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Homework Help: Fourrier Series

  1. Apr 17, 2005 #1
    I am 52, not yet senile and would be greatfulif someone can give me the Fourier series of two isolated pulses, pulswidth "w" spaced "2D" apart or at +- D.

    Thanks in advance
     

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  2. jcsd
  3. Apr 17, 2005 #2

    quasar987

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    I will try it, but don't have a program to check my calculations.

    This function is symetrical. So the coeficients [itex]B_n[/itex] are 0. The period is 2D.

    [tex]A_n = \frac{2}{D}\int_{0}^{D} cos\left( \frac{n\pi x}{D}\right)y(x)dx = \frac{2h}{D}\int_{D-W/2}^{D} cos\left( \frac{n\pi x}{D}\right)dx = \frac{2h}{D}\left[\frac{D}{n\pi}sin\left( \frac{n\pi x}{D}\right)\right]_{D-W/2}^{D}[/tex]

    [tex]A_n = \frac{2h}{n\pi}sin\left( \frac{n\pi W}{2D} - n\pi \right) & n=1,2,...[/tex]

    [tex]A_0 = \frac{Wh}{D}[/tex]

    [tex]\Rightarrow y(x) = \frac{Wh}{2D} + \frac{2h}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}sin\left( \frac{n\pi W}{2D} - n\pi \right) cos\left( \frac{n\pi x}{D}\right)[/tex]

    This makes sense, because the bigest W can be is 2D wide. In this case, the function is y(x) = h, which is correct. (h is the height of the pulse)
     
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3

    saltydog

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    I get:

    [tex]y(x)=\frac{Wh}{2D}+\frac{2h}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}Sin[\frac{n\pi W}{2D}]Cos[\frac{n\pi}{D}(x-c-\frac{W}{2})][/tex]

    with:

    [tex]c=\frac{2D-W}{2}[/tex]

    I've included a plot for:

    W=5
    D=10
    h=1

    for the first 25 terms of the series
     

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  5. Apr 17, 2005 #4

    quasar987

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    Our expressions are equivalent, in case you haven't noticed.

    [tex]-c-\frac{W}{2} = \frac{-2D+W-W}{2} = D[/tex]

    and

    [tex]\cos\left(\frac{n\pi}{D}(x+D)\right) = \cos\left(\frac{n\pi x}{D}+n\pi\right)=-\cos\left(\frac{n\pi x}{D} \right)[/tex]

    So it's cool.


    Edit: Wait, that's not true :grumpy:
     
    Last edited: Apr 17, 2005
  6. Apr 17, 2005 #5

    quasar987

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    Ok I corrected an error in my original post due to this false identity that I had used.

    Now our expressions are equivalent.
     
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