02-25-2007 03:15 AM
02-26-2007 11:16 AM
02-26-2007 11:48 AM
If you are using the Control Design Toolkit, you can modify the example:
\examples\Control Design\Getting Started\Classical Control Design.llb\CDEx PID with Constraints.vi
That helps you enter the Constraint for your design and modify the PID parameters to achieve the desired compensation.
On a side note, I notice that your transfer function already have an integrator in its direct path. So, unless you are trying to follow a ramp, you don't need a Integral Gain for your controller and a simple gain or a PD would work for this plant, which means that you don't need PID for this plant.
Anyway let me know if this answer your question.
Alex Barp - Control and Simulation group - National instruments
02-26-2007 12:22 PM
Thanks for the reply,infact I am using matlab I have made PD controller using graphical root locus .I want to know how to calculate the compensator gain using root locus.ifu have any source of information for this please do send me.
03-01-2007 04:10 AM
You will have to specify if you have the LabVIEW Control Design & Simulation toolkit installed as this makes PID control alot easier.In addition you will have to go into greater detail on how you wish your application to work. Will you be wanting to use your already written MatLab code within LabVIEW also, or simply try to emulate what you have achieved in MatLab with LabVIEW. If you do want to use your MatLab code a good way to integrate this is by navigating to Mathematics > Scripts & Formulas > Script Nodes - MatLab script node.
You will need to post more information about the system you are using and what you want you application to achieve for me to give better guidance how you to approach your programming.
03-01-2007 05:22 AM
Thanks for reply.well for the time being using LAbview or matlab is not an issue infact I am desinging P dcontroller and calcualting the parameter of Kd and Kp.open loop transfer function for my problem is : 31.75/s(s+2.5)(s+0.5) .
you can see there is an integrator in the open loop and theortically it should nt give any steady state error the close loop function obtained from this openloop with feedback is:
---------------------------- the step respons of the close loop function shows a unbounded system ;any comments on it????? should we include 31.75 the numerator of the open l
s^3 + 3 s^2 + 1.25 s + 31.72
loop in the calcualtion of the close loop?
i developed aPD compensator using graphical root locus techniques using the following time doima specification overshoot 10% settling time 10 secod.the desired pole from this specification is -2.-+2,76i which gives from gemoetry the position of the required zero as -1.24 using this zero does not give the required specification so i used z=-.3 (any comments on it)I am also perplexed about the calculation of the Kd :from graphical root locus methos i have used following formula
Kd= product of vector lenght of finte pole to desire pole/(productor of vector length of zero to desired pole * 31.75)
the obtained PD compensator is :
0.2866 s + 0.08599 and coreesponding open loop P Dcompensator is :
9.09 s + 2.727
s^3 + 3 s^2 + 1.25 s
the close loop is obtained as
1.515 s + 0.4545
s^3 + 3 s^2 + 10.34 s + 2.727
this gives settling time 3 second overshoot is 16 % but steady state is .167 i.e about 99 % steady state error.
Looking forrwad for ur reply.
03-09-2007 10:35 AM
I would highly recommend looking at the NI LabVIEW Control Design Toolkit, which will contain all of the functions, examples, and comprehensive documentation on how the Toolkit PID VIs work. To achieve what you want in LabVIEW, especially with little LabVIEW experience will be extremely difficult without the toolkit. I hope this helps.