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Fourier Transform Meaning

  1. Aug 5, 2012 #1
    Hi,

    I am having a little trouble with the physical meaning of a Fourier transform. I will try to pose a concrete example. Mathematically, the fourier transform of an exponential decay results in a Lorentzian function.

    Let's say I have a population that decays exponentially with time. Now, if I Fourier transform an exponential decay, I get some function (a Lorentzian) of frequency. What does this frequency function mean in the context of my exponential decay problem?

    Thank you for your help :)
     
    Last edited: Aug 5, 2012
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  3. Aug 5, 2012 #2

    chiro

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  4. Aug 5, 2012 #3
    Hi,

    Yes, I am aware of the frequency interpretation but in my population decay problem for example, what does the frequency decomposition correspond to physically?
     
  5. Aug 5, 2012 #4

    chiro

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  6. Aug 5, 2012 #5
    Sorry, I still don't get it. Everything is mathematically fine but I have no better physical intuition from knowing that derivatives of the Fourier coefficents have this property.

    Could you please explain, in your own words, what the frequency decomposition corresponds to in an exponential decay? I really don't have any idea how it fits in as of now. Thank you for your help!
     
  7. Aug 5, 2012 #6

    chiro

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    Well in terms of the derivatives, you are seeing how quickly the population is declining and also with regards to the acceleration (and further derivatives) of how it is declining.

    The above PDF links these things (velocity is the first derivative, acceleration is the second and so on) to the derivative (i.e. the fourier co-effecients).

    If you had specific attributes in mind for what you are looking for, then maybe you should make this explicit in this thread.
     
  8. Aug 5, 2012 #7
    chiro, so here's what I've got so far. My function here is an exponential decay in time. The Fourier coefficients of its derivative are related to the Fourier coefficients of the original exponential decay function by the relation in the link, yes? I now have a connection between the Fourier coefficients of the exponential decay function and the Fourier coefficients of its derivative.

    But here's my question, phrased a little differently. Suppose you have a plane wave of the form [itex]e^{i\omega_{0} t}[/itex]. A Fourier transform gives you a delta function [itex]\delta(\omega-\omega_{0})[/itex]. This has a very clear physical interpretation. The Fourier coefficients are telling me that my plane wave has only one frequency, [itex]\omega_{0}[/itex].

    Now, suppose we go back to the exponential decay. What physical property do the Fourier coefficients signify? I am looking for something like the connection in the preceding paragraph.

    I apologize if it hasn't been clear (or if I am taking a long time to understand something really simple) but I can't really see the answer yet. Thank you!
     
  9. Aug 5, 2012 #8

    chiro

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    I'm not sure exactly what you want.

    The PDF I linked gives you the connection of the derivatives of the original function to the fourier coeffecients/frequency information and this will tell you how your population is decaying over time with respect to the rates of change.

    You said above that you wanted to see how the fourier analysis and results related to the physical problem, and I can't really think of a better way than to see how the population decay is affected by its derivatives.

    In terms of the frequency information with the delta, if you want to look at this kind of thinking and approach you should get a decent book on signal processing which will go into the whole "impulse" approach (i.e. this delta function approach) of signal analysis: I'm not an engineer or really that well informed, but that is the place to understand how the paradigm of impulses is used in signal processing.

    One thing that might aid you is to just write down a list of physical attributes that come to your mind that are of particular interest or importance, and then think about how you would see that mathematically in terms of any transform and decomposition: I have mentioned derivatives, you are looking at fourier analysis, but if you can link your ideas to these (and more) then you should move towards answering these kinds of questions.

    So if you want to look at the impulse approach to understanding signals, get a good resource on signal processing.
     
  10. Aug 6, 2012 #9
    I feel like I was slightly without intuition when it comes to the Fourier transform until I looked at the coefficients as "weights" for the various frequencies that go into a particular functional reconstruction.

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

    Look at the second equation, the inverse Fourier transform. Notice that [itex]F(\omega)[/itex] tells you how much "weight" to apply to the waves [itex]e^{i \omega t}[/itex] when reconstructing the time-dependent (original) f(t) function. The weight is different based only on the frequency ([itex]F(\omega)[/itex] is only a function of [itex]\omega[/itex]). Now [itex]F(\omega)[/itex] is the Fourier transform! So the Fourier transform can be interpreted as the weight that each wave at a particular frequency has in reconstructing the time (or position) dependent function. If the Fourier transform is low at a particular frequency range, than the resulting integral over the Fourier transform will contribute less in reconstructing f(t). I think this is a necessary interpretation when understanding the Fourier transform.

    The Fourier transform contains all of the information it takes to know which frequencies have how much importance in reconstructing a particular position (or time dependent) function.

    Your example of plane waves and the delta function exhibited this perfectly.
     
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