10-3 Linear Systems Analysis in the Time Domain III

System Analysis Spring 2015

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10-3

Linear Systems Analysis in the Time Domain III

- Transient Response -

System Analysis Spring 2015

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System Response with Additional Poles

2 2

( )

( ) ( )

 

 

 

  

 

n d

n d

B s C

C s A

s s

2

1

,

2 _{n n}

1

p p     j   

Underdamped System with additional pole at - α

( ) 1

2

cos( )

1



 



 



 



nt

step d

y t e t

System Analysis Spring 2015

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▪ 목적

:

제어시스템을 설계할 때 극점이 과도응답에 미치는 영향을 알아본다

.

▪ 표준

2

차 계단응답에 대한 부가 극점의 효과를 고려

■ 극점 추가의 영향

2

( ) 1 , 0.5

( / 1)[ 2 1]

H s s s s 

 

  

  

System Analysis Spring 2015

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■ 극점 추가의 영향

▪

Simulation Result

2

( ) 1 , 0.5

( / 1)[ 2 1]

H s s s s 

 

  

  

System Analysis Spring 2015

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■ 영점 추가의 영향 (2)

2

0 2 2

/ 1

( ) , 0.5, 1, 1

2 1

1 1

( ) ( ) ( )

2 1 2 1

d

H s s

s s

H s H s H s s

s s s s

  



  

    

 

   

   

System Analysis Spring 2015

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■ 영점 추가의 영향 (2)

0.5, 1

      0.5,    1

▪

Simulation Result

2 0 2 2

/ 1 1 1

( ) , ( ) , ( )

2 1 2 1

^d

2 1

s s

H s H s H s

s s s s s s



   

   

     

System Analysis Spring 2015

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■ 영점 추가의 영향 (1)

2

/ 1

( ) , 0.5

2 1

H s s

s  s

  

 

 

▪Simulation Result

System Analysis Spring 2015

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Non-minimum Phase System

System Analysis Spring 2015

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System Analysis Spring 2015

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System Response with Zeros

System Analysis Spring 2015

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Effects of Nonlinearities

Saturation Deadzone

System Analysis Spring 2015

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Effects of Nonlinearities

Backlash

System Analysis Spring 2015

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1

,

2

, ,

_n

x x  x

system

1 2

m

u u u



1 2

l

y y y





1 2



( ) ( ) ( )

( ) ( ) ( )

( )

_n ^T

x t Ax t Bu t y t Cx t Du t

x t x x x

 

 







Solution of Linear (Time Invariant) State Equation

System Analysis Spring 2015

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Basic Matrix Linear Algebra -Homogeneous Solution

 

( 0)

0 0 0

2

A A

0 0

A

(0) ( ) (0)

( ) 0 i)

( ) 0,

( ) ( ), , ( )

( ) ( )

ii) exp( ) 1

1! 2! !

i) x = A x A : , x : 1

x( ) exp A( ) x( ), ( ) A

ii)

at

at

a t t

k at

t t

t

x C

x t x e

u t x ax

x t Ce t

x t e x t t t x t

at at at

e at

k

n n n

t t t t d e e

dt How to evaluate e











 

 

      

 

  



 



Scalar Function

Homogeneous Solution

System Analysis Spring 2015

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 

   

0

0

A( ) 0

0 0

( )

0 0

2 1 1 0 2 0 0 1 2

2 A 2 A 2

1

x( ) x( )

( ) x( )

( ) exp A( )

1. ( ) ( ) ( ) , ,

2. (0)

3. ( ) ( ) ( ) (2 ) (2 )

( ) ( )

4. ( ) ( )

5. ( )

t t

A t t

t t

g

t e t

t t t

t t e t t

t t t t t t for any t t t

I

t t t t e e t

t gt

t t

t











  

    

      

 

        

  

   





: State transition matrix (STM)

: fundamental matrix of the system Properties of STM

is nonsingular for all finite values of t (inverse exists)

State Transition Matrix

System Analysis Spring 2015

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 

0

( ) 0

0

x A x B , x A x B ,

x A x B , x B ( )

x( ) x(0) B ( )

x( ) x(0) B ( )

( ) x(0) ( ) B ( )

At

At At At At At

t

At A

t

At A t

t

u u

integrating factor e

e e e u d e e u t

dt

e t e u d

t e e u d

t t u d





 

 

  



    

 



   

 

    

  

  

    







 



Complete Solution of the State Equation

System Analysis Spring 2015

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 

⁰

0

0

0 0

0 0

0

, , x( ) x( ) B ( )

x( ) ( ) x( ) ( ) B ( )

x( ) ( ) x(0) ( ) B ( )

t

At At A

t t

t t

for t t e t e t e u d

t t t t t u d

t t t u d



 

  

  





 



      

     







- Zero-input response

- free response

- Zero-state response

- forced response

0

, , 0

0

x( ) ( ) x(0) ( ) B ( )

t

change of variable let t t

t

d d

t t u d

    

 

 

  

    

  

 

     

Complete Solution of the State Equation

System Analysis Spring 2015

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Laplace Transformation Solution of State Equations

 

x  Ax Bu y  Cx  Du

0

1 1

0

X( ) A X( ) B ( ) ,

X( ) ( I A)

^

( I A) B ( )

^

  

   

s s x s U s

s s x s U s

( I A)

X( ) [ (0) B ( )]

det( I A)

  

 adj s

s x U s

s

( )  X( )  ( ) Y s C s DU s

( ) X( )

( )  ( ) 

Y s s

C D

U s U s

( ) ( I A)

( ) det( I A) B

  

       Y s adj s

C D

U s s

( I A) det( I A) det( I A)

  

 

C adj s B D s

s

Laplace transformation 

Transfer function form 

System Poles  Roots of the denominator / Solution of det( I A) s   0

Eigenvalues of A matrix!!

System Analysis Spring 2015

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Time Domain Solution of State Eqns. using Laplace Transform

x   A x B  u

1 1

0

A -1 1 -1

X( ) ( I A) ( I A) B ( )

( I A)

( I A) ( )

det( I A)

 



   

  

 

            

t

s s x s u s

adj s

e s t

L L s

A A( )

0 0

0

( ) B ( )

( ) (0) ( ) B ( )



 

  

 



    





t

t t

t

x t e x e u d

t x t u d

System Analysis Spring 2015

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 

₂ ¹

-1 1 -1 2 3

2 3 4

2 2 3 3

A 2 2 3 3

1 1 1

, , ( ) ( 1) ( )

( 1)!

1 1 1 1

( I A) I A A A

1 1

I A A A

2! 3!

1 1

I A A A

2! 3!

n

n n n

n n

t

t t t f t d F s

s n s ds

s s s s s

t t t

e t t t





   

          

 

 

           

   

     







L L L

L L

Using Laplace transformation table

System Analysis Spring 2015

21 Example1. Find the state transition matrix and solve for x(t).

0 1 0

x( ) x( ) ( )

8 6 1

   

           

 t t u t 1

x(0) 0

    

 

u(t) is a unit step.

System Analysis Spring 2015

22 Example2. Find the state transition matrix using (sI-A)

^-1

0 1 0

x( ) x( ) ( )

8 6 1

   

           

 t t u t 1

x(0) 0

    

 

System Analysis Spring 2015

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End of Lecture 10-3