ECE2260F09_HW5p3csoln

ECE2260F09_HW5p3csoln - 2260 F 09 HOMEWORK #5 prob 3c...

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2260 F 09 HOMEWORK #5 prob 3c solution E X : Find the inverse Laplace transform for the following expression: F ( s ) = s 3 36 s 2 863 s 6900 ( s 2 + 24 s + 400)( s 2 + 24 s + 225) S OL ' N : Factor the denominator to find the "partial fraction" terms needed. (We actually write terms that represent decaying cosines and sines.) F ( s ) = s 3 + 36 s 2 + 863 s + 6900 [( s + 12) 2 + 16 2 ][( s + 12) 2 + 9 2 ] = A ( s + 12) ( s + 12) 2 + 16 2 + B (16) ( s + 12) 2 + 16 2 + C ( s + 12) ( s + 12) 2 + 9 2 + D (9) ( s + 12) 2 + 9 2 or F ( s ) = A ( s + 12) + B (16) ( s + 12) 2 + 16 2 + C ( s + 12) + D (9) ( s + 12) 2 + 9 2 We can use a common denominator and then solve for the coefficients. F ( s ) = [ A ( s + 12) + B (16)][( s + 12) 2 + 9 2 ] [( s + 12) 2 + 16 2 ][( s + 12) 2 + 9 2 ] + [ C ( s + 12) + D (9)][( s + 12) 2 + 16 2 ] [( s + 12) 2 + 16 2 ][( s + 12) 2 + 9 2 ] The multipliers for each power of s are as follows: s 3 : A + C s 2 :12 A + 24 A + 16 B + 12 C + 24 C + 9 D
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ECE2260F09_HW5p3csoln - 2260 F 09 HOMEWORK #5 prob 3c...

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