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8/3/2019 PIC Part-A Student Version
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2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
1
PROCESS INSTRUMENTATIONAND CONTROL
PreparedandCompiledby
Prof.T.K.Ghoshal &Prof.Smita Sadhu
Taughtby
Prof.Smita Sadhu (SS)Deptt.ofElectricalEngg.JadavpurUniversity
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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SyllabusPartI ConceptsofModulatingandSequentialControl.StructureofModulatingControlloops.ProcessControlterminology.ProcessInstrumentationdiagrams.
ControllerImplementation:Electronicanalog,Digital,Pneumatic Controllers.Self-tuningandMultifunctionControllers,ControlValves.ProcessActuators:Electrical,Pneumatic,Hydraulic,Valvepositioners.IndustrialInstrumentationSystems:Components,structure,specification.SelftuningandAdaptivecontrollers.
Supervisorycontrol:ObjectivesandImplementation.
PartII
ConceptofProcessesandUnits:Processstatics,massandenthalpybalance. Modellingofprocessdynamics,ModellingofChemicalprocesses. Singleloopcontrolofstandardfirstorderprocessplants. P-I-Dcontrol,Controllertuning,Ziegler-Nichlol's method,Frequencydomaindesign. Feed-forwardcontrol,Multi-loopandCascadecontrol,Interactionanddecoupling. Non-lineareffectsinplantsandcontrollers. Simulationofprocesscontrolsystems.* BoilerDrumLevelControl.* DiscreteControllers:Selectionofsamplingintervals,stability analysis.*
Books: 1.PrinciplesandPracticeofAutomaticProcessControl- SmithandCorripio 2.PrinciplesofProcessControl- Patranabis 3.AutomaticProcessControl- Eckmann 4.ProcessControlSystems- Shinskey 5.ProcessSystemsAnalysisandControl- Coughanowr &Koppel 6.ChemicalProcessControl- Stephanopoulos 7.Processcontrol- Pollard
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PROCESS in Process Control
DomainofInterest Chemical,thermalandbiotechnical
processes
FormalDefinition: Aphysico-chemicalprocess involves
physicalorchemicalchangeofmatter,orconversionofenergy
Importance: Abovetypesofprocessesproduce
productslikefuels,plastics,cement,metals,adhesives,yarnsforcloth,
foods,beverages,medicines,
Allareimportantinourdailylife.
AChemicalPlant
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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CONTROL
in Process Control Theregulation,manipulation(or modulation)of
variablesinfluencingaprocessinsuchawayastoobtainthedesiredqualityandquantityofproductinasafeandefficientmanner.
Hencemodulatingcontrol
Manualmodulationand
controlAutomatic
control
Controllermodulatessteamflowthrough
theactuator
Operatorreadsthe
temperature
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About the Subject
Caveat!! Nosinglebookmaybegoodenoughforthiscourse
Solution:
RefreshyourknowledgeinControlSystem
Specially,timesresponse,BodePlot,Nyquist plot,stability
margins
Consultthereferencebooks
Solvetheexercisesintheproblemsheet
Readlecturenotesofcurrentandpreviousyears
Applyyourmindandcommonsense
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About the Subject II Historyofintroducingthissubject
Introducedinthelateseventies
Digital(electronic)controlwasthenreplacingtraditionalpneumaticanalogcontrol
Expansionofpetrochemicalindustries expandedjobmarket.
ShrinkageintraditionalEEjobmarket.
Consultingengineeringcompaniesrequired
engineerswiththeseskillsets.
Someoffactorshavechangedinthepastdecades,butthesubjectremainsexciting.
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About the Subject III
Natureofthesubject Concernscontrolofchemicaland bio-chemical
processes Startedbychemicalengineers
Theirownvocabulary.
Obsessionwithfluids&valves.
FormalizedlaterbyElectricalEngineers&Shinskey Intensiveuseofelementarycontrolengg.principles
Requiresdeepappreciationoffreq.domainmethods.
Mixtureoftheoryandpractice
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Nature of chemical processes Flowofliquidandslurries Chemicalandbiochemicalreaction Transferofmassandenthalpy MultiInput- multiOutput. Mathematicalmodelmaybeunavailable,unreliable,
complex,non-linear,etc. Typifiedbyunitoperations(connectedthroughanetwork
ofpipesandvalves). Filtration
Distillation(phasechange)
Fermentation Heatexchangers Boilers
Chemicalreactors
AheatExchanger
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Nature of chemical processes-II
ADistillationColumn AFractionating(Distillation)
ColumnforPetroleum
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Nature of chemical processes-III
(a)
(b)
Acontinuousstirredtankreactor(CSTR)
(a)Schematicwithtemperaturecontrolloop
(b)Photograph
StirrerMotor
Cooling
Coils
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Objectives of Process Control(and justification of investment)
Ensureimprovedsafetyoftheplant bykeepingprocessvariableswithinsafe
boundaries.
Reducecapitalcost bykeepingprocessvariableswithinboundaries,
requiringlesssafetymargins.
Ensurequalityandconsistencyoftheproduct bykeepingprocessvariableswithinclosetolerance
Ensureproductivity&economyoperation. Maintainingprocessvariablesatappropriatevalues
forallthecomponentsoftheplant.
Reduceabuseofenvironment. Bycontrollingandneutralizingpollutingeffluents
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Means of attaining Process control objectivesEnsuringpredictableplantperformance
byfeedback
Ensuringplantcontrollability Choosingfinalcontrolelements(valvesand
actuators)
Structureofcontrolloop.(feedback,feedforward,cascade)
Goodpairingofinput-outputs.
