PID controller in simulink
7 次查看(过去 30 天)
显示 更早的评论
Hi everyone! I am working on PID controller. I am having a model in simulink which I have uploaded at http://tinypic.com/view.php?pic=15cyg2&s=6. The gains of PID are: kp=1,Kd=1,ki=1.Saturation block has limits equal to -30 & +30. The state-space model has A=[0 1 0 0; 0 3.5 -106.5 0.37;0 0 0 1;0 0.21 -6.3 1.59]; B=[0 0;48.06 46664; 0 0; 340.18 47.7]; C=[1 0 1 0]; D=[0 0] and initial condition equal to 0. The system should converge at 20 as input is 20. But the problem is firstly its not converging and secondly the input of PID which is error signal appears as it is; as output with no modification. Need urgent help?
4 个评论
Gul Rukh Khan
2016-9-21
Dear Azizi brother, I have one question, I want to Tune my Simulink Model with PID, but I tried a lot to tune but failed badly. If you kind send your email address, I will send the model, and after Tuned kindly send back to me on this email: grkhan@hec.gov.pk Please, just send me an email with your good name AZIZI and obliged. with Many Thanks
Gul Rukh Khan
采纳的回答
Beenish Mazhar
2012-9-8
1 个评论
Walter Roberson
2012-9-9
Azzi commented "the answer expected was the value of PID parameters Kp Ti and Td"
更多回答(2 个)
Ryan G
2012-9-7
编辑:Ryan G
2012-9-7
I looked at the root locus with the PID and I don't believe this will be enough to adequately stabilize this system. Just for kicks I tried utilizing looptune to stabilize the system and was successful in finding a 4th order TF that would do it. Since I know nothing of your system other than the SS matrices it may or may not make any sense.
If you replace the PID with a Transfer fcn block this should stabilize the system:
num = [-1.19566835199456 0.150677296783384 -4.87470249290078 -5.9378741332245 -3.64167152790914e-05];
den = [1 32.4776016633891 994.732763803663 6558.83645377611 0.0401083722778298];
The resulting controller doesn't do a particularly good job but your system may not be easy to stabilize anyway.
What is the 2nd input and why don't you use it?
3 个评论
Ryan G
2012-9-7
The reason I ask is both inputs are coupled. So if you have a controller on both inputs it is easier to stabilize. As I mentioned previously, it's difficult to determine what makes sense not knowing anything about the system. I would suggest taking a look at slTunable if you have access to the robust control toolbox.
This will allow you to add an arbitrary compensator and have MATLAB automatically place the poles and find the gains based on requirements. However, if the system cannot meet the requirements or cannot be stabilized (which seems to be the case for just PID here) then it won't work.
I would figure out how you want to implement both inputs and tune them together.
Azzi Abdelmalek
2012-9-7
编辑:Azzi Abdelmalek
2012-9-7
eig(A) shows that your system is unstable (positive real(poles)in your system)
eig(A)
ans =
0
2.5582 + 2.3191i
2.5582 - 2.3191i
-0.0264
and if your output is not converging that means your PID parametes are not quite choosen
5 个评论
Azzi Abdelmalek
2012-9-7
We don't use PID to make systems stable, (ofcourse they can do it) . PID are mainly used to control the output, make it below a certain reference in presence of any dsturbance. Now the problem is to find adequat PID parameters, because the ones you gave did'nt work, If I find something I will post it
Azzi Abdelmalek
2012-9-7
编辑:Azzi Abdelmalek
2012-9-9
I guess, since your system is 4rth order and unstable, it's difficult to be stabilized with a PID, maybe you should look for another controller
另请参阅
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!