adaptive control theory: model reference adaptive systems
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Contents
Problem 1 ............................................................................................................................ 2Problem 2 ..........................................................................................................................10
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This is the perfect model following condition.
If two parameters are chosen to be 1 = 10 =
2 =
20 =
Then the input-output relations of the system and the model are the same.
The loss function:() = 12 2()The MIT rule: () = () () 1 () =
() ()2 () = () ()Substitute the control law into the plant model; we obtain the closed-loop dynamic system:
() = () + 1() 2() = ( + 2)() + 11()Define = ()Then, the closed-loop system can be rewritten as:
+ ( + 2)() = 1() () = 1 + ( + 2)()Similarly, for the reference model, we obtain:() = + ()The error between the output y of the closed-loop system and the output ym of the reference
model is
() = () () = 1 + ( + 2)() + () ()1 = + ( + 2)()
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()2 = 1+ ( + 2)2 () = + ( + 2) 1 + ( + 2)()
=
+ (
+
2)()
These formulas cannot be used directly because the process parameters a and b are not
known. Approximations are required. One possible approximation is based on the
observation that: + + 20 = + , when the parameters give perfect model-following. We will therefore use the approximation:
+ + 2 + Hence, the sensitivity derivatives are:
()1 + () = + ()
()2 + () = + ()
Note: + is a low-pass filter, with DC gain = 1.Substitute the sensitivity derivatives into the adjusted law, and then we get the following
equation for updating the controller parameters:
1 () = ()()1 = () + ()= + ()() = + ()()
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2 () = ()()2 = () + () = + ()()=
+
(
)
(
)
Where, = is a design variable. > 0 > 0 > 0 > 0 can be chosen by simulation.Summary:
Plant model transfer function: () = + = 0.65+1.25Reference model transfer function: () = + = 2+2Equations for updating the controller parameters:
1(
) =
+ (
)
(
) =
2
+ 2(
)
(
)
2 () = + ()() = 2 + 2()()Control law:
(
) =
1(
)
2
(
)
Refer to the part 2 for more information.
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Figure 2 Simulation the closed-loop system with some gamma values
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3.Present the results and plot the relation between the two controller parameters. From theresults discuss on the limitation of the gradient method.
Figure 3 Controller parameters theta1 and theta2.
From Fig. 3, we see that the parameters will converge to the true values with increasing
time. The convergence rate increases with increasing gamma.
The parameter estimates are related to each other in a very special way, even they are quite
far from their true values. This is illustrated in Fig. 4, this figure shows that parameters doindeed approach their correct values as time increases. The parameter estimates quickly
approach the line theta2 = theta1 a/b. This line represents parameter values such that the
closed-loop system has correct steady-state gain.
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Figure 4 Relation between controller parameters theta1 and theta2, with gamma = 1, in
increase time to 10 periods (200s).
Discussion on the limitation of the gradient method (MIT rule): look at Fig. 2 and Fig. 5,
we can see that when we try to increase the amplitude of the command input, so the
performances are very different. When gamma increases, the controller parameters
converge quickly but the system becomes unstable. It can be seen clearly when the
amplitude of the command increases. The system will be stable when gamma is small, but
the rate of convergence is very slow. So, there is no guarantee that an adaptive controller
based the gradient method will give a stable closed-loop system.
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Figure 5 ref. output and process output when increase the amplitude of the command signal
to 3 times. With the same gamma = 5.
Problem 2
Consider the plant in P1.
1.Design a model reference adaptive controller for tracking control using the normalizedMIT rule. Parameter values for the reference model are: am = bm = 2.
The design procedure is the same as in Problem 1. A little different is now the equations
for updating the controller parameters change to:
1() = 1() + (12 + 22)
2() = 2(
)
+ (12 + 22)Where,
1 = 1 = + ()
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2 = 2 = + ()2.Simulate the closed-loop system and compare the performance when = 0.001, = 1.
The command is a square wave with amplitude of (0.1, 1, 10) and period of 20 seconds.You can try different .
Figure 6 Block diagram for simulating the normalized MIT rule
ym(t)
5
Out5
4
Out4
3
Out32
Out2
1
Out1
MATLAB
Function
uc(t)
-b
s+am
b
s+am
gamma
s
gamma
s
Theta2
Theta1
bm
s+am
Reference Model
Plant Input
b
s+a
Plant
Output
f(u)Fcn1
f(u)Fcn
Clock y(t)
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Figure 7 Simulation the closed-loop system with some gamma values, and the amplitude of
the command is 0.1.
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Figure 8 Simulation the closed-loop system with some gamma values, and the amplitude of
the command is 1.
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Figure 9 Simulation the closed-loop system with some gamma values, and the amplitude of
the command is 10.
3.Compare the results with those of the MIT rule. In problem 1, we have seen that the choice of adaptation gain is crucial and that the value
chosen depends on the amplitude of the command signal. With gamma = 5 (even gamma =
2) and the amplitude of the command signal = 10, the adaptive system becomes maybe
unstable. But in problem 2, we use the normalized MIT rule, when we try with gamma = 2
and the amplitude of the command signal = 10, the performance of the closed-loop system
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is very good. So, in the normalized MIT rule, the choice of adaptation gain does not
depend on the amplitude of the command signal.