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Getting Harmonics using Fourier Series

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    p(t)={ -1 from -1/220 to -1/330
    0 from -1/330 to -1/660
    1 from -1/660 to 1/660
    0 from 1/660 to 1/330;
    -1 from 1/330 to 1/220 }

    p(t) represents the period of the excess air pressure of a sound wave. Find the harmonics and their intensity.

    2. Relevant equations

    (1) p(t) = A0/2 + Ʃ ( An cos(2.n.pi.f0.t) + Bn sin(2.n.pi.f0.t) )

    (2) An = 2f0 ∫ p(t).cos(2.n.pi.f0.t) dt (from 0 to 1/f0)

    (3) Bn = 2f0 ∫ p(t).sin(2.n.pi.f0.t) dt (from 0 to 1/f0)

    3. The attempt at a solution

    This problem seams quite simple, but I am going crazy with it.

    f0 the fundamental frequency, is the frequency of p(t) wich is 110.

    If you draw p(t) you can easily verify that it is an even function, so you will only need to calculate the coeficients An, using equation (2). This is so, because the integral of p(t) (even) times sin(2.n.pi.f0.t)(odd) yealds zero.

    So I integrate p(t).cos(2.n.pi.f0.t) from zero to 1/220 and multiply by 2 in order to find An, but what I get is An=0, and it just can't be =S

    I tried other equivalent approaches like integrating from -1/220 to +1/220; integrating from -1/220 to 0 and multiply by 2; etc. Always 0.

    Would somebody help?
     
  2. jcsd
  3. Jan 21, 2012 #2

    vela

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    Show us how you're getting 0 when you calculate an.
     
  4. Jan 22, 2012 #3
    I cannot attach images, I press attachments and nothing happens...

    But while I was trying to explain how I did it I noticed my error. At least a big one, cuz I'm not getting exactly the expected result yet.

    The expected result for the relative intensities of the harmonics is:
    1 : 0 : 0 : 0 : 1/25 : 0 : 1/49 ; being 1 the fundamental frequency.

    What I get is: 1 : 0 : 0 : 0 : 1/5 : 0 : 1/7 ; also in the 5th coeficient I'm obtaining a negative value, is it ok for a coefficient to be negative?

    My result for the An formula is as follows: (2/n.pi)(sin(n.pi/3) + sin(2n.pi/3))
    The expected result suggests the 'n' of the first brackets on my An formula should be squared, but it makes no sense through my calculation.

    But it would be nice if I could upload my work, any idea why it does not work?
     
  5. Jan 22, 2012 #4

    vela

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    Yes, a coefficient can be negative.

    You can type in your work using LaTeX. Here's a FAQ on it. It's actually preferable to posting an attachment of your scanned work, which is often hard to read and inconvenient to work with.
     
  6. Jan 22, 2012 #5
    And when I introduce the coefficient in the expanded series shall I put it positive or negative?

    I finally got able to attach files, sorry I'm not using LaTex. I think it is readable, hope you understand it. Thank you for your time.
     

    Attached Files:

  7. Jan 22, 2012 #6

    vela

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    Your work looks fine. Remember power is proportional to the amplitude squared.
     
  8. Jan 22, 2012 #7
    That's it! Thanks a lot =)
     
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