429-2009-HW2key

429-2009-HW2key - Homework 2 Solution Key Problem 1:...

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Homework 2 Solution Key Problem 1: Original equation: d dt x ( t ) = β ( t ) - αx ( t ) (1) In the Laplace domain this becomes: s ˜ x ( s ) = β ( s ) s + α (2) Solving: ˜ x ( s ) = ˜ β ( s ) s + α (3) From the original problem: β ( t ) = β 0 2 (1 + sin ( ω t )) (4) Taking the Laplace transform of (4): ˜ β ( s ) = β 0 2 ± ω s 2 + ω 2 + 1 s ² (5) Substituting (4) into (3), rearranging: ˜ x ( s ) = β 0 2 ± ω ( s 2 + ω 2 ) ( s + α ) + 1 s ( s + α ) ² (6) Taking inverse Laplace transform: β 0 2 π i ˆ i - i ω ( s 2 + ω 2 ) ( s + α ) + 1 s ( s + α ) e st ds (7) This evaluates to: x ( t ) = β 0 ω ( e - αt ( α 2 + ω 2 ) + e iωt ( + α )(2 ) + e - iωt ( - + α )( - 2 ) ) + β 0 α (1 - e - αt ) Or in polar coordinates: x ( t ) = β 0 ω ( e - αt r 2 + e i ( ωt - θ ) 2 irω + e - i ( ωt - θ ) - 2 irω ) + β 0 α (1 - e - αt ) (8) 1
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This simplifies to: β 0 ω r 2 e - αt + β 0 r sin( ωt - θ ) + β 0 α (1 - e - αt ) (9) Problems 2, 3: % Homework 2 Main Matlab file. % Constants:
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429-2009-HW2key - Homework 2 Solution Key Problem 1:...

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