Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Transform justification

  1. Jul 21, 2005 #1
    I'm trying to justify to myself that the inverse Fourier transform of the Fourier transform of a function is the function itself, provided that the FT exists. I can't simplify the double integral that results when this operation is performed, and much to my dismay, nothing at Mathworld, Wikipedia, or Google has the answer. This is the integral which I'm trying to compute:

    [tex] \frac{1}{2 \pi}\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}f(t) e^{-i \omega t}dt) e^{i \omega t}d{\omega} [/tex]

    Is there some sort of coordinate transformation that would help me out here? I have no idea how to begin simplifying this. Thanks.
     
  2. jcsd
  3. Jul 21, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    the t in the e to the iwt shouldn't be a t, since you then have t as a dummy variable (inner integral) and a variable (outer integral), or conversely the inner t's ought not to be t's. the (or a ) correct integral to work out is

    [tex] \frac{1}{2 \pi}\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}f(s) e^{-i \omega s}ds) e^{i \omega t}d{\omega} [/tex]

    which works out as f(t) as required (i think).
     
  4. Jul 21, 2005 #3

    uart

    User Avatar
    Science Advisor

    Hi Manchot, just a quick reply to maybe put you on the right track.

    A significant obstacle to you being able to simplify that equation is that you've used the variable "t" twice in two slightly different contexts. This is a really easy mistake to make with integral transforms. Basically what I'm saying is that your first use of "t" in the inner integral might just as well have been a dummy variable (because it's integrated out anyway.) So replace "t" in the inner integral with [tex]\lambda[/tex] or whatever and you'll suddenly find ways to re-arrange that expression that weren't previously apparent.

    Hope this helps.
     
  5. Jul 21, 2005 #4

    uart

    User Avatar
    Science Advisor

    Sorry Matt, something must be updating slow here today because I didn't see your post before replying with essentually the same answer. o:
     
  6. Jul 21, 2005 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How easily this works out depends on one's perspective. If I were in a physics class, I would blithely interchange the order of integration and do the formal calculation

    $$
    \begin{align}
    \frac{1}{2 \pi}\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty}f(s) e^{i \omega \left( t - s \right) }ds \right) d\omega &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty}f(s) e^{i \omega \left( t - s \right) }d\omega \right) ds\\
    &= \int_{-\infty}^{\infty} f(s) \delta \left( t - s \right) ds\\
    &= f(t),
    \end{align}
    $$

    but if I were in an analysis class, I would proceed somewhat more cautiously!

    Regards,
    George
     
    Last edited: Mar 2, 2016
  7. Jul 21, 2005 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    no need for that level of caution: a mathematician would simply require that f satisfies sufficient criteria for you to be able to do this, whereas a physicist would assume it as being obviously true that one could do this. one of the reasons i never understood physics was i never knew what theyy were assuming.
     
  8. Jul 21, 2005 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I agree. Useful to a mathematician would be something like:

    If [itex]f[/itex] is integrable and piecewise continuous on [itex]\mathbb{R}[/itex], and the Fourier transform [itex]\hat{f}[/itex] is absolutely integrable, then [itex]f[/itex] is continuous and

    [tex]
    f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{it\omega} \hat{f} (\omega) d\omega.
    [/tex]

    But Manchot, in his original post, said

    I interpret this as Why? or How?, rather than When?. The statement I gave above is proved (along with a more general statement) in Gerald Folland's excellent book Fourier analysis and Its Applications. Folland says: "Our task is to investiagte the validity of Manchot's problem. Like the question of whether the Fourier series of a periodic function [itex]f[/itex] converges to [itex]f[/itex], this is not entirely straightforward."

    Regards,
    George
     
  9. Jul 21, 2005 #8
    Thank you all for your help so far. I did know that the inner t was a dummy variable, and had been able to change the order of integration, but the step in George Jones' justification wherein [itex]\frac{1}{2 \pi} \int_{-\infty}^{\infty}f(s) e^{i \omega \left( t - s \right) }d\omega[/itex] is replaced with [itex]\delta (t-s)[/itex] still eludes me. Now, I know that the delta function can be "defined" in this way, but in the class I'm learning this in (an electrical engineering class), the delta function was defined to be the limit of a rectangle centered at zero whose height approached infinity. (I realize that this is not rigorous, but I don't mind that part so much). From there, it was shown that the delta function is the identity element of convolution, and using inverse Fourier transforms, the above statement was taken to be true. So, I guess that my question is now this: using my class' definition, how could one show that [itex]\frac{1}{2 \pi} \int_{-\infty}^{\infty}f(s) e^{i \omega \left( t - s \right) }d\omega = \delta (t-s)[/itex] (without using Fourier transforms, of course)?
     
  10. Jul 21, 2005 #9

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think that at an elementary, non-rigorous level, this representation of the delta function might be hard derive without using Fourier transforms, so, tomorrow, when I'm little less tired, I'll give a non-rigorous proof of the Fourer inversion theorem that doesn't use a delta function.

    This representation of the delta is quite easily derived using the more abstract formalism of tempered distributions.

    Regards,
    George
     
  11. Jul 22, 2005 #10
    Can you show it's true for [tex] x^n [/tex] and then by linearity it's
    true for all functions that have a taylor series?
     
  12. Jul 22, 2005 #11

    uart

    User Avatar
    Science Advisor

    No no no! The f(s) shouldn't be in there, it should be shuffled off to the outer integral because its not a function of [tex]\omega[/tex].

    The inner integral with the "delta identity" should just read,
    [tex]\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i \omega \left( t - s \right) }d\omega = \delta (t-s)[/tex].

    This identity reduces the outer integral to a simple convolution integral of f(s) with [tex]\delta (t-s)[/tex] which of course recovers f(t).
     
  13. Jul 22, 2005 #12

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let [itex]\hat{f}[/itex] be the Fourier transform of [itex]f[/itex]. The key idea used in the proof of the Fourier Inversion Theorem is to take the inverse Fourier transform of [itex] e^{-\epsilon^2 \omega^2} \hat{f} (\omega)[/itex], and, at the end, to let [itex]\epsilon \rightarrow 0[/itex], so that [itex] e^{-\epsilon^2 \omega^2} \hat{f} (\omega) \rightarrow \hat{f} (\omega)[/itex].

    [tex]
    \begin{equation*}
    \begin{split}
    \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{-\epsilon^2 \omega^2} \left( \int_{-\infty}^{\infty}f(s) e^{-i \omega s}ds \right) e^{i t\omega} d\omega &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} f(s) \left( \int_{-\infty}^{\infty} e^{-[\epsilon^2 \omega^2 + i \left( s - t \right) \omega ]} d\omega \right) ds\\
    &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} f(s) \frac{\sqrt{\pi}}{\epsilon} e^{-\frac{\left( s - t \right)^2 }{4\epsilon^2}} ds\\
    &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} 2\sqrt{\pi} f(2u\epsilon + t) e^{-u^2} du
    \end{split}
    \end{equation*}
    [/tex]

    Note: 1) the interchange of the order of integration is justified because the integrand is absolutely convergent; 2) I did the integral in the brackets on the right of the first equals sign was done by completing the square and choosing an appropriate contour; 3) a change of variable [itex]u = (s - t)/(2\epsilon)[/itex] was made.

    Now take the limit as [itex]\epsilon \rightarrow 0[/itex] of both sides of the above equation. Moving the limits through the integral signs (justified by the dominated convergence theorem) gives

    [tex]
    \begin{equation*}
    \begin{split}
    \frac{1}{2 \pi}\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty}f(s) e^{-i \omega s}ds \right) e^{i t\omega} d\omega &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} 2\sqrt{\pi} f(t) e^{-u^2} du\\
    &= f(t).
    \end{split}
    \end{equation*}
    [/tex]

    Regards,
    George
     
  14. Jul 22, 2005 #13

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Here's the proof from the book 'Applied Complex Variables' by Yue Kuen Kwok. It uses Fourier series. The definition is slightly different though. The FT has no minus in the exponent, but since I'd rather have the minus in the FT instead of the inverse FT I'll adjust the proof.
    We wish to establish the relation:

    [tex]u(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}\int_{-\infty}^\infty e^{-i\omega t}u(t)dtd\omega[/tex]

    We start from the Fourier series representation of a periodic function u(x) definied over the finite interval [-L,L]. The Fourier expansion of u(x) takes the form:

    [tex]u(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi x}{L}+\sum_{n=1}^\infty b_n \sin \frac{n\pi x}{L}.[/tex]

    The conditions required for the validity of the above expansion are that f(x) (MY NOTE: he writes f(x), but surely he means u(x)) has only a finite number of finite discontinuities and only a finite number of maxima and minima. These conditions are sufficient, but not necessary and are commonly called the Dirichlet conditions.
    By taking advantage of the orthogonality properties of the component functions over [-L,L], the Fourier coefficients a_n and b_n are given by:
    [tex]a_n=\frac{1}{L}\int_{-L}^L u(t)\cos \frac{n\pi t}{L}dt, \qquad
    n=0,1,...[/tex]
    [tex]b_n=\frac{1}{L}\int_{-L}^L u(t)\sin \frac{n\pi t}{L}dt, \qquad
    n=1,2,...[/tex]

    In full, the Fourier expansion can be written as:
    [tex]u(x)=\frac{1}{2L}\int_{-L}^L u(t)dt+\frac{1}{L}\sum_{n=1}^\infty
    \cos \frac{n \pi x}{L}\int_{-L}^L u(t)\cos \frac{n\pi t}{L}dt[/tex]
    [tex]+\frac{1}{L}\sum_{n=1}^\infty \sin \frac{n \pi x}{L}\int_{-L}^L
    u(t)\sin \frac{n\pi t}{L}dt[/tex]
    [tex]=\frac{1}{2L}\int_{-L}^L u(t)dt+\frac{1}{L}\sum_{n=1}^\infty
    \int_{-L}^L u(t)\cos \frac{n\pi}{L}(t-x)dt[/tex]

