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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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| + | <math>X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt</math> | ||
Revision as of 14:16, 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 $
