1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series, complex coefficents

  1. Sep 30, 2007 #1


    User Avatar
    Science Advisor
    Homework Helper

    1. The problem statement, all variables and given/known data

    Let f be a periodic function with period 2pi


    [tex] g = e^{2it}f(t-3) [/tex]

    Find a relation between f and g's complex fourier coefficents.

    2. Relevant equations

    [tex]y(t) = \sum _{n-\infty}^{\infty} b_n e^{in\Omega t}[/tex]

    [tex]T \Omega = 2\pi[/tex]

    T is period

    [tex]b_n = \dfrac{1}{T}\int _0^{T} y(t)e^{-in\Omega t}dt[/tex]

    3. The attempt at a solution

    g is also periodic with 2pi, since e^{2it} is.

    [tex]f(t) = \sum _{n = -\infty}^{\infty} c_n e^{in t}[/tex]

    [tex]c_n = \dfrac{1}{2\pi}\int _0^{2\pi} f(t)e^{-int}dt[/tex]

    [tex]g(t) = \sum _{m = -\infty}^{\infty} d_m e^{im t}[/tex]

    [tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{2it}e^{-imt}f(t-3)dt[/tex]

    [tex]f(t-3) = \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _0^{2\pi} f(t-3)e^{-int}dt \right) e^{int} [/tex]

    [tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _0^{2\pi} f(t-3)e^{-int}dt \right) e^{int} \right) dt[/tex]

    [tex]r = t-3[/tex]

    Then r = t

    [tex]d_m = \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _{-3}^{2\pi -3} f(t)e^{-in(t+3)}dt \right) e^{int} \right) dt[/tex]

    We are still integrating over an integer multiple of periods, so that we can write:

    [tex]d_m = \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _{0}^{2\pi} f(t)e^{-int}e^{-i3n}dt \right) e^{int} \right) dt[/tex]

    [tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty}c_n e^{-i2n}dt \right) dt[/tex]

    Does this look right to you guys? I don't have an answer to this one, and we have no smiliar problems in our course book =/
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    Do you want more info?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Fourier series complex Date
Help with found Fourier complex series of e^t Feb 28, 2017
Complex Fourier Series Problem Dec 13, 2016
Complex Fourier Series into a Cosine Series Apr 13, 2016
Fourier Series without complex Oct 24, 2014