Revision as of 11:04, 25 September 2008 by Anders89 (Talk)

Part A

CT LTI system:

y(t) = 10x(t-1)

plugging in delta(t) into the system we get:

h(t) = 10delta(t-1)

$ s = j\omega $

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

$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $

$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $

$ H(s) = 10e^{-s}\, $


Part B

$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $

Alumni Liaison

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Ryne Rayburn