Fourier Series Simplification

Discussion in 'Homework Help' started by that_)dude, Nov 5, 2012.

  1. that_)dude

    Thread Starter New Member

    Nov 5, 2012
    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).

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

    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.