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Math Help - Fourier Transform

  1. #1
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    Fourier Transform

    Fourier Transform-vskh4.jpg
    For part i, I found that the answer is
    F(w) = (4sinw - 4wcosw)/w^3

    For part ii, I consider
    [F(x)]^2 which is equal to 16[(xcosx - sinx)^2]/x^6
    and then i tried to use the Convolution relation
    [F(x)]^2 = Fourier Transform of [ f*f] = Fourier Transform of (integrate from -infinity to infinity[f(t-k)f(k)]dk)
    However, the integration is not converge, so I think there must be some mistakes.
    Please gives me some idea on this........
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  2. #2
    MHF Contributor chisigma's Avatar
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    In i) is requested the computation of the Fourier Transform of...

    f(t) = \begin{cases}<br />
0, t < -1 \\[3pt]<br />
1-t^{2}, -1<t<1 \\[3pt]<br />
0, t >1 \end{cases} (1)

    ... and to achieve that we can use the 'little nice formula'...

    \displaystyle \int_{a}^{b} f(t)\ e^{-s t}\ dt = \sum_{n=0}^{\infty} \frac{f^{(n)} (a)\ e^{-s a} - f^{(n)}(b)\ e^{-s b}}{s^{n+1}} (2)

    Applying (2) we find that...

    \displaystyle \mathcal{F} \{f(t)\} = [\int _{-1}^{1} f(t)\ e^{-s t}\ dt]_{s= i \omega}= 4\ [\frac{s\ \cosh s}{s^{2}} - \frac{\sinh s}{s^{3}}]_{s= i \omega}

    \displaystyle = 4\ [\frac{s\ \cosh s - \sinh s}{s^{3}}]_{s=i \omega} (3)

    Kind regards

    \chi \sigma
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