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Lecture3-Laplace TransformsK.J.ÅströmReview of control system analysis1.The Basic Feedback Loopplace T ransforms3.Analysis of Feedback Loops4.Qualitative Understanding of Signals and Systems5.SummaryTheme:Streamline manipulation of equations and block diagrams.Construction of a Block DiagramThe block diagram gives an overview.T o draw a block diagram:•Understand how the system works.•Identify the major components and the relevant signals.•Key questions:Where is the essential dynamics?What are appropriate abstractions?•Describe the dynamics of the blocks in terms of standard models.dt n+a1dn−1ydt n−1+...+b n u is characterized by two polynomialsA(s)=s n+a1s n−1+a2s n−2+...+a n−1s+a nB(s)=b1s n−1+b2s n−2+...+b n−1s+b n•The roots of A(s)are called poles of the system.•The roots of B(s)are called zeros of the system.•The transfer function of the system is G(s)=B(s)Linear Time Invariant Systems(LTI)d n ydt n−1+...+a n y=b1dn−1uInterpretation of the Impulse Responsey(t)=nk=1C k−1(t)eαk t+t(t−τ)u(τ)dτLet the system be initially at rest,i.e.C k=0and let the inputbe an impulse at time0.The output is theny(t)= (t)If the input is a unit step the output(the step response)isy(t)=t(t−τ)dτ=t(τ)dτExperimental determination of step and impulse responses.Recall Cruise ControlProcess modeldvdt2+(0.02+k)dedtThe mathematical tool of Laplace transforms is ideally suitedfor these type of calculations.An essential part of the languageof control.place TransformsConsider a function f defined on 0≤t <∞and a real number σ>0.Assume that f grows slower than e σt for large t .The Laplace transform F =L f of f is defined asL f =F (s )= ∞e −stf (t )dtExample 1:f (t )=1,F (s )=∞e −st dt =−1sExample 2:f (t )=e −at ,F (s )=∞e −(s +a )t dt =−1s +adt=∞e −stf (t )dt =e −st f (t )∞0+s∞e −stf (t )dt =−f (0)+s L fs)dv =f (0)Final value theoremlim s →0sF (s )=lim s →0∞0se −st f (t )dt =lims →0∞e −vf (vPropertiesLinearity:L (a f +b )=a L f +b L Differentiation:L dfs L fTime shift:L f (t −T )=e −sTL fTime stretch:L f (at )=1a ),a >0.Convolution:L t0f (t −τ) (τ)d τ=F (s )G (s )Final value Theorem †:lim s →0sF (s )=lim t →∞f (t )Initial value Theorem †:lim s →∞sF (s )=lim t →0f (t )•†:V alid only if limits exist!A (s )=B (s )s −α1+C 2s −αnC k =lim s →αk(s −αk )F (s )=B (αk )The time function corresponding to the transform isf (t )=C 1e α1t +C 2e α2t +...+C n e αn tParameters αk give shape and numbers C k give magnitudes.Notice that αk may be complex numbers.With multiple roots the constants C k are instead polynomials.Manipulating LTI Systems The differentiation property L dfCruise ControlProcess model:dvsE(s)Pure algebra gives relation between Laplace transforms of slopeΘreference V r and E by eliminating V and Us(s+0.02)+ks+k iE(s)=10sΘ(s)+s(s+0.02)V r(s)2+2ω0s+ω20Θ(s)=10θ0ω0te−ω0tω0te−ω0tThe largest error e max=10θ0e−1occurs for t=1/ωparegraph belowθ0=0.04ζ=1,ω0=0.05(dotted),ω0=0.1Discussion•What do we mean by a solution to a problem?•A historical perspectiveClosed form expressions,tables,curves •The role of computers•The necessity of insight and understanding•The need to check results•What properties can wefind easily using“back of an envelope”calculation.•A perspective on use of Laplace transforms in control engineering•A more general(biased personal)perspective.T echnology changes fast but engineering education changes slowly.U(s)=L yCar Model in Cruise Control Process modeldvU(s)=1U(s)=−10Transfer Function of PID ControllerThe error e is the input and the control signal u is the outputu=ke+k ite(τ))dτ+k ddeE(s)=k+k iTransfer Function of CarA simple model of a car on a horizontal road tells how its position y depends on the throttle.Let the mass be m and assume that the propelling force is proportional to the throttle wefindmd2yU(s)=kTransfer Function of Time DelayConsider a system where the output y is the input u delayed Ttime units.The input output relation isy(t)=u(t−T)and the transfer function isG(s)=Y(s)Transfer Function of Standard Model Consider the systemd n ydt n−1+...+a n y=b1dn−1us n+a1s n−1+a2s n−2+...+a n−1s+a nSolutionIntroduce Laplace transforms and transfer functions.We haveE =R − N +P (D +CE )Solving for E givesE =11+PCN −P5.Qualitative Understanding of Signals andSystemsTime responses can in principle be computed.T ables of Laplace transforms is a help but the work is quite tedious.Time responses are easy to compute using different types of software.•It is a good rule to always make order of magnitude calculations to make sure that results are reasonable whenever you use software.•Much insight can be obtained form very simple calcula-tions (series expansions and factorization).•Some results will be presented.•It will be discussed more in future lecturesA (s )=B (s )s −α1+C 2s −αnC k =lim s →αk(s −αk )F (s )=B (αk )Parameters αk (roots of A (s ))are easy to compute.The signal y (t )has the formy (t )=C 1e α1t +C 2e α2t +...+C n e αn tParameters αk give shape and C k give magnitudes.A (s )=s +5Example...For small s the Laplace transform is Y(s) 2.5/s,which implies that for large t the time function is y(t) 2.5.For larges we have Y(s) 1/s2.This means that for small t the time function isC(s)=s+5Insight from Transfer Functions•Derive transfer function G(s)=B(s)Insight from Transfer Functions•Make a series expansion of G(s)for small s(low fre-quency behavior,large t)G(s)=c−1s+c−2s k+...If c0=0like a static gain.If c−1=0and c−1=0like an integrator.If c−1=c−2=...=c−k+1=0and c k=0like k integrators.ExamplesE XAMPLE 1—PID CONTROLLERThe PID controller has the transfer functionG (s )=k +k is+k +k d sThis is already in series expansion form.For slow signals(small s )it behaves like an integrator and for fast signals (large s )it behaves like a differentiator.E XAMPLE 2—PID CONTROLLER WITHD ERIVATIVEF ILTERG (s )=k 1+11+s T d /NBehaves like a static gain k (1+N )for fast signals (large s ).。
