FT of rectangular pulse and constant; duality

FT of a rectangular pulse

Take a pulse of amplitude $A$ and width $\tau$, centred at $t = 0$:

$$x(t) = A\,\mathrm{rect}(t/\tau) = \begin{cases}A, & |t| < \tau/2\\ 0, & \text{otherwise}\end{cases}$$ $$\begin{aligned} X(\omega) &= \int_{-\tau/2}^{\tau/2}Ae^{-j\omega t}dt = A\left[\frac{e^{-j\omega t}}{-j\omega}\right]_{-\tau/2}^{\tau/2} = \frac{A}{j\omega}\left(e^{j\omega\tau/2} - e^{-j\omega\tau/2}\right) \\ &= \frac{2A}{\omega}\sin\frac{\omega\tau}{2} = A\tau\,\frac{\sin(\omega\tau/2)}{\omega\tau/2} \end{aligned}$$ $$\boxed{X(\omega) = A\tau\,\mathrm{Sa}\!\left(\frac{\omega\tau}{2}\right)}, \qquad \mathrm{Sa}(x) = \frac{\sin x}{x}$$

The peak is $X(0) = A\tau$ (the area of the pulse), and the zeros are at $\omega = \pm 2\pi k/\tau$, $k = 1, 2, \dots$

 x(t)               X(w)
   A  ______          A*tau    _
     |      |                 / \
 ____|      |____    __   _  /   \  _   __
    -tau/2  tau/2      \_/ \/     \/ \_/
                   ---------+---------- w
                  -2pi/tau  0  2pi/tau

FT of a constant A

A constant is not absolutely integrable, so start from the inverse transform of an impulse:

$$F^{-1}\{2\pi A\,\delta(\omega)\} = \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi A\,\delta(\omega)e^{j\omega t}d\omega = A$$ $$\boxed{A \xleftrightarrow{F} 2\pi A\,\delta(\omega)}$$

This is also the limit of the rectangular pulse as $\tau \to \infty$: the sinc becomes taller and narrower, and its area stays $2\pi A$.

Duality property

Statement: if $x(t) \leftrightarrow X(\omega)$, then $X(t) \leftrightarrow 2\pi x(-\omega)$.

Proof: from the inverse transform,

$$2\pi x(t) = \int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega$$

Put $t \to -t$:

$$2\pi x(-t) = \int_{-\infty}^{\infty}X(\omega)e^{-j\omega t}d\omega$$

Then interchange the symbols $t$ and $\omega$:

$$2\pi x(-\omega) = \int_{-\infty}^{\infty}X(t)e^{-j\omega t}dt = F\{X(t)\}$$

Example 1 (from the parts above): $\delta(t) \leftrightarrow 1$. By duality, $1 \leftrightarrow 2\pi\delta(-\omega) = 2\pi\delta(\omega)$. This matches the constant result with $A = 1$.

Example 2 (sinc in time): the rectangle gives $A\,\mathrm{rect}(t/\tau) \leftrightarrow A\tau\,\mathrm{Sa}(\omega\tau/2)$. Since the rectangle is even, duality gives

$$A\tau\,\mathrm{Sa}\!\left(\frac{t\tau}{2}\right) \xleftrightarrow{F} 2\pi A\,\mathrm{rect}(\omega/\tau)$$

A sinc pulse in time has a rectangular (band-limited) spectrum. With $\tau = 2W$ and $A = \frac{1}{2\pi}$, this becomes the ideal low-pass filter pair $\frac{\sin Wt}{\pi t} \leftrightarrow 1$ for $|\omega| < W$ (and 0 otherwise).