b_lecture10 the root locus rules automatic control system

The Root Locus Method

Introduction

The performance of a feedback system can be described in terms of the location of the roots of the characteristic equation in the s-plane.

It is very useful to determine how the roots of the characteristic equation move around the s-plane as we change one parameter.

The root locus method was introduced by Walter R. Evans in 1948.

Introduction

Root-locus is a graphical technique for sketching

the locus of roots in the s-plane as a parameter is

varied in a given range.

Root-locus is based on the poles and zeros of

the open-loop transfer function for determining the

poles of the closed-loop transfer function of a

system.

Root Locus Concept

The characteristic equation

022 Kss

roots Ks 112,1

em.f the systfunction o transfer open-looples of theare the possK 22

01

0

. 1 1 21 otsme real roare the sassK

jsK

jsK

1

11 2

2,1

2,1

K)2(

1

ss

The open-loop transfer function:

)2()(

ss

ksG

0)2(

1)(1

ss

ksG

. 10 2 1 al rootsfferent reare two disandsK

reasing.with k incoots grow n of the rary sectioThe imagin

t.is constanthe roots ection of The real s

x roots.ate comple of conjugare a pairsands 21

Root Locus Concept

According to above discussion: we can sketch the locus of the roots varying with k from 0 to + on the S-plane.

Definition of Root locus:

The root loci is the path of the roots of the

characteristic equation (or the poles of the

systems closed-loop transfer function), traced out in the s-plane as a system parameter is

changed.

Root Locus Concept

Root - root of s polynomial equation(the characteristic equation)

Locus - Set of points (roots)

system.order -second theof 1 case with therespected

becan This poles. real thebeing poles theof because dampedlly exponentia is

response system theofportion transientthe

,10 range in the changes When (2)

k

What information can be got from

the root-loci of the system?

(1) when k changes in the range 0 ~ +, the system is always stable, because the loci of the closed-loop

poles are always in the left-half of

the s-plane.

Root Locus Concept

system.order -

second theof 10 case with therespected

becan This poles.complex conjugate the

being poles theof because sinusoid damped

lly exponentia a is response system theof

portion transient the,1 When (3)

k

Root Locus Concept

constant. being poles theofsection real theof because not varied be willtime

setting theAnd axis. real thefromaway poles theof beause augmented being

with increased be willresponse system theofovershoot the1 when (4)

k

k

... etc.

).0( 0 pole loop-open a of because system typea is system The (5) ks

system. a of eperformanc theanalyze tolocus-root use why weisIt system.

a of loci-root thefromn informatio oflot aget can that weobvious isIt

Root Locus Concept

We usually interested in determining the locus of the roots

as K* varies as K0

)(sG

)(sH

n

j

j

m

i

i

ps

zs

KsHsG

1

1

)(

)(

)()(

The characteristic equation is

0)()(1 sHsG

rearrange the equation, if necessary, so that the parameter of

interest K, appears as the multiplying factor in the form, and

write the polynomial in the form of poles and zeros as follows:

Two Basic Criterion

K*: The root-locus gain, which is proportional to

open-loop gain K. 121

121

22

22

sTsTsTs

sssKsHsG

kkkiv

lllj

(For reasons that will become clear later, this is the definition of the positive or 180

degree locus. Will later define the negative, or 0 degree locus.)

)(

)(

)()(

1

1 eex variabls: a compl

ps

zs

KsHsGn

j

j

m

i

i

criterionMagnitude

ps

zs

Kn

j

j

m

i

i

1

||

||

1

1

criterionAngle kkpszsn

j

j

m

i

i ,2 ,1 ,0 )12()()(11

Two Basic Criterion

0)()(1 sHsG 01)()( jsHsG

the magnitude and angle requirement

for the root locus are:

Two Basic Criterion

1. The number of separate loci is equal to the order of the characteristic equation. Number of loci branches = n

2. The loci are symmetrical about the real axis. The root locus is symmetrical about the real axis since

the roots of 1+G(s)H(s)=0 must either be real or appear

as complex conjugates.

Root Loci Construction Rules

Basic approach: The complete root loci can be constructed

point-by-point finding all points in the s-plane that satisfy the

angle criterion.(usually dont use this approach directly.)

The rapid sketching procedure of the root locus

shown as follows:

0)(

)(

1

1

1

n

j

j

m

i

i

ps

zs

K 0)()(11

m

i

i

n

j

j zsKps

0K njps j ,2,1 , 0)(1

n

j

jps

3. The root loci begin at the open loop poles and end at open zeros. (1) The root loci begin at the open loop poles.


(2) The root loci end at the open loop zeros.

0

)(

)(

1

1

1

n

j

j

m

i

i

ps

zs

K

Kps

zs

n

j

j

m

i

i1

)(

)(

1

1

mizs i ,2,1 , 0)(1

m

i

izsK

Example Second-order system

)4(

)2(

)125.0(

)15.0()()(

*

ss

sK

ss

sKsHsG

zero

poles

-4

-2

0

j


. ,

:

zeropole

symbolsthe withhe s-planezeros in tpoles and open-loop locate the

system physical actual

:zeros loop-open and poles loop-open of loci-root For the

mn

mn

4. Root Loci on the Real Axis


The locus covers the section of the real axis to the left of an odd number of poles and zeros of G(s)H(s).

)4(

)2(

)125.0(

)15.0()()(

*

ss

sK

ss

sKsHsG

Example Second-order system

zero

poles

-4

-2

0

j

s

criterionAngle

kkpszsn

j

j

m

i

i

,2 ,1 ,0 )12()()(11

5. The loci proceed to the zeros at infinity along asymptotes.

These linear asymptotes are centered at a point on the real

axis given by

mn

zpm

i

i

n

j

j

a

11

The angle of the asymptotes with respect to the real axis is

)1,,1,0( )12(

mnk

mn

ka


j

0)3)(1(

)2(1)(1

sss

sKsG

Example: Third-order system

A single-loop feedback control

system has a characteristic

equation follows: -1 -2 -3 0

113

)2()310(

a

)1( 270

)0( 90180

13

120

0

0

k

kka


Sequential Example: A single-loop feedback control system

has a characteristic equation as follows:

0)2)(1(

1)()(1

sss

KsHsG

The intersection of the asymptotes is

13

210

a

The angles of the asymptotes are

)2( 300

)1( 180

)0( 60

1803

)12(

0

0

0

0

k

k

kk

a


-2 -1

j

6. Breakaway points on the root loci.

Breakaway points on the root loci correspond to multiple-

order roots of the equation 1+G(s)H(s)=0. The breakaway

point d can be computed by solving

It is important to point out that the condition for the breakaway

point given in Eq.(above) is necessary but not sufficient. In

other words, all breakaway points on the root loci must satisfy

Eq.(above), but not all solutions of Eq. are breakaway points.

m

i

n

j ji pdzd1 1

11


Necessary conditions only

7. The tangents to the loci at the breakaway points.

In general, due to the phase criterion, the tangents to the

loci at the breakaway point are equally spaced over 3600

Breakaway

point

090

045

045


The two loci at the breakaway

point are spaced 180o apart. The four loci at the breakaway

point are spaced 90o apart.

0)2)(1(

1)()(1

sss

KsHsG

02

1

1

11

ddd

58.1 ,42.0 21 dd


Sequential Example: A single-loop feedback control system

has a characteristic equation as follows:

-2

j

-1

m

i

n

j ji pdzd1 1

11

here no zeros

8. Intersection(the crossing points)of the root loci with the imaginary axis and corresponding values of K*.

0)2)(1(

1)()(1

sss

KsHsG

023 23 Ksss

023 23 Kjj

portion real 03

portionimaginary 022

3

K

6 2

0 0

K

K

The characteristic equation is


To substitute into the characteristic

equation then to solve the value for

js

Sequential Example:

With a little practice, you should be able to sketch root loci very rapidly.

-2

j

-1


Sequential Example: 0)2)(1(

1)()(1

sss

KsHsG

The intersection of the asymptotes is

13

210

a

The angles of the asymptotes are

)2( 300

)1( 180

)0( 60

1803

)12(

0

0

0

0

k

k

kk

a

6 2

0 0

K

K

58.1 ,42.0 21 dd

Breakaway points on the root loci

Intersection of the root loci with the

imaginary axis.

kp

s

)( kps )(lim kps

p psk

k

9. The angles of arrival and departure.

m

i

n

kjj

jkikp ppzpKk1 1

)()()12(


Departure angle

kz

s

)( kzs )(lim kzs

z zsk

k

9. The angles of arrival and departure.

n

j

jk

m

kii

ikz pzzzkk11

)()()12(


Arrival angle

Example: ))(1(

)()(jisjss

KsHsG

jpjpp 1 1 0 321

1

-1

-1 0

j 2p

3p

1p

)()()12( 32122 ppppkp

000 90135180)12(2

kp

0452

p

0453p


m

i

n

kjj

jkikp ppzpKk1 1

)()()12(

Example: )5.15.0)(5.2(

)2)(5.1()(

*

jSss

jssKsG


10. The sum of closed loop poles

0

)(

)(

1

1

1

n

j

j

m

i

i

ps

zs

K

0)()(11

m

i

i

n

j

j zsKps

0)(1

n

i

iss

012

2

1

1

nn

nnn asasasas

constpsan

j

j

n

i

i

11

1

If the order of the denominator of the open loop transfer

function is greater than the numerator by at least 2 ,

then the sum of the closed loop poles is a constant.


)2( mn

constpsn

j

j

n

i

i

11

. 0)((s)1 ofroot theis pole, loop-open theis where sHGsp ij

)2( mn


Example:


In terms of above rules we can rapidly sketch the

root-loci of a control system.

11. Determine the parameter value Kx at any point Sx on the root loci using the magnitude requirement.

xss

zs

ps

K

ps

zs

KSm

i

i

n

j

j

n

j

j

m

i

i

x

1

1

1

1*

||

||

1

||

||

ocriterionMagnitude

ps

zs

Kn

j

j

m

i

i

1

||

||

1

1

xss

psK

ps

KSn

j

jn

j

j

x

1

1

* || 1

||

1 ocriterionMagnitude

ps

Kn

j

j

1

||

1

1

If there is no zeros

Sketch the root-loci for the following open-loop transfer functions:

)4(

)6)(2()( )1(

ss

ssKsG

)2(

)3()( )3(

ss

sKsG

)6(

)4)(2()( )2(

ss

ssKsG

)1(

)3)(2()( )4(

ss

ssKsG

)2(

)136()( )5(

2

ss

ssKsG

)136(

)2)(1()( )6(

2

ss

ssKsG

Re

Im

Re

Im

Re

Im

Re

Im

Re

Im

Re

Im

Examples

(1) (2) (3)

(4) (5) (6)

b_lecture10 the root locus rules automatic control system

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