Converting PID gains of an independent gains equation to a dependent gain equation.

[TConnolly]

I've got a loop that is extreemly well tuned for a CLX PIDE. I also have a SLC5/05 that is controlling an identical piece of equipment at a customer's site over here in Europe where I am now. This loop was tuned manually, but it does not respond nearly as well.

The gains for the independent gains equation are Kp = 3.5697, Ki = 434.447, Kd = .00405. Process gain is .334, process time constant is .425, dead time is .2.

What I have come up wth for the SLC5/05 dependent gains equation is Kc = 3.57, Ti = .008 (will have to be .01), and Td .001 (will have to be 0).

However, I'm not confident that I made the change correctly. I am hoping someone has a trick up their sleeve.

 

[Pandiani]

Hello Alaric,

As I see it, you did well.
Just remember:
Kp+Ki/s+Kds = Kc*(1+1/(Tis)+Tds)
which yields to:
Kp = Kc
Ti = Kc/Ki
Td = Kd/Kc

 

[Peter Nachtwey]

That will translate the gains but the numbers don't look right

http://www.controlguru.com/wp/p71.html
Using the IMC from www.controlguru.com you would get...

Kc=(1/Kp)*(Tp/(Tc+Tdt))
Ti=Tp

assuming Tc=0.1*Tp for aggressive tuning

Ti should be 0.425 which is much slower and Kp should be 5.247

(1/0.334)*(0.425/(0.0425+0.2)=5.427

It is the integrator time constant that doesn't look right for the plant information you have provided. The Kc is OK.
BTW, the Kd or Td isn't really necessary of a one time constant plant.

 

[Ron Beaufort]

the numbers look "fishy" to me too ... have you checked the "timing/triggering" used on both platforms? ... usually that's the part that "skews" the Integral and Derivative settings ... if the timing is "off" then any mathematical conversion that you do won't be accurate ...

post the programs (both please) if you can ... or email them if you'd feel better doing it that way ... I'd be glad to take a look ...

 

[TConnolly]

Peter, I noticed you used .1Tp for Tc, while Doug shows that Tc is the greater of .1Tp or .8Tdt, which in this case .8Tdt would be the greater value, yielding a Kc of 3.467 instead of 5.427.

Is there a reason you chose .1Tp over .8Tdt?

 

[Peter Nachtwey]

I screwed up but you caught it. Good.

Peter, I noticed you used .1Tp for Tc, while Doug shows that Tc is the greater of .1Tp or .8Tdt, which in this case .8Tdt would be the greater value, yielding a Kc of 3.467 instead of 5.427.

Is there a reason you chose .1Tp over .8Tdt?

At least you are double checking me and yourself.

 

[TConnolly]

Today I tried using the gains I derived and none of them performed worth a damn. Finally, scrolling through the program I find someting that doesn't look familiar, its not part of the program I wrote (I had to do an upload when I got here, its one of several changes). Anyways, it looks like someone baked-his-own first order lag filter - at least attempted one, it doesn't look like any way I've seen one done - and dropped it in the program (probably an attempt to compensate for bad tuning). Since this is a very fast responding process, that skews the process time constant and lag time, so I'm working off bad numbers. So I slapped in a JMP-LBL to skip the partially baked filter. All is well now except for deleting the filter once I decipher if thats what it really was.

Thanks Peter, Ron, Pandiani.