---
title: "FT of rectangular pulse and constant; duality"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-rect-pulse-constant-duality"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-rect-pulse-constant-duality
site: BCT Engineering
---

# FT of rectangular pulse and constant; duality

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2080 Asoj, 4+4 marks
- Video: [Fourier Transform of Basic Signals (Rectangular Function)](https://www.youtube.com/watch?v=M10lkIuJmTs) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-rect-pulse-constant-duality
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find the Fourier transform of continuous time rectangular pulse and constant amplitude A. State and prove duality property of continuous time Fourier transform with suitable example.

## Short answer

Rectangular pulse A·rect(t/τ) has FT Aτ Sa(ωτ/2); constant A gives 2πA δ(ω). Duality: if x(t) ↔ X(ω) then X(t) ↔ 2π x(−ω). A sinc pulse in time thus yields a bandlimited rectangular spectrum.

## Answer

#### 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 $\omega = \pm 2\pi k/\tau$, $k = 1, 2, \dots$
```text
 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).

## Check yourself

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

1. The FT of a pulse of amplitude A and width τ centred at 0 is:
   A. A τ Sa(ωτ/2)
   B. A Sa(ωτ)
   C. 2A τ Sa(ωτ)
   D. A τ/ (jω)
   Answer: A. Integrating gives (2A/ω) sin(ωτ/2) = Aτ sin(ωτ/2)/(ωτ/2). (Key point: Rectangular pulse FT)

2. What are the peak value and first zeros of the pulse spectrum?
   A. Peak A, zeros at ±π/τ
   B. Peak Aτ (pulse area), zeros at ±2πk/τ
   C. Peak 2A, zeros at ±τ
   D. Peak Aτ, zeros at ±τ/2π
   Answer: B. X(0) equals the area Aτ, and sin(ωτ/2) = 0 at ω = 2πk/τ. (Key point: Peak and zeros)

3. The Fourier transform of a constant A is:
   A. A δ(ω)
   B. 2π A δ(ω)
   C. A/(jω)
   D. π A δ(ω)
   Answer: B. The inverse transform of 2πA δ(ω) is A, so A <-> 2πA δ(ω). (Key point: FT of constant)

4. Since A rect(t/τ) <-> Aτ Sa(ωτ/2), duality says Aτ Sa(tτ/2) <-> ?
   A. A rect(ω/τ)
   B. 2π A rect(ω/τ)
   C. Aτ rect(ω)
   D. δ(ω)
   Answer: B. Duality gives 2π times the original time function with ω as variable (even pulse), 2πA rect(ω/τ). (Key point: Duality with sinc)

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