MATLAB Answers

0

How can I design a PID Controller to stabllize the plant 1/(s^3+1) ? [beginner]

OHSEONG KWON さんによって質問されました 2018 年 11 月 16 日
最新アクティビティ Aquatris
さんによって コメントされました 2018 年 11 月 17 日
I have a system, it's Transfer function is :
1
------------
s^3 + 1
and I want to design a Discrete PID Controller so discretized the plant at sampling time 0.05s :
0.000019z^2 + 0.00008z + 0.000021
------------------------------------------------- (Sorry, it is not clean..)
z^3 -2.99994z^2 + 3.00006z - 1
This system is unstable and Discrete PID Controller Block's autotuner cannot find Kp, Ki, Kd to stablize the system.
I want to design a PID Controller, not another Controller.
I think it should be added such a filter, derivative filter(Differtiator?) or a Integrator...
How can I design the PID Controller in Matlab and Simulink? It can be designed?
Thanks for read my ask.

  0 件のコメント

サインイン to comment.

タグ

1 件の回答

回答者: Aquatris
2018 年 11 月 16 日
編集済み: Aquatris
2018 年 11 月 16 日
 採用された回答

So you have a transfer function G(s) = 1/(s^3+1) and you want to find gains P,I,D for your controller K(s) = P+D*s+I/s. As commonly known, the closed loop transfer function T(s) can be written as;
T(s) = K(s)*G(s)/(1+G(s)*K(s))
If you plugin G(s) and K(s) functions, you will obtain;
T(s) = (D*s^2 + P*s + I) / (s^4 + D*s^2 + (P+1)*s + I)
From here, it should be clear that a simple controller, K(s) = P+D*s+I/s, cannot stabilize the system (hint: routh-hurwitz stability criterion). That's why your controller needs to have a term that will show up as the s^3 term.
One option is to add a C*s^2 term to your controller. This way your K(s) = P+D*s+I/s+C*s^2, and resulting closed loop transfer function is;
T(s) = (C*s^3 + D*s^2 + P*s + I)/( s^4 + C*s^3 + D*s^2 + (P+1)*s + I )
For instance, imagine you want to place your poles to (-4+0i) (continous time model), then your characteristic equation needs to be;
(s+4)^4 = s^4 + 16*s^3 + +96*s^2 + 256*s + 256 = ( s^4 + C*s^3 + D*s^2 + (P+1)*s + I )
From here, by matching the coefficients, you can find the controller
K(s) = 255+96*s+256/s+16*s^2;
This controller stabilizes G(s), and also places all 4 of the closed loop poles to (-4+0i).
Another method to tune the gains of your controller is to perform Ruth-Hurwitz stability analysis and determine the range of controller gains that achieves closed loop stability.
These calculation are done for continous time however they are valid for discrete time as well, with some modification due to conversion from s to z domain.
Hope this helps.
s = tf('s');
G = 1/(s^3+1);
K = 255+96*s+256/s+16*s^2;
T = minreal(G*K/(1+G*K));
step(T)

  4 件のコメント

Aquatris
2018 年 11 月 17 日
Keep in mind selecting the poles is important since it will determine the performance of your controller, or in other words the response of your closed loop (overshoot,settling time etc.).
Glad this helped :)
OHSEONG KWON 2018 年 11 月 17 日
Thank you but Can I ask you one more question?
I got P, I, D and C what I hoped as you said me, and It works in continuous time model.
But I discretized The plant 1/(s^3+1) at Ts=0.05s :
2.083e-05 z^2 + 8.333e-05 z + 2.083e-05
--------------------------------------- (Now I can use the function c2d :) )
z^3 - 3 z^2 + 3 z - 1
but it do not work for same Controller(PIDC Controller). I use SImulink, but I can't what is wrong.
캡처.PNGsad.PNG
This picture is the continuous and discrete plant and output. In discrete plant, sampling time Ts is set for step input parameter.
What is wrong..?
Aquatris
2018 年 11 月 17 日
I recommend you do do same thing I did for continous system for the discrete system. You know your G(z) and you can parametrize your K(z). Then obtain your T(z) and select some stable poles (z(1,2,3,4) = 0.8 for instance) and try to find controller gains that gives you the same charasteric equation polynomial coefficients.

サインイン to comment.



Translated by