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Fourier Series complex coefficients

  1. Apr 6, 2014 #1
    I have been trying to follow how the complex Fourier coefficients are obtained; the reference I am using is at www.thefouriertransform.com. However I am unable to follow the author's working exactly and wondered if anyone could help me see where I am going wrong.

    First, I understand that the coefficient [itex]c_n=\frac{1}{T}\int^T_0 f(t) e^{\frac{-i2\pi nt}{T}}dt[/itex]

    Given a square wave, of period 2T, and amplitude :

    http://s90.photobucket.com/user/jonnburton/media/SqareWave_zpsbdf193c1.jpg.html

    It can be seen directly from the graph that [itex]c_0=\frac{1}{2}[/itex] (I also managed to calculate this, too.)

    But I can't follow how the author obtained expressions for [itex]c_n[/itex].

    This is what I have done in attempt to obtain [itex]c_n[/itex]:

    [itex]c_n=\frac{1}{T}\int^{\frac{T}{2}}_0 1 \cdot e^{\frac{-i2\pi nt}{T}}dt[/itex]

    [tex]\frac{1}{T}\left[\frac{T e^{\frac{-i2\pi nt}{T}}}{i2\pi n}\right]^{\frac{T}{2}}_0[/tex]

    [tex]\frac{1}{T}\left[\left(\frac{T e^{-in\pi}}{in2\pi}\right)-\left(\frac{T}{in2\pi}\right)\right][/tex]

    [tex]\left(\frac{e^{-in\pi}}{in2\pi}-\frac{1}{in2\pi}\right)[/tex]

    However, the expression the author finds is:

    [tex]\frac{1}{i\pi n}[/tex] for odd n, and zero for even n.
     
    Last edited by a moderator: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2
    You seem to be missing a minus sign in your integral.

    Also, simply for consistency, the limits you originally give (between 0 and T) aren't the limits you say you end up using (between 0 and T/2).

    Lastly, what is [itex] e^{-in\pi} [/itex] for odd/even [itex] n [/itex]? Recall [itex] e^{in\pi} = cos(n\pi) + isin(n\pi) [/itex].
     
  4. Apr 6, 2014 #3
    Thanks for the hints Silversonic. I will go back through it again but I think I can see how this works out now!

    (I just changed the limits because I though it wasn't necessary to integrate beyond [itex]\frac{T}{2}[/itex] because the function there is zero.)
     
  5. Apr 6, 2014 #4
    Okay, the picture won't load for me. That could be an issue on my end.
     
  6. Apr 6, 2014 #5
    Not sure why it wasn't working; I've just typed the URL in again and I think it's OK now.
     
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