(New page: == Specify a signal x(t) == <math>x(t)=cos(8 \pi x)e^{-x^{2}}</math>)
 
 
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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:signals and systems]]
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== Example of Computation of Fourier transform of a CT SIGNAL ==
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A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]]
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----
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== Specify a signal x(t) ==
 
== Specify a signal x(t) ==
  
  
<math>x(t)=cos(8 \pi x)e^{-x^{2}}</math>
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<math>x(t)=cos(8 \pi t)e^{-t^{2}}</math>
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== Fourier Transform of x(t) ==
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:<math>\begin{align} X(\omega)  &=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t-t^2}-e^{-j8\pi t-t^2}}{2}e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty}\frac{e^{t(j8\pi -t)}-e^{-t(j8\pi +t)}}{2}e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}-e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt
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\\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}}{2}e^{-j\omega t}dt  -\int_{-\infty}^{\infty}\frac{e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt
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\end{align}</math>
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----
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 11:25, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


Specify a signal x(t)

$ x(t)=cos(8 \pi t)e^{-t^{2}} $

Fourier Transform of x(t)

$ \begin{align} X(\omega) &=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t-t^2}-e^{-j8\pi t-t^2}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{t(j8\pi -t)}-e^{-t(j8\pi +t)}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}-e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}}{2}e^{-j\omega t}dt -\int_{-\infty}^{\infty}\frac{e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt \end{align} $

Back to Practice Problems on CT Fourier transform

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