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<math>(te^{-4t}\sin{6\pi t})u(t)</math>
+
<math>(t e^{-4t} \sin{6 \pi t}) u(t)</math>
  
  
 
== The Fourier Transform ==
 
== The Fourier Transform ==
 +
 +
 +
<math>X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt</math>
 +
 +
 +
<math>X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt</math>

Revision as of 14:15, 7 October 2008

The Signal

$ (t e^{-4t} \sin{6 \pi t}) u(t) $


The Fourier Transform

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


$ X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn