HOMEWORK_9

# 114248 195122 1 05 04 05 04 fromtable 195122 228496

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Unformatted text preview: he response y ( nΔt ) to a unit step change in x using the partial fraction method. 5 0.6 5 0.6 0.41 1 0.41 5 0.6 1 0.41 1 1 1 5 5 0.6 0.6 0.41 1 0.41 1 5 0.6 0.41 1 9.7561 5.9451 9.7561 5.9451 19.5122 1 0.5 0.4 0.5 0.4 9.7561 5.9451 9.7561 5.9451 19.5122 1 0.5 0.4 0.5 0.4 . 11.4248 . 11.4248 19.5122 1 0.5 0.4 0.5 0.4 From table: ∆ 19.5122 22.8496 0.6403 cos 0.6747 2.5943 (b) Check your answer in part (a) by using long division. 5 0.6 1 0.41 1 1 5 0.6 0.41 1 3 5 0.41 0.41 3 5 2 1.41 0.41 Long division: 2 1.41 5 0.41 √5 5 13 3 10 13 13 21.62 7.05 2.05 7.05 2.05 26 18.33 5.33 5.33 18.95 16.28 18.95 37.9 26.7195 21.62 21.3895 43.24 21.62 21.8505 18.95 7.7695 7.7695 30.4842 21.62 ∆ 8.8642 ∆ ∆ 5 ∆ 13 ∆ 2∆ 18.95 ∆ 3∆ ∆ n 0 0 1 5 2 13 3 18.95 4 21.62 Using part (a) 2.5943 ∆ 19.5122 22.8496 0.6403 cos 0.6747 ∆ n 0 0 1 5 2 13 3 18.95 4 21.62 So, it’s proven. (c) What is the steady‐state value of y? 5 0.6 1 lim ∆ lim 1 0.41 1 1 5 0.6 lim 0.41 1 5 1.6 5 1 0.6 19.51 0.41 1 1 0.41 4∆...
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## This note was uploaded on 02/14/2010 for the course CHE NGN500 taught by Professor Ghaleb during the Spring '10 term at American Dubai.

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