6.003: Signals and SystemsLaplace TransformFebruary 18, 2010Concept Map: Continuous-Time SystemsMultiple representations of CT systems.Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXConcept Map: Continuous-Time SystemsRelations among representations.Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXConcept Map: Continuous-Time SystemsTwo interpretations ofR.Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXConcept Map: Continuous-Time SystemsRelation between System Functional and System Function.Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXA →1sCheck YourselfHow to determine impulse response from system functional?Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXCheck YourselfHow to determine impulse response from system functional?Expand functional using partial fractions:YX=2A22 + 3A + A2=A2(1 +12A)(1 + A)=2A1 +12A−2A1 + ARecognize forms of terms: each corresponds to an exponential.Alternatively, expand each term in aseries:YX= 2A1 −12A +14A2−18A3+ − · · ·− 2A1 − A + A2− A3+ − · · ·Let X = δ(t). ThenY = 21 −12t +18t2−148t3+ − · · ·u(t) − 21 − t +12t2−13!t3+ − · · ·u(t)= 2e−t/2− e−tu(t)Check YourselfHow to determine impulse response from system functional?Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXseriespartialfractionsConcept Map: Continuous-Time SystemsToday: new relations based on Laplace transform.Block DiagramSystem FunctionalDifferential Equation System FunctionImpulse Response+R1+R12X Y− −YX=2A22 + 3A + A22¨y(t) + 3˙y(t) + y(t) = 2x(t)Y (s)X(s)=22s2+ 3s + 1h(t) = 2(e−t/2− e−t) u(t)R˙x(t) x(t)RX AXLaplace 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) =e−tif t ≥ 00 otherwiseX1(s) =Z∞−∞x1(t)e−stdt =Z∞0e−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) > −1Check 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 aboveCheck YourselfX2(s) =Z∞0(e−t− e−2t)e−stdt=Z∞0e−te−stdt −Z∞0e−2te−stdt=1s + 1−1s + 2=(s + 2) − (s + 1)(s + 1)(s + 2)=1(s + 1)(s + 2)These equations converge if Re(s + 1) > 0 and Re(s + 2) > 0, thusRe(s) > −1.−1−2s-planeROC1(s + 1)(s + 2); Re(s) > −1Check 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 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 + 1Left- 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 aboveCheck YourselfX4(s) =Z∞−∞e−|t|e−stdt=Z0−∞e(1−s)tdt +Z∞0e−(1+s)tdt=e(1−s)t(1 − s)0−∞+e−(1+s)t−(1 + s)∞0=11 − s|{z}Re(s)<1+11 + s|{z}Re(s)>−1=1 + s + 1 − s(1 − s)(1 + s)=21 − s2; −1 < Re(s) < 1The ROC is the intersection of Re(s) < 1 and Re(s) > −1.Check YourselfThe Laplace transform of a signal that is both-sided a vertical strip.0tx4(t)x4(t) = e−|t|−1 1s-planeROCX4(s) =21 − s2−1 < Re(s) < 1Check YourselfFind the Laplace transform of x4(t). 20tx4(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 aboveTime-Domain Interpretation of ROCX(s) =Z∞−∞x(t) e−stdttx1(t)s-plane−1tx2(t)s-plane−1−2tx3(t)−1s-plane−1tx4(t)s-plane−1 1Time-Domain Interpretation of ROCX(s) =Z∞−∞x(t) e−stdttx1(t)s-plane−1tx2(t)s-plane−1−2tx3(t)−1s-plane−1tx4(t)s-plane−1 1Time-Domain Interpretation of ROCX(s) =Z∞−∞x(t) e−stdttx1(t)s-plane−1tx2(t)s-plane−1−2tx3(t)−1s-plane−1tx4(t)s-plane−1 1Time-Domain Interpretation of ROCX(s) =Z∞−∞x(t) e−stdttx1(t)s-plane−1tx2(t)s-plane−1−2tx3(t)−1s-plane−1tx4(t)s-plane−1 1Check YourselfThe
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