Revision as of 19:46, 25 September 2008 by Han45 (Talk)

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part A

$ e^{st}\rightarrow h(t)\rightarrow H(S)e^{st} $

$ H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt $

assume

$ h(t) = 5u(t-3) $

$ H(s) = 5\int_3^\infty e^{-st}dt $

$ H(s) = \frac{-5}{s} $

$ 4 to \infty $

$ H(s)= \frac {5}{s} $


part B

$ h(t) = \sum_{-\infty} ^\infty a_k H(s) $

since I got

$ a_1 = \frac{3}{j} $

$ a_2= \frac{-3}{j} $

$ a_3 = 2 $

$ a_4 = 2 $

so,

$ s + s + \frac{s}{j} -\frac{s}{j}= 2s $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood