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Fourier Series Frequency, Period, and Coefficients

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the function 10sin^2(10t)

    Find the fundamental frequency and period.

    Find the exponential and trigonometric coefficients of the Fourier Series.

    2. Relevant equations



    3. The attempt at a solution

    I really have no idea how to start this problem. The sin^2 is what is throwing me off, as im not sure if i am supposed to integrate it first before finding the frequency and period. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 31, 2012 #2

    LCKurtz

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    I would start by looking at the graph of sin(t) and sin2t to see what squaring it does to the period of a sine function.
     
  4. Oct 31, 2012 #3
    Ok, so I did as you said and checked out the difference. The sin^2(t) seems to cut the period in half, so rather than being ∏/5 in my case, it was ∏/10. So, to find my T variable, it would be:

    2∏/(∏/10)

    right?

    Then I should be able to just plug my ω=∏/10, and T into the equations for exponential and trigonometric coefficients I think...

    Sorry, this stuff is really difficult to grasp for some reason.
     
  5. Oct 31, 2012 #4

    LCKurtz

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    In other words, 20.
    Since I don't know what your T represents, I don't know if that is right or not. But you are correct that the period of ##\sin^2(10t)## is ##\pi/10##.
     
  6. Nov 1, 2012 #5
    Ok, so now that I know that the period of this function is ∏/10, the frequency is 20.

    I graphed it first to figure out the period, then solved it mathematically. Using trig identities:

    10sin(10t) = 1/2 - (5cos(20t))/2

    The frequency, ω is the 20 in that equation, then plugging that into the equation for solving the period (this is the T I was referring to in my last post, sorry for being unclear):

    T=2∏/ω

    so T=∏/10.

    Thank you for walking me through that. I understand how that part works. Now, the next part is finding the exponential and trigonometric coefficients of this function.... Which I am again stuck at.
     
    Last edited: Nov 1, 2012
  7. Nov 1, 2012 #6

    LCKurtz

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    Well, there is an easy way and a hard way from here. Here's the easy way. Your function is an even function as you know from graphing it, so its expansion will not have any sine terms. You could note that your trig identity above does express your function as a finite cosine series, so if you go to the work of doing a half range cosine expansion, that is what you will get after a lot of work. And you can substitute for the cosine in terms of complex exponentials and you will have the complex form of the FS, which will also have finitely many nonzero terms.
     
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