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
+
<math>\,x(t)=\frac{1}{2\pi}\left(
+ \int_{-\infty}^{\infty}e^{j(\omega + 5)}\delta(\omega - \pi)e^{j\omega t}\,d\omega\,</math>
+
\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)\,</math>
  
<math>\,x(t)=\int_{-\infty}^{-3}e^{\omega +3}e^{j\omega t}\,d\omega
+
<math>\,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 +
 
+ \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>
+
e^{j5}\int_{-\infty}^{\infty}e^{j(t+1)\omega}\delta(\omega - \pi)\,d\omega
 +
\right)\,</math>
  
<math>\,x(t)=e^{3}\int_{-\infty}^{-3}e^{(jt+1)\omega}\,d\omega
+
<math>\,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^{-3}\int_{-3}^{\infty}e^{(jt-1)\omega}\,d\omega
+ e^{j5}e^{j(t+1)\pi}\,</math>
+
+ e^{j5}e^{j(t+1)\pi}
 +
\right)\,</math>
  
<math>\,x(t)=\frac{e^{3}}{jt+1}\left. e^{(jt+1)\omega}\right]_{-\infty}^{-3}
+
<math>\,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}
 
+ \frac{e^{-3}}{jt-1}\left. e^{(jt-1)\omega}\right]_{-3}^{\infty}
+ e^{j(\pi(t+1)+5)}\,</math>
+
+ e^{j(\pi(t+1)+5)}
 +
\right)\,</math>
  
<math>\,x(t)=\frac{e^{3}}{jt+1}(e^{-3(jt+1)}-0)
+
<math>\,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)})
 
+ \frac{e^{-3}}{jt-1}(0-e^{-3(jt-1)})
+ e^{j(\pi(t+1)+5)}\,</math>
+
+ e^{j(\pi(t+1)+5)}
 +
\right)\,</math>

Revision as of 21:17, 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)\, $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang