how to Design PID controller for second order untable process

조회 수: 8 (최근 30일)
Harshit Gole
Harshit Gole 2012년 2월 10일
편집: Azzi Abdelmalek 2013년 10월 16일
How a PID controller can be designed for a second order SISO process, of the form -a/S^2-b, where a,b>0 .

채택된 답변

Arkadiy Turevskiy
Arkadiy Turevskiy 2012년 2월 13일
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
  댓글 수: 2
Harshit Gole
Harshit Gole 2012년 2월 19일
sir ,
after entering the command
"C=pidtune(plant,'PID');"
it is showing an error
"??? Undefined function or method 'pidtune' for input arguments of type 'tf'."
please guide me.
Walter Roberson
Walter Roberson 2012년 2월 19일
pidtune is new as of R2010b
http://www.mathworks.com/help/toolbox/control/rn/bsllzup-1.html

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 PID Controller Tuning에 대해 자세히 알아보기

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by