Why a Square Wave Has Finite Power

A square wave is a sum of infinitely many sine waves (harmonics). Each harmonic carries some power, so is the total finite? Each harmonic's power is at most a constant over n², and the p-series Σ1/n² converges, so by comparison the harmonics' power adds up to something finite: exactly the square wave's power.

How it works

The comparison test for series

In one sentence

If $0 \le a_n \le b_n$ for all large $n$ , then $\sum b_n$ converging implies $\sum a_n$ converges, and $\sum a_n$ diverging implies $\sum b_n$ diverges.

General formula

{\htmlData{sym=0}{f(t)}} = \frac{4}{\pi}\sum_{{\htmlData{sym=1}{n}}\ \text{odd}}\frac{\sin {\htmlData{sym=1}{n}}{\htmlData{sym=2}{t}}}{\htmlData{sym=1}{{n}}}, \qquad \frac{8}{\pi^2}\sum_{{\htmlData{sym=1}{n}}\ \text{odd}}\frac{1}{{\htmlData{sym=1}{n}}^2} \le \frac{8}{\pi^2}\sum_{{\htmlData{sym=1}{n}}=1}^{\infty}\frac{1}{{\htmlData{sym=1}{n}}^2} < \infty

If every term is smaller than the matching term of a series with a finite total, the smaller series can't add up to more.

What the symbols mean

$f(t)$
Square wave
Switches between +1 and −1.
$n$
Harmonic number
The nth harmonic oscillates n times as fast.
$t$
Time
Measured in radians of the fundamental cycle.

The square wave of height 1 is $\frac{4}{\pi}\left(\sin t + \frac{\sin 3t}{3} + \frac{\sin 5t}{5} + \cdots\right)$ .
A sine of amplitude $b$ carries average power $\frac{b^2}{2}$ , so harmonic $n$ carries $\frac{8}{\pi^2n^2}$ .
Compare with the p-series: $\sum\frac{1}{n^2}$ converges ( $p = 2 > 1$ ), and our terms are smaller, so the total converges.
It converges to exactly 1, the square wave's own power: $\sum_{n\text{ odd}}\frac{1}{n^2} = \frac{\pi^2}{8}$ .

Learn the math

Comparison TestsCalc II · Lesson 30, Sequences and Series Strategy for SeriesCalc II · Lesson 35, Sequences and Series Estimating the Value of a SeriesCalc II · Lesson 36, Sequences and Series

Why a Square Wave Has Finite Power