Root Locus - Control Systems.pdf

Module-4-A-Graphical Tools 1

ECB 3123 CONTROL SYSTEMS

Dr Irraivan Elamvazuthi

MODULE 4-A

GRAPHICAL TOOLS

Root Locus Techniques




In this chapter, you will be able to:

• Sketch a root locus

• Refine the sketch of root locus

• Use root locus to find poles of a closed-loop system

• Use root locus to describe transient response and stability

when system parameter, K is varied




Root Locus: Graphical presentation of closed-loop poles

as system parameter is varied




Root Locus: Graphical presentation of closed-loop poles

as system parameter is varied




Review: Closed-Loop Systems

Closed-loop poles:characteristics equation

0)()()()( sNsKNsDsD HGHG

)()()()(

)()()(

)()(1

)()(

)(

)()(

)(

)()(

sNsKNsDsD

sDsKNsT

sHsG

sKGsT

sD

sNsH

sD

sNsG

HGHG

HG

H

H

G

G




Review: Complex Numbers

1

22

tan

M

MMe

jsj





jasF

ajsF

assF

js

)()(

)()(

)(

jsas

jsasM

js

to connecting line

tohorizontal from angle

to from distance




Example 1: Given F(s) = s+7, find F(s) at point s = 5+j2

43.6312

2tan

16.12212

212)(

7)25()(

25

7)(

1

22

25

M

jsF

jsF

js

ssF

js





js

)())((

)())((

)(

)()(

21

21

n

m

pspsps

zszszs

sD

sNsF

n

j

j

m

i

i

ps

zs

sF

1

1

)(

)(

)(

magnitude

angle

lengthspole

lengthszero

ps

zs

Mn

j

j

m

i

i

1

1

)(

)(

anglespoleangleszero

pszsn

jj

m

ii

11

)()(




Example 2: Evaluation of a complex function via vectors

Given )2(

)1()(

ss

ssF , find F(s) at s = -3+j4

SOLUTION

6.11643.632

4tan

204)2(

42)(

1)43()(

1)(

1

22

43

1

M

jsF

jsF

ssF

js

9.12613.533

4tan

5204)3(

43)(

)(

1

22

43

2

M

jsF

ssF

js

10496.751

4tan

174)1(

41)(

2)(

1

22

43

3

M

jsF

ssF

js

0 -1 -2

j




3.114217.

)0.1049.1266.116(175

20M

6.11620

4.10417

9.1265




)()(1

)()(

sHsKG

sKGsT

Properties of the Root Locus

Root locus:

0)()(1 sHsKG

conditions:

1)()( sHsKG

180)12()()( ksHsKG

)()(

1

sHsGK

Odd multiples of 180




)2)(1(

)4)(3()()(

ss

ssKsHsKG

Test point 1: s = -2+j3

Example 3: Given open loop transfer function as follows, determine whether

the test point is located on the root locus.

55.7043.1089057.7131.56

4321 Not an odd multiple of 180, the

test point is not a point on the root locus.




magnitude:

33.0)22.1)(12.2(

)22.1)(707.0(

zeros from distance

poles from distance

)()(

1

21

43

LL

LLK

sHsGK

Test point 2: )2/2(2 j

Angles do add up to odd multiples of 180,

the test point is a point on the root locus.

Thus, we can say that:

• Given poles and zeros of the open-loop transfer function, KG(s)H(s), a point

in the s-plane is on the root locus for some value of K, if

180)12( kanglespoleangleszero

• Gain K at which angles add up to (2k+1) 180 is

zeros from distance

poles from distance

)()(

1

sHsGK

180

74.1449026.3547.194321




Sketching the Root Locus

1. Write the characteristic equations

0)()(1 sHsKG

2. Open Loop:

Factor and locate poles and zeros of G(s)H(s)

0)())((

)())((1

21

21

n

n

pspsps

zszszsK

3. Real axis:

On the real axis, for K > 0, the root

locus exists on the left of

odd numbered real roots

(poles/zeros).




4. Branches

The number of branches equals

the number of open loop poles.

*Loci always start, K = 0, at a pole

and terminate at a zero, K = .

*Loci that do not terminate at a zero

will approach a point (zero) at infinity .

Root Locus Construction Steps

5. Symmetry

The locus is always symmetrical wrt the real

(horizontal) axis.




6. Asymptotes

Loci proceeding to zeros at will

follow the asymptotic lines with

centroid

angle

zp

ii

Ann

zp

zeros#poles#

zerospoles

3

4

14

)3()421(

1,03

)12(

kforkfor

nn

k

zp

A




7. j crossing

Determine j crossing by finding gain K

that will ensure stability using Routh-Hurwitz

table

KsKsss

sKsT

ssss

sKsHsKG

)8(147

)3()(

)3)(2)(1(

)3()()(

234

s4 1 14 3K

s3 7 8+K

s2 90-K 21K

s1 K

KK

90

720652

s0 21K 0 0

59.1

07.20235.8021)90(

65.9

072065

22

2

js

sKsK

K

KK




82.3,45.1

0)158(

612611

)158(

)23(

1)23(

)158()()(

22

2

2

2

2

2

s

ss

ss

ds

dK

ss

ssK

ss

ssKsHsKG

8. Breakaway and Breakin points

Determine breakaway and breakin points

by determining the maximum/minimum

change in K wrt changes in s

0

1)()(

sds

dK

sHsKG

Breakaway point

Breakin point




Example: Given the following system

1. Find the exact point and gain where the locus crosses

the j-axis

S2

K+1

8+20K

S1

6-4K

0

S0

8+20K

0

6-4K = 0

K = 1.5

(K+1)s2 + (8+20K) = 0

s = j3.9

2. Find the breakaway point on the real axis

204

)86(2

2

ss

ssK

)208()46()1(

)204()(

2

2

KsKsK

ssKsT

dK/ds = 0, b/away point: -2.88

-4 -2

2+j4

2-j4




Thus, sketch of root locus:

Breakaway point,

s = -2.88

j-axis crossing,

s = j3.9




9. Angle of Departure and Arrival

Determine angle of arrival for

complex zeros and angle of

departure for complex poles

180polesother from angle

zerosother from angledep -

180zerosother from angle

polesother from anglearr -




180polesother from angle

zerosother from angledep -

4.1086.251

180)2

1tan90(

1

1tan

180)(

11

423dep

Example: Angle of departure from a complex pole

Documents

Root Locus - Control Systems.pdf