(New page: <math>X(\omega) = 2\pi e^{4j\omega}\,</math> We kenw that the Fourier transformation of <math>x(t - t_o)\,</math> is equal to <math>e^{-j\omega t_o} X(\omega)\,</math> Using this propert...)
 
 
Line 1: Line 1:
 +
[[Category:problem solving]]
 +
[[Category:ECE301]]
 +
[[Category:ECE]]
 +
[[Category:Fourier transform]]
 +
[[Category:inverse Fourier transform]]
 +
[[Category:signals and systems]]
 +
== Example of Computation of inverse Fourier transform (CT signals) ==
 +
A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]]
 +
----
 +
 +
 
<math>X(\omega) = 2\pi e^{4j\omega}\,</math>
 
<math>X(\omega) = 2\pi e^{4j\omega}\,</math>
  
Line 16: Line 27:
  
 
<math>x(t) = \delta(t+4)\,</math>
 
<math>x(t) = \delta(t+4)\,</math>
 +
 +
 +
----
 +
[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 12:42, 16 September 2013

Example of Computation of inverse Fourier transform (CT signals)

A practice problem on CT Fourier transform



$ X(\omega) = 2\pi e^{4j\omega}\, $

We kenw that the Fourier transformation of $ x(t - t_o)\, $ is equal to $ e^{-j\omega t_o} X(\omega)\, $

Using this property, we can solve the question

$ X(\omega) = 2\pi e^{4j\omega} Y(\omega)\, $ where $ Y(\omega) = 1\, $

$ y(t) = \delta(t)\, $

$ x(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}2\pi e^{4j\omega} Y(\omega) dw\, $

$ x(t) = \int^{\infty}_{-\infty}e^{4j\omega} Y(\omega) dw\, $

$ x(t) = \hat{f} (e^{4j\omega } Y(\omega))\, $

$ x(t) = \delta(t+4)\, $



Back to Practice Problems on CT Fourier transform

Alumni Liaison

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

Dr. Paul Garrett