Review ▪ Observer – controller ▪ Reference Input Tracking

Lecture 11-1

▪ Pole Placement – Review

▪ Observer – controller

▪ Reference Input Tracking

Controllable Canonical form

1 2 1

0 1 0 0 0

0 0 1 0 0

0 0 0 0 0

1

n n n

1

u

a a

₋

a

₋

a

   

   

   

   

= +

   

   

 − − − −     

 

x x





 

    



Example Controllability and Control Canonical form

Control :

[

n n 1 1

]

y = b b

₋

 b x

1 1

1

1 1

( ) ( )

n

n

n n

n n

b s b

Y s

u s s a s a s a

−

− −

+ +

= + +  + +



c c c c

c c

A B u y C D u

= +

= +

x x

x



[

1 2

]

c c c c cn c

u = − K x = − K K  K x

Example Controllability and Control Canonical form Control :

u = − K

c

x

c

= − [ K

c1

K

c2

 K

cn

] x

c

( )

( )

1 1 2 1 1

0 1 0

0 0 0

( ) ( ) ( )

c c c c c

c c c c

c

n c n c c

A B K

A B K

a K a

₋

K a K

= + −

= − ⋅

 

 

 

=  

 − + − + − + 

 

x x x

x

x







   



Characteristic Eqn

Characteristic Eqn for the desired pole locations :

1

1 1 2 1

( ) ( ) ( ) 0

n n

n n n

s + a + K s

⁻

+  + a

₋

+ K s + a + K =

1 2

1 2 1

0

n n n

n n

s + α s

⁻

+ α s

⁻

+  + α

₋

s + α =

Then the necessary feedback gains

1

,

2 1 1

, ,

1 1

c n n c n n cn

K = − + a α K = − a

₋

+ α

₋

 K = − + a α

General Form

A Bu y C

= +

=

x x

x



Transformation Matrix

x = T x

c

1 1

c c

c c c c c

c c c

T AT Bu

T AT T Bu A B u y C T C

− −

= +

= + = +

= =

x x

x x x

x x





Controllable

Canonical Form State feedback control law

1

c c c

u = − K x = − K T

⁻

x

Transformation Matrix T

T = CW

2 n 1

C =   B AB A B  A

⁻

B  

1 2 1

2

1

1 0

1

1 0 0 0 0

n n

n

a a a

a W

− −

−

 

 

 

 

=  

 

 

 



 

  

 

Transformation Matrix T

T = CW

2 n 1

C =   B AB A B  A

⁻

B  

1 2 1

2

1

1 0

1

1 0 0 0 0

n n

n

a a a

a W

− −

−

 

 

 

 

=  

 

 

 



 

  

 

Where

1 1 1

T

⁻

= W C

^{− −}

T

c

=

x x

Ackerman’s Formula

Step 1. Converting (A,B) to (A_c, B_c), T Step 2. Solving for the gain, K_c

Step 3. Converting back gain, K = KcT^-1

[ ]

¹

2 1

1 1

0 0 0 1 ( )

[ ]

( )

c n

n n

c n

K C A

C B AB A B A B

A A A I

α

α α α

−

−

−

=

=

= + + +







: " "

i

the coefficient of the desired characteristic polynomial

α

Selection of Pole Location for Good Design 1. Dominant second-order poles

2. Symmetric Root Locus (SRL) - LQR Design

1. Dominant second-order poles

2^nd order systems (p.139 Franklin) Step response

Step input response

2 2 2

2 2 2 2 2 2 2

( ) 2 ( ) (1 ) ( )

n n n

n n n n d

H s s s s s

ω ω ω

ζω ω ζω ω ζ σ ω

= = =

+ + + + − + +

( ) 1

^t

cos(

_d

) sin(

_d

)

d

y t e

^σ

ω t σ ω t ω

−

 

= −  + 

 

Re Im

η ω

n

cos η ζ =

The Desired poles

[ , , , , ]

^T

c n d n d

p = − ζω ω + j − ζω ω − j − α − β − γ

Characteristic eqn :

( s

²

+ 2 ζω

_n

s + ω

_n²

)( s + α )( s + β )( s + γ ) = 0

, ,

_n

α β γ >> ζω

MATALB STATEMENT

A : (4x4) matrix B : (4x1) matrix

pc : desired pole ex) pc=[-0.7+0.7j; -0.7-0.7j; -4;-4;-4]

K2 = acker(A,B,pc)=place(A,B,pc)

A Bu

= +

x  x

·Reference Input Tracking

u = − K x

Introducing the reference input with Full-State Feedback

( )

ss ss

u = u − K x x −

The pole-placement design

Introduce reference input

u = − K x + r

; a non-zero steady-state error to step input New control formula for zero s.s. error to any constant input

