Mathematical Modeling of Dynamic Systems in State Space II

(1)

Mathematical Modeling

of Dynamic Systems in State Space II

(2)

Transfer Function and State Equation

( )( 2) 3 ( ) z s s   u s

( ) 2 ( ) 3 ( ) z t   z t  u t

1 1

1

( ), 2 ( )

2 3 ( )

x z t x z z t

x x u t

y x

   

  

 

  

 

 

1. 

2.

( ) 3

( ) 2

z s u s  s



1 1 1

1 1

1

1 1

1

( ) 5 3 5( 2) 7 7

( ) 2 2 5 2

( ) 5 ( ) 7 ( ) 2

( ) 7 ( ), 2 7

2

2 7

, 5

5

z s s s

u s s s s

z s u s u s

s

let z s u s z z u

s

x x u

y z u

let x z

x u

  

   

  

 



  



  

 

    

  

    

 



(3)

Another way,

( ) 5 3

( ) 2

z s s u s s

 



1 1

1

( ) ( ) 5 3 2 ( )

2 ( )

( ) 5 3

5( 2 ) 3

5 7

, 2

7 5

z s u s s s Q s

q q u t

z t q q

q u q

u q

x x u

let x q

y x u

 

 

 

 

   

 

  

 

       



(4)

2

^nd

order systems

(5)

ex1) Consider a system defined by

 y  6  y  11 y   6 y  6 u

1 2 3

x y







Choose state variables,

Then we obtain, 1 2

2 3

3 1 2 3

,

6 11 6 6

x x

x y x x x u



     



 

 

1 1 1

2 2 2

3 3 3

0 1 0 0

0 0 1 0 , 1 0 0

6 11 6 6

x x x

x x u y x

x x x

         

         

             

            

         



State Equation and Transfer Function

Phase variables (each subsequent state variable is defined to be the derivative of the previous state variable.)

(6)

( ), k b b

my ky b y u y y y u

m m m

       

     

1 2

,

2

k

1

b

2

b

x x x x x u

m m m

    

 

The right side includes term. To explain the reason we should not include differentiation of u, assume

u 

1

,

2

x  y x  y 

Choose state variables,

( ) ( )

u   t unit impulse function

2

k b k ( )

x y dt y t

m m m 

    

ex2) Consider a mechanical system,

includes term. It means and cannot be accepted as a state variable.

x

2

( / ) ( ) k m  t x

₂

(0)  

That’s why the standard form is

x   Ax  Bu y  Cx  Du

x   Ax  Bu

"udisplacement"

(7)

2 1 2

2

1 2

b k b b b k b b

x y u y y u u x x u

m m m m m m m m

k b b

x x u

m m m

   

                  

         

     

1 1

2

2 2

0 1 b

x x m

k b u

x x b

m m

m

 

   

       

                             



term has been eliminated.

u 

To eliminate term,

1

,

2

b

x y x y u

    m

u 

2

k b b b k b

y y y u y u y y

m m m m m m

d b k b b b

y u y y u u

dt m m m m m

        

           

     

     

     

 

So we choose state variables as,

Method 1: Choose a state variable that includes

u

State Equation and Transfer Function

(8)

1 2 0 1 2

y  a y  a y  b u  b u  b u

   

Consider a second-order system,

2

0 1 2

2

1 2

( ) ( )

Y s b s b s b

U s s a s a

 

  

2

0 1 2

2

1 2

( ) 1 ( )

( ) , ( )

Z s Y s

b s b s b U s s a s a Z s

    

 

1 2

0 1 2

z a z a z u y b z b z b z

  

  

 

1 2

, ,

let x  z x  z 

0 1 2 0

(

2 1 1 2

)

1 2 2 1

b z   b z   b z  b  a x  a x   u b x  b x  y

1 2

,

2 2 1 1 2

x x x a x a x u

       

Method 2: Include the input derivates in the output equation

2

1 2

1

s  a s  a b s

⁰ ²

 b s b

¹



²

( )

U s Z s ( ) Y s ( )

2 2 0 1 1 1 0 2 0

( ) ( )

y  b  a b x  b  a b x  b u

1 2

2 2 1 1 2

x x

x a x a x u



   



