state space analysis, eign values and eign vectors

By: Shilpa Mishra

M.E., NITTTR Chandigarh

.

STATE SPACE ANALYSIS Conversion of T.F. models to canonical state variable models and concept of Eignvalues & Eignvectors.

Contents_______________

1. Introduction

2. Need of realization of transfer function into state variable models

3. Realization of transfer function into a state space model or mathematical model in following possible representation:

a. First companion form (controllable form) b. Second companion form (observable form) c. Jordan canonical form

4. Eigenvalues and Eigen vectors

04/09/232 Shilpa Mishra ME IC 122509

Plant

Mathematical Model : Differential equation

Linear, time invariant

Frequency DomainTechnique

Time Domain Technique

Two approaches for analysis and design of control system:

1.Classical Technique or Frequency Domain Technique.(T.F. + graphical plots like root locus, bode etc.)2.Modern Technique or Time Domain Technique (State variable approach).

04/09/233Shilpa Mishra ME IC 122509

Transfer Function form

Need of conversion of transfer function form into state space form:

1.A transfer function can be easily fitted to the determined experimental data in best possible manner. In state variable we have so many design techniques available for system. Hence in order to apply these techniques T.F. must be realized into state variable model.


BuAxx DuCxy

xx

y

uA

B

C

D

= state vector

= derivative of the state vector with respect to time

= output vector

= input or control vector

= system matrix

= input matrix

= output matrix

= Feed forward matrix

State equation

output equation

General State Space form of Physical System


Deriving State Space Model from Transfer Function Model

The process of converting transfer function to state space form is NOT unique. There are various “realizations” possible.

All realizations are “equivalent” (i.e. properties do not change). However, one representation may have some advantages over others for a particular task.

Possible representations:

1. First companion form

2. Second companion form

3. Jordan canonical form


1. First Companion Form (SISO System)

If LTI SISO system is described by transfer function of the form;

Decomposition of transfer function:

.

01223

3

012

2

)(

)(

)(

)(

asasasa

bsbsb

sR

sC

sU

sY


sXbsbsbsCsY 1012

2

101

121

2

2 xbdt

dxb

dt

xdbty

)2........(..........322110)( xbxbxbty

sXasasasasRsU 1012

23

3

)()()()(

101

121

2

231

3

3 txadt

tdxa

dt

txda

dt

txdatu

43322110)( xaxaxaxatu

I.

II.

)1...().........(32211043 tuxaxaxaxa

21 xx 32 xx 1)( xtx 43 xx &

Select state variables like :


21 xx 32 xx 1)( xtx

from equation (1) & (2) and state equation, block diagram realization in first companion form of TF will be

43 xx


Again from equation (1) & (2) complete state model will be ;

)3).....((

3/1

0

0

3

2

1

210

100

010

3

1

3

2

1

,

)(

3

13

3

22

3

11

3

043

)(32211043

tu

ax

x

x

aaaa

x

x

x

or

tua

xa

ax

a

ax

a

axx

tuxaxaxaxa

A B


Equation (3)&(4) combining together gives the complete realization of the given transfer function.

Matrix A has coefficients of the denominator of the TF preceded by minus sign in its bottom row and rest of the matrix is zero except for the superdiagonol terms which are all unity.

In matrix theory matrix with this structure is said to be in companion form therefore this realization is called first companion form of realizing a TF.

)4.......(

3

2

1

210)(

,

322110)(

x

x

x

bbbty

or

xbxbxbty

C


Example :TF to State Space (constant term in numerator)

rcccc 2424269

cx 1 cx 2 cx 3

1. Inverse Laplace

2. Select state variables

21 xx

32 xx

rxxxx 2492624 3213

1xcy

sD

sNsG

sN

sD

numerator

denominator


r

24

0

0

x

x

x

92624

100

010

x

x

x

3

2

1

3

2

1

3

2

1

001

x

x

x

y


Example:


2. Second Companion Form (SISO System)

In second companion form coefficient of the denominator of the transfer function appear in one of the column of the A matrix.

This form can be obtained by the following steps:

Let the transfer function is,

nnnn

nnnn

asasas

bsbsbsb

sU

sYsH

......

........

)(

)()(

22

11

22

110

)5)].......(()([1

....)]()([1

)()(

0)]()([...)]()([)]()([,

)()....()().....(

110

111

0

110

11

sYasUbs

sYasUbs

sUbsY

sUbsYasUbsYassubsYsor

sUbsbsbsYasas

nnn

nnnn

nnn

nnn

On dividing by and solving for Y(s);ns


By equation (5) block diagram can be drawn as follows which is called second companion form of realization:-


To get state variable model, output of each integrator is identified as state variables starting at the left and preceding to the right.

The corresponding differential equations are,

ubn

xy

un

bubn

xn

ax

un

bubn

xn

axx

ububn

xan

xn

x

ububn

xan

xn

x

0

isequation output theand

)0

(1

1)

0(

112

2)

0(

221

1)

0(

11


Now from here state and output equations organized in matrix form are given below:

BuAxx DuCxy

0

011

011

0

1

2

1

;1000

B ;

100

010

001

000

bDC

bab

bab

bab

a

a

a

a

A nn

nn

n

n

n


•In this form of realizing a TF the poles of the transfer function form a string along the main diagonal of the matrix A.

