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Sum to infinity of Heaviside function

  • Thread starter joriarty
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  • #1
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I'm revising my course text for my exam and came across a Fourier series problem finding the Fourier series of the square wave:

[PLAIN]http://img574.imageshack.us/img574/5862/eq1.png. [Broken]

It is then calculated that the complex Fourier coefficients are

[PLAIN]http://img232.imageshack.us/img232/5863/eq2.png [Broken]

with no intermediary calculations provided.

What I don't understand is how they got:

[PLAIN]http://img2.imageshack.us/img2/1776/eq3y.png [Broken]

Could someone please explain? Thanks :)
 
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Answers and Replies

  • #2
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Take a look at the integration range and then at the definition of f again. Then you should see it. Drawing f could also help.
 
  • #3
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I'm still not getting it. Plotting f(t) I get a series of step functions as expected where f(t) = 1 from t = 0 to pi (ie where n = 0), zero from pi to 2*pi (ie where n = 1), one from 2*pi to 3*pi (n = 2) and so on.

But I don't see how this infinite series can be equated to 1 - H(t - pi) which is just equal to 1 from t = 0 to pi and zero elsewhere. That's equivalent to f(t) summed from n = 0 to 0.
 
  • #4
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U do Fourier transform for periodic functions. F is periodic with period 2Pi. So in the integration range f is just 1-H...
 
  • #5
fzero
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I'm still not getting it. Plotting f(t) I get a series of step functions as expected where f(t) = 1 from t = 0 to pi (ie where n = 0), zero from pi to 2*pi (ie where n = 1), one from 2*pi to 3*pi (n = 2) and so on.
If you are integrating from 0 to 2pi, you don't care what happens from 2*pi to 3*pi and so on.

But I don't see how this infinite series can be equated to 1 - H(t - pi) which is just equal to 1 from t = 0 to pi and zero elsewhere. That's equivalent to f(t) summed from n = 0 to 0.
They don't appear to be claiming anything about the entire series, just about it's behavior on [tex][0,2\pi][/tex].
 
  • #6
62
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Ah, I get it now. Thanks guys
 

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