Adaptive control theory: Pole-Placement and Indirect STR

7/31/2019 Adaptive control theory: Pole-Placement and Indirect STR

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NATIONAL CHENG KUNG UNIVERSITY

Department of Mechanical Engineering

ADAPTIVE CONTROL

HOMEWORK 5

Instructor: Ming Shaung Ju

Student: Nguyen Van Thanh

Student ID: P96007019Department: Inst. of Manufacturing & Information Systems

Class: 1001- N164400 - Adaptive Control

November 25, 2011
http://class-qry.acad.ncku.edu.tw/qry/classmate.php?syear=0100&sem=1&co_no=N164400&class_code=http://class-qry.acad.ncku.edu.tw/qry/classmate.php?syear=0100&sem=1&co_no=N164400&class_code=


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Adaptive Control Theory HW5 Page 1

Contents

Problem 1 ...................................................................................................................... 2

Problem 2 .................................................................................................................... 13

Problem 3 .................................................................................................................... 37


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Problem 1

Consider a process

Transfer function of the process:

( ) = 0+ 12+ 1+ 2=( ) ( )

Transfer function of the desired closed-loop system:

( ) = 02+ 1+ 2=( )

( )

Where,

0= 0.17611= 1.32052= 0.4966

A general linear controller:

( ) = ( ) ( )

Our job is to define three polynomials: R, S and T.

(a) Design a controller in which the stable zero of process model is canceled.

Step 1 :


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We can see that

( ) = + 100 + 10= 0 = 10= 0.081180.1172 = 0.6927 | | = 10< 1 = += + 100

+= + 10

=

0

deg ( ) = 2, deg ( ) = 1 Let deg ( ) = 2 deg ( ) 1 = 22 1 = 3 deg ( ) = deg ( ) deg ( ) = 3 2 = 1

deg ( )

deg ( ) ,deg ( ) = deg (

) 1

deg ( ) = 1

deg ( 0) = deg ( ) deg ( +) 1 = 2 1 1 = 0 Choose 0( ) = 1 = +,deg ( ) = deg ( ) = 1 deg ( ) = 0 = 1

( ) = +( ) = + 1

0= + 0.6927

Step 2 :

The Diophantine equation:

+ = ++ += 0 + + = 0


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+ 0= ( 2+ 1+ 2) + 0( 0+ 1) = ( 2+ 1+ 2)

2+ (

1+

00) +

2+

01=

2+

1+

2

By comparing coefficients of this equation, we obtain

0= 110 = 1.06911= 220 = 1.4556

Step 3 :

= + 0 += = 0 = 0= 000 = 00

=

0 = 1

00

= 1.5026

The controller:

( ) = ( ) ( )

is thus characterized by the polynomials:

( ) = + 0.6927

( ) = 1.0691 1.4556 ( ) = 1.5026 Plug these polynomials into the controller:

( ) = ( ) ( )


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( + 0.6927) ( ) = 1.5026 ( ) (1.0691 1.4556) ( ) (1 + 0.6927 1) ( ) = 1.5026 ( ) (1.0691 1.4556 1) ( )

( ) + 0.6927 (

1) = 1.5026 ( ) 1.0691 ( ) + 1.4556 (

1)

( ) = 0.6927 ( 1) + 1.5026 ( ) 1.0691 ( ) + 1.4556 ( 1) (b) Design a controller in which no zero of process model is canceled. Choose a properdelay d0.

Step 1 :

No zero of process model is canceled

= += 1

deg(

+)

= 0 = = 0+ 1 = = = (deg ( ) = deg ( ) = 1)


( ) =

0+

12+ 1+ 2=

( )

( )

Parameter is chosen such that ( = 1 ) = 1 = 1+++ = 0.8877 deg ( ) = 2, deg ( ) = 1

Let deg ( ) = 2 deg ( ) 1 = 22 1 = 3

deg ( ) = deg (

) deg (

) = 3 2 = 1

deg ( ) deg ( ) ,deg ( ) = deg ( ) 1 deg ( ) = 1 deg ( 0) = deg ( ) deg ( +) 1 = 2 0 1 = 1

