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I'm trying to calculate the Fourier series for:

$g(t) =$ t if $-\pi/2 ≤ t ≤ \pi/2 $

$g(t) =$ π - t if $ π/2 ≤ t ≤ 3π/2$

The function is odd, so its Fourier Series should contain only sin terms. However, if you consider $g(t)$ as the integral of square wave function ie. sq(t+π/2), where

$ sq(t) = -1$ if $-π ≤ t ≤ 0 $

$ sq(t) = 1$ if $ 0 ≤ t ≤ π $

The Fourier series would contain only cosine terms. How is this possible?

My guess is that I can't consider g(t) as the integral of sq(t+π/2) because we would have -t instead of π - t. I have two questions:

  1. Why exactly can't I use this method?

  2. Can I use sq(t) at all for this question?

I referred to this question and everything seems fine.

user1752323
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1 Answers1

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Note: If $f$ is defined on an interval $I$ then speaking literally it makes no sense to say that $f$ is odd or even unless $I$ is centered at the origin. Here when I say such a function is odd or even I mean that the $2\pi$-periodic extension of $f$ to $\Bbb R$ is odd or even.

Yes, $sq$ is odd. But the function $sq(t+\pi/2)$ is even.

David C. Ullrich
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