Fall 2010 1ME451: Control SystemsME451: Control SystemsProf. Clark Radcliffe, Prof. Prof. Clark Radcliffe, Prof. JongeunJongeunChoiChoiDepartment of Mechanical EngineeringDepartment of Mechanical EngineeringMichigan State UniversityMichigan State UniversityLecture 2Lecture 2Laplace transformLaplace transformFall 2010 2Course roadmapCourse roadmapLaplace transformLaplace transformTransfer functionTransfer functionModels for systemsModels for systems••electricalelectrical••mechanicalmechanical••electromechanical electromechanical Block diagramsBlock diagramsLinearizationLinearizationModelingModelingAnalysisAnalysisDesignDesignTime responseTime response••TransientTransient••Steady stateSteady stateFrequency responseFrequency response••Bode plotBode plotStabilityStability••RouthRouth--HurwitzHurwitz••((NyquistNyquist))Design specsDesign specsRoot locusRoot locusFrequency domainFrequency domainPID & LeadPID & Lead--laglagDesign examplesDesign examples((MatlabMatlabsimulations &) laboratoriessimulations &) laboratoriesFall 2010 3Laplace transformLaplace transformOne of most important math tools in the course!One of most important math tools in the course!Definition: For a function f(t) (f(t)=0 for t<0),Definition: For a function f(t) (f(t)=0 for t<0),We denote Laplace transform of f(t) by F(s).We denote Laplace transform of f(t) by F(s).f(tf(t))tt00F(sF(s))(s: complex variable)Fall 2010 4Example of Laplace transformExample of Laplace transformStep functionStep function00f(t)f(t)tt55Remember L(u(t)) = 1/sf (t) = 5u(t) =5 t ≥ 00 t < 0⎧⎨⎪⎩⎪F(s) = f (t)e−stdt0∞∫= 5e−stdt0∞∫= 5 e−stdt0∞∫= 5 −1se−st⎡⎣⎤⎦0∞⎡⎣⎢⎤⎦⎥= 51s⎡⎣⎢⎤⎦⎥=5sFall 2010 5Integration is HardIntegration is HardTables are EasierTables are EasierFall 2010 6Laplace transform tableLaplace transform table(Table B.1 in Appendix B of the textbook)(Table B.1 in Appendix B of the textbook)Inverse Laplace TransformInverse Laplace TransformFall 2010 7Properties of Laplace transformProperties of Laplace transformLinearityLinearityEx.Ex.Proof.Proof.Fall 2010 8Properties of Laplace transformProperties of Laplace transformDifferentiationDifferentiationEx.Ex.Proof.Proof.tt--domaindomainss--domaindomainFall 2010 9Properties of Laplace transformProperties of Laplace transformIntegrationIntegrationProof.Proof.tt--domaindomainss--domaindomainFall 2010 10Properties of Laplace transformProperties of Laplace transformFinal value theoremFinal value theoremEx.Ex.if all the poles of if all the poles of sF(ssF(s) are in ) are in the left half planethe left half plane(LHP)(LHP)Poles of Poles of sF(ssF(s) are in LHP) are in LHP, so final value , so final value thmthmapplies.applies.Ex.Ex.Some poles of Some poles of sF(ssF(s) are not in LHP) are not in LHP, so final value , so final value thmthmdoes does NOTNOTapply.apply.Fall 2010 11Properties of Laplace transformProperties of Laplace transformInitial value theoremInitial value theoremEx.Ex.Remark: In this theorem, it does not matter if Remark: In this theorem, it does not matter if pole location is in LHP or not. pole location is in LHP