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Thread: Fourier Transforms

  1. #1
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    Fourier Transforms

    1. Obtain from first principles, the Fourier transform of the function:

    f(t) = 1, |t|<1 and 0, |t|>1


    - I'm unsure of where to begin with this question, is it by using the formula (picture attached) and I don't know how to use the fact that |t|<1 or |t|>1

    2. Find the fourier transform of

    f(t) = u(t)e^(-at), a>0


    - I know this can be read easily off tables but I tried using first principles and I found this statement (picture in red). A little off-topic but why is this statement true?

    Any help would be greatly appreciated! Thanks in advance
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  2. #2
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    Quote Originally Posted by blueturkey View Post
    1. Obtain from first principles, the Fourier transform of the function:

    f(t) = 1, |t|<1 and 0, |t|>1


    - I'm unsure of where to begin with this question, is it by using the formula (picture attached) and I don't know how to use the fact that |t|<1 or |t|>1
    Yes, the formula F(\omega) = \int_{-\infty}^\infty\!\!f(t)e^{-i\omega t}dt is the place to start. The function f is 1 in the interval -1\leqslant t\leqslant1, and zero outside that interval. So the integral becomes F(\omega) = \int_{-1}^1\!\!e^{-i\omega t}dt, which you should be able to do.

    Quote Originally Posted by blueturkey View Post
    2. Find the fourier transform of

    f(t) = u(t)e^(-at), a>0


    - I know this can be read easily off tables but I tried using first principles and I found this statement (picture in red). A little off-topic but why is this statement true?
    Not sure about this, but I suspect that u(t) is meant to be the Heaviside function, defined by u(t) = 1 if t>0, and u(t) = 0 if t<0. If so then this function f(t) is zero when t<0, so the integral over negative values of t is going to be zero.
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  3. #3
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    thanks!
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