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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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----
 
==signal ==
 
==signal ==
 
assume
 
assume
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== answer ==
 
== answer ==
  
<math>x(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt</math>
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<math>X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt</math>
  
 
<math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt</math>
 
<math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt</math>
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<math>=\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+jw)}</math>
 
<math>=\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+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 12:31, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


signal

assume

$ x(t) = e^{-5t}u(t) + e^{-4(t-1)}u(t-3)\! $


answer

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

$ =\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt $

$ =\int_{0}^{\infty}e^{-5t} e^{-jwt}dt + \int_{3}^{\infty}e^{-4(t-1)}e^{-jwt}dt $

$ =\frac{e^{-(5+jw)t}}{-(5+jw)}\bigg]^{\infty}_{0}+ \int_{3}^{\infty}e^{-4t}e^{4}e^{-jwt}dt $

$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4*\int_{\infty}^{3}e^{-(4+jw)t}dt $

$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4 * \frac{e^{-(4+jw)t}}{-(4+jw)}\bigg]_3^\infty $

$ =\frac{1}{(5+jw)}+ e^4 * \frac{e^{-(12+3jw)}}{(4+jw)} $

$ =\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+jw)} $


Back to Practice Problems on CT Fourier transform

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Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin