---
title: "Energy of e^(jω₀n)u[n]; period of cos(41πn/7)"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:energy-power-period-cos41"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-period-cos41
site: BCT Engineering
---

# Energy of e^(jω₀n)u[n]; period of cos(41πn/7)

Signals and Systems · Chapter 1: Signals

- Asked in: 2071 Magh, 6 marks
- Video: [Energy and Power of Discrete-Time Signals](https://www.youtube.com/watch?v=QafVncdLd-g) — 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/energy-power-period-cos41
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Find the energy and power of the signal x[n] = e^(jω₀n)u[n], u[n] is unit step function. What is the period of signal x[n] = cos(41πn/7)?

## Short answer

For x[n]=e^(jω0 n)u[n], energy is infinite and average power P=1/2, so it is a power signal. For cos(41πn/7), ω0=41π/7, ω0/(2π)=41/14 rational, so fundamental period N=14 samples.

## Answer

#### Energy and power of $x[n] = e^{j\omega_0 n}u[n]$
Formulas for a discrete-time signal:

$$
E = \lim_{N\to\infty} \sum_{n=-N}^{N} |x[n]|^2, \qquad P = \lim_{N\to\infty} \frac{1}{2N+1}\sum_{n=-N}^{N} |x[n]|^2
$$

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

**Energy:**

$$
E = \lim_{N\to\infty} \sum_{n=0}^{N} 1 = \lim_{N\to\infty}(N+1) = \infty
$$

**Power:**

$$
\begin{aligned}
P &= \lim_{N\to\infty} \frac{1}{2N+1}\sum_{n=0}^{N} 1 \\
&= \lim_{N\to\infty} \frac{N+1}{2N+1} = \lim_{N\to\infty}\frac{1 + 1/N}{2 + 1/N} = \frac{1}{2}
\end{aligned}
$$

**Answer:** $E = \infty$, $P = 0.5$ W. Since the power is finite and non-zero, $x[n]$ is a **power signal**.

#### Period of $x[n] = \cos(41\pi n/7)$
A discrete-time sinusoid $\cos(\omega_0 n)$ is periodic only if $\omega_0/2\pi$ is rational, and then $N = 2\pi m/\omega_0$ with the smallest integer $m$ that makes $N$ an integer.

$$
\begin{aligned}
\omega_0 &= \frac{41\pi}{7} \\
\frac{\omega_0}{2\pi} &= \frac{41}{14} = \frac{m}{N} \quad \text{(rational, already in lowest terms)}
\end{aligned}
$$

So $m = 41$ and $N = 14$. Check: $\omega_0 N = \frac{41\pi}{7}\times 14 = 82\pi = 41 \times 2\pi$.

**Answer:** the signal is periodic with fundamental period $N = 14$ samples.

## Check yourself

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

1. What is the energy of e^(jω0 n)u[n]?
   A. Infinite
   B. 1
   C. 0.5
   D. Zero
   Answer: A. |x[n]|^2 = 1 for all n ≥ 0 and the sum diverges. (Key point: Energy of x[n])

2. What is the average power of e^(jω0 n)u[n]?
   A. 1
   B. 0.5
   C. 0
   D. Infinite
   Answer: B. lim (N+1)/(2N+1) = 1/2. (Key point: Power of x[n])

3. How is e^(jω0 n)u[n] classified?
   A. Energy signal
   B. Neither
   C. Power signal
   D. Periodic signal
   Answer: C. P is finite and non-zero, while E is infinite. (Key point: Classification)

4. What is the fundamental period of cos(41πn/7)?
   A. 7
   B. 41
   C. 82
   D. 14
   Answer: D. ω0/2π = 41/14 is in lowest terms, so N = 14 (m = 41). (Key point: Period of cos(41πn/7))

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