3
$\begingroup$

I am trying to understand how a PID controller moves the poles and zeros of an transfer function. I've been playing a bit with it, but aren't able to see some kind of connection.

I mean that P and I rises the Overshoot which would mean that the damping ratio gets smaller, thereby should away from the real axis.

and D should should do the opposite, but it doesn't seem to be true with the examples i've used.. am i doing something wrong??

Well i kind of just want a general knowlegde of how it affect second order systems.

$\endgroup$
  • $\begingroup$ Can you give some more details on the examples you're trying? $\endgroup$ – ryan0270 May 17 '14 at 2:31
  • $\begingroup$ Well i kind of just want a general knowlegde of how it affect second order systems. $\endgroup$ – Control May 17 '14 at 9:18
0
$\begingroup$

So imagine this scenario:

You are driving you car towards a red stop light. As a human, you would naturally adjust your speed so that you smoothly stop in front of the stop line.

Here is how a PID controller would do it:

P: The proportional part of the controller would measure the exact distance from the car to the stop light, and drive slower and slower as it approaches the light. The P-controller will drive super slow when getting close to the light, eventually stop without actually reaching it.

I: The integral part of the controller would go fast towards the light, pass it, realise it had gone too far, drive back, drive back too far, then go forwards again, drive pass the light again, and so on. The I-controller might, in fact, never stop on the line.

D: The derivative part of the controller would know exactly when to start breaking, so that it does not break to early - like the P-controller - nor does it drive to far - like the I-controller. Instead, it would start breaking exactly when it needs to, giving a firm but smooth stop in front of the stop light.

A properly tuned PID would combine these characteristics: The P would measure the exact distance to the light, and adjust the initial speed according to this distance. The I would be more aggressive, and enforce that no breaking is done until the the stop light is close, while the D would take over breaking when it realises the car is going way too fast towards the stop light, and break firmly.

Depending on the weighting of Kp, Ki and Kd, these characteristics will be more or less visible in the final system. Of course, the example contains some simplifications. For a general tuning guide, have a look at the Ziegler-Nichols method; the first paragraph in this wiki pretty much explains it all. Note that this tuning method is merely a guide, and that you will most likely have to manually fine tune the system afterwards.

Hope that helps!

$\endgroup$
  • $\begingroup$ thanks for you answer, it cleared some things, but i am still missing an answer on how it affect the root locus of an system. $\endgroup$ – Control May 19 '14 at 8:34
  • $\begingroup$ All right. I'm a bit rusty on this, so I might be wrong, but this is how I remember it: Poles in the right half plane gives an unstable system. The further out in the left half plane the poles are placed, the faster the system will stabilize. Finally, the rate of oscillation will increase as their distance from the real axis increases (and the angle between the poles and the imaginary axis decreases). Was that more of what you were aiming for? :) $\endgroup$ – Tormod Haugene May 19 '14 at 14:06
  • $\begingroup$ Well.. more how increasing Ki,Kp and Kd moves the poles and zeroes for an transfer function. $\endgroup$ – Control May 19 '14 at 14:25
  • $\begingroup$ Well, I believe there is no single answer to that. If you, for instance, by increaseing Kp, the poles can both move to the left or to the right, depending on the system. The same can be said for the Ki and Kd parameters. $\endgroup$ – Tormod Haugene May 19 '14 at 14:30
  • 1
    $\begingroup$ Never mind.. I've realized it, it because of the increasing and decreasing damping ratio. $\endgroup$ – Control May 20 '14 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.