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Given the inverse rectangular function: $p(t) = \begin{cases}1&\mbox{ if }|t| > a,\\ 0 &\mbox{ if } |t| < a,\end{cases}$

where $a>0$ is real. And using the Fourier transform defined by:

$$F(p(t)) = P(\omega) = \int_{-\infty}^{\infty} p(t) e^{-i\omega t} d\omega,$$

with inverse: $$F^{-1}(P(\omega)) = p(t) = \frac{1}{2 \pi}\int_{-\infty}^{\infty} P(\omega) e^{i\omega t} d\omega.$$

Find and then plot the FT of the inverse rectangular function, $p(t)$, noting the stationary points and intercepts.


My approach has been to use a couple properties and results:

$p(t) = 1 - s(t)$

where $s(t)$ is the normal rectangular pulse function given by: $$s(t) = \begin{cases}0&\mbox{if }|t| > a,\\ 1 &\mbox{if }|t| < a.\end{cases} $$ Using the above definition of the FT, we get: $F([1]) = 2 \pi \delta (\omega)$

where $\delta (\omega)$ is a delta impulse function centered at origin, and

$F([s(t)]) = 2a\cdot \frac{\sin(\omega a)}{\omega a}$

Also, the FT is linear, hence $F([p(t)]) = F[(1)] - F([s(t)])$

Hence:

$F([p(t)]) = 2 \pi \delta (\omega) - 2a\cdot \frac{\sin(\omega a)}{\omega a}$.

Is this okay so far? Is there a simplification of this? how does one plot something like this? What exactly are the 'stationary points' of a function - are they just the points where the derivative WRT $t$ is $0$?

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    $\begingroup$ Your result is correct. You cannot (easily) plot this because the Fourier transform is in fact not a real function but a distribution. $\endgroup$ – Fabian Sep 5 '12 at 15:22
  • $\begingroup$ I see! I guess, without computer simulation, I could still analytically note the points of intercept with horizontal axis - which would just be those points where $\omega a$ is a multiple of $\pi$. Could also note the horizontal axis values of when the function has 0 slope - though, I still have to work out how to find that.. $\endgroup$ – confused Sep 5 '12 at 15:40
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Seems I answered my own question.

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