Results 1 to 2 of 2

Math Help - About a basic property of the Laplace Transform...

  1. #1
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    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...

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

    ... then is...

    \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...

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

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




    In this case...

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

    ... so that is...

    \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...

    \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 t \rightarrow \infty. In this case is...

    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...

    \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

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    A carefull search has pemitted me to discover the mistake in my procedure: the correct diction of the 'Final value theorem' is...

    If...

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

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

     \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 \sigma axis.

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

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basic Laplace Transform of a 1st order Differential Equation
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: October 14th 2011, 12:57 AM
  2. Z transform help (work included, maybe property help)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: October 9th 2011, 11:57 AM
  3. Laplace transform and Fourier transform what is the different?
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: December 29th 2010, 11:51 PM
  4. Need help proving basic property of tensors
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 24th 2009, 11:53 AM
  5. Laplace Transform
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 7th 2009, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum