Department of AUTOMATIC CONTROL Exam in FRT Systems Engineering and FRTN5 Process Control June 5, 5, 4: 9: Points and grades All answers must include a clear motivation. Answers may be given in English orswedish.thetotalnumberofpointsisforsystemsengineeringand5for Process Control. The maximum number of points is specified for each subproblem. Preliminary grading scales: Systems Engineering: Grade3: points 4: 4points 5: 7points Process Control: Grade3: points 4: 7points 5: points Acceptable Aid Authorized Formelsamling i reglerteknik/ Collection of Formulae. Standard mathematical tables like TEFYMA. Pocket calculator. Results Thesolutionsarepostedonthecoursehomepage,andtheresultswillbetransferredtoLADOK.Dateandlocationfordisplayofthecorrectedexamswillbe posted on the course home page.
Exam FRT/FRTN5, 5-6-5 u FIC y FT LIC y LT Buffer tank FT y 3 Figure ThelevelcontrolsysteminProblem s to Exam in Systems Engineering/Process Control 5-6-5. FigureshowsaP/Idiagramforlevelcontrolofabuffertank.Theprocess hasonecontrolsignal,u,andthreemeasurementsignals,y,y,andy 3. a. Two of the measurement signals are used for feedback which? One of the measurement signals is used for feedforward which? Explain! ( p) b. Thecurrentsolutionisdeemedtooexpensivebyyourboss,andyouhave been ordered to remove two of the sensors(transmitters). Which sensors could you remove and still have a working system? Explain! ( p) a. y andy areusedforfeedbackcontrolinacascadedstructure.themaster level controller(lic) calculates the flow setpoint for the slave flow controller(fic).y 3 measurestheoutflow,whichcanbeseenasadisturbance. The level controller uses feedforward from this signal to compensate for the disturbance. b. If y isremoved,therewillbenofeedbackfromthetanklevel,whichis thequantitywearetryingtocontrol.hence,y shouldnotberemoved.removingy (andfic)woulddisabletheinnerloopofthecascadecontroller, andremovingy 3 woulddisablethefeedforward.bothoftheseactionswould decrease the control performance but still leave a working system.. Consider the nonlinear dynamical system ẋ = 8x 3 + u ẋ =x y=x wherex andx arethestatevariables,uistheinputandyistheoutput.
Exam FRT/FRTN5, 5-6-5 a. Determinethestationarystate(x,x )andthestationaryoutputy,when theinputsignalisconstant,u =.Thenlinearizethesystemaroundthis stationary point. ( p) b. Determine the transfer function G(s) that describes the relation between the input and the output of the linearized system. Determine the poles of the system and comment on its stability properties. ( p) a. Instationaritywehaveẋ=,whichyields= 8(x )3 + x =.5. Further,x =.Finally,wehavey =(x ) =.5.Let f (x,x,u)= 8x 3 + u f (x,x,u)=x (x,x,u)=x The partial derivatives are f = x f = x = x f = 4x x f = x =x x f u = u f u = u = Insertingthestationary valuesandintroducingthevariables x = x x, u=u u and y=y y,wegetthelinearizedsystem x= ( ) ( ) 6.5 x+ u y=( ) x b. The transfer function is calculated as G(s)=C(sI A) B=( ) ( ) ( ) s 6.5 s =.5 s +6 Thetwopolesarelocatedin± 6i.Hence,thesystemismarginallystable (i.e., stable but not asymptotically stable). 3. Figureshowsthepole-zeromapsoffoursystems,andFigure3showsthe step responses of the same systems, but not necessarily in the same order. Thetransferfunctionofeachsystemcanbewrittenintheform G(s)= ω s +ζ ω s+ ω Match the pole-zero maps 4 with the corresponding step responses A D. Motivate! (p) 3
Exam FRT/FRTN5, 5-6-5 Pole-zero map Pole-zero map Imaginary Axis (seconds - ) - - - - Real Axis (seconds - ) Imaginary Axis (seconds - ) - - - - Real Axis (seconds - ) Pole-zero map 3 Pole-zero map 4 Imaginary Axis (seconds - ) - - - - Real Axis (seconds - ) Imaginary Axis (seconds - ) - - - - Real Axis (seconds - ) Figure Pole-zero maps for Problem 3.5 Step response A.5 Step response B Amplitude.5 Amplitude.5 5 Time (seconds) 5 Time (seconds).5 Step response C.5 Step response D Amplitude.5 Amplitude.5 5 Time (seconds) 5 Time (seconds) Figure 3 Step responses for Problem 3 4
Exam FRT/FRTN5, 5-6-5 Larger ω givesfasterstepresponse,andpolesfurtherawayfromtheorigin. Larger ζ gives a more damped step response, and smaller angle between thepolesandthenegativerealaxis(seepage45inthecoursebook).hence, the correct matching is D, C, 3 B, 4 A. 4. A process is described by the following differential equation, where u is the inputandyistheoutput. 3ẏ(t)+y(t) u(t.)= a. Calculate the transfer function of the process. What are the static gain, the time constant, and the deadtime of the process? ( p) b. Calculate the impulse response of the process. ( p) a. Laplace transformation of the equation yields 3sY(s)+Y(s) e.s U(s)= SolveforY(s)toobtain Y(s)= e.s 3s+ U(s) Now, the transfer function can be identified as G(s)= e.s 3s+ The system is asymptotically stable, so the static gain is given by G()=. ThetimeconstantisidentifiedasT=3. ThedeadtimeisidentifiedasL=.. b. WehaveU(s)=,andhenceY(s)= e.s 3s+.InverseLaplacetransformation gives 5. Consider an ideal PD controller, t<. y(t)= 3 e (t.)/3, t. G c (s)=k(+st d ) a. Calculatethegainofthecontrollerforasinusoidalinpute(t)=sinωt. (p) 5
Exam FRT/FRTN5, 5-6-5 b. Whathappenstothecontrollergainforlargefrequencies,andwhyisthis a problem? How can the problem be handled in a practical implementation of the controller? ( p) a. Thegainiscalculatedas G(iω) = K(+iωT d ) =K +(ωt d ) b. Verylargeinputfrequencygivesaverylargegain,andthecontrolleris very sensitive to high-frequency noise. This can be coped with by using the following approximation of the derivative term, where the derivative gain islimitedbyn: D(s)= skt d +skt d /N 6. Consider the block diagram in Figure 4, where the transfer functions are assumed to be ( G (s)=3 + ) 5s G (s)= s 5s+ a. WhattypeofcontrollerisG (s)?motivate! (p) b. CalculatethetransferfunctionG yr (s)fromrtoy.whatareitspolesand zeros? (p) c. Determine,ifpossible,thestaticgainofG yr (s).assumingaconstantreferencevaluer,whatdoesthisimplyforthestaticerrore( )? (p) a. ThisisaPI-controller,sinceitisintheform ( G (s)=k + ) st i The s in the denominator corresponds to time integration, and forms the integral part together with the control parameters. r l l e u Σ G (s) Σ G (s) Σ y Σ n Figure4 BlockdiagraminProblem6(andProblem). 6
Exam FRT/FRTN5, 5-6-5 b. Settingl =l =n=,weobtain Y(s)=G (s)g (s)(r(s) Y(s)) Y(s)= G (s)g (s) R(s) +G (s)g (s) }{{} =G yr (s) Inserting the given transfer functions, we calculate G yr (s)=.5(s+.)(s ) (s+.)(s+3) =.5(s ) s+3 Thezeroishencelocatedin+,andthepoleislocatedin 3. (Theprocesspolein.iscancelledbythecontrollerzerointhesame location this is typical for Lambda tuning.) c. Since the system is asymptotically stable(pole strictly in left half-plane), thestaticgaincanbefoundasg()=.thisimpliesthattherewillbeno static error, since we will have y(t) = r(t) in stationarity(again assuming l =l =n=). 7. TheNyquistplotofalinearsystemG(s)isshowninFigure5.Youmay assumethatallpolesofthesystemarelocatedinthelefthalf-plane..5 Nyquist Diagram Imaginary Axis -.5 - -.5 -.5.5.5 Real Axis Figure5 NyquistplotofG(s)inProblem7. a. Estimate the phase margin, the amplitude margin, and the static gain of the system. (.5 p) b. LetthesystembefeedbackinterconnectedasshowninFigure6withK. Letd=beaconstantdisturbance.Calculatethestationaryerrore( )as afunctionofk.whatisthesmalleststationaryerrorthatmaybeachieved? (.5 p) 7
Exam FRT/FRTN5, 5-6-5 d Σ G(s) e K Figure 6 Feedback system in Problem 7. a. Thephase-marginis9degasseenbytheintersectionoftheunit-circle and the Nyquist curve. The Nyquist curve intersects the negative real axis atabout.(theactualvalueis 6 ),whichgivesanamplitudemargin of 5(respectively 6). ThestaticgainisgivenbyG(),whichisabout.6(theactualvalueis 5 3 ). b. IntheLaplacedomain,wehave E(s)=G(s)(D(s) KE(s)) E(s)= WithD(s)= s,wehave e( )=lim s se(s)= G(s) +KG(s) D(s) G() +KG() = 5/3 +5K/3 Thestationary errordecreaseswithlarger K,andforstabilityweneed K<6.Hence,thesmalleststationaryerrorthatispossibleis.5. 8. Only for FRTN5 Process Control. You have designed a sequential controller for a batch reactor and used GRAFCET to describe the logic, see Figure 7. The controller works as follows: ThesequencestartswhentheuserpushestheStartbutton(Startbecomes true). Thereactorisfilledwithreactantbythepump(action P).Whenthe desiredlevelisreached(l becomestrue),thepumpisstopped. Theheatingofthereactorisnowstarted(action Q).Theheatingstops when the temperature reaches the desired value(t becomes true). Thereactorisnowwaitingforthereactiontofinish.Whentheoperator pushes the Empty button(empty becomes true), the output valve is opened(action V).Whenthereactorisempty(L becomesfalse),the valveisclosedandtheprogramreturnstotheinitialstep. Modify the program so that the following new features are implemented: Filling and heating should be done in parallel. Heating may however notstartuntilthelowlevelsensor L becomestrue. Itshouldbepossibletoaborttheparallelfillingandheatingatany time by pressing the Empty button. (3 p) 8
Exam FRT/FRTN5, 5-6-5 Start P L T Q Empty V L Figure7 GRAFCETinProblem8 One solution is shown in Figure 7. More elaborate solutions are also possible. 9. OnlyforFRTN5ProcessControl. Therelativegainarrayfora4 4 model of a distillation column has been computed as follows:.93.5.8.64..49.86.54 Λ=.35 3.34.7.9.5.3.9.99 Usingfoursimplecontrollers,whichoutputsignaly... y 4 shouldbefed backtowhatinputsignalu... u 4?Motivate! (p) Foreachrowin Λ (eachcorrespondingtoanoutput),oneshouldfinda column(each corresponding to an input) with a non-negative element close toone.inrow,thefirstelementis.93.inrow,thefourthelementis.54.inrow3,theonlypositiveelementisthesecondone,3.34.inrow4, thethirdelementis.9.asuitablepairingishencey u,y u 4,y 3 u, y 4 u 3.. Only for FRTN5 Process Control. Again consider the block diagram infigure4,butnowassumethatg (s)andg (s)arearbitrarymultivariable systems of matching dimensions. Calculate the multivariable transfer function from the reference signal vector r to the control signal vector u. (p) 9
Exam FRT/FRTN5, 5-6-5 Start P L Q T or Empty P L or Empty Empty V L Figure8 GRAFCETinProblem8 The control signal vector is given by SolvingforU(s)weobtain U(s)=G (s)(r(s) G (s)u(s)) (I+G (s)g (s))u(s)=g (s)r(s) U(s)=(I+G (s)g (s)) G (s)r(s) Themultivariabletransferfunctionfromrtouishencegivenby G ur (s)=(i+g (s)g (s)) G (s)