# HW2s - AE 353 HW 2 Solutions 1 a F s = 2 s s 2 = 1 s 1 s 2...

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Unformatted text preview: AE 353: HW 2 Solutions 1 a F ( s ) = 2 s ( s + 2) = 1 s- 1 s + 2 = ⇒ f ( s ) = 1( t )- 1( t ) . e- 2 t b F ( s ) = 10 s ( s + 1)( s + 10) = A s + B s + 1 + C s + 10 A = sF ( s ) | s = = 1 B = ( s + 1) F ( s ) | s =- 1 =- 10 9 C = ( s + 10) F ( s ) | s =- 10 = 1 9 f ( t ) = 1( t )- 10 9 e- t + 1 9 e- 10 t c We use formula 19 and 20 of table A.2 in text F ( s ) = 3 s + 2 s 2 + 4 s + 20 = 3 s + 2 ( s + 2) 2 + (4) 2- 4 ( s + 2) 2 + (4) 2 f ( t ) = 3 e- 2 t cos 4 t- e- 2 t sin 4 t d F ( s ) = 3 s 2 + 9 s + 12 ( s + 2)( s 2 + 5 s + 11) = A s + 2 + Bs + C s 2 + 5 s + 11 A = ( s + 2) F ( s ) | s =- 2 = 6 5 Using the fact that A = 6 5 we equate numerators on both sides to get 3 s 2 + 9 s + 12 = 6 5 ( s 2 + 5 s + 11) + ( Bs + C )( s + 2) from which we find B = 9 5 and C =- 3 5 . Therefore F ( s ) = 6 5 1 s + 2 + 1 5 9 s- 3 s 2 + 5 s + 11 = 6 5 1 s + 2 + 3 5 3 s- 1 ( s + 2 . 5) 2 + ( √ 4 . 75) 2 = 6 5 1 s + 2 + 9 5 s + 2 . 5 ( s + 2 . 5) 2 + ( √ 4 . 75) 2- 25 . 5 5 * √ 4 . 75 √ 4 . 75 ( s + 2 . 5) 2 + ( √ 4 . 75) 2 f ( t ) = 6 5 e- 2 t + 9 5 e- 2 . 5 t cos( √ 4 . 75 t )- 25 . 5 5 * √ 4 . 75 e- 2 . 5 t sin( √ 4 . 75 t ) Page 1 of 6 AE 353: HW 2 Solutions e F ( s ) = 1 s 2 + 4 = 1 2 2 s 2 + 2 2 Use 17 from table A.2 f ( t ) = 1 2 sin(2 t ) f F ( s ) = 2( s + 2) ( s + 1)( s 2 + 4) = A s + 1 + B s s 2 + 2 2 + C 2 s 2 + 2 2 A = ( s + 1) F ( s ) | s =- 1 = 2 5 Using A value, we equate numerators on both sides to get 2( s + 2) = 2 5 ( s 2 + 4) + ( Bs + 2 C )( s + 1) from which we find...
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HW2s - AE 353 HW 2 Solutions 1 a F s = 2 s s 2 = 1 s 1 s 2...

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