Ensuringappropriateoperatingconditions choiceofsetpoints.
EnsuringCorrectsequenceofoperation SequentialControl(PLC)
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Components of A PROCESS CONTROL LOOP
Controller
Transducer/
Transmitter
Process
FeedforwardElement
FCE/ValveActuator
PVSPCO
Load Disturbance
Feedforward
Feedbacksignal
FCE:FinalControlElementCO:ControllerOutputPV:ProcessVariableSP:Setpoint
A physical quantity orproperty that can bemeasured and controlled
An elementwhich receives
signals in one form andconverts it to anotherform
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Process Variable and SET POINTMostfeedbackcontrolloopsattempttomaintain
aprocessvariable(e.g.,temperature,pressure,pH,etc.)atadesiredvalue.
AninputvariablethatsetsthedesiredvalueofcontrolledvariableiscalledtheSetPoint
AControlloopisdesignedsuchthattheprocessvariable isnearlyequaltothesetpoint.
Thesetpointmaybeassigned manuallythroughlocaloperation
Manuallybutfromadistance(remotecontrol) suppliedautomaticallybyanothercontroller(cascade
control).
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Actuator & FCE
ProcessvariablesinChemicalprocessesaremostoftenmanipulatedandcontrolledbymodulatingsomeflow
Theflowismodulatedbycontrolvalves
ValvesarethereforecalledtheFinalControlElement(FCE) ApneumaticallyactuatedValve
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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Valves
Manually
operated
Butterfly
Valve
AplugorGlobe
Valve
Diaphragmvalve
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Actuator & FCE-II
Inorderthatvalvesmaybemanipulatedbyacontroller,valveactuatorsorvalvemotorsarerequired.
Valvemotorscanbe pneumaticor Electric
Valvepositionersarespecialactuatingsystemswithfeedbackofvalveposition
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Loads and DisturbancesAprocessoraplantisdesignedtosupplyor
sustainload Ageneratorshould supplyadequatecurrent Aboilershould supplyadequateflowofsteam (loadispartofduty,notdisturbance)
ADisturbanceisexternalinfluencebutnot aload,thatsomehowinfluencethevariabletobecontrolled Oftenthedisturbanceschangewithtime.
Speedchangeintheprimemoverisadisturbancetoadcgenerator
Pressurefluctuationofinletsteamisadisturbancetoaturbine
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Loads and disturbances in simple processes
Process Schematicdiagram PV Load Disturbance1. Flow
throughpipe
Flowrate
Downstreampressure
Upstreampressure/fouling
2. Heatexchanger
Processfluido/ptemp
Processfluidflow
Steamtemp/steamflow
3. Levelcontrol
Level Outletflowrate
Inletflowpressure
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Transmitter AControlLooprequiresthat
ProcessVariableandsomeloads,bemeasuredandthevaluestransmittedtothecontrollerforappropriateaction.
ATransmitter Measuresthevariablewitha
sensor, Convertsthesensorsignaltoa
standardlevelwhichmaybeused(remotely)byanindicatinginstrument,acontrollerorarecorder.
Thesensingelement,mayormaynotbephysicallyinsidethetransmitter.
Atemperaturetransmitterwithoutthe
temperaturesensor
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Transmitter-II
StandardoutputsignalofTransmitters DCCurrentLoop(most
common) 4 20mA 0 20mA
DCVoltage 0 10V 0 5V 15V
Pneumatic(nearlyobsolete)
0.21.0Atmosphericpressure
TransmittersaregenerallyspecifiedwithFullScale(range)
Transmitterreadingisusuallyexpressedaspercentageorperunitoffullscale.
Example: Witharangeof1250C
and40%reading, thecurrentoutputfora4-
20mAtransmitterwouldbe4+0.4*16=10.4ma,
correspondingto1250*0.4=500C
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Transmitter-IIIPreferredStandardtransmitteroutput:
4-20mA(currentloop,suppressedzero)
Advantagesofcurrentloopsignals
Nodangerfromshortcircuit
Novoltagedropconcerns,lowcablecost
Advantageofsuppressedzero
Detectionofcircuitbreak(openloop) Possibilityofsupplyingcurrenttothe
transmitter(twowiretransmitter)
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Transmitter-IV
Transmitter
Transducer
DCPowerSource
CurrentLoopSignal
ToController
AFourWireTransmitter
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Transmitter
TransducerDCPowerSource
Transmitter-V
ATwoWireTransmitter
CurrentLoopSignalToControllerTransmitter
Transducer
Transmitterdrawsa
constant4macurrentforown
powering
Transmitter-V
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Transmitter-VI
ATwoWireTransmitterwithSensorandController
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Transmitter-VIIARosemount
FlowTransmitterconnectedtoatubesection.
Note:
(i)Local Indicator
(ii)pressuretappings
(iii)connectingflanges
FlowPipe
Connecting
Flanges
Pressure
Taps
Pressure
TapsPower
In
Signalout
LocalIndicator
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Process Instrumentation Diagram
FT
FC
PV
SP
CO
Flow
Actuator
Valve
FT:FlowTransmitter
FC:FlowController
AlsocalledPipingandInstrumentationDiagram
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Example (PI Diagram) Identifytransmitters,controllersandFCEinthediagrambelow
Whatisthis,
BTW?
FlowIndicatingController
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Recap-Control Loop
Wehavestudiedessentialaspectsofacontrolloop
LoopStructure
Setpoint,ProcessVariable
Valve,Actuator,FinalControlElementandValvepositioners
Transmitter
ProcessInstrumentationdiagram
Comingup:
Controller,ControlLaw,FeedbackandFeedforward
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Controller Error=SP-PV
Transmitteroutputshouldmatchthecontrollerinputsphysically Type(saycurrentloop)
Range
Controlleroutputandactuatorinputalsoshouldmatchphysically
Signalisonesided,i.e.noreversalofsignals Biasisoftenusedtogenerate
someCOwithnoerror
MaximumCOvoltagecapabilitylimitstheelectricalload
ControlLaw Thedefiningfunctionthat
generatesCOfromallinputs
Controller
Bias FF
SP
PV
CO
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Industrial Controllers
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Control Laws: Feedback & feed forwardControllawdefineshowthecontroller
output(CO)isgenerated.
Infeedback,CO=f1 (PV,SP)
Infeed-forward,CO=f2 (LOAD)
Incombined,CO=f3(PV,SP,LOAD)
ControllerPlant&Actuator
PVSP
CO
Feedback
FeedForward Load
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Control Laws:-II
Feedbackapproach: AdjustCOsuchthatPV SPnotwithstanding
loadanddisturbance slowbutgoodadjustmentcapability
Cantolerateinaccurateplantdescription(robustness)
Feedforwardapproach: AdjustCOtonullifytheeffectofloadonPV
fastbut
notaccuratewhenplantdescriptionisapproximate Usedasanauxiliarytofeedback
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Controllability(in the sense of process control)
Aprocesstobecontrolledmustbecontrollable
Thetermcontrollability hasaspecialmeaninginprocesscontrolcontext.
Thisissometimescalledinput-outputcontrollability
InthesimplestSISOcasethereisonlyoneprocessvariableofinterest.
Forprocesscontrollabilityinthesimplecase: Theremustexistaninputvariable(usuallyaflow),whichmay
bemanipulatedindependently.
Theprocessvariableshouldchangeinresponsetochangeintheinputvariable.
Forastepchangeininput,theoutputshouldeithersettleataconstantvalueorkeeponincreasing(typeonehighertype)
Donotconfuse
withstatecontrollability
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Fly-ball governor- The First Feedback Controller
JamesWattsDrawing
Principle
Amoremodernversion(1930)
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Controllability-II(in the sense of process control)
Formulti-outputsystems,therewouldbemorethanone(saymnumberof)processvariablesofinterest.
Forinput-outputcontrollabilityofmulti-outputcase Theremustbeatleastmnumberofinputvariableswhichcan
beindependentlymanipulated.
Itwouldbepossibletoformmnumberofinput outputpairs,
eachwithanuniqueinput(thatisinputvariablescannotberepeated)and
eachofwhichiscontrollable.
Notethat
Theremaybemorethanoneschemeofpairing Somepairingschemecanbebetterthantheother.
Responseofaprocessvariabletoitsinputpairshouldbemoresignificantthantheresponsetounpairedinput.
Thisiscalleddiagonaldominance.
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Modelling ElementaryProcesses
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Modelling of elementary processFlowthroughOrifice,Venturi andPipe
FlowthroughControlValve
Dyeinjection
Heattransferinanoven
LiquidlevelSystem
Linearization
LessonslearntfromModelling
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Flow Through Orifice
q=cd (/4)D22 [2(p1 - p2)/(1- d
4)]1/2
Whereq=volumetricflow,cd isaconstant
D2
=orifice,venturi ornozzleinsidediameter
D1=upstreamanddownstreampipediameter
d=D2 /D1 diameterratio,=densityp1 ,p2=upstreamanddownstreampressures
q=cf(P)
P=p1-p2p1 p2
D1
D2
Usedforsmalltomediumflow
measurement
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Venturi Tube
Usedformeasurementofmedium(100litres/minute)tolargeflow
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Common Flow equation
Thiscommonflowequationmaybeusedfororifice,venturi andlongpipesections
Theflowrateq isvolumetric
Flowisassumedtobeturbulent
Theflowcoefficientcf isstronglydependentondensityandgeometry
andmildlyonReynoldsnumber(1%-2%)
Theflowcoefficientcf isstronglydependentontheunitschosenfor
flowandpressure
Whenpressureisexpressedashead,densityeffectgetscancelled.
Formassflowrate,q istobemultipliedbydensity
q=cf(P)
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Common Flow equation-II
Theflowequationisstronglynonlinear Forlinear,smallsignalapproximation,localslopemay
beused.
(2%changeinpressuredrop=>1%changeinflow)
qP
Pq
PPPc
PcPPcq
PPPqqq
Pcq
f
ff
f
~~
2
1
~)1
~/1(
~~
~;~
+=
+=
+=+=
=
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Control Valve Characteristics
Flowthroughvalvealsoobeysthegeneralflowequation
P indicatespressuredropacrossthevalveTheflowcoefficientcf() becomesafunctionof
valveopening Letthevalveopeningbeexpressedbya
variable ,where =1indicatesfullopening =0indicatesthatthevalveisfullyclosed Theflowequationisvalidwithintheinterval
0.05
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Control Valve Characteristics-III
Smallsignallinearequationscanbederivedasabove
Theoperatingpointshouldbeknown
PPcP
Pcq
PcPPcqPcq
PPPqqq
Pcq
ff
fff
f
+
=
++==
+=+=+==
})(){(})(){(
~)(
~)~(;
~)~(~
~;~
;~)(
)~
,
~
,~( Pq
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Control Valve Characteristics-IVAcontrolvalveisusedinapipingcircuitwith
differenttypesofload
Theflowisdeterminedbythepumpcharacteristics,valveopeningandflowcharacteristicsofthepipingandtheload.
Forefficientflowcontrolinapipingcircuit acontrolvalveisgenerallyconnectedinseries when
aconstantpressure pumpisused
acontrolvalveisgenerallyconnectedinparallelorshunt whenaconstantflow pumpisused
Exercise:Drawdiagramsandwithanalogyfromelectriccircuits,explaintheabove
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Control Valve Characteristics-V
a=quickopeningb=linear
c=squareroot
d,e =equalpercentage
f=hyperbolic
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.0 0.2 0.4 0.6 0.8 1.0
X/Xmax
F/Fmax
a b
c d
e f
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Dye injection
v: Averagevelocityoffluid
L:Lengthofpipesection
Delay=L/v
q1Colourless
Fluidflowrate
vq1+q2
L
Dyeq2
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
Input
IdealresponseActualresponse
T=L/v
(Delay) Percentagecomposition
=(q2/(q1+q2))*100 (q2/q1)*100=y (say)
q2
y
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Dye injection-IIStep Response
q2 Blue:input
Red:idealresponseGreen:actualresponse
Percentagecomposition
=(q2/(q1+q2))*100
(q2/q1)*100=y(say)
y
T=L/v
(Delay)
time
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
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Dye injection-IIIDyeinjectionisusedroutinelyforcolouring
subsidizedkerosenetobesoldinFairPriceShops.
Processessimilartodyeinjectionarefairlycommon
Thedelayinresponseforthedyeinjectionandsimilarsystemsiscalledtransportdelay.
Transportdelayisacommonphenomenoninchemicalprocesseswhichinvolvepipelinesand
fluidtransportation.
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Heat transfer in an oven
H
watts
Insulation
Charge
m:heatedmass
c:specificheatA:Externalsurfacearea
hAdissipatedHeat
ambienaboveeTemperatur:
Insteadystate,Heatdissipated=Heatinput
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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Heat transfer in an oven- Equations:
Considerheatbalanceequations
mc
H
mc
hA
dt
dor
differenceetemperaturinchangehAdtdcmHor
gssurroundinand
ovenbetweendifferenceetemperaturthAcmtH
fisitactuallythough
etemperaturtoalproportionbetoassumedistransferheatsimplicityFor
CmwatttcoefficientransferHeathcmabsorptionHeat
timetJthAPO
JtHPI
n
=+
=+=
=+=
==
==
=
,
..,
....
)],([
,
/..
)(./
)(./
2 o
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Heat transfer in an oven-Transfer function
t
H Hss
t
ss
( )( )
sHsH
sor
hAHconsttimetheishA
mcwhere
hAHmc
H
mc
hA
dt
d
ss
ss
ssss
+=
==
=+=+
1
1
/;
/&
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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Heat transfer in an oven-
non-dominant poles Ifonecarriesoutanexperimentonarealoven,the
responsecouldbeabitdifferent.
Thisisbecausesomeeffectshavebeenneglectedinthesimplemodel.
Forexample: Theovenitselfwouldhaveitsowntimeconstant,whichcanbe
muchmorethanthetimeconstantofthecharge.
Thetemperaturemeasuringdevicemayhaveitsowntimeconstant
Whentheaboveeffectsaremodelled properly,onewouldgetahigherordertransferfunction.
Sometimesonetimeconstantmaybesignificantlylargerthattheother.Thistimeconstantiscalleddominantandtheothersnon-dominant.
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Heat transfer in an oven-non-dominant poles-II
( )( ) )1)(1)(1(
1321
sssHsH
sss
ss
+++=
t
ss
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.5 2 2.5 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Effectofnon-dominantpoles
appearsasadelay
1st orderresponse
3rdorderresponse
A3rd ordermodelwithtwonon-
dominanttimeconstants
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Liquid Level Process-AA:Areaofcross-sectionofcontainer
Q1:Inletflowrate
Q2:Outletflowrate
H:Heightoffluid
CylindricalTank,LinearOutputFlowEquation
Q2
A
Q1
H
Q2 dependsonthe
pressureatthebottom-
whichisproportionaltotheheightoftheliquid
column
Inthisspecialcase,outletflow
IsassumedtobelinearwithH.(say); 22 HCQHQ f=
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Liquid Level Process-A-II
Insteadystate
AQAHCdt
dH
QHCdt
dHAQQ
dt
dHA
HAtQtQ
f
f
//)( 1
112
21
=+
=+=+
+=
)/~(~~~~
~;
~~
121
122
ff CQHHCQQ
HHQQQ
===
===
DifferentialEquation
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
58
Liquid Level Process-A-IIITransferFunction
( )( )
sAs
QH
sC
Q
sQ
sH
QHACAwhere
C
QH
dt
dh
C
QH
dt
dhCA
f
f
f
f
f
+=
+=
+=
==
=+
=+
1
1.
1
)~/~(
1
1
]~
/~
/[
)/(
11
1
1
1
1
Timeconstant=timerequiredto
fillthetanktothesteadystate
height!!
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Liquid Level Process-B
A:Areaofcross-sectionofcontainer
Q1:Inletflowrate
Q2:Outletflowrate
H:Heightoffluid
CylindricalTank,non- LinearOutputFlowEquation
ThistimeoutletflowrateIsassumed
tobeproportionaltosquarerootofHClosertoempiricalresult.
Pressure squareofflowrate
(say); 22 HCQHQ f=Q2
A
Q1
H
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
60
Liquid Level Process-B-II
Insteadystate
1
12
21
QHCdt
dHA
QQdt
dHA
HAtQtQ
f =+
=+
+=
HQC
CQHHCQQ
HHQQQ
f
ff
~/
~
)/~
(~~~~
~;
~~
1
2
121
122
=
===
===
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61
Liquid Level Process-B-III
Forsmallchangesaboutsteadystate
dt
dh
dt
dH
hHH
qQQ
qQQ
=
+=
+=
+=
~
~
~
222
111
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
62
Liquid Level Process-B-IV Hencetheprocessequationsare:
1
1
1
11
11
11
11
~2
~
,
~2
,
~~
2
~,
~~
21
~,
~~1
~,
~)
~(,
qhH
Q
dt
dhAor
qH
ghC
dt
dhAor
qQH
ghCgHC
dt
dhAor
qQH
hgHC
dt
dhAor
qQgH
hHC
dt
dhAor
qQghHCdt
dHAor
f
ff
f
f
f
=+
=+
+=++
+=
++
+=
++
+=++
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63
Liquid Level Process-B-V
( )
( )
sQ
H
sq
shor
QHAwhere
Q
qHh
dt
dhor
Q
qHh
dt
dhQHA
A
qh
HA
Q
dt
dh
+
=
=
=+
=+=+
1
1.~
~2
,
]~
/~
2,[
~~
2,
~~
2)~
/~
2(~2
~
11
1
1
1
1
11
11
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
64
Liquid Level Process-C
Q1A:Areaoffreeliquidsurface
Q1:Inletflowrate
Q2:Outletflowrate
H:Heightoffluid
R:RadiusCylinder
HorizontalCylindricalTank,non- LinearOutputFlowEquation
(say); 22 HCQHQ f=
Q2
L
H
R
LHRHLHRRbLA222
22)(22 ===
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65
Liquid Level Process-C-II
dt
dh
dt
dH
hHH
qQQ
qQQ
=
+=
+=
+=
~
~
~ changesmallgConsiderin
222
111
2
1
2
1
21
122
~~22
~
~/
~
)/~
(~
~~~
;~
;~~
statesteadyIn
HHRLAA
HQC
CQH
HCQQ
HHQQQ
f
f
f
==
=
=
==
===
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
66
Liquid Level Process-C-III
DeriveTransferfunctionfromabove
1
11
11
111
11
112
21
~
~~2where;~
~2
~2
~~~~
/1(~~
~/1(
~~~~
~~
~
QAHq
QHh
dtdh
qhH
Q
dt
dhAqQHhQ
dt
dhA
QHhHCdt
dhAQhHC
dt
dhA
QHCdt
dhAQQ
dt
dHA
HAtQtQ
ff
f
==+
=++=++
=++=++
=+=+
+=
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67
Liquid Level Process-D
Q1:Inletflowrate
Q2:Outletflowrate
H:Heightoffluid A:Areaoffreeliquid
surface=R2;R:Radius
(D/B)=(2R/H)
ConicalTank,non- LinearOutputFlow
(say); 22 HCQHQ f=
(say)
})/)(4/{()2/(
2
2222
H
HBDBDHRA
=
===
Q1
D
H
A
B
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
68
Liquid Level Process-D-II
dt
dh
dt
dH
hHH
qQQ
qQQ
=
+=
+=
+=
~
~
~changesmallgConsiderin
222
111
2
1
2
1
21
122
~~
~/
~
)/~
(~
~~~
;~
;~~
statesteadyIn
HAA
HQC
CQH
HCQQ
HHQQQ
f
f
f
==
=
=
==
===
2
1
2
1
21
122
~~
~/
~
)/~
(~
~~~
;~
;~~
statesteadyIn
HAA
HQC
CQH
HCQQ
HHQQQ
f
f
f
==
=
=
==
===
2
1
2
1
21
122
~~
~/
~
)/~
(~
~~~
;~
;~~
statesteadyIn
HAA
HQC
CQH
HCQQ
HHQQQ
f
f
f
==
=
=
==
===
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69
Liquid Level Process-D-III
DeriveTransferfunctionfromabove
1
3
1
11
11
111
11
112
21
~)
~(2
~
~~2
where;~
~2
~2
~~~~
/1(~~
~/1(
~~~~
~~
~
Q
H
Q
AHq
Q
Hh
dt
dh
qhH
Q
dt
dhAqQHhQ
dt
dhA
QHhHCdt
dhAQhHC
dt
dhA
QHCdt
dhAQQ
dt
dHA
HAtQtQ
ff
f
===+
=++=++
=++=++
=+=+
+=
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
70
Linearization
Nonlinearfirstorderdifferentialequationscanbelinearizedatequilibriumpoints(steadystatevalues)
vV
fy
Y
fVYf
dt
dy
dt
dYvVVyYY
VYf
VVYYdt
dYVYf
dt
dY
VVYYVVYY)|()|()
~,
~(
~;
~point,nominalabouttheonperturbatismallagConsiderin
0)~,
~(
~;
~;0state,steadyin;),(
~;
~~;
~====
+
+==
+=+=
=
====
0
(Taylorseriesexpansion)
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Linearization-II
GeneralCase
HQf
f
hAHCAQH
qAQQdt
dh
AHCQAQQdt
dH
~,
~111
1
121
1
|])}/)/(({)}/([{
/)(/)(
+
=
==
HQCCQH
HCQQHHQQQ
ff
f
~/
~)/
~(
~
~~~;
~;
~~ statesteadyIn
1
2
1
21122
==
=====
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
72
Linearization-III
Comparewithpreviousresult
hHQ
Hq
hHHH
QHQ
H
q
hHH
CHQH
q
hHHCHQHA
q
dt
dh
HQf
HQf
3
1
2
1
2
13
12
1
~,
~2/33
12
1
~,
~22
11
~2
~
~
}~~2/3
~
~~~
2{1
~
|})(~~
2{1
~
|])}/()/(({~
1
1
=
=
+=
+=
2~~:DProcessFor HA =
1
3
~)
~(2
Q
H =
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Lessons learnt from Modelling
Simplechemicalprocessesmaybemodelledwiththehelpofunderlyingphysics.
Manychemicalprocessesaredescribedbynonlineardifferentialequations. Thenonlinearequationscanbelinearizedabout
operatingpoints.
Manychemicalprocessesmayexhibitdelayedresponse.
Thedelayscouldbedueto
Transportoffluidoverlongpipes Contributionofsmallernon-dominanttimeconstants(moreaboutitlater)
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
74
Step Response of ProcessPlants
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Step response of process plants
SelfLimiting1st order
SelfLimitinghigherorder
0 0.5 1 1 .50
0.5
1
S te p Re sp o n se
Time(sec)
Amplitude
0 0.5 1 1 .50
0.5
1
S te p Re sp o n se
Time(sec)
Amplitude
0 0.5 1 1 .50
0.5
1
S te p Re sp o n se
Time(sec)
Amplitude
0 0.5 1 1 .50
0.5
1
S te p Re sp o n se
Time(sec)
Amp
litude
AlsoknownasType0system
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
76
Step response of process plants-II SelfLimitingOscillatory
withovershoot
SelfLimitingOscillatorywith
overshootandUndershoot
22
2
)(2
)()(
nn
n
sssG
++=
22
2
)(2
)(
2/1
2/1)(
nn
n
ssTs
TssG
++
+
=
Non-minimum
phasezerocauses
undershoot
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Step response of process plants-III
Non-SelfLimiting(Type-1):Ramp
Non-SelfLimiting(Type-2):
Parabola
)1()(
ss
KsG
+=
)1(
)(2
ss
KsG
+
=
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
78
Step response of process plants-IV
SelfLimitingwithtransportdelay
)1()(
s
esG
sT
+=
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Response of non-oscillatory higher order processes
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.50
0.5
1
StepResponse
Time(sec)
Amplitude
1/(0.1s+1)2
1/(0.1s+1)3
1/(0.1s+1)4
1/(0.1s+1)5
1/(0.1s+1)6
1/(0.1s+1)7
1/(0.1s+1)8
1/(0.1s+1)9
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
80
Frequency Response
Revisit
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Frequency Response Intro
Frequencyresponsemethodisaveryconvenienttoolforanalyzinglinearsystems,because Frequencyresponsecanbeexperimentallyobtained
CanbeeasilysketchedfromTransferFunction
StabilitymarginsandapproximatespeedofresponsecanbeeasilyobtainedfromOPENLOOPfrequencyresponse
InthiscoursewewouldfrequentlyuseBodePlotandoccasionallyNyquist plot.
Bode:Magnitude,Phaseagainstfreq Nyquist:Phasor plotincomplexplane
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
82
Frequency Response Intro-II InBodePlot
Frequencyaxisislogarithmic
Magnitudemaybeplotted,eitherinabsolutevalueandusingalogarithmicscale
OrinDecibelvalueusinglinearscale
Wewouldoftenusetheasymptoticplotforaqualitativeanalysis Slopesandcornerswouldbeimportant
consideration
Meaningofunitslopeindecibelscale: 20dB/decade
Meaningofunitslopeinabsolutevaluescale: Onedecadeperdecade
Iffrequencyischanged(n=)10times,themagnitudewouldchange(n=)10times
Example:unitnegativeslope,frequencyincreased7.4times,magnitudewouldbe(1/7.4times)
100
100
1
100
100
1
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Frequency Response Intro-III
Meaningofdoublepositive slopeinabsolutevaluescale: Increasefrequencyntimes,the
magnitudewouldbecomen-squaretimes.
Meaningofdoublenegative slopeinabsolutevaluescale: Iffrequencyisincreasedbyntimes,the
magnitudewouldbecome(1/n-square)times
Example:M=100@20r/s,at30r/s,themagnitudewouldbe100X(20/30)2 =44.44
Example:M=100@20r/s,themagnitudewouldbe250at20X((100/250)=
12.65r/s
100
2030
44.44
Doublenegativeslope
12.65
250
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
84
Bode plot of Integrator
10-1
100
101
102
Magnitude(abs)
100
101
102
103
-91
-90
-89
Phase
(deg)
Frequency(rad/sec)
( ) deg90==
K
j
KjG
GainCrossoveratK(r/s)
Unitnegative
slope
GainatunityFreq=K
Phase
constantat- 90deg
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Double Integrator
100
101
102
10-2
10-1
100
101
102
Magnitude(abs)
BodeDiagram
Frequency(rad/sec)
( ) deg180)( 22
==
Kj
KjGDouble
negativeslope
GainCross
overatK(r/s)
Gainatunity
Freq=K
WhatisthePhaseplot??
