Line 8: Line 8:
 
<math>\,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega \,</math>
 
<math>\,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega \,</math>
  
<math>\,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\,</math>
+
<math>\,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\,</math>
  
<math>\,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\,</math>
+
<math>\,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\,</math>
  
<math>\,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}\,</math>
+
<math>\,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}\,</math>
 +
 
 +
<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}
 +
+ e^{j(\pi(t+1)+5)}\,</math>

Revision as of 21:10, 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)}\, $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett