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Apply Lyaponove function
to design the observer
Advisor: Ming-Shyan Wang
Presenter: Hanh Nguyen Thi
1
Contents
Principle of method
Apply for servo motor
Simulation results
2
Principle of method
Theorem : Lyaponove function
x  f x 
V x   0
V x   0x  0
(1)
(2)
If (2) is satisfied, then (1)
is asymtotically stable
system
3
Principle of method
 The system is affected by the unknown disturbance.
 x  A x  bu    

T
y  c x
 Unknown disturbance
 Given value
z  Sx
u  u   
~
1 ~
A  SAS , b  S b
~
~
z  A z  b u 
Canonical control form
4
Principle of method
State feedback controller for canonical control form
w
u
~ ~
z  A z  b u
z
R
Block diagram of State feedback controller
u  w  R z
u  u   
u  w  Rz 
State feedback
controller
 is the unknown disturbance
̂ Is the known disturbance
ˆ
u  w  Rz 
ˆ 
u  w  R z   
~
~
z  A z  b u 


~ ~
ˆ 
z  A z  b w  R z   
~
ˆ
 Am z  b w     
~ ~
( Am  A  b R )


~
z m  Am z m  b w
5

Principle of method
Method:
~
ˆ
  
Lyaponove function:
~
~ ~
V (e, )  e T Pe   T 
e  z  zm
~ ~
de

 Am e  b 
dt
6
Principle of method
~ T
~


dV  d e 
d
e
d

d

~
~
T
   T

   Pe  e P
 
dt  dt 
dt  dt 
dt
T
~
d

~
~
~
T
 A m e  b T (x)(t ) Pe  e P A m e  b T (x)(t )  2T
dt
~


d

~
~
T
T
T
T


 e ( A m P  PA m )e  2  b Pe 



dt 

Q
~


d

~
T

 e Qe  2T  bT Pe 

dt


~
 H


T


7
Principle of method
~
 d
~


H

 bT Pe

ˆ
d
d
 dt
T
ˆ


H



H



b
Pe

ˆ
dt
dt
 d  d  H (  
ˆ )  bT Pe
 dt
dt
Choose:
Satisfy:
ˆ
d
ˆ  b T P K e  K edt  K d e 
  H
p
i
d
dt
dt 

T
Ki

b
Pe
ˆ

(K p 
 K d s)
sH
s
dV
~T ~
T
 e Qe  2 H  0, e  0
dt
8
Apply for servo motor
 :angular position
 :speed of motor.
 d :disturbance torque
Mechanical model:
 d
 dt  
 d
Bm
1
1


  d  e
 dt
Jm
Jm
Jm
Bm
x1   , x2   ,   
Jm

 e :motor torque.
J m :moment of inertia.
B m :the friction coefficient.
 :unmodelled uncertainty
 ,  :parameter variations
1
, u   e ,        d  
Jm
 dx1
 dt  x 2

 dx 2  x   (u  )
2
 dt
z
1

x
 dz1
 dt  z 2

 dz 2  x  (u  )
2
 dt
Canonical
control form
9
Apply for servo motor
Controller:


s
ˆ
u  w  Rz  

x

R
x

ˆ

 1
ˆ
u 
 r1  r2   
 
ˆ 

bT Pe
(K p 
1   1  r1

ˆ
u 
   r2   
 s s

Ki
  p3e1  p4 e2 
K
 K d s) 
( K p  i  K d s)
s
sh
s
sH
 1

  p3  p4 e2
s
 ( K  K i  K s)
 
p
d
sh
s
Observer
10
Apply for servo motor
 Determine A matrix: use pole placement method.
 0
Am  
  a0
a0  s1s2
1 
 a1 
a1  s1  s2 
 Determine controller parameter:
1 
 0
~ ~
Am  A  b R  


a

a
1
 0
~ 0 1  ~  0 
A
, b  1, R  r1 r2 
0



 
 r1  a0 , r2  a1  
11
Apply for servo motor
Determine Q matrix:
dV
~ ~
T
 e Qe  2T H
dt
 q1 q2   e1 
~2
 e1 e2 

2
H

 
q2 q4  e2 
~
  q1e12  q4 e22  2q2 e1e2  2 H 2


dV/dt the larger, the better.
Choose q2=0 and q1,q4, H big enough
12
Apply for servo motor
Determine P matrix:
1 
a12   0
a
Am   11


 
a21 a22   a0  a1 
 p p2 
 q1 q2 
P 1
,
Q


q q 
p
p
4
4
 3
 3
Am P  PAm  Q
T
 a11
a
 12
a21   p1
a22   p3
 p3 
p4 
p2   p1

p4   p3
p2   a11 a12 
 q1
 
p4  a21 a22 
q3
q2 
q4 
q1
2a0
q1
q
 4
2a0 a1 2a1
13
Simulation
1
s
1  r1

  r2 
s

u

1

*
1
s 
x2

 1

 p3  p4 
 s

1
~
s  a1 s  a~0
z2
1
s h
PID
̂
2
Block diagram for the disturbance observer
14
Simulation
Parametter for simulation:
Bm=8e-5;
Jm=6.45e-5;
h=500;
s1=-50;
%the pole
s2=-400;
%the pole
q1=0.5;
%the element of Q matrix
q4=2;
%the element of Q matrix
Kp=100;
% Kp,Ki,Kd of PID (disturbance)
Ki=1500;
Kd=2
15
Simulation
Disturbance profile
16
Simulation
Speed response at 2 rpm
17
Thank you for your attention !
18
Model of canonical control form :
 x  A x  bu

T
y

c
x  du

 0
 0

A 

 0
 a0
1
0
0
1


0
0
 a1  a2

0 
0 
0 

0 
 

 , b    

 

1 
0 
1
  an 1 
19