# 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} )\$

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.

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. You got the transform of the cosine right. However, regarding the time shift property, you still have some mistake in it. Try first to identify what is $t_0$ equal to in this case.
Instructor's hint: this has to do with cascading a time shift and a time scaling. Recall that the order of the operation is relevant.-pm