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traianus is offline
Aug8-05, 03:54 PM
P: 80
Hello everybody,
I am new in this forum and I have a problem. First of all, I am not a mathematician, I am an engineer that is using the Laplace transform. Considering this, please do not be too formal in the answers...


The LAPLACE transform of a function [tex]f(t)[/tex] is
[tex]\mathcal{L}\left[f(t)\right][/tex]. Now we let the first derivative of [tex]f(t)[/tex] be called with [tex]\dot{f}(t)[/tex] and the second derivative of [tex]f(t)[/tex] be called with [tex]\ddot{f}(t)[/tex]. It is well known that the LAPLACE transforms (the unilateral) for the first and second derivatives are:

First derivative ---> [tex]\mathcal{L}\left[\dot{f}(t)\right] = s\mathcal{L}\left[{f}(t)\right] - f(0)\qquad (1)[/tex]

Second derivative ----> [tex]\mathcal{L}\left[\ddot{f}(t)\right] = s^2\mathcal{L}\left[{f}(t)\right] -sf(0) - \dot{f}(0)\qquad (2)[/tex]
Inverse laplace transform: I indicate it by using the symbol [tex]\mathcal{L}^{-1}\left[f(t)\right][/tex].
Now, from equation [tex](1)[/tex], the inverse Laplace transform of the term
[tex]s\mathcal{L}\left[{f}(t)\right] [/tex] is:

[tex]\mathcal{L}^{-1}\left[s\mathcal{L}\left[{f}(t)\right]\right] = \mathcal{L}^{-1}\left[ \mathcal{L}\left[\dot{f}(t)\right] + f(0)\right] = \dot{f}(t) + \mathcal{L}^{-1}\left[f(0)\right]\qquad (3)[/tex]

Remembering that the inverse Laplace transform of 1 is Dirac's delta function (indicated with [tex]\delta(t)[/tex]), equation [tex](3)[/tex] becomes:

[tex]\fbox{$\mathcal{L}^{-1}\left[s\mathcal{L}\left[{f}(t)\right]\right] = \dot{f}(t) + f(0)\delta(t)$}\qquad (4)[/tex]

Similarly, from equation [tex](2)[/tex], I can obtain the inverse LAPLACE transform of the term [tex]s^2\mathcal{L}\left[{f}(t)\right][/tex]:

[tex]\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[{f}(t)\right]\right] = \mathcal{L}^{-1}\left[ \mathcal{L}\left[\ddot{f}(t)\right] +s f(0) + \dot{f}(0)\right] \qquad (5)[/tex]


[tex]\fbox{$\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[{f}(t)\right]\right] = \ddot{f}(t) +\dot{\delta}(t) f(0) + \delta(t)\dot{f}(0)$} \qquad (6)[/tex]

where [tex]\dot{\delta}(t))[/tex] is the derivative of the Dirac distribution.


Suppose that I like to take a "stupid" but equivalent approach using the CONVOLUTION theorem. Suppose that the goal is to obtain equations (4) and (6) using that theorem. Let's start with equation (4).
The inverse Laplace transform of the Laplace variable s is:

[tex]\mathcal{L}^{-1}\left[s\right] = \dot{\delta}(t)\qquad (7)[/tex]

The inverse of [tex]\mathcal{L}\left[f(t)\right][/tex] is just [tex]f(t)[/tex]. Using the convolution theorem I can find the inverse of the product [tex]s\mathcal{L}[f(t)][/tex] :

[tex]\mathcal{L}^{-1}\left[s\mathcal{L}\left[f(t)\right]\right] = \int\limits_0^t\dot{\delta}(t-u)f(u)\mathrm{d}u\qquad (8)[/tex]

How can I obtain from equation [tex](8)[/tex] equation [tex](4)[/tex]? Should I integrate by parts? I tried it, but I have some theoretical problems with the DIRAC function. Please, help me!

Similar problem if I use the convolution theorem in order to obtain equation [tex](6)[/tex]:

[tex]\mathcal{L}^{-1}\left[s^2\mathcal{L}\left[f(t)\right]\right] = \int\limits_0^t\ddot{\delta}(t-u)f(u)\mathrm{d}u\qquad (9)[/tex]

how do I obtain equation [tex]

If you have some answers (but not too technical because I am not a mathematician) please let me know:
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