(9)

 

1 1 1

2 2 0 1 1 0 0

2 2 1 2 2

0 1 0

1 ,

x x x

u y b a b b a b b u

x a a x x

         

     

             

        

( ) ( 1)

1 1

( ) ( 1)

0 1 1

n n

y a y a y a y

b u b u b u b u



 

   

    

 

1 1

2 2

1 1

1 2 1

0 1 0 0 0

0 0 1 0 0

,

0 0 0 1 0

1

n n

n n n n n

x x

u

x x

x a a a a x

 

       

       

         

       

             

     

 

      

 

 

1 2

0 1 1 0 1 1 0 0

n n n n

n

x

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

x

 

   

      

   

 

   

 N-th order differential equation,

Method 2: Include the input derivates in the output equation

(10)

( ),

my     ky b y u    my   by   ky  bu 

2 2

( ) ( ) 1 ( )

, ,

( ) ( ) ( )

Y s b s Z s Y s

U s  ms bs k U s  ms bs k Z s  bs

   

( ms

2

 bs  k Z s ) ( )  U s ( ), bs Z s ( )  Y s ( )

ex2) Consider this mechanical system again,

2

1

( ) ( ) b ( ) k ( )

s Z s U s s Z s Z s

m m m

  

¹ ²

2 1 2

2

1 x x

k b

x x x u

m m m

y b x



   





State variables,

1

,

2

x  z x  z 

0 1

k b

m m

 

 

    

 

A

0 1 m

   

  

  B

 0 b 

 C

  0

D 

Draw Block Diagram,

System Matrices:

State Equation and Transfer Function

2 2

( ) ( ) ( )

, ( )

( ) ( )

Y s bs Y s u s

cU s  ms bs k b s  ms  Q s

 

(11)

ex) Consider a spring-mass-damper system

2 2

( )

d y dy du

m b k y u

dt dt dt

d y dy du

or m b ky b ku

dt dt dt

 

      

 

   

2

( ) ( )

Y s bs k

Transfer Function

U s ms bs k

  

 

1 1 1

2 2 2

0 1

0 u, 1

x x k b x

k b y

x x m m x

m m

 

              

                 

       



2 2 0

1 1 0

0 0

k k k

b a b

m m m

b b b

b a b

m m m

    

m

0 1 2

1 2

0, ,

,

b b b b k

b k

a a

m m

  

 

State variables:

State Eq:

Output Eq:

Rewrite equations:

(12)

0 1 2 3 4 5 6 7 8 0

0.5 1 1.5

Unit-Step Response (Method 2)

t(sec)

Output y(m)

t=0:0.02:8;

A=[0 1;-10 -2];

B=[0;1];

C=[10 2];

D=[0];

sys=ss(A,B,C,D);

[y,t]=step(sys,t);

plot(t,y) grid

title('Unit-Step Response (Method 2)','Fontsize',15) xlabel('t(sec)','Fontsize',15)

ylabel('Output y(m)','Fontsize',15)

If, m=10kg, b=20N-s/m, k=100N/m

 

1 1 1

2 2 2

0 1 0

u, 10 2

10 2 1

x x x

x x y x

         

  

               

     



Matlab Example

(13)

Transformation of Mathematical Models with MATLAB

( ) ( )

Y s numerator polynomial in s num U s  denominator polynomial in s  den

MATLAB command, [A, B, C, D] = tf2ss(num,den) ex) Consider,

3 2

( )

( ) 14 56 160

Y s s

U s  s s s

  

 

1 1

2 2

3 3

1 2 3

14 56 160 1

1 0 0 0

0 1 0 0

0 1 0 [0]

x x

x x u

x x

x

y x u

x

 

       

         

       

       

   

   

   



One of many possible state-space representations is,

gives a state space representation.

>> num=[0 0 1 0];

>> den=[1 14 56 160];

>> [A,B,C,D]=tf2ss(num,den) A =-14 -56 -160

1 0 0 0 1 0 B =1

00

C =0 1 0 D =0

(14)

x = Ax+ Bu y = Cx+ Du



can be written as,

ˆ ˆ

Px = AP x+ Bu y = CP x+ Du ˆ



or

x = P A P x+ P Bu ˆ

^-1

ˆ

^-1

y = CP x+ D u ˆ



- Since infinitely many n x n matrices can be a transformation matrix P, there are infinitely many state-space models for a given system.

