Results 1 to 2 of 2

Math Help - Fourier Series: Triangular waveform, Ao calc going wrong somewhere

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    4

    Fourier Series: Triangular waveform, Ao calc going wrong somewhere

    The question asked is calculate Fourier series for waveform
     f(t) = 5 + \frac{5}{\pi}\theta for -\pi < \theta < 0
     f(t) = 5 - \frac{5}{\pi}\theta for -\pi < \theta < 0

    I calculate Ao to be 5 but as you are given the fourier series with the question I can see that that is incorrect and it should be \frac{5}{2}

    Here is my calculation as I have done it, could someone please tell me where I am making a mistake? Thanks Ed.

    A_o = \frac{1}{2\pi}<br />
\int_{-\pi}^{\pi} \! f\theta \, d\theta

    A_o = \frac{1}{2\pi}<br />
\int_{-\pi}^{0} \! 5 + \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta + \frac{1}{2\pi}\int_{0}^{\pi} \! 5 - \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta

    A_o = \frac{1}{2\pi}\displaystyle \left(\int_{-\pi}^{0} \! 5 d\theta + \frac{5}{\pi} \int_{-\pi}^{0} \! \theta \, d\theta + \int_{0}^{\pi} \! 5 d\theta - \frac{5}{\pi} \int_{0}^{\pi} \! \theta \, d\theta \right)

    A_o = \frac{1}{2\pi}\displaystyle \left( 5[\theta]^0_{-\pi} + \frac{5}{2\pi} [\theta^2]^0_{-\pi} \, + 5[\theta]^\pi_0 - \frac{5}{2\pi}[\theta^2]^\pi_o \right)

    A_o = \frac{1}{2\pi}\displaystyle \left( 5\pi + 5\pi \right)

    The \frac{5}{2\pi}[\theta^2] \ cancelling \ each \ other \ out

    A_o = \frac{10\pi}{2\pi} = 5
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by emmv View Post
    The question asked is calculate Fourier series for waveform
     f(t) = 5 + \frac{5}{\pi}\theta for -\pi < \theta < 0
     f(t) = 5 - \frac{5}{\pi}\theta for -\pi < \theta < 0

    I calculate Ao to be 5 but as you are given the fourier series with the question I can see that that is incorrect and it should be \frac{5}{2}

    Here is my calculation as I have done it, could someone please tell me where I am making a mistake? Thanks Ed.

    A_o = \frac{1}{2\pi}<br />
\int_{-\pi}^{\pi} \! f\theta \, d\theta

    A_o = \frac{1}{2\pi}<br />
\int_{-\pi}^{0} \! 5 + \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta + \frac{1}{2\pi}\int_{0}^{\pi} \! 5 - \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta

    A_o = \frac{1}{2\pi}\displaystyle \left(\int_{-\pi}^{0} \! 5 d\theta + \frac{5}{\pi} \int_{-\pi}^{0} \! \theta \, d\theta + \int_{0}^{\pi} \! 5 d\theta - \frac{5}{\pi} \int_{0}^{\pi} \! \theta \, d\theta \right)

    A_o = \frac{1}{2\pi}\displaystyle \left( 5[\theta]^0_{-\pi} + \frac{5}{2\pi} [\theta^2]^0_{-\pi} \, + 5[\theta]^\pi_0 - \frac{5}{2\pi}[\theta^2]^\pi_o \right)

    A_o = \frac{1}{2\pi}\displaystyle \left( 5\pi + 5\pi \right)

    The \frac{5}{2\pi}[\theta^2] \ cancelling \ each \ other \ out

    A_o = \frac{10\pi}{2\pi} = 5
    I did not read all of your stuff. But, whenever I have seen Fourier expansions it is \frac{a_0}{2}

    Fourier series - Wikipedia, the free encyclopedia

    Could that be the reason? I hope you don't take this as an insult to your intelligence.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 18th 2011, 08:24 PM
  2. [SOLVED] Fourier series to calculate an infinite series
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 4th 2010, 01:49 PM
  3. Complex Fourier Series & Full Fourier Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2009, 05:39 AM
  4. matlab changing time scale on a waveform
    Posted in the Math Software Forum
    Replies: 3
    Last Post: July 27th 2009, 09:06 AM
  5. from fourier transform to fourier series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum