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Laplace and Fourier Transforms

  1. May 29, 2009 #1
    Well, not sure where to begin. I have a couple questions that I have been unable to find answers to from professors/books/etc... I feel like I'm making a little progress in filling in the gaps but not completely.

    First of all, I just don't understand what makes the Laplace transform work so well. All of my books on differential equations just give the definition and then results and such. But how is the integral derived? I shudder to think that someone just "noticed" it works.

    On to Fourier.. I have not formally studied the Fourier transform yet, but I have studied the Fourier series. My first reaction to seeing the transform is that is looks so similar to the Laplace transform that maybe the Laplace is a specific case of the Fourier transform (kind of doubt that). I am still not sure exactly how the Fourier transform comes out of the Fourier series. How does this work? There seems to be a resemblance between the series in exponential form and the integral definition, but I am still not sure how the series goes into the integral definition.

    Does the Fourier transform work pretty much the same way as the Laplace transform only it takes care of complex functions?

    Also, are these two things linked somehow? Or should I stop with the bad habit of assumption and start to think of these as two completely independent things?

    Sorry if this type of thing has been posted before. I used the search option but didn't find much.
     
  2. jcsd
  3. May 29, 2009 #2

    chroot

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    The variable s, used in the Laplace transform, is a complex number usually written in the form [itex]s = \sigma + i \omega[/itex]

    The Fourier transform is obtained from the Laplace transform by restricting s to only the imaginary axis. In other words, if you fix [itex]s = i \omega[/itex], you get the Fourier transform. The Fourier transform is a special case of the Laplace transform.

    The imaginary parameter, [itex]\omega[/itex], indicates frequency. The real parameter indicates damping. Since the Fourier transform considers only pure tones (undamped (co)sine waves which oscillate forever), it makes sense that the Fourier transform considers only the imaginary parameter.

    The history of these transforms involved a number of people and a great many years, and I'm not well-versed in it, so take what follows with a grain of salt. I believe Fourier's work came first, in describing functions as sums of sines and cosines. It was later discovered that Fourier's transform is a type of integral transform, and people like Laplace began studying other kinds of integral transforms, including those which can represent functions in terms of both damped and undamped (co)sine waves.

    - Warren
     
  4. May 30, 2009 #3
    Ok thank you! That helps a lot. I reviewed the course description for PDE I'll be taking soon, I think I will wait until that class before continuing to study the subject since I'll be learning about Fourier series in that class. So hopefully that will help bridge some gaps.

    The problem I have with studying topics on my own is I get sidetracked to other topics so easily..
     
  5. May 30, 2009 #4

    Mute

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    That statement should probably be qualified by saying the Fouirer Transform is a special case of the bilateral laplace transform, whose limits are the whole real line. The usual Laplace transform just covers the half real line, so the Fourier transform is generally not a special case of it.

    There's brief bit of history about the transform on wikipedia's page:

    http://en.wikipedia.org/wiki/Laplace_transform

    Also, note that there are some cases in which the Fourier transform exists but the bilateral laplace transform doesn't, as the integral doesn't converge (or converges, but only for a limited range of the real part of s). A typical example are the characteristic function and moment generating function of a probability distribution, the first being the Fourier transform, which always exists, and the second being the bilateral laplace transform, which does not necessarily exist for all or any s.
     
    Last edited: May 31, 2009
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