Find if a function is the Laplace transform of a periodic function

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libelec
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Homework Statement



Could [tex]\[F(s) = \frac{1}{{s(1 - {e^{ - s}})}}\][/tex] be the Laplace transform of some periodic function? Why? If so, find that periodic function

The Attempt at a Solution



If it was the Laplace transform of some periodic function, the the Laplace transform of the first wave should be 1/s, and the period T should be 1. The function whose Laplace transform is 1/s is H(t). Then the periodic function should be the stairs function, [tex]\[f(t) = \left\{ \begin{array}{l}<br /> n,x \in [nT,(n + 1)T] \\ <br /> n + 1,x \in [(n + 1)T,(n + 2)T] \\ <br /> \end{array} \right.,n \in N + \left\{ 0 \right\}\][/tex]

Now, the stairs function is of exponential order, since it's smaller or equal than the function t+1 for all t, and that is an exponential order function. Then it has a Laplace transform.

So far, so good. If the stairs function transforms to F(s) there, then F(s) is the Laplace transform of some periodic function.

But I don't think that's what the exercise asks me to do, since a latter question asks me to find the periodic function. I think there's something I have to prove through F(s) that allows me to say that it is the Laplace transform of a periodic function.

But I don't know what is that.

Any ideas?
 
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Allright, it's true.

But then why couldn't it be periodic? Is it because the inverse Laplace transform of 1/s is H(t), and that isn't a periodic function? What does F(s) has to have to check if it belongs to a periodic function or not?
 
libelec said:
Allright, it's true.

But then why couldn't it be periodic? Is it because the inverse Laplace transform of 1/s is H(t), and that isn't a periodic function? What does F(s) has to have to check if it belongs to a periodic function or not?

Well, I didn't check your steps, but I assume you have the correct inverse with your staircase function. It isn't periodic and the FT is a 1-1 transform so doesn't that settle it?
 
Yes, but I thought there was another argument to say that. I mean, since the question seems to be theorical.