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ROOT

LOCUS

DESIGN

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Designcascadecompensatorstoimprovethe

steadystateerror

Designcascadecompensatorstoimprovethe

transientresponse

Designcascadecompensatorstoimprovebothsteadystateerrorandtransientresponse

Designfeedbackcompensatorstoimprovethe

transientresponse

Realizethedesignedcompensatorphysically

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Introduction

Thelimitationofchangingthegain

Settingthegainataparticularvalueyieldsthe

transientresponsedictatedbythepolesatthat

pointontherootlocus,thuswearelimitedto

thoseresponsesthatexistalongtherootlocus

Onewaytosolvethelimitation:

Augmentorcompensatethesystemwith

additionalpolesandzerossothatthe

compensatedsystemhasarootlocusthatgoes

throughthedesiredpolelocationforsomevalue

ofgain

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Introduction(cont.)

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Introduction(cont.)

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Effects of Addition of Poles and Zeros

Figure: Effect of Adding Zero

1.Addition of poles pulls the root locus to the right

2.Additional zero pulls the root-locus to the left

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ImprovingSteadyStateError

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ImprovingSSerror(PI)

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ImprovingSSerror(PI)(cont.)

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Drawrootlocus

withoutcompensator

Drawastraightlineof

damping

ratio EvaluateKfromthe

intersectionpoint

FromK,findthelast

pole(at 11.61)

Calculatesteadystateerror

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Findinganintersectionbetweendampingratio

lineandrootlocus

Dampingratiolinehasanequation:

wherea=realpart,b=imaginarypartofthe

intersectionpoint,

Summationofanglefromopenlooppolesand

zerostothepointis180degrees

mab =

18010

tan2

tan1

tan 111 =

+

+

ab

ab

ab

))(tan(cos 1 =m

TakenfromProf.Poj Tangmamchit andProf.Benjamas Panomrattanarugs lecturenotes(KMUTT)

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Arctanformula

+

=

AB

BABA 1tan)(tan)(tan

111

+=+

AB

BABA

1tan)(tan)(tan 111


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Usetheformulatogettherealandimaginary

partoftheintersectionpointandget

0.6936-1.5893,=a

3.9255-=b

1

321 pppK= No open loop zero

( ) ( ) ( )53.164

1

1021 222222

=+++

= bababa

K


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Drawrootlocuswithcompensator

(systemorderisupby1from3rd to

4th) Needscomplexpolescorresponding

todampingratioof0.174(K=158.2)

FromK,findthe3rd and4th poles(at

11.55and 0.0902)

Poleat 0.0902candophase

cacellationwithzeroat 1(3thorder

approx.)

Compensatedsystemanduncompensatedsystemhavesimilar

transientresponse(closedloop

polesandKareaprrox.Thesame)

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ImprovingSSerror(Lag)(cont.)

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ImprovingSSerror(Lag)(cont.)

Uncompensated systemWith lag compensation

(root locus remains the same)

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ImprovingSSerror(Lag)cont.

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Example(Lag)

With damping ratio of 0.174, add lag

Compensator to improvesteady-state error

by a factor of 10

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ExampleLag(cont.)

Step I: find an intersection of root locus and

damping ratio line (-0.694+j3.926 with K=164.56)

Step II:

find

Kp = lim G(s) as s0 (Kp=8.228)

Step III:steady-state error =1/(1+Kp)= 0.108

Step IV:

want to decrease error down to

0.0108

[Kp = (1 0.0108)/0.0108 = 91.593]

Step V: require a ratio of compensator zero to poleas 91.593/8.228 = 11.132

Step VI: choose apole at 0.01, the corresponding

Zero will be at 11.132*0.01 = 0.111

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ExampleLag(cont.)

3rd

order approx. for lag compensator(= uncompensated system) making

Same transient response but 10 times

Improvement in ss response!!!

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Example(Lag)

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Example(Lag)

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PhysicalRealizationofPIandLag

Active

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Active

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Active

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PassiveComponentRealization

Apure

integrator

(orPI)

controllercannotbe

realized

using

simple

passivecomponent

s

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SSImprovementConclusion

CanbedoneeitherbyPIcontroller(poleat

origin)orlagcompensator(poleclosedto

origin). ImprovingSSerrorwithoutaffectingthe

transientresponse.

Nextstepistoimprovethetransientresponse

itself.

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ImprovingTransientResponse

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ImprovingTransientResponse(PD)(a)Uncompensatedsystem,(b)compensatorzeroat 2(d)compensatorzeroat 3,(d)compensatorzeroat 4

Indicate settl ing time

Indicate peak time

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Improving Transient Response (PD)

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ImprovingTransientResponse(PD)

Example

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PDExamplecont.

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PDExamplecont.

StepI:calculateacorrespondingdampingration(16%

overshoot=0.504dampingratio)

StepII:searchalongthedampingratiolineforanodd

multipleof180(at 1.205j2.064)andcorrespondingK(43.35)

Step

III:

find

the

3

rd

pole

(at

7.59)

which

is

far

away

fromthedominantpoles 2nd orderapprox.works!!!

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PDExamplecont.

StepIV:evaluateadesiredsettlingtime:

StepV:getcorrespondingrealandimagine

numberofthedominantpoles

(3.613and 6.193)

sec107.13

320.3:systemdcompensate

sec320.3205.1

44

:systemteduncompensa

==

===

s

ns

T

T

613.3107.1

44 ===sT

193.6))504.0(tan(cos613.3 1 == d

PD E l t

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PDExamplecont.

Location of polesas desired is at

-3.613j6.192

PD Example cont

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PDExamplecont.

StepVI:summationofanglesatthedesiredpolelocation,

275.6,isnotanoddmultipleof180(notontherootlocus)

needtoaddazerotomakethesumof180.

StepVII:theangularcontributionforthepointtobeonroot

locusis +275.6180=95.6 putazerotocreatethedesired

angle

006.3

)607.95180tan(613.3

192.6

=

=

oo

PD Example cont

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PDExamplecont.

Compensator: (s+3.006)

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PD Compensator

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PDCompensator

ImprovingTransientResponse(Lead

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Compensator)

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ImprovingTransientResponse(Lead

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Compensator)

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Lead Compensator Example

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LeadCompensatorExample

StepI:%OS=30%equivalentto

dampingratio=0.358,=

69.02

StepII:Searchalongthelinetofindapointthatgives180

degree(1.007j2.627)

StepIII:FindacorrespondingK( )

StepIV:calculatesettlingtimeofuncompensatedsystem

StepV:Twofoldreductioninsettlingtime(Ts=3.972/2=

1.986),correspoding realandimaginarypartsare:014.2

986.1

44===

sT

253.5))358.0(tan(cos014.2 1 == d

sec972.3007.1

44===

n

sT

212.63=K

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Improving Transient Response (Lead)

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ImprovingTransientResponse(Lead)

Note:checkifthe2nd orderapprox.isvalidforjustifyourestimatesof

percentovershootandsettlingtime

Searchfor3rd and4th closedlooppoles

(43.8, 5.134)

43.8ismorethan20timestherealpartofthedominantpole

5.134isclosetothezeroat 5

Theapprox.isthenvalid!!!

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PhysicalRealizationofPDandLead

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y

Passive

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PIDCompensator

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p

PIDDesign

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g

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PIDExample

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sec297.057.10

===

d

pT

Step1

%OS=20% dampingratio=0.456 =62.87

Searchalongtherlinetofindapointof180

degree(5.415j10.57)

FindacorrespodingK=121.51

Thenfindthepeaktime

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PIDExample(cont.)

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Decreasepeaktimebyafactorof2/3 get

imaginarypointofacompensatorpole:

Tokeepadampingratioconstant,realpartofthe

polewillbeat

Thecompensatorpoleswillbeat 8.13j15.867

867.15)297.0)(3/2(

===

p

dT

13.8)87.62tan( == o

d

Step2

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LeadLagCompensatorDesign

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SameproceduresasindesigningPID:

Beginwithdesigningleadcompensatortogetthe

desiredtransient

response

designlagcompensatortoimprovesteadystate

error

PartofHW3(doyourself)

PhysicalRealizationofPIDandLead

Lag

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g

Active

R1

R2 C2

Vi(s) Vo(s)

C1

2 1 1 22 1

1 2

1

( )c

R C R CG s R C s

R C s

= + + +

PID Compensator

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PhysicalRealizationofPIDandLead

Lag

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Passive Lead-Lag Compensator

Conclusions

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Documents

Week_15_16