# About a basic property of the Laplace Transform...

• Sep 17th 2009, 01:18 AM
chisigma
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...

http://digilander.libero.it/luposabatini/LTtest1.bmp

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...

http://digilander.libero.it/luposabatini/LTtest2.bmp

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! (Nerd) ...

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

$\chi$ $\sigma$
• Sep 17th 2009, 10:01 AM
chisigma
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 (Headbang) ...

Kind regards

$\chi$ $\sigma$