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Fourier Series Expansion

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    f(t)= -1 if -∏ < t ≤ 0
    1 if 0 < t ≤ ∏

    f(t+2∏) = f(t)

    question asks to compute first 3 non-zero terms in Fourier series expansion of f(t)
    2. Relevant equations



    3. The attempt at a solution
    since this is an odd function i used the fourier sine series formula

    f(t)=

    Ʃ (bn) sin(nwt)
    n=1

    (bn)= (2/L)*
    L
    ∫ f(t)sin(nwt)
    0
    this is just integral from 0 to L cause i dont know how to use the subscipts on the forum

    i got L=∏ since the period,T=2∏
    w=1

    so my (bn)=(-2/n∏) [cos(n∏)-1]
    so as a refult my fourier expansion becomes (bn) sin(nwt)
    and i get (-2/n∏) [cos(n∏)sin(nt) - sin(nt)]

    and whatever n value i get cos(n∏)=1 so it will be 0 for every n value. im pretty sure i did something wrong here since the answer is
    4/∏ [sin(t) + 1/3sin(3t) + 1/5sin(5t)]
     
  2. jcsd
  3. Nov 20, 2011 #2

    I like Serena

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    Welcome to PF, kiwifruit! :smile:

    You have (bn)=(-2/n∏) [cos(n∏)-1].

    Let's try to fill in a couple of values for n.
    What do you get for n=1?
    Since I do not get zero.
     
  4. Nov 20, 2011 #3
    thank you. i got it now. i confused cos∏=1 when it should be -1
     
  5. Nov 20, 2011 #4

    I like Serena

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