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Signals - is this problem supposed to be so long?

  1. Jan 28, 2008 #1
    Ok, I am working through these review problems and I'm coming up short here. I have a TON of problems to do, and I just feel like I am spending way to much time on this one. Could someone tell me if I am on the right track, or is there an easier way to do this.

    Thank you.

    Determine the Fourier series of the output of the RC circuit with input shown below:

    Image Link (here)

    So in the previous part of the problem I found the frequency response (given conditions that RC=1) to be: [tex] H(\omega) = \frac{1}{1+j \omega} [/tex].

    So my thought at the solution is to use the fact that [tex] Y(\omega) = X(\omega)H(\omega) [/tex], and then from this I could find [tex] y(t) [/tex]. Thus, an outline of the solution is:

    Outline of possible solution
    1: Find [tex] X(\omega) [/tex]
    2: Calculate [tex] Y(\omega) [/tex]
    3: Calculate [tex] y(t) [/tex]
    4: Find coefficients of y(t)

    So here are my steps thus far.

    1: One period of x(t) is [tex] T = 2 [/tex]. I can therefore write the function as:
    x(t) = -2t+1 for t=[0,1]
    x(t) = 2(t-2)+1 for 1=[1,2]

    The series coefficients are found by:
    [tex] a_k = \frac{1}{T} \int_T x(t) e^{-jk \omega_0 t} dt [/tex]

    After a lot of simplifcation I arrive at:
    [tex] a_k = \frac{4}{k^2 \pi^2}, \,\,\, k=odd [/tex]
    [tex] a_k = 0, \,\,\, k=even [/tex]

    [tex] X(\omega) = 2 \pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega -k \pi) [/tex]

    2: [tex] Y(\omega) = \left( \frac{1}{1+j\omega} \right) \left( 2 \pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega -k \pi) \right) [/tex]

    I stopped here... am I on the right track?

    Thanks !
    Last edited: Jan 28, 2008
  2. jcsd
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