1

Root Locus Analysis

Root Locus Analysis

The transient response of a closed-loop system is completely

determined by the location in the s-plane of the closed-loop

system poles and zeros. This shows if the system is stable and

also whether there is any oscillatory behaviour in the time

response. Therefore, it is worthwhile to determine how the

roots of the characteristic equation as a system parameter is

varied. The root locus method is proposed by Evans in 1948.

It is a graphical method for system analysis and design

2

Root Locus Concept

G(s)+-

E(s)R(s)k

Y(s)C(s)

1

)()()(

1

1sDsNsG =

)()()(

2

2sDsNsC =

)()()()()()(

)()(1)()()(

2121

21sDsDsNskN

sNsNsCskG

sCskGsT+

=+

=

3

The poles of T(s) = zeros of )()(1 sCskG+ = zeros of kN s N s D s D s1 2 1 2( ) ( ) ( ) ( )+

Let )()()()(

)()()()(

1

1

n

mpspszszsk

sPskZsCskG

++++

==LL

LL

then the closed-loop poles are the roots of P s kZ s( ) ( )+ = 0

Therefore

zeros loopopen system poles loopopen system

0)(0)(0

⇔=⇔=

⇒∞→⇒→

sZsP

kk

zeros loop-open the poles loop-closed thepolesloop-openthepolesloop-closed the0

→→

⇒∞→⇒→

kk

For certain k, to find the corresponding closed-loop poles isto find the roots of )()(1 sCskG+ = 0, that is todetermine s, such that 1)()( −=sCskG .

For k > 0Find s, such that(1) 1)()( =sCskG , (Magnitude condition)

(2) arg π)21()}()({ rsCskG += (Phase condition)

r Z∈

4

Root Locus Construction

1. Loci branchThe branches of the locus are continuous

curves that start at each of n poles of G(s)C(s),for k > 0. As k → +∞ , the locus branchesapproach the m zeros of G(s)C(s). Locusbranches for excess poles extend infinitely farfrom the origin; for excess zeros, locus segmentextends from infinity.

Example

Consider )84)(2(

)1()()( 2 ++++

=sss

ssCsG , the corresponding

root locus branch, for k = [0, 10] are shown below.

-8 -6 -4 -2 0-4

-3

-2

-1

0

1

2

3

4

5

2. Real-axis locusThe root locus on those portion of the real axis for which thesum of poles and zeros to the right is an odd (even) number,for k > 0 (for k < 0).

3. Locus end pointspoles ⇒ zeros (finite or infinite) for k → ∞

4. Asymptotes of locus as s → ∞The angles of the asymptotes of the root locus branches,which end at infinity, are given by:

φ asyr

n m=

+−

( )1 2 180o , k > 0

φ asyrn m

=⋅−

2 180o , k < 0

Note: For s → ∞ ,

)(

)(lim)()(lim

in

i

jm

j

ss ps

zsksCskG

−

−

=∞→∞→

π

π

n

m

s ssk

)()(

limσσ

−

−≅

∞→

mns sk

−∞→ −=

)(lim

σ

= -1 ⇒ ks mn −=− −)( σ

zrmn

rjks mn ∈−+

+= − ),)12(exp(1 πσ

})arg{()21(180}arg{ mnsrk −−=+=− σo ∴ − = +( ) arg{ } ( )n m s r180 1 2o

Therefore, φ asyr

n m=

+−

( )1 2 180o , for k > 0.

6

ExampleConsider the following system

+-

E(s)R(s) Y(s)

s(s+2)k

1

kG s ks s

( )( )

=+ 2

T s ks s k

( ) =+ +2 2

The poles of T(s) ⇒ the roots of s2 + 2s + k = 0

⇒ − ± −1 1 k

For k ≤ 1, the roots are real within [-1, 0].For k > 1, the roots are complex conjugates with real part = -1.

