6.003: Signals and Systems Lecture 7 October 1, 200916.003: Signals and SystemsLaplace and Z TransformsOctober 1, 2009Mid-term Examination #1Wednesday, October 7, 7:30-9:30pm, Walker Memorial.No recitations on the day of the exam.Coverage: DT Signals and SystemsLectures 1–5Homeworks 1–4Homework 4 will include practice problems for mid-term 1.However, it will not collected or graded. Solutions will be posted.Closed book: 1 page of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Review sessions during open office hours.Conflict? Contact [email protected] before Friday, October 2, 5pm.Last TimeMany continuous-time systems can be represented with differentialequations.Example: leaky tankr0(t)r1(t)h1(t)Differential equation representation:τ ˙r1(t) = r0(t) − r1(t)Last time we considered two methods to solve differential equations:• solving homogeneous and particular equations• singularity matchingSolving Differential Equations with Laplace TransformThe Laplace transform provides a particularly powerful method ofsolving differential equations — it transforms a differential equationinto an algebraic equation.Method (where L represents the Laplace transform):differentialalgebraicalgebraic differential equation−→↓ solve−→differentialequationalgebraicequationalgebraicanswersolution todifferential equationL−→↓ solveL−1−→Laplace Transform: DefinitionLaplace transform maps a function of time t to a function of s.X(s) =Zx(t)e−stdtThere are two important variants:Unilateral (18.03)X(s) =Z∞0x(t)e−stdtBilateral (6.003)X(s) =Z∞−∞x(t)e−stdtBoth share important properties — will discuss differences later.Laplace TransformsExample: Find the Laplace transform of x1(t):0tx1(t)x1(t) =Ae−σtif t ≥ 00 otherwiseX1(s) =Z∞−∞x1(t)e−stdt =Z∞0Ae−σte−stdt =Ae−(s+σ)t−(s + σ)∞0=As + σprovided Re{s + σ} > 0 which implies that Re{s} > −σ.−σs-planeROCAs + σ; Re{s} > −σ6.003: Signals and Systems Lecture 7 October 1, 20092Check Yourself0tx2(t)x2(t) =e−t− e−2tif t ≥ 00 otherwiseWhich of the following is the Laplace transform of x2(t)?1. X2(s) =1(s+1)(s+2); Re{s} > −12. X2(s) =1(s+1)(s+2); Re{s} > −23. X2(s) =s(s+1)(s+2); Re{s} > −14. X2(s) =s(s+1)(s+2); Re{s} > −25. none of the aboveRegions of ConvergenceLeft-sided signals have left-sided Laplace transforms (bilateral only).Example:tx3(t)−1x3(t) =−e−tif t ≤ 00 otherwiseX3(s) =Z∞−∞x3(t)e−stdt =Z0−∞−e−te−stdt =−e−(s+1)t−(s + 1)0−∞=1s + 1provided Re{s + 1} < 0 which implies that Re{s} < −1.−1s-planeROC1s + 1; Re{s} < −1Left- and Right-Sided SignalsWe can concisely express left- and right-sided signals by multiplica-tion with step functions.0tx1(t)x1(t) =e−tif t ≥ 00 otherwise≡ e−tu(t)tx3(t)−1x3(t) =−e−tif t ≤ 00 otherwise≡ −e−tu(−t)Left- and Right-Sided ROCsLaplace transforms of left- and right-sided exponentials have thesame form (except −); with left- and right-sided ROCs, respectively.0te−tu(t)time functionLaplace transform−1s-planeROC1s + 1t−e−tu(−t)−1−1s-planeROC1s + 1Check YourselfFind the Laplace transform of x4(t).0tx4(t)x4(t) = e−|t|1. X4(s) =21−s2; −∞ < Re{s} < ∞2. X4(s) =21−s2; −1 < Re{s} < 13. X4(s) =21+s2; −∞ < Re{s} < ∞4. X4(s) =21+s2; −1 < Re{s} < 15. none of the aboveSolving Differential Equations with Laplace TransformsSolve the following differential equation:˙y(t) + y(t) = δ(t)Take the Laplace transform of this equation.L {˙y(t) + y(t)} = L {δ(t)}The Laplace transform of a sum is the sum of the Laplace transforms(prove this as an exercise).L { ˙y(t)} + L {y(t)} = L {δ(t)}What’s the Laplace transform of a derivative?6.003: Signals and Systems Lecture 7 October 1, 20093Laplace transform of a derivativeAssume that X(s) is the Laplace transform of x(t):X(s) =Z∞−∞x(t)e−stdtFind the Laplace transform of y(t) = ˙x(t).Y (s) =Z∞−∞y(t)e−stdt=Z∞−∞˙x(t)e−stdt= x(t)e−st∞−∞−Z∞−∞x(t)(−se−st)dtThe first term must be zero since X(s) converged. ThusY (s) = sZ∞−∞x(t)e−stdt = sX(s)Solving Differential Equations with Laplace TransformsBack to the previous problem:L { ˙y(t)} + L {y(t)} = L {δ(t)}Let Y (s) represent the Laplace transform of y(t).Then sY (s) is the Laplace transform of ˙y(t).sY (s) + Y (s) = L {δ(t)}What’s the Laplace transform of the impulse function?Laplace transform of the impulse functionLet x(t) = δ(t).X(s) =Z∞−∞δ(t)e−stdt=Z∞−∞δ(t) e−stt=0dt=Z∞−∞δ(t) 1 dt= 1Sifting property: δ(t) sifts out the value of the integrand at t = 0.Solving Differential Equations with Laplace TransformsBack to the previous problem:sY (s) + Y (s) = L {δ(t)} = 1This is a simple algebraic expression. Solve for Y (s):Y (s) =1s + 1We’ve seen this Laplace transform previously.y(t) = e−tu(t) (why not y(t) = −e−tu(−t) ?)Notice that we solved the differential equation ˙y(t) + y(t) = δ(t) with-out computing homogeneous and particular solutions and withoutsingularity matching.Solving Differential Equations with Laplace TransformsSummary of method.Start with differential equation:˙y(t) + y(t) = δ(t)Take the Laplace transform of this equation:sY (s) + Y (s) = 1Solve for Y (s):Y (s) =1s + 1Take inverse Laplace transform (by recognizing form of transform):y(t) = e−tu(t)Solving Differential Equations with Laplace TransformsRecognizing the form ...Is there a more systematic way to take an inverse Laplace transform?Yes ... and no.Formally,x(t) =12πjZσ+j∞σ−j∞X(s)estdsbut this integral is not generally easy to compute.This equation can be useful to prove theorems.We will find better ways (e.g., partial fractions) to compute inversetransforms for common systems.6.003: Signals and Systems Lecture 7 October 1, 20094Solving Differential Equations with Laplace TransformsExample 2:¨y(t) + 3 ˙y(t) + 2y(t) = δ(t)Laplace transform:s2Y (s) + 3sY (s) + 2Y (s) = 1Solve:Y (s) =1(s + 1)(s + 2)=1s + 1−1s + 2Inverse Laplace transform:y(t) =e−t− e−2tu(t)These forward and inverse Laplace transforms are easy if• differential equation is linear with constant coefficients, and• the input signal is an impulse function.Properties of Laplace TransformsThe use of Laplace Transforms to solve differential equations de-pends on several important properties.Property x(t) X(s) ROCLinearity
View Full Document
Unlocking...