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Thread: About a basic property of the Laplace Transform...

  1. #1
    MHF Contributor chisigma's Avatar
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    About a basic property of the Laplace Transform...

    I suppose is well known to you the so called 'Final value theorem' , connected to the Laplace Trasform:

    If...

    $\displaystyle f(s) = \mathcal{L} \{\varphi (t)\}$

    ... then is...

    $\displaystyle \lim _{t \rightarrow \infty} \varphi (t) = \lim_{s \rightarrow 0} s\cdot f(s) $

    ... provided that both limits exist.

    Let's take an example and consider the function...

    $\displaystyle \varphi (t) = \sum_{n=0}^{\infty} \mathcal {U} (t-n)$ (1)

    ... where $\displaystyle \mathcal {U}$ is the so called 'Haeviside step function'. This function is represented here...




    In this case...

    $\displaystyle f(s) = \frac{1}{s\cdot (1-e^{-s})}$ (2)

    ... so that is...

    $\displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} \frac{1}{1-e^{-s}} = + \infty$ (3)

    ... and it seems to be all right. Let' try now with another function...

    $\displaystyle \varphi (t) = \sum_{n=0}^{\infty} \mathcal {U} (t-n) - t$ (4)

    ... that is represented here...



    It is quite obvious that such type of function doesn't have limit for $\displaystyle t \rightarrow \infty$. In this case is...

    $\displaystyle f(s)= \frac{s -1 + e^{-s}}{s^{2}\cdot (1-e^{-s})}$ (5)

    ... and if we try to apply the 'Final value theorem', using three times the L'Hopital rule, we obtain...

    $\displaystyle \lim_{s \rightarrow 0} s\cdot f(s) = \frac{1}{2}$ (6)

    ... and that's is a little surprising! ...

    At this point the questions for You are...

    a) did I make some mistrake?...
    b) if not how to interpret all that?...

    Any help will be gratefully accepted!...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  2. #2
    MHF Contributor chisigma's Avatar
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    A carefull search has pemitted me to discover the mistake in my procedure: the correct diction of the 'Final value theorem' is...

    If...

    $\displaystyle f(s) = \mathcal{L} \{\varphi (t)\} $

    ... $\displaystyle \varphi(t)$ and $\displaystyle \varphi^{'}(t)$ are both Laplace transformable, and $\displaystyle s\cdot f(s)$ has no singularities of the $\displaystyle i\omega$ axis and in the right half plane, then ...

    $\displaystyle \lim_{ t \rightarrow \infty} \varphi(t) = \lim_{ s \rightarrow 0} s\cdot f(s)$

    .... where the limit on the right is taken along the positive $\displaystyle \sigma$ axis.

    In both my examples $\displaystyle s\cdot f(s)$ did have infinite poles on the $\displaystyle i\omega$ axis so that ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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