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Fourier Series (Sawtooth)

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex] f(t)=\sum_{i=1}^\infty (-1)^{n+1} \frac{Sin(n\omega t)}{n} [/tex]
    represents the sawtooth function.
    2. Relevant equations

    Fourier Series Equations:
    [tex] f(t)=\frac{a_0}{2}+\sum_{i=0}^\infty \left(a_n Cos(nt)+b_n Sin(nt) \right) [/tex]
    [tex] a_n=\frac{2}{T}\int_{t}^{t+T} dt cos(nt) f(t) [/tex]
    [tex] b_n=\frac{2}{T}\int_{t}^{t+T} dt sin(nt) f(t) [/tex]
    [tex] a_0=\frac{2}{T}\int_{t}^{t+T} dt f(t) [/tex]
    3. The attempt at a solution
    I see from wikipedia:http://en.wikipedia.org/wiki/Fourier_series that they get
    [tex] f(t)=2 \sum_{i=1}^\infty (-1)^{n+1} \frac{Sin(n\omega t)}{n} [/tex]
    Is this 2 out in front because of their series is twice the period (T) of the one I'm given?

    Anyways. The way I go about this is to first take out the [tex]a_n[/tex] since this is an odd function. The part I am stuck on is the limits of integration. The period is the time it takes to travel from the bottom of the sawtooth wave to the top of one right? So should T be (in the case of my given equation) [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex]?
  2. jcsd
  3. Sep 29, 2008 #2


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    Multiplying by two changes the amplitude of a function, not it's period. What's your definition of the sawtooth exactly?
  4. Sep 29, 2008 #3
    If you refer to the picture in the link that is saw tooth. My professor gave me a mathematica notebook with that exact picture; plotted with the equation I am suppose to prove. He graphed it with [tex]\omega=2\pi[/tex]. The amplitude appears greater then 1 though.
  5. Oct 1, 2008 #4
    So the amplitude just needs to be divided by 2?
    I start off with:

    b_n=\frac{1}{T}\int_{t}^{t+T} dt sin(nt) f(t)

  6. Oct 2, 2008 #5


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    No. The original formula you gave is correct, T=2pi. The 2 in the wikipedia comes from doing the integration. I don't think there is anything wrong with the definitions, you just aren't calculating something correctly.
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