Root Locus Explanation

8/10/2019 Root Locus Explanation

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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 byEvans in 1948.

It is a graphical method for system analysis and design


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Root Locus Concept

G(s)+

-

E(s)R(s)k

Y(s)

C(s)

1

)(

)()(

1

1

sD

sNsG =

)(

)()(

2

2

sD

sNsC =

)()()()(

)()(

)()(1

)()()(

2121

21

sDsDsNskN

sNsN

sCskG

sCskGsT

+=

+=


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Root Locus Construction1. Loci branch

The 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 branches

approach the m zeros of G(s)C(s). Locus

branches for excess poles extend infinitely farfrom the origin; for excess zeros, locus segment

extends 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


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2. Real-axis locus

The root locus on those portion of the real axis for which the

sum of poles and zeros to the right is an odd(even) number,

fork> 0(for k< 0).

3. Locus end points

poles 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:

asyr

n m=

+

( )1 2 180o, k> 0

asyr

n m=

2 180o, k < 0

Note:

For s ,

)(

)(

lim)()(lim

i

n

i

j

m

j

ssps

zs

ksCskG

=

n

m

s s

sk

)(

)(lim

mns s

k

=

)(lim

= -1 ks mn = )(

zrmn

rjks mn

++= ),

)12(exp(

1

})arg{()21(180}arg{ mnsrk =+= o

= +( ) arg{ } ( )n m s r 180 1 2o

Therefore, asyr

n m=

+

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


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Example

Consider the following system

+

-

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

s(s+2)

k

1

kG s k

s s

( )

( )

=

+ 2T s

k

s 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

1

2

1 2 180+ = o (phase Condition)


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Since

By polynomial parameter comparison, the

common point at which all asymptotes intercept the

real axis is given by

=

==

Re( ) Re( )p z

n m

i j

j

m

i

n

11, 2 mn

Note:A root locus branch may cross its asymptote.

zrmn

rjks mn

++= ),

)12(exp(

1

6. Break-away/ Break-in point on the real axis

The break-away point for the locus between two

poles on the real-axis occurs when the value of k is a

maximum. The break-in point for the locus between

two zeros on the real-axis occurs where the value k

is a minimum.

k= k=

k= 0k= 0

kmax

kmin


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)()()}()({ 1

sZsPsCsGk ==

k

s= 0

10

2Z s

P s dZ s

dsZ s

dP s

ds( )( )

( )( )

( )

=

(1)

(2) Find the roots of )]()([ sCskGds

d= 0

The roots of )]()([ sCskG

ds

d= 0 are the

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

Formula:d

dsf f

d

dsf= ln

Hint:ds

df

ds

df

fff

ds

df ==

1ln

=

=

)()(ln

)()(

)()()]()([

sPsZ

dsd

sPsZ

sPsZ

dsdsHskG

dsd

=

ds

sdP

sPds

sdZ

sZsP

sZ )(

)(

1)(

)(

1

)(

)(

=

ds

sdPsZ

ds

sdZsP

sP

)()(

)()(

)(

12


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ExampleConsider

ss

sksCskG

)2(

)4()()(

+

+= . Using the formula above,

it is obtained that1

4

1 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


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

=

ionconsideratunderpolethetotoanglezero)()(+

ionconsideratunderpolethetotoanglepole)()(other180

sCsG

sCsGdo

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

o180

ionconsideratunderpolethetotoanglezero)()(other

ionconsideratunderpolethetotoanglepole)()(

=

sCsG

sCsGa

Example

-4 -3 -2 -1 0-2

-1

0

1

2

210

120

90

o

ooo

210

12090180

=

+=d

o

ooo

180

18000

=

=a


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Imaginary axis crossing point

The value of k that cause a change of sign in the

Routh Array, is that value for which the locus crosses

into 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 2

s2 3 6k

s1

2 - 2k

s0

6k

1= k 063 2 =+ s

js 2=


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Non-intersection or intersection of root locus branches

The angle between two adjacent approaching branches is

=

360o

where denotes the number of branches

approaching and leaving the intersection point.

The angle between a branch leaving and an

adjacent branch that is approaching the samepoint is given by

=

180o

Example

-4 -3 -2 -1 0-2

-1

0

1

2

leaving branch

approaching branch

= 180

= 90


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Grants Rule

For system rank 2, Grants rule state that the sum o

the (unity-feedback) closed-loop system poles is equal

to the sum of the open-loop system poles.

Note:

P s kZ s( ) ( )+ = 0

s a s a s a s an nn

n

n+ + + + + =

1

1

2

2

1 0 0LL ,

where an1 is independent of kalso

an = 1 poles


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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 : 3Root locus on the real axis ]2,( and ]0,1[


3

)12(

+=

kk

k=0,1,2

Centroid of the asymptotes

1

3

)2()1(0=

++=

3. Imaginary axis crossing point

The characteristic equation is

0)2)(1( =+++ Ksss 023 23 =+++ Ksss

and the corresponding Routh table is


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ExamplePlot the root locus for the system with

)22)(2(

1)()(

2 +++

+=

sss

ssHsG

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[


213

)12(

=

+=

kk

Centroid of the asymptotes

2

3

13

)1()1()1()2(=

+++=

jj

3. Angle of deparatured

)12()( 21 +=++ kdppz

)12(4

+= kd

For 4

3,1 d ==k


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Example

Consider the system with

)1(

1)()(

+=

sssHsG

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


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Additional zero

Root locus with additional zero -2

Example

)1(

)1()(

2

2

+

+=

ss

sKsG

Consider a negative unity feedback system has a

plant transfer function

(a) Sketch the root locus for K > 0. (b) Find thegain 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=


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j

K

sss

ssKsKKss

sssKs

n

nnnn

nn

22,5723.0:Roots

5723.0

619.4

87.2

matchingtsCoefficien

0)414.1()414.1(

0)2)((0)12(

0)1()1()(

1Method

2223

2223

22

=

=

=

=+++++

=+++=++++

=+++=

Matlabby96.196.1,58.0:Roots j

967.1967.1,584.0:Roots j

584.0

07382.7)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,magnitudeFrom

1.967j1.967-arerootsconjugatethe1.967x

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

2r)180(1135))x

1x(tan180)

x

1x(tan(180)

1x

xtan2(180

2Method

23

223

222222

2222

1-1-1-

111

=

=+++++

=+++=++++

=

=+++

++

=

=

+=++

+

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 attractclosed-loop poles and alter the root locus location. It is

also very useful to applied stable pole-zero cancellation

for improving system performance.


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Exercise 1

-10 -8 -6 -4 -2 0

-20

-10

0

10

20

-2.5

)10(

)5()()(

2 +

+=

ss

sksCskG

Exercise 2


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Exercise 3

Exercise 4


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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

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Root Locus Explanation