NATIONAL CHENG KUNG UNIVERSITY Department of Mechanical Engineering
ADAPTIVE CONTROL HOMEWORK 6
Instructor:
Ming – Shaung Ju
Student:
Nguyen Van Thanh
Student ID:
P96007019
Department:
Inst. of Manufacturing & Information Systems
Class:
1001- N164400 - Adaptive Control
December 19, 2011
Contents
Problem 1 .................................. ................................................... .................................. .................................. .................................. ................................. ................ 2 Problem 2 .................................. ................................................... .................................. .................................. .................................. ............................... .............. 17 Problem 3 .................................. ................................................... .................................. .................................. .................................. ............................... .............. 23
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Problem 1 Consider a process:
a. Derive the pulse transfer function for the sampling period h = 0.1 s, and with zero order hold and determine a reference model. By using Matlab, we can write a script to solve this problem: clear all; close all; clc; %% process % in continuous-time num_s = 316.54; den_s = [1 19.49 452.20]; Gs = tf(num_s, den_s); % convert to discrete-time h = 0.1; % sampling time Hq = c2d(Gs, h, 'zoh'); % H(q) = B(q)/A(q) a1 = 0.2369; a2 = 0.1424; b0 = 0.6536; b1 = 0.3119; % choose d0 = deg(A) - deg(B) = 2 - 1 = 1; %% Reference model (desired closed-loop) % in continuous-time d = 0.7071; % damping ratio wn = 19.49/2/d; % natural frequency Gcls = tf(wn^2, [1 2*d*wn wn^2]); % converto to discrete-time Hmq = c2d(Gcls, h, 'zoh'); am1 = -0.4238; am2 = 0.1424; bm0 = 0.4758; bm1 = 0.2427;
We obtain: -
The pulse transfer function:
-
The desired closed-loop system:
+ In continuous-time:
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+ In discrete-time:
b. Design a direct STR in which the stable zero of process model is canceled Now, let the desired closed-loop system be:
let A0(q) = 1.
Step1: estimate the coefficients of two polynomials R(q) and S(q) in the model
Where,
̂ ̂ ̂ ̂ ̂ ̂ ̂ ̂ ̂
Using the RLS algorithm, we can obtain: Step 2: compute the control signal
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̂̂ ̂̂ ̂̂ ̂̂ ̂ Repeat step 1 and 2 at each sampling period. c. Design a direct STR in which no zero of process model is canceled. Let the desired closed-loop system be:
|| |||| || ( )
Parameter is chosen such that
The poles of Am(q):
Process model:
Where,
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Step 1: estimate the coefficients of the polynomials R1 and S1 in the model below by
the RLS algorithm.
Where,
{ Step 2: cancel possible common factors in R1 and S1 to obtain R and S.
̂ ̂ ̂ ̂ ̂ ̂
Step 3: calculate the control signal from:
Repeat step 1, 2 and 3 at each sampling period.
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d. 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 tracking performance. You may try step with different magnitude. d.1 Direct STR with stable zero cancellation:
Figure 1.1 Command, output & control input
Figure 1.2 Estimated parameters
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Figure 1.3 Command, output, reference output & open-loop output (t = 0:20)
Figure 1.4 Command, output, reference output & open-loop output (t = 20:end)
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Figure 1.5 Command, output & control input, with step disturbance at t = 60
Figure 1.6 Estimated parameters, with step disturbance at t = 60
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Figure 1.7 Command, output, reference output & open-loop output, with step disturbance at t = 60 d.2 Direct STR without stable zero cancellation:
Figure 1.8 Command, output & control input
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Figure 1.9 Estimated parameters
Figure 1.10 Command, output, reference output & open-loop output (t = 0:30)
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Figure 1.11 Command, output, reference output & open-loop output (t = 30: end)
Figure 1.12 Command, output & control input, with step disturbance at t = 60
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Figure 1.13 Estimated parameters, with step disturbance at t = 60
Figure 1.14 Command, output, reference output & open-loop output (t = 30: end), with step disturbance at t = 60
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e. 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. -
Tracking performance: the no zero cancellation is better than the other
-
Ringing of input: the control input signal with stable zero cancellation design has a severe oscillation (ringing) compare to that of no zero cancellation design
-
Disturbance rejection: with zero cancellation is better than the other
-
In a practical case the time delay and the order of the process that we would like to control are not known. It is therefore natural to consider these variables as design parameters that are chosen by the user. The parameter d0 is of particular importance for a direct algorithm. We can find the interesting and surprising result that cancellation of the process zero can be avoided by increasing the parameter d0. It is depend so much on experience.
Matlab code: clear all; close all; clc; h = Tuc fuc k = t = inp
0.1; % sampling time = 25*h; % period of uc = 1/Tuc; % frequency of uc 4; % number of cycles 0: h: k*Tuc; = square(2*pi*fuc*t);
% tt = 0: length(inp) - 1; % figure(1); % plot(tt,inp); hold on; v = [zeros(1,60) 1*ones(1, length(inp) - 60)]; % disturbance % real system a_1 = 0.2369; a_2 = 0.1424; b_0 = 0.6536; b_1 = 0.3119; % reference system am_1 = -0.4238; am_2 = 0.1424; bm_0 = 0.7186; oput(1) = 0; oput(2) = b_0*inp(1); oputf(1) = 0;
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oputf(2) = oput(2); % STR with zero cancellation % RLS % initial condition p = [150 0 0 0; 0 150 0 0; 0 0 3 0; 0 0 0 3]; estheta = [0 0; 0 0; 0 0.01; 0 0.2]; ctr_u = [0 0]; ctr_uf = [0 0]; lambda = 1; % fogetting factor % update for i = phi p = k =
3: length(inp) = [oputf(i-1); oputf(i-2); ctr_uf(i-1); ctr_uf(i-2)]; p - p*phi*inv(lambda + phi'*p*phi)*phi'*p/lambda; p*phi;
oput(i) = -a_1*oput(i-1) - a_2*oput(i-2) + b_0*(ctr_u(i-1)+v(i-1))... +b_1*(ctr_u(i-2)+v(i-2)); oputf(i) = -am_1*oputf(i-1) - am_2*oputf(i-2) + oput(i); estheta(:,i) = estheta(:,i-1) + k*(oputf(i) - phi'*estheta(:,i-1)); ests_0(i) = estheta(1,i); ests_1(i) = estheta(2,i); estr_0(i) = estheta(3,i); estr_1(i) = estheta(4,i); estt_0(i) = 1 + am_1 + am_2; ctr_u(i) = -estr_1(i)/estr_0(i)*ctr_u(i-1) + estt_0(i)/estr_0(i)*inp(i)... - ests_0(i)/estr_0(i)*oput(i) - ests_1(i)/estr_0(i)*oput(i-1) + 0.1*v(i); ctr_uf(i) = -am_1*ctr_u(i-1) - am_2*ctr_u(i-2) + ctr_u(i); end figure(2); subplot(211); stairs(inp,'linewidth',2);hold on; plot(oput,'r','linewidth',2); xlabel('Normalized time'); ylabel('uc(t) & y(t)'); legend('Command', 'output'); subplot(212); stairs(ctr_u,'linewidth',1); xlabel('Normalized time'); ylabel('u(t)'); legend('Control input signal'); hold on; figure(3); subplot(411); stairs(t,estr_1./estr_0,'r','linewidth',2); legend('r1hat/r0hat'); subplot(412); stairs(t,ests_0./estr_0,'g','linewidth',2); legend('s0hat/r0hat'); subplot(413); stairs(t,ests_1./estr_0,'b','linewidth',2); legend('s1hat/r0hat'); subplot(414); stairs(t,estt_0./estr_0,'m','linewidth',2); legend('t0hat/r0hat'); xlabel('Time'); hold on; % STR without zero cancellation % initial condition pd = [5000 0 0 0 0 0; 0 5000 0 0 0 0; 0 0 5000 0 0 0; 0 0 0 1000 0 0;
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0 0 0 0 1000 0; 0 0 0 0 0 1000]; a_0 = 0.6; oputd(1) = 0; oputd(2) = b_0*inp(1); oputd(3) = -a_1*oputd(2) - a_2*oputd(1) + b_0*inp(2) + b_1*inp(1); oputdf(1) = 0; oputdf(2) = oputd(2); oputdf(3) = -(a_0 + am_1)*oputdf(2) -(am_2 + a_0*am_2)*oputdf(1) + oputd(3); esthetad = [0 0 0; 0 0 0; 0 0 0; 0 0 0.1; 0 0 0.2; 0 0 0.3]; ctr_ud = [0 0 0]; ctr_udf = [0 0 0]; % update for ii = 4:length(inp) phid = [oputdf(ii-1); oputdf(ii-2); oputdf(ii-3); ctr_udf(ii-1);... ctr_udf(ii-2); ctr_udf(ii-3)]; pd = pd - pd*phid*inv(lambda + phid'*pd*phid)*phid'*pd/lambda; kd = pd*phid; oputd(ii) = -a_1*oputd(ii-1) - a_2*oputd(ii-2) + b_0*(ctr_ud(ii-1)... + v(ii-1))+b_1*(ctr_ud(ii-2)+v(ii-2)); oputdf(ii) = -(a_0 + am_1)*oputdf(ii-1) - (am_2 + am_2*a_0)*oputdf(ii-2)... - a_0*am_2*oputdf(ii-3)+ oputd(ii); esthetad(:,ii) = esthetad(:,ii-1) + kd*(oputd(ii) - phid'*esthetad(:,ii-1)); ests_10(ii) = esthetad(1,ii); ests_11(ii) = esthetad(2,ii); ests_12(ii) = esthetad(3,ii); estr_10(ii) = esthetad(4,ii); estr_11(ii) = esthetad(5,ii); estr_12(ii) = esthetad(6,ii); % cancel common factors iin R1 & S1 to obtaiin R & S ests_0d(ii) = ests_10(ii)/estr_10(ii); estb_1d(ii) = (ests_11(ii) - estr_11(ii)*ests_12(ii)/estr_12(ii))... /(ests_0d(ii) - ests_12(ii)/estr_12(ii)); estr_0d(ii) = (estr_11(ii) - estb_1d(ii))/(estr_10(ii)); ests_1d(ii) = estr_0d(ii)*ests_12(ii)/estr_12(ii); beta(ii) = (1 + am_1 + am_2)/(estr_10(ii) + estb_1d(ii)); ctr_ud(ii) = -estr_0d(ii)*ctr_ud(ii-1) + beta(ii)*inp(ii)... + beta(ii)*a_0*inp(ii-1) - ests_0d(ii)*oputd(ii-1)... - ests_1d(ii)*oputd(ii-2) + 0.1*v(ii); ctr_udf(ii) = -(a_0 + am_1)*ctr_udf(ii-1) - (am_2 + am_2*a_0)*ctr_udf(ii-2)... - a_0*am_2*ctr_udf(ii-3)+ oputd(ii); end % % % % % % % % % % % % % % % %
figure(4); subplot(211); stairs(inp,'linewidth',2);hold on; plot(oputd,'r','linewidth',2); xlabel('Normalized time'); ylabel('uc(t) & y(t)'); legend('Command', 'output'); subplot(212); stairs(ctr_ud,'linewidth',1); xlabel('Normalized time'); ylabel('u(t)'); legend('Control input signal'); hold on; figure(5) subplot(311);stairs(t,estr_0d,'r','linewidth',2); legend('r0hat'); subplot(312);stairs(t,ests_0d,'g','linewidth',2);
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% % % %
legend('s0hat'); subplot(313);stairs(t,ests_1d,'b','linewidth',2); legend('s1hat'); xlabel('Time');
% reference output ref_oput(1) = 0; ref_oput(2) = bm_0*inp(1); for j = 3: length(inp) ref_oput(j) = -am_1*ref_oput(j-1) - am_2*ref_oput(j-2) + bm_0*inp(j-1); end % open-loop output op_oput(1) = 0; op_oput(2) = bm_0*inp(1); for jj = 3:length(inp) op_oput(jj) = -a_1*ref_oput(jj-1) - a_2*ref_oput(jj-2) + b_0*inp(jj-1)... + b_1*inp(jj-2); end figure(6) stairs(inp(1:20),'linewidth',2);hold on; plot(oput(1:20),'r','linewidth',2); plot(ref_oput(1:20),'g--','linewidth',2); plot(op_oput(1:20),'k--','linewidth',2); xlabel('Time'); legend('Command', 'Output', 'Reference oput', 'Open-loop output'); hold on; figure(7) stairs(inp(20:end),'linewidth',2);hold on; plot(oput(20:end),'r','linewidth',2); plot(ref_oput(20:end),'g--','linewidth',2); plot(op_oput(20:end),'k--','linewidth',2); xlabel('Time'); legend('Command', 'Output', 'Reference oput', 'Open-loop output'); hold on; figure(8) stairs(inp(1:30),'linewidth',2);hold on; plot(oputd(1:30),'r','linewidth',2); plot(ref_oput(1:30),'g--','linewidth',2); plot(op_oput(1:30),'k--','linewidth',2); xlabel('Time'); legend('Command', 'Output', 'Reference oput', 'Open-loop output'); hold on; figure(9) stairs(inp(30:end),'linewidth',2);hold on; plot(oputd(30:end),'r','linewidth',2); plot(ref_oput(30:end),'g--','linewidth',2); plot(op_oput(30:end),'k--','linewidth',2); xlabel('Time'); legend('Command', 'Output', 'Reference oput', 'Open-loop output');
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Problem 2 a. Consider the direct STR in P1. Modify your simulation program so that the parameters of the process are time-varying. Investigate experimentally how well the adaptive system can adapt to 20% of parameter variations. (a1 = a1o+a1 sint, a2 = a2o +a2 sint, b0 = b0o+b0 sint, b1 = b1o +b1 sint, where a1o is the nominal value of a1, try different % of parameter variations with different frequencies). a.1 Direct STR with stable zero cancellation
Figure 2.1 Command, output & control input with delta = 20%, w = 1 rad/s
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Figure 2.1 Command, output & control input with delta = 30%, w = 2 rad/s a.2 Direct STR without stable zero cancellation
Figure 2.3 Command, output & control input with delta = 20%, w = 1 rad/s
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Figure 2.4 Command, output & control input with delta = 30%, w = 2 rad/s Investigate experimentally how well the adaptive system can adapt to 20% of parameter variations: from Problem1 and through simulation in Problem 2, they are small difference, so the adaptive system can adapt to 20% of parameter variations, but not well.
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b. Introduce the forgetting factor and compare the performance with that of the un-weighted designs. b.1 Direct STR with stable zero cancellation
Figure 2.5 Command, output & control input with delta = 20%, w = 1 rad/s, lambda = 0.98 b.2 Direct STR without stable zero cancellation
Figure 2.6 Command, output & control input with delta = 20%, w = 1 rad/s, lambda = 0.98 From figure 2.1 and figure 2.5, we can see that with lambda = 0.98 is better than lambda = 1.0. Matlab code: clear all; close all; clc;
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h = Tuc fuc k = t = inp
0.1; % sampling time = 25*h; % period of uc = 1/Tuc; % frequency of uc 4; % number of cycles 0: h: k*Tuc; = square(2*pi*fuc*t);
% tt = 0: length(inp) - 1; % figure(1); % plot(tt,inp); hold on; v = [zeros(1,60) 0*ones(1, length(inp) - 60)]; % disturbance % real system a_1 = 0.2369; a_2 = 0.1424; b_0 = 0.6536; b_1 = 0.3119; delta = 0.2; % percentage of variation w = 1; % frequency rad/s a_1 = a_1 + delta*a_1*sin(w*t); a_2 = a_2 + delta*a_2*sin(w*t); b_0 = b_0 + delta*b_0*sin(w*t); b_1 = b_1 + delta*b_1*sin(w*t); % reference system am_1 = -0.4238; am_2 = 0.1424; bm_0 = 0.7186; oput(1) = 0; oput(2) = b_0(1)*inp(1); oputf(1) = 0; oputf(2) = oput(2); % STR with zero cancellation % RLS % initial condition p = [150 0 0 0; 0 150 0 0; 0 0 3 0; 0 0 0 3]; estheta = [0 0; 0 0; 0 0.01; 0 0.2]; ctr_u = [0 0]; ctr_uf = [0 0]; lambda = 0.98; % fogetting factor % update for i = 3: length(inp) phi = [oputf(i-1); oputf(i-2); ctr_uf(i-1); ctr_uf(i-2)]; p = (p - p*phi*inv(lambda + phi'*p*phi)*phi'*p)/lambda; k = p*phi; oput(i) = -a_1(i)*oput(i-1) - a_2(i)*oput(i-2) + b_0(i)*ctr_u(i-1) + b_1(i)*ctr_u(i-2); oputf(i) = -am_1*oputf(i-1) - am_2*oputf(i-2) + oput(i); estheta(:,i) = estheta(:,i-1) + k*(oputf(i) - phi'*estheta(:,i-1)); ests_0(i) = estheta(1,i); ests_1(i) = estheta(2,i); estr_0(i) = estheta(3,i); estr_1(i) = estheta(4,i); estt_0(i) = 1 + am_1 + am_2; ctr_u(i) = -estr_1(i)/estr_0(i)*ctr_u(i-1) + estt_0(i)/estr_0(i)*inp(i)... - ests_0(i)/estr_0(i)*oput(i) - ests_1(i)/estr_0(i)*oput(i-1) + 0.1*v(i); ctr_uf(i) = -am_1*ctr_u(i-1) - am_2*ctr_u(i-2) + ctr_u(i);
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end
figure(2); subplot(211); stairs(inp(20:end),'linewidth',2);hold on; plot(oput(20:end),'r','linewidth',2); xlabel('Normalized time'); ylabel('uc(t) & y(t)'); legend('Command', 'output'); subplot(212); stairs(ctr_u(20:end),'linewidth',1); xlabel('Normalized time'); ylabel('u(t)'); hold on; figure(3); subplot(411); stairs(t(20:end),estr_1(20:end)./estr_0(20:end),'r','linewidth',2); legend('r1hat/r0hat'); subplot(412); stairs(t(20:end),ests_0(20:end)./estr_0(20:end),'g','linewidth',2); legend('s0hat/r0hat'); subplot(413); stairs(t(20:end),ests_1(20:end)./estr_0(20:end),'b','linewidth',2); legend('s1hat/r0hat'); subplot(414); stairs(t(20:end),estt_0(20:end)./estr_0(20:end),'m','linewidth',2); legend('t0hat/r0hat'); xlabel('Time'); hold on; % STR without zero cancellation % initial condition pd = [5000 0 0 0 0 0; 0 5000 0 0 0 0; 0 0 5000 0 0 0; 0 0 0 1000 0 0; 0 0 0 0 1000 0; 0 0 0 0 0 1000]; a_0 = 0.6; oputd(1) = 0; oputd(2) = b_0(i)*inp(1); oputd(3) = -a_1(i)*oputd(2) - a_2(i)*oputd(1) + b_0(i)*inp(2) + b_1(i)*inp(1); oputdf(1) = 0; oputdf(2) = oputd(2); oputdf(3) = -(a_0 + am_1)*oputdf(2) -(am_2 + a_0*am_2)*oputdf(1) + oputd(3); esthetad = [0 0 0; 0 0 0; 0 0 0; 0 0 0.1; 0 0 0.2; 0 0 0.3]; ctr_ud = [0 0 0]; ctr_udf = [0 0 0]; % update for ii = 4:length(inp) phid = [oputdf(ii-1); oputdf(ii-2); oputdf(ii-3); ctr_udf(ii-1);... ctr_udf(ii-2); ctr_udf(ii-3)]; pd = (pd - pd*phid*inv(lambda + phid'*pd*phid)*phid'*pd)/lambda; kd = pd*phid; oputd(ii) = -a_1(ii)*oputd(ii-1) - a_2(ii)*oputd(ii-2) + b_0(ii)*(ctr_ud(ii1)... + v(ii-1))+b_1(ii)*(ctr_ud(ii-2)+v(ii-2)); oputdf(ii) = -(a_0 + am_1)*oputdf(ii-1) - (am_2 + am_2*a_0)*oputdf(ii-2)... - a_0*am_2*oputdf(ii-3)+ oputd(ii); esthetad(:,ii) = esthetad(:,ii-1) + kd*(oputd(ii) - phid'*esthetad(:,ii-1)); ests_10(ii) = esthetad(1,ii); ests_11(ii) = esthetad(2,ii);
