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P-only controller via root locus P-only controller via root locus

P-only controller via root locus - PowerPoint Presentation

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Uploaded On 2018-11-09

P-only controller via root locus - PPT Presentation

by Frank Owen PhD PE polyXengineering Inc San Luis Obispo California Ponly controller simple gain adjustment K P 0 K I 0 K D 0 Controller does not change root locus of existing G ID: 725629

order system controller pole system order pole controller criterion poles polyxengineering locus root magnitude desired cont draw vector location

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Slide1

P-only controllervia root locus

by Frank Owen, PhD, PE

polyXengineering, Inc.

San Luis Obispo, CaliforniaSlide2

P-only controller: simple gain adjustment

K

P

≠ 0, K

I

= 0, K

D

= 0

Controller does not change root locus of existing G

OL

K

P

-adjustment moves poles along existing root locus to a point that gives desired performanceSlide3

Example 1: 1

st

-order system

A 1

st

-order system has a steady-state gain of 1 and a time constant of 2 seconds. Install a P-only controller to double the speed of the system.

 

Re

Im

-1/2

Pole at -1/T

-1/1

Need to move pole to hereSlide4

Example 1: 1

st

-order system, cont.

Re

Im

-1/2

-1/1

M

p1

= 0.5

Draw vector from OL pole to desired pole location

Use graphical method to solve:

q

p1

= 180

°

There is only one such vector, so angle criterion met

 

To meet the magnitude criterion, K

P

= 0.5Slide5

Example 2: 2

nd

-order system

Re

Im

-3

A 2

nd-order system has open-loop poles at s = -6 and s = -3. KOL without KP = 1. A controller is added with the goal of minimizing the time to peak value but without exceeding 10% overshoot.

Design this controller.

-6

 

10% overshoot means:

z

= 0.591 means:

 

53.8

°

Put closed-loop pole hereSlide6

Example 2: 2

nd

-order system, cont.

Re

Im

-3

Again, the graphical solution:

-6

53.8

°

M

p2

M

p1

q

p1

q

p2

Draw vectors from the OL poles to the desired location of the CL pole.

q

p2

Since

s

a

is halfway between the two OL poles, the red vectors form an isosceles triangle. Note that the two bottom interior angles are

q

p2

.

So

q

p1

+

q

p2

= 180

°

…and the angle criterion is satisfied.Slide7

Example 2: 2

nd

-order system, cont.

Re

Im

-3

Now use K

P to satisfy the magnitude criterion:

-6

53.8

°

M

p2

M

p1

q

p1

q

p2

q

p2

 

 

 

 

 

To meet the magnitude criterion, K

P

= M

p1

·M

p2

=39.9Slide8

That’s all folks!

Fin

©

polyXengineering, Inc.

San Luis Obispo, California

www.polyxengineering.com