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# 102_1_lecture15_student - UCLA Fall 2011 Systems and...

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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

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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 + Oscillation component A decay/growth component σ Also defined the corresponding complex exponential e st For which s does f ( t ) e st 0 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

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Notice that Fourier transform is just the special case of this: F ( ) = F ( s ) | s = 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 0 - f ( t ) e - st dt The lower limit 0 - indicates that we include impulses at the origin. A bilateral Laplace transform can correspond to different signals (causal, anti-causal, 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 anti-causal 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

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Example. Consider the Laplace transform of f ( t ) = e - at u ( t ) : F ( s ) = Z 0 e - at e - st dt = Z 0 e - ( a + s ) t dt = 1 s + a provided we can say e - ( s + a ) t 0 as t → ∞ . If < ( s + a ) = σ + a > 0 : e - ( s + a ) t = e - ( σ + + 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 + a EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7

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Example : What is the Laplace transform of unit step signal? (this expression will be different from Fourier transform of u ( t ) !) What is the Laplace transform of δ ( t ) ? What is the Laplace transform of cos( ω 0 t ) ? EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
Solution: EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 9

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We’ve studied many properties of Fourier series and Fourier Transform (shift, time-stretch, reversal, etc). Same relationships also hold for Laplace Transform.
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