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How's Fourier series modified for function f(t)= f(2Pi t)?

  1. May 15, 2016 #1
    1. The problem statement, all variables and given/known data
    How are the coefficients of the Fourier series modified for a function with a period 2πT?

    2. Relevant equations
    a0 = 1/π ∫π f(x) dx
    an = 1/π ∫π f(x) cos(nx) dx
    bn = 1/π ∫π f(x) sin(nx) dx

    3. The attempt at a solution
    I tried letting x= t/T
    so dx = dt/T and the limits x = ± π, x = ± πT
    and I was going to plug this into the integrals but I don't think it's right.

    Any help would be greatly appreciated, thanks!
     
  2. jcsd
  3. May 15, 2016 #2

    Charles Link

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    Homework Helper

    Usually the function f(t) is periodic over period T. (not ## 2 \pi ## T). The integration over t will go from t=0 to t=T. You can readily google the topic to get the precise form of the coefficients, etc.
     
  4. May 15, 2016 #3

    LCKurtz

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    Gold Member

    For general period ##2p## the Fourier coefficients are$$
    b_n = \frac 1 p \int_{-p}^p f(x) \sin(\frac{n\pi x}{p})~dx $$ $$
    a_n =\frac 1 p \int_{-p}^p f(x) \cos(\frac{n\pi x}{p})~dx $$ $$
    a_0 = \frac 1 {2p}\int_{-p}^p f(x)~dx$$
    ##a_0## may or may not have the ##2## in the denominator depending on whether you start the FS with ##a_0## or ##\frac {a_0} 2##.
     
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