Inverse FT of a rectangular spectrum

Given: $X(j\omega) = 1$ for $|\omega| < \omega_c$, and $0$ otherwise (an ideal low-pass spectrum).

Inverse FT formula:

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

Working:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega = \frac{1}{2\pi}\left[\frac{e^{j\omega t}}{jt}\right]_{-\omega_c}^{\omega_c} \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} \\ &= \frac{1}{2\pi}\cdot\frac{2j\sin(\omega_c t)}{jt} = \frac{\sin(\omega_c t)}{\pi t} \end{aligned}$$

Answer:

$$x(t) = \frac{\sin(\omega_c t)}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\left(\frac{\omega_c t}{\pi}\right)$$

Properties of the result:

• At $t = 0$ (by L'Hospital's rule): $x(0) = \frac{\omega_c}{\pi}$, which equals $\frac{1}{2\pi}\times$ (area of the spectrum $2\omega_c$).
• Zeros at $t = \pm\frac{\pi}{\omega_c}, \pm\frac{2\pi}{\omega_c}, \dots$; main lobe width $\frac{2\pi}{\omega_c}$.
• A wider band ($\omega_c$ larger) gives a taller, narrower sinc pulse.
• $x(t)$ is non-zero for $t<0$, so the ideal low-pass filter (whose impulse response this is) is non-causal and cannot be built exactly.

 X(jw)                  x(t)
 1 +-------+       wc/pi ^
   |       |            .|.
---+---+---+--> w       . | .
 -wc   0  wc      .  . .  |  . .  .
                ----'-----+-----'-----> t
                    -pi/wc   pi/wc