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Fouier series from the sign function

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    If I have the standard sign function defined on R is the corresponding fourier series the standard fourier series with plus/minus in front respectively?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 4, 2009 #2

    jbunniii

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    Can you write down exactly what you mean by "standard sign function"? To me, it means

    [tex]f(x) = \left\{\begin{array}{ll}1, & \textrm{if }x > 0 \\
    -1, & \textrm{if }x < 0 \end{array}\right.[/tex]

    and [itex]f(0)[/itex] is typically defined as either 0 or 1, though its value doesn't make any difference with regard to Fourier series.

    There is no Fourier series defined for this function unless you restrict the domain to a finite interval. And the series will depend on which interval you choose.
     
  4. Oct 4, 2009 #3
    Is says find the fourier series

    [tex]f(x) = sign(x)[/tex] where [tex]x \in ]-\pi,\pi[[/tex]

    But since my interval is finite. How do I then go about finding the fourier series for that f(x)??
     
  5. Oct 4, 2009 #4

    jbunniii

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    Can you write down the formula for finding the Fourier coefficients of [itex]f(x)[/itex], and then try plugging in this particular [itex]f(x)[/itex], so we can see how far you are able to get on your own?
     
  6. Oct 4, 2009 #5
    Yes if f(x) is even defined on the interval [-n,n]

    then the fourier series is [tex]n_0 + \sum_{k=1}^{\infty} n_k cos(k\pi x / n)[/tex]


    and

    [tex]n_0 = \frac{1}{n} \int_{0}^{n} f(x) dx[/tex]

    [tex]n_k = \frac{1}{n} \int_{0}^{n} f(x) \cdot cos(k \pi x /n) dx[/tex]


    if f(x) is odd

    then [tex]f(x) = \sum_{k=1}^{\infty} n_k \cdot sin(k \pi x /n)[/tex]

    where [tex]b_k = \frac{2}{n} \int_{0}^{n} f(x) \cdot sin(k \pi x /n)[/tex]

    Is this what you mean?
     
    Last edited: Oct 4, 2009
  7. Oct 4, 2009 #6

    jbunniii

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    Yes, let's work with that. Your function is clearly odd, so use the formula for [itex]b_k[/itex]. (By the way, there's a typo in your formula for [itex]f(x)[/itex]: surely [itex]n_k[/itex] should be [itex]b_k[/itex].)

    Now plug in [itex]f(x)[/itex] for this particular function. What is [itex]n[/itex] in this case?
     
  8. Oct 4, 2009 #7
    am I stupid or is n = 1 ??

    and I then end up with two version of [tex]b_k[/tex]

    where [tex]b_{k_1} = \frac{2}{n=1} \int_{0}^{1} sin(kx) dx [/tex]

    and [tex]b_{k_2} = \frac{2}{n=-1} \int_{0}^{-1} sin(-kx) dx [/tex] ???
     
  9. Oct 4, 2009 #8

    jbunniii

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    No, [itex]n[/itex] is the (one-sided) interval over which the function is defined.

    Your function is defined on [itex][-\pi, \pi][/itex], so [itex]n = \pi[/itex].

    The formula for [itex]b_k[/itex] tells you to integrate only over [itex][0,\pi][/itex]. Your function is equal to ONE over that entire interval, so you simply end up with

    [tex]b_k = 2 \int_{0}^{\pi} sin(kx) dx[/tex]

    You don't have to worry about the negative half of the function, because this formula for [itex]b_k[/itex] assumes that the function is odd.

    There is a more general formula for Fourier coefficients which does not assume that the function is even OR odd:

    [tex]a_k = \int_{-\pi}^{\pi} f(x) \cos(kx) dx[/tex]
    [tex]b_k = \int_{-\pi}^{\pi} f(x) \sin(kx) dx[/tex]

    Note that in the special case where [itex]f(x)[/itex] is odd, [itex]a_k = 0[/itex] (can you see why?), and the formula for [itex]b_k[/itex] can be simplified as follows:

    [tex]\begin{align*}b_k &= \int_{-\pi}^{\pi} f(x) \sin(kx) dx \\
    &= \int_{-\pi}^0 f(x) \sin(kx) dx + \int_{0}^{\pi} f(x) \sin(kx) dx \\
    &= 2 \int_{0}^{\pi} f(x) \sin(kx) dx \end{align*}[/tex]

    which matches the formula you are using. Thus if the function is odd, this simplified formula allows you to integrate only over the positive half, [itex][0,\pi][/itex].
     
  10. Oct 4, 2009 #9
    So basicly my fourier series for the sign function over the finite is with that b_k above?
     
  11. Oct 4, 2009 #10

    LCKurtz

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  12. Oct 4, 2009 #11
    Hi I had some family troubles so I didn't get arround to finishing it.

    I did the intermediate calculations for the b_k but how was suppose to compare that result to find the some for the second series that I didn't?

    Is it some obvious theorem that I overlooked maybe? Compare two series?
     
  13. Oct 4, 2009 #12
    Did you mean "sign" or --sine--?
     
  14. Oct 4, 2009 #13
    sign function!
     
  15. Oct 4, 2009 #14

    LCKurtz

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    Reply in that other thread and we can continue there.
     
  16. Oct 4, 2009 #15

    jbunniii

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    Yes, that's right. Just evaluate the integral and simplify and that gives you the [itex]b_k[/itex] values. Then you can plug those into the formula for the Fourier series.
     
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