1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Transformations: Rewriting a F.T.

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I have the following Fourier transformation:

    [tex]
    u(x,t) = \sqrt {\frac{2}{\pi }} \int_0^\infty {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}
    \over f} _s (\omega )\,e^{ - c^2 \omega ^2 t} \sin \omega x\,d\omega },
    [/tex]
    where fs is the sine-transform of a function f. I am supposed to rewrite this into the form:

    [tex]
    u(x,t) = \frac{1}{{2c\sqrt {\pi t} }}\int_0^\infty {f(s)\left[ {\exp \left( { - \frac{{(x - s)^2 }}{{4c^2 t}}} \right) - \exp \left( { - \frac{{(x + s)^2 }}{{4c^2 t}}} \right)} \right]} \,ds
    [/tex]

    I can see I have to start working with the sine transform of f. But I don't know what to write this as?

    Thanks in advance,

    sincerely Niles.
     
  2. jcsd
  3. Oct 10, 2008 #2

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    Hint: look at the convolution theorem...
     
  4. Oct 11, 2008 #3
    Ok, I have that:

    [tex]
    \widehat u(\omega ,t) = \widehat f_s (\omega )e^{ - c^2 \omega ^2 t},
    [/tex]

    where the exponential function is the Fourier transform of the heat kernel (Gauss' kernel). But I do not know how the convolution theorem works, when it is the sine-transform of a function.

    If it works the regular way, then I do not get the expression above.
     
  5. Oct 11, 2008 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    If [itex]\widehat{f}_s(\omega)[/itex] is the Foruier sin transform of f(x), then [itex]\widehat{f}_s(\omega)=\mathcal{F}_s[f(x)][/itex] and

    [tex]\widehat{u}(\omega ,t) = \widehat{f}_s (\omega )e^{ - c^2 \omega ^2 t}=\mathcal{F}_s[f(x)]e^{ - c^2 \omega ^2 t}[/tex]

    Can you write [itex]e^{ - c^2 \omega ^2 t}[/itex] as the Fourier transform of a function [itex]g(x)[/itex]?
     
    Last edited: Oct 11, 2008
  6. Oct 11, 2008 #5
    Hmm, I can always find the inverse Fourier sine transform of [tex]e^{ - c^2 \omega ^2 t}[/tex], but that seems tedious.

    Also I can't find the function in a table of Fourier sine transforms. Is it the Gauss' kernel?

    Thanks for replying so fast.
     
  7. Oct 11, 2008 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    It's the Gauss' kernel, but it is the full Fourier transform (or a cosine Fourier transform) of a function.
     
  8. Oct 11, 2008 #7
    How can I use convolution in this context? Now I have:

    [tex]
    \widehat{u}(\omega ,t) = \mathcal{F}_s[f(x)]\mathcal{F}[g_G(x)]
    [/tex]

    where g_G is Gauss' kernel. The Fourier transform of a convolution is the product of each Fourier transformed part of the convolution - in the above expression I have both sine and regular F.T.?
     
  9. Oct 11, 2008 #8
    Ok, so the convolution theorem does not care whether we are looking at Fourier sine, cosine or "ordinary" transforms?

    Can you give me a hint of how I can rewrite the convolution to the expression in my very first post?

    Thanks again.
     
  10. Oct 11, 2008 #9

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    The convolution theorem says

    [tex]\mathcal{F}[(f*g)(x)]= \sqrt{2\pi} \hat{f}(\omega)\hat{g}(\omega)[/tex]

    [tex] \Rightarrow u(x,t)=\frac{1}{\sqrt{2\pi}}(f*g)(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(s)g(x-s)ds[/tex]

    Use the fact that f must be odd for its FS to be a sine series, f(-s)=-f(s) to rewrite this as:

    [tex] u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} f(s)[g(x-s)-g(x+s)]ds[/tex]

    Then substitute your g(x).
     
    Last edited: Oct 11, 2008
  11. Oct 11, 2008 #10
    Ok, I am a little uncertain about this. Here are my thoughts:

    The inverse Fourier transform (i.e. the complex form of the Fourier Integral Representation) is the analogous of the Fourier series, correct? So the inverse Fourier sine transform is the analogous of the Fourier sine representation, and hence f(s) must be odd? Am I correct about this?

    And if my thoughts are correct, then the inverse Fourier cosine transformation of a function f means that f is even?

    EDIT: In my book there is a theorem explaining the relationships between odd and even transforms. Perhaps that has something to do with this? The theorem is written in this PDF: http://www.math.ualberta.ca/~thillen/math300/assign5.pdf

    It is the comments after question 9 and 10 (at page 3). What is meant by restriction?
     
    Last edited: Oct 11, 2008
  12. Oct 13, 2008 #11

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    Yes.

    Yes again.

    Sorry, I can't access your link.
     
  13. Oct 17, 2008 #12
    I think the document is accessible now (it is to me, atleast).

    Thanks for taking a look at it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier Transformations: Rewriting a F.T.
  1. Fourier transform (Replies: 4)

  2. Fourier transform (Replies: 1)

  3. Fourier transform (Replies: 1)

  4. Fourier transformation (Replies: 1)

Loading...