control systems - chibum lee · chibum lee -seoultech control systems root locus stability and...

Control Systems

Root locus

[email protected]

Control SystemsChibum Lee -Seoultech

Outline

Concept of Root Locus

Constructing root locus


Root Locus

Stability and transient response is closely related

with the location of poles in s-plane

How the poles of a given system migrate about

the s-plane as the parameters are varied.

Root locus method (Evans in 1948)


Example

DC motor with P control

• Open Loop Transfer Function:

• Closed Loop Transfer Function:

)(

1

)()()(

2

BJssT(s)

θ(s)

sTsBssJs

TBJ

y (=)

KBsJs

K

2

)( BJss

K


Example

Closed loop poles are

• Closed loop poles

• Natural frequency & Damping ratio

JK

B

J

Kn

2,

22

2

2 2)(

)(

nn

n

ssKBsJs

K

sR

sY

12

4, 2

2

21

nnJ

JKBBss


Example

For a given plant, (ie. J,B is fixed), how K will affect the location of poles?

Re

Im

roots conjugatecomplex 04

roots double04

roots real two04

2

4,

2

2

2

2

21

JKB

JKB

JKB

J

JKBBss

JK

B

J

Kn

2, cf)


Root Locus

Path of poles traced out in s-plane

as a system parameter varies from 0 to

• Characteristic equation is

• Closed loop poles satisfies

)(1

)()(

sKG

sKGsT

plane)-s(in 1)( sKG

0)(1)( sKGs

)12()(

1)(

ksKG

sKG


Root Locus

Magnitude condition

Angle condition

),2 ,1 ,0( )12()( kksKG

1)( sKG

Re

Im

,5 ,3 ,)( sKG

-1

1)( sKG


Recall Complex Functions


Example

Ex.

• Open-loop transfer function

• Characteristic Eqn.

• Closed-loop poles

02

0)2(

111)(

2

Kss

ssKKGs

)2(

1

ssK

Closed-loop pole

0

1

2

5

0 2

,

1

,

1

1

1

j

1

1

2j

j

K

Ks 11


Example


• At poles

0)2(

111)(

ssKKGs

)12()2(

1

1)2(

1

kss

K

ssK

pp

pp

)2(

)2(

1

2)2(

pp

pp

pppp

ssKss

K

ssssK


Example

)2()2(

1

2)2(

pp

pp

pppp

ssKss

K

ssssK

generalized method?


Constructing Root Locus

Step0: Prepare the characteristic equation

• Assume the parameter of interest is the gain K (0 K )

• Ex.

)( )())((

)())((1)(1)(

21

21

mnpspsps

zszszsKsKGs

n

m

parameter) locusroot :(0

1

21

0)2(1202

11

2

2

s

s

sssss

parameter) locusroot :()2)(1(

)1(10

)2)(1(

11 KK

ss

sK

ss

sK



Step1: Locate the poles & zeros of KG(s) on the complex plane

• the root locus branches start from the open loop poles (K=0) and terminate at the zeros (K=)

0)()(

)()()()(:0

1

1 1

n

mn

psps

zszsKpspsK

Open Loop Poles Starting point

Open Loop Zeros Terminating point

0)()(

)()()()(1

0)()()()(:

1

1

1

1

1

m

mn

mn

zszs

zszspspsK

zszsKpspsK



• # of separate loci = # of OL poles

• Starting point of root loci (K=0) Open loop poles (n)

• Terminating point of root loci (K=)

Finite zeros (m) : Open loop zeros

Infinite zeros (n-m)



Step2: Locate the segments of the real axis that are root loci

• The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros. (K>0)

• Root loci must be symmetrical with respect to the real axis.


Example

Ex.


• Step1:

• Step2:

01

11)(2

41

21

ss

sKKGs

0)4(

)2(21

ss

sK

Poles: s = 0,-4Zero: s = -2

s1 = -1



Step3: Find the center and the angle of asymptotes

• Root loci emanate from real axis along asymptotes centered at A

and proceed to the zeros at with angle of A

• #of distinct asymptotes is N=n-m (i.e., # of zeros at )

n = # of finite poles of KG(s)

m = # of finite zeros of KG(s)

n-m = Relative degree of system

• The center of asymptotes: (on the real axis)

• The angle of the asymptotes:

mn

zzzppp

mn

zpmn

m

i

i

n

j

j

A

)()(

212111

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

mnk

mn

kA


Example

Ex.

open-loop poles: 0,-2,-4,-4

open-loop zeros: -1

n=4, m=1

• Center of asymptotes:

2)4)(2(

)1(1)(1

sss

sKsGH

314

)1()4(2)2(

A


Example

3

5,,

314

)12(

kA

• Angles of asymptotes:


Step4: Find the breakaway point on the real axis (if any)


0 such that Find

)(

)(')()()('

0)()(

0)(

)(1)(1

2

ds

dKs

sY

sYsXsYsX

ds

dK

sKYsX

sX

sYKsKG


Ex.

• Open-loop

• Rewritten


0)4)(2(

1)(1)(

ss

KsGs

)4)(2()(

ss

KsG

)4)(2(

0)4)(2(

ssK

Kss


Example

Ex.

0)3)(2(

)1(1)()(1

sss

sKsHsG

)1(

)3)(2(

s

sssK

0)1(

61082

)1(

)3)(2(

2

23

s

sss

s

sss

ds

d

ds

dK

js 79.077.0,46.2


Step5: Find the crossover point where the locus cross the

imaginary axis

• Routh-Hurwitz criterion

• Let 𝑠 = 𝑗𝜔 in Δ 𝑠



Example

Ex. Find the crossover point of following open loop

transfer function G(s) and corresponding K

• Characteristic eqn.

• Find crossover point

- 25 -

)5)(3()(

sss

KsG

0)5)(3( Ksss

0158 23 Ksss

js

0)15()8()(15)(8)( 2223 jKKjjj

120

1200

CK

K

15

Routh Array

15



Step6: Find the departure angle (@ poles) and

the arrival angle (@ zeros)

• Angle of departure from a complex pole pj

D = π - angles of vectors to pj from other poles

+ angles of vectors to pj from zeros

• Angle of arrival at a complex zero zi

A = π - angles of vectors to zi from other zeros

+ angles of vectors to zi from poles

• Only need to deal with finite zeros

( the zeros will be taken care of by asymptotes)

),2 ,1 ,0( )12()( kksKG



211)()( sHsG

: angle from zero z1

1: angle from pole p1

2: angle from pole p2

121 D

))((

)(

2

)()()(

21

1

22

1

psps

zsK

ss

zsKsHsG

nn


Ex.

Example

)1)(1(

)2()()(

jsjs

sKsHsG

o

ooo

D j

135

4590180)1(

2- @ zerofrom angle

j-1- @ polefrom angle

135o

90o

45o

-2

j

-j


Example

Ex.

)1(

))(()()(

ss

jsjsKsHsG

ooooo

A j 225459090180)(

1- @ polefrom angle

0 @ polefrom angle

j- @ zerofrom angle

90o

45o

-1

j225o

90o



Step7: Complete the sketch and

locate closed loop poles of interest

• From K to find closed loop pole

Plug in numerical value of K to find s=+j

• From closed loop pole to find K

Use magnitude criterion

*1

1

ss

M

i

i

n

j

j

zs

ps

K

0)(

)(1)()(1

sD

sNKsHsG

( )

( )

D sK

N s



MATLAB command

• rlocfind

• rlocus

• rltool


Rules for Constructing Root Locus

Note: If the numerator and the denominator of the

open-loop transfer function have common factors, they

will cancel each other.

• Pole-zero cancellation.

The root locus plot of G(s)H(s) will then not show all

the roots of the characteristic equation.

• Only those that haven’t been cancelled are shown.


)2()2)(1(

)1()()(

ss

K

sss

sKsHsG

Cancellation of Poles of G(s) with Zeros of H(s)

If the denominator of G(s) and the numerator of H(s)

have common factors, they will cancel each other. (Pole-

zero cancellation.)

• Ex.

)(sY )(sY


Outline

Root locus examples


Example

Ex

• Step1: Open loop poles: 0,0,-2

Open loop zeros: -1

• Step2: Real axis portions (0,0), (-1,-2)

)0()2(

)1(1)()(1

2

K

ss

sKsHsG


Example

• Step3:

Center of asymptotes

Angle of asymptotes

5.013

)1()2(11

mn

zpm

j

j

n

i

i

A

2

3,

2)12(

mnkA


Example

Ex.

• Step1: Open loop poles and zeros

• Step2: Real axis portions

• Step3: Center and angle of asymptotes

• Step4: Breakaway point

• Step5: Crossover point

• Step6: Angle of departure and arrival

)0(32

)2(1)()(1

2

K

ss

sKsHsG

)(sY


Example


Example

Ex.







)0()2)(1(

1)()(1

Ksss

KsHsG

)(sY


Example

Ex

• Step1: Open loop poles: 0, -4, –4j4

• Step2: Real axis portions: (0, -4)

• Step3:

Center of asymptotes:

Angles of asymptotes:

)0(1286412

1)()(1234

Kssss

KsHsG

34

)444440(

jjA

4

7,

4

5,

4

3,

44)12(

kA




s4 1 64 K

s3 12 128

s2 53.33 K

s1 128- 0.225K

s0 K

0<K<568.89 for stability

crossover point : j3.266

Characteristic Eqn.01286412)( 234 Ksssss

Routh

Array

5533.27117.3,5767.1

)128128364(

)1286412(

23

234

js

sssds

dK

ssssK


Example

• Step6:

angle of departure

angle of arrival

4

3

4

4tan

4

3

22

1

3

33

4321

D


Example

• Step7:


Example

Ex.







20

𝑠(𝑠 + 1)(𝑠 + 4)

)(sR

+-

)(sY

1 + 𝐾𝑠


Example

Ex.







)(sR

+-

)(sY

)3(

)1(

ss

sk


Outline

Typical root locus configuration

Conditional stability

Non-minimum phase system


Typical Root Locus Configurations


Typical Root Locus Configurations

Caution: the relative position of the open-loop poles and zeros.


Conditional Stability

There may be ranges of K over which the system is

stable/unstable. Conditional Stability

• Ex.)(sY


Non-minimum Phase Systems

If a system has at least one zero in the RHP

Non-minimum phase system

• Gain should be limited


Outline

Addition of open-loop poles

Addition of open-loop zeros

Root locus of multivariables


Addition of Open-loop Poles

Adding poles to open loop transfer function.

Original: 𝝉𝒑𝟏 = 𝟎, 𝝉𝒑𝟐 = 𝟎 Add one: 𝝉𝒑𝟏≠ 𝟎, 𝝉𝒑𝟐 = 𝟎 Add two: 𝝉𝒑𝟏≠ 𝟎, 𝝉𝒑𝟐 ≠ 𝟎

)(sY


Addition of Open-loop Zeros

Adding zeros to open loop transfer function

Original: 𝝉𝒛𝟏 = 𝟎, 𝝉𝒛𝟐 = 𝟎Add one: 𝝉𝒛𝟏≠ 𝟎, 𝝉𝒛𝟐 = 𝟎

Add two: 𝝉𝒛𝟏≠ 𝟎, 𝝉𝒛𝟐 ≠ 𝟎

)(sY


Addition of Open-loop Poles and Zeros

Adding a pole to the LHP “pushes” part of the locus into

the RHP.

Adding a zero to the LHP “pulls” part of the locus into

the LHP.


Root locus of multivariables

Root locus for variable K, a

• Characteristic eqn.: 𝑠2 + 𝑎𝑠 + 𝐾 = 0

𝑎 = 0 , K=variable 𝐾 = 𝑐𝑜𝑛𝑠𝑡 , 𝑎 = variables

1 +𝐾

𝑠2= 0 1 +

𝑎𝑠

𝑠2+𝐾= 0

𝐾

𝑠(𝑠 + 𝑎))(sR

+-

)(sY

control systems - chibum lee · chibum lee -seoultech control systems root locus stability and...

Documents