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Math Help - Direction of the Laplace transform

  1. #1
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    Direction of the Laplace transform

    Hello,
    the Laplace transform is typically interpreted as a transformation from the time-domain to the frequency-domain:

    F(s) = L{f(t)}.

    However, if I measure a time-dependent (nonexponential) decay function, I interprete this as a weighted sum of (infinitively many) monoexponential decay components. But then you have to perform the Laplace transform with s and t swapped (from the frequency-domain to the time-domain):

    f(t) = L{g(s)}.

    To find the distribution g(s), being given a time-dependent decay function, one has to apply an inverse Laplace transform.

    Why is the definition exactly the other way round? When is the Laplace transform applied this way and how can you (geometrically?) interprete the the image F(s)?

    Alex
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  2. #2
    Senior Member yeKciM's Avatar
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    for which purpose you need this ?

    so it can be answered leaning to that direction

    anyway what is at most time needed is that you can see and understand region of convergence, so sometimes it's needed to show on "s-plane" (real and imaginary axis) how and where does it converge (exist) because if it diverges than doesn't exist Laplace transform of that signal
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  3. #3
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    Thank you for your answer.
    In my case, I measure (often nonexponential) luminescence decay process. In many cases, one can assume that the nonexponential process is a superposition of a very large number of monoexponential processes. And I want to find out the distribution of the decay times for these.
    This is very similar to Laplace DLTS, where a current transient is analyzed to deternmine the decay rates for several types of impurities.

    In both cases, an inverse Laplace transform is needed to convert the time dependent transient to the corresponding frequency spectrum (which is in contrast to the definition, where a forward Laplace transform is applied on the time-domain function to find the function int the frequency-domain).

    Alex
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  4. #4
    Senior Member yeKciM's Avatar
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    interesting

    perhaps I'm just lost while translating... one use Laplace transform to go to  S= \sigma +j \omega region or often called frequency region, and use Inverse Laplace transformation to go back from that region.

    but in your case you say it's another way around

    please post some example of what do you do, and how do you do that perhaps that way it will be more clear of what is the trouble there

    in most cases i'll used Laplace transformation for solving equation of state of some electrical circuits... it's just being used to simplify calculations, so that you don't have to solve integral/differential equations but simple algebraic equations... and after solving those just go reverse to time domain using inverse Laplace transformation ...

    as for that interpretation of images F(S), you understand that is just another function in S-plane so you can do that (for those regions for which your function is converge)

    please post some work, now i'm really interested
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  5. #5
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    Sorry that it took a while.
    One sample:

    I measure a power law decay of the form

    Code:
    f(t) = 1/(1+t/t0)^beta.
    As this model is empirical, the parameters are more or less meaningless.
    However, you can interprete this function as a sum of exponential functions:

    Code:
    f(t) = int(1/s*(t0*s)^beta*exp(-t0*s)/GAMMA(beta) * exp(-t*s),s=0..infinity)
    Where

    Code:
    invlaplace(f(t),t,s) = g(s) = 1/s*(t0*s)^beta*exp(-t0*s)/GAMMA(beta)
    is the distribution of decay components. In my case, I sometimes want the distribution of decay times, so I simply replace in the integral s by 1/tau and ds by dtau:

    Code:
    f(t) = int(1/tau*(t0/tau)^beta*exp(-t0/tau)/GAMMA(beta) *exp(-t/tau),tau=0..infinity)
    with the distribution

    Code:
    h(tau) = 1/tau*(t0/tau)^beta*exp(-t0/tau)/GAMMA(beta)
    Using h(tau), I can also define some mean decay time

    Code:
    m_tau = int(h(tau)*tau,tau=0..infinity) = t0/(beta-1).
    Alex
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