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Integral want same answer as book

  1. Aug 10, 2012 #1
    Hi Guys,

    I'm doing some work on signals and power spectrums.

    The book I am using says the spectrum for this light scattering signal is given by:

    [tex]I(q,\omega ) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {\exp ( - i\omega t)} S(q,t)dt[/tex]

    In the book an S term is given as:

    [tex]S(q,t) = N\left\langle {{\gamma ^2}} \right\rangle \exp [ - ({q^2}{D_T} + 6{D_R})t][/tex]

    So what I want to do is work out the spectrum manually (using mathemtica or matlab) and get the same answer that they do. I need this because soon I will be using equations that aernt in the book and need to make sure the spectrum comes out correct.

    But when I try to do it in mathematic I get an error saying the integral does not converge.

    Matlab gives an answer but it looks strange and wrong.

    The book then says that the spectrum is therefore (this is what I want to arrive at via matlab or mathematica):

    [tex]I(q,\omega ) = \frac{{N\left\langle {{\gamma ^2}} \right\rangle }}{\pi }\frac{{{q^2}{D_T} + 6{D_R}}}{{{\omega ^2} + {{({q^2}{D_T} + 6{D_R})}^2}}}[/tex]

    any ideas?

    Thanks
     
  2. jcsd
  3. Aug 10, 2012 #2

    uart

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    Science Advisor

    I think it's pretty obvious that S(q,t) is diverging badly for negative t. Are you sure there's not an absolute value or a square ([itex]t^2[/itex]) or something in the [itex]({q^2}{D_T} + 6{D_R})t[/itex] term that you're omitting?
     
    Last edited: Aug 10, 2012
  4. Aug 10, 2012 #3
    Hey, thanks for your answer. For these experiments, t is a delay time, so it cant really go negative. t spans from several decades from say e-7 to e5 seconds. So the exp term will be 1 at the shortest of times, and 0 at longest. But that is the equation exactly in the book. I dont get why I cant just put it in and watch it spin. How do the authors do these equations lol
     
    Last edited: Aug 10, 2012
  5. Aug 10, 2012 #4

    uart

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    To get the answer in the book you need to use the absolute value of "t". It must have been "implied" that S(q,t) as written was for positive "t" and that it was symmetric for negative "t".

    [tex]S(q,t) = N\left\langle {{\gamma ^2}} \right\rangle \exp [ - ({q^2}{D_T} + 6{D_R}){\bf|t|}] [/tex]

    BTW. This is just Fourier transform. You can look up that integral as the FT of [itex]e^{-a|x|}[/itex] here: http://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms
     
  6. Aug 10, 2012 #5
    Wow, thanks so much! When I use Abs(t) it gets the exact answer. I guess it makes sense because time cant be negative. Huge help thanks!
     
  7. Aug 12, 2012 #6
    Hey,

    You've been such a great help may I extend my question now?

    I have some real data that I want to turn into the power spectrum. The data is in the form of x y where x is t, and y is S(q,t).

    So I can fourier transform the theoretical equation, to obtain the answer for how we expect the power spectrum to turn out. Now... how do I fourier transform my experimental data? Its just a set of numbers.

    So for the first delay time, S(q,t) will be 1. Do I just try

    [tex]I(q,\omega ) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {\exp ( - i\omega t)} \times1dt[/tex]

    I dont get what goes in ω, or what limits I use?

    Thanks for the help! Right now the machine I use Pumps out the autocorrelation function (S(q,t)), but I want the power spectrum.
     
  8. Aug 12, 2012 #7

    uart

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    Hi Steve. For a set of numbers like that you use the DFT ( http://en.wikipedia.org/wiki/Discrete_Fourier_transform ). Matlab has a built in function "fft" to perform this operation.
     
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