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EE3TP4_11c_CTConvolutionExamples_Lecture 15

EE3TP4_11c_CTConvolutionExamples_Lecture 15 - Continuous...

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Unformatted text preview: Continuous Time Convolution Examples These overheads were originally developed by Mark Fowler at Binghamton University, State University of New York. Example 1: h ( t ) Zero ICs ) ( t f ) ( t y Given: f t = e t u − t h t =− δ t 2 e − t u t Find: Zero − state response : y t = f t ∗ h t y t = f t ∗ h t = f t ∗ [ − δ t 2 e − t u t ] Solution : First we’ll use properties of convolution to break this down into sub-problems. By the distributive property we have: ) ( 2 ) ( ) ( ) ( ) ( t u e t f t t f t y t- +- = δ Call this convolution y 1 ( t ) call this h 1 ( t ) Use Property of Convolution with impulses ) ( ) ( ) ( 1 t y t f t y +- = f τ 1 τ h 1 τ 2 f − τ 1 f t − τ 1 τ L , = τ L ,t = t Flipped Flipped & Shifted Write as functions of τ : How many regions are there? τ τ τ h 1 τ 2 τ 1 f t − τ t τ Only two cases for product h 1 τ f t − τ : ⇒ y 1 t = ∫ ∞ [ e t − τ ] 2 e − τ dτ = 2 e t ∫ ∞ e − 2τ dτ = 2 e t [ − 1 2 e − 2τ ] ∞ =− e t [ − 1 ] y 1 t = e t for t RI: t < 0 ) ( 1 τ h τ ) ( τ- t f τ t RII: t ≥ 0 ) ( 1 τ h τ...
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EE3TP4_11c_CTConvolutionExamples_Lecture 15 - Continuous...

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