    Suppose we let L approach infinity so that [-L,L] becomes [itex](-\infty,\infty)[/itex]. Further, we set:[itex]\frac{n\pi}{L}=\omega[/itex] and [itex]\pi/L=\Delta \omega[/itex].
    As [itex]L\to \infty[/itex], the first term in the above expansion
    vanishes since the value of the integral is finite given that u(t) is absolutely integrable. The second term becomes the series:
    [tex]\frac{1}{\pi}\sum_{i=1}^{\infty}\Delta
    \omega\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt[/tex]
    In the limit [itex]\Delta \omega \to 0[/itex], the above infinite sum is replaced
    by an integral with integration variable [itex]\omega[/itex].
    We thus formally obtain:
    [tex]u(x)=\frac{1}{\pi}\int_{0}^{\infty}\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt d\omega[/tex]
    To arrive at the form shown in the beginning, we observe that [itex]\cos \omega(t-x)[/itex] is an even function of [itex]\omega[/itex] and [itex]\sin \omega(t-x)[/itex] is an odd function of [itex]\omega[/itex]. We then have:
    [tex]u(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(t)\cos \omega(t-x)dt d\omega \quad (1)[/tex]
    and
    [tex]0=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(t)\sin \omega(t-x)dt d\omega \quad (2)[/tex]

    Adding (1) and -i times (2) together, the result is finally established.
    ------

    The proof is not overly rigorous, but it does show nicely that the FT follows more or less empirically from the Fourier series.
     
    Last edited: Jul 22, 2005
  15. Jul 22, 2005 #14
    I thank all of you for your help. It does not go unappreciated.
     
    Last edited: Jul 22, 2005
  16. Jun 4, 2010 #15
    Referring to post 12,we define the 'inner integral' as the fourier transform and after that,'outer one' as the inverse fourier transform...so does that mean that the whole idea of a fourier transform (and its inverse)is to make it easier for us to find the fourier integral?(it seems to be loop:we start off with f(x),then we find the transform by finding the inner integral,then we seem to get back f(x) by doing the outer integral...pretty strange.)

    Also,(since the fourier transform seems to have some kind of relation with the fourier integral)is the concept of fourier transform applicable for only non periodic functions?
     
  17. Jun 6, 2010 #16
    Anyone?
     
  18. Jun 9, 2010 #17

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The integral that defines the Fourier transform is

    [tex]\int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt[/tex]

    Since

    [tex]\left|f(t) e^{-i\omega t}\right| = \left| f(t) \right|[/tex]

    it follows that the definition makes sense (the integral converges absolutely) if and only if [itex]f(t)[/itex] is absolutely integrable.

    This in general cannot be true for a periodic function, UNLESS the function is zero almost everywhere, in which case the Fourier transform is simply zero.

    Periodic functions are handled most naturally via Fourier series.

    However, it is possible to enlarge the class of "functions" for which Fourier transforms exist. One way to do this is via the theory of distributions (aka generalized functions). There's a lot of machinery needed to make this rigorous, but the outcome is that periodic functions DO have Fourier transforms (in the sense of distributions), consisting of an evenly spaced sequence of Dirac "delta functions" with "complex amplitudes" equal to the Fourier series coefficients.

    You should be able to find a non-rigorous treatment of this in many applied books that use Fourier transforms/series - e.g., a differential equations text, or an engineering "digital signal processing" book or the like. See also Kammler, "A First Course in Fourier Analysis" and Strichartz, "A Guide to Distribution Theory and Fourier Transforms."
     
  19. Jun 10, 2010 #18
    Thanks,jbunniii,so your post confirms,basically,that fourier transforms are used specially for non periodic functions,but can be made to work on periodic functions also,provided we manipulate the function(using dirac delta functions and all that) to make it non periodic.
    Could you also throw some light on my other question:is the whole idea of a fourier transform (and its inverse)to make it easier for us to find the fourier integral?

    (it seems to be loop:we start off with f(x),then we find the transform by finding the inner integral,then we seem to get back f(x) by doing the outer integral...pretty strange.)
     
  20. Sep 28, 2010 #19
    I think I can make some sense of the fourier transform now!!

    I think the fourier transform is basically the coefficient of the fourier series when the interval 'l' tends to infinity(fourier integral) whereas we get back the function f(x) when we do fourier inverse transform because the inverse transform is nothing but the fourier series of the function....is this right?
    I've seen the derivation for the fourier transform in Kreyzig...that's where I got this concept from.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier Transform justification
  1. Fourier Transform (Replies: 2)

  2. Fourier transform? (Replies: 3)

  3. Fourier Transformations (Replies: 10)

Loading...