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I need help understanding the Fourier components of a square wave

  1. May 20, 2013 #1
    In my physics book there is an example of making a square wave by "simply" summing up a few cosine waves. The book says these first three waves are the first three Fourier components of a square wave, yet when I sum the three wave functions up, I get something way off; as does my calculator.

    For example, if we take the easiest case of x = 0, we get the sum of 1, 1/3, and 1/5 equals 1.53m. However, when I look at the plot for the sums, the amplitude seems to be at about 0.9m. That is nowhere near the sum of the three wave functions at zero. This means that I am missing something fundamentally important here. What is it?

    Here's a link to the graphs and the example problem. Thanks for your help.

  2. jcsd
  3. May 20, 2013 #2


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    Staff: Mentor

    They forgot a minus sign: the amplitude of ##D_2## should be ##-\frac{1}{3} D_M##.
  4. May 20, 2013 #3
    Excellent. Thank you very much, DrClaude.
  5. May 20, 2013 #4
    Here is how to relate the amplitude of the cosine waves to the amplitude A of the square wave. For this example, [itex]A = \frac{\pi}{4}[/itex].

    [itex]D_1 = \frac{4}{\pi}A cos(kx)[/itex]
    [itex]D_2 = -\frac{4}{3\pi}A cos(3kx)[/itex]
    [itex]D_2 = \frac{4}{5\pi}A cos(5kx)[/itex]


    If are curious how this can be found:
    [itex]a_n = \frac{2}{T} \int_{-T/2}^{T/2} f(x)cos\left(2\pi \frac{n}{T} x\right)dx[/itex]
    Where f(x) is the square wave. We observe that the square wave is symmetric across x=0, so we integrate from 0 to T/2, and multiply by 2.
    [itex]a_n = \frac{4}{T} \int_{0}^{T/2} f(x)cos\left(2\pi \frac{n}{T} x\right)dx[/itex]
    We split this up into two integrals for the positive an negative portions. The square wave is +A in the positive portion and -A in the negative portion.
    [itex]a_n = \frac{4}{T}\int_{0}^{T/4} A cos\left(2\pi \frac{n}{T} x\right)dx - \frac{4}{T}\int_{T/4}^{T/2} A cos\left(2\pi \frac{n}{T} x\right)dx [/itex]

    [itex]a_n = \left. \frac{4}{T}\frac{T}{2\pi n} A sin\left(2\pi \frac{n}{T} x\right) \right|_{0}^{T/4} - \left. \frac{4}{T}\frac{T}{2\pi n} A sin\left(2\pi \frac{n}{T} x\right)\right|_{T/4}^{T/2} [/itex]
    [itex]a_n = \frac{2}{\pi n}A \left[sin \left( \frac{\pi n }{2}\right) - 0 - sin\left(\pi n\right) + sin\left(\frac{\pi n }{2}\right) \right] = \frac{4}{\pi n}Asin \left( \frac{\pi n }{2}\right) - \frac{2}{\pi n}Asin\left(\pi n\right) [/itex]
    The second term is always zero, since n is an integer, and thus [itex]sin\left(\pi n\right) = 0[/itex]
    [itex]a_n = \frac{4}{\pi n}A sin \left( \frac{\pi n }{2}\right) [/itex]
    [itex]a_n = 0, \frac{4A}{\pi}, 0, -\frac{4A}{3\pi}, 0, \frac{4A}{5\pi}, 0 , ... [/itex]

    Notice that only the terms with odd n are non-zero. Also, the sign alternates every odd term.

    Edit: I have fixed this to show a square wave for any amplitude A. For the square wave in this example, [itex]A = \frac{\pi}{4}[/itex]
    Last edited: May 20, 2013
  6. May 20, 2013 #5


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    Staff: Mentor

    If you look carefully at the figure, you will see that the square wave being approximated has an amplitude of ##\pi/2##, so there is no missing ##4/\pi## factor.
  7. May 20, 2013 #6
    Oops, somehow I assumed it was supposed to have an amplitude of [itex]D_M[/itex]. The depicted square wave does seem to have an amplitude of about ##\pi/4##, or in other words a ##\pi/2## peak-to-peak amplitude.
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