# PID setpoint problem

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Moritz on 16 Apr 2012
Commented: OSAMA YOUNAS on 21 Nov 2017
Hi,
for my bachelor project I'm designing a controller for an inverted pendulum problem. The first simulation I have created in Labview but after purchasing MATLAB R2012a I wanted to try to 'translate' my simulation into simulink. It all works nicely except of one big problem. When I use a normal continues PID controller, the control works fine and it is controlled to the zero (equilibrium) point. Hence, the automatic set point of the PID(s) block seems to be zero (?). I have used the following values: P: 500, I:0.01, D:1, N:1. The form is ideal. Now, since I need to be able to set my setpoint myself I replaced the PID(s) block by the PID(s) 2DOF block (is this right?). Here I used: P: 500, I:0.01, D:1, N:1, b:1, c:1. I set the form to ideal and connected a 0 constant to the reference input. This (to my judgment) should be the same controller as the PID(s) was. However, the system now is highly unstable. What am I doing wrong? Where have I gone wrong? I only want the PID(s) block and set my own setpoint.
I hope this all makes sense!
Thank you so much in advance!
Regards

Arkadiy Turevskiy on 17 Apr 2012
Edited: Arkadiy Turevskiy on 2 Oct 2017
PID Controller block does not assume anything about the setpoint value, it operates on the error signal which is the difference between the setpoint and measured output.
If you want to control your pendulum around some non-zero angle, you do not need 2DOF PID. All you have to do is have a block for your setpoint (constant, for example). Set the value of this constant as you wish. Then substract measured angle to form the error signal, and feed this error signal to the PID Controller.
Take a look at this demo to see how to do this in a Simulink model.
More demos and other relevant info is available on PID Control with MATLAB and Simulink page.
HTH.
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OSAMA YOUNAS on 21 Nov 2017
Thanks alot! I was wasting my time finding a Setpoint block but couldn't find it

R2012a

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