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unit step disturbance at a time of 40 seconds

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MUHAMED SIDIBEH
MUHAMED SIDIBEH il 11 Dic 2022
Modificato: Sam Chak il 12 Dic 2022
How do I generate a unit step disturbance at a time of 40 seconds for my PID controller
clear all;
s=tf('s');
load time.dat;
load response.dat;
K = 2;
Tau = 3;
dt = 0.9;
g = ((2/(3*s + 1))*(exp(-dt*s)))
[gy,tg]=step(g,20)
plot(time, response,'k',tg,gy,'r','LineWidth',2)
hold on
% Cohen - Coon PI tuning
Kc = (1/K)*(Tau/dt)*(0.9 + (dt/(12*Tau)));
Tau_1 = (dt)*((30 + 3*(dt/Tau))/(9+20*(dt/Tau)));
P = Kc;
I = Kc/Tau_1;
cont = ((s*P + I)/s);
gol = cont*g
gcl = feedback(gol,1)
step(gcl,20)
hold on
grid on
Thanks

Risposte (1)

Sam Chak
Sam Chak il 12 Dic 2022
Modificato: Sam Chak il 12 Dic 2022
The closed-loop transfer function for Y(s)/D(s) is given by
The following shows how to compute the comparison between the step responses for 40 seconds. Let us know if this is the technical answer that you are looking for.
s = tf('s');
% Parameters
K = 2;
Tau = 3;
dt = 0.9;
% Plant
Gp = ((2/(3*s + 1))*(exp(-dt*s)));
% Cohen - Coon PI tuning
Kc = (1/K)*(Tau/dt)*(0.9 + (dt/(12*Tau)));
Tau_1 = (dt)*((30 + 3*(dt/Tau))/(9 + 20*(dt/Tau)));
Kp = Kc;
Ki = Kc/Tau_1;
Gc = ((Kp*s + Ki)/s);
% Response to the Unit Step Reference
Gol = Gc*Gp;
Gcl = feedback(Gol, 1); % closed-loop transfer function Y(s)/R(s)
step(Gcl, 40)
hold on
% Response to the Unit Step Disturbance
Gcd = feedback(Gp, Gc); % closed-loop transfer function Y(s)/D(s)
step(Gcd, 40)
legend('Step Reference Response', 'Step Disturbance Response')
hold off, grid on

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R2022b

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