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
86
Magnitude plot of first order system( )
( )22
11
+=
+=
K
j
KjG
BodeDiagram
Frequency (rad/ sec)
100
101
102
100
101
Magnitude(abs)
Corner
frequency1/
K
Asymptotic
plot
ActualplotUnitnegative
slope
/1)/1( 2 KKgcf =
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Phase lag of first order term
BodeDiagram
Frequency(rad/sec)
100
101
102
-90
-60
-30
0
Phase(deg)
( )
( )
( )
( )
1
2,/1
45,/1
,/1
tan1
2
2
2
1
2
+>>
==
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More General TF-2
( ) )1.1(
)02.01(
ss
sKsG +
+
=
100
101
102
103
104
-60
-40
-20
0
20
40
60
Magnitude(dB)
Frequency(rad/sec)
Unitnegativeslope
1st corner
@1r/s
Doublenegative
slope
Unitnegative
slope
2nd corner@50r/s
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
90
Standard Process Model
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Standard type zero process model
Usuallychemicalprocessesarenotoscillatorybynature Manyplantsexhibit1 st order-likeselfregulating
response Havetransportdelayorhigherorderfastpoleswhich
canbeapproximatedasdelay Sometextbookscallthedelayas lag!! Separatemodelrequiredfornon- selfregulatingplants
( )s
eKsG
sT
p+
=
1
1Time
Constant
DelayProcessGain
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
92
Step Response of Standard Type Zero Process Model
( )s
eKsG
sT
p+
=
1
1
0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
StepResponse
Time(sec)
Amplitude
T
K1
( ) )()1( /)(1 TtueKtyTt =
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Fitting empirical step response in the standard process model
y
K
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
time
0.26K
0.67K
t1 t2
u
t
u
u
yK
=
Att1=T+0.3
y=K(1-e-0.3)=0.26K
Att2=T+y=K(1-e-1)=0.67K
Wemaydetermine andTfromt1 andt2
ProcessReaction
Curve
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
94
Step Response of Standard Type Zero Process Model
( )s
KesG
sT
p
+=
1
0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
StepResponse
Time(sec)
Amplitude
T
K
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Unit step response for
0 0.5 1 1.5 2 2.5 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
System:sys
Time(sec):0.0964
Amplitude:0.1
System:sys
Time(sec):0.511
Amplitude:0.6
StepResponse
Time(sec)
Amplitude
( )( )( )ss
sGp05.015.01
1
++=
0427.0;511.0
),2(&)1(
)2(..........916.0511.0,
6.01
)1........(105.00964.0,
1.01
511.0
0964.0
==
=
=
=
=
T
From
Tor
e
Tor
e
T
T
Asmallnondominanttime
constantcanbeapproximatedas
adelay
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
96
Unit step response for
StepResponse
Time(sec)
Amplitude
0 0.5 1 1.5 2 2.5 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
System:sys
Time(sec):0.1
Amplitude:0.0512
System:sys
Time(sec):0.6
Amplitude:0.628
( )( )( )205.015.01
1
sssG
p ++
=
07.0;535.0
),2(&)1(
)2(..........988.06.0,628.01
)1(..........0525.01.0,
0512.01
6.0
1.0
==
==
=
=
T
From
Tore
Tor
e
T
T
5.0=
( )
1.0205.0
05.012
+
T
Fairlygood
approximationofparameters!
Twonon-dominantand
smalltimeconstants
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Alternative method: Graphical method
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
time
T
Tangenttothe
highestslope
FinalsteadyValue
Smoothentheplotfirst
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
98
Normalized modelIfthesteadystateO/Pwith100%inputis
alsodefinedas100%O/P,thegainKbecomesunity.
Thenormalizedprocessmodelbecomes:
( )s
esGsT
p
+=
1
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Frequency response of plant
( )
( ) ( )
( ) ( )
( ) ( ) =
+===
=+
===
+==
+=
j
KjGejG
j
KjGejG
s
KsGandesGLet
s
KesG
Tj
Tj
sT
sT
p
1
1
1
1
21
21
21
??
?? ??
??
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
100
Magnitude contribution of delay( ) TjejG =1
=1forall
frequencies
1sincos == TjT
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101
Phase contribution of delay
( ) TTT == cossintan1
-T
( ) TjejG = 1
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
102
Bode plot for delay usinglinearscale for frequency
10-1
100
101
Magnitude(abs)
0 50 100 150 200 250 300 350 400 450-540
-450
-360
-270
-180
-90
0
Phase(deg
)
BodeDiagram
Frequency(rad/sec)
-T
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Magnitude contribution of first order term
( )( )
22
11
+=
+=
K
j
KjG
BodeDiagram
Frequency (rad/ sec)
100
101
102
100
101
Magnitude(abs)
Cornerat
1/
K
Asymptoticplot
Actualplot
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104
Phase contribution of first order term
BodeDiagram
Frequency(rad/sec)
100
101
102
-90
-60
-30
0
Phase(deg)
( )
( )
( )
( )
1
2,/1
45,/1
,/1
tan1
2
2
2
1
2
+>>
==
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Frequency Response of Standard TypeZero Process Model
sTesG=)(1
BodeDiagram
Frequency(rad/sec)
10-1
100
101
Magnitude(abs)
100
101
102
-90
-45
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
10-1
100
101
Magnitude(abs)
101
102
103
-540
-450
-360
-270
-180
-90
0
Phase(deg)
1/s
sG+
=1
1)(2
Whyistheshapelikethis?
