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 )\ $

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Answer 1

Guess $ \chi(\omega)=2\pi\delta(\omega-2\pi k)\, $ such that $ \mathfrak{F}^{-1}=e^{jk2\pi t} $


$ \mathfrak{F}(2\pi\delta(\omega-2\pi k))=\frac{1}{2\pi}\int_{-\infty}^\infty2\pi\delta(\omega-2\pi k)e^{j\omega t}d\omega=e^{jk2\pi t} $

Therefore,$ \chi(\omega)=2\pi\delta(\omega-2\pi k)\, $

--Cmcmican 20:47, 21 February 2011 (UTC)

Instructor's comments: Take a look at your answer: it depends on k. However, the input does not depend on k. By the way, you can use the "mathcal" font to produce the curly X. Like this: $ {\mathcal X} $. And if you use the inline class, it is aligned with the line like this: $ {\mathcal X} $. -pm

Answer 2

I'll try this again, using the formula for Fourier transform of a periodic signal.

$ \mathfrak{F}\Bigg(\frac{1}{2}e^{j2\pi t}+\frac{1}{2}e^{-j2\pi t}\Bigg)=\frac{2\pi}{2}\delta(\omega-2\pi)+\frac{2\pi}{2}\delta(\omega+\pi) $

Therefore $ \mathcal X (\omega) =\pi\delta(\omega-2\pi)+\pi\delta(\omega+2\pi) $

--Cmcmican 17:38, 23 February 2011 (UTC)

Answer 3

Write it here.

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