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Finding the Fourier Coefficients for mechanics homework

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier Coefficients for the triangular wave equation shown in this picture:
    Fourier_Series.png
    2. Relevant equations
    ##f(t)= a_0 + \sum_{n=1}^\infty a_{n}cos(n{\omega}t) + \sum_{n=1}^\infty b_{n}sin(n{\omega}t)##
    ##a_0 = \frac{1}{\tau}\int_{-\tau/2}^{\tau/2} f(t) \, dt##
    ## \omega = \frac{2\pi}{\tau}##
    ##a_n = \frac{2}{\tau}\int_{-\tau/2}^{\tau/2}f(t)cos(n\omega t) \, dt##
    ##b_n = \frac{2}{\tau}\int_{-\tau/2}^{\tau/2}f(t)sin(n\omega t) \, dt = 0, \forall n##, because it's an even function
    3. The attempt at a solution
    ## \tau = 2##
    ##a_0 = \frac{1}{2}\int_{-1}^{1} 1-|t| \, dt= \frac{1}{2}[\int_{-1}^{0} 1+t \, dt+\int_{0}^{1} 1-t \, dt]=\frac{1}{2} [\frac{1}{2}+\frac{1}{2}]=\frac{1}{2} .##

    ##a_n = \frac{2}{2}\int_{-1}^{1}f(t)cos(n\omega t) \, dt= \int_{-1}^{0}(1+t)cos(n\omega t) \, dt + \int_{0}^{1}(1-t)cos(n\omega t) \, dt ,##
    Which, when computed through many tiresome intermediate steps, gives:
    ##a_n= \frac{2}{n^2\pi^2} ##, which differs from the answer in the textbook, ##a_n= \frac{4}{n^2\pi^2}, n=1,3,5,7,...; a_n=0, n=2,4,6,8,10,...##

    So my question is, have I set up the problem correctly and made a computational mistake somewhere along the way in my integrals? Or am I missing something in the setup of the problem and missing the point entirely? It's a *** problem in our textbook(aka the most difficult level) and it's also a "computer problem", so I'm sure my prof won't expect me to do a problem this laborious on a test, but I'd still like to know where I went sour.
     
    Last edited: Mar 10, 2015
  2. jcsd
  3. Mar 11, 2015 #2

    DrClaude

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    Staff: Mentor

    Are you sure about these equation? If yes, what is ##\omega##?
     
  4. Mar 11, 2015 #3
    Well, these are the generic formula for the Fourier Series for a function. ##\omega = \frac{2\pi}{\tau}=\frac{2\pi}{2}=\pi##, I guess I wasn't as explicit about it but I considered it trivial.
     
  5. Mar 11, 2015 #4

    DrClaude

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    Staff: Mentor

    There can be different conventions, so I just wanted to make sure that I understood.


    You'll have to give the intermediate steps, because the first line is correct, but the last equality doesn't follow.
     
  6. Mar 11, 2015 #5
    Oh, I see what you were asking. Sorry.:smile:

    ##a_n = \frac{2}{2}\int_{-1}^{1}f(t)cos(n\omega t) \, dt= \int_{-1}^{0}(1+t)cos(n\omega t) \, dt + \int_{0}^{1}(1-t)cos(n\omega t) \, dt ,##
    ##=\int_{-1}^{0}cos(n\omega t) \, dt +\int_{-1}^{0}tcos(n\omega t) \, dt +\int_{0}^{1}cos(n\omega t) \, dt -\int_{0}^{1}tcos(n\omega t) \, dt ##
    ##= \int_{-1}^{1}cos(n\omega t) \, dt +\int_{-1}^{0}tcos(n\omega t) \, dt -\int_{0}^{1}tcos(n\omega t) \, dt##
    ##= \frac{sin(n{\pi}t)}{n\pi}|_{-1}^1+\int_{-1}^{0}tcos(n\omega t) \, dt -\int_{0}^{1}tcos(n\omega t) \, dt##
    ##= 0 +[\frac{tsin(n{\pi}t)}{n\pi}+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{-1}^0-[\frac{tsin(n{\pi}t)}{n\pi}+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{0}^1##
    ##=0 +[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{-1}^0-[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{0}^1##
    ##= 0 + \frac{1}{n^2\pi^2}+\frac{1}{n^2\pi^2}-\frac{1}{n^2\pi^2}+\frac{1}{n^2\pi^2}##
    ##=\frac{2}{n^2\pi^2}##:wideeyed:

    BTW, thanks a bunch for helping out DrClaude.
     
  7. Mar 11, 2015 #6
    I think I see where I went wrong :oldbiggrin:
    Looks like it was just a silly case of the negative signs cancelling each other.
    Should be:
    ##=0 +[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{-1}^0-[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{0}^1##
    ##=\frac{cos(n\pi)}{n^2\pi^2}+\frac{cos(n\pi)}{n^2\pi^2}+\frac{cos(n\pi)}{n^2\pi^2}+\frac{cos(n\pi)}{n^2\pi^2} ##
    ##= \frac{4cos(n\pi)}{n^2\pi^2}##
    ##=\frac{4(-1)^n}{n^2\pi^2} ##
     
    Last edited: Mar 11, 2015
  8. Mar 12, 2015 #7

    DrClaude

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    Staff: Mentor

    Getting closer, but that second line doesn't follow from the first.
     
  9. Mar 12, 2015 #8
    ##=0 +[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{-1}^0-[0+\frac{cos(n{\pi}t)}{n^2\pi^2}|_{0}^1##
    ##=\frac{cos(0)}{n^2\pi^2} - \frac{cos(-n\pi)}{n^2\pi^2}-\frac{cos(n\pi)}{n^2\pi^2}+\frac{cos(0)}{n^2\pi^2}##
    ##=\frac{1}{n^2\pi^2} - \frac{cos(-n\pi)}{n^2\pi^2}-\frac{cos(n\pi)}{n^2\pi^2}+\frac{1}{n^2\pi^2} ##
    ##=\frac{1}{n^2\pi^2} - \frac{cos(n\pi)}{n^2\pi^2}-\frac{cos(n\pi)}{n^2\pi^2}+\frac{1}{n^2\pi^2}##
    ##=\frac{1}{n^2\pi^2} +\frac{1}{n^2\pi^2}+\frac{1}{n^2\pi^2}+\frac{1}{n^2\pi^2}## when n is odd
    and ##=\frac{1}{n^2\pi^2} - \frac{1}{n^2\pi^2}-\frac{1}{n^2\pi^2}+\frac{1}{n^2\pi^2}=0## when n is even
    So the Fourier series for the function would be:
    ##f(t)= a_0 + \sum_{n=1}^\infty a_{n}cos(n{\omega}t) + \sum_{n=1}^\infty b_{n}sin(n{\omega}t)##
    ##=\frac{1}{2}+\sum_{n=1}^\infty \frac{4}{(2n-1)^2\pi^2}cos((2n-1){\pi}t)##

    That looks right. DrClaude, I definitely owe you one.
     
  10. Mar 12, 2015 #9
    victory.jpg

    Case closed.
     
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