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Experiments on Convolution - Linear Dynamic Systems and Signals - Lab Handout, Exercises of Electronic Circuits Analysis

Main points are: Experiments on Convolution, Convolution of Signals, Arbitrary Input, Discrete-Time Convolutions, Distributivity Properties, Associativity Properties, Finite Sum of Form, Convolution Integral

Typology: Exercises

2012/2013

Uploaded on 04/16/2013

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Download Experiments on Convolution - Linear Dynamic Systems and Signals - Lab Handout and more Exercises Electronic Circuits Analysis in PDF only on Docsity! 6.6 MATLAB Laboratory Experiments on Convolution Purpose: In this section we design two experiments dealing with continuous- and discrete-time convolutions and their applications to linear continuous- and discrete-time dynamic systems. The purpose of the first experiment is to present the convolution operator, and to demonstrate some of its properties in both continuous- and discrete-time domains. By writing and modifying the corresponding MATLAB programs, students will master every step of the convolution process. In the second experiment, the convolution method will be used to determine the zero-state responses of both continuous- and discrete-time linear dynamic systems by using the famous formula that states that the response of a linear system at rest due to an arbitrary input is the convolution of that input with the system impulse response. 6.6.1. Convolution of Signals In this experiment, students must verify the commutativity and distributivity properties of continuous- and discrete-time convolutions. Part 1. Continuous-Time Signals The convolution of two signals is defined by        ! #" "$ (6.30) Consider the continuous-time signals %'&( )* +%,.-/ )*+0124365782957%2: ; Write a MATLAB program to perform and verify the continuous-time convolution commutativity and associativity properties:  1 $ 1    % <= ?>' 0 @A   ?>'  ? 0 % While writing the program you must discretize the convolution integral given in (6.30); that is the integral in (6.30) must be approximated by a finite sum of the form  =BC!? DBEC!FGC HI J%K H)L  MC! 36=BNOM1=CP; (6.31) Docsity.com
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