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Fourier Series

  1. Apr 16, 2008 #1
    [SOLVED] Fourier Series

    1. The problem statement, all variables and given/known data
    I know this may be a simple problem but im just beginning to understand this subject and this question has confused me.

    Expand the following as a whole-range fourier series:

    f(x) = 1 -pi < x < 0
    f(x) = x 0 < x < pi

    I can solve FS with one eqn, but what do you do with two, do you integrate twice or something?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 16, 2008 #2

    Hootenanny

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    Welcome to PF Shomy,

    The first thing to decide is whether the function is even, odd or neither.
     
  4. Apr 16, 2008 #3
    Thanks for the welcome!
    The first part is even and the second is odd
     
  5. Apr 16, 2008 #4

    Hootenanny

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    I suspect that the function is written thus,

    [tex]f(x) = \left\{\begin{array}{cr}1 & -\pi<x<0 \\ x & 0<x<\pi\end{array}\right.[/tex]

    Which means it isn't actually two functions, but one piecewise function.
     
  6. Apr 16, 2008 #5
    Yeah
     
  7. Apr 16, 2008 #6

    Hootenanny

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    So now, is the function odd or even?
     
  8. Apr 16, 2008 #7
    The first part is even and the second is odd
     
  9. Apr 16, 2008 #8

    Hootenanny

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    So overall the function is...?
     
  10. Apr 16, 2008 #9
    Odd?
     
  11. Apr 16, 2008 #10

    Hootenanny

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    Correct! What does this tell you about the Fourier series coefficients?
     
  12. Apr 16, 2008 #11
    One of the coefficients (cos) is zero
     
  13. Apr 16, 2008 #12

    Hootenanny

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    Correct, all the coefficients of cosine are zero. So what is the Euler formula for calculating the coefficients of sine?
     
    Last edited: Apr 16, 2008
  14. Apr 16, 2008 #13
    Well i meant that one of the coefficients (ie the cos ) is zero because i am using sum notation.

    an = 1/pi integral f(x)cosmx dx from pi to -pi. Do I integrate both functions and sum them or what?
     
  15. Apr 16, 2008 #14

    Hootenanny

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    Haven't we just agreed that the cosine coefficients are zero?
     
  16. Apr 16, 2008 #15
    Yeah sorry
     
  17. Apr 16, 2008 #16

    Hootenanny

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    No problem :smile:. So what about the coefficients of sine?
     
  18. Apr 16, 2008 #17
    bm = 1/pi integral f(x)sinmx dx

    So do you integrate the first part from pi to 0 and the second part from 0 to -pi
     
  19. Apr 16, 2008 #18

    Hootenanny

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    I believe it's the other way around. Overall the coefficient is given by,

    [tex]b_n=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\sin(nx)dx[/tex]

    However, since f(x) is defined piecewise we must write,

    [tex]b_n = \frac{1}{\pi}\left[\int_{-\pi}^{0}1\cdot\sin(nx)dx + \int_{0}^{\pi}x\cdot\sin(nx)dx\right][/tex]

    Is that what you meant?
     
  20. Apr 16, 2008 #19
    Yeah thats what was holding up. Thanks so much!
     
  21. Apr 16, 2008 #20

    Hootenanny

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    No problem :smile:
     
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