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| − | <math>X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt</math> | + | <math>\sout{X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt}</math> |
<math>X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt</math> | <math>X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt</math> | ||
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<math>X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt</math> | <math>X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt</math> | ||
Revision as of 14:41, 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 $
$ \sout{X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt} $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt $
