(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Computing the Fourier Transform of a Continuous-time Signal = Compute the Fourier transform o...)
 
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=== Answer 1  ===
 
=== Answer 1  ===
Write it here.
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Guess <math>\chi(\omega)=2\pi\delta(\omega-2\pi k)\,</math> such that <math>\mathfrak{F}^{-1}=e^{jk2\pi t}</math>
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check:
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<math>\mathfrak{F}(2\pi\delta(\omega-2\pi k))=\frac{1}{2\pi}\int_{-\infty}^\infty2\pi\delta(\omega-2\pi k)e^{j\omega t}d\omega=e^{jk2\pi t}</math>
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Therefore,<math>\chi(\omega)=2\pi\delta(\omega-2\pi k)\,</math>
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--[[User:Cmcmican|Cmcmican]] 20:47, 21 February 2011 (UTC)
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=== Answer 2  ===
 
=== Answer 2  ===
 
Write it here.
 
Write it here.

Revision as of 16:47, 21 February 2011


Practice Question on Computing the Fourier Transform of a Continuous-time Signal

Compute the Fourier transform of the signal

$ x(t) = \cos (2 \pi t )\ $


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Answer 1

Guess $ \chi(\omega)=2\pi\delta(\omega-2\pi k)\, $ such that $ \mathfrak{F}^{-1}=e^{jk2\pi t} $

check:

$ \mathfrak{F}(2\pi\delta(\omega-2\pi k))=\frac{1}{2\pi}\int_{-\infty}^\infty2\pi\delta(\omega-2\pi k)e^{j\omega t}d\omega=e^{jk2\pi t} $

Therefore,$ \chi(\omega)=2\pi\delta(\omega-2\pi k)\, $

--Cmcmican 20:47, 21 February 2011 (UTC)


Answer 2

Write it here.

Answer 3

Write it here.


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