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 $0$?
Best Answer
Seems I answered my own question.