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Varification needed for small trigonometrical Fourier series, PRESSING

  1. Nov 24, 2012 #1
    I have x(t) = 1/2 + cos(t) + cos(2t)

    so I can see that a0 = 1/2
    and that it is an even function so there is no bn
    Also that T = 2pi so
    an = 2/2pi ∫02pi x(t).cos(nω0t) dt

    but when I integrate this I get an = 0 yet I've been told that the answer is

    x(t) = 1/2 + Ʃn = 12 cos(nω0t)

    which would mean that an = 1, but I'm not sure how.


    Thanks heaps!

    EDIT: I just found another related problem I'm struggling with if anyone's interested:
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 26, 2012 #2

    rude man

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    This has been adequately addressed already. Your x(t) is already a Fourier series. If you computed the coefficients to be different you made a math error.
  4. Nov 26, 2012 #3
  5. Nov 26, 2012 #4


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