---
title: "DT periodicity; energy/power of e^(jω₀n)u[n]"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:dt-periodicity-energy-power-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-periodicity-energy-power-exponential
site: BCT Engineering
---

# DT periodicity; energy/power of e^(jω₀n)u[n]

Signals and Systems · Chapter 1: Signals

- Asked in: 2076 Baisakh, 3+4 marks
- Video: [Shortcut for Energy, Power & NENP Signals](https://www.youtube.com/watch?v=pa8BtYyZcrY) — 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-periodicity-energy-power-exponential
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Derive necessary condition for a discrete time signal to be periodic. Find the energy and power of the discrete time signal x[n] = e^(jω₀n)u[n], where u[n] is unit step function.

## Short answer

A DT signal e^{jω0 n} is periodic iff ω0/(2π) is rational, because ω0 N = 2π m for some integers N,m. For x[n]=e^{jω0 n}u[n], |x[n]|=1 for n≥0; total energy E = Σ 1 = ∞; average power P = lim_{N→∞}(N+1)/(2N+1)=1/2 (power signal).

## Answer

#### Necessary condition for periodicity
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.

#### Energy and power of $x[n] = e^{j\omega_0 n}u[n]$

$|x[n]|^2 = |e^{j\omega_0 n}|^2 u[n] = u[n]$, i.e. 1 for $n \ge 0$ and 0 otherwise.

**Energy:**

$$
E = \sum_{n=-\infty}^{\infty}|x[n]|^2 = \sum_{n=0}^{\infty}1 = \infty
$$

**Power:**

$$
\begin{aligned}
P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2 = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=0}^{N}1 \\
&= \lim_{N\to\infty}\frac{N+1}{2N+1} = \frac{1}{2}
\end{aligned}
$$

**Answer:** $E = \infty$, $P = 1/2$; $x[n]$ is a **power signal**.

## Check yourself

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

1. For e^(jω0 n) to be periodic, what must hold?
   A. ω0 is any real number
   B. ω0 is a rational number
   C. ω0/π is irrational
   D. ω0/2π is a rational number
   Answer: D. e^(jω0 N) = 1 needs ω0 N = 2πm, so ω0/2π = m/N must be rational. (Key point: Periodicity condition)

2. Why does a DT exponential have a stricter periodicity condition than a CT one?
   A. The period N must be an integer because n takes only integer values
   B. DT signals have no frequency
   C. DT signals are always aperiodic
   D. DT signals are measured in different units of energy
   Answer: A. CT e^(jω0 t) is periodic for every ω0 ≠ 0, but in DT the period must be an integer number of samples. (Key point: CT vs DT periodicity)

3. What is the energy of x[n] = e^(jω0 n)u[n]?
   A. 1
   B. Infinite
   C. 1/2
   D. Zero
   Answer: B. |x[n]|^2 = 1 for every n ≥ 0, so the sum of ones diverges. (Key point: Energy of the signal)

4. What is the average power of x[n] = e^(jω0 n)u[n]?
   A. 1
   B. 0
   C. 1/2
   D. Infinite
   Answer: C. P = lim (N+1)/(2N+1) = 1/2, so the signal is a power signal. (Key point: Power of the signal)

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