Tutorial on Root Locus

7/27/2019 Tutorial on Root Locus

1/30

Chapter 7:

The Root Locus Method

Tutorial Session

EE4314 - Summer 2008

Prof. K. Alavi

TA: An Vo

Content

Characteristic Equation

Definition of Root-locus

Angle and Magnitude Condition

Sketching Root Locus

Change In Pole-zero Configuration


2/30

Closed loop system

consider a general form of closed loop system:

)(sR )(sC)(sE )(sG

)(sH

with transfer function:)()(1

)(

)(

)(

sHsG

sG

sR

sC

+=

Characteristic equation

0)()(1 =+ sHsG

or:

1)()( =sHsG


3/30

Root locus

The root locus is the path of the roots of

the characteristic equation in the s-plane as

a system parameter is changed.

Example : system definitionconsider a closed loop system:

)(sR )(sC)(sE1K

)10(

2

+ss

K

after reduction:

)(sR )(sC

K

K

++102

21KKK=


4/30

Example : table of pole positions

Norman S. Nisse (2004).

Control Systems Engineering, 4th edition

John Wiley & Sons

0102 =++ Kss

Example : poles in s-plane



John Wiley & Sons


5/30

Example : root locus



John Wiley & Sons

Angle and magnitude condition

because

1)()( =sHsG

)()( sHsG is complex

amplitude condition:

,...2,1,0)12(180)()( =+= kksHsG

angle condition:


6/30

Angle and magnitude condition

The values ofs that fulfill both the angle andmagnitude conditions are the roots of the

characteristic equation (closed-loop poles).

Root locus property

A locus of the points in the complex plane

satisfying the angle condition alone is the

root locus.


7/30


8/30

Example : vector representation of G(s)

consider a closed loop system:

)(sR )(sC

)2)(1(

)4)(3(

++

++

ss

ssK

Example : angle and magnitude

=

++

++=

)2)(1(

)4)(3()(

ss

ssKsG

)12(180)2()1()4()3( +=+++++= kssss

amplitude condition:

angle condition:

1)2)(1(

)4)(3()( =

++++=

ss

ssKsG


9/30

Example

Check, if pointbelongs to root locus

32 js +=

+==+=+

180)12(55.70

43.1089057.7131.564321

k

Norman S. Nisse(2004).


John Wiley & Sons

NO

Example

Check, if point belongs to root locus2

22 js +=

=+ 1804321

33.0)22.1)(22.1(

)22.1(

2

2

21

43=== LL

LLK

For which value ofK?

YES


10/30

Sketching the root-locus plot

1. Locate the poles and zeros of G(s)H(s) on the s plane.

The root locus is symmetrical about the real axis

The number of branches of the root locus equals the number of

the closed-loop poles

The root locus begins at the finite and infinite poles of G(s)H(s)

and ends at the finite and infinite zeros of G(s)H(S)

Example : construction of root locus


)(sR )(sC

)2)(1(

)4)(3(

++

++

ss

ssK

for this system system:

1)( =sHand)2)(1(

)4)(3()(

++

++=

ss

ssKsG


11/30

Example



John Wiley & Sons

Example



John Wiley & Sons


12/30

Sketching the root-locus plot

2. Determine the asymptotes of the root loci on the

real axis.

The root locus approaches straight lines as asymptotes as the

locus approaches infinity. The equation of the asymptotes is

given by the real-axis intercept, a, and angle, a, as follows

zerosfinite#polesfinite#

zerosfinitepolesfinite

=a

,...2,1,0zerosfinite#polesfinite#

)12(

=

+

= kk

a

Example: finding asymptotes


)(sR )(sC

)4)(2)(1(

)3(

+++

+

ssss

sK

real axis intercept:

34

14)3()421( =

=a

angles of the lines:

3

)12(

+=

ka

33

5,,

3

==a


13/30

Example : finding asymptotes



John Wiley & Sons

sketching the root-locus plot3. Find the points where the root loci may cross the

imaginary axis.

a) Use the Rouths stability criterion

b) Let s=j in characteristic equation, equate both the real part andimaginary part to zero, and then solve for and K.


14/30

Example :j-axis crossing


)(sR )(sC

)4)(2)(1(

)3(

+++

+

ssss

sK

closed-loop transfer function of the system:

KsKsss

sKsT

3)8(147

)3()(

234

+++++

+=


Routh table:

65.90720652 ==+ KKK

59.107.20235.8021)90( 22 jssKsK ===

-74.6456 9.6456


15/30




John Wiley & Sons

1.59

Closed loop system is stable

if K


16/30


17/30

Example : break-away point



John Wiley & Sons

-0.435

sketching the root-locus plot5. Taking a series of test points in the broad

neighborhood of the origin of the s plane, sketch the

root loci.

if accurate shape of the root-loci is needed MatLab


18/30

Example num = [1];den = [ 1 1];rlocus(num, den)

Examplenum = [1];

den = conv([1 1],[1 2]);

rlocus(num, den)


19/30


20/30

Example

num = [1];

den = conv([1 0],[1 1]);

den = conv(den,[1 3]);

rlocus(num, den)

Addition of zeros

num = [1 4];

den = conv([1 0],[1 1]);


rlocus(num, den)


21/30

Addition of zeros

num = [1 2];

den = conv([1 0],[1 1]);


rlocus(num, den)

Addition of zeros

num = [1 0.5];

den = conv([1 0],[1 1]);


rlocus(num, den)


22/30

Root-locus plot: MatLab

% open-loop system

clear;

num = [0 0 4];

den = [1 2 0];

% root locus plot

rlocus(num,den)

v=[-5 1 -3 3]; axis(v); axis('square')

grid

title('root-locus plot')

% r: complex root locations, K: gai vector[r K] = rlocus;


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

-2

-1

0

1

2

30.160.340.50.640.76

.

0.94

0.985

0.160.340.50.640.760.86

0.94

0.985

1234

root-locus plot of the system

Real Axis

ImaginaryAxis


23/30


% open-loop system

clear;

num = [0 1 1];

den = [1 2 3 4];

% root locus plot

sys = tf(num, den);

rltool(sys)


-1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-5

-4

-3

-2

-1

0

1

2

3

4

5Root Locus Editor for Open-Loop 1 (OL1)

Real Axis

ImagAxis


24/30

example :

construction of root locusconsider a closed loop system:)(sR )(sC

32

)2(2 ++

+sK

G(s) has a pair of complex conjugated poles at:

21 js +=and

21 js=

change in pole-zero configuration

A slight change in the pole-zero

configuration may cause significant

changes in the root-locus configurations.


25/30




26/30




27/30




28/30


29/30


Problems

Prob 9 and 10: E7.26 & 7.27 Plot root locus

using matlab

Prob 11: See section 7.6/page 444 and

7.7/447

Example 7.11/ page 452


30/30

impulse(sys)

step(sys)

References

Lecture notes, Dr. Darek Korzec

Tutorial on Root Locus

Documents

Transcript of Tutorial on Root Locus