Duality property of CTFT

Statement

If $x(t) \xleftrightarrow{F} X(\omega)$, then a time function with the same shape as $X$ has a transform with the shape of $x$:

$$X(t) \xleftrightarrow{F} 2\pi\,x(-\omega)$$

For an even $x(t)$ this simplifies to $X(t) \leftrightarrow 2\pi x(\omega)$.

Proof

Start from the inverse transform:

$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \;\Rightarrow\; 2\pi x(t) = \int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega$$

Replace $t$ by $-t$:

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

Now exchange the names of the variables $t$ and $\omega$:

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

This proves $X(t) \leftrightarrow 2\pi x(-\omega)$.

Example

A rectangular pulse $x(t) = 1$ for $|t| < T_1$ has $X(\omega) = \frac{2\sin\omega T_1}{\omega}$. By duality, with $x$ even:

$$\frac{2\sin T_1 t}{t} \xleftrightarrow{F} 2\pi\,x(\omega) = \begin{cases}2\pi, & |\omega| < T_1\\ 0, & |\omega| > T_1\end{cases}$$

Dividing by $2\pi$ and writing $W$ for $T_1$:

$$\frac{\sin Wt}{\pi t} \xleftrightarrow{F} \begin{cases}1, & |\omega| < W\\ 0, & |\omega| > W\end{cases}$$

A sinc pulse in time is an ideal low-pass spectrum. Similarly, $\delta(t) \leftrightarrow 1$ gives $1 \leftrightarrow 2\pi\delta(\omega)$.