---
title: "DT complex exponential periodic if rational"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:dt-exponential-rational-frequency"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-exponential-rational-frequency
site: BCT Engineering
---

# DT complex exponential periodic if rational

Signals and Systems · Chapter 1: Signals

- Asked in: 2082 Kartik, 4 marks
- Video: [Periodic Discrete Time Signals (Solved Problems)](https://www.youtube.com/watch?v=dD_67ZmfPTc) — 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/dt-exponential-rational-frequency
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Prove that discrete time complex exponential is periodic if its frequency is rational.

## Short answer

For x[n]=e^(jω0 n) to be periodic with period N, we need e^(jω0 N)=1 ⇒ ω0 N=2πm (m integer). Thus ω0/(2π)=m/N is a rational number. Conversely, if rational, N exists, so the DT exponential is periodic if and only if ω0/(2π) is rational.

## Answer

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$
\begin{aligned}
x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N}
\end{aligned}
$$

For $x[n+N] = x[n]$ we need

$$
e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}
$$

$$
\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}
$$

So **$e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number.** The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for **every** $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

**Conversely**, if $\omega_0/2\pi = m/N$ is rational, then $e^{j\omega_0 (n+N)} = e^{j\omega_0 n}e^{j2\pi m} = e^{j\omega_0 n}$, so the signal is periodic. Hence the DT complex exponential is periodic **if and only if** its frequency (in cycles/sample, $f_0 = \omega_0/2\pi$) is rational. Hence proved.

**Example:** $e^{j(2\pi/5)n}$: $f_0 = 1/5$, periodic with $N = 5$. $e^{jn}$: $f_0 = 1/2\pi$, irrational, not periodic.

## Check yourself

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

1. If w0/2pi = m/N, then e^(j w0 (n+N)) equals:
   A. -e^(j w0 n)
   B. e^(j w0 n) e^(j 2 pi m) = e^(j w0 n)
   C. e^(j w0 n)/N
   D. 0
   Answer: B. w0 N = 2 pi m so the extra factor is e^(j2 pi m) = 1. (Key point: Rational implies periodic)

2. The result established is that e^(j w0 n) is periodic:
   A. Only if w0 is an integer
   B. If and only if w0/2pi is rational
   C. For every w0
   D. Only for w0 = pi
   Answer: B. Necessity and sufficiency together give an iff statement. (Key point: If and only if)

3. The fundamental period is found by:
   A. Largest m
   B. Smallest m making N = 2 pi m / w0 an integer, with m and N coprime
   C. Taking N = 2 pi / w0 always
   D. Using m = 0
   Answer: B. Choose the smallest integer m giving an integer N. (Key point: Fundamental period)

4. e^(j(2 pi/5)n) has fundamental period:
   A. 2
   B. 5
   C. 10
   D. Not periodic
   Answer: B. f0 = 1/5, so N = 5. (Key point: Example)

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