| Line 22: | Line 22: | ||
Therefore, | Therefore, | ||
| − | <math>\ | + | <math>\mathcal X (\omega)=e^{j\omega \frac{\pi}{12}}2\pi \delta(\omega-2\pi k)</math> |
--[[User:Cmcmican|Cmcmican]] 20:52, 21 February 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 20:52, 21 February 2011 (UTC) | ||
| Line 29: | Line 29: | ||
=== Answer 2 === | === Answer 2 === | ||
| − | + | I'll try this again, using my new answer from the previous problem, and correcting my time shifting property. | |
| + | |||
| + | <math>\mathfrak{F}\Bigg(s(t-t_0)\Bigg)=e^{-j\omega t_0}\mathfrak{F}\Bigg(x(t)\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> | ||
| + | |||
=== Answer 3 === | === Answer 3 === | ||
Write it here. | Write it here. | ||
---- | ---- | ||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
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) $
Answer 3
Write it here.
