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| − | = Practice Question on Computing the Fourier Transform of a Continuous-time Signal = | + | = [[:Category:Problem_solving|Practice Question]] on Computing the Fourier Transform of a Continuous-time Signal = |
Compute the Fourier transform of the signal | Compute the Fourier transform of the signal | ||
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=== Answer 2 === | === Answer 2 === | ||
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| + | I'll try this again, using the formula for Fourier transform of a periodic signal. | ||
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| + | <math>\mathfrak{F}\Bigg(\frac{1}{2}e^{j2\pi t}+\frac{1}{2}e^{-j2\pi t}\Bigg)=\frac{2\pi}{2}\delta(\omega-2\pi)+\frac{2\pi}{2}\delta(\omega+\pi)</math> | ||
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| + | Therefore <math class="inline">\mathcal X (\omega) =\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi)</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 17:38, 23 February 2011 (UTC) | ||
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=== Answer 3 === | === Answer 3 === | ||
Write it here. | Write it here. | ||
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Latest revision as of 10:25, 11 November 2011
Contents
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 )\ $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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)
- Instructor's comments: Take a look at your answer: it depends on k. However, the input does not depend on k. By the way, you can use the "mathcal" font to produce the curly X. Like this: $ {\mathcal X} $. And if you use the inline class, it is aligned with the line like this: $ {\mathcal X} $. -pm
Answer 2
I'll try this again, using the formula for Fourier transform of a periodic signal.
$ \mathfrak{F}\Bigg(\frac{1}{2}e^{j2\pi t}+\frac{1}{2}e^{-j2\pi t}\Bigg)=\frac{2\pi}{2}\delta(\omega-2\pi)+\frac{2\pi}{2}\delta(\omega+\pi) $
Therefore $ \mathcal X (\omega) =\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi) $
--Cmcmican 17:38, 23 February 2011 (UTC)
Answer 3
Write it here.
