Root-locus Technique for

Control Design

2

INTRODUCTION

Stability and the characteristics of transient response of closed-loop systems

Locations of the closed-loop poles

Problems to solve characteristic equation : 1. Difficult for a system of third or higher order. 2. Tedious for varying parameters.

3

VARYING THE LOOP GAIN K

In many systems, simple gain adjustment may move the closed-loop poles to desired locations.

Then the design problem may become the selection of an appropriate gain value.

It is important to know how the closed-loop poles move in the s plane as the loop gain K is varied.

The open loop gain K is an important parameter that can affect the performance of a system

R(s)K

Y(s)

H(s)

G (s)－

4

Root Locus:

the locus of roots of the characteristic equation of the closed-loop system as a specific parameter (usually, gain K) is varied form 0 to ∞.

The advantages of RL approach:

1. Avoiding tedious and complex roots-solving calculation

2. Clearly showing the contributions of each loop poles or zeros to the location of the closed-loop poles.

3. Indicating the manner in which the loop poles and zeros should be modified so that the response meets system performance specifications.

5

START FROM AN EXAMPLE

Consider a second-order system shown as follows:

)1( sskR(

s) -

Y(s)

The roots of CE change as the value of k changes.

Closed-loop TF:

kss

ks

2)(

Characteristic equation (CE): 02 kss

Roots of CE: kk

s 412

1

2

1

2

4112,1

When k changes from 0 to ∞, how will the locus of the roots of CE move?

6

1,2

1 11 4 , : 0

2 2s k k

1/ 4k＝ 2/121 ss

01 s 1s2 k＝ 0

0 1/ 4k As the value of k increases, the two negative real roots move closer to each other.

A pair of complex-conjugate roots leave the negative real-axis and move upwards and downwards following the line s=-1/2.

1/ 4 k

－ 1/2－ 1 0

On the s plane, using arrows to denote the direction of characteristic roots move when k increases, by numerical value to denote the gain at the poles.

0k 0k

1

4k

k

k

7

By Root loci, we can analyze the system behaviors

(2)Steady-state performance: there ’s an open-loop pole at s=0, so the system is a type I system. The steady-state error is 0 under step input signal R/Kv under ramp signal v0t

∞ under parabolic signal.

(1)Stability: when Root loci are on the left half plane, then the system is definitely stable for all k>0.

(3)Transient performance: there’s a close relationship between root loci and system

behavior

on the real-axis: k<0.25 underdamped;

k=0.25 critically damped

k>0.25 underdamped.

However, it’s difficult to draw root loci directly by closed-

loop characteristic roots-solving method.

The idea of root loci : by loop transfer function, draw closed-loop root loci directly.

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RELATIONSHIP BETWEEN ZEROS AND POLES OF G(S)H(S) AND

CLOSED-LOOP ONES

G(s)

H(s)

-

R(S) Y(s)

Forward path TF: Closed-loop TF:( )G s

)()(1)(

)(sHsG

sGs

Feedback path TF: ( )H s

10

Suppose that G(s)H(s) has m zeros (Zi) and n poles （ Pi), the above equation can be re-written as

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

K* varies

from 0 to ∞

To draw Closed-loop root locus is to solve the CE

1 ( ) ( ) 0G s H s That is

( ) ( ) 1G s H s RL equation

11

Magnitude equation (ME)

1

1

1

m

iin

ii

K s Z

s P

Angle equation (AE)Angle equation (AE)

1 1

( ) ( )

(2 1) , 0, 1, 2,

m n

i ii i

s Z s P

l l

Since G(s)H(s) is function of a complex variable s, the root locus equation can be described by the following two equations:

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

Magnitude equation is related not only to zeros and poles of G(s)H(s), but also to RL gain ；

Angel equation is only related to zero and poles of G(s)H(s).

Use AE to draw root loci and use ME to determine the value of K on root loci.

RL equation:

12

2 21 2 2

'

1

( 1)( ) ( )

( 1)( 2 1)

( 1/ )

( 1/ )( )( )n d n d

K sG s H s

s T s T s T s

K s

s s T s j s j

' 21 2/K K TT

21/n T 2

2d T1

/11 z

1 4

1 11 1

1 1 2 3 4

( ) ( ) ( ) ( )

( )

i ii i

G s H s s z s p

Poles of G(s)H(s) （ × ）

Zeros of G(s)H(s) （〇）

Angel is in the direction of anti-clockwise

For a point s1 on the root loci, use AE

Use ME

11

4131211'

zs

pspspssK

Example 1

2 11/p T dn4,3 jp 1 0p

13

Unity-feedback transfer function:s

K)s(G

Test a point s1 on the negative real-axis

11 1

1 1

( ) ( ) |

180

m n

i i si i

s z s p

s p

Test a point outside the negative real-axis s2=-1-j

21 1

2 1

( ) ( ) |

135

m n

i i si i

s z s p

s p

All the points on the negative real-axis are on RL.

All the points outside the negative real-axis are not on RL.

Example 2

One pole of G(s)H(s): 1 0p

No zero.

1. Find all the points that satisfy the Angle Equation on the s-plane, and then link all these points into a smooth curve, thus we have the system root locus when k* changes from 0 to ∞;2. As for the given k, find the points that satisfy the Magnitude Equation on the root locus, then these points are required closed-loop poles.

However, it’s unrealistic to apply such “probe by each point” method. W.R. Evans (1948) proposed a set of root loci drawing rules which simplify the our drawing work.

Probe by each point:

15

4-3 RULES TO DRAW REGULAR ROOT LOCI

(suppose the varying parameter is open-loop gain K )

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2 Number of Branches on the RL

3 Symmetry of the RL

4 Root Loci on the real-axis

5 Asymptotes of the RL

6 Breakaway points on the RL

7 Departure angle and arrival angle of RL

8 Intersection of the RL with the imaginary axis

9 The sum of the roots and the product of the roots of the closed-loop characteristic equation

PROPERTIES OF ROOT LOCI1 and points of Root Loci0K K

17

Root loci originate on the poles of G(s)H(s) (for K=0) and terminates on the zeros of G(s)H(s) (as K=∞).

MagnitudeEquation:

1

1

n

iim

ii

s PK

s Z

＝

0K Root loci start from poles of G(s)H(s)

K Root loci end at zeros of G(s)H(s).

1

1

1

m

iin

ii

K s Z

s P

1 and points0K K

is P

is Z

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:

18

2 Number of branches on the RLnth-order system, RL have n starting points and RL have n branches

For a real physical system, the number of poles of G(s)H(s) are more than zeros ， i.e. n > m.

m root loci end at open-loop zeros （ finite zeros) ；（ n － m ） root loci end at (n － m) infinite zeros.

The order of the characteristic equation is n as K varies from 0 to ∞ ,n roots changen root loci.

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:

n root loci end at open-loop zeros （ finite zeros) ；

19

3 Symmetry of the RL

The RL are symmetrical with respect to the real axis of the s-plane.

The roots of characteristic equation are real or complex-conjugate.

Therefore, we only need to draw the RL on the up half s-plane and on the real-axis, the rest can be obtained by plotting its mirror image.

20

On a given section of the real axis, RL for k>0 are found in the section only if the total number of poles and zeros of G(s)H(s) to the right of the section is odd.

4 RL on the Real Axis

zero ： z1 poles ： p1 、 p2 、 p3 、 p4 、 p5

Pick a test point s1 on [p2 ， p3]

The sum of angles provided by every pair of complex conjugate poles are 360°;

The angle provided by all the poles and zeros on the left of s1 is 0°.

The angle provided by all the poles and zeros on the right of s1 is 180°;

1 5

1 1 1 11 1

( ) ( ) ( ) ( )i ii i

G s H s s z s p

1 1( ) ( ) (2 1)180G s H s l ＝ ？

jw

21

Consider the following loop transfer function

2 2

( 1)( 4)( 6)( ) ( )

( 2)( 3)

K s s sG s H s

s s s

On the right of [-2 ， -1] the number of real zeros and poles=3.

On the right of [-6 ， -4], the number of real zeros and poles=7.

Example

Determine its root loci on the real axis.

Repeated poles:S-plane

0-1-2-3-4-5-6

Poles: Zeros:

22

5 Asymptotes of RL

(2 1)180a

i

n m

mn

zpm

1ii

n

1ii

a

＝

When n ≠ m, there will be 2|n-m| asymptotes that describe the behavior of the RL at |s|=∞.

The angles between the asymptotes and the real axis are（ i= 0 ， 1 ， 2 ，… ,n-m-1 ） ：

The asymptotes intersect the real axis at ：

1

1

( )( ) ( ) 1

( )

m

iin

ii

K s ZG s H s

s P

RLEquation:

23

*(0.25 1) ( 4)( ) ( )

( 1)(0.2 1) ( 1)( 5)

K s K sG s H s

s s s s s s

The angles between the asymptotes and the real axis are

3 poles ： 0 、 -1 、 -5

1 zero ： -4n-m = 3 -1 = 2

The asymptotes intersect the real axis at

113

)4()5()1()0(

mn

zpm

1ii

n

1ii

a

＝

(2 1)180, 0,1a

ii

n m

90 ,270a

Example Consider the following loop transfer function

Determine the asymptotes of its root loci.

24

2

( 1)( ) ( )

( 4)( 2 2)

K sG s H s

s s s s

4 poles ： 0 、 -1+j 、 -1-j 、 -4

1 zero ： -1

n-m=4-1=3

The asymptotes intersect the real axis at

1 1

(0) ( 1 ) ( 1 ) ( 4) ( 1) 5

4 1 3

n m

i ii i

a

p z

n mj j

＝

The angles between the asymptotes and the real axis are

(2 1)180, 0,1,2a

ii

n m

60 ,180 ,300a

Example Consider the following loop transfer function

Determine the asymptotes of its root loci.

