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Therefore <math class="inline">\mathcal X (\omega) =e^{-j\omega \frac{\pi}{12}}\Bigg(\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi)\Bigg)</math> | Therefore <math class="inline">\mathcal X (\omega) =e^{-j\omega \frac{\pi}{12}}\Bigg(\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi)\Bigg)</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 17:43, 23 February 2011 (UTC) | ||
=== Answer 3 === | === Answer 3 === | ||
Revision as of 12:43, 23 February 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+\frac{\pi}{12} )\ $
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
Use answer to previous practice problem and the time shifting property.
$ \mathfrak{F}\Bigg(s(t-t_0)\Bigg)=e^{j\omega t_0}\mathfrak{F}\Bigg(x(t)\Bigg) $
Therefore,
$ \mathcal X (\omega)=e^{j\omega \frac{\pi}{12}}2\pi \delta(\omega-2\pi k) $
--Cmcmican 20:52, 21 February 2011 (UTC)
- TA's comments: In the time shift property of the Fourier transform that you provided, it should be $ e^{-j\omega t_0} $ and not $ e^{j\omega t_0} $. Another thing is that the transform of a cosine should yield only two deltas in the frequency domain.
Answer 2
I'll try this again, using my new answer from the previous problem, and correcting my time shifting property.
$ \mathfrak{F}\Bigg(s(t-t_0)\Bigg)=e^{-j\omega t_0}\mathfrak{F}\Bigg(x(t)\Bigg) $
Therefore $ \mathcal X (\omega) =e^{-j\omega \frac{\pi}{12}}\Bigg(\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi)\Bigg) $
--Cmcmican 17:43, 23 February 2011 (UTC)
Answer 3
Write it here.
