---
title: "Inverse FT of a rectangular spectrum"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:inverse-ft-rectangular-spectrum"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/inverse-ft-rectangular-spectrum
site: BCT Engineering
---

# Inverse FT of a rectangular spectrum

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2072 Magh, 6 marks
- Video: [Signals and Systems - Inverse Fourier Transform](https://www.youtube.com/watch?v=mq3bR4E-QU8) — UConn HKN
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/inverse-ft-rectangular-spectrum
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find inverse Fourier transform of the rectangular pulse X(jω) = 1 for −ω_c < ω < ω_c; = 0 otherwise.

## Short answer

Apply the inverse FT integral: x(t) = (1/2π)∫_{-ωc}^{ωc} e^{jωt} dω = sin(ωc t)/(π t). The result is x(t) = (ωc/π) sinc(ωc t/π).

## Answer

**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.

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives x(t) from X(jω)?
   A. ∫X(jω)e^(-jωt)dω
   B. (1/2π)∫X(jω)e^(-jωt)dω
   C. 2π∫X(jω)e^(jωt)dω
   D. (1/2π)∫X(jω)e^(jωt)dω
   Answer: D. This is the inverse Fourier transform. (Key point: Formula)

2. What is x(t)?
   A. sin(ωc t)/(π t)
   B. sin(ωc t)/t
   C. 2 sin(ωc t)/(π t)
   D. sin(ωc t)/(2π t)
   Answer: A. (1/2π)(2j sin ωc t)/(jt) = sin(ωc t)/(πt). (Key point: Result)

3. What is x(0)?
   A. ωc
   B. ωc/π
   C. 1/π
   D. 2ωc/π
   Answer: B. Limit of sin(ωc t)/(πt) is ωc/π, equal to the spectrum area 2ωc divided by 2π. (Key point: Value at zero)

4. What does the result say about the ideal low-pass filter?
   A. It is causal
   B. It is unstable but causal
   C. It is non-causal since x(t) is nonzero for t < 0
   D. It is time varying
   Answer: C. The impulse response sin(ωc t)/(πt) extends to negative time, so the filter cannot be built exactly. (Key point: Causality)

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