CHEE 434/821Process Control IISome Review MaterialWinter 2006

Instructor:M.Guay

TA:V. Adetola

IntroductionIn the chemical industry,

the design of a control system is essential to ensure:Good Process OperationProcess SafetyProduct QualityMinimization of Environmental Impact

IntroductionWhat is the purpose of a control system?

To maintain important process characteristics at desired targets despite the effects of external perturbations.ControlPlantProcessingobjectives

SafetyMake $$$Environment...Perturbations

MarketEconomyClimateUpsets...

IntroductionPlantControlWhat constitutes a control system?

Combination of process sensors, actuators and computer systems designed and tuned to orchestratesafe and profitableoperation.

IntroductionProcess Dynamics:

Study of the transient behavior of processes

Process Control

the use of process dynamics for the improvement of process operation and performance

or

the use of process dynamics to alleviate the effect of undesirable (unstable) process behaviors

IntroductionWhat do we mean by process?

A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT

INPUT: (u) Something that you can manipulate DISTURBANCE: (d) Something that comes as a result of some outside phenomenon OUTPUT: (y) An observable quantity that we want to regulate

udyPInformation Flow

ExamplesStirred tank heater

MTin, wQT, wTinwQTProcessInputsOutput

ExamplesThe speed of an automobileForce ofEngineFrictionInputsOutputFriction

EngineSpeedProcess

Examples

e.g. Landing on Mars

Examples

e.g. Millirobotics

Laparoscopic Manipulators

IntroductionProcess

A process, P, is an operation that takes an INPUT or a DISTURBANCE and gives an OUTPUT

INPUT: (u) Something that you can manipulate DISTURBANCE: (d) Something that comes as a result of some outside phenomenon OUTPUT: (y) An observable quantity that we want to regulate

udyPInformation Flow

ControlWhat is control?

To regulate of a process output despite the effect of disturbances e.g.Driving a carControlling the temperature of a chemical reactorReducing vibrations in a flexible structure

To stabilize unstable processes e.g.Riding a bikeFlight of an airplaneOperation of a nuclear plant

Benefits of ControlEconomic BenefitsQuality (waste reduction)Variance reduction (consistency)Savings in energy, materials, manpower

Operability, safety (stability)PerformanceEfficiencyAccuracyroboticsReliabilityStabilizabilitybicycleaircraftnuclear reactor

Control

A controller is a system designed to regulate a given processProcess typically obeys physical and chemical conservation lawsController obeys laws of mathematics and logic (sometimes intelligent)

e.g. - Riding a bike (human controller)- Driving a car- Automatic control (computer programmed to control)

ProcessControllerWhat is a controller?

Block representationsBlock diagrams are models of the physical systemsProcessSystem Physical BoundaryTransfer offundamental quantitiesMass, Energy and MomentumInput variablesOutput variablesPhysicalOperationAbstract

ControlA controlled process is a system which is comprised of two interacting systems:

e.g. Most controlled systems are feedback controlled systems

The controller is designed to provide regulation of process outputs in the presence of disturbances

ProcessControllerOutputsDisturbancesActionObservationmonitorintervene

IntroductionWhat is required for the development of a control system?1. The Plant (e.g. SPP of Nylon)WaterSteamGas Make-upVentNylonBlowerDehumidifierReheaterReliefPotHeater

IntroductionWhat is required?

1. Process UnderstandingRequired measurementsRequired actuatorsUnderstand design limitations2. Process InstrumentationAppropriate sensor and actuator selectionIntegration in control systemCommunication and computer architecture3. Process ControlAppropriate control strategy

ExampleCruise ControlControllerFrictionProcessSpeedEngineHuman or Computer

Classical ControlControl is meant to provide regulation of process outputs about a reference, r, despite inherent disturbances

The deviation of the plant output, e=(r-y), from its intended reference is used to make appropriate adjustments in the plant input, u

ProcessControllerClassical Feedback Control Systemdyure+-

ControlProcess is a combination of sensors and actuators

Controller is a computer (or operator) that performs the required manipulations

e.g. Classical feedback control loop

yreACPMdComputerActuatorProcessSensor-+

ExamplesDriving an automobileyeACPMDriverAutomobile-+SteeringrVisual and tactile measurementDesired trajectoryrActual trajectoryy

ExamplesStirred-Tank HeaterTin, wQT, wHeaterTCThermocoupleyeACPMControllerTank-+HeaterThermocoupleTin, wTR

ExamplesMeasure T, adjust Q

Controller:Q=K(TR-T)+QnominalwhereQnominal=wC(T-Tin)

Q: Is this positive or negative feedback?Tin, wTeACPMControllerTank-+HeaterThermocoupleFeedback controlTR

ExamplesMeasure Ti, adjust QAPCMTiQiDQQ++Feedforward Control

Control NomenclatureIdentification of all process variables

Inputs(affect process) Outputs(result of process)

Inputs

Disturbance variablesVariables affecting process that are due to external forcesManipulated variablesThings that we can directly affect

Control NomenclatureOutputsMeasuredspeed of a carUnmeasuredacceleration of a carControl variablesimportant observable quantities that we want to regulatecan be measured or unmeasured

ControllerManipulatedDisturbancesProcessControlOther

ExampleTLTwi, Tiwc, Tciwc, Tcowo, TohVariables

wi, wo:Tank inlet and outlet mass flows Ti, To:Tank inlet and outlet temperatures wc:Cooling jacket mass flow Pc:Position of cooling jacket inlet valve Po:Position of tank outlet valve Tci, Tco:Cooling jacket inlet and outlet temperatures h:Tank liquid levelPoPc

ExampleVariables InputsOutputs

Disturbances Manipulated Measured Unmeasured ControlwiTiTciwchwoToPcPo

Task: Classify the variables

Process Control and Modeling

In designing a controller, we mustDefine control objectivesDevelop a process modelDesign controller based on modelTest through simulationImplement to real processTune and monitor

ModelControlleryuredProcessDesignImplementation

Control System DevelopmentDefine ObjectivesDevelop a process modelDesign controller based on modelTest bySimulationImplement and TuneMonitorPerformance

Control development is usually carried out following these important stepsOften an iterative process, based on performance we may decide to retune, redesign or remodel a given control system

Control System Development ObjectivesWhat are we trying to control?

Process modelingWhat do we need?Mechanistic and/or empirical

Controller designHow do we use the knowledge of process behavior to reach our process control objectives?What variables should we measure?What variables should we control?What are the best manipulated variables?What is the best controller structure?

Control System DevelopmentImplement and tune the controlled processTest by simulationincorporate control strategy to the process hardwaretheory rarely transcends to realitytune and re-tune

Monitor performanceperiodic retuning and redesign is often necessary based on sensitivity of process or market demandsstatistical methods can be used to monitor performance

Process ModelingMotivation:

Develop understanding of processa mathematical hypothesis of process mechanismsMatch observed process behavioruseful in design, optimization and control of process

Control:

Interested in description of process dynamicsDynamic model is used to predict how process responds to given inputTells us how to react

Process ModelingWhat kind of model do we need?

Dynamic vs. Steady-state

Steady-stateVariables not a function of timeuseful for design calculationDynamicVariables are a function of timeControl requires dynamic model

Process ModelingWhat kind of model do we need?

Experimental vs Theoretical

ExperimentalDerived from tests performed on actual processSimpler model formsEasier to manipulateTheoreticalApplication of fundamental laws of physics and chemistrymore complex but provides understandingRequired in design stages

Process ModelingDynamic vs. Steady-state

Step change in input to observe Starting at steady-state, we made a step changeThe system oscillates and finds a new steady-stateDynamics describe the transitory behavior

050100150200250300404550556065OutputTimeSteady-State 1Steady-State 2

Process ModelingEmpirical vs. Mechanistic modelsEmpirical Modelsonly local representation of the process(no extrapolation)model only as good as the dataMechanistic ModelsRely on our understanding of a processDerived from first principlesObserving laws of conservation of MassEnergyMomentumUseful for simulation and exploration of new operating conditionsMay contain unknown constants that must be estimated

Process Modeling

Empirical vs Mechanistic modelsEmpirical models do not rely on underlying mechanisms Fit specific function to match processMathematical French curve

Process ModelingLinear vs NonlinearLinearbasis for most industrial controlsimpler model form, easy to identifyeasy to design controllerpoor prediction, adequate controlNonlinearrealitymore complex and difficult to identifyneed state-of-the-art controller design techniques to do the jobbetter prediction and controlIn existing processes, we really onDynamic models obtained from experimentsUsually of an empirical natureLinearIn new applications (or difficult problems)Focus on mechanistic modelingDynamic models derived from theoryNonlinear

Process ModelingGeneral modeling procedure

Identify modeling objectivesend use of model (e.g. control)

Identify fundamental quantities of interestMass, Energy and/or Momentum

Identify boundaries

Apply fundamental physical and chemical lawsMass, Energy and/or Momentum balances

Make appropriate assumptions (Simplify)ideality (e.g. isothermal, adiabatic, ideal gas, no friction, incompressible flow, etc,)

Write down energy, mass and momentum balances (develop the model equations)

Process ModelingModeling procedure

Check model consistencydo we have more unknowns than equations

Determine unknown constantse.g. friction coefficients, fluid density and viscosity

Solve model equationstypically nonlinear ordinary (or partial) differential equationsinitial value problems

Check the validity of the modelcompare to process behavior

Process ModelingFor control applications:

Modeling objectives is to describe process dynamics based on the laws of conservation of mass, energy and momentum

The balance equation

1. Mass Balance (Stirred tank)2. Energy Balance (Stirred tank heater)3. Momentum Balance (Car speed)

Rate of Accumulationof fundamental quantityFlowInFlowOutRate ofProduction=-+

Process ModelingApplication of a mass balanceHolding Tank

Modeling objective: Control of tank level

Fundamental quantity: Mass

Assumptions: Incompressible flow

hFFin

Process ModelingTotal mass in system = rV = rAhFlow in = rFinFlow out = rF

Total mass at time t = rAh(t)Total mass at time t+Dt = rAh(t+Dt)Accumulation rAh(t+Dt) - rAh(t) = Dt(rFin-rF ),

Process ModelingModel consistencyCan we solve this equation?

Variables: h, r, Fin, F, A5Constants: r, A2

Inputs: Fin, F2

Unknowns:h1

Equations1

Degrees of freedom0

There exists a solution for each value of the inputs Fin, F

Process ModelingSolve equation

Specify initial conditions h(0)=h0 and integrate

Process ModelingEnergy balance

Objective:Control tank temperatureFundamental quantity: EnergyAssumptions:Incompressible flowConstant hold-up

MTin, wQT, w

Process ModelingUnder constant hold-up and constant mean pressure (small pressure changes)Balance equation can be written in terms of the enthalpies of the various streams

Typically work done on system by external forces is negligible

Assume that the heat capacities are constant such that

Process ModelingAfter substitution,

Since Tref is fixed and we assume constant r ,Cp

Divide by r CpV

Process ModelingResulting equation:

Model Consistency

Variables: T, F, V, Tin, Q, Cp, r7

Constants: V, Cp, r 3Inputs: F, Tin, Q3Unknown: T1

Equations1

There exists a unique solution

Process ModelingAssume F is fixed

where t=V/F is the tank residence time (or time constant)

If F changes with time then the differential equation does not have a closed form solution.

Product F(t)T(t) makes this differential equation nonlinear.

Solution will need numerical integration.

Process ModelingA simple momentum balance

Objective:Control car speedQuantity:MomentumAssumption:Friction proportional to speed

MomentumOut=Sum of forces acting on systemMomentumInRate of Accumulation-+Force ofEngine (u)FrictionSpeed (v)

Process ModelingForces are:Force of the engine = uFriction = bv

Balance:

Total momentum = Mv

Model consistency

Variables:M, v, b, u4Constants:M, b2Inputs:u1Unknownsv1

Process ModelingGravity tank

Objectives: height of liquid in tankFundamental quantity: Mass, momentumAssumptions:Outlet flow is driven by head of liquid in the tankIncompressible flowPlug flow in outlet pipeTurbulent flowhLFFo

Process ModelingFrom mass and momentum balances,

A system of simultaneous ordinary differential equations results

Linear or nonlinear?

Process ModelingModel consistency

VariablesFo, A, Ap, v, h, g, L, KF, r9

Constants A, Ap, g, L, KF, r6

Inputs Fo1

Unknownsh, v2

Equations2

Model is consistent

Solution of ODEsMechanistic modeling results in nonlinear sets of ordinary differential equations

Solution requires numerical integration

To get solution, we must first:specify all constants (densities, heat capacities, etc, )specify all initial conditionsspecify types of perturbations of the input variables

For the heated stirred tank,

specify r, CP, and Vspecify T(0)specify Q(t) and F(t)

Input SpecificationsStudy of control system dynamicsObserve the time response of a process output in response to input changes

Focus on specific inputs

1. Step input signals2. Ramp input signals3. Pulse and impulse signals4. Sinusoidal signals5. Random (noisy) signals

Common Input Signals1. Step Input Signal: a sustained instantaneous change

e.g. Unit step input introduced at time 1

Common Input Signals2. Ramp Input: A sustained constant rate of change e.g.

