Chapter 5: Root LocusChapter 5: Root Locus

Nov. 13-15, 2012

Two Conditions for Plotting Root Locus

1

1

( )1

( )

m

g ii

k n

jj

K s zG s

s p

Given open-loop transfer function Gk(s)

Characteristic equation

m

ii

n

jj

g

zs

ps

K

1

1

|)(|

|)(|

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

kkpszsn

jj

m

ii

Magnitude Condition and Argument Condition

Content Rules1 Continuity and Symmetry Symmetry Rule

2

Starting and end points

Number of segments

n segments start from n open-loop poles, and end at m open-loop zeros and (n-m) zeros at infinity.

3 Segments on real axis On the left of an odd number of poles or zeros

4 Asymptote n-m segments：

5 Asymptote

3

Rules for Plotting Root Locus

mn

zpn

j

m

iij

1 1

)()(

,2,1,0,)12(

kmn

k ＝

4

7Angle of

emergence and entry

Angle of emergence

Angle of entry

8 Cross on the imaginary axis

Substitute s = j to characteristic equation and solve

Routh’s formula

n

j

m

zii

ijz k1 1

)12(

m

i

n

pjj

jip k1 1

)12(

6Breakaway

and break-in points

0

][

ds

sFd 0 sZKsPsF g

0 sZsPsZsP

m

i

n

j ii pz1 1

11

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jBreak-in point

jBreakaway point

Rule 6: Breakaway and Break-in Points on the Real Axis

Use the following necessary condition

0 sZsPsZsP

0

d

dor 0

1

d

dor 0

d

d

s

K

sHsGss

sHsG g

ji pszs

11

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3.3.Symmetry 5.Asymptote6.Breakaway and break-in points4. Segments on real axis1.Draw the open-loop poles and zeros

-1-2-3

2. Two segments

polepolezero

18012

)12(180

k

0

12

3)21(1 1

mn

zpn

j

m

iij

breaka

wayBreak-

in23a

aaa

2

1

1

1

3

1

Example 5.3.1: Given the open-loop transfer function, please draw the root locus.

)2)(1(

)3()()(

ss

sksHsG

3

21

1

1

a

aa

zs

ps

k m

ii

n

jj

=0.172k =5.818k

2

* )1()(

s

sKsGk

Example 5.3.2:

j

0

s1

s2

Conclusion: For the open-loop transfer function with one zero and two poles, the root locus of characteristic equation is probably a circle in the complex plane.

please prove that the root locus in the complex plane is a circle.

Given the open-loop transfer function

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Example 5.3.3:

)2)(1()()(

sss

KsHsG

1. Open-loop poles and zeros

-1-2

2. Segments on real axis3. Asymptotes

103

210

180,6003

)12(180

k

4. Breakaway and break-in points

58.142.0 0]263[ 0

21

2

sssolutionssyieldssZsPsZsP

-0.42

5. Points across the imaginary axis

Ks

Ks

Ks

s

Ksss

0

1

2

3

23

3

63

21

023

K=6063 2 s

j1.414K=6

K=6-j1.414

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3. Symmetry4. No segments on real axis5. Asymptote6. Breakaway and break-in points7.The point where the locus across the imaginary axis

Example 5.3.4:

1. Find poles and zeros

-1

2. 4 segments

14

11111 1

mn

zpn

j

m

iij

135454

)12(180或

k

311010123

0)]181)(181()1[(

3,21

23

2

jssSolutionssssyields

ds

jsjssd

Breakaway

ks

ks

ks

s

ks

kssss

19

014

44841914

0404

19241

01940244

0

1

2

3

4

234

)181)(181()1()()(

2 jsjss

ksHsG

484-4 =0 get =121k k

014014 2 s

16.3js =121-j3.16k

j3.16 =121k

Example 5.3.5

s

K

15.0

5.0

ss

sR sC

2

Please sketch the root locus with respect to K=[0,+∞).

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Extension of Root Locus

Parameter Root Locus

Zero-degree Root Locus

Canonical form1

)(

)()(

1

1

n

jj

m

ii

gk

ps

zsKsG

Root locus gain

1. How to sketch the root locus for other parameters?

2. How to sketch if Gk(s)=1

ConventionalRoot Locus

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Example 5.3.6: Ks is a ramp feedback

gain, please sketch the root locus with respect to Ks=[0,+∞).

010)102(2 sKs s

01102

102

ss

sK s

0)(')('1 sHsG

102

10)(')('

2

ss

sKsHsG s

1. Break-in and breakaway point2. Angle of emergence 198)10890(180

3.412.3

65.365.3

10

12.310

0

21

1

1

zs

psps

zs

ps

K

s

ds

sdF

m

ii

n

jj

s

90o

108o

198o

0101022 sKss s

5.3.2 Parameter Root Locus

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Example: Sketch the root loci of the system with the open loop transfer function:

0 : )1(

)1()()( 1

1 Kss

sKsHsG

5.3.2 Zero-degree Root Locus

)1(

)1(

)1(

)1()()( 11

ss

sK

ss

sKsHsGAnalysis:

For this kind of systems, the characteristic equations are like as:

1)()( 0)()(1 111111 sHsGKsHsGK

0 : 1

)( gg

k KK

sG

,2 ,1 ,0 ,2)( kksGk

Magnitude equation

Argument equation

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1 1

Root locus by using the sketching rules with the following modification:Real-axis: Left of even number of zeros or polesAsymptoteAngles of emergence and entry

1

1

K

js

)16.0(

3.0

1

K

s

)17.6(

3.3

1

K

s

For Kg varying from ∞－ → 0 together with Kg =[0,+∞) simultaneously, the root loci are named the “complete root loci”.

m

i

n

pjj

jip k1 1

2

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5-4 Application of Root Locus

Insert a zero )4)(1(1

)()(

)4)(1(1)(

sss

asKsG

sss

KsG g

kg

k

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Add a pole to the open-loop transfer function

))(2(

)3()(

)2(

)3()( 1

21

1 asss

sKsGH

ss

sKsGH

Im

Re

Im

5

Re

Im

23 23

23

Re

Re

23

Im

1

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Generally, adding an open zero in the left s-plane will lead the root loci to be bended to the left. The more closer to the imaginary axis the open zero is, the more prominent the effect on the system’s performance is.

Generally, adding an open pole in the left s-plane will lead the root loci to be bended to the right. The more closer to the imaginary axis the open pole is, the more prominent the effect on the system’s performance is.

5.4.1 The effects of Zeros and Poles

Attracting effect

Repelling effect

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Example 5.4.1:

How to enlarge the stability region?)12)(15)(110(

6.3)(

sss

KsG c

k

)5.0)(2.0)(1.0(

036.0)(

sss

KsG c

k

)5.0)(2.0)(1.0(

sss

K

60o

j0.2ω1=0.17

Kc=0.778

-j0.2ω2=-0.17Kc=0.778

ω3=0Kc=0.278

)5.0)(2.0)(1.0(

)4.0()(

sss

sKsGk

ω1=0Kc=0.278

-0.06-0.027

278.0for 0axisimaginary with Intercept

06.0 :point Breakaway90 1.0:Asymptote

1

12

cK

a

278.0for 0778.0for 17.0

:axisimaginary with Intercept 0.027a:pointBreakaway

180,60 2.0:Asymptote

3

12

312

c

cK

K

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5.4.2 Performance Analysis Based on Root Locus

Example 5.4.2: Given the open-loop transfer function

)15.0()(

ss

KsGk

please analyze the effect of open-loop gain K on the

system performance.

Calculate the dynamic performance criteria for K ＝5.

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k¡ú¡

Þ

0

k £½0

k £½2

k £½0

k £½3

k £½10

k £½3

k £½2

3 j

k¡ú¡

Þ

[s ]

k £½1£­ 2 £­ 1

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It is observed from the root locus that the system is

stable for any K.

For 0 ＜ K ＜ 0.5(0 ＜ k ＜ 1), there are two different

negative real roots.

For K=0.5(k=1), there are two same negative real roots.

For K ＞ 0.5(k ＞ 1), there are a pair of conjugate

complex poles.

For K=5( k=10), the closed-loop poles are 311 2

12 jjs nn

316.016.3

1,16.310 n

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The criteria for transient performance can be given by

%35%100%100 05.11/ 2

eep

Peak time stn

p 05.11 2

Settling time %)5(33

stn

s

Overshoot

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1. We can get the information of the system’s stability by checking whether or not the root loci are on the left half s-plane with the system’s parameter varying.

2. We can get some information of the system’s steady-state error in terms of the number of the open-loop poles at the origin of the s-plane.

3. We can get some information of the system’s transient performance in terms of the tendencies of the root loci with the system’s parameter varying.

4. If the root loci in the left half s-plane move to somewhere far away from the imaginary axis with the system’s parameter varying, the system’s response decays more rapidly and the system is more stable, vice versa.

Complement Use Matlab to sketch root locus

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2

12

1

( ) +7 +12+3 +4= =

+1 +2 +3 +2( )

m

g iggi

k n

jj

K s z K s sK s sG s

s s s ss p

In Matlab:

num=[1 7 12]

den=[1 3 2]

rlocus(num,den)

sys=tf[num,den]

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