I A 0

  

 Eigenvalues of an n x n matrix A are the roots of the characteristic equation.



The eigenvalues are also called the characteristic roots.

ex)

0 1 0

A 0 0 1 ,

6 11 6

 

 

  

    

 

3 2

1 0

I A 0 1 6 11 6

6 11 6

( 1)( 2)( 3) 0



    



  



      



    

Transformation of a State-Space Models into Another One

(15)

1 2 1

0 1 0 0

0 0 1 0

A

0 0 0 1

n n n

a a _ a _ a

 

 

 

 

    

 



   



Diagonalization of State Matrix A

Consider an n x n state matrix A :

1 2

2 2 2

1 2

1 1 1

2 2

1 1 1

x = Pz, P

n n

n n n

n

where

  



^



^



^

 

 

  

 

 



  



If matrix A has distinct eigenvalues and the state vector x is transformed into another state vector z by use of a transformation matrix P ,

1 1 2

0 P AP

0

_n





 

 

  

 

 



is a canonical matrix and each column of P is an eigenvector of matrix A

P AP

^1

(16)

Jordan Canonical Form

3 2 1

0 1 0

A 0 0 1

a a a

 

 

  

    

 

If matrix A involves multiple eigenvalues, diagonalization is not possible but matrix A can be transformed into a Jordan Canonical Form.

Consider the 3 x 3 matrix A :

1

,

2

,

3

where

1 2 3

       

Assume that matrix A has eigenvalues

1 1 2 3 1 2 3

( I   A) v  0 A : 3 3  matrix ,    , , v v v , ,

1 1 1

,

2 1 2

,

3 3 3

Av   v Av   v Av   v



1

I A  1 rank   



1 2 3

 

1 2 3



¹ 1 ^-1 ¹ 1

3 3

0 0 0 0

A 0 0 , P AP 0 0

0 0 0 0

v v v v v v

 

   

   

     

   

   

can determine two eigenvectors .

v v

1

,

2

Case 1.

(17)

Jordan Canonical Form

1 3

2 2

1 1 3

1 0 1

x S z S 1

2

where  

  

 

 

   

 

 

will yield

1 1

1 3

1 0

S AS 0 0 J

0 0





 

 

   

 

 

This is in the Jordan Canonical Form.



1

I A  2 rank   

1 1 1 3 3 3

A v   v , A v   v

1 2 1

A   I v  v



1 2 3

 

1 2 3



¹ 1 3

1 0

A 0 0

0 0

v v v v v v



 

 

  

 

 

can determine one eigenvector .

v

1

Case 2.

2 1 1 2

A v   v  v

Find such that

v

2

(18)

Canonical Forms

Case 3. Complex Roots

• Diagonal (or Jordan) Canonical Form

(Partial Fraction Expansion)



Complex case의 diagonalization

방법 이용 x x bu

y Cx

  





0 0

j

 

  

    

1

2 2

1

2 2

j

K j

  

 

  

 

 

1 2 2 2

1 1 2 2 j j

K j



 

 

  

 

 

 

K KJ

 

J

 

 

 

  

Note : Complex Roots, Complex State x

J  K

^1

 K

1 1

K P APK

^ ^



AP   P

(19)

Canonical Forms

Case 3. Complex Roots

• Diagonal (or Jordan) Canonical Form

 

1 2

1 0

z z b u

b

y z

 

 

   

      



Ex)



( ) Y s



1 / s

1 / s b

1

( ) u s

b

2

( )

u s 







(20)

Canonical Forms

Case 3. Complex Roots

• Diagonal (or Jordan) Canonical Form

x   Ax  Bu

let

: diagonal

1 1

x P

P AP P Bu



^



^





 

 

let

1 1 1

J

Kz

z K K z K P Bu



  



  

 

1 2

z b u b

 

 

 

      

1 1 1

J

x P PKz

z K K z K P bu y CPKz



  

  

    

 

  

  

Step 1 Step 2

Step 3

(21)

End of Lecture 5(II)