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Complex fourier series

  1. Apr 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve [tex]\frac{1}{1}[/tex] [tex]\int^{0}_{-1}[/tex] -t e[tex]^{-j2\pi*nt}[/tex]dt

    2. Relevant equations
    So I use integration by parts
    u = -t and dv = e[tex]^{-j2\pi*nt}[/tex] , du= -1 and v = [tex]\frac{1}{-j2\pi*n}[/tex]e[tex]^{-j2\pi*nt}[/tex]

    3. The attempt at a solution
    after integration by parts I get:
    [tex]\frac{e^-j2\pi*nt}{j2pi*n}[/tex] + [tex]\frac{1}{(j2pi*n)^{2}}[/tex] [1 - e[tex]^{-j2\pi*nt}[/tex]]

    So basically im suppose to end up with [tex]\frac{1}{j2pi*n}[/tex] but it doesnt work out like it should.
    Any help would be good thanks.
    Last edited: Apr 7, 2008
  2. jcsd
  3. Apr 7, 2008 #2


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    First, you certainly do not get anything involving "t" when you have evaluated at t= 0 and -1! Did you mean
    [tex]\frac{e^{2\pi i n}}{2\pi i n}+ \frac{1- e^{2\pi i n}}{(2\pi i n)^2}[/tex]

    What is [itex]e^{-2\pi i nt}[/itex]?
    (Sorry, but I just can't force my self to use "j" instead of "i"!)
  4. Apr 7, 2008 #3
    woops my bad, those t's arent spose to be in there. But the rest is right.

    And yes it is what you typed out.

    Im leading to think I shouldnt get the same answer as what I got using the Trig-Complex Relationship equation. Which is why im getting something different.

    The exponential part is what I integrate with to evaluate the complex fourier series. See the equation. -t is the function.
  5. Apr 7, 2008 #4


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    Science Advisor

    Once again, what is
    [tex]e^{-2\pi i n}[/tex]?
  6. Apr 7, 2008 #5
    In trig FS, you evaluate each component a0, an, bn. Well the exponential replaces that I think.

    im guessing you use Euler's Rules (i think) to evaluate it which says

    sin theta = 1 / 2i [ e ^ i*theta - e ^ -i*theta]
    cos theta = 1 /2 [ e ^ i*theta + e ^ -i*theta]

    Hopefully that answered that.
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