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CISE302_ch 7 Al-Amer2008 ١

CISE302: Linear Control Systems

Dr. Samir Al-Amer

Chapter 7: Root locus

Term 081

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Learning Objectives

Understand the concept of root locus and its role in control system designBe able to sketch root locus and use MATLAB to plot the root locusRecognize the role of root locus in parameter design and sensitivity analysisBe able to design controllers using root locus

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Outlines

The concept of root locus

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Root locus

The location of the roots of the characteristic equation determines the systems response.Modifying one or more of the system’s parameter cause the roots of the characteristic equation to change.Root locus is a graphical method that determines the location of the roots as one parameter change.

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Root locusRoot locus was developed by Evans in 1948.Used to analyze and design systemsWe use it to ensure that the roots of the characteristic equation are in a desirable regionCan be used to study sensitivity of the pole location w.r.t changes in the parameter.

Desirableregion

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Root Locus Using MATLAB

rlocus(1,[1 2 0])

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Example 1

KrKr

Kr

arerootsKs

−−−=−+−=

−±−=

=++

11,112

442

:02s equation sticcharacteri the

21

2,1

2 )(sR E(s)Y(s)

_ K )2(1+ss

XX

Root Locus

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Root locus can be used to study sensitivity of system w.r.t variations in the parameter

K=0:0.1:10; G=rlocus(1,[1 2 0],K); plot(G,’.’)

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Mathematics of root locus c

)(sR E(s)Y(s)

_ K )(sG

integer:360180)(;1

01)(0)(1

kksKGKG

jsKGKGsKG

±=∠=⇒

+−=∠

=+

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Question c

)(sR E(s)Y(s)

_ K )(sG

? and 0between variesK as change0)(1 equation

sticcharacteri theof roots thedoes How

∞=+ sKG

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Example

200)2(0

12)s(s

K)(

−==→=+→=

=+

=

sorsssKWhen

sKG

)(sR E(s)Y(s)

_ K )2(1+ss

The locus of the roots of 1+KG(s)=0 starts at the poles of G(s)

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Root Locus Procedure

Initial Step:Express the characteristic equation in the form

1+K P(s) = 0 Factor P(s) and rewrite the equation in the form

0)(

)(1

1

1 =−

−+

∏

∏

=

=n

ii

m

ii

ps

zsK

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Example 1

Apply the initial step of the root locus procedure for the system

sssK

23 23 ++_

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Example

Step 1: Express the characteristics as

023

11

0)(1

23 =++

+

=+

sssK

sPK

sssK

23 23 ++_

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Example

Step 2: Factor P(s)

0)2)(1(

11

0)(1

=++

+

=+

sssK

sPK

sssK

23 23 ++_

n = 3

m = 0

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Root Locus Procedure Locate the open loop poles and zeros of P(s)

Locate the poles and zeros of P(s) on the s-plane. Use X for poles and O for zeros

0)2(

)1(1 =+−

+sssK

XX-2 1

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The root locus starts at the poles of P(s) and terminates at the zeros of P(s) or at infinity

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Root Locus Procedure # of separate root loci

The number of separate root loci = The number of poles of P(s) Note that ( # poles ≥ # zeros)

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Example 2

How many separate root loci does the root locus of the system has?

sssK

23 23 ++_

0)2)(1(

11

0)(1

=++

+

=+

sssK

sPK

# of poles =3Three separate root loci

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Root Locus ProcedureSegments of real line that is part of root locus

2: Segments of real line:A segment of real line is a part of root locus if sum of # of poles plus # of zeros to right of that segment is odd.

XXX

3 2 1 0

Yes No Yes No

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ExampleA segment of real line is a part of root locus if sum of # of poles

plus # of zeros to right of that segment is odd.

XXX

6 5 3 2 1 0

No Yes Yes No Yes No

X

Double poles

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ExampleA segment of real line is a part of root locus if sum of # of poles

plus # of zeros to right of that segment is odd.

XXX

5 4 3 2 1 0 Yes No Yes No Yes No

X

X

Complex poles or zeros do not

affect the count

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Example

Step 2: Segments of real axis that are parts of the root locus

0)2)(1(

11 =++

+sss

K

XXX

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Asymptotes

If n > m # of poles of P(s) ># of zeros

The root locus approaches infinity at n-mdirections.

