# Thread: Inverse Fourier Transform Hellp

1. ## 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!

2. If:

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

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)
$

and this is the same results obtained by the integration above.