A Bu y C Du

= +

= +

x x

x



The system state equation

0

_{ss ss}

ss ss ss

A Bu

y C Du

= +

= +

x x

In the constant steady state

ss ss

y = r

We want For any value of rss

To do this, we make ss x ss

ss u ss

N r u N r

=

= x

Then,

0

_{x ss u ss}

ss x ss u ss

AN r BN r r CN r DN r

= +

= +

0 1

x u

N A B C D N

 

   

⇒     =  

     

This eqn can be solved for Nx and Nu

1

0

1

x u

N A B

N C D

     

−

   =   

   

 

With this values

( ) ( ) ( )

( )

u x u x

u N r K N r K N KN r a

K Nr b

= − − = − + +

= − +

x x

x

(a) is better in practical implementation

(a) Zero s.s. error even if the gain K is slightly in error (b) nonzero s.s. error if gains do not match exactly

R

N

u

Plant

K

N

x

u y y

R

K

Plant N u

+

−

− +

+

−

( ) a ( ) b

+ N −

r u

K

C y

A Bu

= +

x x

x

( A BK ) GNr y C

= − +

=

x x

x



The closed-loop zeros,

( )

det 0 det 0

0 0

sI A B sI A BK BN

C C

− −

 − − −  = ⇒   =

   

 

 

(By elementary row and column operations)

When full state feedback is used as

The zeros remain unchanged by the feedback

(only closed loop poles have been changed by the state feedback) Conclusions

u − − K x + Nr

( r ≠ 0)

Tracking Integral Control

A Bu y C

= +

=

x x

x



(for simplicity, assume D=0)

( ) ( )

ss ss u x

u = u − K x x − = N r − K x − N r

1

0

0 1

x u

N A B

N C

     

−

   =   

   

 

(

_{u x}

) ( ) (

_{u x}

)

N

A B N r K KN r A BK B N KN r

= + − + = − + +

x  x x x 

1

1

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

sI A BK X s BNr s X s sI A BK BNr s

y s CX s C sI A BK BNr s

−

−

− + =

= − +

= = − +

( )

1

( ) : (closed loop DC gain)

( )

Y s C sI A BK BN R s

= − +

−

: Parameter dependent

u x

N = N + KN

Parameter error → Steady state error

The closed loop DC gain

: the ratio of the output of a system to its input

(presumed constant) after all transients have decayed State Feedback

A Bu

y C

= +

=

x x

x

 u = u

_ss

− K ( x x −

_ss

)

1 0

lim lim ( ) 1 ( )

t s

y sG s C A BK BN

r s

−

→∞

=

→

= − +

( A BK ) BNr y C

= − +

=

x x

x



1

( ) ( ) ( )

Y s = C sI − + A BK

⁻

BNR s

DC gain ;

1

0

1

0 1

x u

N A B A

N C

α β γ

−



−



     

= =  

     

   

   

u x

N = N + KN

DC gain is not 1if there exist

parametr errors or parameter variations

State space design :

Plant parameter variations → s.s. error

⇒ Integral Control and Robust Tracking Integral control

The system

Augment the plant state with the extra state x_I which obeys the differential equations

A Bu B u

1

y C

= + +

=

x x

x



x 

I

= C x − = r e

thus

0 t

x

I

= ∫ e dt

The augmented state equations become

The feedback law

1

0

0 0 1

0 0

I I

x C x

u r w

A B B

   =      + −   +    

         

   x      x    



[

0

]

I I

u K K   x

= −  

  x

Plugging into the eq

[ ]

[ ]

0 1

0

I I I

I

A B K x K B w x C r

y C C x

= + − − +

= −

= =    

 

x x x

x

x x





0 1

( ) ( ) ( ) ( )

( ) ( ) ( )

I I

I

sI A BK X s BK X s B w s sX s CX s R s

− + = − +

= −

I

( ) X s

( )

0 1

( sI A BK X s ) ( ) BK

_I

1 CX s ( ) R s ( ) B W S ( )

− + = − s − +

( )

( ) [ ]

( )

0 1

1

0 1

1 0

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

I I

I I

I I

s sI A BK BK C X s BK R s sB W s X s s sI A BK BK C BK R s sB W s Y s C s sI A BK BK C BK

R s

−

−

 − + +  = +

 

 

=  − + +  +

 

=  − + + 

[ ] { (

⁰

) }

¹

1 1

( ) ( ) ( ) 1 ( )

I I I

X s CX s R s C s sI A BK BK C BK R s

s s



−



= − =   − + + −  

D.C. gain =1 for any KI, K0

r − K

_I

u Plant

K

o

− e

+ y

−

+

1 s

+

1 s

− e

r + − K

_I

K

o

−

Plant

Estimator +

+ y

Integral Control

Integral Control with Estimator

End of Lecture Note 11