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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 13:43, 23 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+\frac{\pi}{12} )\ $


Share your answers below

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.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang