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ME430_HW1_solution

# ME430_HW1_solution - 2 = 0 K 2 = ± 1 6 K 1 K 3 = 0 K 3 =...

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ME430 / ECE451 (Assigned: Thursday 1/7/2010 Quiz: Thursday 1/14/2010): Homework #1 Solutions Problem 1 Page 1 of 4

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ME430 / ECE451 (Assigned: Thursday 1/7/2010 Quiz: Thursday 1/14/2010): Homework #1 Solutions Problem 1 Problem 2 Page 2 of 4
ME430 / ECE451 (Assigned: Thursday 1/7/2010 Quiz: Thursday 1/14/2010): Homework #1 Solutions Problem 2 Problem 3 Problem 4 Page 3 of 4

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ME430 / ECE451 (Assigned: Thursday 1/7/2010 Quiz: Thursday 1/14/2010): Homework #1 Solutions Problem 4 Problem 5 a) 25 s ( s 2 +6 s +25) = K 1 s + K 2 s + K 3 s 2 +6 s +25 Multiply both sides by s : 25 s 2 +6 s +25 = K 1 + K 2 s 2 + K 3 s s 2 +6 s +25 Let s ! 0: 25 25 = K 1 = 1 Now use algebraic method, multiplying both sides of original equation by full denominator: 25 = K 1 ( s 2 + 6 s + 25) + K 2 s 2 + K 3 s 25 = ( K 1 + K 2 ) s 2 + (6
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Unformatted text preview: 2 = 0 ) K 2 = ± 1 6 K 1 + K 3 = 0 ) K 3 = ± 6 Which gives us: F ( s ) = 1 s ± s +6 s 2 +6 s +25 = 1 s ± s +3+3 ( s 2 +3) 2 +4 2 = 1 s ± s +3 ( s 2 +3) 2 +4 2 ± 3 ( s 2 +3) 2 +4 2 We know that: L [ Ae ± at cos !t ] = A ( s + a ) ( s + a ) 2 + ! 2 and L [ Be ± at sin !t ] = B! ( s + a ) 2 + ! 2 from the constant multiplication and frequency shift properties. SO: ) f ( t ) = 1 ± e ± 3 t cos 4 t ± 3 4 e ± 3 t sin 4 t b) s +4 s 2 +8 s +160 = s +4 s 2 +8 s +16+144 = s +4 ( s +4) 2 +12 2 Glancing at the above problem, we can write the solution by inspection: ) f ( t ) = e ± 4 t cos 12 t Page 4 of 4...
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