t i o n

x x ß u

x x ß u

= +

= +

&

&

M22

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1 1

1 1 2 1

n n n

n n n n n

x x ß u

x a x a x a x ß u− −

−

= +

= − − + +

M

&

& L

1 1 1

2 2 2

Combining the above set of equations

0 1 0 00 0 1 0

x x x x

u

β β

= +

& L& L

M O O M MM M


[ ]

1 1

1

2

1 01 0 0

n n nn n

n

a a a x x

x

x

y x ß u

x

β −− − −

= = +

& L

L M

Note that

•The derivatives of the control

only affect B matrix

•A and C matrices are unchanged

•Matrix D is non-zero in this case

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Decomposition of Transfer functionConsider the following transfer function

( )( )

n n n n n n n n

n

Y s b s b s b s b s b U s s a s a s a

− − −

− −+ + + + +=

+ + + +

1 2 3

0 1 2 3

1 2

1 2

⋯

⋯

Control System Design-II by Dr. A.Q. Khan24

Steps:1. Express the transfer function in –ve power of s. Multiply

numerator and denominator by n s −

( )( )

n n n

n

Y s b b s b s b s b s U s a s a s a s

− − − −

− − −+ + + + +=

+ + + +

1 2 3

0 1 2 3

1 2

1 21

⋯

⋯

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2. Multiply numerator and denominator by a dummy variableX(s)

Decomposition of Transfer function ctd…

( )( )

( )( )

n

n n

n

Y s X s b b s b s b s b s

U s a s a s a s X s

− − − −

− − −

+ + + + +=

+ + + +

1 2 3

0 1 2 3

1 2

1 21

⋯

⋯


3. Rewrite numerator and denominator as

( ) ( )

( )

( ) ( ) ( )

n

n

n

n

Y s b b s b s b s X s

U s a s a s a s X s

− − −

− − −

= + + + +

= + + + +

1 2

0 1 2

1 2

1 21

⋯

⋯

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4. Construct the state diagram in the above two equations.Re-arrange the last equation as

5. Substituting as( ) ( ) ( ) ( )n n X s U s a s a s a s X s − − −= − + + +1 21 2 ⋯



( ) ( )( ) ( )

( )

,

,

n n

n

n n

n

n n n n n

n n n n n

s X s x s X s x x

s X s x x s X s x x

X s x x a x a x a x a x u

y b x b x b x b x b x

− − +

− + −−

− −

− −

→ → =← ← → →= =← ←

→← = − − − − − +

= + + + + +

1

1 2 1

2 1

3 2 1

1 1 2 2 3 1

1 1 2 2 3 1 0

ℓ ℓ

ℓ

ɺ

ɺ ɺ⋯

ɺɺ ⋯

ɺ⋯

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substituting

into n n n n n

n n n n n

x a x a x a x a x u

y b x b x b x b x b x

− −

− −

= − − − − − +

= + + + + +

1 1 2 2 3 1

1 1 2 2 3 1 0

ɺ ⋯

ɺ⋯



( )( ) ( )

( )

n n n n

n n n n

n n n n

n

As

y b x b x b x b x

b a x a x a x a x u b b a x b b a x

b b a x b u

− −

− −

− −

= + + + +

+ − − − − − += − + − + +

− +

1 1 2 2 3 1

0 1 1 2 2 3 1

0 1 1 0 1 2

1 0 1 0

⋯

⋯

⋯

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where

x Ax Bu

y Cx Du

= +

= +

ɺ


Control System Design-II by Dr. A.Q. Khan28 ( ) ( ) ( )

, ,

n n n

n n n n

A B D b

a a a a

C b b a b b a b b a

− −

− −

= = =

− − − −

= − − −

0

1 2 1

0 1 0 1 1 0 1

0 1 0 0 00 0 1 0 0

0 0 0 0 0

1

⋯⋯

⋯

⋮ ⋮ ⋮ ⋯ ⋮ ⋮

⋯

⋯

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Matlab functions ss2tf: Transforms state-space to transfer function

[num,den] = ss2tf(A,B,C,D)

G(s) = num/den = Y(s)/U(s) For multi-input systems, the above function still works; that is, [num,den] = ss2tf(A,B,C,D,iu)


iu: ith input and should be an integer value, that is, 1,2,…. For example [num,den] = ss2tf(A,B,C,D,1) returns transfer

function from first input.

tf2ss: Transforms the transfer function to state-space [A,B,C,D]=tf2ss(num,den) This function is only for SISO

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State space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted Pendulum

Control SystemControl SystemControl SystemControl System


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State space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted Pendulum

Control SystemControl SystemControl SystemControl System----Free body diagramFree body diagramFree body diagramFree body diagram


ɺɺ

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( )

( )

( )

M m x m u

I m m x mg or

M M m g u

Mx u mg

Let

θ

θ θ

θ θ

θ

+ + =

+ + =

= + −

= −

2

ɺɺɺɺ ℓ

ɺɺ ɺɺℓ ℓ ℓ

ɺɺℓ

ɺɺ

ɺ ɺ


, , , ,

x x

M m x gx u

M M

x x

m x gx u

M M

= = = =

=

+= −

=

= − +

1 2 3 4

1 2

2 1

3 4

4 1

1

1

ɺ

ɺℓ ℓ

ɺ

ɺ

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x x M m g

x x M M

u x x

x x m g

+ −

= + −

1 1

2 2

3 3

4 4

0 1 0 0 0

10 0 0

0 0 0 1 0

10 0 0

ɺ

ɺℓ ℓ

ɺ

ɺ


x

x y

x

x

=

1

2

3

4

1 0 0 0

0 0 1 0

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Canonical forms 4 canonical forms are important

Controllable canonical form

Observable canonical forms

Diagonal canonical form


Given a nth order linear differential equation

and the transfer function

1 1 .

1 0 1 1

n n n n

n n n y a y a y b u b u b u b u− −

−+ + + = + + + +L L

( )

1

0 1 1

1

1 1

n n

n n

n n

n n

b s b s b s bG s

s a s a s a

−

−−

−

+ + + +=

+ + + +

L

L34

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Controllable Canonical forms The controllable canonical form for the system shown is

{

1 1

2 2

1 1

1 2 1

0 1 0 0 0

0 0 1 0 0

0 0 1 0

1

n n

n n nn n

x x

x x

u x x

a a a a x x

− −

− −

= + − − − −

& L

& L

M M O O O M M M& L L

& L144444 444443


[ ]0 1 1 0 1 1 0

cc

c

B A

n n n n

C

y b a b b a b b a b− −= − − −L L1444444442444444 3 {

1

2

0

1 c D

n

n

x

x

b u

x x

−

+

M44

•Very important in design of state-feedback

controller

•Pole placement35

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Observable Canonical forms The observable canonical form for the system shown is1 1 0

2 2 1 1 01

1 1 2 2 02

0 0 0

1 0 0

0 0

n nn

n nn

n n

x x b a ba

x x b a ba

u

x x b a ba

− −−

− −

−− −−

= +

−−

& L

& L

M M MM O O O M

& L L

&


[ ]

1 1 01

0 0 1

o b o b

o b

n n

A B

C

aa x x

y

−−

=

L

1 4 4 442 4 4 4 43 1 442 4 43

L L1 4 442

1

2

0

1n

n

x

x

b u

x

x

−

+

M4 4 43

•Very important in design of

state observers

•

Pole placement•Note also that

,

,

T T

ob c ob c

T

ob c ob c

A A B C

C B D D

= =

= =

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Diagonal canonical form (DCF) Consider a transfer function having distinct poles, that is,

The diagonal canonical form is given as

( )

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

1

0 1 1 1 10

1 2 1 1

n n

n n n

n n

Y s b s b s b s b cc cb

U s s p s p s p s p s p s p

−

−+ + + += = + + + ++ + + + + +

LL

L

1 11 0 0 0 1

0 0 0 1

x x p

x x

− −

& L

& L


{

[ ]

1 1

1

2

1 2 0

1

0 0 0 1

0 0 0 1

d d

d

n n

nn n

B A

n

C n

n

u x x

p x x

x

x

y c c c b u

x

x

− −

−

= +

−

= +

M M O O O M M M& L L

& L1 4 4 4 442 4 4 4 4 43

L L M1 4 4 4 2 4 4 4 3

37

J d C i l f

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Jordan Canonical form

Suppose a system transfer function has multiple poles ofmultiplicity m, the proceeding diagonal can be modified. Thismodified representation is known as Jordan Canonical form.

( )( ) ( ) ( ) ( )

1

0 1 1

3

1 2

31 2 40 3 2

n n

n n

n

n

Y s b s b s b s bU s s p s p s p

c cc c cb

−

−+ + + +=+ + +

= + + + + + +

LL

L


1 21 1 ns p s p s ps p s p+ +

[ ]

1

1

1

4

5

6

1 2 2 n 0

p 1 0 0 0 0 0

0 p 1 0 0 0 0

0 0 p 0 0 0 0x x u

0 0 0 p 0 0 1

0 0 0 0 p 0

0 0 0 0 0 p 1

y c c c c x b u

− − −

= + −

−

−

= +

&

M

L L

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An Example( )

( )1 2

2 2

1 2

0 1 2 0 1 2

Y s b s bs 3

U s s 3s 2 s a s a

Obtain CCF, OCF, DCF

Here b 0, b 1, b 3; a 1,a 3,a 2

0 1 0CCF : A , B ,C 3 1 , D 0

++= =

+ + + +

= = = = = =

= = = =


[ ]T T T0 c 0 c 0 c 0

0 2 3OCF : A A , B C ,C 0 1 B , D 0

1 3 1

− −

− = = = = = = = −

[ ]D D D D1 0 1

DCF : A , B ,C 2 1 , D 00 2 1

− = = = − = −

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Similarity TransformationSimilarity TransformationSimilarity TransformationSimilarity Transformation

State-space representation is not unique Often desirable to work with some especial form of state-

space models


A transformation exists from one canonical form to anothercanonical form

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x Ax Bu

y Cx Du

= +

= +

ɺConsider

Letx Pz =

Where is non-sin ular matrix.n n ×


z P x −= 1

The transformed equation is

z P x P Ax P Bu

P APz P Bu

CPz Du η

− − −

− −

= = +

= +

= +

1 1 1

1 1

ɺ ɺ

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z Az Bu

Cz Du η

= +

= +

ɺ

where, ,A P AP B PB C CP D D −= = = =1


Invariance Properties:• Characteristic equation• Eigenvalues•

Eigenvectors• Transfer function/Transfer function matrix (MIMO Systems)

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Controllability canonical formConsider the dynamic system

to get the CCF, we choose

where

x Ax Bu

y Cx Du

= +

= +

ɺ

P MW =

n −2 1


n n

n n

a a a

a a

W

a

− −

− −

=

1 2 1

2 3

1

1

1 0

1 0 0

1 0 0 0

⋯

⋯

⋯

⋮ ⋮ ⋯ ⋮ ⋮

⋯

⋯

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and are found as

Then

s a

n n n n n sI A s a s a s a s a

− −−− = + + + + =

1 2

1 2 1 0⋯

0 1 0 0 0

0 0 0 0 0

⋯

⋯


,

,

n n n

A P AP B PB

a a a a

C CP D D

−

− −

= = = = − − − −

= =

1 2 1

0

0 0 0 1

1

⋮ ⋮ ⋯ ⋮ ⋮

⋯ ⋮

⋯

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Observable canonical formConsider the dynamic system

to get the OCF, we choose

where

x Ax Bu

y Cx Du

= +

= +

ɺ

( )P WN −

= 1


,

n n

n n

n

C a a a CA a a

N W CA

a

CA

− −

− −

−

= =

1 2 1

2 3

2

1

1

1

1 0

1 0 0

1 0 0 0

⋯

⋯

⋮ ⋮ ⋯ ⋮ ⋮

⋮ ⋯

⋯

s

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and are found as

Then

s a

n n n n n sI A s a s a s a s a

− −−− = + + + + =

1 2

1 2 1 0⋯

n

n

a

a −−

− − = = −

1

1

0 0 0

1 0 0

⋯

⋯

⋯


[ ]

n

a

C CP

−

−

= =

10 0 0

0 0 0 1

⋮ ⋮ ⋯ ⋮ ⋮

⋯

⋯

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Diagonal canonical form

In DCF, the system matrix A is transformed into Diagonalmatrix,

The entries on the diagonal are the eigenvalues of the A

The transformation matrix is obtained as


where are the eigenvectors corresponding to theeigenvalues

If the matrix A is in CCF and has distinct eigenvalues, Then Pmatrix has especial structure.

n n

e e e e −

=1 2 1

⋯

i λi e

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Diagonalization of AAAA Let a matrix A has distinct eigenvalues and is

represented as1 2 n, ,λ λ λL

0 1 0 0

0 0 1 0

A

=

L

L

L L L L L

L L L L L


To diagonalize A, let us use some transformation; that is,

With

n n 1 1a a a−

− − − L L x Pz=

1 2 3 n

2 2 2 21 2 3 n

n 1 n 1 n 1 n 1

1 2 3 n

1 1 1 1

P

− − − −

λ λ λ λ

= λ λ λ λ λ λ λ λ

M

M

M

M M M M M

M

48

Vandermondematrix

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Then

Case: If A has multi le ei envalues, then the dia onalization

Diagonalization of AAAA1

2

1

3

n

0 0 0

0 0 0

P AP 0 0 0

0 0 0

−

λ λ = λ λ

L

L

L

L L L L LL


is not straight forward. For instance A has eigenvaluesThen P can be constructed as

1 2λ =λ =λ

-11 1

P with P does not exists.

= λ λ This implies that diagonalization cannot be achieved this way

49

Is there any way possible??

An ExampleAn ExampleAn ExampleAn Example

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An ExampleAn ExampleAn ExampleAn Example

Diagonalise the following system dynamics

----------- (a)

[ ]

0 1 0 0

x 0 0 1 x 0 u

6 11 6 6

y 1 0 0 x

= +

− − − =

&


Step-1: Find eigenvalues of A . In this case the eigenvaluesare

Step-2: Form matrix P1 2 3

1, 2, 3λ = − λ = − λ = −

1 2 3

2 2 2

1 2 3

1 1 1 1 1 1P 1 2 3

1 4 9

= − − − = λ λ λ λ λ λ

50

Apply transformation to (a)x Pz=

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pp y ( )

[ ]

1 1

0 1 0 0 0 1 0 0

Pz 0 0 1 Pz 0 u z P 0 0 1 Pz P 0 u

6 11 6 6 6 11 6 6

y 1 0 0 Pz

1 0 0 3

− − = + ⇒ = + − − − − − −

=

−

& &


[ ]z 0 2 0 z 6 u, y 1 1 1 x

0 0 3 3

⇒ = − + − = −

&

• Transformation matrix P modifies the coefficient matrix of z into

diagonal form• All the states are decoupled from one another• The eigenvalues remain unchanged

51

Fact: The eigenvalues are unchanged during similarity

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transformation.

Proof:

( )

1 1 1

1

1

Consider

I P AP P P P AP

= P I A P

= P I A P

− − −

−

−

λ − = λ −

⇒ λ −

⇒ λ −


( )

( )

( )

1

1

= P P I A

= P P I A

= I A

−

−

⇒ λ −

⇒ λ −

⇒ λ −Hence proved that the eigenvalues are invariant under linear transformation

52

Example

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Example Diagonalize the following

Step 1: The eigenvalues are -1,-3,-3[ ]

0 1 0 0

x 0 0 1 x 0 u

9 15 7 3y 1 0 0 x

= +

− − − =

&


Step 2: Form P matrix1

1 2 3

2 2 2

1 2 3

1 1 1 1 1 1

P 1 3 3 P does not exist.

1 9 9

−

= λ λ λ = − − − ⇒ λ λ λ

Diagonalization ?

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Generalized Eigenvalues

A has multiple eigenvalues The formula for computing eigenvectors

does not hold to com ute all ei envectors( )i i I A e λ − = 0


Let a q be distinct eigenvalues among n eigenvalues of A. Theequation hold for computing qeigenvectors.

( )i i I A e λ − = 0

Let be the eigenvalues of mth order i e; Thejλ m n q≤ −

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Let be the eigenvalues of mth order. i.e; . The

corresponding eigenvectors are called generalized eigenvectorsand can be determined using the following formula

j λ m n q ≤

( )

( )

j n q

j n q n q

I A e

I A e e

λ

λ

− +

− + − +

− =

− =−

− =−

1

2 1

0


( )

n q n q

j n q m n q m I A e e λ

− + − +

− + − + −− =−

2

1

⋮

An Example

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An Example

2 3

1

Diagnolize

; ; ; ;

:

Eigenvector associated to is

A

Solution

λ λ λ

λ

− = = = =

=

−

1

0 6 5

1 0 2 2 1 1

3 2 4

2


( ) ( )1

I A e I A e e e

e e e

λ − = − = − − = ⇒ − − −

1 21 21 21

31 31 31

2 1 2 2 0

3 2 4

( ) ( )2

e e e

I A e I A e e e

e e e

λ

= − − − − = − = − − = ⇒ = −

− − − −

11 12 12

2 21 22 22

31 32 32

1

2

11 6 5

31 1 1 2 0

73 2 35

7

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( ) ( )2

;

e e

I A e I A e e e e e

e

e e

λ

−

− = − = − − = − − − −

⇒ = − −

11 12

3 21 22 2

31 32

13

23

33

1 6 5

1 1 1 2

3 2 3

1

22

49

46

49


;P

A P AP −

= − − − − − −

= 1

2 1 1

3 221

7 49

5 462

7 49

=

2 0 0

0 1 1

0 0 1

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Transfer function State space-Decomposition of transfer functionsDecomposition of transfer functionsDecomposition of transfer functionsDecomposition of transfer functions Direct decomposition :

Applied to a transfer function which is not in factored form. It canbe conducted in two ways1. Direct decomposition to CCF


2. rect ecompos t on to

Cascade decompositionApplied to transfer functions that are written as product of simplefirst order or second order components

Parallel decompositionApplied to transfer functions whose denominator is in factoredform and partial fraction expansion can be obtained

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Direct decomposition to CCFConsider the following transfer function

( )( )

n n n

n n n n

n

Y s b s b s b s b U s s a s a s a

− − −

− −+ + + += + + + +

1 2 3

1 2 3

1 2

1 2

⋯⋯


Steps:

1. Express the transfer function in –ve power of s. Multiplynumerator and denominator by n s −

( )( )

n

n n

n

Y s b s b s b s b s U s a s a s a s

− − − −

− − −+ + + += + + + +

1 2 3

1 2 3

1 2

1 21

⋯⋯

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2. Multiply numerator and denominator by a dummy variableX(s)

Direct decomposition to CCF ctd…

( )( )

( )( )

n n

n

n

Y s X s b s b s b s b s

U s a s a s a s X s

− − − −

− − −

+ + + +=

+ + + +

1 2 31 2 3

1 2

1 21

⋯

⋯


3. Rewrite numerator and denominator as

( ) ( ) ( )

( ) ( ) ( )

n

n

n

n

Y s b s b s b s b s X s

U s a s a s a s X s

− − − −

− − −

= + + + +

= + + + +

1 2 3

1 2 3

1 2

1 21

⋯

⋯

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Direct decomposition to CCF ctd…4. Construct the state diagram in the above two equations.

Re-arrange the last equation as

5. Substituting as( ) ( ) ( ) ( )n n X s U s a s a s a s X s − − −= − + + +1 21 2 ⋯


( ) ( )( ) ( )

( )

,,

n n

n

n n

n

n n n n n

n n n n

s X s x s X s x x s X s x x s X s x x

X s x x a x a x a x a x u

y b x b x b x b x

− − +

− + −−

− −

− −

→ → =← ←

→ →= =← ←

→←

= − − − − − +

= + + + +

1

1 2 1

2 1

3 2 1

1 1 2 2 3 1

1 1 2 2 3 1

ℓ ℓ

ℓ

ɺ

ɺ ɺ⋯

ɺ

ɺ ⋯

⋯

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where

Direct decomposition to CCF ctd…

x Ax Bu

y Cx Du

= +

= +

ɺ

Control System Design-II by Dr. A.Q. Khan62 [ ]

,

,

n n n

n n n

A B

a a a a

C b b b b D

− −

− −

= = − − − −

= =

1 2 1

1 2 1

0 1 0 0 0

0 0 1 0 0

0 0 0 0 0

1

0

⋯

⋯

⋯

⋮ ⋮ ⋮ ⋯ ⋮ ⋮

⋯

⋯

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Direct decomposition to OCFConsider the following transfer function

( )( )

n n n

n n n n

n

Y s b s b s b s b U s s a s a s a

− − −

− −+ + + += + + + +

1 2 3

1 2 31 2

1 2

⋯

⋯


Steps:

1. Express the transfer function in –ve power of s. Multiplynumerator and denominator by n s −

( )( )

n

n n

n

Y s b s b s b s b s U s a s a s a s

− − − −

− − −+ + + += + + + +

1 2 3

1 2 3

1 2

1 21

⋯⋯

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2. Expand the equation asDirect decomposition to OCF ctd…

( ) ( )

( ) ( )

n

n

n

n

a s a s a s Y s

b s b s b s b s U s

− − −

− − − −

+ + + +

= + + + +

1 2

1 2

1 2 3

1 2 3

1 ⋯

⋯


or

( ) ( ) ( )

( ) ( )

n

n

n

n

Y s a s a s a s Y s

b s b s b s b s U s

− − −

− − − −

= − + + +

+ + + + +

1 2

1 2

1 2 3

1 2 3

⋯

⋯

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The output of the integrators are assigned as state variable

Finally we get

Direct decomposition to OCF ctd…

x A x B u

y C x D u

= +

= +

ɺ


where

[ ]

,

,

n n

n n

n n

a b

a b

A B a b

a b

C D

− −

− −

= = −

= =

1 1

2 3

1 1

0 0 0

1 0 0

0 1 0

0 0 1

0 0 0 1 0

⋯

⋯

⋯

⋮ ⋮ ⋮ ⋯ ⋮ ⋮

⋯

⋯

Cascade Decomposition

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p

( )( )

1 2

1 2

1 2 1 2

where, , , and are real constant. Each of the first

+ += + +

Y s s b s bK U s s a s a

a a b b


1 1 2

2

decomposition.

