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Fourier Transform of an equation representing a sound

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    http://dl.dropbox.com/u/11341635/IntegrationProblem.jpg [Broken]

    2. Relevant equations

    http://dl.dropbox.com/u/11341635/Fourier%20Transform%20Equations.jpg [Broken]

    3. The attempt at a solution

    http://dl.dropbox.com/u/11341635/1st%20part%20of%20attempt.png [Broken]
    http://dl.dropbox.com/u/11341635/2nd%20part%20of%20attempt.JPG [Broken]
    http://dl.dropbox.com/u/11341635/3rd%20part%20of%20attempt.JPG [Broken]

    The 3rd part of the attempt is only the top half of the page.

    Any help would be greatly appreciated.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 25, 2012 #2

    vela

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    You made a mistake in the very beginning. You can't pull ##\cos\omega_0 t## out like that.

    You had
    \begin{align*}
    x(t) &= e^{-(t/t_0)^2}\cos [\omega_0t(1-qt)] \\
    &= e^{-(t/t_0)^2}\left[\frac{e^{j\omega_0t(1-qt)} + e^{-j\omega_0t(1-qt)}}{2}\right]
    \end{align*}So far, so good. When you multiply it out, you get two terms. Looking at just the exponential in the first term, you have
    $$e^{-(t/t_0)^2}e^{j\omega_0t(1-qt)} = \exp \left[-\left(\frac{t}{t_0}\right)^2 + j\omega_0t(1-qt)\right]$$Now collect powers of t in the exponent to get:
    $$e^{-(t/t_0)^2}e^{j\omega_0t(1-qt)} = \exp \left[-\left(\frac{1}{t_0^2} + j\omega_0q\right)t^2 + j\omega_0 t\right] = \exp \left[-\alpha^2\left(t^2 - \frac{j\omega_0 }{\alpha^2} t \right)\right]$$ where ##\alpha^2 = \frac{1}{t_0^2} + j\omega_0q##. What you want to do now is to complete the square in the exponent so you'll end up with an integral that looks like the one given in the identity. Obviously, you'll have to do a similar calculation for the second term.
     
    Last edited: Jan 25, 2012
  4. Jan 25, 2012 #3
    Thankyou so much - I've just watched a quick refresher on YouTube for completing the square so we'll see how I get on now I guess.
     
  5. Jan 25, 2012 #4
    I've done the majority of the work I think. It'd be great if you'd be willing to give it a quick look.

    http://dl.dropbox.com/u/11341635/2nd%20attempt.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Jan 25, 2012 #5

    vela

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    Be careful with the signs, otherwise it looks good so far. You also need a factor of m in front.
     
  7. Jan 25, 2012 #6
    a solution of sorts

    Well I'm done. The end equation looks pretty weighty, so I've know idea if it's right or not. Thanks again for your help so far.

    http://dl.dropbox.com/u/11341635/Part%201.png [Broken]

    http://dl.dropbox.com/u/11341635/Part%202.png [Broken]

    http://dl.dropbox.com/u/11341635/Part%203.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Jan 25, 2012 #7
    Sorry if this is a bit OT, but does anybody know where I could learn about this kind of stuff? It really interests me. (keep in mind I'm only in calculus in high school right now)
     
  9. Jan 25, 2012 #8
    Actually I've been finding YouTube to be a pretty good place for tutorials on even some pretty hefty maths. Check out Matlab - it's like Excel but a million times better. Also check out Mathematica and their demos project - http://demonstrations.wolfram.com/

    I wouldn't recommend Wikipedia as a place to start with Maths - unless it's incredibly simple stuff. It's not that it's unreliable - it's actually pretty reliable. It's more that you get way way too much information without any nice ways of explaining things.
     
  10. Jan 26, 2012 #9

    vela

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    Whoops! We should have multiplied by e-jωt right from the start. Also, I still see numerous algebra errors I alluded to before.

    Why are you integrating from -0.1 to 0.1? For the Fourier transform, you're supposed to integrate from -∞ to +∞.
     
    Last edited: Jan 26, 2012
  11. Jan 26, 2012 #10
    Well I checked the plus and minus signs quite a few times already. I'm not sure about the factor of m - where does that come in?

    I'm integrating from -0.1 to 0.1 because the original graph dies off towards -0.1 and 0.1
     
    Last edited: Jan 26, 2012
  12. Jan 26, 2012 #11

    vela

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    You could carry on from where you are (after you fix the errors that are already there). You'll end up completing the square yet again.

    If you start with the factor e-jωt right from the start, you only have to complete the square once. It's actually not that hard to fix. Going back to what I wrote in post #2, when you include the exponential from the Fourier transform, you get

    $$e^{-(t/t_0)^2}e^{j\omega_0t(1-qt)}e^{-j\omega t} = \exp \left[-\left(\frac{1}{t_0^2} + j\omega_0q\right)t^2 + j\omega_0 t\right]e^{-j\omega t} = \exp \left[-\alpha^2\left(t^2 - \frac{j(\omega_0-\omega)}{\alpha^2} t \right)\right]$$So only the coefficient of the linear term changes slightly.
     
  13. Jan 26, 2012 #12

    vela

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    Look at the coefficient of t2 before and after you completed the square. They should be equal, but they're not in your work. Also, the sign of the quadratic term has to be negative, otherwise the integral won't converge.

    The problem is coming up because of the way you're completing the square. You don't need to set anything to 0, and you can't just arbitrarily decide to divide out m or p. Look at it this way:
    $$ax^2+bx = a\left(x^2+\frac{b}{a}x\right)$$Now complete the square of the quadratic inside the parentheses to get
    $$ax^2+bx = a\left(x^2+\frac{b}{a}x\right) = a\left[x^2+\frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] = a\left(x+\frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2$$
     
  14. Jan 26, 2012 #13
    Okay - got it now. Many many thanks.
     
    Last edited: Jan 26, 2012
  15. Jan 26, 2012 #14
    New version of the solution

    Not sure who's around - have been working on this for a bit - would be grateful for a hand - my solution seems too hefty. The aim was to do an analytical Fourier transform using a couple of trigonometric identities.

    http://dl.dropbox.com/u/11341635/1st%20bit.png [Broken]

    http://dl.dropbox.com/u/11341635/2nd%20bit.png [Broken]

    http://dl.dropbox.com/u/11341635/3rd%20bit.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  16. Jan 26, 2012 #15

    vela

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    Looks good. You can simplify it a bit by getting the imaginary terms out of the denominators, but the answer doesn't look like it's going to come out nice and neat like you want.
     
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