Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


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 $


$ 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 $


$ X(\omega)=\int_{0}^{\infty} \frac {t e^{t(j(6 \pi - \omega)-4)}}{2 j} - \frac {t e^{t(-j(6 \pi + \omega)-4)}}{2 j}dt $


$ X(\omega)= \frac{(t (j(6 \pi - \omega)-4) - 1) e^{t(j(6 \pi - \omega)-4)}}{2 j (j(6 \pi - \omega)-4)} - \frac{(t (-j(6 \pi + \omega)-4) - 1) e^{t(-j(6 \pi + \omega)-4)}}{2 j (-j(6 \pi + \omega)-4)}\bigg]_0^\infty $


$ X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)} $



Comments/questions

  • A faster/easier way to solve this problem would be to use the Multiplication Property

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Questions/answers with a recent ECE grad

Ryne Rayburn