---
title: "DTFT is periodic with period 2π"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-periodicity-2pi"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-periodicity-2pi
site: BCT Engineering
---

# DTFT is periodic with period 2π

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2076 Baisakh, 4 marks
- Video: [Periodicity of DTFT](https://www.youtube.com/watch?v=xobwqgYmnvY) — Kishore Kashyap
- 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-periodicity-2pi
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Explain with necessary derivation that the discrete time Fourier Transform is periodic in frequency with period 2π.

## Short answer

X(e^{j(ω+2π)}) = Σ x[n] e^{-jωn} e^{-j2πn}. Since e^{-j2πn}=1 for integer n, this equals Σ x[n] e^{-jωn}=X(e^{jω}). Hence the DTFT is periodic with period 2π, because discrete frequencies separated by 2π are indistinguishable.

## Answer

The DTFT of a sequence $x[n]$ is

$$
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n}
$$

**Derivation:** Replace $\omega$ by $\omega + 2\pi$:

$$
\begin{aligned}
X(e^{j(\omega+2\pi)}) &= \sum_{n=-\infty}^{\infty} x[n]\,e^{-j(\omega+2\pi)n} \\
&= \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n}\,e^{-j2\pi n}
\end{aligned}
$$

Since $n$ is an integer, $e^{-j2\pi n} = \cos(2\pi n) - j\sin(2\pi n) = 1$. Therefore

$$
X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n} = X(e^{j\omega})
$$

More generally $X(e^{j(\omega+2\pi k)}) = X(e^{j\omega})$ for any integer $k$. So the DTFT is periodic in $\omega$ with period $2\pi$.

**Why it happens:** The discrete-time exponentials $e^{j\omega n}$ and $e^{j(\omega+2\pi)n}$ are the same sequence, because $n$ takes only integer values. Frequencies $2\pi$ apart cannot be told apart in discrete time. This is unlike continuous time, where $e^{j\omega t}$ are all different for different $\omega$.

**Consequences:**

- Only one period, usually $-\pi \le \omega < \pi$ (or $0 \le \omega < 2\pi$), needs to be computed or plotted.
- Low frequencies are near $\omega = 0, \pm 2\pi, \dots$; the highest frequency is $\omega = \pm\pi$.
- The inverse DTFT integrates over one period only: $x[n] = \frac{1}{2\pi}\int_{2\pi} X(e^{j\omega})e^{j\omega n}d\omega$.

**Example:** For $x[n] = a^n u[n]$, $X(e^{j\omega}) = \frac{1}{1-ae^{-j\omega}}$, which repeats every $2\pi$ since $e^{-j\omega}$ does.

## Check yourself

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

1. In the derivation, which fact makes X(e^(j(ω+2π))) = X(e^(jω))?
   A. e^(-j2πn) = 1 because n is an integer
   B. x[n] is real
   C. The sum is finite
   D. e^(-jω) has unit magnitude
   Answer: A. The extra factor e^(-j2πn) equals 1 for integer n, leaving the original sum. (Key point: Derivation step)

2. Physically, why is the DTFT periodic with period 2π?
   A. Because x[n] is periodic
   B. Discrete exponentials e^(jωn) and e^(j(ω+2π)n) are the same sequence
   C. Because the DTFT is continuous
   D. Because of aliasing in time
   Answer: B. With integer n, frequencies 2π apart cannot be distinguished. (Key point: Why periodic)

3. Because of periodicity, only one period is needed, and the highest frequency is at:
   A. ω = 0
   B. ω = ±π
   C. ω = ±2π
   D. ω = π/2
   Answer: B. Low frequencies are near 0, ±2π, ..., and the highest discrete-time frequency is ω = ±π. (Key point: Consequence for plotting)

4. The inverse DTFT integrates over:
   A. All ω from -∞ to ∞
   B. Any one period of length 2π
   C. Only ω >= 0
   D. One sample
   Answer: B. x[n] = (1/2π)∫ over 2π of X(e^(jω))e^(jωn)dω, since the spectrum repeats. (Key point: Inverse DTFT)

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