Effacer les filtres
Effacer les filtres

control design for marginally stable system

6 vues (au cours des 30 derniers jours)
Mohamed
Mohamed le 31 Mar 2020
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]

Réponses (1)

Sam Chak
Sam Chak le 23 Jan 2024
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)
P = -2.56 s^4 - 8287 s^3 + 1.374e07 s^2 + 9.281e05 s - 3250 -------------------------------------------------------------- s^5 + 4444 s^4 + 390.8 s^3 + 3.653e05 s^2 + 1.989e04 s + 23.19 Continuous-time transfer function.
%% Controller
[C, info] = pidtune(P, 'PIDF')
C = 1 s Kp + Ki * --- + Kd * -------- s Tf*s+1 with Kp = -0.00745, Ki = -6.76e-06, Kd = 1.55, Tf = 209 Continuous-time PIDF controller in parallel form.
info = struct with fields:
Stable: 1 CrossoverFrequency: 0.0013 PhaseMargin: 64.8902
%% Closed-loop system
% Gcl = minreal(feedback(C*sys, 1));
Gcl = minreal(feedback(C*P, 1));
zcl = zero(Gcl) % closed-loop zeros
zcl = 4×1
1.0e+03 * 1.2073 -0.0001 0.0000 -0.0000
pcl = pole(Gcl) % closed-loop poles
pcl =
-0.0083 + 9.0655i -0.0083 - 9.0655i -0.0539 + 0.0000i -0.0016 + 0.0026i -0.0016 - 0.0026i -0.0006 + 0.0000i
%% 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')

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by