PID controller, difference when graphing step function with PID control block in matlab and simulink
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hoang
le 2 Juil 2024 à 2:55
Commenté : Sam Chak
le 3 Juil 2024 à 6:03
Hi everyone,
please tell me, why is there a difference when graphing step function with PID control block in matlab and simulink
s =tf('s');
g = 1.883e5/(s*(s^2+4466*s+6.43e6));
kp = 60;
ki = 63000;
kd = 3;
gpid = pid(kp,ki,kd);
gsys = feedback(g*gpid,1);
step(gsys)
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Sam Chak
le 2 Juil 2024 à 4:16
Hi @hoang
Both responses look similar if you implement them correctly as shown.
s = tf('s');
Gp = 1.883e5/(s^2 + 4466*s + 6.43e6);
kp = 60;
ki = 63000;
kd = 3;
Gc = pid(kp, ki, kd);
Gcl = feedback(Gc*Gp, 1);
step(Gcl), grid on, ylim([-0.1, 1.1])
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Sam Chak
le 3 Juil 2024 à 6:03
Hi @hoang
I'm glad to hear that. If you find the explanation and code helpful, please consider clicking 'Accept' ✔️ on my Answer. Additionally, you can show your appreciation by voting 👍 for other answers by Paul and Umar as a token of support for their knowledge sharing. Your support is greatly appreciated!
Plus de réponses (2)
Paul
le 2 Juil 2024 à 3:28
Hi hoang,
The posted code doesn't result in the same figure as in the question
s =tf('s');
g = 1.883e5/(s*(s^2+4466*s+6.43e6))
kp = 60;
ki = 63000;
kd = 3;
gpid = pid(kp,ki,kd);
gsys = feedback(g*gpid,1);
step(gsys)
Note also that g in the Matlab code has an integrator in the denominator that is not present in the plant model in the Simulink diagram.
What is inside the subsystem block in the Simulink diaggram?
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Umar
le 2 Juil 2024 à 3:00
Hi Hoang,
My suggestion would be tuning the PID controller gains appropriately. You can use tools like the PID Tuner in Simulink or MATLAB's Control System Toolbox to adjust the controller parameters for better system performance. Try using the pidtune function to tune the PID controller. For more information regarding this function, please refer to
https://www.mathworks.com/help/control/ref/dynamicsystem.pidtune.html
Let me know if you need further assistance.
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