2695: ECE Adaptive Control (3 Credits, Fall 2008)
Lecture 6: Model Reference Adaptive Control (II)
October 6, 2008 Instructor: Zhi-Hong Mao Assistant Professor of ECE and Bioengineering University of Pittsburgh, Pittsburgh, PA
1
Outline
Announcements Homework problems Review of last lecture Examples of the MIT rule (contd) Stability of the MIT rule Lyapunov design (design based on Lyapunov stability theory)
2
Announcements
Midterm exam will be given on October 14 (Tuesday); October 13 (Monday) is Fall Break
Please pay attention to my lecture notes Please pay attention to the homework problems
Extra office hour session for this week
Friday 4:00 pm6:00 pm
Possible personal emergency
3
1
Homework problems
No homework problems for this week Past homework problems
Problem 1.10(b) Problem 0.3
4
Homework problems
No homework problems for this week
Past homework problems
1 Problem 4.1: Consider the plant P( s ) = s( s + a ) where a is an unknown parameter. Determine a controller that can give the closed-loop system 2 M ( s) = 2 . s + 2 s + 2
Problem 4.2: Control an integrator P ( s ) = b / s with a controller u = 1 r 2 yp . The desired response model is given by
M ( s ) = bm /( s + am ).
Derive the MIT rule for parameter update.
5
Review of last lecture
What is the MIT rule? Mixture of time-domain and frequency-domain notations
Exercise: Rewrite the following equation using timedomain notation only
b e0 = k m r. s + am
6
2
Review of last lecture
M
r
ym e0
P
yp
Question: Is this a linear system? Exercise: Write down the parameter-update equations for the above system.
7
Review of last lecture
Plant: Model:
dy p dt = ay p + bu
r
M
ym
e0
P
yp
dy m = am y m + bm r dt
Controller: u = 1 r 2 yp
e0 = y p ym e0 b = r 1 s + a + b 2 e0 b 21 b = r= yp ( s + a + b 2 )2 s + a + b 2 1
MIT rule:
s + a + b 2 s + a + b 2* = s + am
r e0 am d 2 = y p e0 dt s + am
8
a d1 = m dt s + am
Question: Why do we use am here?
Example 3
System dynamics
dc
d1
J + c + mgd c sin = d1T
T mg
(s) d1 = T ( s) Js 2 + cs + mgd c
(s)
T ( s) = 1.89 s 2 + 0.0389s + 10.77
9
3
M
ym
Example 3
Form of controller
r u
P
e0 yp
u = 1r 2 y p e0 = y p ym = Pu Mr
1.89 y p = Pu = 2 (1r 2 y p ) s + 0.0389s + 10.77 1.891 yp = 2 r s + 0.0389 s + 10.77 + 1.89 2
10
M
ym
Example 3
Gradients
e0 =
r u
P
e0 yp
1.891 r Mr s 2 + 0.0389 s + 10.77 + 1.89 2
e0 1.89 = r 1 s 2 + 0.0389 s + 10.77 + 1.89 2 e0 1.89 21 = 2 ( s 2 + 0.0389s + 10.77 + 1.89
yp =
2
)
2
r=
1.89 yp s 2 + 0.0389 s + 10.77 + 1.89 2
1.891 r s 2 + 0.0389 s + 10.77 + 1.89 2
11
Example 3
Approximation of the gradients
e0 1.89 = r 1 s 2 + 0.0389 s + 10.77 + 1.892 e0 1.89 = 2 yp 2 s + 0.0389 s + 10.77 + 1.89 2
yp = 1.891 r s 2 + 0.0389 s + 10.77 + 1.89 2
s 2 + 0.0389s + 10.77 + 1.89 2 s 2 + a1m s + a0 m e0 1.89 = r 1 s 2 + a1m s + a0 m e0 1.89 = 2 yp 2 s + a1m s + a0 m
12
4
M
ym
Example 3
r u
P
e0 yp
Adjustment mechanism
e d1 a0 m = ' 0 e0 = 2 r e0 dt 1 s + a1m s + a0 m d 2 e a0 m = ' 0 e0 = 2 y p e0 dt 2 s + a1m s + a0 m
13
Example model
bm 3
Reference s 2 + a1m s + a0m
ym
Plant
1.89 s 2 + 0.0389s + 10.77
_
r
1
+ _
yp
+
e0
2
s
Example 3
0
200
400
600
Remark: The plant almost goes unstable. This response is largely due to the near instability of the open loop system. Tuning of gamma and changing the reference model did not alleviate this 15 problem.
a0m s2 + a1m s + a0m
s
a0m s 2 + a1m s + a0 m
14
ym ym g=.0001 g=.0001
800
1000
1200
5
Example 3
Remark: For this specific example, introducing PD control may stabilize the system dynamics.
16
Example 3
17
Example 3
18
6
Stability of the MIT rule
k k , M ( s) = 0 P( s) = s +1 s +1 r = sin( t )
Remark: In this case, for small enough , the MIT rule is stable for all g, while for small enough g, it is stable for all . Otherwise, the dynamic behavior of the algorithm is quite complex, with interleaving of the stable and unstable regions.
19
Stability of the MIT rule
The case where
P( s ) and M ( s )
do not share the same set of poles and zeros
20
Lynapunov design
Example 4
Question: How to determine the update law for ?
21
7
Lynapunov design
Example 4
Question: Can we guarantee the convergence of e0? Can we guarantee the convergence of ?
22
Lynapunov design
Example 4
The MIT rule
Rule derived from Lyapunov stability theory
23
Lynapunov design
Example 5
Plant: dy p = a p y p + k pu dt Model:
r u
M
ym
P
yp
e0
dym = am ym + k m r dt
Controller:
u = c0 r + d 0 y p
Question: What are the nominal values of c0 and d0?
24
8
Lynapunov design
Example 5
The system
= a p y p + k pu dt dym = am y m + k m u dt u = c0 r + d 0 y p dy p
Parameters
e0 = y p ym r c c* = = 0 0* y d 0 d 0
* c0 =
a am km * , d0 = p kp kp
Exercise: Derive equation for e0.
25
Lynapunov design
Example 5
Lyapunov function
v ( e0 ,r , y ) =
e0 2 k p 2 + (r + y 2 ) 2 2g
Exercise: Desig...