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Math Help - Fourier Transform

  1. #1
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    Fourier Transform

    Hi there,

    Just wondering if anyone can give me some clues as to how to go about getting the fourier transform using the formula integral from - infinity to + infinity f(t) e^-iwt dt

    f(t) = { 1 - t/2 : 0 =< t <= 2

    1 + t/2 : -2 =< t <= 0

    0 : |t| > 2.


    i have got as far as splitting it into four integrals and started integration by parts on 1/2 * integral between -2 and 0 of t.e^-iwt dt. it gets messy from there.. Just wondering if integration by parts is the way to go. I think i should end with an answer containing a few sinc functions.

    Many thanks
    Jack
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  2. #2
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    Quote Originally Posted by JackLondon View Post
    Hi there,

    Just wondering if anyone can give me some clues as to how to go about getting the fourier transform using the formula integral from - infinity to + infinity f(t) e^-iwt dt

    f(t) = { 1 - t/2 : 0 =< t <= 2

    1 + t/2 : -2 =< t <= 0

    0 : |t| > 2.


    i have got as far as splitting it into four integrals and started integration by parts on 1/2 * integral between -2 and 0 of t.e^-iwt dt. it gets messy from there.. Just wondering if integration by parts is the way to go. I think i should end with an answer containing a few sinc functions.

    Many thanks
    Jack
    I can't imagine why that would "get messy". To integrate \int_{-2}^0 t e^{i\omega t}dt let u= t, dv= e^{i\omega t} so that du= dt and v= \frac{-i}{\omega}e^{i\omega t} so the integral becomes \frac{-2i}{\omega}e^{-2i\omega}+ \int_{-2}^0 e^{i\omega t}dt = \frac{-2i}{\omega}e^{-2i\omega}- \frac{i}{\omega}(1- e^{-2i\omega})
    Last edited by mr fantastic; February 13th 2009 at 07:18 PM. Reason: Fixed the latex tags
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I can't imagine why that would "get messy". To integrate \int_{-2}^0 t e^{i\omega t}dt let u= t, dv= e^{i\omega t} so that du= dt and v= \frac{-i}{\omega}e^{i\omega t} so the integral becomes \frac{-2i}{\omega}e^{-2i\omega}+ \int_{-2}^0 e^{i\omega t}dt = \frac{-2i}{\omega}e^{-2i\omega}- \frac{i}{\omega}(1- e^{-2i\omega})

    Thanks HallsofIvy

    It was 1/2 * integral between -2 and 0 of t.e^-iwt dt. You missed the minus sign.

    Thanks anyway. dv = e^-iwt dt I was getting v = e^-iwt / -iw

    Correct me if I am wrong.

    So I assume integration by parts is the best way to get the Fourier Transform. The Maths Lecturer mentioned Partial Fractions.

    Thanks again
    Jack
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