What is the result of the PID vi. in Labview?

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• The transfer of the PID block function does not show the derivator.

  Hello

  I try to use the vi of PID, but I when I try the box by itself, it does not behave as a 'classic' PID should behave. The main problem is that I can't see the derivator. In the attached vi, I compare the transfer function of the PID vi with the function of transfer of a PID built by me. My version shows should have all what a PID: section integral (with an amplitude decreasing and phase-90), Center (with a constant amplitude and phase 0) and area derived (with the magnitude and the phase of + 90).

  

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  Kind regards

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  Nathand dear,

  Thanks for your reply, I tried the change and you are right. This shows that's not a PID standard, since this means (and in fact I can see it now by checking within the vi) is not using the derivative of the error, but the derivative of the process variable. I know that it is sometimes used to prevent the effect of fast changing set points, but it's a shame that they don't how about it in the help, and that's not a selectable feature. Do you know if there is a way to edit these pre-packaged vi?

  Another question, about the style of labview. I've included the footage just for the terms of the Group and to make the code more readable separate the integrator of the derivator and not having a knot of tangled signals. Especially when I run the cleaning diagram, it becomes very tangled. I searched for another way to do this (container boxes, groups). Would be nice if you could suggest me something that I can do for her.

  Thanks again,

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  ____________

  Blog: LucD notes

  Twitter: lucd22

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