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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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Suppose we have a signal:
 
Suppose we have a signal:
  
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:<math>X(w) = \frac{e^{-jw}}{2+jw} \,</math>
 
:<math>X(w) = \frac{e^{-jw}}{2+jw} \,</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 11:31, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


Suppose we have a signal:

$ e^{-2(t-1)}u(t-1)\, $

The formula of Fourier Transform is:

$ X(w) = \int_{-\infty}^{ \infty} x(t)e^{-jwt}dt\, $

Substituting:

$ X(w) = \int_{-\infty}^{ \infty} e^{-2(t-1)}u(t-1)e^{-jwt}dt\, $

From the step function, the range becomes 1 to $ \infty $, so the equation becomes:

$ X(w) = \int_{1}^{ \infty} e^{-2(t-1)}e^{-jwt}dt\, $
$ X(w) = \int_{1}^{ \infty} e^{2-(2+jw)t}dt\, $

Integrating yields:

$ X(w) = {\left. -\frac{e^{2-(2+jw)t}}{2+jw} \right]_{1}^{\infty}}\, $
$ X(w) = 0 - -\frac{e^{2-(2+jw)}}{2+jw} \, $
$ X(w) = \frac{e^{2-2-jw}}{2+jw} \, $
$ X(w) = \frac{e^{-jw}}{2+jw} \, $

Back to Practice Problems on CT Fourier transform

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn