BARU - EPM2036 Chapter 4 Notes _condensed

8/7/2019 BARU - EPM2036 Chapter 4 Notes _condensed

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CHAPTER 4: ROOT LOCUS METHOD

4.1 INTRODUCTION

What? Find the positions of the closed-loop poles when a

certain system parameter varies.

Why? Positions of closed-loop poles determine important

properties such as stability, damping, etc.

How? By using root locus method.

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A root locus of a system is a plot of the roots of the system characteristic

equation (poles of the closed-loop transfer function) as some parameters

of the system are varied.

Each root locus starts at an open-loop pole with K= 0 and terminates at

infinity as K.

Each root locus gives one characteristic root (closed-loop pole) for aspecific value ofK.

Note that information about damping can be obtained from the root

locus.

4.3 BASIC PROPERTIES OF ROOT LOCI

4

KG(s)


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Closed-loop transfer function: )()(1)(

)(

)(

sHsKG

sKG

sR

sY

+=

1 + KG(s)H(s) = 0

G(s)H(s) = -1/K

Any point on the root locus (with K> 0) must satisfy

Magnitude criterion: |G(s)H(s)| = 1/K

Angle criterion: G(s)H(s) = 180o

5

KG(S)


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Let )())((

)())(()()(

21

21

n

m

pspsps

zszszsKsHsKG

+++

+++=

Magnitude criterion:( )

( ) Kps

zs

sHsGn

j

j

m

i

i1

)()(

1

1 =

+

+

=

=

=

Angle criterion:o

j

n

j

i

m

i

pszssHsG 180)()()()(11

=++= ==

Example: Find ifs = -1 and s = -3 are points on the root locus. If so, find

the value(s) ofKat these points.

6

-2

-1+j

-1-j

Re(s)

Im(

s)


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Fors = -1 Fors = -3

7

-2

-1+j

-1-j

a

b

c = 0o

-2

-1+j

-1-j

d

ef

-3


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Angle criterion:o

j

n

j

i

m

i

pszssHsG 180)()()()(11

=++= ==

Total phase = c a b = 0o Total phase = f d e = -180o

Point s = -1 is not on root locus Point s = -3 is on root locus

Find Kat s = -3.

8

-2

-1+j

-1-j

p

q

r-3


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Magnitude criterion:

( )

( ) Kps

zs

sHsGn

j

j

m

i

i 1)()(

1

1 =

+

+

=

=

=

Kqp

r 1=

K= 5

4.4 PROPERTIES AND CONSTRUCTION OF ROOT LOCI

RULES FOR CONSTRUCTION OF ROOT LOCI

Rule 1: Number of Root Loci (Branches)

The root locus plot consists of n root loci (branches) as K varies from 0to . The loci are symmetric with respect to the real axis.

Characteristic equation: 0)()()(11

=+++= ==

m

i

i

n

j

j zsKpss

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Degree n n roots

Complex roots occur in complex conjugate pairs

Root loci symmetrical about the real axis

Rule 2: Starting and Ending Points of Root Loci

As K increases from 0 to , each root locus starts from an open-loop

pole with K = 0 and ends on an open-loop zero with K = . The number

of root loci ending atequals the number of open-loop zeros at.

Example: For a system with )1)(1(2

)(++

+=

ss

ssG and 1)( =sH , find the starting and

ending points of the root loci.

Characteristic equation:0)2()1)(1( =++++ sKss

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Starting points: K= 0, s = -1, -1 (open-loop poles)

Ending points: K= , s = -2, (open-loop zeros)

Rule 3: Asymptotes to Root Loci (Behaviour at Infinity)

The (n-m) root loci which tend to do so along straight line asymptotes

radiating out from a single point s= - a on the real axis (called thecentroid) where

mna

=

)zerosloop-openofpartreal()polesloop-openofpartreal(

These (n-m) asymptotes have angles

)1(,,1,0;180)12( 0

=

+= mnq

mn

qa

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( )

( )0

)(11

1

1 =+

++

++

=

=

mn

a

n

j

j

m

ii

s

K

ps

zsK

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( )

( )

mn

am

i

i

n

j

j

s

zs

ps

=

= ++

+

)(

1

1

...)(...11

11

++=+

+

==

mn

a

mnmnm

i

i

n

j

j

mn smnsszps

mn

zpn

j

m

i

ij

a

=

= =1 1

)()(

From angle criterion:,1,0;180)12()(

0=+= qqmn

)1(,,1,0;180)12(

0

=

+= mnqmn

qa

Example: Find the centroid and asymptotes of the root locus for

)4)(3)(41)(41(

)2()()(

+++++

+=

ssjsjs

sKsHsKG .

Rule 1: 4 branches.

Rule 2: Starting points are open-loop poles s = -1 4j, -3, -4.

Ending points are open-loop zeros s = -2, , , .Rule 3: 3 asymptotes.

Centroid, 37

14

)2(4311=

= a

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Angles of asymptotes

)1(,,1,0;180)12( 0

=

+= mnq

mn

qa

2,1,0;14

180)12(0

=

+= q

qa a= 600, 1800, and 3000

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Rule 4: On-locus Segments on the Real Axis

A point on the real axis lies on the locus if the number of open-loop polesplus zeros on the real axis to the right of this point is odd.

Take a point s0 on the real axis.

(i)Poles and zeros on the real axis to the right of this point contribute an

angle of 1800 each.(ii) Poles and zeros to the left of this point contribute angle of 00 each.

(iii) The net angle contribution of a complex conjugate pole or zero

pair is always zero.

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G(s)H(s) = (mr+ nr)180

0

mr= number of open-loop zeros on the real axis to the right ofs0

nr= number of open-loop poles on the real axis to the right ofs0

From angle criterion: G(s)H(s) = 1800 = (2q + 1)1800, q = 0, 1, 2,

Angle criterion is satisfied if (nr+ mr) is odd.

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Rule 5: On-Locus Points of the Imaginary Axis

The intersections (if any) of root loci with the imaginary axis can be

determined by use of the Routh criterion.

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Stability limits Imaginary axis crossings

Example: Find the imaginary axis crossings for the root locus plot of

)4)(3)(41)(41()2()()(

++++++=

ssjsjssKsHsKG .

Characteristic equation02204)143(439

234=++++++ KsKsss

Routh Table:

s4 1 43 204+2K

s3 9 143+K

s2 (244 K)/9 204+2K

s

1

(18368 61KK

2

)/(244 K)s0 204+2K

244 K> 0 K< 244

18368 61KK2 > 0 -169.4 < K< 108.4

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204 + 2K> 0 K> -102

Hence, 0 < K< 108.4

ForK= 108.4, all the coefficients in s1

row are zero.Auxiliary equation formed from the coefficients ofs2 row.

0)2204(9

244 2=++

Ks

K s = j5.28

Root loci intersect the imaginary axis at s = j5.28, corresponding value ofK

is 108.4.

Rule 6: Angle of Departure from Complex PolesThe angle of departure, p, of a locus from a complex open-loop pole is

given by p = 1800+ , where is the net angle contribution at thispole of all other open-loop poles and zeros.

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Net angle contribution of all other open-loop poles and zeros at s0 is)( 4312 ++=

Total phase G(s)H(s) at s0 = - p.

From angle criterion, - p = 1800

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Angle of departure from the complex open-loop pole, p = 1800 + .

For this example,

1 = 90076)1/4(tan

1

2 ==

63)2/4(tan1

3 ==

53)3/4(tan1

4 ==

0

4312130)( =++=

p = 1800

+ = 500

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Rule 7: Angle of Arrival at Complex Zeros

The angle of arrival, z, of a locus at a complex zero is given by

z = 1800 - , where is the net angle contribution at this zero of allother open-loop poles and zeros.

Example:Let us consider the characteristic equation 0)2()1(

12

=+

++

ss

sK.

Rule 1: 2 branches.

Rule 2: Starting points are open-loop poles s = 0, -2.

Ending points are open-loop zeros s = j1.

Rule 3: Not applicable.

Rule 4: On-locus segments on the real axis between 0 and -2.

Rule 5: Root locus touches imaginary axis at s =

j1.Rule 6: Not applicable.

Rule 7:

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Points at which multiple roots of the characteristic equation occur

(breakaway/break-in points of root loci) are the solutions of 0=dsdK

where( )

( )

=

=

+

+

= m

i

i

n

jj

zs

ps

K

1

1

).

Example: Draw the root locus plot for the characteristic equation0

)1(

)3)(2(1 =

+

+++

ss

ssK

.

Rule 1: 2 branches.


Ending points are open-loop zeros s = -2, -3.


Rule 4: On-locus segments on the real axis between 0 and -1, and

between -2 and -3.Rule 5: Not applicable.


Rule 7: Not applicable.26


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Rule 8:

65)3)(2(

)1(2

2

++

+=

++

+=

ss

ss

ss

ssK

0)65(

)52)(()12)(65(22

22

=

++

+++++=

ss

ssssss

ds

dK

03622

=++ ss

s = -0.634 (breakaway point), s = -2.366 (break-in point).

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Condition for breakaway/break-in point (as derived above) is necessary

but not sufficient.

4.5 COMPLETE EXAMPLES

Question 1: Consider a feedback system with the characteristic equation0;0

)2)(1(1 =

+++ K

sss

K

. Plot the root locus for this system. Find the dominant

closed-loop poles for = 0.5.

Solution:

Rule 1: 3 branches.

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Rule 2: Starting points are open-loop poles s = 0, -1, -2.

Ending points are open-loop zeros s = , , .

Rule 3: 3 asymptotes.

Centroid, 103 12 == a Angles, )1(,,1,0;

180)12( 0=

+= mnq

mn

qa

2,1,0;03

180)12( 0=

+= q

qa a= 600, 1800, and 3000

Rule 4: On-locus segments on the real axis between 0 and -1, and

between -2 and -.

Rule 5: Characteristic equation: 023 23 =+++ Ksss

Routh Table:

s3 1 2

s2 3 K

s1 (6-K)/329


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s0 K

For stability, K> 0, (6 - K)/3 > 0

Crosses imaginary axis when K= 6.Auxiliary equation: 0633 22 =+=+ sKs 2js =



Rule 8: K= - (s3 + 3s2 + 2s).0)263(

2=++= ss

ds

dK

s = -0.4226 (breakaway point), -1.5774 (not valid)

If two loci break away from a breakaway point, their tangents will be

1800 apart.

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In general, if r loci break away from a breakaway point, then their

tangents will be 3600/rapart, i.e., the tangents will equally divide 3600.

= 0.5 line makes an angle cos-1 = 600 with the negative real axis.

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Points of intersection: s = -0.33 j0.58.

From magnitude criterion, { } 04.121 58.033.0 =++= += jssssK .

Question 2 (past year exam question 2004/2005): Sketch the root locus

of a unity feedback system with forward path transfer function G(s) givenas follows:

)54()(

2++

=sss

KsG

Solution:

Rule 1: 3 branches.Rule 2: Starting points are open-loop poles s = 0, -2+j, -2-j.

Ending points are open-loop zeros s = , , .


Centroid, 34

03

22=

=

a

Angles, )1(,,1,0;180)12( 0

=

+= mnq

mn

qa

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2,1,0;

03

180)12( 0=

+= q

qa a= 600, 1800, and 3000

Rule 4: On-locus segments on the real axis between 0 and -.

Rule 5: Characteristic equation: 05423

=+++ Ksss Routh Table:

s3 1 5

s2 4 K

s1 (20-K)/4

s0 K

For stability, K> 0, (20-K)/4 > 0

Crosses imaginary axis when K= 20.

Auxiliary equation: 02044 22 =+=+ sKs 5js =

Rule 6: For pole -2+j,

= -153.43 - 90 p = 1800 + = -63.430

For pole -2-j,

Due to symmetry properties: p = 63.430

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Rule 8: Characteristic equation: K= - (s3 + 4s2 + 5s).0)583(

2=++= ss

ds

dK

s = -1 (breakaway point), -1.667 (break-in point)

From the characteristic equation,21 ==sK , 852.1667.1 ==sK

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4.6 EFFECTS OF ADDING POLES AND ZEROS TO G(s)H(s)

Controller design may be treated as an investigation of the effects to root

loci when poles and zeros are added to the loop transfer functionKG(s)H(s).

Addition of Poles:

Has the effect of pushing the root loci toward the right-half s-plane.

Example: Consider the loop transfer function given by )2()()( += ssK

sHsKG .

Rule 1: 2 branches.


Ending points are open-loop zeros s = , .


Centroid, 10220

=

= a

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Angles, )1(,,1,0;

180)12( 0=

+= mnq

mn

qa

1,0;02

180)12( 0=

+= q

qa a= 900, 2700

Rule 4: On-locus segments on the real axis between 0 and -2.Rule 5: Not applicable.



Rule 8: 0)22( =+= sdsdK

s = -1 (breakaway point)

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Stable for all K.

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Let us introduce a pole at s = -b (b > 2).

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

++=

++=

sss

K

bsss

KsHsKG , with b = 3

The centroid of asymptotes changes from -1 to -(2+b)/3.

The angles of asymptotes change from 90 to 60and 180. Root loci bend towards right-half s-plane unstable at large K.

Addition of Zeros

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Has the effect of pushing the root loci toward the left-half s-plane.

Let us introduce a zero at s = -3. )2()3(

)()(+

+=

ss

sKsHsKG .

Relative stability is improved.

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BARU - EPM2036 Chapter 4 Notes _condensed

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