| Line 21: | Line 21: | ||
<math>\,x(t)=\frac{e^{3}}{jt+1}\left. e^{(jt+1)\omega}\right]_{-\infty}^{-3} | <math>\,x(t)=\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} | + \frac{e^{-3}}{jt-1}\left. e^{(jt-1)\omega}\right]_{-3}^{\infty} | ||
| + | + e^{j(\pi(t+1)+5)}\,</math> | ||
| + | |||
| + | <math>\,x(t)=\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)}\,</math> | + e^{j(\pi(t+1)+5)}\,</math> | ||
Revision as of 20:14, 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)=\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\, $
$ \,x(t)=\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\, $
$ \,x(t)=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}\, $
$ \,x(t)=\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)}\, $
$ \,x(t)=\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)}\, $
