Root Locus Analysis

ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems

Root Locus AnalysisRoot Locus AnalysisRoot Locus AnalysisRoot Locus Analysis

ME2142/ME2142E Feedback Control Systems1

Root Locus AnalysisRoot Locus Analysis

Consider the closed-loop system RG

C

-

+ E

B

The transient response, and stability, of the closed-loop system is determined by the values of the roots of the characteristic

HB

yequation or, in other words, the location of the closed-loop poles on the s=plane.

The open-loop transfer function can be written in the form

0)()(1 sHsG

where K is an adjustable gain, the z’s and p’s are the zeros

)())(()())((

)()(21

21

n

m

pspspszszszs

KsHsG

j g , pand poles of the open-loop transfer function. As the gain K changes, the values of the closed-loop poles will change and thus the transient response, and stability.Th t l l t i l t f th l i f th l d l l


The root locus plot is a plot of the loci of the closed-loop poles on the s-plane as the gain K varies from 0 to infinity.

Why Root Locus?Why Root Locus?

1

Consider the closed-loop system for which K is a proportional controller:Consider the closed-loop system for which K is a proportional controller:

)2)(1(1

sss

Questions:Questions:

1)How will the system respond as K is varied?

2)Can the system ever be unstable for some values of K?

3)How should K be adjusted to give a “good” response?

Depends upon how changes in K changes the values of the

closed-loop poles, or the roots of the characteristic

Depends upon how changes in K changes the values of the

closed-loop poles, or the roots of the characteristic 1


equation:equation: 0)2)(1(

111

sss

KGH


ExampleExample

Characteristic Eqn: 0)2)(1( Ksss

0)2)(1(

1sss

K

Root Locus Plot

Closed-loop Poles vs KClosed-loop Poles vs K

)2)(1( sss

Closed loop Poles vs KClosed loop Poles vs KK P1 P2 P3 0 0 -1 -2

0.1 -0.054 -0.90 -2.05 0 2 -0 12 -0 79 -2 09 K=0.2

K=1

K=7

0.2 -0.12 -0.79 -2.090.4 -0.42+j0.09 -0.42-j0.09 -2.16 0.7 -0.38+j0.41 -0.38-j0.41 -2.25 1 -0.34+j0.56 -0.34-j0.56 -2.32 2 -0.24+j0.86 -0.24-j0.86 -2.522 0.24+j0.86 0.24 j0.86 2.524 -0.10+j1.9 -0.10-j1.9 -2.80 7 0.04+j1.50 0.04-j1.50 -3.09

10 0.15+j1.73 0.15-j1.73 -3.31

K=0K=0.4 K=1

K=7


3 poles, thus 3 loci.

Why Root Locus?Why Root Locus?

0)2)(1(

11

sss

KHow do the roots of change as K varies?

This can be easily

seen, K=7

seen,

graphically,

from a Root K=0.2

K=1

Locus for

)2)(1()()(

sssKsHsG

K=0.4 K=1)2)(1( sssK=0

K=7


Plotting the Root LociPlotting the Root Loci

The root loci are plotted either

Manually, or

The root loci are plotted either

Manually, oru y, o

Using a computer program such as OCTAVE(easy if you have the program and knows how to use it.)

u y, o

Using a computer program such as OCTAVE(easy if you have the program and knows how to use it.)


Manual plotting – Root Locus ConceptsManual plotting – Root Locus Concepts

The Characteristic Equation is first written in the formThe Characteristic Equation is first written in the form

0)(1)()(1 sKFsHsG

where K>0 is a constant gain.where K>0 is a constant gain.

0)(1)()(1 sKFsHsG

Th t f th h t i ti ti th l The roots of the characteristic equation are the values of s which satisfy the equation

for 18011)( nsKF ,5,3,1 n

Or when 180)( nsF ,5,3,1 n

The magnitude condition is satisfied by having The magnitude condition is satisfied by having

1)( sKF giving )(

1sF

K


Manual plotting – Root Locus ConceptsManual plotting – Root Locus Concepts

Consider

Determining the phase angle for F(s)Determining the phase angle for F(s)

))()()(()(

)()()(4321

1

pspspspszsK

sKFsHsG

)(4321

1

4321

14321

1

4321

1

j

j

jjjj

j

ePPPPeKZ

ePePePePeKZ

Then

where Im

(s+p2) 2Im

(s+p2) 2

180)( 43211 nsF ,5,3,1 n

)( 11 zs

-p2s

s

( p2)

-p2s

s

( p2))( 11 ps

)( 22 ps

0 Re-z

s(s+z1)

1

0 Re-z

s(s+z1)

1)( 33 ps

)( 44 ps


0 Rez1 0 Rez1Note that 360n,3,2,1n

Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines

1) Locate the poles and zeros of the open-loop transfer, G(s)H(s), function on the s plane.

1) Locate the poles and zeros of the open-loop transfer, G(s)H(s), function on the s plane.

2) There are as many loci, or branches, as poles of the G(s)H(s).

2) There are as many loci, or branches, as poles of the G(s)H(s).

3) Each branch starts from a pole of G(s)H(s) and ends in a zero. If there are no zeros in the finite region, then the zeros are at infinity.

3) Each branch starts from a pole of G(s)H(s) and ends in a zero. If there are no zeros in the finite region, then the zeros are at infinity.

Reason: 0)()( sKNsD

)(

0)()(1)()(1

sDsNKsHsG

When K=0, and roots are roots of D(s).0)( sD

When , and roots are roots of N(s).1K 0)( sN



4) The loci exist on the real axis only to the left of an odd number of poles and/or zeros. Complex poles and/or zeros have no effect because for a point on the real

4) The loci exist on the real axis only to the left of an odd number of poles and/or zeros. Complex poles and/or zeros have no effect because for a point on the real zeros have no effect because, for a point on the real axis, the angles involved are equal and opposite. zeros have no effect because, for a point on the real axis, the angles involved are equal and opposite.

ImConsider a test point, s, on the real axis as shown.

j1

j2

-p1

1Every pair of complex conjugate poles (or zeros) will contribute a pair of

Every pair of complex conjugate poles (or zeros) will contribute a pair of

0 1-1-2-3

-j1

Re-p3

-z1

3s

will contribute a pair of angles, and such that

They can thus be ignored

will contribute a pair of angles, and such that

They can thus be ignored

1 2 36021

-j2-p2

2They can thus be ignored.They can thus be ignored.



4) The loci exist on the real axis only to the left of an odd number of poles and/or zeros. Complex poles and/or zeros have no effect because for a point on the real

4) The loci exist on the real axis only to the left of an odd number of poles and/or zeros. Complex poles and/or zeros have no effect because for a point on the real zeros have no effect because, for a point on the real axis, the angles involved are equal and opposite. zeros have no effect because, for a point on the real axis, the angles involved are equal and opposite.

ImConsider a test point, s, on the real axis as shown.

j1

j2

-p1

1Each real pole or zero to the left of point s does not contribute to the angle sum

Each real pole or zero to the left of point s does not contribute to the angle sum

0 1-1-2-3

-j1

Re-p3

-z1

3s

contribute to the angle sum and thus can be ignored.contribute to the angle sum and thus can be ignored.

Each pole/zero to the right of point s contributes an

-j2-p2

2180

180n531

angle of . An odd number of them will thus contribute a total of

h


,5,3,1 nwhere

Locus exists to left of odd No. of zeros/polesLocus exists to left of odd No. of zeros/poles

Example: Consider the characteristic equationExample: Consider the characteristic equation

)22)(22(

)3(jsjss

sKGH

)())((

)(

21

sKFpspss

zsK

Im

-pOne zero at s=-z=-3;

21

j1

j2-p1

One zero at s=0;

0 1-1

j

-2-3 Re

p0-zComplex conjugate poles at s=-p1=-2-j2, and s=-p2=-2+j2Complex conjugate poles at

s=-p1=-2-j2, and s=-p2=-2+j2

-j1

j2


-j2-p2



)22)(22(

)3(jsjss

sKGH

)())((

)(

21

sKFpspss

zsK

Im

-pConsider a test point, S, on the real

i h l f f ONE l 0

Im

-p

21

j1

j2-p1axis to the left of ONE pole at s=0.

j1

j2

2

ss+p1

p1

This test point will be part of theroot locus if 180)( nsF

0 1-1

j

-2-3 Re

p0-z0 1-1

j

-2-3 Re

1 s

sroot locus if 180)( nsF

Or 180321 n

-j1

j2

-j1

j23

s+p2

s+zNote that ,

and

0 1801

032


-j2-p2-j2

-p2Point S forms part of the root locus.



)22)(22(

)3(jsjss

sKGH

)())((

)(

21

sKFpspss

zsK

Im

-p

21

Consider a test point, S, on the reali h l f f h 3

Im

-p

j1

j2-p1axis to the left of the zero at s=-3.

j1

j2

2

s

s+p1

-p1

Note that , 032

0 1-1

j

-2-3 Re

p0-z0 1-1

j

-2-3 Re

1s

sand

Thus

1801 0)(sF

-j1

j2

Point S is not part of of the rootlocus. Note S is to the left of an

-j1

j23

s+p

s+z


-j2-p2even No. of zero/pole-j2-p2

s+p2


Consider the system with ExampleExample

)2)(1()()(

sssKsHsG

0)2)(1( Ksss 01 GH 0)2)(1( Ksss

Or )2)(1( sssK

01 GH

Consider s=-0.5:

)25.0)(15.0)(5.0( K3750)51)(50(50 375.0)5.1)(5.0(5.0

Consider s=-1.5:

)25.1)(15.1)(5.1( K

375.0)5.0)(5.0(5.1


But K needs to be positive.


Consider the system with ExampleExample

)2)(1()()(

sssKsHsG

0)2)(1( Ksss 01 GH 0)2)(1( Ksss

Or )2)(1( sssK

01 GH

Similarly, any point swhich does not satisfy the angle condition

will 180)( nsF will result in K being complex and not a

l l

180)( nsF


pure real value.


5) Because complex roots must occur in conjugate pairs, i.e. symmetrical about the real axis, the root-locus plot is symmetrical about the real axis.

5) Because complex roots must occur in conjugate pairs, i.e. symmetrical about the real axis, the root-locus plot is symmetrical about the real axis.

Im6) Loci which terminates at infinity approach asymptotes i d i

-p2

in doing so.For a test point S at infinity, each pole/zero contributes an equal phase angle.

0 Re-p1 -z1

Thus

for

Z/P=No of zeros/poles

180)( nPZ ,5,3,1 n

-p3

Z/P=No. of zeros/poles

PZn

180



7) All the asymptotes start from a point on the real axis with coordinate

Im

ZPzp m

in

ia

11

axis with coordinate

j2

Example: For

)2)(1(1)()( sHsG 0 1-1

j1

-2-3 Re

)2)(1()()(

sss

Then30

180180

nPZ

n 300,180,60

-j1

j2

13

03

a

30 PZ -j2



8) Break-in and breakaway points (on real axis)

Break-in pt.

At a break-in point, value of K increases as the loci moves onto the real axis and away from the break-in pointpoint.At a breakaway point, values of K increases along the real axis from both sides and reach maximum at the b k i tbreakaway point.

sAlong the real axis, .

Thus, at the break-in or breakaway points points,

0sds

Kd


Breakaway pt.provided K>0 and exists on the root loci.


9) Imaginary Axis Crossing

Two approaches:Two approaches:Im Axis Crossing

a) Use Routh Criteria to determine the value of K at which the system is critically stable. This is system is critically stable. This is indicated by a value of zero in the first column but with no sign change in the first column of the R th ARouth Array.

b) Since the roots are on the imaginary axis, by letting in the characteristic equation and

js in the characteristic equation and solve for and K. This is done by equating both the real and imaginary parts of the


imaginary parts of the characteristic equation to zero.


10) Angle of Departure from complex poles and Angle of Arrival from complex of Arrival from complex zeros.

Angle of Departure



Angle of Arrival



Im

Im

These angles are determined by taking a test point very close to the complex pole, or

d l i h l

-p2

2-p2

2zero, and applying the angular criteria.

11For diagram on right,

180)( 211 n0 Re-z1 0 Re-z1,3,1

)( 211

n

O 180

-p1

1-p1

1Or 180112 n


11


Angle of DepartureSuppose Suppose 60,90 11

Then using 180112 n2

21015018090602 n

112

1

210or150

1


Example Root Locus PlotsExample Root Locus Plots


022

1 23

sss

K22 sss

Rewriting as 0)(1 sKF

0)11)(11(

11)22(

1 2

jsjss

Ksss

K

3 poles ; No zeros at

)11();11(;0 321 jpjpp



3 poles at

;01p

)11(

);11(

3

2

jp

jp



4. The loci exist on the real axis only to the left of an odd number of poles

d/and/or zeros.



4. The loci exist on the real axis only to the left of an odd number of poles

d/and/or zeros.

3. Each branch starts from an open-loop pole and p p pends at an open-loop zero.



6. 3 poles. No zeros. Therefore 3 asymptotes.

30180180

n

PZn

300,180,60

7 A t t i t t 7. Asymptote intercept on real axis.

zp mi

ni 11

ZPa 11

20)2(


303


8. No break-in or break-away point.

9. Imaginary axis crossing.

022

1 23 K

22 23 sss

022 23 Ksss

0)(2)(2)( 23 Kjjj

0)2()2( 23 Kjj 0)2()2( Kjj

023 2


02 2 K 42 2 K


10. Angle of departure

180321 n

18090135 n 18090135 2 n

180135902 n1for45

2

n



10. Angle of departure

180321 n

18090135 n 18090135 2 n

180135902 n1for45

2

n



OCTAVE Program

>> h tf([1] [1 2 2 0])>> gh=tf([1],[1 2 2 0])

Transfer function:1

-----------------s^3 + 2 s^2 + 2 s

>> rlocus(gh)>> rlocus(gh)

Poles at o es at

11,0 js



OCTAVE Program

>> h ([ 3 4] [0 1] [1])>> h=zp([-3 -4],[0 -1],[1])

Zero/pole/gain:(s+3) (s+4)-----------s (s+1)

>> rlocus(h)>> rlocus(h)>>

Poles at s=0, -1o es at s 0,

Zeros at s=-3, -4



OCTAVE Program

>> gh=zp(-2,[0 -1 -1+2i 1 2i] 1)-1-2i],1)

>> rlocus(gh)

N t if i lNote: specifying a pole orzero as a+bi meanslocating that pole or zero at a+jb.

Poles atPoles at 21,1,0 js


Zero at Zero at 2s


OCTAVE Program

>> gh=zp([],[0 -4 -1+2i 1 2i] 1)-1-2i],1)

>> rlocus(gh)

Poles atPoles at 21,4,0 js


End


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Root Locus Analysis