Root locus - Department Of Electrical...

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Root locus Control system by: Nafees Ahmed 1 Root locus

Transcript of Root locus - Department Of Electrical...

Page 1: Root locus - Department Of Electrical Engineeringeedofdit.weebly.com/.../7326910/notes_tee602_root_locus.pdfRoot locus construction (i i) L oci Branches: Each locus starts from an

Root locus

Control system by: Nafees Ahmed 1

Root locus

Page 2: Root locus - Department Of Electrical Engineeringeedofdit.weebly.com/.../7326910/notes_tee602_root_locus.pdfRoot locus construction (i i) L oci Branches: Each locus starts from an

What is Root Locus ?

H(s)

The characteristic equation of the closed-loop system is

1 + K G(s)H(s) = 0

The root locus is essentially the trajectories of roots of thecharacteristic equation as the parameter K is varied from 0 toinfinity.

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A simple example

A camera control system:

How the dynamics of the camera changes as K is varied ?

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A simple example (cont.) : pole locations

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A simple example (cont.) : Root Locus

(a) Pole plots from the table. (b) Root locus.

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Root locus

k G(s)

H(s)

+-

)(ty)(tr

Control system by: Nafees Ahmed 6

)()(1

)(

)(

)()(

sHskG

skG

sR

sysT

0)()(1 sHskG poles

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)12()()(

1)()(

1)()(

0)()(1

nsHskG

sHskG

sHskG

sHskGOpen loop transfer function

Using open loop transfer function + system parametersto analyze the closed-loop system response

Control system by: Nafees Ahmed 7

Using open loop transfer function + system parametersto analyze the closed-loop system response

0k

Draw the s-plan root locus

zerossHsGk

polessHsGk

)()(,

)()(,0

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)1.01)(5.01()()(

sss

ksHskG

Example: Draw Root locus

Sol:In Matlab>>n=[1];>>d=[0.05 0.6 1 0];>>rlocus(n,d)

Control system by: Nafees Ahmed 8

Sol:In Matlab>>n=[1];>>d=[0.05 0.6 1 0];>>rlocus(n,d)

Page 9: Root locus - Department Of Electrical Engineeringeedofdit.weebly.com/.../7326910/notes_tee602_root_locus.pdfRoot locus construction (i i) L oci Branches: Each locus starts from an

Control system by: Nafees Ahmed 9

Page 10: Root locus - Department Of Electrical Engineeringeedofdit.weebly.com/.../7326910/notes_tee602_root_locus.pdfRoot locus construction (i i) L oci Branches: Each locus starts from an

Root locus construction

(ii) Loci Branches: Each locus starts from an open pole (K=0) andterminates at either on an open loop zero oron infinity (K=∞)

P=No of finite polesZ=No of finite zeros

(i) Root Locus is symmetrical about real axis

No of branches in root loci =P if P>Z=Z if Z>P

Let

Control system by: Nafees Ahmed 10

if ZP

if togoesbranchesZP

zerostofromcomesbranchesPZ

=>Poles are more

=>Zeros are more

No of branches in root loci =P if P>Z=Z if Z>P

PZ

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(iii) Existence of Root loci on real axis

Poles + zeros = odd

0180

Note: Only real axis Poles+ Zeros not complex

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Poles + zeros = even

01801)()( sHskG

If Poles + Zeros on RHS= Odd=> That portion will be in root locusIf Poles + Zeros on RHS= Even=> That portion will not be in root locus

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(iv) Asymptotic angles,2,1,0,

)12(

kZP

kk

0454

1802,6 ZPif

Reference for angle

Control system by: Nafees Ahmed 12

045

045

Total no of asymptotes =P-Z

Reference for angle

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(v) Centroid of the asymptotes