Common Input Signals3. Pulse: An instantaneous temporary change

e.g. Fast pulse (unit impulse)

Common Input Signals3. Pulses:

e.g. Rectangular Pulse

Common Input Signals4. Sinusoidal input

Common Input Signals5. Random Input

Solution of ODEs using Laplace TransformsProcess Dynamics and Control

Linear ODEsFor linear ODEs, we can solve without integrating by using Laplace transforms

Integrate out time and transform to Laplace domain

MultiplicationY(s) = G(s)U(s)Integration

Common TransformsUseful Laplace Transforms1. Exponential

2. Cosine

Common TransformsUseful Laplace Transforms3. Sine

Common TransformsOperators1. Derivative of a function f(t)

2. Integral of a function f(t)

Common TransformsOperators3. Delayed function f(t-t)

Common TransformsInput Signals1. Constant

2. Step

3. Ramp function

Common TransformsInput Signals4. Rectangular Pulse

5. Unit impulse

Laplace TransformsFinal Value Theorem

Limitations:

Initial Value Theorem

Solution of ODEsWe can continue taking Laplace transforms and generate a catalogue of Laplace domain functions. See SEM Table 3.1

The final aim is the solution of ordinary differential equations.

ExampleUsing Laplace Transform, solve

Result

Solution of Linear ODEsStirred-tank heater (with constant F)

taking Laplace

To get back to time domain, we mustSpecify Laplace domain functions Q(s), Tin(s)Take Inverse Laplace

Linear ODEsNotes:The expression

describes the dynamic behavior of the process explicitly

The Laplace domain functions multiplying T(0), Tin(s) and Q(s) are transfer functions

+++Tin(s)Q(s)T(0)T(s)

Laplace TransformAssume Tin(t) = sin(wt) then the transfer function gives directly

Cannot invert explicitly, but if we can find A and B such that

we can invert using tables.

Need Partial Fraction Expansion to deal with such functions

Linear ODEsWe deal with rational functions of the form r(s)=p(s)/q(s) where degree of q > degree of p

q(s) is called the characteristic polynomial of the function r(s)

Theorem:Every polynomial q(s) with real coefficients can be factored into the product of only two types of factorspowers of linear terms (x-a)n and/orpowers of irreducible quadratic terms, (x2+bx+c)m

Partial fraction Expansions1. q(s) has real and distinct factors

expand as

2. q(s) has real but repeated factor

expanded

Partial Fraction ExpansionHeaviside expansion

For a rational function of the form

Constants are given by

Note: Most applicable to q(s) with real and distinct roots. It can be applied to more specific cases.

Partial Fraction Expansions3. Q(s) has irreducible quadratic factors of the form

where

Algorithm for Solution of ODEs

Take Laplace Transform of both sides of ODESolve for Y(s)=p(s)/q(s)Factor the characteristic polynomial q(s)Perform partial fraction expansionInverse Laplace using Tables of Laplace Transforms

Transfer Function Modelsof Dynamical ProcesseProcess Dynamics and Control

Transfer FunctionHeated stirred tank example

e.g. The block is called the transfer function relating Q(s) to T(s)

+++Tin(s)Q(s)T(0)T(s)

Process ControlTime DomainTransfer function Modeling, Controller Design and AnalysisProcess Modeling,Experimentation and ImplementationLaplace DomainAbility to understand dynamics in Laplace and time domains is extremely important in thestudy of process control

Transfer functionOrder of underlying ODE is given by degree of characteristic polynomiale.g. First order processes

Second order processes

Steady-state value obtained directlye.g. First order response to unit step function

Final value theorem

Transfer functions are additive and multiplicative

Transfer functionEffect of many transfer functions on a variable is additive+++Tin(s)Q(s)T(0)T(s)

Transfer FunctionEffect of consecutive processes in series in multiplicative

Transfer Function U(s)Y2(s)Y1(s)

Deviation VariablesTo remove dependence on initial conditione.g.

Remove dependency on T(0)

Transfer functions express extent of deviation from a given steady-state

ProcedureFind steady-stateWrite steady-state equationSubtract from linear ODEDefine deviation variables and their derivatives if requiredSubstitute to re-express ODE in terms of deviation variables

ExampleJacketed heated stirred tank

Assumptions: Constant hold-up in tank and jacketConstant heat capacities and densitiesIncompressible flowModelF, TinFc, TcinFc, TcF, Th

Nonlinear ODEsQ: If the model of the process is nonlinear, how do we express it in terms of a transfer function?

A: We have to approximate it by a linear one (i.e.Linearize) in order to take the Laplace.f(x0)f(x)xx0

Nonlinear systemsFirst order Taylor series expansion

1. Function of one variable

2. Function of two variables

3. ODEs

Transfer functionProcedure to obtain transfer function from nonlinear process modelsFind steady-state of processLinearize about the steady-stateExpress in terms of deviations variables about the steady-stateTake Laplace transformIsolate outputs in Laplace domainExpress effect of inputs in terms of transfer functions

First order ProcessesExamples, Liquid storagehFFi

First Order ProcessesExamples: Speed of a Car

Stirred-tank heater

Note:

First Order ProcessesLiquid Storage Tank

Speed of a car

Stirred-tank heater Kpt

r/b rA/b

M/b1/b

1/rCpF V/FFirst order processes are characterized by:

1. Their capacity to store material, momentum and energy2. The resistance associated with the flow of mass, momentum or energy in reaching theircapacity

First order processesLiquid storage:Capacity to store mass : rAResistance to flow : 1/b

Car:Capacity to store momentum: MResistance to momentum transfer : 1/b

Stirred-tank heaterCapacity to store energy: rCpVResistance to energy transfer : 1/ rCpF

Time Constant = t = (Storage capacitance)*(Resistance to flow)

First order processStep response of first order process

Step input signal of magnitude M

0.632y(t)/KpMt/t

First order processWhat do we look for?

Process Gain: Steady-State Response

Process Time Constant:

t =

What do we need?

Process at steady-stateStep input of magnitude MMeasure process gain from new steady-stateMeasure time constant

Time Required to Reach 63.2% of final value

First order processRamp response:

Ramp input of slope a00.511.522.533.544.5500.511.522.533.544.55tat/ty(t)/Kpa

First order ProcessSinusoidal response

Sinusoidal input Asin(wt) 02468101214161820-1.5-1-0.500.511.52ARfy(t)/At/t

First order Processes

10-210-110010110210-210-1100AR/KptpwBode Plots10-210-1100101102-100-80-60-40-200ftpwHigh Frequency AsymptoteCorner FrequencyAmplitude RatioPhase Shift

Integrating ProcessesExample: Liquid storage tank

Process acts as a pure integratorhFFi

Process ModelingStep input of magnitude M

OutputTimeInputTimeSlope = KM

Integrating processesUnit impulse response

OutputTimeInputTimeKM

Integrating ProcessesRectangular pulse responseOutputTimeInputTime

Second Order ProcessesThree types of second order process:

1. Multicapacity processes: processes that consist of two or more capacities in seriese.g. Two heated stirred-tanks in series

2. Inherently second order processes: Fluid or solid mechanic processes possessing inertia and subjected to some acceleratione.g. A pneumatic valve

3. Processing system with a controller: Presence of a controller induces oscillatory behaviore.g. Feedback control system

Second order ProcessesMulticapacity Second Order ProcessesNaturally arise from two first order processes in series

By multiplicative property of transfer functionsU(s)Y(s)Y(s)U(s)

Second Order ProcessesInherently second order process:e.g. Pneumatic Valve

Momentum Balancexp

Second order ProcessesSecond order process:Assume the general form

whereKP = Process steady-state gaint = Process time constantx = Damping Coefficient

Three families of processes

x1Overdamped

Note: Chemical processes are typically overdamped or critically damped

Second Order ProcessesRoots of the characteristic polynomial

Case 1) x>1: Two distinct real rootsSystem has an exponential behavior

Case 2) x=1:One multiple real rootExponential behavior

Case 3) x

Second order ProcessesStep response of magnitude M01234567891000.20.40.60.811.21.41.61.82x=2x=0x=0.2

Second order processObservations

Responses exhibit overshoot (y(t)/KM >1) when x

Second order processesExample - Two Stirred tanks in series

MTin, wQT1, wMQT2, wResponse of T2 toTin is an example of anoverdamped second order process

Second order ProcessesCharacteristics of underdamped second order process

1. Rise time, tr2. Time to first peak, tp3. Settling time, ts4. Overshoot:

5. Decay ratio:

Second order Processes-5%+5%abctrtsPtp

Second Order ProcessSinusoidal Response

where

Second Order ProcessesBode Plotsx=1x=0.1x=1x=0.1

More Complicated processesTransfer function typically written as rational function of polynomials

where r(s) and q(s) can be factored as

s.t.

Poles and zeroesDefinitions:the roots of r(s) are called the zeros of G(s)

the roots of q(s) are called the poles of G(s)

Poles: Directly related to the underlying differential equation

If Re(pi)0 then there is at least one term of the form epit - y(t) does not vanish

Polese.g. A transfer function of the form

withcan factored to a sum of

A constant term from sA e-t/t from the term (t1s+1)A function that includes terms of the form

Poles can help us to describe the qualitative behavior of a complex system (degree>2)The sign of the poles gives an idea of the stability of the system

PolesCalculation performed easily in MATLAB

Function ROOTSe.g.

ROOTS([1 1 1 1])ans =

-1.0000 0.0000 + 1.0000i 0.0000 - 1.0000i MATLAB

PolesPlotting poles in the complex plane

Roots: -1.0, 1.0j, -1.0j

PolesProcess Behavior with purely complex poles

PolesRoots: -0.4368, -0.4066+0.9897j, -0.4066-0.9897j

PolesProcess behavior with mixed real and complex poles

PolesRoots: -0.7441, -0.3805+1.0830j, -0.3805-1.0830j, 0.2550

PolesProcess behavior with unstable pole

ZerosTransfer function:

Let t1 is the dominant time constant

ZerosObservations:

Adding a zero to an overdamped second order process yields overshoot and inverse response

Inverse response is observed when the zeros lie in right half complex plane, Re(z)>0

Overshoot is observed when the zero is dominant ()

Pole-zero cancellation yields a first order process behavior

In physical systems, overshoot and inverse response are a result of two process with different time constants, acting in opposite directions

ZerosCan result from two processes in parallel

If gains are of opposite signs and time constants are different then a right half plane zero occursU(s)Y(s)

Dead TimeTime required for the fluid to reach the valveusually approximated as dead timehFiControl loopManipulation of valve does not lead to immediate change in level

Dead timeDelayed transfer functions

e.g. First order plus dead-time

Second order plus dead-timeU(s)Y(s)

Dead timeDead time (delay)

Most processes will display some type of lag time Dead time is the moment that lapses between input changes and process response

tDStep response of a first order plus dead time process

Dead TimeProblemuse of the dead time approximation makes analysis (poles and zeros) more difficult

Approximate dead-time by a rational (polynomial) functionMost common is Pade approximation

Pade ApproximationsIn general Pade approximations do not approximate dead-time very well

Pade approximations are better when one approximates a first order plus dead time process

Pade approximations introduce inverse response (right half plane zeros) in the transfer function

Limited practical use

Process ApproximationDead timeFirst order plus dead time model is often used for the approximation of complex processes

Step response of an overdamped second order process01234567800.10.20.30.40.50.60.70.80.91- First Order plus dead timeo Second Order

Process Approximation Second order overdamped or first order plus dead time?

Second order process model may be more difficult to identify-- First order plus dead time- Second order overdampedo Actual process

Process ApproximationTransfer Function of a delay system

First order processes

Second order processes

G(s)Y(s)U(s)

Process ApproximationMore complicated processesHigher order processes (e.g. N tanks in series)

For two dominant time constants t1 and t2 process well approximated by

For one dominant time constant t1, process well approximated by

Y(s)U(s)

Process ApproximationExample

Empirical ModelingObjective:

To identify low-order process dynamics (i.e., first and second order transfer function models)Estimate process parameters (i.e., Kp, t and x)

Methodologies:

1. Least Squares Estimationmore systematic statistical approach2. Process Reaction Curve Methodsquick and easybased on engineering heuristics

Empirical ModelingLeast Squares Estimation:Simplest model form

Process Description

where y vector of process measurementxvector of process inputsb1, b0process parameters

Problem:Find b1, b0 that minimize the sum of squared residuals (SSR)

Empirical ModelingSolutionDifferentiate SSR with respect to parameters

These are called the normal equations. Solving for parameters gives:

where

Empirical ModelingCompact formDefine

Then

Problemfind value of b that minimize SSR

Empirical ModelingSolution in Compact FormNormal Equations can be written as

which can be shown to give

or

In practiceManipulations are VERY easy to perform in MATLABExtends to general linear model (GLM)

Polynomial model

Empirical ModelingControl Implementation:previous technique applicable to process model that are linear in the parameters (GLM, polynomials in x, etc)

i.e. such that, for all i, the derivatives are not a function of b

typical process step responsesfirst order

Nonlinear in Kp and toverdamped second order

Nonlinear in Kp, t1 and t2

nonlinear optimization is required to find the optimum parameters

Empirical ModelingNonlinear Least Squares required for control applicationssystem output is generally discretized

or, simply

First Order process (step response)

Least squares problem becomes the minimization of

This yields an iterative problem solution best handled by software packages: SAS, Splus, MATLAB (function leastsq)

Empirical ModelingExample

Nonlinear Least Squares Fit of a first order process from step response data

Model

Data

Empirical ModelingResults:Using MATLAB function leastsq obtained

Resulting Fit

Empirical ModelingApproximation using delayed transfer functionsFor first order plus delay processes

DifficultyDiscontinuity at q makes nonlinear least squares difficult to apply

Solution1. Arbitrarily fix delay or estimate using alternative methods2. Estimate remaining parameters3. Readjust delay repeat step 2 until best value of SSR is obtained

Empirical ModelingExample 2Underlying True Process

Data

Empirical ModelingFit of a first order plus dead time

Second order plus dead time

Empirical ModelingProcess reaction curve method:based on approximation of process using first order plus delay model

1. Step in U is introduced2. Observe behavior ym(t)3. Fit a first order plus dead time model

GpGcGsM/sD(s)Y(s)Ym(s)Y*(s)U(s)Manual Control

Empirical ModelingFirst order plus dead-time approximations

Estimation of steady-state gain is easyEstimation of time constant and dead-time is more difficulttKMq

Empirical ModelingEstimation of time constant and dead-time from process reaction curvesfind times at which process reaches 35.3% and 85.3%

Estimate tt1t2

Empirical ProcessExampleFor third order process

Estimates:

Compare:Least Squares FitReaction Curve

Empirical ModelingProcess Reaction Curve Method

based on graphical interpretationvery sensitive to process noiseuse of step responses is troublesome in normal plant operationsfrequent unmeasurable disturbancesdifficulty to perform instantaneous step changesmaybe impossible for slow processesrestricted to first order models due to reliabilityquick and easy

Least Squaressystematic approachcomputationally intensivecan handle any type of dynamics and input signalscan handle nonlinear control processesreliable

Feedback ControlSteam heated stirred tank

Feedback control system: Valve is manipulated to increase flow of steam to control tank temperature

Closed-loop process: Controller and process are interconnected

TTTCIPPsCondensateSteamFin,TinF,TIPLTLC

Feedback ControlControl Objective:maintain a certain outlet temperature and tank level

Feedback Control:

temperature is measured using a thermocouplelevel is measured using differential pressure probesundesirable temperature triggers a change in supply steam pressurefluctuations in level trigger a change in outlet flow

Note:level and temperature information is measured at outlet of process/ changes result from inlet flow or temperature disturbances inlet flow changes MUST affect process before an adjustment is made

ExamplesFeedback Control:requires sensors and actuators

e.g. Temperature Control Loop

Controller:software component implements math hardware component provides calibrated signal for actuatorActuator:physical (with dynamics) process triggered by controllerdirectly affects processSensor:monitors some property of system and transmits signal back to controller

Tin, FTeACPMControllerTank-+ValveThermocoupleTR

Closed-loop ProcessesStudy of process dynamics focused on uncontrolled or Open-loop processes

Observe process behavior as a result of specific input signals

In process control, we are concerned with the dynamic behavior of a controlled or Closed-loop process

Controller is dynamic system that interacts with the process and the process hardware to yield a specific behaviour

GpY(s)U(s)GcGmGpGv+-++controlleractuatorprocesssensorR(s)Y(s)D(s)

Closed-Loop Transfer FunctionBlock Diagram of Closed-Loop Process

Gp(s)- Process Transfer Function

Gc(s)- Controller Transfer Function

Gm(s)- Sensor Transfer Function

Gv(s)- Actuator Transfer Function

GcGmGpGv+-++controlleractuatorprocesssensorR(s)Y(s)D(s)

Closed-Loop Transfer FunctionFor control, we need to identify closed-loop dynamics due to:- Setpoint changesServo- DisturbancesRegulatory

1. Closed-Loop Servo Responsetransfer function relating Y(s) and R(s) when D(s)=0

Isolate Y(s)

Closed-Loop Transfer Function2. Closed-loop Regulatory Response

Transfer Function relating D(s) to Y(s) at R(s)=0

Isolating Y(s)

Closed-loop Transfer Function2. Regulatory Response with Disturbance Dynamics

Gd(s)Disturbance (or load) transfer function

3. Overall Closed-Loop Transfer FunctionRegulatoryServo

PID ControllersThe acronym PID stands for:P- ProportionalI- IntegralD- Derivative

PID Controllers: greater than 90% of all control implementationsdates back to the 1930svery well studied and understoodoptimal structure for first and second order processes (given some assumptions)always first choice when designing a control system

PID controller equation:

PID ControlPID Control Equation

PID Controller ParametersKcProportional gaintIIntegral Time ConstanttDDerivative Time ConstantuRController Bias

Proportional ActionIntegralActionDerivativeActionControllerBias

PID ControlPID Controller Transfer Function

or:

Note:

numerator of PID transfer function cancels second order dynamicsdenominator provides integration to remove possibility of steady-state errors

PID ControlController Transfer Function:

or,

Note:

Many variations of this controller existEasily implemented in SIMULINKeach mode (or action) of controller is better studied individually

Proportional FeedbackForm:

Transfer function:

or,

Closed-loop form:

Proportional FeedbackExample:Given first order process:

for P-only feedback closed-loop dynamics:

Closed-LoopTime Constant

Proportional FeedbackFinal response:

Note:for zero offset response we require

Possible to eliminate offset with P-only feedback (requires infinite controller gain)

Need different control action to eliminate offset (integral)Tracking ErrorDisturbance rejection

Proportional FeedbackServo dynamics of a first order process under proportional feedback

- increasing controller gain eliminates off-set

Kcy(t)/KMt/t

Proportional FeedbackHigh-order processe.g. second order underdamped process

increasing controller gain reduces offset, speeds response and increases oscillation

y(t)/KM

Proportional FeedbackImportant points:proportional feedback does not change the order of the systemstarted with a first order processclosed-loop process also first orderorder of characteristic polynomial is invariant under proportional feedback

speed of response of closed-loop process is directly affected by controller gainincreasing controller gain reduces the closed-loop time constant

in general, proportional feedbackreduces (does not eliminate) offsetspeeds up responsefor oscillatory processes, makes closed-loop process more oscillatory

Integral ControlIntegrator is included to eliminate offset

provides reset actionusually added to a proportional controller to produce a PI controllerPID controller with derivative action turned offPI is the most widely used controller in industryoptimal structure for first order processes

PI controller form

Transfer function model

PI FeedbackClosed-loop response

more complex expressiondegree of denominator is increased by one

PI FeedbackExamplePI control of a first order process

Closed-loop response

Note:offset is removedclosed-loop is second order

PI FeedbackExample (contd)effect of integral time constant and controller gain on closed-loop dynamics

natural period of oscillation

damping coefficient

integral time constant and controller gain can induce oscillation and change the period of oscillation

PI FeedbackEffect of integral time constant on servo dynamicsy(t)/KM0.010.10.51.0Kc=1

PI FeedbackEffect of controller gain

affects speed of responseincreasing gain eliminates offset quicker

y(t)/KM0.10.51.05.010.0tI=1

PI FeedbackEffect of integral action of regulatory response

reducing integral time constant removes effect of disturbancesmakes behavior more oscillatoryy(t)/KM

PI FeedbackImportant points:

integral action increases order of the system in closed-loop

PI controller has two tuning parameters that can independently affectspeed of responsefinal response (offset)

integral action eliminates offset

integral actionshould be small compared to proportional actiontuned to slowly eliminate offsetcan increase or cause oscillationcan be de-stabilizing

Derivative ActionDerivative of error signalUsed to compensate for trends in outputmeasure of speed of error signal changeprovides predictive or anticipatory actionP and I modes only response to past and current errorsDerivative mode has the form

if error is increasing, decrease control actionif error is decreasing, decrease control action

Always implemented in PID form

PID FeedbackTransfer Function

Closed-loop Transfer Function

Slightly more complicated than PI form

PID FeedbackExample:PID Control of a first order process

Closed-loop transfer function

PID FeedbackEffect of derivative action on servo dynamics2.01.00.50.1y(t)/KM

PID FeedbackEffect of derivative action on regulatory response

increasing derivative action reduces impact of disturbances on control variableslows down servo response and affects oscillation of process

2.01.00.50.1

Derivative ActionImportant Points:

Characteristic polynomial is similar to PIderivative action does not increase the order of the systemadding derivative action affects the period of oscillation of the processgood for disturbance rejectionpoor for tracking

the PID controller has three tuning parameters and can independently affect,speed of responsefinal response (offset)servo and regulatory responsederivative actionshould be small compared to integral actionhas a stabilizing influencedifficult to use for noisy signalsusually modified in practical implementation

Closed-loop StabilityEvery control problem involves a consideration of closed-loop stability

General concepts:

BIBO Stability:

An (unconstrained) linear system is said to be stable if the output response is boundedfor all bounded inputs. Otherwise it is unstable.

