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laplace transform of impulse function

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Dec 1, 2005
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laplace transform impulse

Hi All,
I have a bad problem with laplace transform.
a signal should be absolutely integrable, with finite extrema points in a finite region and some other properties to have a laplace transform.
but all books take the laplace transform of δ(t), δ'(t), ... which are not supported. and I am now confused. help please.

laplace transform impulse function


I don't understand what you mean by 'not supported'. If you are implying that there is a problem with absolute integrablity then you should note that the Dirac delta function, δ(t), is in fact absolutely integrable. It's integral will yield 1.

∫δ(x) dx=1 w/ integral limits from -∞ yo ∞


laplace impulse


You should look at the strict definition of δ(t), its a limiting case, where its area approaches 1 and its height approaches infinity causing its area to be 1 by definition, also, its integral is defined in this way, it has special properties, other than an simply infinity at a point.

I hope you got what I'm trying to say.

That is a tricky problem, isn't it? I ran into a similar problem yesterday. Here is what happened. Yesterday, I wanted to learn how to build a boat. I went to the bookshelf and pulled out a book. It turned out to be a philosophy book and, therefore, I was immediately getting furious. How could it be ... "philosophy"? It does not talk about "boat" at all. I started throwing chairs, dishes, ..., and I almost set the whole house on fire. Suddenly, my 4 - year old son came in and said :"Daddy, put down that wrong book and find the right one, ok?" ...Oh, yeah, ok, ...problem solved ...

So, got a good signal, huh? Not too big and staying in a finite range? And want to take the laplace transform? Unfortunately, the books you got only talk about the laplace transforms of δ(t), δ'(t) functions? So .... may I have my baby's words for you:"put down those wrong books and find the right one, ok?"

laplace of impulse function

∫δ(t).exp(-st) dt = ∫δ(t) dt = 1

laplace impulse function

impulse function is integrable.The definition itself says that "it is defined by the eqn

∫δ(t)dt=1 for 0- <t<0+

laplace transform of impulse

There is no problem with the integral at all
\[ \int_{0^-}^{\infty} \delta (t) e^{-st} dt \]
Due to the sifting property (see for example ), the value of the integral is just \[e^{st}\] evaluated at where the impluse occurs which in this case is \[t=0\]. So the result is \[e^0=1\]. In general, the sifting property says that \[ \int_{t_a}^{t_b} \delta (t-t_0) f(t) dt = f(t_0)\], as long as \[ t_a \le t_0 \le t_b \]. All this means is that the delta function must occur within the limits of integration. If the impulse occurs outside these limits the result of the integral is simply zero.

Best regards,
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