The transfer function of the PID block doesn't show the derivator.

Question

Hello,
I am trying to use the PID vi, but I when I try the box by itself, it doesn't behave as a "clasic" PID should behave. The main problem is that I don't manage to see the derivator. In the attached vi I compare the transfer function of the PID vi with the transfer function of a PID built by me. My version shows all what a PID should have: integral section (with decreasing magnitude and -90 phase), center area (with constant magnitude and 0 phase), and derivative area (with increasing magnitude and +90 phase).
The PID vi only shows the integral part.
You can also select a step input, and see the output directly. If you choose a large enough derivative time (100 times bigger than the integrator time), and you look closely to the first part of the output, you will see the pick due to the derivator in my version, but not in the PID.vi version.
Does anyone knows what am I doing wrong? 
Kind regards,
Pablo Estevez
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Attachments:
TestPID.vi ‏31 KB

Answer 1

Dear Nathand,
Thanks for your answer, I tried the change and you are right. That shows that this not a standard PID, since that means (and actually I can see it now by checking inside the vi) that it is not using the derivative of the error but the derivative of the process variable. I know that this is used sometimes to prevent the effect of fast changing set-points, but it is a shame that they do not comment on it in the help, and that this is not a selectable feature. Do you know if there is a way to edit these pre-packaged vi's? 
One more question, about the labview style. I included the sequences just to group terms and make the code more readable to separate the integrator from the derivator and not have a knot of entangled signals. Specially when I run the clean-up diagram, it gets very entangled. I have been looking for another way to do that (container boxes, groups). It would be nice if you could suggest me something I can do for it.
Thanks again,
Pablo