Choose 0( ) = + 0( | 0| < 1)


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Poles of the closed-loop system:

( ) = 0 2+ 1+ 2= 0 1= 0.6603 + 0.24632= 0.6603 0.2463

| 1| = | 2| = 0.7047 0.7047 | 0| < 1 Again, 0( ) = + 0(0.7047 | 0| < 1)

= += 1 = = + 1( ) =

0+

1

Step 2 :


+ = ++ += 0 + + = 0 ( 2+ 1+ 2)( + 1) + ( 0+ 1)( 0+ 1) = ( + 0)( 2+ 1+ 2)

3+ ( 1+ 1+ 00) 2+ ( 11+ 2+ 01+ 10) + ( 21+ 11)= 3+ ( 0+ 1) 2+ ( 2+ 10) + 20


1+ 1+ 00= 0+ 111+ 2+ 01+ 10= 2+ 1021

+

11=

20

1+ 00= 0 1+ 111+ 10+ 01= 2+ 2+ 1021+ 11= 20


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1= 0.8802 0+ 0.08300= 1.0222 0+ 0.36111= 1.1169 00.68200.7047 | 0| < 1

Step 3 :

( ) = 0= ( + 0) = 0.8877 + 0.8877 0(0.7047 | 0| < 1) The controller:

( ) = ( ) ( )


( ) = + 1 ( ) = 0+ 1

( ) = ( + 0) = + 0 Where

1= 0.8802 0+ 0.08300= 1.0222 0+ 0.36111= 1.1169 00.68200.7047 | 0| < 1= 0.8877 Plug these polynomials into the controller:

( ) = ( ) ( )

( + 1) ( ) = ( + 0) ( ) ( 0+ 1) ( ) (1 + 11) ( ) = ( + 01) ( ) ( 0+ 11) ( ) ( ) + 1( 1) = ( ) + 0( 1) 0( ) 1( 1)


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( ) = 1( 1) + ( ) + 0( 1) 0( ) 1( 1) (c) Simulate above designs when command is a square wave of amplitude 1 and period

of 25 sampling periods. Test the effect of step disturbance on the tracking

performance. You may try step with different magnitude. ( for simulating let a 0 = 0.8 )

Figure 1. Block diagram of sys with stable cancellation

Figure 2. Command & output, with stable zero cancellation

1

Out1

t0.z

z+r1

T/R

s0.z+s1

z+r1

S/R

Disturbance

MATLABFunction

CommandCmd & OptClock

b0.z+b1

z +a1.z+a22

B/A


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Figure 3. Command & output, with stable zero cancellation, and step disturbance

Figure 4. Command & output, with no zero cancellation


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Figure 5. Command & output, with no zero cancellation, and step disturbance

(d) Compare the tracking performance, ringing of input and disturbance rejection of the

two designs. Discuss on the effect of time delay on the performance of the adaptive

controller. Plot y(t), u c(t) vs. t and u(t), e(t) vs. t.

Plot y(t), u c(t): we can see through Fig. 2 to Fig. 5. Two controllers have very same

performance (tracking and disturbance).

Discuss on the effect of time delay on the performance of the adaptive controller: we

should choose the time delay as small as possible. When the time delay is large that

means the output signal lags phase to the command signal. Hence, the performance of

the system is not good.

From Fig. 6 and Fig.8, the control input signal with stable zero cancellation design has a

severe oscillation (ringing) compare to that of no zero cancellation design


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Figure 6. Control input signal with stable zero cancellation design

Figure 7. Error between u c(t) and y(t), with stable zero cancellation design


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Figure 8. Control input signal with no zero cancellation design

Figure 9. Error between u c(t) and y(t), with no zero cancellation design


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Problem 2

Consider a process

(a) Design an indirect STR in which the stable zero of process model is canceled.

Transfer function of the process:

( ) = 0+ 12+ 1+ 2=( ) ( )

Where, parameters {a1, a2, b0, b1} are unknown constants.


( ) = 02+ 1+ 2=( ) ( )

Where,

0= 0.17611= 1.32052

= 0.4966

A general linear controller:

( ) = ( ) ( )


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Step 1 : Estimate the coefficients of the polynomials A and B i.e. {a1, a2, b0, b1}, by

using the RLS:

( ) = 1(

1) 2(

2) + 0(

1) + 1(

2) = (

1)

( 1) = [ ( 1) ( 2) ( 1) ( 2) ] = [ 1 2 0 1] The RLS with exponential forgetting is given by

( ) = ( 1) + ( ) ( ) ( ) = ( ) (

1) (

1)

( ) = ( 1) ( 1) + ( 1) ( 1) ( 1) 1 ( ) = ( ( ) ( 1)) ( 1)/

Step 2 :

With the stable zero of process model is canceled (this step for only define the degree of

the polynomials)

( ) = + 100 + 10= 0 = 10 | | = 10< 1 | 1| < | 0| = += + 100 += + 10= 0


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deg ( ) = 2, deg ( ) = 1 Let deg ( ) = 2 deg ( ) 1 = 22 1 = 3

deg ( ) = deg (

) deg (

) = 3 2 = 1

deg ( ) deg ( ) ,deg ( ) = deg ( ) 1 deg ( ) = 1 deg ( 0) = deg ( ) deg ( +) 1 = 2 1 1 = 0 Choose 0( ) = 1

= +,deg ( ) = deg ( ) = 1 deg ( ) = 0 = 1 ( ) = +( ) = + 10= + 1


+ = ++ += 0 + + = 0

+ 0=

( 2+ 1+ 2) + 0( 0+ 1) = ( 2+ 1+ 2) 2+ ( 1+ 00) + 2+ 01= 2+ 1+ 2 By comparing coefficients of this equation, we obtain

0= 1101= 220

= + 0 += = 0


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= 0= 000 = 00 =

0 = 1

00

=

0

( ) = +( ) = + 10= + 1 Step 3 : Calculate the control law

( ) = ( ) ( )

( ) +

1(

1) =

0( )

0( )

1(

1)

( ) = 1( 1) + 0( ) 0( ) 1( 1) Repeat steps 1, 2 and 3 at each sampling period.

(b) Design an indirect STR in which no zero of process model is canceled. Choose a

proper delay d 0.

Step 1 : Estimate the coefficients of the polynomials A and B i.e. {a1, a2, b0, b1}, by

using the RLS:

( ) = 1( 1) 2( 2) + 0( 1) + 1( 2) = ( 1) ( 1) = [ ( 1) ( 2) ( 1) ( 2) ]

= [

1 2 0 1]

The RLS with exponential forgetting is given by

( ) = ( 1) + ( ) ( ) ( ) = ( ) ( 1) ( 1)


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( ) = ( 1) ( 1) + ( 1) ( 1) ( 1) 1 ( ) = ( ( ) ( 1)) ( 1)/

Step 2 :

No zero of process model is canceled = += 1deg ( +) = 0 = = 0+ 1 = = = (deg ( ) = deg ( ) = 1)


( ) = 0+ 12+ 1+ 2=( ) ( )

Parameter is chosen such that ( = 1 ) = 1 = 1+++ deg ( ) = 2, deg ( ) = 1

Let deg (

) = 2 deg (

) 1 = 2

2 1 = 3

deg ( ) = deg ( ) deg ( ) = 3 2 = 1 deg ( ) deg ( ) ,deg ( ) = deg ( ) 1 deg ( ) = 1

deg ( 0) = deg ( ) deg ( +) 1 = 2 0 1 = 1 Choose

0( ) = +

0( |

0| < 1)

Poles of the closed-loop system:

( ) = 0 2+ 1+ 2= 0 1= 0.6603 + 0.24632= 0.6603 0.2463 | 1| = | 2| = 0.7047


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0.7047 | 0| < 1 Again, 0( ) = + 0(0.7047 | 0| < 1)

=

+= 1

= = +

1( )

= 0+ 1 The Diophantine equation:

+ = ++ += 0 + + = 0 ( 2+ 1+ 2)( + 1) + ( 0+ 1)( 0+ 1) = ( + 0)( 2+ 1+ 2)

