Revision as of 12:47, 24 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} )\ $

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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)

TA's comments: You're almost there. Be careful with the sign when using the time shift property of the Fourier transform. Here $ t_0=-\frac{\pi}{12} $. So this will yield a multiplication by $ e^{j\omega \frac{\pi}{12}} $ in the frequency domain and not $ e^{-j\omega \frac{\pi}{12}} $.

Answer 3

Write it here.

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