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Complex Fourier Series

  1. Sep 28, 2009 #1
    Having some trouble with this, any help is appreciated

    1. The problem statement, all variables and given/known data
    Give the complex fourier series for [tex]f(x) = 2 - x, -2<x<2[/tex]


    2. Relevant equations
    [tex]f(x) = \sum_{n=-\infty}^\infty C_ne^{\frac{i n \pi x}{l}[/tex]

    [tex]C_n=\frac{1}{2l} \int_{-l}^l f(x)e^{\frac{-i n \pi x}{2}} dx [/tex]

    3. The attempt at a solution
    [tex]f(x) = 2 - x, l = 2[/tex]

    [tex]f(x) = \sum_{n=-\infty}^\infty \frac{1}{4} \int_{-2}^2 ((2-x)e^{\frac{-i n \pi x}{2}}dx) e^{\frac{i n \pi x}{2} [/tex]


    now here, I don't know which steps I should take next. Should I take this integral for general n? Or should I break it up into the case for when n=0, n>0, and n<0?

    I've kind of tried both to no avail :(
     
  2. jcsd
  3. Sep 29, 2009 #2

    lanedance

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    from a quick look here's a ew helpful (hopefully) tips:

    the sum over negative & positive n values in the complex exponent, is equivalent to summing only over positive n's with both sin & cos terms, & this one might be easier to evaluate in terms of sin's & cos's

    also you should find a proof of the integral of sin(2.pi.n.x/L).sin(2.pi.n.x/L) over L = 0 for integers m not equal p and similar for cosines & mixes, this will allow you to write each coefficient in terms of single integral

    now consider your function
    g(x) = 2-x

    first consider the constant part, 2. Any integral of a sinusoid over an integer times period will be zero, real or complex so only the n= 0 (or constant cos part will contribute)

    now -x is an odd function, the cos integrals will be zero, (assuming it is just repeated),so only sin's will contribute...

    so you should be able to reduce the integrals to required to ~x.sin(n.pi.x/2)
     
  4. Sep 29, 2009 #3
    yeah, i think that was part a), they had us find the fourier series in sin/cos form, now they want us to re-derive it with complex exponentials

    as a side note, i didnt think about splitting up the 2 - x into both 2 and the -x part, that makes sense now and seems quicker for part a :)
     
  5. Sep 29, 2009 #4

    lanedance

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    not sure if you are allowed, but you could always write your complex exponential as:
    [tex] e^{i x } = cos(x) + i sin(x) [/tex]
    then use what you know about sin & cos integrals - evaluating the integrals will essentially be the same then...

    and the negative n ones will just be complex conjugates of the poistives
     
  6. Sep 29, 2009 #5
    i think they want it purely in the e^ix form :(

    just stuck on whether to worry about positive/negative cases and whether i do or dont how to simplify it to something reasonable
     
  7. Sep 29, 2009 #6

    lanedance

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    hmm.. by separating 2 & -x should be able to integrate *2 easy,for n = 0

    then simple int by parts on the *(-x) for when n is not equal to zero

    knowing what is going to happen to the terms based on what you have already done (sin & cos), i would attempt it for a generic n & -n, then show what happens when you sum the result... so express you results as a sum from 0 to n
     
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