2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Rafael Wagner Feb 19 '17 at 16:43
4
$\begingroup$

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 $L[\delta(x-a)] = 0$.

$\endgroup$
  • $\begingroup$ Oh wow that makes sense, thank you! $\endgroup$ – firesage123 Mar 9 '16 at 22:29
  • 2
    $\begingroup$ @kibble The Dirac Delta is not a function and it therefore is nonsensical to write $\delta(t-a)=0$. $\endgroup$ – Mark Viola Mar 12 '16 at 20:41
  • $\begingroup$ @Mark Viola $\delta (t-a)=0,$if $t\not=a$ $\endgroup$ – Dhamnekar Winod Oct 28 '18 at 13:49
  • $\begingroup$ @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. $\endgroup$ – Mark Viola Oct 28 '18 at 14:02
  • $\begingroup$ @Mark Viola, Here you can find the definition of $\delta(x)$.en.wikipedia.org/wiki/Dirac_delta_function $\endgroup$ – Dhamnekar Winod Oct 28 '18 at 14:11
1
$\begingroup$

In this answer and this one, I provided primers on the Dirac Delta. Here, we present a simple 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

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

Then, we can write

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.