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Fourier Sawtooth

  1. Nov 8, 2004 #1

    cj

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    I can derive the Fourier series for a regular
    sawtooth wave.

    A different kind of sawtooth is represented by:

    [tex] f(x)=\left\{\begin{array}{cc}-\frac{1}{2}(\pi +x),&\mbox{ if }
    =-\pi \leq x < 0\\+\frac{1}{2}(\pi -x),& \mbox{ if } 0 < x \leq \pi\end{array}\right. [/tex]

    For the life of me I can't figure out how
    to derive the series for this, which is:

    [tex]f(x)=\sum_{n=1}^{\infty} sin (nx/n)[/tex]
     
  2. jcsd
  3. Nov 8, 2004 #2

    Galileo

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    Probably a typo, but:
    [tex]f(x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}[/tex]

    It's obtained the usual way. The function is odd, so all the cosine coefficients are zero.
    Now just get:

    [tex]a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx[/tex]

    where [itex]a_n[/itex] is the coefficient of sin(nx)
     
  4. Nov 8, 2004 #3

    cj

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    Yes, you're right -- there was a typo.

    Does the integration breaks down
    into

    [tex]a_n=\frac{1}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx + \frac{1}{\pi}\int_{-\pi}^{0}f(x)\sin(nx)dx \text { ??}[/tex]

    Also, does an [itex]a_0[/itex] term need to be
    determined? I'm not sure when,
    or when not, to include an [itex]a_n[/itex].

    Thanks a lot.

     
  5. Nov 8, 2004 #4

    Galileo

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    No. There's an easy way to remember/see it. If n=0, then sin(nx)=0.
    So the zeroth coeff. of the sine is always zero.
    For the cosine: cos(nx)=1 if n=0.

    That's correct, so that's all there's to it.
    Both integrals are equal though, since f(x) and sin(nx) are odd, f(x)sin(nx) is even.
     
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