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Fourier Series Simplification

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Two similar problems, but once I find out how to do the first one, I can figure out how to do the second. My signals book tells me the answers to the following "Dn"s are:

    First problem: Dn = (1/∏) ∫ sin(t) * e^(-j2nt) dt = 2/(∏ (1-4n^2) )

    if x(t) = rectified sin(t) wave, period T = ∏.

    Second problem: Dn = ∫(t/2∏) * e^(-jn*ωnaught*t) = 1/(2∏n).

    2. Relevant equations

    Formula for Dn = (1/period) ∫x(t)*e^(-(j*n*ωnaught*t))

    3. The attempt at a solution

    See, when I pop the first one into Wolfram, I get
    e^(-2*j*n*t) * (cos(t) + 2*j*n*sin(t))
    /
    ∏(4n^2 - 1).

    Since we integrate over 0 to ∏ in the first one, I understand Euler' identity is used to get the e^(-2*j*n*t) term to 1, and the cos(∏) goes away, so i'm left with 2jn / ∏(4n^2 - 1). How did they simplify that to get what I put up top as the answer?

    First post so I'm a noob.
    Thanks
     
  2. jcsd
  3. Nov 5, 2012 #2
    whoops, double post... sorry... can't figure out how to delete the post...
     
    Last edited: Nov 5, 2012
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