how to Design PID controller for second order untable process
6 次查看(过去 30 天)
显示 更早的评论
How a PID controller can be designed for a second order SISO process, of the form -a/S^2-b, where a,b>0 .
0 个评论
采纳的回答
Arkadiy Turevskiy
2012-2-13
Not sure if that is what you are looking for, but here is how you could do it in Control System Toolbox:
>>a=1; % change to desired value
>>b=4; % change to desired value
>>s=tf('s');
>>plant=-a/(s^2-b);
>>C=pidtune(plant,'PID'); design pid controller
>>step(feedback(C*plant,1)); plot the step response of the closed-loop system
>>pidtool(plant); % open the GUI for interactive tuning
HTH. Arkadiy
2 个评论
Walter Roberson
2012-2-19
pidtune is new as of R2010b
http://www.mathworks.com/help/toolbox/control/rn/bsllzup-1.html
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 PID Controller Tuning 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!