Assignment 2-Solutions - ECE3085 - Homework #1 Due: Jan....

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ECE3085 - Homework #1 Due: Jan. 21, 2009 Problem 1. (5 pts) What is the inverse Laplace transform of F ( s ) = 3 s 2 + 2 s - 3 Solution 1. First, F ( s ) needs to be broken up into the the contribution of several simpler rational polynomials: F ( s ) = 1 ( s - 3)( s + 1) = A 1 s - 3 + A 2 s + 1 . We’ve actually seen this in class as an example, already in the factored form. What we did in the example, was to solve for the coefficients, A 1 = ( s - 3) F ( s ) | s =3 = 1 / 4 , A 2 = ( s + 1) F ( s ) | s = - 1 = - 1 / 4 , to get F ( s ) = 1 4 1 s - 3 - 1 4 1 s + 1 , whose inverse Laplace transform is f ( t ) = p 1 4 e 3 t - 1 4 e - t P 1( t ) . Problem 2. (5 pts) What is the inverse Laplace transform of F ( s ) = 10 s ( s + 1)( s + 10) Solution 2. The same procedure is applied to this problem and those that follow, F ( s ) = 10 s ( s + 1)( s + 10) = A 1 s + A 2 s + 1 + A 3 s + 10 The coefficients are A 1 = sF ( s ) | s =0 = 10 / 10 = 1 , A 2 = ( s + 1) F ( s ) | s = - 1 = 10
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This note was uploaded on 05/26/2010 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.

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Assignment 2-Solutions - ECE3085 - Homework #1 Due: Jan....

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