(System Function)
(Response to a signal)
 
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===Response to a signal===
 
===Response to a signal===
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:<math> x(t) = 2sin(2\pi t) + cos(\pi t). </math>
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:<math> x(t) = \frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt}) </math>
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:<math> y(t) = H(s)e^{-st}x(t) </math>
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:<math> y(t) = 5e^{-st}(\frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt})) </math>
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:<math> y(t) = \frac{5e^{-st}}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{5e^{-st}}{2}(e^{\pi jt}+e^{-\pi jt}) </math>

Latest revision as of 08:43, 25 September 2008

System

y(t) = 5x(t)

Unit Impulse Response

$ x(t) = \delta(t) $ $ h(t) = 5\delta(t) $

System Function

$ y(t) = \int^{\infty}_{-\infty} h(\tau) * x(\tau) d\tau, $
$ y(t) = \int^{\infty}_{-\infty} 5\delta(\tau) * e^{-jw(t -\tau)} d \tau, $
$ y(t) = e^{jwt} \int^{\infty}_{-\infty} 5 \delta(\tau) e^{-jw\tau} d\tau $
$ H(s) = \int^{\infty}_{-\infty} 5 \delta(\tau) e^{-jw\tau} d\tau $
$ H(s) = 5 e^{-jw0} $
$ H(s) = 5\, $

Response to a signal

$ x(t) = 2sin(2\pi t) + cos(\pi t). $
$ x(t) = \frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt}) $
$ y(t) = H(s)e^{-st}x(t) $
$ y(t) = 5e^{-st}(\frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt})) $
$ y(t) = \frac{5e^{-st}}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{5e^{-st}}{2}(e^{\pi jt}+e^{-\pi jt}) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood