Revision as of 18:40, 25 September 2008 by Han45 (Talk)

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

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva