(New page: Let x(t)= <math>cos(t)</math> Then <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega</math> <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(t)e^{j\omega...)
 
 
(2 intermediate revisions by one other user not shown)
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]]
 +
----
 +
 
Let x(t)= <math>cos(t)</math>
 
Let x(t)= <math>cos(t)</math>
  
Line 8: Line 18:
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(t)e^{j\omega t}d\omega</math>
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(t)e^{j\omega t}d\omega</math>
  
−
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{j\omega t}d\omega</math>
+
<math>x(t)=\frac{1}{2\pi}cos (t)\int_{-\infty}^{\infty}e^{j\omega t}d\omega</math>
  
 +
<math>x(t)=\frac{1}{2\pi}cos (t){\left.\frac{e^{j\omega t}}{jt}\right]_{-\infty}^{\infty}}</math>
 +
 +
<math>x(t)=\frac{1}{2j\pi t}cos (t){\left.e^{j\omega t}\right]_{-\infty}^{\infty}}</math>
  
  
 
----
 
----
−
 
+
[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]
−
 
+
−
 
+
−
 
+
−
<math>x(t) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt</math>
+
−
 
+
−
<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
+
−
 
+
−
<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
+
−
 
+
−
<math>X(\omega)={\left. \frac{e^{jt(1-\omega)}}{j(1-\omega)}\right]_{-\infty}^{\infty}} + {\left. \frac{e^{-jt(1+\omega)}}{-j(1+\omega)}\right]_{-\infty}^{\infty}}</math>
+
−
 
+
−
<math>X(\omega)={\left.\frac{(1+\omega)e^{jt(1-\omega)}-(1-\omega)e^{-jt(1+\omega)}}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math>
+
−
 
+
−
<math>X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math>
+
−
 
+
−
<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\infty}^{\infty}}</math>
+
−
 
+
−
<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\pi}^{\pi}}</math>
+
−
 
+
−
<math>X(\omega)=0</math>
+

Latest revision as of 11:50, 16 September 2013

Example of Computation of inverse Fourier transform (CT signals)

A practice problem on CT Fourier transform

Let x(t)= $ cos(t) $


Then

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

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

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

$ x(t)=\frac{1}{2\pi}cos (t){\left.\frac{e^{j\omega t}}{jt}\right]_{-\infty}^{\infty}} $

$ x(t)=\frac{1}{2j\pi t}cos (t){\left.e^{j\omega t}\right]_{-\infty}^{\infty}} $


Back to Practice Problems on CT Fourier transform

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=HW5.3_Xujun_Huang_ECE301Fall2008mboutin&oldid=58496"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn