1 Modern Control Theory Digital Control Lecture 4.

1

Modern Control TheoryDigital Control

Lecture 4

2

OutlineThe Root Locus Design Method

Introduction Idea, general aspects. A graphical picture of how changes of

one system parameter will change the closed loop poles. Sketching a root locus

Definitions 6 rules for sketching root locus/root locus characteristics

Selecting the parameter value

3

Introduction

0)()()(1

)()()(1

)()()(

)(

)(

sHsGsD

sHsGsD

sGsDsT

sR

sYClosed loop transfer function

Characteristic equation,roots are poles in T(s)

4

Introduction

)(,8.1

))((2)(

1

1)(

2

22

2

2

1

pnr

dd

n

nn

n

Mt

jsjssssH

ssH

Dynamic features depend on the pole locations.

For example, time constant , rise time tr, and overshoot Mp

(1st order)

(2nd order)

5

Introduction

Root locus Determination of the closed

loop pole locations under varying K.

For example, K could be the control gain.

Ksa

sbK

sL

sKbsa

sa

sbK

sKL

sHsGsD

1

)(

)(

1)(

0)()(

0)(

)(1

0)(1

0)()()(1

The characteristic equation can be written in various ways

6

Introduction

Polynomials b(s) and a(s)

n

iim

mnn

m

iim

mmm

pspspsps

asassa

zszszszs

bsbssb

sbsasa

sbsL

121

11

121

11

)()())((

)(

)()())((

)(

als1)polynomi power highest thent to(coefficie

monic are )( and )(,)(

)()(

7

Definition of Root Locus

Definition 1 The root locus is the values of s for which 1+KL(s)=0

is satisfied as K varies from 0 to infinity (pos.).

Definition 2 The root locus is the points in the s-plane where the

phase of L(s) is 180°. Def: The angle to the test point from zero number i is i.

Def: The angle to the test point from pole number i is i. Therefore,

In def. 2, notice,

integeran is l ),1(360180 lii

positive and realK for

180)/1(,/1)( KKsL

8

Root locus of a motor position control (example)

2

41

2

1,

0)()(

of roots of grapha is locusRoot

)(,1)(

,)1(

1)(,

have, We

)1()(

)(

)(

)(

)( positionmotor -DC

21

2

2

Krr

KsssKbsa

sssasb

sssLAK

ss

AsG

sU

sY

sV

s

a

m

9

Root locus of a motor position control (example)

0.5) (here, somefor calculated be can

21part real ,conjugatedcomplex :),(

,41For

[-1;0] interval thein real :),(

,410For

1

0

2

1

2

1),(

loop)-(open,0For 2

41

2

1),(

21

21

21

21

K

rr

K

rr

K

rr

K

Krr

break-away point

10

IntroductionSome root loci examples (K from zero to infinity)

11

Sketching a Root LocusHow do we find the root

locus?

We could try a lot of test points s0

Sketch by hand (6 rules)

Matlab

s0

p1

12

110

10

100

0

1

)(

where

1

)(

1)(

spoint Test

)(

1

)(

1)(

1

ps

Rps

eRpssG

psassG

j

Sketching a Root Locus

s0

p1

What is meant by a test point? For example

13

)(exp

)()(

)()()(general, In

1121

21

)()(1

)()(1

010

0100 1

1

nmaa

bb

jan

ja

jbm

jb

n

m

jrr

rr

erer

erer

psps

zszssG

n

m


s0

p

p*

integeran is l ),1(360180 :definition locusRoot lii

A inconviniet method !

14

Root Locus characteristics(can be used for sketching)

Rule 1 (of 6) The n branches of the locus start at the poles L(s) and

m of these branches end on the zeros of L(s).

3) Rule()(

1) Rule)((0)(,(end) if

(poles) 0)(,(start)0 if

1

)(

)(0)()(

,order of is )( and ,order of is )(

Notice,

sa

zerossbK

saK

Ksa

sbsKbsa

mnmsbnsa

15

Root Locus characteristics

Rule 2 (of 6) The loci on the real axis (real-axis part) are to the left

of an odd number of poles plus zeros.

Notice, if we take a test point s0 on the real axis :

• The angle of complex poles cancel each other.• Angles from real poles or zeros are 0° if s0 are to the right.• Angles from real poles or zeros are 180° if s0 are to the left.• Total angle = 180° + 360° l

16


Rule 3 (of 6) For large s and K, n-m branches of the loci are

asymptotic to lines at angles l radiating out from a point s = on the real axis.

mn

zp

mnlmn

l

ii

l

,,2,1,)1(360180

l

For example

17


180 havemust weNotice,

,0)(

11

pages),next (see ionApproximat

01

for ionapproximat an derive can we largeFor

3) Rule()(

1) Rule(0)(0)(, If

,/1)()(

)(have, We

1

11

11

n-m

ii

mn

nnn

mmm

α)(s

Rs

K

asas

bsbsK

s

sa

sbsLK

mnKsLsa

sb

s0

°

Example,n-m = 4

Thus, l°

18


)1()()()(

)(

so ,for small are )( and )( Notice,

)()(

)())((

2 that assuming , find usLet

,)()()(

)()()(

)(

)(

)()()()( Notice,

,)(

)(1

)(

)(

11

11

2

111111

21

11

11

1

11

11

11

11

sqssqssQsb

sa

ssRsO

baqqba

sRsOsqsbssas

sRsbssqssas

n-mQ(s)

ssqssQsb

sRsbsQ

sb

sa

sRsbsQsa

mnKsb

sa

Ksa

sb

mnmnmn

n

nnnnnn

mmmnmnnn

mnmn

19


0)(

11

1

)(

1 have weSo,

)( where,

)(

)()(11

1)(

)(

have weNow,

0 around,1)1(Taylor notice First,

11

11

1

11

1

mnmn

mn

mnmn

mn

r

sK

Ks

mn

q

mn

qsK

mn

qs

mns

qs

s

qsK

s

qsK

sb

sa

xxrx

20


mn

zp

mn

ba

mn

qα

zzzzb

s

zszszsbsbssb

ppppa

s

pspspsasassa

ii

im

m

nmmm

in

n

nnnn

)(

have weNow,

: is oft coefficien The

)())(()(

: is oft coefficien The

)())(()(

Notice, .at look usLet

111

211

1

211

1

211

1

211

1

= Starting point for the asymptotes

21


n-m=1 n-m=2 n-m=3 n-m=4

mn

zp

mnlmn

l

ii

l

,,2,1

,)1(360180

3 Rule

For ex.,n-m=3

22

Root Locus characteristicsRule 4 (of 6)

The angle of departure of a branch of the locus from a pole of multiplicity q is given by (1)

The angle of departure of a branch of the locus from a zero of multiplicity q is given by (2)

)1(360180)2(

)(

)()(

)(

)( mult. of Zero

)1(360180)1(

)(

)(

)(

1

)(

)( mult. of Pole

,

0

,

0

lq

sa

sbzs

sa

sbq,

lq

sa

sb

pssa

sbq,

liiidepl

q

liiidepl

q

Notice, the situation is similar to the approximation in Rule 3

23


arrldepl

liiarrli

liidepli

ii

qq

q

q

l

,,

,

,

and find can weSo,

Notice,

)1(360180

have weBecause

s0

Departure and arrival?For K increasing from zero to infinity poles go towards zeros or infinity. Thus,• A branch corresponding to a pole departs at some angle.• A branch corresponding to a zero arrives at some angle.

In general, we must find as0 such thatangle(L(s))=180°

Vary s0 untilangle(L(s))=180°

24

Root Locus characteristicsRule 5 (of 6)

The locus crosses the j axis at points where the Routh criterion shows a transition from roots in the left half-plane to roots in the right half-plane.

Routh: A system is stable if and only if all the elements in the first column of the Routh array are positive => limits for stable K values.

0)Row(s

3)Row(s

2)Row(s

1)Row(s

1)Row(s

0

3213-n

3212-n

5311-n

42n

cccn

bbbn

aaan

aan

21

31

11

51

4

12

31

2

11

det1

1det

1

1det

1

bb

aa

bc

aa

a

ab

aa

a

ab

25


Rule 6 (of 6) The locus will have multiplicative roots of q at points

on the locus where (1) applies. The branches will approach a point of q roots at

angles separated by (2) and will depart at angles with the same separation.

q

l

ds

dba

ds

dab

)1(360180)2(

0)1(

26

Root Locus characteristicsThe locus will have multiplicative roots of q in the point s=r1

)()()()( 11 sfrssbKsa q

01

rsds

dba

ds

dab

We diffrentiate with respect to s

10

)()()(...)()( 12

111 rs

qq

ds

sdfrsrsrsq

ds

dbK

ds

da

1)(

)(1 rssb

saK

27

Root Locus sketchingExampleRoot locus for double integrator with P-control

Rule 1: The locus has two branches that starts in s=0. There are no zeros. Thus, the branches do not end at zeros.

Rule 3: Two branches have asymptotes for s going to infinity. We get

02

0

270

90

2

)1(360180)1(360180

mn

zp

l

mn

l

ii

l

Eq.)(Char.01

1)(,1

)(22

skksD

ssG pp

28


Rule 2: No branches that the real axis.

Rule 4: Same argument and conclusion as rule 3.

Eq.)(Char.01

1)(,1

)(22

skksD

ssG pp

Rule 5: The loci remain on the imaginary axis. Thus, no crossings of the j-axis.

Rule 6: Easy to see, no further multiple poles. Verification:

0021)(

10

1)(,)(

22

2

ssds

ds

ds

sd

ds

dba

ds

dab

sbssa

29


Eq.)(Char.0)1(

101

)1(1

1and

Eq.)(Char.01

)(1

)(,1

)(

22

2

2

s

sK

ss

K

k

kkK

sskk

skksDs

sG

D

pD

Dp

Dp

ExampleRoot locus for satellite attitude control with PD-control

30


90 areat 0 pole double thefrom departure of angles The

:4 Rule

axis real negative thealong asymptote one is there,1 Because

:3 Rule

locus theon is 1 ofleft the toaxis real The

:2 Rule

infinity. approachesother The

.1at zero theon s terminateOne

.0at start that branches twoare There

:1 Rule

1)1(0

)1(1

22

s

mn

s

s

s

Ks

s

s

sK

31


2,002)1(

2,1,)(,1)(

from found are roots multiple of points The :6 Rule

1

axis.imaginary thecrossnot does locus theThus,

below.array thefind wecriterion, s Routh'Applying :5 Rule

Eq.)(Char.0)1(

1

2

2

2

issssds

dba

ds

dab

sds

da

ds

dbssassb

K

K

K

s

sK

32


33

Selecting the Parameter ValueThe (positive) root locus

A plot of all possible locations of roots to the equation 1+KL(s)=0 for some real positive value of K.

The purpose of design is to select a particular value of K that will meet the specifications for static and dynamic characteristics.

For a given root locus we have (1). Thus, for some desired pole locations it is possible to find K.

LK

sLK

1

)(

1)1(

34

Selecting the Parameter Value

657.7*1.2*4)()()(

1

))((

1

then,5.0 want say we usLet

, ,at Poles

)44)(44(

1

)16)4((

1

302000

32000

321

2

ssssssL

K

sssss)L(s

sss

jsjss

ssL(s)

Example

35

Selecting the Parameter ValueMatlab

A root locus can be plotted using Matlab rlocus(sysL)

Selection of K [K,p]=rlocfind(sysL)

1 Modern Control Theory Digital Control Lecture 4.

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Transcript of 1 Modern Control Theory Digital Control Lecture 4.