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HW2_solution

# HW2_solution - MAE107 Homework#2 Solution Prof M'Closkey...

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MAE107 Homework #2 Solution Prof. M’Closkey Problem 1 1. The ODE’s are, - K 4 - J ˙ Ω+ bJ = M 4 - 1 - R 2 C 3 ˙ v 3 + v 3 = v 8 - v 1 ( t ) M 4 ( t ( t ) v 8 ( t ) v 3 ( t ) 2. Using a transfer function approach to calculating a single ODE relating v 3 and v 1 , - K 4 - 1 Js + b - 1 - 1 R 2 C 3 s +1 - v 1 ( t ) M 4 ( t ( t ) v 8 ( t ) v 3 ( t ) The overall transfer function is v 3 v 1 = 1 R 2 C 3 s +1 · 1 · 1 + b · K 4 = K 4 JR 2 C 3 1 ( s + b/J )( s /R 2 C 3 ) . This transfer function is the same as the one in Problem 4, part 5 in Homework #1. Hence, the ODE is the same as the ODE from Problem 4, Homework #1 (replacing v 8 with v 3 ). 3. Considering the block diagram in the present problem, v 8 ( t ) = 0 because v 1 ( t )=0and Ω(0) = 0 (thus Ω( t ) = 0). Thus, we need to solve an IVP in which the input is zero: R 2 C 3 ˙ v 3 + v 3 =0 ,v 3 (0) = 1 . The solutions is calculated to be v 3 ( t )= e 1 R 2 C 3 t ,t 0 . (1) Now we consider the arrangement of subsystems according to Homework #1, 1

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- R 2 C 3 ˙ v a + v a = v 1 - K 4 - J ˙ Ω+ b Ω= M 4 - 1 - v 1 ( t ) v a ( t ) M 4 ( t ( t ) v 8 ( t ) Considering the same initial conditions on the subsystems we fnd v a ( t )= e 1 R 2 C 3 t ,t 0 . Thus, M 4 ( t K 4 e 1 R 2 C 3 t 0 , which is the “input” into the motor block. Assuming 1 /R 2 C 3 6 = b/J , Ω( t Z t 0 1 J e b J ( t τ ) M 4 ( τ ) = Z t 0 1 J e b J ( t τ ) K 4 e 1 R 2 C 3 t = K 4 R 2 C 3 bR 2 C 3 J ± e 1 R 2 C 3 t e b J t ² 0 . Hence, v 8 ( t K 4 R 2 C 3 bR 2 C 3 J ± e 1 R 2 C 3 t e b J t ² 0 . (2) Note that (2) is not equal to (1) so the order oF the blocks does matter when considering initial conditions! Problem 2 Given the input, δ ( t 1 2 c 3 t 2 e ct μ ( t ) ,c> 0 , we assume a particular solution oF the Form, x p ( t α 1 e ct + α 2 te ct + α 3 t 2 e ct . Substituting this into the non-homogeneous ODE yields, α 1 = bc 3 ( c + a ) 3 α 2 = bc 3 ( c + a ) 2 α 3 = 1 2 bc 3 c + a 2
0.1 0.08 0.06 0.04 0.02 0 0.02 0.04 0.06 0.08 0.1 0 0.2 0.4 0.6 0.8 1 c=100 c=500 c=1000 Figure 1: The solution of the IVP with di±erent impulse durations.

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HW2_solution - MAE107 Homework#2 Solution Prof M'Closkey...

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