25

6 Breakaway points on the RLBreakaway points on the RL correspond to multi-order roots of the RL equation.

j

4p

3p

1p2p

A

B

0

[ s]

The breakaway points on the RL are determined by finding the roots of dK/ds=0 or dG(s)H(s)=0.

26

The poles and zeros of G(s)H(s) are shown in the following figure, determine its root loci.

Rule 1 、 2 、 3RL have three branches ， starting from poles 0 、－ 2 、－ 3 ， ending at on finite zero － 1 and two infinite zeros. The RL are symmetrical with respect to the real axis.

Ruel 4The intersections [-1,0] and[-3,-2] on the real axis are RL.

Rule 5The RL have two asymptotes(n － m ＝ 2)

(2 1)18090 ,270

0,1

a

i

n mi

22

)1()3()2(0

mn

zpm

1ii

n

1ii

a

＝

Rule 6The RL have breakaway points on the real axis (within[-3,-2])

3

1

2

1

0

1

1

1

＋＋

＋＋＝

＋ bbbb 47.2b

0-1-2-3

jω

σ

Example

27

2( ) ( )

( 4)( 4 20)

KG s H s

s s s s

Rule 1 、 2 、 3 、 4 n=4,m=0the RL are symmetrical with respect to the real axis;the RL have four branches which start from poles 0,-4 and -2±j4 ;the RL end at infinite zeros;the intersection [-4,0] on the real-axis is RL

0-2

-j4

-4

jω

σ

j4

(2 1)180( 0,1,2,3)

45 135 , 225 315

a

ll

n m

， ，

Rule 5 The RL have four asymptotes.

24

)2()2()4(0

mn

zpm

1ii

n

1ii

a

＋

＝

Example

28

21 ( ) ( ) 1 0

( 4)( 4 20)

KG s H s

s s s s

2

4 3 2

( 4)( 4 20)

( 8 36 80 )

K s s s s

s s s s

3 2(4 24 72 80) 0dK

s s sds

Rule 6the breakaway point of the RL

21 b

45.223,2 jb

2( ) ( )

( 4)( 4 20)

KG s H s

s s s s

0-2

-j4

-4

jω

σ

j4

Example

29

7 Angles of departure and angles of arrival of the RL

The angle of departure or arrival of a root locus at a pole or zero, respectively, of G(s)H(s) denotes the angle of the tangent to the locus near the point.

30

Angle of Departure:

1 1

(2 1) ( ) ( ), 0, 1,m n

pj j i j ii i

i j

l p z p p l

＝ ＋

Pick up a point s1 that is close to p1

Applying Angle Equation (AE)

1 1 1 1 1 2 1 3( ) ( ) ( ) ( )

(2 1)

s z s p s p s p

l

s1p1 )( 11 ps angle of departure θp1

1 1 1 1 2 1 3(2 1) ( ) ( ) ( )p l p z p p p p ＝ ＋

31

1 1

(2 1) ( ) ( ), 0, 1,n m

zj j i j ii i

i j

l z p z z l

＝ ＋

Angle of Arrival:

32

2( ) ( )1 1

KsG s H s

s s

3 poles P1,2=-1(repeated poles) P3=1 ； 1 zero Z1=0 ， n-m=2 。3 branches ， 2 asymptotes

1 1 1 1 1 00.5

3 1

n m

i ii i

a

P Z

n m

(2 1) 3,

2 2a

l

n m

3(2 1) ,

2 2 2pl l

Angle of departure:

Example Consider the following loop transfer function

Determine its RL when K varies from 0 to ∞.

-1

j

-0.5 1

33

8 Intersection of the RL with the Imaginary Axis

Intersection of the RL with Im-axis?

Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K.

The characteristic equation have roots on the Im-axis and the system is marginally stable.

Method 2

js 1 ( ) ( ) 0G j H j

0)()(1Im

0)()(1Re

jHjG

jHjG

1 ( ) ( ) 0G s H s

34

( ) ( )( 1)( 2)

KG s H s

s s s

Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K

3

2

1

0

1 2

3

60

3

s

s K

Ks

s K

Marginally stable: K =6

Auxiliary equation:

063 2 s

2js Intersection point:

Example Consider the following loop transfer function

Determine the intersection of the RL with Im-axis.

Routh’s Tabulation:

Method 1

35

3 2 2 3( ) 3( ) 2( ) ( 3 ) (2 ) 0j j j K K j

2 3 0K Real part: ＝

3Imaginary part: 2 0 ＝2 6K

( ) ( )( 1)( 2)

KG s H s

s s s

js 1 ( ) ( ) 0G j H j → Closed-loop CE

Example Consider the following loop transfer function

Determine the intersection of the RL with Im-axis.

Method 2

Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K