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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 s :225 A + 24(16) B + 24(12) A + 400 C + 24(9) D + 24(12) C 1:(12 A + 16 B )225 + (12 C + 9 D )400 Equating the above values with the numerator coefficients in the original F ( s ), we work through the four equations. A + C = 1 36( A + C ) + 16 B + 9 D = 36 225 A + 24(16) B + 24(12)( A + C ) + 400 C + (24)9 D = 863 12(225) A + 225(16) B + 12(400) C + 400(9) D = 6900 or

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A + C = 1 36 + 16 B + 9 D = 36 225 A + 24(16) B + 24(12) + 400 C + (24)9 D = 863 12(225) A + 225(16) B + 12(400) C + 400(9) D = 6900 or A + C = 1 16 B + 9 D = 0 225 A + 24(16 B + 9 D ) + 400 C = 863 24(12) 12(225) A + 225(16) B + 12(400) C + 400(9) D = 6900 or A + C = 1 16 B + 9 D = 0 225 A + 400 C = 575 12(225) A + 225(16) B + 12(400) C + 400(9) D = 6900
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ECE2260F09_HW5p3csoln - 2260 F 09 HOMEWORK#5 prob 3c...

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