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Asymptotes

Step 3: # of asymptotes = n – m =3

300,180,60

2,1,018012

103

}0{}210{

=Φ

=−+

=Φ

−=−

−−−=

−−

= ∑ ∑

A

A qmn

qmn

zerospolescentroid

XXX60

-2 -1

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Example

How many asymptotes?

0)1(

11 =+

+ss

K

XX

n=2, m=0, n-m=22 asymptotes

Centroid =(0+(-1))/2

=-0.5

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Intersection with Imaginary axis

To find intersection with imaginary axis use Routh Hurwitz method

Ks

KKs

Kss

Kssssss

K

0

1

2

3

23

23

6onIntersecti3

63

21023

023

11equation sticCharacteri

=−

=+++

=++

+

XXX60

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Intersection with Imaginary axis

263

s Equation006321

2

1

2

3

jss

Auxiliarys

Kss

±=⇒+

=

XXX60

Intersection with imaginary axis

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Breakaway points

Breakaway points are points at which the root loci breakaway from real axis or the root loci return to real axis.At breakaway points

Solve the above equation to determine the breakaway point. Select solution that are in segments of real axis that is part of root locus

0)(=

dssdP

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Breakaway points

To find breakaway points

( )( )

422604226057741 pointsaway break

023

26310

023

1

223

2

23

.Select. and.

sssss

sssdsd

−−−

=++

++−

=⎟⎠⎞

⎜⎝⎛

++ XXX-1-2

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Symmetry of root locus

The root locus is symmetric with respect to the real axis.

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Angle of departure

Sum of angle contributions of poles and zeros (measured with standard reference) = 180+360kIn the example all poles/zeros are realNo need to do this

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Angle criteria

At all points of the root loci : Sum of angle contributions of poles and zeros (measured with standard reference) = 180+360k

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Example 8Draw root locus for the following system

2)1(

++

ssK

_

1,21)(

0211

:

==++

=

=++

+

mnsssP

ssK

EquationsticCharacteri

X

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02

)1(1 =++

+ssKofLocusRoot

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Example 9Draw root locus for the following system

)3)(2)(1()5(+++

+sss

sK_

1,3,)3)(2)(1(

5)(

0)3)(2)(1(

51

:

==+++

+=

=+++

++

mnsss

ssP

ssssK

EquationsticCharacteri

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Example 9

21

13)5(321

270,90213#

5:,3,2,1:

1,3,)3)(2)(1(

5)(

−=

−−−−−−

=

=−=−−−−

==+++

+=

Centroid

Anglesymptotesof

ZerosPoles

mnsss

ssP

XXX

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0)3)(2)(1(

51 =+++

++

ssssK

oflocusroot

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Example10

How do we draw root locus for this system?

Ksss +++ 231

23_

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Example10

How do we draw root locus for this system?

Ksss +++ 231

23_

Initial Step: Express the characteristics as

?)(0)(1

sPisWhatsPK =+

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Example10

Initial Step: Express the characteristics as

( )

1231)(

0123

11

1230123

023

11:Equation sticCharacteri

23

23

2323

23

+++=⇒

=+++

+

++++==++++

=+++

+

ssssP

sssK

KsssKsssKsss

Continue the Root Locus Procedure

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Example11

How do we draw root locus for this system?

)2(2

2 Kssss

+++

_

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Example11

Initial Step: Express the characteristics as

( )

22)(

022

1

222202)2(

0)2(

21:Equation sticCharacteri

23

23

23

232

2

+++=⇒

=+++

+

++++

++++==++++

=++

++

sssssP

ssssK

KsssssKssssKsss

Kssss

Continue the Root Locus Procedure

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Controller Design Based on Root Locus

The idea here is to select the parameter K so that all the poles in the desired location.

Desiredregion

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The desired region is obtained to satisfy the given specificationsDraw the desired region on the root locusSelect the gain K such that all poles of the closed loop system are in the desired region

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Desiredregion

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What can you tell about behavior of this system?

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What can you tell about this system?

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Keywords

Root locusBreakaway pointsAsymptotesCentroidAngle of departureAngle of arrival