IENG3004 Lecture 6-11-12 S1 Root Locus

8/3/2019 IENG3004 Lecture 6-11-12 S1 Root Locus

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IENG3004IENG3004Control Systems TechnologyControl Systems Technology

Lecture 6Lecture 6

Root LocusRoot Locus


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Root Locus IntroductionRoot Locus Introduction

We know that the transient response of a system isgoverned by the roots of the characteristic equation

It is frequently necessary to analyse the variation intransient response due to changes in the values ofsystem parameters

2IENG3004 Control Systems Technology Lecture 6 Root Locus

cumbersome

Requires a graphical method to describe variation in transientresponse (Root Locus)

Root Locus traces the movement of the closed looproots as an OLTF parameter (usually gain) is changed

Roots can be real or complex, so the s-plane has real andcomplex dimensions


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Poles & ZerosPoles & Zeros

Zero:

A value of s which reduces the Transfer Function tozero

(a root of the numerator)

Pole


va

ue o s w c sen s e rans er unc on oinfinity

(a root of the denominator)

The poles of a Transfer Function are the same as the

roots of a the characteristic equation


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Poles & ZerosPoles & Zeros

Example: Determine the poles and zeros of theT.F.:

)22)(1(

)2(2 +++

+

ssss

s

ZEROS: s + 2 = 0 s = -2


POLES: s = 0 s = 0s + 1 = 0 s = -1

s2

+ 2s + 2 = 0

s = -1 j


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Example: First Order SystemExample: First Order System

)3(

1

+s


)3()3(1

)3(

R(s)

C(s)

Ks

K

sK

sK

++

=

++

+=

From the block diagram, the CLTF reduces to:

Location of the closed-loop pole is: s = -(3+K)


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Root locus has one pole on the real axis:



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Simple example shows that: For K = 0, closed loop pole is located at -3

As K increases, the pole moves to the left for

increasing values of K As K increases, the transient response decays more

rapidly and thus steady-state is achieved morerapidly


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Root Locus Construction StepsRoot Locus Construction Steps

by way of an example

Kc(t)r(t)

+-)1(

1

+

ss

1


+s

Step 1: Obtain Open Loop Poles & Zeros

(a) Root loci start at open-loop poles

(b) Root loci end at open-loop zeros or infinity

(c) The number of separate loci is equal to the number ofopen-loop poles

(d) Loci are symmetrical about the real axis


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)2)(1(OLTF ++== sss

K

KGH

Open loop poles at: 0, -1, -2

Step 1: Obtain Open Loop Poles & Zeros (cont)


Real & Imaginary Axes Poles as x Zeros as o Locus is plotted

Step 2: Sketch Root Loci on Real Axis (if existing)


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Step 3: Derive asymptotes of Root Loci

The branches of a root locus approach a set of straight-

line asymptotes at large distances from the origin.(a) The asymptotes emanate from a point on the real

axis called the centre of asymptotes, given by:


mn

zpm

i

i

n

i

i

c

=

== 11

where n = number of open-loop polesm = number of open-loop zerospi = location of i

th polezi = locaiton of i

th zero

103

)0()210(=

=c


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Step 3: Derive asymptotes of Root Loci (cont)

(b) The angle made by the asymptotes and the real

axis is given by:

k+=

)21( k = 0, 1, 2, 3


mn

35,,

3

03

)21(

=

+=k


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Step 4: Determine breakaway points, b

A breakaway point (b) occurs on the real axis where two

or more branches depart from, or arrive at, the real axis.They occur where:

1= from characteristic=

dK


0=dsdK

)(sGH

equation)

0)2)(1(

1 =++

+sss

K023 23 =+++ Ksss

0263 2 =+++dsdKss

if

1.58or42.06

24366=

=s

ds

then:

x

No RootLocus inthis area


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Step 5: Find K for marginal stability and c