ZP

zerospoles

)186)(2(

3)()(

2

sss

ssHsG

Zero : 0Poles: -2, -3+j3, -3-j3 4

13

0)33332(

jj

example

Control system by: Nafees Ahmed 13

Zero : 0Poles: -2, -3+j3, -3-j3 4

13

0)33332(

jj

09013

180

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(vi) Breakaway and entry points 0ds

dk

example)2)(1(

sss

kkGH

01 kGH The characteristic function of closed loop system

0)2)(1(

231

23

sss

kssskGH

Control system by: Nafees Ahmed 14

0)2)(1(

231

23

sss

kssskGH

577.1,423.0

0263

)23(

2

23

s

ssds

dk

sssk

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(vii) Angle of departure and arrival

example)1)(1(

)2(

jsjs

skkGH

Angle of departure from the pole: js 1

)()(1800 sHsGA

)()(1800 sHsGD Angle of departure from a complex pole

Angle of arrival at a complex zero

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0

0

0

0

225

)11()21(180

180)11()21(

180)1()2(

180)1()1()2(

D

D

D

D

jjj

jjj

jss

jsjss

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Angle of approach to the zero:

example )1(

))((

ss

jsjskkGH

js

0

0

0

0

0

135

)1()(180

180)1()(

180)1()(

180)1()()(

A

A

A

A

jjjj

jjjj

ssjs

ssjsjs

Control system by: Nafees Ahmed 16

0

0

0

0

0

135

)1()(180

180)1()(

180)1()(

180)1()()(

A

A

A

A

jjjj

jjjj

ssjs

ssjsjs

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(viii) Intersection with Img axis: By Routh Hurwitz criterion

example)22)(3( 2

ssss

kkGH

The characteristic function of closed loop system:

0685

0)22)(3(234

2

kssss

kssss

ks

ks

ks

s

ks

0

1

2

3

4

34

252045

3465

81

16.8

034

25204

k

k

rowzerogetTo

Control system by: Nafees Ahmed 17

ks

ks

ks

s

ks

0

1

2

3

4

34

252045

3465

81

16.8

034

25204

k

k

rowzerogetTo

095.1

05

34 2

js

equauxilaryks

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)1.01)(5.01( sss

k

+ -

)(sC)(sR

)1.01)(5.01()()(

sss

ksHskG

,,

10,2,0

zeros

poles

05.7,945.0

0)6.005.0(

06.005.0

0)1.01)(5.01(

21

23

23

ss

sssds

d

ds

dk

ksss

ksss(i)

Example

Control system by: Nafees Ahmed 18

,,

10,2,0

zeros

poles

6003

180

403

0)10()2(0

k

05.7,945.0

0)6.005.0(

06.005.0

0)1.01)(5.01(

21

23

23

ss

sssds

d

ds

dk

ksss

ksss(i)

(ii)

(iii)

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ks

ks

ks

s

ksss

6.0

05.06.06.0

105.0

06.005.0

1

2

3

23

5.4

0126.0

122

js

s

k

)12(5.4 kj

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)0(

0

k)0(

2

k)0(

10

k

)12(5.4 kj

945.0s4

060

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Determination of K on Root loci

point the toG(s)H(s)ofzeros thefromdrawnlengths vectorallofProduct

point the toG(s)H(s)ofpoles thefromdrawnlengths vectorallofProductK

A

BxCK

Control system by: Nafees Ahmed 20

AB

C

Root Loci

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)1.01)(5.01()(

sss

kskGH

MATLAB method

ssss

skskGH

15

)93()(

234

gh=zpk([],[0 –2 -10],[1])rltool(gh)

Example 1

Example 2

Control system by: Nafees Ahmed 21

n=[-3 -9]m=[1 –1 –1 –15 0]gh=tf(n,m)rltool(gh)

ssss

skskGH

15

)93()(

234

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Example 3:

1)(,)84(

)(2

sHsss

KsG

>> n=[1];>> d=[1 4 8 0];>> rlocus(n,d)

Ord may be written as

d=conv([1 0],[1 4 8]);

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>> n=[1];>> d=[1 4 8 0];>> rlocus(n,d)

Ord may be written as

d=conv([1 0],[1 4 8]);

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Thanks?

Control system by: Nafees Ahmed 23

Thanks?