•In Jordan canonical form state space model will be like:-

3. Jordan canonical form (Non-repeated roots)


nnnn

nnnn

asasas

dsdsdsd

sU

sYsH

......

........

)(

)()(

22

11

22

110

Let general transfer function;

)()(

)(

)(

1

21

1

ssxdt

tdxL

xxdt

tdx

xtx

if


urxλx

.

.

.urxλx

urxλx

nnnn

2222

1111

nxxxudty ......)( 210

& output equation;

With the help of above equations jordan canonical state model can be obtained as;

State equation:


Jordan Canonical Form(Non-repeated roots): Block Diagram


Jordan canonical form: Example (Non repeated roots)


Jordan canonical form: Example (repeated roots):


Eignvalues and Eignvector

Definition :

Given a linear transformation A, a non-zero vector x is defined to be an eigenvector of the transformation if it satisfies the eigenvalue equation

for some scalar λ. In this situation, the scalar λ is called an eigenvalue of A corresponding to the eigenvector x.

It indicates that vector x has the property that its direction is not changed by the transformation A, but that it is only scaled by a factor of λ.

Only certain special vectors x are eigenvectors, and only certain special scalars λ are eigenvalues.

The eigenvector must be non-zero because the equation A0 = λ0 holds for every A and every λ.

λxAx


A acts to stretch the vector x, not change its direction, so x is an eigenvector of A.

The eigenvalue λ is simply the amount of "stretch" or "shrink" to which a vector is subjected when transformed by A.If λ = 1, the vector remains unchanged (unaffected by the transformation). A transformation I under which a vector x remains unchanged, Ix = x, is defined as identity transformationIf λ = −1, the vector flips to the opposite direction; this is defined as reflection.


Computation of eigenvalues & the characteristic equation

When a transformation is represented by a square matrix A, the eigenvalue equation can be expressed as;

Ax = λx

(A − λI)x = 0

As x must not be zero, this can be rearranged to;

det(A − λI) = 0.

which is defined to be the characteristic equation of the n × n matrix A. Expansion of the determinant generates a polynomial of degree n in λ and may be

written as;

here λ1, λ2……λn are called eignvalues. For each eign value there exist eign vector x.


Example 1. find the eigenvalues and eigenvectors of the matrix

75.075.0

5.13][A

Solution

75.075.0

5.13][][ IA

0)5.1)(75.0()75.0)(3()det( IA

0125.1375.025.2 2 0125.175.32

)1(2

)125.1)(1(4)75.3()75.3( 2

2

092.375.3

3288.0,421.3 So the eigenvalues are 3.421 and 0.3288.


3288.0,421.3 21 Let

2

1][x

xX

be the eigenvector corresponding to 421.31

Hence 0]])[[]([ 1 XIA

010

01421.3

75.075.0

5.13

2

1

x

x

0

0

671.275.0

5.1421.0

2

1

x

x

.421x1 +1.5x2 = 0 .75x1 +2.671x2 = 0


If sx 1

then

sx

xs

2808.0

05.1421.0

2

2

The eigenvector corresponding to 421.31 then is

s

sX

2808.0][

2808.0

1s

The eigenvector corresponding to 421.31

is

2808.0

1

Similarly, the eigenvector corresponding to 3288.02

is

781.1

1


Example 2. Find the eigenvalues and eigenvectors of

005.0

5.05.05.0

105.1

][A

Solution

The characteristic equation is given by 0])[]det([ IA

0

05.0

5.05.05.0

105.1

det

0)]5.0)(5.0()0)(5.0)[(1()]0)(5.0())(5.0)[(5.1(

025.12 23 The roots of the above equation are

0.1,5.0,5.0 Note that there are eigenvalues that are repeated. Since there are only two distinct eigenvalues, there are only two eigenspaces. But, corresponding to = 0.5 there should be two eigenvectors that form a basis for the eigenspace


To find the eigenspaces, let

3

2

1

][

x

x

x

X

Given 0][)][( XIA

then

0

0

0

05.0

5.05.05.0

105.1

3

2

1

x

x

x

For 5.0 ,

0

0

0

5.005.0

5.005.0

101

3

2

1

x

x

x

Solving this system gives axbxax 321 ,


So

a

b

a

x

x

x

3

2

1

0

0

0 b

a

a

0

1

0

1

0

1

ba

So the vectors

1

0

1

and

0

1

0

form a basis for the eigenspace for the eigenvalue 5.0 .

For 1 ,

0

0

0

105.0

5.05.05.0

105.0

3

2

1

x

x

x

Solving this system gives axaxax 5.0,5.0, 321

The eigenvector corresponding to 1 is

5.0

5.0

1

5.0

5.0 a

a

a

a

;

Hence the vector

5.0

5.0

1

is a basis for the eigenspace for the eigenvalue of 1 .


Thanks

for

Attentio

n04/09/2337 Shilpa Mishra ME IC 122509

state space analysis, eign values and eign vectors

Engineering