---
title: "Periodicity of 3e^(j3π(n+1/2)/5)"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:periodicity-complex-exponential-example"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/periodicity-complex-exponential-example
site: BCT Engineering
---

# Periodicity of 3e^(j3π(n+1/2)/5)

Signals and Systems · Chapter 1: Signals

- Asked in: 2076 Bhadra, 4+4 marks
- Video: [Signals & Systems - Checking of Periodicity](https://www.youtube.com/watch?v=JePnh5ZygdA) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/periodicity-complex-exponential-example
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Derive necessary condition for a discrete time signal x[n] = e^(jω₀n) to be periodic. Determine whether the discrete time signal x[n] = 3e^(j3π(n + 1/2)/5) is periodic or not.

## Short answer

A discrete-time exponential e^(jω₀n) is periodic iff ω₀/(2π) is rational; then period N is the smallest positive integer such that Nω₀ is a multiple of 2π. For the given signal, ω₀ = 3π/5, ω₀/(2π) = 3/10 is rational, so it is periodic with fundamental period N = 10.

## Answer

#### Necessary condition
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.

#### Is $x[n] = 3e^{j3\pi(n + 1/2)/5}$ periodic?

Separate the constant phase:

$$
x[n] = 3e^{j3\pi/10}\,e^{j\frac{3\pi}{5}n}
$$

$3e^{j3\pi/10}$ is a constant complex amplitude, so periodicity depends on $\omega_0 = \frac{3\pi}{5}$:

$$
\frac{\omega_0}{2\pi} = \frac{3\pi/5}{2\pi} = \frac{3}{10} \quad\text{(rational)}
$$

So the signal is **periodic**. Fundamental period:

$$
N = \frac{2\pi m}{\omega_0} = \frac{10m}{3} \;\Rightarrow\; N = 10 \text{ (with } m = 3\text{)}
$$

**Check:** $x[n+10] = x[n]\,e^{j\frac{3\pi}{5}\cdot 10} = x[n]e^{j6\pi} = x[n]$.

**Answer:** periodic with fundamental period $N = 10$ (fundamental frequency $2\pi/10 = \pi/5$ rad/sample).

## Check yourself

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

1. What is ω0/2π for x[n] = 3e^(j3π(n+1/2)/5)?
   A. 3/5
   B. 5/3
   C. Irrational
   D. 3/10
   Answer: D. Rewriting, ω0 = 3π/5, so ω0/2π = 3/10, which is rational. (Key point: Rational test)

2. What role does the 1/2 inside (n + 1/2) play in periodicity?
   A. It gives a constant complex amplitude and does not affect periodicity
   B. It makes the signal aperiodic
   C. It doubles the period
   D. It changes ω0 to 3π/10
   Answer: A. The factor e^(j3π/10) is a constant; only the coefficient of n sets ω0. (Key point: Constant phase)

3. What is the fundamental period of the signal?
   A. 5
   B. 10
   C. 3
   D. 30
   Answer: B. N = 2πm/ω0 = 10m/3; the smallest m making N an integer is m = 3, giving N = 10. (Key point: Fundamental period)

4. Which step in the derivation gives the periodicity condition?
   A. e^(jω0 N) = 0
   B. ω0 N = π
   C. e^(jω0 N) = 1, so ω0 N = 2πm
   D. ω0 = N
   Answer: C. Requiring x[n+N] = x[n] forces e^(jω0 N) = 1, which is true only for multiples of 2π. (Key point: Necessary condition derivation)

## Close variants set in other exams

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