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 = t 2 e t u t Find: Zero state response : y t = f t h t y t = f t h t = f t [ t 2 e t u t ] Solution : First well 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- +- = 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 1 h 1 2 f 1 f t 1 L , = L ,t = t Flipped Flipped & Shifted Write as functions of : How many regions are there? h 1 2 1 f t t Only two cases for product h 1 f t : y 1 t = [ e t ] 2 e d = 2 e t e 2 d = 2 e t [ 1 2 e 2 ] = e t [ 1 ] y 1 t = e t for t RI: t < 0 ) ( 1 h ) ( - t f t RII: t 0 ) ( 1 h...
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This note was uploaded on 02/01/2011 for the course ECE 3TP4 taught by Professor Drterrencetodd during the Fall '10 term at McMaster University.

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EE3TP4_11c_CTConvolutionExamples_Lecture 15 - Continuous...

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