Chapter 8 Root Locus and Magnitude-phase Representation

§ 8.1 Root Locus in Control Problem

§ 8.2 Magnitude-phase Representation

§ 8.3 Evan’s Root-locus Method

§ 8.4 Root-locus Method in Design

• Proportional Control Problem:

§ 8.1 Root Locus in Control Problem (1)

KG(s)Y(s)R(s)

+

H(s)

r(t) y(t)

b(t)

K: Can be controller parameter or plant parameter

)s(N)s(KN)s(D)s(D

)s(D)s(KNT(s)

)s(D

)s(NH(s) ,

)s(D

)s(NG(s) Let

G(s)H(s)K1

KG(s)T(s)

HGHG

HG

H

H

G

G

C-L System:

Zeros of T(s): Consist of the zeros of G(s) and poles of H(s)

Poles of T(s): Changes with K

Problem: Investigate the effect of varied K on controlling system dynamics

through the poles of closed-loop system.

m

lg

Kr

u(t)

b(t)

controller plant and sensor

• Fundamental of Root Locus: Definition of root locus:

§ 8.1 Root Locus in Control Problem (2)

Usage of root locus:

Graphical representation of the path of the closed-loop poles as one or more parameters of the open-loop transfer function are varied (usually ).

Provide relationship among open-loop system parameter, closed-loop pole location, and output transient response.

0

Root locus:

• Root Locus of 2nd-order Closed-loop System: § 8.1 Root Locus in Control Problem (3)

R(s) + Y(s)

)2s(s n

2n

1 ,i1

is 1,

- ,s 1,

1s 1, :Roots

s2s)s(T

2ndddn

nn

2nn

2nn

2

2n

parameter varied a as , fixed For (1) n parameter varied a as , fixed For (2) n

j

1

0

0

1

0 0

1

j

0n

ncos

n 1

• Vector Representation of Complex Numbers: § 8.2 Magnitude-phase Representation (1)

2 2 -1 bs = -a+bj= a +b -(tan ), a, b>0a

Cartesian

Coordinate

Polar Coordinate

s s

b

b

a 0

j

1s =-a+bj

2s =-a-bj

Part)y j(ImaginarPart) (Reals

Coord. Cartesian

0

1s

2s

(phase))(Magnitudes

Coord. olarP

12

12

ss

ss

1s2s1s

2s

1s

• Magnitude-phase Representation(1) Absolute magnitude and phase

sjessss

§ 8.2 Magnitude-phase Representation (2)

(2) Relative magnitude and phase As a free vector

s s

o

s

s s

s

Any horizontal reference line

a

a 0

ja

b

a a

a a

c

point a: - +jpoint b: - -jpoint c: -

a

c1

1

1

1

c

a

ac

a

b

a1

1

1

1

c0

j

a

c

c/a

1j1

11

aac

caa

e

j)(

)()j( :c to relative a point

)j(1

1o

111

1e

)360( or :c to relative b point

Ex: Find residues in partial fraction expansion for

Y(s) 1.25T(s)

R(s) s(s 2.5)(s 4.5)

§ 8.2 Magnitude-phase Representation (3)

by pole-zero representation.

5.4s

K

5.2s

K

s

K)s(T 321

Sol:

9

1

)05.4)(05.2(

25.1

)5.4s)(5.2s(

25.1s)s(TK

0s0s1

4

1

)02)(1805.2(

25.1

)5.4s(s

25.1)5.2s()s(TK

5.2s5.2s2

9

25.1

)1802)(1805.4(

25.1

)5.2s(s

25.1)5.4s()s(TK

5.4s5.4s3

pole-zero representation

5.4s9

25.1

5.2s4

1

s9

1)s(T

j

5.25.4

05.2

05.4

)K( 1

j

5.4

1805.2

02

)K( 2

j

1805.4

1802

)K( 3

Ex: If s is a complex variable, solve 0)1s(s

25.01

§ 8.2 Magnitude-phase Representation (4)

(1) 1)1s(s

25.00

)1s(s

25.01

Sol:

poles: s=0, s=-1In complex plane

From eq.(1)

3, 1,n ,e11

)e1s)(es(

25.0 o180nj)1s(jsj

3, 1,n ,180n))1s(s(:Phase

)a2(11ss

25.0:Mag

o

)b2(1801)s(s ,1n If o

0

j

11

j

j

)e( )0(j

)e( )270(j

)e( )90(j

)e( )180(j

je

circle Unit

2

)1s(

1s

1s

21s 3

seq.(1)satisfy to ?s

Where is the “s” to satisfy eqs. (2a) and (2b) in complex plane?

§ 8.2 Magnitude-phase Representation (5)

(A) From phase relationship (2b)

(3)

180

180

21

32

31

21

5.0

25.0

(eq.3)

125.0

21

2

2

2

1

21

21

(B) From magnitude relationship (2a)

i.e. s is on the bisection line

Conclusions: From (A) and (B) s=-0.5, -0.5

0

j

s

1 0

j

s

1 21

§ 8.2 Magnitude-phase Representation (6)Ex: Find the root locus of a DC-servo position control system with varied K.

Sol: C-L system,

Ku+

-

R(s) Y(s)

)1s(s

1

plant

controller

K

r

Controller

y

u

y

Plant

0K ,

Kss

K

GH1

G

R

C2

characteristic eq. 0Kss2 (A) Direct solution method

K25.05.0s pole-zero representation

Root locus with varied K

C-L system is stable with varied K.

C-L system is underdamping when K>0.25.

0

j

1 5.0

5.0

0.1

0.1

5.0

25.1K

5.0K

25.0K 0K 0K

5.0K

25.1K

Gain, K Poles, s

0

0.25

1.25

0.5

1 ,0

5.0 ,5.0

j5.05.0

j5.0

O-L System

§ 8.2 Magnitude-phase Representation (7)(B) Use magnitude-phase representation

characteristic eq. 0Kss2

180))1(s(0)(s :Phase

K1ss :Mag1

1)s(s

K or

K

1

ss

1 .e.i

2

From the example in section 8.2:

(a) Phase relationship s is on the bisection line between 0 and –1.

(b) Magnitude relationship

2

1 i.e. ,5.05.0K

5.025.0K

2

21

j0.5 0.5s 0.5,K

0.5 0.5,s 0.25,K (b) and (a) From

0

j

1

5.0 0K 0K

25.0K

5.0K

5.0K

Root locus

§ 8.2 Magnitude-phase Representation (8)Step response

Key parametric values of K in root locus:

(1) Intersection between root locus and -axis

Critical K of Instability

(2) Intersection between root locus and -axis

Critical K of Oscillation

j

1

r(t)

t

K=1.25

K=0.5

K=0.25K=0.25

Start

Rotation and then stop

shaft pointer

K=0.5

Start

Oscillation and then stop

§ 8.3 Evan’s Root-locus Method (1)

1. Parametrilization of closed-loop system with K varied as:

• MethodEvan's R-L

MethodOpen-loop

pole-zero diagramClosed-loop

pole locations

Adjust parameterK

2. Obtain characteristic equation of the closed-loop poles: 1+KG(s)H(s)=0

3. From magnitude-phase representation to obtain criteria for poles and zeros:

KG(s)+

H(s)

4. Sketch shape with scale and parameter K:

The shape of root locus is determined entirely by the phase criterion.

The magnitude criterion is used only to assign scale and parameter K of the

locus.

1KG(s)H(s) :criterion Magnitude

mn ,asasas

bsbsbsG(s)H(s)

011n

1nn

011m

1mm

The highest power in both the numerator and denominator of G(s)H(s) are

normalized to unity.

2, 1, 0,k ,180)1k2(360k180KG(s)H(s) :criterion Phase

§ 8.3 Evan’s Root-locus Method (2)• Terminologies and Symbols About Root Locus

mn ,)ps()ps)(ps(

)z(s)z)(szK(sKG(s)H(s)

c)a 0,K ,0bi)-abi)(sac)(s(s

1K1 :Ex(

n21

m21

zeros of values the of sum Algebraic:z zeros of number :z#

poles of values the of sum Algebraic:p poles of number :p#

i

i

j

Breakaway point

Branch(1)

Branch(2)

Branch(3)

Break-in point

Asymptotic line

Angles ofasymptotes

k

Centroid ofasymptotes

Angles ofdeparture

d

o

§ 8.3 Evan’s Root-locus Method (3)• Five Rules for Sketching the Root Locus

2 1180

i i0

k

Equation of asymptotes

( p z )

#p # z

( k ) , k 0, 1, 2,

#p # z

1. Number of branches: The number of branches of the locus is equal to

the order of the characteristic polynomial.

2. Symmetry: The locus is symmetrical about the real axis.

3. Real-axis segments: For K>0, the locus on the real axis exists to the

left of an odd number of real-axis open-loop poles

plus zeros.

4. Starting and ending points: The locus begins at the poles of G(s)H(s)

with K=0 and terminate with , either at

the zeros of G(s)H(s) or at infinity.

5. Behavior at infinity: The locus approaches straight line asymptotes as the

locus approaches infinity.

K

§ 8.3 Evan’s Root-locus Method (4)• Refining the Sketch

Points of imaginary-axis crossings:

Found by using Routh-Hurwitz criterion or substituting into the equation of root locus and solving the equations.

Angles of departure (arrival):

Obtain by choosing an arbitrary point infinitesimally close to the pole (zero)

and applying the angle criterion.

Root locus calibration:

Breakaway (Break-in) points:

K attains local maximum (minimum) on the real axis.

t t t

t i

t i

1 G(s )H(s ) for points s on the locus

Ks p

Ks z

condition.necessary for 0ds

dK .e.i

tss

js

§ 8.3 Evan’s Root-locus Method (5)Ex: Find root locus for a DC motor position servo as

+

)1s(s

K

Sol: C-L poles:K

1 0, s(s 1)+k=0, Two branchess(s 1)

O-L poles: s=0, s=-1, # p=2

zeros: none, # z=0

Real-axis segment: lies between the poles of s=0 and s=-1.

,2 ,1 ,0k ,

02

180)1k2(kAsymptotes:

Breakaway point:4

1K

2

1s ,0

ds

dK

270 ,90 10

2

1

02

)0())1(0(0

§ 8.3 Evan’s Root-locus Method (6)

The closed-loop position servo is stable for any positive proportional gain.

j

1s 2

1

0K 0K

4

1K 0s

90

270

Breakaway point

Asymptotes

real axis segment

§ 8.3 Evan’s Root-locus Method (7)Ex: Find stability conditions for various K in the characteristic equation:

Sol: Standard form 02)1)(ss(s

K1

C-L poles:

O-L poles: s=0, s=-1, s=-2, # p=3

zeros: none, # z=0

Real-axis segment: lies between the poles of s=0 and s=-1.

lies to the left of pole s=-2.

,2 ,1 ,0k ,

03

180)1k2(kAsymptotes:

0K ,0Ks2s3s 23

loci of branches Three ,0Ks2s3s 23

Imaginary axis crossing:

js

23 )s2s3s(K js set ⇒

0)-(2j part,Imaginary 3K part, Real

2

2

6K ,2 0K ,0

300 ,180 ,60 210

103

)0()210(0

§ 8.3 Evan’s Root-locus Method (8)

Breakaway point:

)feasible Not( 58.1s

385.0K 423.0s 0

ds

dK

Root locus and stability conditions

Stability conditions

(1)

three negative real roots

stable

(2) 0.385<K<6

two complex roots, one negative real root

stable

(3) K>6

two complex roots in RHP, one negative

real root

unstable

385.0K0

0-1-2K=0

K=0

K=0

-0.423K=0.385

Asymptotes

6K ,2

6K ,2

60180

300

j

§ 8.3 Evan’s Root-locus Method (9)Ex: Consider a system includes lightly damped flexible modes near

imaginary axis, find the root for 0)s(H)s(KG1

)36)1.0s((s

250.1)(sG(s)H(s) (1)

2

2

)25)1.0s((s

360.1)(sG(s)H(s) (2)

2

2

Sol: (1) (2)

18090

90

0k ,k360180

zero) to close (s arrival of angle Find

12

321

21321

t1

090

90

0k ,k360180

pole) to close (s departure of angle Find

21

321

21321

t

j

j

j

ts

j

Stability insensitive

to varied KStability sensitive

to varied K

§ 8.4 Root-locus Method in Design (1)• Design Problems

• Configurations of Compensation

Adjust parameter and / or introduce the pole(s) and zero(s) of the dynamic

compensator to alter the root locus so that the performance specifications

can be satisfied.

Performance specs: Stability Margin, Transient Response, Steady State Error.

G(s)+

Hc(s)

Gc(s)+

G(s) Gc(s)+

G(s)

Hc(s)

Cascade Compensation Feedback Compensation Cascade and Feedback Compensation

O-L Transfer Function:

Static Compensator: pc

c

K)s(G

)s(G)s(G

bc

c

K)s(H

)s(H)s(G

bcpc

cc

K)s(H ,K)s(G

)s(H)s(G)s(G

§ 8.4 Root-locus Method in Design (2)• Dynamic Compensators

Passive compensator --- Mainly RC type network

Active compensator --- Mainly OP and RC circuit

Require external power

Basic Gain and Phase Compensation Passive Compensator Active Compensator

PD controller PI controller

Phase lead Phase lag

PID controller or lag-lead compensator can be used to improve transient

response, s.s. error and trade off stability margin.

p

zGain DC ,

ps

zs)s(D

sKK)s(G DPc

s

KK)s(G I

Pc

p z

j

,pz00)j(D

pz

j

,zp0

0)j(D

CompensatorEffects Passive Active

LeadNetwork

LagNetwork

PDController

PIController

Improve transient speedIncrease stability margin

Improve s.s.errorReduce stability margin

§ 8.4 Root-locus Method in Design (3)• Design Constraints

Feasible region of 1st / 2nd-order dominant poles

Optimal region of 2nd-order dominant poles

j

1cos

j

1 4

sst

j

%a.s.o % )ordernd2(

s sSettlingTime (T ) t)ordernd2/st1(

s s

%o.s. a%

T t

j

°45

stable butoscillatiory

5%<o.s. % ,7.0=

stable butsluggish

optimalregion

§ 8.4 Root-locus Method in Design (4)• Addition of Poles to G(s)H(s)

The effect of adding a pole to G(s)H(s) is to push the root loci toward the

RHP.

:)s(H)s(G )2s)(1s(s

K

)j2s)(j2s)(1s(s

K

j

12

1 0

0K 0K

K

K

0-1-2

K=0 K=0 K=0

j

K

K

)1s(s

K

0-1-2

K=0

K=0 K=0

j

K

K

K=0-1

1

K

K

§ 8.4 Root-locus Method in Design (5)• Addition of Zeros to G(s)H(s)

The effect of adding a left-half plane zero to G(s)H(s) is to move and bend

the root loci toward the LHP.

:)s(H)s(G )1s(s

)2s(K

)1s(s

)j2s)(j2s(K

j

12

1 0

0K 0K

K

K

)1s(s

K

j

0K

-2 -1K

0K K

j

0K

-2 -1

K

0K

K

0

§ 8.4 Root-locus Method in Design (6)• Practical Compensator Realized by OP:

1)(

rCompensato Lead :CRCR (1) 2211

1)(

rCompensato Lag :CRCR (2) 1122

j

22CR

1

11CR

1

ts 1

j

22CR

1

11CR

1

ts 1

+

--

+

R1

R2

R3

R4

Eo(s)Ei(s)

C1

C2

E(s)

T1

s

T1

sK

CR1

s

CR1

s

CR

CR

)s(E

)s(Ec

22

11

23

14

i

o

23

14c2211 CR

CRK ,CRT ,CRT

§ 8.4 Root-locus Method in Design (7)• Root Sensitivity Robust K of characteristic roots

Breakaway (breakin) point:

dK

ds

s

K

KK

ss

SsK

0ds

dK

sKs

Avoid selecting the value of K to operate at the breakaway (breakin) points.

Root locus is less sensitive to changes in gain at the lower value of K.

Root sensitivity provides information of changes in both magnitude and direction of specific root for designer.

For characteristic eq. 1+KG(s)H(s)=0,

n

1ii

m

1ii

)ps(

)zs(K)s(H)s(KG

K ss

Ksensitivity of C-L pole at s - : s GH s

§ 8.4 Root-locus Method in Design (8)

From ramp input

Characteristic eq.

)2s(s

)sK1(K)s(H)s(G 21

20K1.0)s(H)s(sG

1lime 1

0sss

Ex: For a DC-servo with position and tacho feedbacks, find K1 and K2 to satisfy the

control specs: (1) settling time sec, (2) dominant poles with ,

(3) ramp-input steady state error 3 707.0

%10

Y(s)R(s) + Ea(s)

sK1 2

)2s(sK1

sK1 2

Sol:

0KsKK2ss0,G(s)H(s)1 1212

1212 Kb ,KKa ,0bass2s or

From setting time

3

41 ,

4

3 sec,34Ts

Root locus for b=K1=20

020s2s

as1

2

§ 8.4 Root-locus Method in Design (9)

0.215K20K4.3a

:line 0.707ξ and locus root of onIntersecti

22

Feasible region and root locus

3.151

3.15j,-3.15s

point onIntersecti

d

sec27.14T

time settling Check

s

215.0K

20K select

2

1

-6 -5 -4 -3 -2 -1

-5

-4

-3

-2

-1

0

1

2

3

4

5

-4.36

4.36

707.0

ds3.4a

34