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ECE2260F09_HW5p3asoln

# ECE2260F09_HW5p3asoln - ratio of polynomials and verify...

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2260 F 09 HOMEWORK #5 prob 3a solution E X : Find the inverse Laplace transform for the following expression: F ( s ) = 6 s 3 19 s 2 + 42 s + 72 s 4 + 7 s 3 + 12 s 2 S OL ' N : Factor the denominator to find the partial fraction terms needed. F ( s ) = 6 s 3 19 s 2 + 42 s + 72 s 2 ( s + 3)( s + 4) = A s 2 + B s + C s + 3 + D s + 4 we can find all but B by the pole cover-up method. A = s 2 F ( s ) s = 0 = 72 3 4 = 6 C = ( s + 3) F ( s ) s = 3 = 6( 3) 3 19( 3) 2 + 42( 3) + 72 ( 3) 2 ( 3 + 4) = [ 6( 3) 19]( 3) 2 54 9 = 63 9 = 7 D = ( s + 4) F ( s ) s = 4 = 6( 4) 3 19( 4) 2 + 42( 4) + 72 ( 4) 2 ( 4 + 3) = [ 6( 4) 19]( 4) 2 96 16 = 16 16 = 1 An easy way to find B 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 B .) Here, we use s = 1. F ( s = 1) = 6 19 + 42 + 72 1 2 (1 + 3)(1 + 4) = 6 1 2 + B 1 + 7 1 + 3 + 1 1 + 4 or F ( s = 1) = 89 20 = 6 + B 1 + 7 4 + 1 5 = B + 120 7(5) + 4 20 = B + 89 20 or B = 0

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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 ). 6 s 2 + s + − 7 s + 3 + 1 s + 4 = 6( s + 3)( s + 4) + − 7 s 2 ( s + 4) + s 2 ( s + 3) s 2 ( s + 3)( s + 4) = − 6 s 3 − 19 s 2 + 42 s + 72 s 4 + 7 s 3 + 12 s 2 = F ( s ) (result is verified √ ) Now we take the inverse transform of the partial fraction terms to get our final result. f ( t ) = [6 t − 7 e − 3 t + e − 4 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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