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!

Homework Help: Obtain a fourier series equation from a given graph

  1. Dec 15, 2012 #1
    The problem statement:
    Obtain a Fourier Series Expression Form from the above graph:

    I can't post the graph, so I will describe it. It's a periodic function with period 1 and magnitude 5. The equation is the following: [itex]f(x) = -x, -1/2<x<1/2[/itex]

    I'm really stuck at trying to obtain a series expression. I really don't know how to start this problem. If anyone can tell me how to approach it I would appreciate it.
  2. jcsd
  3. Dec 15, 2012 #2
    You've identified the period so you should be able to find the fundamental frequency. The fourier series will represent the signal as a sum of sines and cosines at the fundamental frequency and its harmonics. The coefficients of the sinusoids are found with the fourier integral.

    There is a shortcut in this problem. The function is odd... do you know what that tells you about the sines and cosines in the fourier representation?
  4. Dec 15, 2012 #3
    Since the function is odd the Fourier representation will only contain sines...
  5. Dec 15, 2012 #4
    I think I found the solution. On this page (http://www.sosmath.com/fourier/fourier2/fourier2.html) it has the Fourier representation of the sine series. But on there the intergral for bn goes from 0 to pi, is that because they are assuming that that is the half of the period of the sine function?

    If that's true, in my case I just have to solve for the intergral from 0 to 1/2 correct? And then plug bn back into the series expansion formula....?
  6. Dec 15, 2012 #5
    Yes that's right. Any cosines added to the result would destroy the odd symmetry so no cosines can be involved in representing the original function.

    On that page they've assumed functions have had their periods normalized to 2pi and so that they are periodic from -pi to pi. They integrate over half the period (0 to pi) instead of the whole period (-pi to pi) because the integrand is the same from -pi to 0 as from 0 to pi so they just integrate from 0 to pi and double it.

    Don't be afraid to apply the equations yourself :)

    [itex]b_{n}=\frac{2}{T}\int_{T}x(t)sin(n\omega_{o}t) dt[/itex]
    Last edited: Dec 16, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook