# Fourier Transform

• Mar 24th 2011, 06:31 AM
progrocklover
Fourier Transform
Attachment 21247
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........
• Mar 25th 2011, 06:38 AM
chisigma
In i) is requested the computation of the Fourier Transform of...

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

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

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

Applying (2) we find that...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$