PI design via root locus - PowerPoint Presentation

Slide1

PI design via root locus

By Frank Owen, PhD, PE

polyXengineering, Inc.

San Luis Obispo, CaliforniaSlide2

Purpose of integral control

The primary purpose of using integral control is to reduce or eliminate steady-state error

In Controls, usually you don’t get something for nothing

Placing a pole right on the border of stability will make a branch of the root locus start thereThis tends to make the system less stableThere is a conventional strategy for placing a pole at the originWe like the dynamics of the existing system, just not the fact that it has non-zero essSo add pole at origin without disrupting existing system dynamics

Let’s see how to do this…Slide3

Example system: 2

nd

–order system

As an example, take a 2nd-order system with OL poles at s = -2, -4 System has two asymptotes with

s

a

= -3The system operates with a P-only controller at 10% overshoot: , thus z·wn = 3 , so For completeness, let KOL = 1 without the controllerThus, , KOL-pz = 8See video on PD design for unit step response of this raw system

Aim:Eliminate essPreserve these system dynamicsSlide4

Put a pole at the origin

…in fact, now, if we were to turn up the gain, the system would go unstable.

Put the pole at the origin

…but this changes the root locus completely…Slide5

How not to change the existing root locus

The answer to this is to put a zero near the pole at the origin. It needs to be far enough away from the pole not to cancel it.

q

p1

q

z1

Recall the formula for evaluating GOL at the closed-loop pole: 

M

p1

M

z1

M

p2

M

p3

Now draw the vectors to the closed-loop pole that are part of evaluating it graphically.Slide6

Effect of the zero near the pole at the origin

q

p1

q

z1

M

p1Mz1Mp2Mp3

But M

z1

≈ M

p1

But M

z1

≈ M

p1

But M

z1

≈ M

p1

…and

q

z1

≈

q

p1

So adding the pole and the zero near each other did not affect the root locus very much.Slide7

How close do we locate the zero?

The question always arises, how close to put the zero next to the pole at the origin. This depends on the scale of the system. In this example, the largest pole is 4 units away from the origin. So let’s choose to put the zero at 1/10

th

of this distance. So we put the zero at s = -0.4 . The resulting system can be checked using Matlab to view its response.Slide8

Now calculate K

P

and K

IThe controller we have added with the pole and zero does not change the gain of the P-only controller used to put the closed-loop pole originally at s = -3 ± j·4.09 .

Our controller is now a gain, a pole at the origin, and a zero at -0.4 . That is

As stated, KOL-pz of the system without the controller is 8 . Using the graphical solution procedure, to meet the magnitude criterion. Note that we have used the pz form here, which is required when using the graphical solution for the root locusSlide9

Now calculate K

P

and K

IBut

So

,

, and thus We compare this with the standard form of a PI controller to get KP and KI…Slide10

Now calculate K

P

and K

I

From this we can see that

,

Now let’s check this via Matlab…Slide11

Matlab commands to compare P and PI

Enter these Matlab commands to compare the system with P-only control and with PI control.Slide12

Comparison of P vs. pi

The integrator continues to act until

e

ss = 0

P-only controller has

e

ss ≠ 0PI controller gets rid of essOur analysis is approximate because we simply used the same gain we had with the P-only controller. But it is evident that the dynamics of the system is very similar to the P-only controller but with the integral action superimposed on it.Slide13

That’s all folks!

Fin

©

polyXengineering, Inc.San Luis Obispo, Californiawww.polyxengineering.com