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<math> = \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\,</math>
 
<math> = \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\,</math>
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 +
<math> = \int_{-5}^{5}e^{3jt -j\omega t}dt + \int_{0}^{\infty}e^{-2t -j\omega t}dt\,</math>

Revision as of 12:20, 7 October 2008

Signal

$ x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}*u(t)\, $


Transformed

$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\, $

$ = \int_{-\infty}^{\infty}e^{3jt}*(u(t+5) - u(t-5))e^{-j\omega t}dt + \int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt\, $

$ = \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\, $

$ = \int_{-5}^{5}e^{3jt -j\omega t}dt + \int_{0}^{\infty}e^{-2t -j\omega t}dt\, $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn