control design for marginally stable system
2 次查看(过去 30 天)
显示 更早的评论
Hello dears
this system is a marginly stable system, I am trying to design a PID controller for it but the problem is that it has a big pole value that causes the system keep osilating for ever. please any body has an idea how to fix the pole location to design the controller
thank you
Mohamed
A =[ 0 0 -2.2444 0 0.01995
0 0 0.81081 -1.3514 -0.0054054
29.508 -19.672 -0.0018769 0 0
0 222.22 0 -4444.4 0
-0.2623 0.13115 0 0 0]
B =[ 104.74
0
-2.56
0
0]
C =[ 0 0 1 0 0]
D=0
sys1=ss(A,B,C,D)
egv=eig(sys1)
egv=[-4444.4 + 0i
-0.0074954 + 9.0655i
-0.0074954 - 9.0655i
-0.053263 + 0i
-0.0011919 + 0i]
0 个评论
回答(1 个)
Sam Chak
2024-1-23
Hi @Mohamed
Since no performance requirements are specified, the pidtune() command can be used as the initial step in designing the PID controller. Additionally, it is possible to cancel out the undesired pole in the closed-loop transfer function, as demonstrated in this case.
%% Original system in State-space
A = [ 0 0 -2.2444 0 0.01995
0 0 0.81081 -1.3514 -0.0054054
29.508 -19.672 -0.0018769 0 0
0 222.22 0 -4444.4 0
-0.2623 0.13115 0 0 0];
B = [104.74
0
-2.56
0
0];
C = [0 0 1 0 0];
D = 0*C*B;
sys = ss(A, B, C, D);
%% Plant
P = tf(sys)
%% Controller
[C, info] = pidtune(P, 'PIDF')
%% Closed-loop system
% Gcl = minreal(feedback(C*sys, 1));
Gcl = minreal(feedback(C*P, 1));
zcl = zero(Gcl) % closed-loop zeros
pcl = pole(Gcl) % closed-loop poles
%% Plot the results
subplot(2,1,1)
step(sys), grid on, legend('Original system', 'location', 'east')
subplot(2,1,2)
step(Gcl, 6000), grid on, legend('Compensated system', 'location', 'east')
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 PID Controller Tuning 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!