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An understanding of Laplace Transform.

  1. Sep 15, 2010 #1
    Currently enrolled in an Electrical Engineering program, i naturally had to come across Laplace Transformation.

    Maths and specially calculus isn't my favourite :)

    What I understand is that Laplace Transform is more or less a tool used to simplify matters by defining a time based quantity in terms of frequency. Hence in terms of signals, breaking down a complex signal into separate components based on frequency.

    The equation being :



    here is the part i need a little help with...

    Why e^-st is involved in the equation, i tried looking at a derivation but couldn't understand much from what i found. What is e^-st really signifying here, in physical or mathematical terms...

    Also, F(s), s here is basically frequency right, but...how do i phrase this...what does F(s) actually represent. I know it doesn't make sense but thats the best i could word it :P

    Any help would be great, thanks...
  2. jcsd
  3. Sep 15, 2010 #2
    The Laplace transform is quite similar to the Fourier transform. Both transformation can, in some sense, be viewed as a "change of basis".

    Normally a function f(x) has an input x and an ouput f(x). This can be viewed as the "coordinate x has an amplitude f(x)". The whole function is then a bunch of amplitudes associated to coordinates.

    If we take the Fourier transform we're making a change of basis. Instead of assigning amplitudes to the coordinates, we switch to an interpretation in which we assign amplitudes to frequencies. The idea behind this is that any function can be written as a sum over (complex) waves: exp(iw). (In turn, you can also interpret this as a sum over sines and cosines). The amplitude F(w) we assign to a frequency w "measures" the contribution of the wave with that frequency to the original function f(x).

    For the Laplace transformation the same idea applies: instead of representing the function as amplitudes assigned to coordinates (or time), we switch bases and assign amplitudes to moments. Moments are yet another way of representing the "shape" of a function f(x).

    To be a bit more specific: The k'th moment of the function f(x) can be found by evaluating the k'th derivative of the function F(s) at s=0. Since the function F(s) is completely determined by all its derivatives at s=0, each function F(s) correspond to a different set of moments.

    So what is a moment? A moment is a measure of "how do points deviate from the mean value". If a signal f(x) is a delta peak at t=0, then it does not deviate from its mean value. In turn, the Laplace transform is a constant: F(s) = 1. All its derivatives are at s=0 are zero.

    To be more specific, the first moment is nothing but the mean value of the function f(x), namely:

    [tex]dF(0)/ds = \int_0^\infty x f(x) dx \equiv \mu [/tex].

    The second moment is related to the variance.

    [tex]dF(s)/ds (s=0) = \int_0^\infty x^2 f(x) dx [/tex].

    I say related, because the true variance is actually calculated relative to the mean value: [tex]\int_0^\infty (x-\mu)^2 f(x) dx[/tex]

    Higher moments are given by:

    [tex] \int_0^\infty (x-\mu)^k f(x) dx [/tex].

    And knowledge of all moments completely specifies the signal f(x)!
  4. Sep 15, 2010 #3
    @xempa, good reply :)
  5. Sep 15, 2010 #4
    Thanks a lot...

    Really helped !
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