Comments:Stability is much easier to prove than unstabilityThis is just one type of stability

Closed-loop StabilityClosed-loop dynamics

if GOL is a rational function then the closed-loop transfer functions are rational functions and take the form

and factor as

GOL

Closed-loop stabilityGeneral Stability criterion:

A closed-loop feedback control system is stable if and only if all roots of the characteristic polynomial are negative or have negative real parts. Otherwise, the system is unstable.

Unstable region is the right half plane of the complex plane.

Valid for any linear systems.

Underlying system is almost always nonlinear so stability holds only locally. Moving away from the point of linearization may cause instability.

Closed-loop StabilityProblem reduces to finding roots of a polynomial

Easy (1990s) way : MATLAB function ROOTS

Traditional:1. Routh array:Test for positivity of roots of a polynomial2. Direct substitutionComplex axis separates stable and unstable regionsFind controller gain that yields purely complex roots3. Root locus diagram Vary location of poles as controller gain is variedOf limited use

Closed-loop stabilityRouth array for a polynomial equation

is

where

Elements of left column must be positive to have roots with negative real parts

Example: Routh ArrayCharacteristic polynomial

Polynomial Coefficients

Routh Array

Closed-loop system is unstable

Direct SubstitutionTechnique to find gain value that de-stabilizes the system.

Observation: Process becomes unstable when poles appear on right half plane

Find value of Kc that yields purely complex poles

Strategy:Start with characteristic polynomial

Write characteristic equation:

Substitute for complex pole (s=jw)

Solve for Kc and w

Example: Direct SubstitutionCharacteristic equation

Substitution for s=jw

Real PartComplex Part

System is unstable if

Root Locus DiagramOld method that consists in plotting poles of characteristic polynomial as controller gain is changed

e.g. Kc-0Kc-01

Stability and PerformanceGiven plant model, we assume a stable closed-loop system can be designed

Once stability is achieved - need to consider performance of closed-loop process - stability is not enough

All poles of closed-loop transfer function have negative real parts - can we place these poles to get a good performance

S: Stabilizing Controllers for a given plantP: Controllers that meet performanceSPCSpace of all Controllers

Controller TuningCan be achieved byDirect synthesis : Specify servo transfer function required and calculate required controller - assume plant = model

Internal Model Control: Morari et al. (86) Similar to direct synthesis except that plant and plant model are concerned

Tuning relations:Cohen-Coon - 1/4 decay ratiodesigns based on ISE, IAE and ITAE

Frequency response techniquesBode criterionNyquist criterion

Field tuning and re-tuning

Direct SynthesisFrom closed-loop transfer function

Isolate Gc

For a desired trajectory (C/R)d and plant model Gpm, controller is given by

not necessarily PID forminverse of process model to yield pole-zero cancellation (often inexact because of process approximation)used with care with unstable process or processes with RHP zeroes

Direct Synthesis1. Perfect Control

cannot be achieved, requires infinite gain

2. Closed-loop process with finite settling time

For 1st order Gp, it leads to PI controlFor 2nd order, get PID control

3. Processes with delay q

requiresagain, 1st order leads to PI control2nd order leads to PID control

IMC Controller TuningClosed-loop transfer functionIn terms of implemented controller, Gc

Gpm

Gp

R

-

+

+

-

+

+

D

C

IMC Controller Tuning1. Process model factored into two parts

where contains dead-time and RHP zeros, steady-state gain scaled to 1.

2. Controller

where f is the IMC filter

based on pole-zero cancellationnot recommended for open-loop unstable processesvery similar to direct synthesis

ExamplePID Design using IMC and Direct synthesis for the process

Process parameters: K=0.3, t=30, q=9

1. IMC Design: Kc=6.97, tI=34.5, td=3.93Filter

2. Direct Synthesis: Kc=4.76, tI=30Servo Transfer function

ExampleResult: Servo ResponseIMC and direct synthesis give roughly same results

IMC not as good due to Pade approximationy(t)tIMCDirectSynthesis

ExampleResult: Regulatory response

Direct synthesis rejects disturbance more rapidly (marginally)y(t)tIMCDirect Synthesis

Tuning RelationsProcess reaction curve method:based on approximation of process using first order plus delay model

1. Step in U is introduced2. Observe behavior ym(t)3. Fit a first order plus dead time model

GpGcGs1/sD(s)Y(s)Ym(s)Y*(s)U(s)Manuel Control

Tuning RelationsProcess response

4. Obtain tuning from tuning correlationsZiegler-NicholsCohen-CoonISE, IAE or ITAE optimal tuning relationstKMq

Ziegler-Nichols Tunings- Note presence of inverse of process gain in controllergain- Introduction of integral action requires reduction incontroller gain- Increase gain when derivation action is introduced

Example:

PI:Kc= 10tI=29.97PID: Kc= 13.33tI=18 tI=4.5

Controller

Kc

Ti

Td

P-only

PI

PID

_944383096.unknown

_944383334.unknown

_944383335.unknown

_944383336.unknown

_944383174.unknown

_944383025.unknown

ExampleZiegler-Nichols Tunings: Servo responsey(t)t

ExampleRegulatory Response

Z-N tuningOscillatory with considerable overshootTends to be conservative

Cohen-Coon Tuning RelationsDesigned to achieve 1/4 decay ratiofast decrease in amplitude of oscillation

Example:

PI:Kc=10.27tI=18.54Kc=15.64tI=19.75td=3.10

Controller

Kc

Ti

Td

P-only

PI

PID

_944384254.unknown

_944384407.unknown

_944384527.unknown

_944384577.unknown

_944384330.unknown

_944384201.unknown

Tuning relationsCohen-coon: Servo

More aggressive/ Higher controller gainsUndesirable response for most cases

Tuning RelationsCohen-Coon: Regulatory

Highly oscillatoryVery aggressivey(t)t

Integral Error Relations1. Integral of absolute error (IAE)

2. Integral of squared error (ISE)

penalizes large errors3. Integral of time-weighted absolute error (ITAE)

penalizes errors that persist

ITAE is most conservativeITAE is preferred

ITAE RelationsChoose Kc, tI and td that minimize the ITAE:

For a first order plus dead time model, solve for:

Design for Load and Setpoint changes yield different ITAE optimum

Type of Input

Type of Controller

Mode

A

B

Load

PI

P

0.859

-0.977

I

0.674

-0.680

Load

PID

P

1.357

-0.947

I

0.842

-0.738

D

0.381

0.995

Set point

PI

P

0.586

-0.916

I

1.03

-0.165

Set point

PID

P

0.965

-0.85

I

0.796

-0.1465

D

0.308

0.929

ITAE RelationsFrom table, we getLoad Settings:

Setpoint Settings:

Example

ITAE RelationsExample (contd)Setpoint Settings

Load Settings:

ITAE RelationsServo Response

design for load changes yields large overshoots for set-point changes

ITAE RelationsRegulatory response

Tuning relations are based GL=Gp Method does not apply to the processSet-point design has a good performance for this case

Tuning RelationsIn all correlations, controller gain should be inversely proportional to process gain

Controller gain is reduced when derivative action is introduced

Controller gain is reduced as increases

Integral time constant and derivative constant should increase as increases

In general,

Ziegler-Nichols and Cohen-Coon tuning relations yield aggressive control with oscillatory response (requires detuning)

ITAE provides conservative performance (not aggressive)

CHE 446Process Dynamics and ControlFrequency Response ofLinear Control Systems

First order ProcessResponse to a sinusoidal input signal

Recall: Sinusoidal input Asin(wt) yields sinusoidal output caharacterized by AR and f02468101214161820-1.5-1-0.500.511.52ARfy(t)/At/t

First order Processes

10-210-110010110210-210-1100AR/KptpwBode Plots10-210-1100101102-100-80-60-40-200ftpwHigh Frequency AsymptoteCorner FrequencyAmplitude RatioPhase Shift

Second Order ProcessSinusoidal Response

where

Second Order Processes

Bode Plotx=1x=0.1x=1x=0.1Amplitude reachesa maximum atresonance frequencyARfw

Frequency ResponseQ: Do we have to take the Laplace inverse to compute the AR and phase shift of a 1st or 2nd order process?

No

Q: Does this generalize to all transfer function models?

Yes

Study of transfer function model response to sinusoidal inputs is called Frequency Domain Response of linear processes.

Frequency ResponseSome facts for complex number theory:

i) For a complex number:

It follows that where

such thatReImwqab

Frequency ResponseSome facts:ii) Let z=a-bj and w= a+bj then

iii) For a first order process

Let s=jw

such that

Frequency ResponseMain Result:

The response of any linear process G(s) to a sinusoidal input is a sinusoidal.

The amplitude ratio of the resulting signal is given by the Modulus of the transfer function model expressed in the frequency domain, G(iw).

The Phase Shift is given by the argument of the transfer function model in the frequency domain.i.e.

Frequency ResponseFor a general transfer function

Frequency Response summarized by

where is the modulus of G(jw) and j is the argument of G(jw)

Note: Substitute for s=jw in the transfer function.

Frequency ResponseThe facts:

For any linear process we can calculate the amplitude ratio and phase shift by:

i) Letting s=jw in the transfer functionG(s)

ii) G(jw) is a complex number. Its modulus is the amplitude ratio of the process and its argument is the phase shift.

iii) As w, the frequency, is varied that G(jw) gives a trace (or a curve) in the complex plane.

iv) The effect of the frequency, w, on the process is the frequency response of the process.

Frequency ResponseExamples:

1. Pure Capacitive Process G(s)=1/s

2. Dead Time G(s)=e-qs

Frequency ResponseExamples:

3. n process in series

Frequency response of G(s)

therefore

Frequency ResponseExamples.

4. n first order processes in series

5. First order plus delay

Frequency ResponseTo study frequency response, we use two types of graphical representations

1. The Bode Plot:Plot of AR vs. w on loglog scalePlot of f vs. w on semilog scale

2. The Nyquist Plot:Plot of the trace of G(jw) in the complex plane

Plots lead to effective stability criteria and frequency-based design methods

Bode PlotPure Capacitive Process

Bode PlotG1G2G3G

Bode Plot

Example: Effect of dead-timeG=GdGGd

Nyquist PlotPlot of G(jw) in the complex plane as w is varied

Relation to Bode plot

AR is distance of G(jw) for the originPhase angle, j , is the angle from the Real positive axis

Example First order process (K=1, t=1)

j

Nyquist PlotDead-time

Second Orderw

Nyquist PlotThird Order

Effect of dead-time (second order process)

Che 446: Process Dynamics andControlFrequency DomainController Design

PI ControllerARjw

PID ControllerARjw

Bode Stability CriterionConsider open-loop control system

1. Introduce sinusoidal input in setpoint (D(s)=0) and observe sinusoidal output2. Fix gain such AR=1 and input frequency such that f=-1803. At same time, connect close the loop and set R(s)=0

Q: What happens if AR>1?GpGcGsD(s)Y(s)Ym(s)R(s)U(s)Open-loop Response to R(s)+-++

Bode Stability Criterion A closed-loop system is unstable if the frequency of the response of the open-loop GOL has an amplitude ratio greater than one at the critical frequency. Otherwise it is stable.

Strategy:1. Solve for w in

2. Calculate AR

Bode Stability CriterionTo check for stability:1. Compute open-loop transfer function2. Solve for w in f=-p 3. Evaluate AR at w4. If AR>1 then process is unstable

Find ultimate gain:

1. Compute open-loop transfer function without controller gain2. Solve for w in f=-p 3. Evaluate AR at w4. Let

Bode CriterionConsider the transfer function and controller

- Open-loop transfer function

- Amplitude ratio and phase shift

- At w=1.4128, f=-p, AR=6.746

Ziegler-Nichols TuningClosed-loop tuning relation

With P-only, vary controller gain until system (initially stable) starts to oscillate.Frequency of oscillation is wc,

Ultimate gain, Ku, is 1/M where M is the amplitude of the open-loop systemUltimate Period

Ziegler-Nichols Tunings

PKu/2PI Ku/2.2 Pu/1.2PIDKu/1.7 Pu/2 Pu/8

Nyquist Stability Criterion

If N is the number of times that the Nyquist plot encircles the point (-1,0) in the complex plane in the clockwise direction, and P is the number of open-loop poles of GOL that lie in the right-half plane, then Z=N+P is the number of unstable roots of the closed-loop characteristic equation.

Strategy1. Substitute s=jw in GOL(s)2. Plot GOL(jw) in the complex plane3. Count encirclements of (-1,0) in the clockwise direction

Nyquist CriterionConsider the transfer function

and the PI controller

Stability ConsiderationsControl is about stability

Considered exponential stability of controlled processes using:Routh criterionDirect SubstitutionRoot LocusBode Criterion (Restriction on phse angle)Nyquist Criterion

Nyquist is most general but sometimes difficult to interpret

Roots, Bode and Nyquist all in MATLAB

MAPLE is recommended for some applications.

Polynomial (no dead-time)

CHE 446Process Dynamics andControlAdvanced Control Techniques:1. Feedforward Control

Feedforward ControlFeedback control systems have the general form:

where UR(s) is an input bias term.

Feedback controllersoutput of process must change before any action is takendisturbances only compensated after they affect the process

GpGcGsY(s)Ym(s)R(s)U(s)+++GDGv++D(s)UR(s)

Feedforward ControlAssume that D(s) can be measured before it affects the processeffect of disturbance on process can be described with a model GD(s)Feedforward Control is possible.

Feedback/Feedforward ControllerStructure

GpGcGsY(s)Ym(s)R(s)U(s)+++GDGvGf++D(s)FeedforwardController

Feedforward ControlHeated Stirred Tank

Is this control configuration feedback or feedforward?How can we use the inlet stream thermocouple to regulate the inlet folow disturbancesWill this become a feedforward or feedback controller?TTTC1PsCondensateSteamF,TinF,TTT

Feedforward ControlA suggestion:

How do we design TC2?TTTC1PsCondensateSteamF,TinF,TTC2++TT

Feedforward ControlThe feedforward controller:

Transfer Function

Tracking of YR requires that

GpY(s)U(s)++GDGvGf++D(s)UR(s)

Feedforward ControlIdeal feedforward controller:

Exact cancellation requires perfect plant and perfect disturbance models.

Feedforward controllers:very sensitive to modeling errorscannot handle unmeasured disturbancescannot implement setpoint changes

Need feedback control to make control system more robust

Feedforward ControlGpGcGsY(s)Ym(s)R(s)U(s)+++GDGvGf++D(s)What is the impact of Gf on the closed-loopperformance of the feedback control system? Feedback/Feedforward Control

Feedforward ControlRegulatory transfer function of feedforward/feedback loop

Perfect control requires that (as above)

Note:Feedforward controllers do not affect closed-loop stabilityFeedforward controllers based on plant models can be unrealizable (dead-time or RHP zeroes)Can be approximated by a lead-lag unit or pure gain (rare)

Feedforward ControlTuning: In absence of disturbance model lead-lag approximation may be good

Kf obtained from open-loop data

t1 and t2from open-loop data

from heuristics

Trial-and-error

Feedforward ControlExample:

Plant:

Plant Model:

Feedback Design from plant model: IMC PID tunings

Feedforward ControlPossible Feedforward controllers:

1. From plant models:

Not realizable

2. Lead-lag unit

3. Feedforward gain controller:

Feedforward ControlFor Controller 2 and 3

Some attenuation observed at first peakDifficult problem because disturbance dynamic are much faster

Feedforward ControlUseful in manufacturing environments if good models are availableoutdoor temperature dependencies can be handle by gain feedforward controllersscheduling issues/ supply requirements can be handled

Benefits are directly related to model accuracyrely mainly on feedback control

Disturbances with different dynamics always difficult to attenuate with PID may need advanced feedback control approach (MPC, DMC, QDMC, H4-controllers, etc)

Use process knowledge (and intuition)

CHE 446:Process Dynamics and ControlAdvanced Control Techniques2. Cascade Control

Cascade ControlJacketed Reactor:

Conventional Feedback Loop:operate valve to control steam flowsteam flow disturbances must propagate through entire process to affect outputdoes not take into account flow measurementTTTC1PsCondensateSteamF,TinF,TTTFT

Cascade ControlConsider cascade control structure:

Note:TC1 calculates setpoint cascaded to the flow controllerFlow controller attenuates the effect of steam flow disturbancesTTTC1PsCondensateSteamF,TinF,TTTFTFC

Cascade ControlCascade systems contain two feedback loops:

Primary Loopregulates part of the process having slower dynamicscalculates setpoint for the secondary loope.g. outlet temperature controller for the jacketed reactor

Secondary Loopregulates part of process having faster dynamicsmaintain secondary variable at the desired target given by primary controllere.g. steam flow control for the jacketed reactor example

Cascade ControlBlock Diagram

Cascade ControlClosed-loop transfer function

1. Inner loop

2. Outer loop

Characteristic equation

Cascade ControlStability of closed-loop process is governed by

Example

Cascade ControlDesign a cascade controller for the followingsystem:

1. Primary:

2. Secondary:

Cascade Control1. PI controller only

Critical frequency

Maximum gain

ARjBode PlotsCascade Controljln(w)

Cascade Control2. Cascade Control

Secondary loop

no critical frequency gain can be largeLet Kc2=10.

Primary loop

Cascade ControlClosed-loop stability:

Bode

Maximum gain Kc1=10.44Secondary loop stabilizes the primary loop.

Cascade ControlUse cascade when:conventional feedback loop is too slow at rejecting disturbancessecondary measured variable is available whichresponds to disturbanceshas dynamics that are much faster than those of the primary variablecan be affected by the manipulated variable

Implementation

tune secondary loop firstoperation of two interacting controllers requires more careful implementationswitching on and off

CHE 446Process Dynamics and ControlAdvanced Control Techniques3. Dead-time Compensation

Dead-time CompensationConsider feedback loop:

Dead-time has a de-stabilizing effect on closed-loop systemPresence of dead-time requires detuning of controllerNeed a way to compensate for dead-time explicitlyGcGpe-qsRCD

Dead-time CompensationMotivation

0.10.750.50.25

Dead-time CompensationUse plant model to predict deviation from setpoint

Result:Removes the de-stabilizing effect of dead-timeProblem:Cannot compensate for disturbances with just feedback (possible offset)Need a very good plant modelGcGpe-qsRCDGpm

Dead-time CompensationClosed-loop transfer function

Characteristic Equation becomes

Effect of dead-time on closed-loop stability is removedController is tuned to stabilize undelayed process modelNo disturbance rejection

Dead-time Compensatione-0.5sRCD

Dead-time CompensationInclude effect of disturbances using model predictions

Adding this to previous loop givesGcGpe-qsRCDGpmGpme-qs+++-+++-

Dead-time CompensationClosed-loop transfer function

Characteristic Equation

Effect of dead-time on stability is removed Disturbance rejection is achievedController tuned for undelayed dynamics

Fast DynamicsSlowDynamics

Dead-time Compensatione-0.5sRCDe-0.5s+++-+++-

Dead-time CompensationAlternative form

Reduces to classical feedback control system with

called a Smith-PredictorGcGpe-qsRCDGpm(1-e-qs)+++++-

Dead-time compensationSmith-Predictor Design

1. Determine delayed process model

2. Tune controller Gc for the undelayed transfer function model Gpm

3. Implement Smith-Predictor as

4. Perform simulation studies to tune controller and estimate closed-loop performance over a range of modeling errors (Gpm and qm)

Dead-time CompensationEffect of dead-time estimation errors:

e-0.5sRCDe-ts+++-+++-t