Transcript Model Reference Adaptive Control Survey of Control Systems (MEM 800) Presented by Keith Sevcik.

Model Reference
Adaptive Control
Survey of Control Systems (MEM 800)
Presented by
Keith Sevcik
Concept
Model
ymodel
Adjustment
Mechanism
Controller Parameters
uc
Controller


u
Plant
yplant
Design controller to drive plant response to mimic ideal
response (error = yplant-ymodel => 0)
Designer chooses: reference model, controller structure,
and tuning gains for adjustment mechanism
MIT Rule
Tracking error:
e  y plant  ymodel

Form cost function:

Update rule:
1 2
J ( )  e ( )
2
d
J
e
 
 e
dt



sensitivity
derivative
– Change in  is proportional to negative gradient of J
MIT Rule


Can chose different cost functions
EX:
J ( )  e( )
d
e
 
sign (e)
dt

 1, e  0

where sign (e)   0, e  0
 1, e  0


From cost function and MIT rule, control law can be
formed
MIT Rule

EX: Adaptation of feedforward gain
Reference Model
Gm ( s)  koG( s)
ymodel
Adjustment Mechanism
-

s
θ
Π
+
Plant
uc
Π
u
G p ( s)  k G( s)
yplant
MIT Rule
Y ( s)
 kG( s) where k is unknown
 For system
U ( s)
Y (s)
 koG ( s)
 Goal: Make it look like
U c ( s)
using plant Gm (s)  koG(s) (note, plant model
is scalar multiplied by plant)
MIT Rule

Choose cost function:

Write equation for error:
1 2
d
e
J ( )  e ( ) 

 e
2
dt

e  y  ym  kGU  GmU c  kGU c  koG U c


Calculate sensitivity derivative:
Apply MIT rule:
e
k
 kGUc 
ym

ko
d
k
   ' ym e   ym e
dt
ko
MIT Rule

Gives block diagram:
Reference Model
Gm ( s)  koG( s)
ymodel
Adjustment Mechanism
-

s
θ
Π
+
Plant
uc


Π
u
G p ( s)  k G( s)
yplant
considered tuning parameter
MIT Rule

NOTE: MIT rule does not guarantee error
convergence or stability

 usually kept small

Tuning  crucial to adaptation rate and
stability.
MRAC of Pendulum

System
J  c  mgdc sin   d1  T
 (s)
d1
 2
T ( s) Js  cs  mgdc
d2
dc
d1
T
 ( s)
1.89
 2
T ( s) s  0.0389s  10.77
MRAC of Pendulum

Controller will take form:
Model
ymodel
Adjustment
Mechanism
Controller Parameters
uc
Controller
u
1.89
2
s  0.0389s  10.77
yplant
MRAC of Pendulum

Following process as before, write
equation for error, cost function, and
update rule:
e  y plant  ymodel
1 2
J ( )  e ( )
2
d
J
e
 
 e
dt


sensitivity
derivative
MRAC of Pendulum

Assuming controller takes the form:
u  1uc   2 y plant
e  y plant  ymodel  G p u  Gmuc
1.89


y plant  G p u   2
1uc   2 y plant 
 s  0.0389s  10.77 
1.891
y plant  2
uc
s  0.0389s  10.77  1.89 2
MRAC of Pendulum
1.891
e 2
uc  Gmuc
s  0.0389 s  10.77  1.89 2
e
1.89
 2
uc
1 s  0.0389 s  10.77  1.89 2
e
1.89 2 1

uc
2
2
 2
s  0.0389 s  10.77  1.89 2


1.891
 2
y plant
s  0.0389 s  10.77  1.89 2
MRAC of Pendulum

If reference model is close to plant, can
approximate:
s  0.0389s  10.77  1.89 2  s  a1m s  a0 m
2
2
a1m s  a0 m
e
 2
uc
1 s  a1m s  a0 m
a1m s  a0 m
e
 2
y plant
 2
s  a1m s  a0 m
MRAC of Pendulum

From MIT rule, update rules are then:
 a1m s  a0 m

d1
e
 
e    2
uc e
dt
1
 s  a1m s  a0 m 
 a1m s  a0 m

d 2
e
 
e    2
y plant e
dt
 2
 s  a1m s  a0 m

MRAC of Pendulum

Block Diagram
Reference Model
bm
s  a1m s  a0 m
ymodel
2
-
uc
Π
θ1
+
1.89
s 2  0.0389s  10.77
-
a1m s  a0 m
s  a1m s  a0 m
2
Π
θ2

s
Π
+
e
Plant
Π

s
yplant
a1m s  a0 m
s  a1m s  a0 m
2
MRAC of Pendulum

Simulation block diagram (NOTE: Modeled
to reflect control of DC motor)
omega^2
s+am
ym
Reference Model
Error
Saturation
35
4.41
2/26
Step
s2 +.039s+10.77
Degrees
Degrees
to Volts
Plant
180/pi
y
Radians
to Degrees
-gamma
s
Theta1
gamma
Theta2
s
am
s+am
am
s+am
MRAC of Pendulum

Simulation with small gamma = UNSTABLE!
150
ym
g=.0001
100
50
0
-50
-100
0
200
400
600
800
1000
1200
MRAC of Pendulum

Solution: Add PD feedback
omega^2
s+am
ym
Reference Model
1.5
du/dt
Error
D
Saturation
35
1
4.41
2/26
Step
Degrees
s2 +.039s+10.77
P
Degrees
to Volts
Plant
180/pi
y
Radians
to Degrees
-gamma
s
Theta1
gamma
Theta2
s
am
s+am
am
s+am
MRAC of Pendulum

Simulation results with varying gammas
45
ym
g=.01
g=.001
g=.0001
40
3.56
ym  2
s  2.67 s  3.56
35
30
25
Designed such that :
Ts  3 sec
  .707
20
15
10
5
0
0
500
1000
1500
2000
2500
LabVIEW VI Front Panel
LabVIEW VI Back Panel
Experimental Results
Experimental Results
PD feedback necessary to stabilize system
 Deadzone necessary to prevent updating
when plant approached model
 Often went unstable (attributed to
inherent instability in system i.e. little
damping)
 Much tuning to get acceptable response

Conclusions
Given controller does not perform well enough
for practical use
 More advanced controllers could be formed from
other methods

– Modified (normalized) MIT
– Lyapunov direct and indirect
– Discrete modeling using Euler operator

Modified MRAC methods
– Fuzzy-MRAC
– Variable Structure MRAC (VS-MRAC)