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Unformatted text preview: UCLA Fall 2011 Systems and Signals Lecture 15: Inversion of Laplace Transform November 21, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1 Review Last class we introduced Laplace transform: • Generalizes Fourier transform • Allows handling of growing signals, unstable systems • Simplifies analysis of LCCDEs (converts differential equations into algebraic equations) • Important for systems with feedback, control EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2 We defined a complex frequency s = σ + jω • Oscillation component jω • A decay/growth component σ Also defined the corresponding complex exponential e st For which s does f ( t ) e st → as t → ∞ ? Defined bilateral Laplace transform: F ( s ) = Z ∞∞ f ( t ) e st dt. with the inverse: f ( t ) = 1 2 πj Z c + j ∞ c j ∞ F ( s ) e st ds EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3 Notice that Fourier transform is just the special case of this: F ( jω ) = F ( s )  s = jω ℜ ℑ ℜ ℑ Fourier Transform Laplace Transform s = j ω s = σ + j ω σ f ( t ) = 1 2 π j Z c + j ∞ c j ∞ F ( s ) e st ds f ( t ) = 1 2 π Z ∞ ∞ F ( j ω ) e j ω t d ω Important: Laplace transform is not unique! F ( s ) is usually specified along with region of convergence (region in complex plane for which F ( s ) does not blow up) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4 We are primarily interested in causal signals (for which f ( t ) = f ( t ) u ( t ) ), so we defined a unilateral Laplace transform: F ( s ) = Z ∞ f ( t ) e st dt The lower limit indicates that we include impulses at the origin. A bilateral Laplace transform can correspond to different signals (causal, anticausal, or infinite extent) depending on the region of convergence. The e σt factor that makes the integral converge for a causal signal can make the integral for an anticausal signal blow up. If we restrict ourselves to the unilateral transform the Laplace transform is (almost) unique, and we can ignore the region of convergence. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5 Example. Consider the Laplace transform of f ( t ) = e at u ( t ) : F ( s ) = Z ∞ e at e st dt = Z ∞ e ( a + s ) t dt = 1 s + a provided we can say e ( s + a ) t → as t → ∞ . If < ( s + a ) = σ + a > : e ( s + a ) t = e ( σ + jω + a ) t = e jωt  {z } =1 e ( σ + a ) t = e ( σ + a ) t The region of convergence is then σ > a , or < s > a . The Laplace transform pair is e at ⇔ 1 s + a EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6 This is very similar to the Fourier transform relationship: e at ⇔ 1 jω + a EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7 Example : • What is the Laplace transform of unit step signal?...
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 Fall '08
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 Jin Hyung Lee

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