The state space model is

b= − −

&&

x a a x

[ ]

2 1

2 2

1 2 2 2

+0

b b

−

= − − +

x K ua x K

y a a x Ku

Parallel Decomposition

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p

( )

( )

( )

( ) ( )

( )

1 2

1 2

w h e r e

i s a p o l y n o m i a l o f o r d e r l e s s t h a n 2

a n d , a r e r e a l c o n s t a n t .

=+ +

Y s Q s

U s s a s a

Q s

a a


( )( ) ( ) ( )

1 2

1 2

1

2

T h e s t a t e s p a c e m o d e l i s g i v e n a s

0 10 1

= ++ +

−

= + − &

Y s K K U s s a s a

a x xa

[ ]1 2

=

u

y K K x

Relationship b/w various methodRelationship b/w various methodRelationship b/w various methodRelationship b/w various method

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

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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


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 ab

12b ab a b b a b , Similarly

2

1b a

6

−

−

= + + +

+ + + = + + + + +

=

= = ⇒ =

=

& L

L L L

3 k

0 0 k 0

1 1b a b , b a b

3 k = =LL

( ) 2 0

As

1x t 1 at at b

= + + + L

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( )

( )

( ) ( ) ( )

0

0

2 2 at

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

2Now consider

x Ax (6)

+ + +

=

=

= + + + =

=

L

&


( ) 20 1 2

Analogy with scalar, we have

x 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.

Referring to equation (9), the term is of particulari t t

Ate

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interest.

It is referred to as matrix exponential and analogously, it can berepresented as

This matrix exponential for an converges absolutely forall finite time.

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 Aedx

e e e if AB BA

e e e if AB BA e e = I

+

+

−

• =

• = =

• ≠ ≠•

C id i (6) d ki L l T f

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( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

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

orx t t x 0

where

−

−−

= ⇒ = −

= − ⇒ =

= φ

L


( )t : n n m atrix and is the unique solution

: State transition m atr

φ ×

( )

( ) ( )-1 -At

ix. It has all the inform ation abou t

the free motion o f the system define by x Ax

0 I

Also note t =e = t

=

φ =

φ φ −

&

Case-I: If be distinct eigenvalues of A, thenwill contain exponentials

1 2 n, ,λ λ λL ( )tφ

1 2 nt t te ,e , eλ λ λL

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will contain exponentials

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

e ,e , e

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 to1 1 1 2 n, , , ,λ λ λ λ λL

( )tφ 1 1t t2te , t eλ λ

1 2 nt t te ,e , eλ λ λL

Properties of ( )tφ

( ) AtAs t eφ =

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( )

( )

( ) ( ) ( ) ( ) ( )

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

( ) ( )

1 2 1 2

A(0)

1 1At At 1

A t t A t A t1 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 −

0 1For x Ax, where A=

2 3

= − −

&

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( ) ( )( ) ( ){ }

( )

-1

1At

1

Obtain t and t

Solution : As e t sI A

s 1Note that sI A

2 s 3

s 3 11sI A

−

−

φ φ

= φ = −

− − =

+ +

− = −

-1L


( )( ) ( )

Using partial fraction expansion

1t

s 1 s 2φ =

+ +

-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 1

s 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

Solution to Homogenous State equationsConsider

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n 1

n 1

n n

n p

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 side

Integrating both sides

de x Ax d e Bu d e x

d

− − −

− τ − − τ

− =

− τ = τ τ ⇒ ττ∫ ∫

&

& ( )

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

( ) ( ) ( ) ( ) ( )

t t

A

0 0

t t

A 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 − − − − − − − −

∫ ∫

∫ ∫

∫

Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach

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

t

A 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

∫

∫

An ExampleAn ExampleAn ExampleAn ExampleConsider

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( ) ( )

( )

x Ax Bu,

where

0 1 0 0 0 for t

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Laplace Transform method:

Method of Diagonalization of A:

( ) ( ){ }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 1 D t

e P e P

−

=

cds-ii (2)

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