3+ ( 1+ 1+ 00) 2+ ( 11+ 2+ 01+ 10) + ( 21+ 11)= 3+ ( 0+ 1) 2+ ( 2+ 10) + 20


1+ 1+ 00= 0+ 111+ 2+ 01+ 10= 2+ 1021

+

11=

20

1+ 00= 0 1+ 111+ 10+ 01= 2+ 2+ 1021+ 11= 20 1010.7047

| 0| < 1

( ) = 0= ( + 0) (0.7047 | 0| < 1) Step 3 :The controller:

( ) = ( ) ( )


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( ) = + 1 ( ) =

0+

1

( ) = ( + 0) = + 0 Plug these polynomials into the controller:

( ) = ( ) ( )

( + 1) ( ) = ( + 0) ( ) ( 0+ 1) ( )

(1 + 11) ( ) = ( + 01) ( ) ( 0+ 11) ( ) ( ) + 1( 1) = ( ) + 0( 1) 0( ) 1( 1) ( ) = 1( 1) + ( ) + 0( 1) 0( ) 1( 1) Repeat step 1, 2 and 3 at each sampling period.

(c) Simulate above designs using following process parameters [a1, a2, b0, b1] = [-

1.3735, 0.7006, 0.1231, 0.07712] and initial guess of above parameter vector [0, 0,

0.01, 0.2]. The command is a square wave of amplitude 1 and period of 25 sampling

periods. You may test the effect of disturbance on the tracking performance. Either

stochastic or deterministic disturbance can be used.

c.1. Stable zero cancellation:

From the true model, we can find the true R, S and T.

0= 110 = 0.43051= 220 = 1.6572


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= 0 = 100 = 0= 1.4305 ( ) = +( ) = + 1

0= +

1= + 0.6265

( ) = 1( 1) + 0( ) 0( ) 1( 1) ( ) = 0.6265 ( 1) + 1.4305 ( ) 0.4305 ( ) + 1.6572 ( 1) A function is written in Matlab can solve this problem:

function [estmtd_theta,yhat,ue,e,re1,se0,se1,te0] =

adtvCtrolHw5_STR1(uc,u,y,lambda)

am1 = -1.3205; am2 = 0.4966;

bm0 = 0.1761;

numOfCoeffs = 4; % number of coefficents

alpha = 100;

t = length(y);

j=1;

phi_1(:,j)=[0;u(1:t-1,j)];

phi_2(:,j)=[0;0;u(1:t-2,j)];

phi_3(:,j)=[0;-y(1:t-1,j)];

phi_4(:,j)=[0;0;-y(1:t-2,j)];

phi = [phi_1(:,j) phi_2(:,j) phi_3(:,j) phi_4(:,j)];

[M,N] = size(phi);

P = alpha*eye(N); % initial condition for projection matrix

estmtd_theta(1,:) = [0.01 0.2 0 0]; % inital condition for [b0 b1 a1

a2]

yhat = zeros(1,t); % estimated value

re1 = zeros(1,N);


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se0 = zeros(1,N);

se1 = zeros(1,N);

te0 = zeros(1,N);

% initial conditions for three polynomials

re1(1) = estmtd_theta(1,2)/estmtd_theta(1,1);

se0(1) = (am1 - estmtd_theta(1,3))/estmtd_theta(1,1);

se1(1) = (am2 - estmtd_theta(1,4))/estmtd_theta(1,1);

te0(1) = bm0/estmtd_theta(1,1);

e = zeros(1,t);

ue(1) = 0;

for i = 2:M-numOfCoeffs+1;

z = P*phi(i,:)';

K = z/(lambda + phi(i,:)*z);

yhat(i) = phi(i,:)*estmtd_theta(i-1,:).';

e(i) = y(i) - phi(i,:)*estmtd_theta(i-1,:).';

estmtd_theta(i,:) = estmtd_theta(i - 1,:) + e(i)*K.';

