ECE2260F09_HW5p3bsoln

ECE2260F09_HW5p3bsoln - ratio of polynomials and verify...

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2260 F 09 HOMEWORK #5 prob 3b solution E X : Find the inverse Laplace transform for the following expression: F ( s ) = 4 s 2 + 17 s + 16 s 3 + 4 s 2 + 4 s S OL ' N : Factor the denominator to find the partial fraction terms needed. F ( s ) = 4 s 2 + 17 s + 16 s 3 + 4 s 2 + 4 s = A s + B ( s + 2) 2 + C s + 2 we can find A and B by the pole cover-up method. A = sF ( s ) s = 0 = 16 4 = 4 B = ( s + 2) 2 F ( s ) s = 2 = 4( 2) 2 + 17( 2) + 16 2 = 1 An easy way to find C is to choose a convenient value for s and equate F ( s ) with the partial fraction expression. (This works because the only coefficient left to be found is C .) Here, we use s = 1. (Note: we must avoid using any roots.) F ( s = 1) = 4 + 17 + 16 1 + 4 + 4 = 37 9 = 4 1 + 1 (1 + 2) 2 + C 1 + 2 = 36 + 1 + 3 C 9 or C = 0 To check our answer, we convert the partial fraction expression into a
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Unformatted text preview: ratio of polynomials and verify that we get F ( s ). 4 s + 1 ( s + 2) 2 + s + 2 = 4( s + 2) 2 + s s ( s + 2) 2 = 4 s 2 + 17 s + 16 s 3 + 4 s 2 + 4 s = F ( s ) (result is verified ) We take the inverse transform of the partial fraction terms to get our final result. f ( t ) = [4 + te 2 t ] u ( t ) N OTE : We multiply by u ( t ) as a reminder that we are uncertain of the value of f ( t ) for t < 0....
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ECE2260F09_HW5p3bsoln - ratio of polynomials and verify...

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