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Sang–Chul Lee
DCASL
2009.03.11
 POLE
PLACEMENT
 STATE
OBSERVER

EFFECTS OF THE ADDITION OF THE OBSERVER ON
A CLOSED-LOOP SYSTEM

INTRODUCTION OF THE DISTURBANCE OBSERVER
Open-loop system
closed-loop control system with u = -Kx
(Regulator system)
What is the difference between two figures ??
x  Ax  Bu
y  Cx
Where
choose the Control signal ->
(1)
x = state vector (n-vector)
u = control vector (sclar)
A = n x n constant matrix
B = n x 1 constant matrix
C = 1 x n constant matrix
u  Kx
(2)
Substituting Eq (2) into Eq (1) gives
x(t )  ( A  BK ) x(t )
x(t )  e( A BK )t x(0)
If the eigenvalues of matrix (A-BK) are asymptotically
stable, x(t) approaches 0 as t approaches infinity
The problem of placing the closed-loop poles
at the desired location is called
a pole-placement problem
So, What is the pole-placement technique ??
Define T(transformation matrix) by
M is the controllability matrix
T  MW
M  B AB
A n1B 
And
Refer to the notes please. Page. 1
The
ai ' s
are coefficients of the characteristic polynomial.
Define a new state vector x̂ by
Eq (1) can be modified to
(Controllable canonical form)
Where
,
Refer to the notes please. Page.1
Let us choose a set of the desired eigenvalues as 1 , 2 ,
then the desired characteristic
( s  1 )( s  2 )
, n .
Eq. becomes
( s  n )  s n  1s n 1 
  n 1s   n  0 (3)
Let us write
when
is used to control system
The characteristic Eq. is
(4)
This characteristic Eq. is the same as the desired characteristic
Eq. for the system, when ‘u = -Kx’ is used as the control signal.
We can rewrite the Eq. (4) as
 s n  (a1  1 ) s n 1 
 (an 1   n 1 ) s  (an   n )  0
This is the characteristic Eq. for the system with
state feedback. Therefore, it must be equal to Eq.(3)
So we get
And then, we obtain
 What
is the STATE OBSERVER ??
- A state observer estimates the state variables
based on the measurements of the output
and control variables.
 Why
we use the STATE OBSERVER ??
- Not all state variables are available for feedback!
x  Ax  Bu
(5)
y  Cx
x  Ax  Bu  K e ( y  Cx)
 ( A  K eC) x  Bu  K e y (6)
Model of the state observer
Block diagram of the system and full-order state observer
Subtract Eq.(6) from Eq.(5). So we can obtain the observer error Eq.
x  x  Ax  Ax  K e (Cx  Cx)
 ( A  K eC)( x  x)
Let’s Define
Then Eq.(7)becomes
(7)
e  xx
e  (A  K eC)e
The dynamic behavior of the error vector is determined by
the eigenvalues of matrix (A  K eC) .
thus the problem here becomes the same as the pole-placement problem.
It’s very similar to the pole-placement technique
Refer to the notes please.
Consider the system defined by
x  Ax  Bu
y  Cx
u  Kx
And, the observer Eq. is
The laplace transform of u is
x  ( A  K eC) x  Bu  K e y
U (s)  KX(s)
(8)
The laplace transform of observer Eq. is
sX( s )  ( A  K eC) X ( s)  BU ( s)  K eY ( s )
If x(0) = 0, then we can obtain
X( s)  (Is  A  K eC  BK ) 1 K eY ( s)
By substituting this Eq. into (8) , we get
U ( s)  K (Is  A  K eC  BK ) 1 K eY ( s)
Block diagram representation of system
U ( s)  K (Is  A  K eC  BK ) 1 K eY ( s)
Note that the transfer function acts as a controller for the system.
The application of disturbance observer to practice
J : inertia
Kt : torque coefficient of electric motor
Tl : load torque
When the disturbance ‘Tl’ is generated ??
How can we use the estimated value ‘Tdis’ ??
Disturbance observer in motion control
(designed by Gopinath’s method)
Thank you!