---
title: "Energy and power: 4sin(2πt), cos(3n/4)"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:energy-power-4sin-cos"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-4sin-cos
site: BCT Engineering
---

# Energy and power: 4sin(2πt), cos(3n/4)

Signals and Systems · Chapter 1: Signals

- Asked in: 2083 Bhadra (new course), 5 marks
- 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-4sin-cos
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Determine the energy and power of the following signal. (i) f(t) = 4sin(2πt) (ii) x[n] = cos(3n/4)

## Short answer

f(t)=4sin(2πt) has infinite energy and 8 W average power, so it is a power signal. x[n]=cos(3n/4) also has infinite energy and 0.5 W average power, making it a power signal.

## Answer

#### (i) $f(t) = 4\sin(2\pi t)$
Periodic with $T = 1$ s.

**Energy:**

$$
E = \int_{-\infty}^{\infty}16\sin^2(2\pi t)\,dt = \infty
$$

(each period contributes the same positive amount, and there are infinitely many periods).

**Power:**

$$
\begin{aligned}
P &= \frac{1}{T}\int_0^T 16\sin^2(2\pi t)\,dt = \int_0^1 8\left[1 - \cos(4\pi t)\right]dt \\
&= 8 - 0 = 8\ \text{W}
\end{aligned}
$$

**Answer:** $E = \infty$, $P = 8$ W, a **power signal** ($A^2/2 = 16/2$).

#### (ii) $x[n] = \cos(3n/4)$
$\omega_0 = 3/4$, $\omega_0/2\pi = 3/(8\pi)$ is irrational, so $x[n]$ is **not periodic**; use the limit definitions.

**Energy:**

$$
E = \sum_{n=-\infty}^{\infty}\cos^2\frac{3n}{4} = \infty
$$

**Power:**

$$
\begin{aligned}
P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\cos^2\frac{3n}{4} \\
&= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1}{2}\left[1 + \cos\frac{3n}{2}\right] \\
&= \frac{1}{2} + \lim_{N\to\infty}\frac{1}{2(2N+1)}\sum_{n=-N}^{N}\cos\frac{3n}{2} = \frac{1}{2} + 0
\end{aligned}
$$

(the cosine sum is bounded, so the second term goes to zero).

**Answer:** $E = \infty$, $P = 0.5$ W, a **power signal**.

## Check yourself

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

1. What is the average power of 4 sin(2πt)?
   A. 16 W
   B. 4 W
   C. 8 W
   D. 2 W
   Answer: C. P = A^2/2 = 16/2 = 8 W. (Key point: Power of 4 sin 2πt)

2. Why is the energy of 4 sin(2πt) infinite?
   A. Because the amplitude is large
   B. Because it is odd
   C. Because P = 0
   D. Each period adds the same positive energy over infinitely many periods
   Answer: D. A non-decaying periodic signal accumulates unbounded energy. (Key point: Energy of periodic signal)

3. Is x[n] = cos(3n/4) periodic?
   A. No, since ω0/2π = 3/(8π) is irrational
   B. Yes, with N = 8
   C. Yes, with N = 4
   D. Yes, with N = 3
   Answer: A. No integer N satisfies 3N/4 = 2πm. (Key point: Periodicity of cos(3n/4))

4. What is the power of x[n] = cos(3n/4)?
   A. 1 W
   B. 0.5 W
   C. 0
   D. Infinite
   Answer: B. cos^2 = (1 + cos(3n/2))/2; the cosine sum is bounded, so P = 1/2. (Key point: Power of cos(3n/4))

## Close variants set in other exams

- [Energy or power: 3cos(4πt), sin(n/6)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aenergy-power-3cos-sin) — Signals and Systems, chapter 1, asked 2×

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