Finding c will give us the intersection of the root locus

with the Imaginary Axis Use Routh-Hurwitz stability criterion to find this

(characteristic equation)023 23 =+++ Ksss


Ks

Ks

Ks

s

0

1

2

3

03

3

213

21

Ks

Ks

Ks

s

0

1

2

3

03)6(

3

21

Therefore stability for 0 < K < 6


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Step 5: Find K for marginal stability and c (cont)

Form the Auxiliary Equation to find c

For the auxiliary equation (s2 row):

3s2 + K = 0


3s2 + 6 = 0

s = j2


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Step 6: If there are no complex poles, plot Root Locus

2

jSymmetrical about the Real axis


j

Re

XXX-2-3

-1

60o

-60o


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Root Locus Second Order ApproximationRoot Locus Second Order Approximation

Dominant Poles

The transient response of a system may be

equated to a sum of exponential terms Contribution of each term is determined by one of the

roots of the characteristic equation


Some roots dominate the response

Example

)203)(4)(8()( 2 ++++=

ssss

K

sG

Gives a transient response in the form:

)2.4sin(.)(5.148

+++=

tCeBeAetyttt

Constantsdetermined byinitial conditions


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)2.4sin(.)( 5.148 +++= tCeBeAety ttt

Pole Locations


Transient response

Total response is

DOMINATED bydecaying sinusoidal

roots for this termmuch closer toImaginary axis


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Since dominating roots are second order

the system may be approximated by an

equivalent second order system This gives us a system with well documented

time res onses.


Performance given in normalised form in terms ofdamping ratio () and natural frequency (n)

The higher order system has been reduced to a

second order approximation, making it possible to useavailable performance data to effect a successfuldesign


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DeterminingDetermining andand nn of an Equivalent Secondof an Equivalent Second--OrderOrder

SystemSystem

Consider normalised form of the second order transferfunction:

22

2

2 nn

n

i

o

ssXX

++=


The poles of this transfer function are located at:

)1( 2 = nn js


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SystemSystem

X

j

Pn

One pole from the complex conjugate pair is located at P,shown in this diagram:


Re

n

0

d = n

cos


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SystemSystem

From the geometry of the diagram:

X

j

n

Pn

d = nnn

nn

OP

OP

+=

+=

)1(

)1()(22222

2222


Re

0

cos

n=

==

n

ncos

In the figure, cos defines the damping line and thedistance of point P from the origin is equal to the naturalfrequency of the system


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DeterminingDetermining Gain KGain K

From the characteristic equation [ 1 + KGH (s) = 0 ]:

)(

1

sGHK

=)(

1

sGHK

=


L

L

))()((

))()((

321

321

pspsps

zszszsKG(s)

+++

+++=

Then, from above, assuming unity feedback:

L

L

321

321

zszszs

pspspsK

+++

+++=


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DeterminingDetermining Gain KGain K

If the location of the closed-loop pole is known, then Kmay be determined graphically:

Q is the closedloop pole


If the real and imaginary axes of the root locus diagramhave been drawn to the same scale then the lengths ofvectors P1Q, P2Q, P3Q, and ZQ can be measured

.

... 321

ZQ

QPQPQP

K=


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ExampleExample

)2)(1(

)(

++

=

sss

KsG

Determine the value of K whichwould give an equivalent

Sketch the root locus of the system with an open looptransfer function:


o60)5.0cos( =

.

8.076.188.052.0

)2(.)1(.)(

1

==

++==

K

ssssGH

K


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Root Locus with Complex PolesRoot Locus with Complex Poles

In some circumstances the open-loop system hascomplex poles:

)2(1

22

nnsss ++


In order to sketch the root locusfor the closed loop system, weneed to determine the angle of

emergenceof the locus from thecomplex poles

X

j

Re

is angle of

emergenceThe angle of emergencecan be determined usingthe angle condition


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Angle Condition

Key concept: if a point lies on the root locus, the argument

G(s) will be a multiple of 180o

.o

321 180ofmultipleofsum=++


n)( 21.180321 +=

Since poles are on denominator,angles will be negative


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2s

+++

+

Angle Condition (cont)

If there is a zeroin the Transfer Function, then the zero will

be associated with a POSITIVE angle.e.g.:


So the angle condition becomes:

32121.180

PPPZn)( =+

Or:)(arg180 i

o

D pHG +=

)(arg180 io

A zHG=


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Angle ConditionAngle Condition Example 1Example 1

Take an OLTF with four poles, two of which are complex,

and no zeros

Select a point just to the right

From Dunn Root Locus Tutorial (http://www.freestudy.co.uk)


enough that angle 3 = 0

Measure or calculateother angles


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P1 = -10 P2 = -1P3 = -4 + 4j P4 = -4 4j

1 = tan-1 (4/6) = 33.7o

2 = 180 tan-1 (4/3) = 126.9o

= 90o


180 = -33.7-126.9-90- 33 = -430.6

= -70.6

Apply:

432121.180 PPPPZn)( =+

30IENG3004 C l S T h l L 6 R L


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Similar OLTF as in Example 1, but introduce 1 zero.

P1 = 0 P2 = -10P

3= -4 + 4j P

4= -4 4j

Z1 = -1

1 = 180 tan-1 (4/4) = 135o

-1 o


2 .

4 = 90o = 180 tan-1 (4/3) = 126.9o

180 = 126.9-135-22.7-90- 3

3 = -300.8 60

Apply:

32121.180 PPPZn)( =+

31IENG3004 C t l S t T h l L t 6 R t L


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P1 = 0P

2= -4 + 4j P

3= -4 4j

1 = 180 tan-1 (4/4) = 135o

o

Consider a 3-pole example:


3

180 = 135-90- 3

3 = -120.8

Apply:

321

21.180PPPZ

n)( =+



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Consider the following system:

52

12

++ sss

1

X jP1


P1 = -1 + 2j P2 = -1 2jP3 = 0

Re

Q

X

P3

P3

X

P1

P2P2

Set Q to be just to the right of P1

2 = 90o

3 = 180 tan-1 (2/1) = 117o

180 = -387- 1

1 = -27



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Resulting Root Locus plot:


32

03

0)11(00=

=

= ==mn

ZP

m

ii

n

ii

c

35,,

3

03

)21(

=

+=k



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Effect of Gain on Second Order ApproximationEffect of Gain on Second Order Approximation

Kc(t)

+

-

r(t)

s

1

Controller Hydraulic

Actuator

Mass-spring-damper

(load)

22

2

1

nnss ++

Consider the following generalised third order system:


Looking at gain where < 1 (i.e.underdamped)

Varying K (0 K ), the closedsystem root locus looks like:

Is the second order approximation valid

for K giving pole set A? What about B?

Re

Xj

XP3

P1

P2 X

Dynamics

ActuatorPole set A

Pole set B


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Effect of adding Poles on Root LocusEffect of adding Poles on Root Locus

)()()(ass

KsHsG

+= 0>a

Take the function:


This gives two poles, one atthe origin and the other at a.

The following slides show theeffect of adding additionalpoles and zeros to the shapeof the Root Locus

From Glonaraghi & Kuo, Automatic Control Systems, 9th Edition, 2010, pp385-388



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Add a pole at -b Add a pole at -c

y gy



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Add a complexconjugate pair

y gy



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Effect of adding Zeros on Root LocusEffect of adding Zeros on Root Locus

Add a zero at -b



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Effect of adding Zeros on Root LocusEffect of adding Zeros on Root Locus

Add a complexzero with b realparts



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With three polesand one zero

42


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Open-Loop Pole-ZeroConfigurations & theCorresponding Root Loci

From Ogata, Modern ControlEngineering, 5th Edition, p289

T i l Q iT i l Q i43IENG3004 Control Systems Technology Lecture 6 Root Locus


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Tutorial QuestionTutorial Question

Plot the Root Locus by hand for both of these:

1.

2.



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IENG3004 Lecture 6-11-12 S1 Root Locus