control design for marginally stable system
11 views (last 30 days)
Show older comments
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 Comments
Answers (1)
Sam Chak
on 23 Jan 2024
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 Comments
See Also
Categories
Find more on PID Controller Tuning in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!