Difference between revisions of "HW5.3 Jeff Kubascik ECE301Fall2008mboutin" - Rhea

Revision as of 20:27, 5 October 2008

Compute the inverse Fourier transform of the following signal using the integral formula:

$ \,\mathcal{X}(\omega)=e^{-|\omega +3|} + e^{j(\omega + 5)}\delta(\omega - \pi)\, $

Answer

$ \,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega \, $

$ \,x(t)=\frac{1}{2\pi}\left( \int_{-\infty}^{\infty}e^{-|\omega +3|}e^{j\omega t}\,d\omega + \int_{-\infty}^{\infty}e^{j(\omega + 5)}\delta(\omega - \pi)e^{j\omega t}\,d\omega \right)\, $

$ \,x(t)=\frac{1}{2\pi}\left( \int_{-\infty}^{-3}e^{\omega +3}e^{j\omega t}\,d\omega + \int_{-3}^{\infty}e^{-\omega -3}e^{j\omega t}\,d\omega + e^{j5}\int_{-\infty}^{\infty}e^{j(t+1)\omega}\delta(\omega - \pi)\,d\omega \right)\, $

$ \,x(t)=\frac{1}{2\pi}\left( e^{3}\int_{-\infty}^{-3}e^{(jt+1)\omega}\,d\omega + e^{-3}\int_{-3}^{\infty}e^{(jt-1)\omega}\,d\omega + e^{j5}e^{j(t+1)\pi} \right)\, $

$ \,x(t)=\frac{1}{2\pi}\left( \frac{e^{3}}{jt+1}\left. e^{(jt+1)\omega}\right]_{-\infty}^{-3} + \frac{e^{-3}}{jt-1}\left. e^{(jt-1)\omega}\right]_{-3}^{\infty} + e^{j(\pi(t+1)+5)} \right)\, $

$ \,x(t)=\frac{1}{2\pi}\left( \frac{e^{3}}{jt+1}(e^{-3(jt+1)}-0) + \frac{e^{-3}}{jt-1}(0-e^{-3(jt-1)}) + e^{j(\pi(t+1)+5)} \right)\, $

$ \,x(t)=\frac{1}{2\pi}\left( \frac{1}{jt+1}e^{-j3t} - \frac{1}{jt-1}e^{-j3t} + e^{j(\pi(t+1)+5)} \right)\, $

$ \,x(t)= \frac{e^{-j3t}}{2\pi} \left(\frac{1}{jt+1} - \frac{1}{jt-1} \right) + \frac{e^{j(\pi(t+1)+5)}}{2\pi} \, $

$ \,x(t)= \frac{e^{-j3t}}{2\pi}\frac{jt-1-(jt+1)}{-t^2-1} + \frac{e^{j(\pi(t+1)+5)}}{2\pi} \, $

$ \,x(t)= \frac{e^{-j3t}}{2\pi}\frac{2}{t^2+1} + \frac{e^{j(\pi(t+1)+5)}}{2\pi} \, $

$ \,x(t)= \frac{e^{-j3t}}{\pi(t^2+1)} + \frac{e^{j(\pi(t+1)+5)}}{2\pi} \, $