-3 -2 -1 0 1-2

-1

0

1

2

φ1φ2

φ φ1 2 180+ =o (phase Condition)

7

Since

By polynomial parameter comparison, thecommon point at which all asymptotes intercept thereal axis is given by

σ =−

−==

∑∑ Re( ) Re( )p zn m

i jj

m

i

n

11 , 2≥− mn

Note: A root locus branch may cross its asymptote.

zrmn

rjks mn ∈−+

+= − ),)12(exp(1 πσ

6. Break-away/ Break-in point on the real axisThe break-away point for the locus between two

poles on the real-axis occurs when the value of k is amaximum. The break-in point for the locus betweentwo zeros on the real-axis occurs where the value kis a minimum.

k = ∞k = ∞k = 0k = 0

kmax

kmin

8

)()()}()({ 1

sZsPsCsGk −=−= −

∂∂ ks

= 0 ⇒ 1 02Z sP s dZ s

dsZ s dP s

ds( )( ) ( ) ( ) ( )−⎛⎝⎜

⎞⎠⎟ =

(1)

(2) Find the roots of )]()([ sCskGdsd = 0

The roots of )]()([ sCskGdsd = 0 are the

break-in/break-away points for all k R∈

Formula:dds

f f dds

f= ln

Hint: dsdf

dsdf

fff

dsdf == 1ln

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

)()(ln

)()(

)()()]()([

sPsZ

dsd

sPsZ

sPsZ

dsdsHskG

dsd

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

dssdP

sPdssdZ

sZsPsZ )(

)(1)(

)(1

)()(

⎟⎠⎞

⎜⎝⎛ −=

dssdPsZ

dssdZsP

sP)()()()(

)(1

2

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Example

Consider ss

sksCskG)2()4()()(

++

= . Using the formula above,

it is obtained that1

41 1

2s s s+= +

+ ⇒ s = -6.83, or -1.17

-10 -8 -6 -4 -2 0-3

-2

-1

0

1

2

3

-6.83

-1.17

K > 0

-5 0 5-5

0

5

K < 0

10

7. Angles of departure and approachThe angle of departure φd of a locus branch from a complex pole is given by

∑∑−=

ionconsideratunder pole the to toangle zero )()(+

ionconsideratunder pole the to toangle pole )()(other 180

sCsG

sCsGdoφ

The angle of approach φa of a locus branch from a complex zero is given by

o180

ionconsideratunder pole the to toangle zero )()(other

ionconsideratunder pole the to toangle pole )()(

−

−

=

∑∑

sCsG

sCsGaφ

Example

-4 -3 -2 -1 0-2

-1

0

1

2210

120

90

o

ooo

210

12090180

=

+−=dφ

o

ooo

180

18000

=

−−=aφ

11

Imaginary axis crossing pointThe value of k that cause a change of sign in the

Routh Array, is that value for which the locus crossesinto the right half s-plane.

Note:point of crossover s xj= → =phase 180o .

Example

Consider )2)(1(

6)()(++

=sss

ksCskG . The Routh array for

the unity-feedback closed-loop system is

s3 1 2s2 3 6ks1 2 - 2ks0 6k

1=→ k 063 2 =+⇒ s

js 2±=

12

Non-intersection or intersection of root locus branchesThe angle between two adjacent approaching branches is

αλλ

= ±360o

where λ denotes the number of branchesapproaching and leaving the intersection point.The angle between a branch leaving and anadjacent branch that is approaching the samepoint is given by

βλλ

= ±180o

Example

-4 -3 -2 -1 0-2

-1

0

1

2

leaving branch

approaching branch

αλ = 180

βλ = 90

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Grant’s Rule

For system rank ≥ 2, Grant‘s rule state that the sum ofthe (unity-feedback) closed-loop system poles is equalto the sum of the open-loop system poles.Note:

P s kZ s( ) ( )+ = 0⇒ s a s a s a s an n

nn

n+ + + + + =−−

−−

11

22

1 0 0LL ,where an−1 is independent of k also

an− = −∑1 poles

我可能要來杯咖啡才行

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Example

Plot the unity feedback closed-loop root locus for

)2)(1(1)()(

++=

ssssHsG

Solution

1. Open loop poles are : 0、-1、-2

Number of root-locus : 3

Root locus on the real axis ]2,( −−∞ and ]0,1[−

2. Asymptotes of locus as s → ∞

3)12( πθ += kk ，k=0,1,2。

Centroid of the asymptotes

13

)2()1(0−=

−+−+=σ

3. Imaginary axis crossing point

The characteristic equation is

0)2)(1( =+++ Ksss 023 23 =+++⇒ Ksss

and the corresponding Routh table is

15

Ks

KsKs

s

0

1

2

3

36

321

−

From the Routh table, the system will be stable for

60

16

Example

Plot the root locus for the system with

)22)(2(1)()( 2 +++

+=

sssssHsG

Solution:

1. Open-loop poles : j±−− 1,2 ，open-loop zero: -1

Number of the locus branches : 3

Locus on the real axis ]1,2[ −−

2. Asymptotes of locus as s → ∞

213)12( πθπθ =⇒

−+

=k

k ，

Centroid of the asymptotes

23

13)1()1()1()2(

−=−

−−−−++−+−=

jjσ

3. Angle of deparature： φd

πφφφφ )12()( 21 +=++− kdppz ππφ )12(4

+−−=⇒ kd

For πφ43 ,1 d =−=k 。

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Example

Consider the system with

)1(1)()(+

=ss

sHsG

Plot the root locus of the following cases.

(i)with additional pole at –2

(ii)with additional zero at -2

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Root locus without additional pole and zero

Additional pole

Root locus with additional pole -2

19

Additional zero

Root locus with additional zero -2

Example

)1()1()( 2

2

++

=sssKsG

Consider a negative unity feedback system has a plant transfer function

(a) Sketch the root locus for K > 0. (b) Find the gain K when two complex roots have a damping ratio and calculate all three roots. (c) Find the entry point (break-in point) of the root locus at the real axis.

707.0=ζ

20

j

K

sss

ssKsKKss

sssKs

n

nnnn

nn

22 ,5723.0 :Roots5723.0619.487.2

matching tsCoefficien0)414.1()414.1(

0)2)((0)12(

0)1()1()(

1Method

2223

2223

22

±−−===

=+++++

=+++⇔=++++

=+++=

α

ω

αωαωωωα

ωζωα

δ

Matlabby 96.196.1 ,58.0 :Roots j±−−

967.1967.1 ,584.0 :Roots j±−−

584.007382.7)934.37382.7()934.3(

0)7382.7934.3)((0518.4)1518.42(518.4

518.4

1967.2967.1967.1967.1)967.1967.0(

)967.1967.0()967.1967.0(

condition, magnitude From1.967j1.967- are roots conjugate the1.967x

-45180/π*1)))-(x/(xtan*2-1)/x)+((xtan+1)/x)-((x(tan

2r)180(1135))x

1x(tan180)x

1x(tan(180)1x

xtan2(180

2 Method

23

223

222222

2222

1-1-1-

111

=⇒=+++++

=+++⇔=++×++

=

=+×+×+

+×+×

±⇒==

+=++

−+−

−−−

− −−−

αααα

α

sss

ssssss

K

K

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Conclusions

(1). The system will tend to be unstable with additional poles (increasing the system rank).

(2). The system will tend to be stable with additional zeros.

In many design exercises, zeros can be introduced to attract closed-loop poles and alter the root locus location. It is also very useful to applied stable pole-zero cancellation for improving system performance.

22

Exercise 1

-10 -8 -6 -4 -2 0-20

-10

0

10

20

-2.5)10(

)5()()( 2 ++

=sssksCskG

Exercise 2

23

Exercise 3

Exercise 4

24

Control System Design by Root Locus Method

1. Determine the desired dominant pole locations using the performance requirements.

2 . Calculate the phase of the desired pole location corresponding to the uncompensated system G(s), and determined the required phase change.

3. Determine the pole and zero of the compensator C(s), such that the phase of the desired pole location corresponding to the compensated system is 180.

4. Determine the value of K, such that is satisfied.

5. Confirm the result by time domain simulation.1)()( =sCsKG