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
106
Frequency Response of Standard Type ZeroProcess Model
-20
-15
-10
-5
0
5
Magnitude(dB)
10 20 30 40 50 60 70 80 90 100
-180
-135
-90
-45
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
Frequencyscaleislinearforphaseplot
1/
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Frequency Response of Standard Type ZeroProcess Model
-20
-15
-10
-5
0
5
Magnitude(dB)
10-1
100
101
102
-180
-135
-90
-45
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
1/
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108
Frequency Response of Standard Type Zero Process Model
Comments Ordinaryfirstordersystemhas90degree
PhaseMarginandinfinitegainmargin.Suchplantsarealwaysstableinclosedloopwithconstantgain.
Standardprocessplantisnotordinarybecauseofthedelay
Delayhasgotthecapabilityofprovidinginfinitephaselag
Delayed1st orderplantsareliabletobecomeunstableinclosedloop
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Effect of gain on magnitude plotBodeDiagram
Frequency(rad/sec)
10-1
100
101
102
103
10-1
100
101
Magnitude(abs)
K=1
K=2
K=3
gcf2
gcf3
Unitygain
1/
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110
GM=1/|G(jwp
)|
PM=180deg+/G(jwg)
10-1
100
101
Magnitude(abs)
BodeDiagram
Frequency(rad/sec)
100
101
102
-180
-135
-90
-45
0
Phase(deg)
pcf=wp
gcf=wg
PhaseMargin
Gain
Margin
gcfdecreaseswithgain
pcf doesnot
depend
upongain
PMincreases
withdecrease
ingain
GMdecreases
withincreasein
gain
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Nyquist plot
Resultantplot
sTesG
=)(1 ssG
+= 11)(2
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112
Pad Approximation forDelay
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Pad Approximation
Thedelaytermisdifficulttohandlewhilecomputingclosedloopresponse.
FrenchmathematicianHenriPad (1863-1953),formulatedthe"best"approximationofagivenfunctionbyarationalfunction ofspecifiedorder.
UsingtheabovemethodtheLaplacetransformexp(-sT)ofdelay,maybeapproximatedbytransferfunctionsofanygivenorder.
Normally,formanualcomputation1st orsecondorderapproximationsareused.
Forcomputersimulation5
th
to10th
orderapproximationsmaybeemployed.
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Pad Approximation -II
( ) ( )
( ) ( )...
!3
2/
!2
2/2/1
...!3
2/
!2
2/2/1
32
32
2/
2/
2/2/
++++
++==
sTsTsT
sTsTsT
e
eeee
sT
sT
sTsTsT
Thesetermsarechangedslightlytoaccommodatethe
effectoftruncationofhigherorder
terms
Pade approximationmaybeappreciatedfromthis
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First order Pad approximation
Frequencyresponse
Unitstepresponse
TimeResponseofprocessplantwithfirst
orderPad appx.fordelay
( )2/1
2/11
sT
sTesG
sT
+
=
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Frequency response of First order Padapprox. for unit delay
BodeDiagram
Frequency(rad/sec)
10-2
10-1
100
101
102
-180
-90
0
Phase
(deg)
10-1
100
101
Magnitude(abs)
Maxphaselagis-180degree
Unitygain
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Unit step response of 1st order Padapprox. for delay
0 0.5 1 1.5 2 2.5 3-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1StepResponse
Time(sec)
Amplitude
100%Negativekick
CausedbyNonMinimumPhase
term
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
118
Time Response of standard plant with 1st order Pad
apprx. for delay
0 0.5 1 1.5 2 2.5 3-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1StepResponse
Time(sec)
Amplitude
Negative
kick reducedduetoplanttimeconstant
( )2/1
2/1
1 sT
sT
s
KsGp
+
+=
OpenLoop
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Second order Pad approximation
Frequencyresponse
Unitstepresponse
TimeResponseofprocessplantwith
secondorderPad appx.fordelay
( )12/2/1
12/2/122
22
1
TssT
TssTesG
sT
++
+=
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120
Frequency response of Second order Padapprox. for delay
10-1
100
101
Magnitude(abs)
10-1
100
101
102
-360
-270
-180
-90
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
Maxphaselag
is-360degree
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Unit step response of 2nd order Pad approx.for delay
StepResponse
Time(sec)
Amplitude
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Smallernegative
kick for2nd
order
positivekick
2010PICLecture2010 Prof.T.K.Ghoshal &Prof.Smita Sadhu
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Time Response of standard plantwith 2nd order Pad appx. for delay
StepResponse
Time(sec)
Amplitude
0 1 2 3 4 5 6-0.2
0
0.2
0.4
0.6
0.8
1
1.2 Open
Loop
Kickssubstantially
reducedduetothetimeconstant
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Higher the order, better the approximationUnit step response
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1
-0.5
0
0.5
1
1.5
time
Firstorder
Secondorder
Thirdorder
Fourthorder
Puredelay
Higher the order, better the approximation
Frequency response
10-2
10-1
100
101
102
-540
-360
-180
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
10-1
100
101
102
-540
-360
-180
0
Phase(deg)
BodeDiagram
Frequency(rad/sec)
10-1
100
101
102
-540
-360
-180
0
Ph
ase(deg)
BodeDiagram
Frequency(rad/sec)
0
Firstorder
Secondorder
Thirdorder
Phasekeepsonincreasingastheorder
becomeshigher
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