P = (P - K*z.')/lambda;

re1(i) = estmtd_theta(i,2)/estmtd_theta(i,1);

se0(i) = (am1 - estmtd_theta(i,3))/estmtd_theta(i,1); se1(i) = (am2 - estmtd_theta(i,4))/estmtd_theta(i,1);

te0(i) = bm0/estmtd_theta(i,1);

ue(i) = - re1(i)*ue(i-1) + te0(i)*uc(i) - se0(i)*yhat(i) -

se1(i)*yhat(i-1);

end

end


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Figure 10. Output

Figure 11. Error between process output and estimated output


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Figure 12. Output with step disturbance



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c.2. No zero cancellation:

0( ) = + 0(0.7047 | 0| < 1) 1

+

1+

00=

0+

111+ 2+ 01+ 10= 2+ 1021+ 11= 20

1+ 00= 0 1+ 111+ 10+ 01= 2+ 2+ 1021+ 11= 20 ( ) = 0= ( + 0) ( = 0.8795, 0.7047 | 0| < 1)

Controller:

( ) = ( ) ( )

( ) = 1( 1) + ( ) + 0( 1) 0( ) 1( 1) A function in Matlab can solve this problem:

function [estmtd_theta,yhat,ue,e,re1,se0,se1,te0] =adtvCtrolHw5_STR2(uc,u,y,lambda)

numOfCoeffs = 4; % number of coefficents

beta = 0.8795;

a0 = 0.8;

am1 = -1.3205; am2 = 0.4966;

t = length(y);

j=1;

phi_1(:,j)=[0;u(1:t-1,j)];

phi_2(:,j)=[0;0;u(1:t-2,j)];

phi_3(:,j)=[0;-y(1:t-1,j)];

phi_4(:,j)=[0;0;-y(1:t-2,j)];


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phi = [phi_1(:,j) phi_2(:,j) phi_3(:,j) phi_4(:,j)];

[M,N] = size(phi);

alpha = 100;

P = alpha*eye(N); % initial condition for projection matrix estmtd_theta(1,:) = [0.01 0.2 0 0]; % inital condition for [b0 b1 a1

a2]

yhat = zeros(1,t); % estimated value

re1 = zeros(1,N);

se0 = zeros(1,N);

se1 = zeros(1,N);

te0 = zeros(1,N);

te1 = zeros(1,N);

% initial conditions for three polynomials

re1(1) = estmtd_theta(1,2)/estmtd_theta(1,1) - ((estmtd_theta(1,2) -

a0*estmtd_theta(1,1))*(am2*estmtd_theta(1,1)^2 -

am1*estmtd_theta(1,1)*estmtd_theta(1,2) + estmtd_theta(1,2)^2))/ ...

(estmtd_theta(1,1)*(estmtd_theta(1,4)*estmtd_theta(1,1)^2 -

estmtd_theta(1,3)*estmtd_theta(1,1)*estmtd_theta(1,2) +

estmtd_theta(1,2)^2)); se0(1) = (estmtd_theta(1,1)*(a0*estmtd_theta(1,4) -

estmtd_theta(1,3)*estmtd_theta(1,4) - a0*am2 + estmtd_theta(1,4)*am1)

- estmtd_theta(1,2)* ...

(estmtd_theta(1,4) - am2 + a0*estmtd_theta(1,3) - a0*am1 +

estmtd_theta(1,3)*am1 -

estmtd_theta(1,3)^2))/(estmtd_theta(1,4)*estmtd_theta(1,1)^2 -


estmtd_theta(1,2)^2);

se1(1) = (estmtd_theta(1,1)*(estmtd_theta(1,4)*am2 -

estmtd_theta(1,4)^2 - a0*estmtd_theta(1,3)*am2 +

a0*estmtd_theta(1,4)*am1) ...

- estmtd_theta(1,2)*(a0*estmtd_theta(1,4) -

estmtd_theta(1,3)*estmtd_theta(1,4) - a0*am2 +


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estmtd_theta(1,4)*am1))/(estmtd_theta(1,4)*estmtd_theta(1,1)^2 -


estmtd_theta(1,2)^2);

te0(1) = beta; te1(1) = beta*a0;

e = zeros(1,t);

ue(1) = 0;

for i = 2:M-numOfCoeffs+1;

z = P*phi(i,:)';

K = z/(lambda + phi(i,:)*z);

yhat(i) = phi(i,:)*estmtd_theta(i-1,:).';

e(i) = y(i) - phi(i,:)*estmtd_theta(i-1,:).';

estmtd_theta(i,:) = estmtd_theta(i - 1,:) + e(i)*K.';

P = (P - K*z.')/lambda;

re1(i) = estmtd_theta(i,2)/estmtd_theta(i,1) -

((estmtd_theta(i,2) - a0*estmtd_theta(i,1))*(am2*estmtd_theta(i,1)^2

- am1*estmtd_theta(i,1)*estmtd_theta(i,2) + estmtd_theta(i,2)^2))/ ... (estmtd_theta(i,1)*(estmtd_theta(i,4)*estmtd_theta(i,1)^2 -

estmtd_theta(i,3)*estmtd_theta(i,1)*estmtd_theta(i,2) +

estmtd_theta(i,2)^2));

se0(i) = (estmtd_theta(i,1)*(a0*estmtd_theta(i,4) -

estmtd_theta(i,3)*estmtd_theta(i,4) - a0*am2 + estmtd_theta(i,4)*am1)

- estmtd_theta(i,2)* ...

(estmtd_theta(i,4) - am2 + a0*estmtd_theta(i,3) - a0*am1 +

estmtd_theta(i,3)*am1 -

estmtd_theta(i,3)^2))/(estmtd_theta(i,4)*estmtd_theta(i,1)^2 -


estmtd_theta(i,2)^2);


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se1(i) = (estmtd_theta(i,1)*(estmtd_theta(i,4)*am2 -

estmtd_theta(i,4)^2 - a0*estmtd_theta(i,3)*am2 +

a0*estmtd_theta(i,4)*am1) ...

- estmtd_theta(i,2)*(a0*estmtd_theta(i,4) -

estmtd_theta(i,3)*estmtd_theta(i,4) - a0*am2 +

estmtd_theta(i,4)*am1))/(estmtd_theta(i,4)*estmtd_theta(i,1)^2 -


estmtd_theta(i,2)^2);

te0(i) = beta;

te1(i) = beta*a0;

ue(i) = - re1(i)*ue(i-1) + te0(i)*uc(i)+ te1(i)*uc(t-1) -

se0(i)*yhat(i) - se1(i)*yhat(i-1);

end

end

Figure 14. Output


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Figure 16. Output with step disturbance


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Figure 17. Error between process output and estimated output with step disturbance

(d) Plot y and u c vs. t, u vs. t and estimated process parameters vs. t. Compare the

tracking performance, ringing of input and disturbance rejection of the two designs.

Discuss on the effect of time delay on the performance of the adaptive controller.

d.1. Stable zero cancellation:


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Figure 18. u c and y_estimated vs time

Figure 19. u (control input is computed from controller law) vs time


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Figure 20. Estimated process parameters


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Figure 21. u c and y_estimated vs time with step disturbance

Figure 22. u (control input is computed from controller law) vs time with step

disturbance


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d.2. No zero cancellation:

Figure 24. u c and y_estimated vs time


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Figure 25. u (control input is computed from controller law) vs time



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Figure 27. u c and y_estimated vs time with step disturbance

Figure 28. u (control input is computed from controller law) vs time with step

disturbance


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Figure 29. Estimated process parameters with step disturbance

From Fig. 19 and Fig.25, the control input signal with stable zero cancellation design

has a severe oscillation (ringing) compare to that of no zero cancellation design

Two controllers have very same performance (tracking and disturbance).

Discuss on the effect of time delay on the performance of the adaptive controller: we

should choose the time delay as small as possible. When the time delay is large that

means the output signal lags phase to the command signal. Hence, the performance of

the system is not good.


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Problem 3

Solution :

I will take only one case in P2: the stable zero of the process model is canceled

Matlab code:

clear all ; close all ; clc;

% define command signal

Ts = 0.5; Tuc = 25*Ts;

fuc = 1/Tuc;

t = 0:Ts:100;

uc = square(2*pi*fuc*t);

w = 1; % frequency

% variations

delta_a1 = 0.13735;

delta_a2 = 0.07006;

delta_b0 = 0.01231;

delta_b1 = 0.007712;


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% define a1, a2, b0, b1

a1 = -1.3735 + delta_a1*sin(w*t);

a2 = 0.7006 + delta_a2*sin(w*t);

b0 = 0.1231 + delta_b0*sin(w*t);

b1 = 0.07712 + delta_b1*sin(w*t);

% parameters of the desired closed-loop system

am1 = -1.3205; am2 = 0.4966;

bm0 = 0.1761;

% Rq = q + r1

r1 = b1./b0;

% S(q) = s0*q + s1

s0 = (am1 - a1)./b0;

s1 = (am2 - a2)./b0;

% T(q) = t0*q

t0 = bm0./b0;

% define input u(t) and process output y(t)

u(1) = 0; u(2) = 0; y(1) = 0; y(2) = 0;

n = length(t);

for i = 3:n

y(i) = -a1(i)*y(i-1) - a2(i)*y(i-2) + b0(i)*u(i-1) + b1(i)*u(i-

2);

u(i) = -r1(i)*u(i-1) + t0(i)*uc(i) - s0(i)*y(i) - s1(i)*y(i-1);

end ;

% call function

lambda = 1;

[estmtd_theta,yhat,ue,e,re1,se0,se1,te0] ...

= adtvCtrolHw5_STR1(uc',u',y',lambda);


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figure(1);

plot(t(1:198),y(1:198),t(1:198),yhat(1:198));

xlabel( 'Time, sec' );

ylabel( 'Output' );

legend( 'Process output' , 'Estimated output' );

figure(2);

plot(e); grid on ;


ylabel( 'Error' );

figure(3);

plot(t(1:198),uc(1:198), '--' ,t(1:198),yhat(1:198));


ylabel( 'uc(t) & y-estimated(t)' ); legend( 'uc(t)' , 'y-estimated(t)' );

figure(4);

plot(ue);


ylabel( 'Control input signal' );

figure(5);

plot(t(1:198),estmtd_theta(:,1),t(1:198),b0(1:198), '--' , ...

t(1:198),estmtd_theta(:,2),t(1:198),b1(1:198), '--' , ...

t(1:198),estmtd_theta(:,3),t(1:198),a1(1:198), '--' , ...

t(1:198),estmtd_theta(:,4),t(1:198),a2(1:198), '--' );

grid on ;


ylabel( 'Process parameters' );

legend( 'b0-estimated' , 'b0' , 'b1-estimated' , 'b1' , 'a1-

estimated' , 'a1' , ... 'a2-estimated' , 'a2' );


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Figure 30. Output (delta = 10% of true value, w = 1 rad/sec)

Figure 31. Error between process and estimated output (delta = 10% of true value, w =

1 rad/sec)


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Figure 32. Command signal and estimated output (delta = 10% of true value, w = 1

rad/sec)

Figure 33. Control input signal (delta = 10% of true value, w = 1 rad/sec)


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Figure 34. True and estimated parameters (delta = 10% of true value, w = 1 rad/sec)

Figure 35. Output (delta = 15% of true value, w = 2 rad/sec)


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2 rad/sec)


rad/sec)


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Figure 40. Output (delta = 10% of true value, w = 1 rad/sec, lambda = 0.98)


1 rad/sec, lambda = 0.98)


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rad/sec, lamda)

Figure 43. Control input signal (delta = 10% of true value, w = 1 rad/sec, lambda = 1)


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Figure 44. True and estimated parameters (delta = 10% of true value, w = 1 rad/sec,

lambda = 0.98)

From all of figures above, we can see, it is very similar between weighted and un-

weighted design in this case. Because, the process parameters change as a sine function.

When frequency and variation increase the performance of the system is not good, even

maybe not stable.

Adaptive control theory: Pole-Placement and Indirect STR

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Transcript of Adaptive control theory: Pole-Placement and Indirect STR