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ests_12(ii) = esthetad(3,ii); estr_10(ii) = esthetad(4,ii); estr_11(ii) = esthetad(5,ii); estr_12(ii) = esthetad(6,ii); % cancel common factors iin R1 & S1 to obtaiin R & S ests_0d(ii) = ests_10(ii)/estr_10(ii); estb_1d(ii) = (ests_11(ii) - estr_11(ii)*ests_12(ii)/estr_12(ii))... /(ests_0d(ii) - ests_12(ii)/estr_12(ii)); estr_0d(ii) = (estr_11(ii) - estb_1d(ii))/(estr_10(ii)); ests_1d(ii) = estr_0d(ii)*ests_12(ii)/estr_12(ii); beta(ii) = (1 + am_1 + am_2)/(estr_10(ii) + estb_1d(ii)); ctr_ud(ii) = -estr_0d(ii)*ctr_ud(ii-1) + beta(ii)*inp(ii)... + beta(ii)*a_0*inp(ii-1) - ests_0d(ii)*oputd(ii-1)... - ests_1d(ii)*oputd(ii-2) + 0.1*v(ii); ctr_udf(ii) = -(a_0 + am_1)*ctr_udf(ii-1) - (am_2 + am_2*a_0)*ctr_udf(ii-2)... - a_0*am_2*ctr_udf(ii-3)+ oputd(ii); end figure(4); subplot(211); stairs(inp(40:end),'linewidth',2);hold on; plot(oputd(40:end),'r','linewidth',2); xlabel('Normalized time'); ylabel('uc(t) & y(t)'); legend('Command', 'output'); subplot(212); stairs(ctr_ud(40:end),'linewidth',1); xlabel('Normalized time'); ylabel('u(t)'); hold on; figure(5) subplot(311);stairs(t(40:end),estr_0d(40:end), 'r','linewidth',2); legend('r0hat'); subplot(312);stairs(t(40:end),ests_0d(40:end), 'g','linewidth',2); legend('s0hat'); subplot(313);stairs(t(40:end),ests_1d(40:end), 'b','linewidth',2); legend('s1hat'); xlabel('Time');
Problem 3 Consider a process:
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a. Use the indirect STR you have designed in which no zero of process model was canceled in. Simulate the closed-loop system with a step disturbance applied at t ≥ 40. Using following process parameters [a1, a2, b0, b1] = [-1.6065, 0.6065, 0.1065, 0.0902] and initial guess of above parameter vector [0, 0, 0.01, 0.2], lambda = 0.98. The command is a square wave of amplitude 1 and period of 50 sampling periods.
Figure 3.1 Command, output & control input signal
Figure 3.2 Estimated parameters
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b. Design a STR with modified estimator and controller by using the internal model principle and verify effectiveness of the method via simulation. Process model:
Where,
( ) Control law:
Compute uf (t):
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( ) ( ) ( ) And
c. If the disturbance is a sinusoidal v(t) = 0.5 sint, re-design a STR as requested in part (b). You may change the frequency of the disturbance and discuss on the effect of frequency on the performance of your controller.
( ) ( ) {
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