---
title: "DTFT of a rectangular pulse"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-rectangular-pulse"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-rectangular-pulse
site: BCT Engineering
---

# DTFT of a rectangular pulse

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Chaitra, 4 marks
- Video: [DT Fourier Transform-Rectangular Pulse](https://www.youtube.com/watch?v=jE07a66rkRQ) — Darryl Morrell
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-rectangular-pulse
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find the Fourier transform of the given signal. x[n] = 1 for −N₁ ≤ n ≤ N₁; = 0 elsewhere

## Short answer

The DTFT of the rectangular pulse x[n]=1 for |n|≤N₁ is X(e^{jω}) = sin(ω(N₁+½)) / sin(ω/2). It is real, even, periodic 2π, with peak 2N₁+1 at ω=0. Derived by summing geometric series from −N₁ to N₁.

## Answer

The signal is a rectangular pulse of $2N_1 + 1$ samples, all equal to 1, centred at $n = 0$.

#### DTFT
$$
X(e^{j\omega}) = \sum_{n=-N_1}^{N_1}e^{-j\omega n}
$$
Put $m = n + N_1$ ($m = 0$ to $2N_1$) and sum the geometric series:
$$
\begin{aligned}
X(e^{j\omega}) &= e^{j\omega N_1}\sum_{m=0}^{2N_1}e^{-j\omega m} = e^{j\omega N_1}\frac{1 - e^{-j\omega(2N_1+1)}}{1 - e^{-j\omega}} \\
&= \frac{e^{j\omega(N_1+\frac{1}{2})} - e^{-j\omega(N_1+\frac{1}{2})}}{e^{j\omega/2} - e^{-j\omega/2}}
\end{aligned}
$$
The second line comes from multiplying the numerator and denominator by $e^{j\omega/2}$. Using $e^{j\theta} - e^{-j\theta} = 2j\sin\theta$:
$$
\boxed{X(e^{j\omega}) = \frac{\sin\left(\omega\left(N_1 + \frac{1}{2}\right)\right)}{\sin(\omega/2)}}
$$
At $\omega = 0$ the value is $2N_1 + 1$ (by L'Hôpital, or as the number of samples).

#### Remarks
- $X(e^{j\omega})$ is real and even, because $x[n]$ is real and even.
- It is the DT counterpart of the sinc function and is periodic with period $2\pi$.
- The first zeros are at $\omega = \pm\frac{2\pi}{2N_1+1}$. A wider pulse gives a narrower main lobe.
- **Example:** for $N_1 = 2$, $X(e^{j\omega}) = \frac{\sin(2.5\omega)}{\sin(0.5\omega)}$, with peak 5 and first zeros at $\pm 2\pi/5$.

## Check yourself

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

1. How is the DTFT of the pulse x[n] = 1 for -N1 <= n <= N1 found?
   A. Summing a geometric series of 2N1+1 terms e^(-jωn)
   B. Integrating e^(-jωt) over the pulse width
   C. Using the time-shift property only
   D. Using Parseval's relation
   Answer: A. The DTFT is a finite sum of e^(-jωn), a geometric series after substituting m = n + N1. (Key point: Method)

2. The DTFT of the pulse is:
   A. sin(ωN1)/sin(ω)
   B. sin(ω(N1 + 1/2))/sin(ω/2)
   C. sin(ωN1)/ω
   D. sin(ω(2N1+1))/sin(ω)
   Answer: B. Summing the series and multiplying numerator and denominator by e^(jω/2) gives sin(ω(N1+1/2))/sin(ω/2). (Key point: Result)

3. What is X(e^(jω)) at ω = 0?
   A. N1
   B. 2N1
   C. 2N1 + 1
   D. 0
   Answer: C. At ω = 0 the sum counts the 2N1+1 samples, each equal to 1. (Key point: Value at zero)

4. For this pulse, what happens to the main lobe as the pulse gets wider?
   A. It gets wider
   B. It gets narrower, with first zeros at ±2π/(2N1+1)
   C. It stays unchanged
   D. It becomes imaginary
   Answer: B. Larger N1 moves the first zero 2π/(2N1+1) closer to the origin, so the main lobe narrows. (Key point: Width and zeros)

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