(Fourier Transform)
(Fourier Transform)
Line 5: Line 5:
 
<math>X(j \omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math>
 
<math>X(j \omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math>
  
<math> = \int_{-\infty}^{\infty} e^{3|t-1|} e^{-j\omega t} dt \!</math>
+
<math> = \int_{-\infty}^{\infty} e^{2|t-1|} e^{-j\omega t} dt \!</math>
 +
 
 +
<math> = \int_{1}^{\infty} e^{2|t-1|} e^{-j\omega t} dt \!</math> + <math> \int_{-\infty}^{1} e^{2|t-1|} e^{-j\omega t} dt \!</math>

Revision as of 16:27, 7 October 2008

Fourier Transform

Signal: x(t) = $ e^{3|t-1|} $

$ X(j \omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \! $

$ = \int_{-\infty}^{\infty} e^{2|t-1|} e^{-j\omega t} dt \! $

$ = \int_{1}^{\infty} e^{2|t-1|} e^{-j\omega t} dt \! $ + $ \int_{-\infty}^{1} e^{2|t-1|} e^{-j\omega t} dt \! $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman