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I've seen everywhere that that the Laplace Transform of Dirac Delta function is:

$$L[\delta(t-a)] = e^{-sa} \text{ when } a > 0$$

But they never explain what happens when $a < 0$. Can I assume that the Laplace transform in the case where $a < 0$ is still the same? Because we're just essentially working in the negative half of the coordinate system? Am I right in thinking this way?

Kibble
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firesage123
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  • you should mark an answer of your questions as the best, i.e., the one that most helped you to comprehend the topic. This is important so that the question will not remain as unanswered. – R.W Feb 19 '17 at 16:43

2 Answers2

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The Laplace transform is defined as

$$L[f(t)] = \int_0^\infty f(t) e^{-st}{\rm d} t$$

If $a<0$ then $f(t) = \delta(t-a) = 0$ for all $t\in[0,\infty)$ so we simply have $L[\delta(x-a)] = 0$.

Kibble
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  • Oh wow that makes sense, thank you! – firesage123 Mar 09 '16 at 22:29
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    @kibble The Dirac Delta is not a function and it therefore is nonsensical to write $\delta(t-a)=0$. – Mark Viola Mar 12 '16 at 20:41
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    @Mark Viola $\delta (t-a)=0,$if $t\not=a$ – Win_odd Dhamnekar Oct 28 '18 at 13:49
  • @dhamnekar What on earth are you talking about. The Dirac Delta is NOT a function. It is a distribution and it does not take on any value at any point. – Mark Viola Oct 28 '18 at 14:02
  • @Mark Viola, Here you can find the definition of $\delta(x)$.https://en.wikipedia.org/wiki/Dirac_delta_function – Win_odd Dhamnekar Oct 28 '18 at 14:11
  • @dhamekar And what does that article tell you in its opening passage? It say that the Dirac Delta is a GENERALIZED FUNCTION (or a DISTRIBUTION) Do you know what that means? – Mark Viola Oct 28 '18 at 14:14
  • @Mark Voila, Generalised functions are useful in making discontinuous function as smooth function – Win_odd Dhamnekar Oct 28 '18 at 14:24
  • @dhamenkarwinod No. A Generalized Function is NOT a function. I suggest that you do some research before you continue writing nonsensical comments. – Mark Viola Oct 28 '18 at 14:30
  • @Mark Voila, I don't have a PhD or MBA from any university, so i shall consider your suggestion. Thanks. – Win_odd Dhamnekar Oct 28 '18 at 14:43
  • Hello, don't argue like this Indeed generalized function is a function for but I think Sir Mark Viola mess it up with relations rather than functions. Indeed what we call a geralized relation is quiet a thinkable point. But it's okay!!! Do, rude stay polite!!! And when things don't go with you just Google. –  Nov 28 '22 at 19:32
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In this answer and this one, I provided primers on the Dirac Delta. Here, we present a simple heuristic way to evaluate the Laplace Transform of the Dirac Delta.

We use the definition of the unit step function $u(t)$ for right-continuous functions as given by

$$u(t)=\begin{cases}1&t\ge0\\\\0&,t<0\end{cases}$$

The function $e^{-st}u(t)$ is not a suitable test function due to the discontinuity at $t=0$. However, for $a\ne0$, we can exploit the fact that the Dirac Delta $\delta_a$ has support $\{0\}$ around $a$. Therefore, we can write

$$\begin{align} \mathscr{L}\{\delta_a\}(s)&=\int_0^\infty \delta(t-a)e^{-st}\,dt\\\\ &=\int_{-\infty}^\infty \delta(t-a)e^{-st}u(t)\,dt\\\\ &=e^{-sa}u(a)\\\\ &=\begin{cases} e^{-sa}&,a> 0\\\\ 0&,a<0 \end{cases} \end{align}$$

where the notation $\delta_a$ is the Dirac Delta $\delta(t-a)$.

We can interpret the Laplace transform of $\delta_0$ as the right-sided limit

$$\mathscr{L}\{\delta_0\}=\lim_{a\to 0^+} \mathscr{L}\{\delta_a\}=1$$

Mark Viola
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  • I have read some books that claim that it is necessary to extend the lower limit of integration of the Laplace transform definition to 0- in order to include delta functions at the origin. Given the way you did, without extending to 0-, why they made that statement? – Vinicius ACP Mar 18 '21 at 17:35
  • Linear System Theory by Zadeh and Desoer: "In many fo the applications that are of interest to us the function f(t) may contain delta functions at t=0. In such cases [...] one (alternative) is to extend the lower limit of integration to $0_-$ in order to include the delta functions in question within the limits of integration..." – Vinicius ACP Mar 18 '21 at 17:36
  • Linear Systems by Thomas Kailath: "The major reason for our use of $\mathscr{L}_-$ is that this enable us to handle initial conditions and impulsive inpults at the origin in a direct manner." – Vinicius ACP Mar 18 '21 at 17:36
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    Well, first of all, the Dirac Delta is not a function. And the symbol $\int_0^\infty \delta(t-a) e^{-st},dt$ is not an integral. Rather, it is a linear functional $\langle \delta_a, f\rangle$ where $f(t)=e^{-st}u(t)$. Note that $f$ is not in the class of functions $C_C^\infty$ since it is not smooth at $t=0$. But inasmuch as the Dirac Delta has support ${0}$, we can interpret the functional as $ \langle \delta_a , g\rangle$ where $g=e^{-st}\phi(t)$, where $\phi \in C_C^\infty$ and and $\phi=1$ in a neighborhood of $0$. – Mark Viola Mar 18 '21 at 21:42