Hw4Soln

# Hw4Soln - ME 375 Homework 4 Solution October 2 2009 Problem...

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Unformatted text preview: ME 375 Homework 4 : Solution October 2, 2009 Problem 1 Part A Problem 3.14(b), Palm textbook, p. 151. Given: 4 ˙ x = 3 e- 5 t x (0) = 4 Solution: The Laplace-transformation of equation :4 ˙ x = 3 e- 5 t is 4 sX ( s )- 4 x (0) = 3 1 s + 5 4 sX ( s )- 16 = 3 1 s + 5 X ( s ) = 3 4 s ( s + 5) + 4 s Using Partial Fractions, we get: X ( s ) = A s + B s + 5 + 4 s 1 Where A = 3 20 and B =- 3 20 X ( s ) = 3 20 1 s- 3 20 1 s + 5 + 4 s X ( s ) = 83 20 1 s- 3 20 1 s + 5 If we look in the Table for Inverse Laplace-transformations, we see the fol- lowing equation: L- 1 1 s = U s L- 1 1 s + a = e- at Solution in time-domain : x ( t ) = 83 20 U s ( t )- 3 20 e- 5 t (1) Part B Problem 3.15(c), Palm textbook, p. 151. Given: 5 ˙ x + 7 x = 15 x (0) = 4 Solution: The Laplace-transformation of equation : 5 ˙ x + 7 x = 15 is 5 sX ( s )- 5 x (0) + 7 X ( s ) = 15 1 s (5 s + 7) X ( s ) = 15 1 s + 20 X ( s ) = 15 1 (5 s + 7) s + 20 1 5 s + 7 2 X ( s ) = 3 1 ( s + 7 5 ) s + 4 1 s + 7 5 Using Partial Fractions, we get:...
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Hw4Soln - ME 375 Homework 4 Solution October 2 2009 Problem...

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