or not. if the limits exist.if the limits exist.Ex.Ex.Fall 2010 12Properties of Laplace transformProperties of Laplace transformConvolutionConvolutionIMPORTANT REMARKIMPORTANT REMARKConvolutionConvolutionL−1F1(s)F2(s)()≠ f1(t) f2(t)Fall 2010 13An advantage of Laplace transformAn advantage of Laplace transformWe can transform an ordinary differential We can transform an ordinary differential equation (ODE) into an algebraic equation (AE).equation (ODE) into an algebraic equation (AE).ODEODEAEAEPartial fraction Partial fraction expansionexpansionSolution to ODESolution to ODEtt--domaindomainss--domaindomain112233Fall 2010 14Example 1Example 11st Order ODE with input and Initial Condition1st Order ODE with input and Initial ConditionTake Laplace TransformTake Laplace TransformSolve for Solve for Y(s)Y(s) 5&y(t) +10y(t) = 3u(t)y(0) = 15 sY (s) − y(0)[]+10 Y(s)[]= 31s⎡⎣⎢⎤⎦⎥5s +10()Y(s) = 5y(0) + 31s⎡⎣⎢⎤⎦⎥Y(s) =55s +10()+3s 5s +10()=1s + 2()+0.6ss+ 2()(Initial Condition) + (Input)Fall 2010 15Example 1 (cont)Example 1 (cont)Use table to Invert Use table to Invert Y(s)Y(s)term by term to find term by term to find y(t)y(t)From the Table:From the Table:So thatSo thatY(s) =1s + 2()+0.6ss+ 2()L1s + a⎛⎝⎜⎞⎠⎟= e−at⇒ L−11s + 2⎛⎝⎜⎞⎠⎟= e−2tLass+ a()⎛⎝⎜⎞⎠⎟= 1− e−at()⇒ L−10.3()2ss+ 2()⎛⎝⎜⎞⎠⎟= 0.3 1− e−2t()y(t) = e−2t+ 0.3 1− e−2t()(Initial Condition) + (Input)Fall 2010 16Properties of Laplace transformProperties of Laplace transformDifferentiation (review)Differentiation (review)tt--domaindomainss--domaindomainFall 2010 17Example 2Example 2s2Y(s) − s −1− Y(s) =1s2s2−1()Y(s) = (s +1) +1s2Y(s) =(s+1)s2−1()+1s2s2−1()Y(s) =As −1()+Bs +1()+Cs2(Find A, B and C)Y(s) =3/2()s −1()+−1/2()s +1()+(−1)s2y(t) = 3/2()et+−1/2()e−t+−1()tHow do we do that???Fall 2010 18Partial fraction expansionPartial fraction expansionMultiply both sides by (sMultiply both sides by (s--1) & let s 11) & let s 1Similarly,Similarly,unknownsunknownsY(s) =(s + 1)s2−1()+1s2s2−1()=As −1()+Bs +1()+Cs2s −1()Y(s)s→1= s − 1()As −1()+Bs +1()+Cs2⎡⎣⎢⎤⎦⎥= Aso A = s − 1()(s + 1)s2−1()+1s2s2−1()⎧⎨⎪⎩⎪⎫⎬⎪⎭⎪s→1= 1+1s2s +1()⎧⎨⎩⎫⎬⎭= 1+12=32B = s +1()Y(s )s→−1=−12C = s2()Y(s)s→0=−1Example 2 (contExample 2 (cont’’d)d)Fall 2010 19Example 3Example 3ODE with initial conditions (ICs)ODE with initial conditions (ICs)Laplace transformLaplace transform(This also isn’t in the table…)Fall 2010 20Inverse Laplace transformInverse Laplace transformIf we are interested in only the final value of y(t), apply If we are interested in only the final value of y(t), apply Final Value Theorem:Final Value Theorem:Example 3 (contExample 3 (cont’’d)d)Fall 2010 21Example: NewtonExample: Newton’’s laws lawWe want to know the trajectory of We want to know the trajectory of x(tx(t). By Laplace transform,). By Laplace transform,MM(Total response)(Total response)= = (Forced response)(Forced response)+ + (Initial condition response)(Initial condition response)Fall 2010 22EX. Air bag and accelerometerEX. Air bag and accelerometerTiny MEMS accelerometerTiny MEMS
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