Line 10: Line 10:
 
We can proceed to compute its inverse
 
We can proceed to compute its inverse
  
<math> x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} (\delta(\omega - 1)e^{j\omega t} + \delta(\omega - 3)e^{j\omega t} d\omega \ </math>
+
<math> x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - 1)e^{j\omega t} + \delta(\omega - 3)e^{j\omega t} d\omega \ </math>
  
 
<math> x(t) = \frac{1}{2\pi}[e^{jt}+ e^{3jt}]</math>
 
<math> x(t) = \frac{1}{2\pi}[e^{jt}+ e^{3jt}]</math>

Revision as of 18:11, 8 October 2008

INVERSE FOURIER TRANSFORM

$ X(\omega) = \delta(\omega - 1) + \delta(\omega - 3) $


Knowing the formula for the Inverse Fourier transform

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

We can proceed to compute its inverse

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

$ x(t) = \frac{1}{2\pi}[e^{jt}+ e^{3jt}] $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn