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Math Help - Inverse Fourier Transform Hellp

  1. #1
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    Inverse Fourier Transform Hellp

    I'm trying to find the inverse fourier transform of:


    P(f)=
    1/2*W for 0<=|f|<=f1

    (1/(4*W))*[1+cos((pi/(2*W*rho))*(|f|-W*(1-rho)))] for f1 <= |f|< (2*W-f1)

    0 for |f| >= (2*W - f1)

    How would I attempt to begin this?

    Thanks!
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
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    If:

    f(s)=\begin{cases} f_1(s) & |s|\leq r_1 \\<br />
                     f_2(s) & r_1\leq |s|\leq r_2 \\<br />
                     0 & |s|> r_2 \end{cases}

    Then would the inverse Fourier Transform be:

    F^{-1}\left\{f(s)\right\}=\frac{1}{\sqrt{2 \pi}}\left(\int_{-r_2}^{-r_1} f_2(s)e^{- i x s}ds+\int_{-r_1}^{r_1} f_1(s)e^{- i s x}ds+\int_{r_1}^{r_2} f_2(s)e^{- i s x} ds\right)

    I think so because Mathematica can calculate inverse transforms. I used the code below with the indicated values for the parameters:

    Code:
    In[23]:= \[Rho] = 1; 
    w = 3; 
    f1 = 1; 
    p[f_] := Piecewise[
           {{(1/2)*w, Abs[f] < f1}, 
             {(1/(4*w))*(1 + Cos[Pi/(2*w*\[Rho])]*
                      (Abs[f] - w*(1 - \[Rho]))), 
               f1 <= Abs[f] <= 2*w - f1}, 
             {0, Abs[f] > 2*w - f1}}]; 
    FullSimplify[InverseFourierTransform[
         p[f], f, x]]
    and obtained:

    \frac{1}{6 \sqrt{2 \pi } x^2}
    \sin x \left(2 \left(9+\sqrt{3}\right) x+\left(2+5 \sqrt{3}\right) x (\cos (2 x)+\cos(4 x))-\sqrt{3} (\sin(2 x)+\sin(4x))\right)<br />

    and this is the same results obtained by the integration above.
    Last edited by shawsend; September 27th 2009 at 12:17 PM. Reason: added Mathematica code
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