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<math>X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt</math>
 
<math>X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt</math>
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 +
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<math>X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt</math>

Revision as of 15:16, 7 October 2008

The Signal

$ (t e^{-4t} \sin{6 \pi t}) u(t) $


The Fourier Transform

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


$ X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt $


$ X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang