(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...)
 
 
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[[Category:problem solving]]
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[[Category:ECE301]]
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[[Category:ECE]]
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[[Category:Fourier transform]]
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[[Category:inverse Fourier transform]]
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[[Category:signals and systems]]
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== Example of Computation of inverse Fourier transform (CT signals) ==
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A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]]
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Let x(t)= <math>cos(t)</math>
 
Let x(t)= <math>cos(t)</math>
  
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<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>
  
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<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{j\omega t}d\omega</math>
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<math>x(t)=\frac{1}{2\pi}cos (t)\int_{-\infty}^{\infty}e^{j\omega t}d\omega</math>
  
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<math>x(t)=\frac{1}{2\pi}cos (t){\left.\frac{e^{j\omega t}}{jt}\right]_{-\infty}^{\infty}}</math>
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<math>x(t)=\frac{1}{2j\pi t}cos (t){\left.e^{j\omega t}\right]_{-\infty}^{\infty}}</math>
  
  
 
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]
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<math>x(t) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt</math>
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<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
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<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
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<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>
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<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>
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<math>X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math>
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<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\infty}^{\infty}}</math>
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<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\pi}^{\pi}}</math>
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<math>X(\omega)=0</math>
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Latest revision as of 12:50, 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

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