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Fourier series question

  1. May 23, 2005 #1
    I got this question out of a book, but I can't get the book's answer. Since I can't draw, I'll just describe the graph given. Express q(t) as a Fourier series expansion.

    The charge q(t) on the plates of a capacitor at time t is shown as a saw-tooth wave with period [itex]2\pi[/itex] and its peak is at [itex]t = \pi[/itex], where q(t) = Q.

    At first, I tried to use the expressions for open boundary conditions (as q(t) is 0 at t = 0 and t = 2 pi). I got stuck, so I started again, with

    [tex]q(t) = \frac{a_{0}}{\sqrt{L}} + \sum_{n=1}^\infty a_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} + \sum_{n=1}^\infty b_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L}[/tex]


    [tex]a_{0} = \int_{0}^{L} \left(\frac{1}{L}\right)^\frac{1}{2} q(t) dt[/tex]

    [tex]a_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} q(t) dt[/tex]

    [tex]b_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L} q(t) dt[/tex]

    Now, since the saw-tooth is symmetric, there was no need to calculate [itex]b_{n}[/itex], since they'd have been 0 anyway.

    I represented q(t) using top hat functions:

    [tex]q(t) = \left[\theta(t) - \theta(t - \frac{L}{2})\right]\frac{2tQ}{L} + \left[\theta(t - \frac{L}{2}) - \theta(t - L)\right]\frac{2(L - t)Q}{L}[/tex]

    Then I used that to calculate [itex]a_{0}[/tex] and [itex]a_{n}[/itex], using [itex]L = 2\pi[/itex]

    After pages of working out, I got:

    [tex]a_{0} = Q\pi \left(\frac{1}{2\pi}\right)^\frac{1}{2}[/tex]

    [tex]a_{n} = \left(\frac{1}{\pi}\right)^\frac{1}{2} \frac{Q}{\pi} \frac{2}{n^2} (\cos n\pi - 1)[/tex]


    [tex]q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi - 1)}{n^2}\right][/tex]

    The book's answer is [tex]q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right][/tex]

    I don't know where I've gone wrong. I can't post all my working, as it's a lot and would take forever to LaTeX.
  2. jcsd
  3. May 23, 2005 #2
    Well, your results are equivalent :)) I'm poor at LaTeX, but I'll try to explain.

    First, there is a typo in a final result (I suspect it is a typo), there should be

    [tex]q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt (\cos (n\pi)-1)}{n^2}\right][/tex]

    (notice the braces in second cos), but you got it right in a_n expression.

    OK, now whenever n is even, expression under sumation is equal to 0 because [tex](\cos (n\pi)-1) = 0[/tex] for even values of n.

    So, we have to sum only if n is odd, otherwise written as [tex]n = 2k-1[/tex], for k going from 1 to inf.

    For odd numbers [tex](\cos (n\pi)-1) = -2[/tex]

    If you put it all together you can write
    [tex]\sum_{k=1}^\infty \frac{-2\cos (2k-1)t}{(2k-1)^2}\right][/tex]

    you can take the "-2" out of summation, and then you get the book's answer.

    Hope my explanation was clear,
  4. May 24, 2005 #3
    Yay, thanks very much. Yes, I did make a typo with the brackets in my post.

    Thanks again :D.
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