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I was given an assignment by my instructor where i had to write the function

$$ f(t) = \begin{cases} 1-t & 0\leq t < 1 \\ t-1 & 1 \leq t < 2 \end{cases}\\ f(t + 2) = f(t) $$

as a complete Fourier series with the hint (you should only get cosine terms)

Now i realized this function was just a triangle wave and the even extension of 1-t on the interval 0 to 1 would be the given function, so i decided to expand it by simply writing it as the Fourier cosine series of 1-t from 0 to 1

would i have gotten an equivalent Fourier series if i expanded the piece-wise function above from 0 to 2 ? I get awfully confused about the limits and normalization factors, but from what i understand the even extension of the function i expanded should represent exactly the same periodic function?

Quazi
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1 Answers1

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If the function is periodic to begin with, then expanding it on any interval equal to an integer multiple of its periods will give you the same Fourier series.

For example, if the function is defined by the rules

$$ f(t) = \begin{cases} 1 & 0\leq t < 1 \\ 0 & 1 \leq t < 2 \end{cases}\\ f(t + 2) = f(t) $$

then you will get the same Fourier series whether you expand on $[0,2]$ or $[0,4]$ or $[0,6]$, etc. If you expand the series over a different interval (e.g. $[0,1]$ or $[0,2.7]$ or whatever), then you will get a different result. Try thinking about drawing the appropriate extensions, and this will be easier to see.

In your case, extension after expanding on $[0,1]$ will be a "sawtooth" because you are only capturing and extending the behavior between $0$ and $1$.

enter image description here

To get a triangular wave, you also have to use an interval that includes the part where the function "goes up again" (i.e. from $1$ to $2$)

enter image description here

BaronVT
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  • i rewrote my question, the function i wrote was getting truncated for some reason – Quazi Oct 07 '14 at 18:36
  • what is the original interval on which you calculated things? your post still just says "on the interval 0" – BaronVT Oct 07 '14 at 19:34
  • sorry, look like the (greater/less than) symbol was messing things up, i'm not familiar with how math.stackexchange handles formatting. i've fixed the post yet again. i expanded it on 0 to 1 , so the extensions should be the same as the actual function – Quazi Oct 07 '14 at 19:58
  • The expansion on $[0,1]$ will not be the same as the original function, since it ignores the behavior of the function on $[1,2]$. See my edited answer above for more detail. – BaronVT Oct 07 '14 at 20:15
  • i was wrong in saying [0,1], i had just adjusted the normalization factor so in reality i was expanding [-1,1], this should capture the behavior properly shouldn't it? – Quazi Oct 08 '14 at 03:28
  • Ok then, yes: in this case they are the same since they both capture exactly one period of the original function. – BaronVT Oct 08 '14 at 15:48