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I am having a problem understanding dirac-delta distribution. Why the strength of the pulse is equal to the area under it?

Is the strength the pick value of the delta distribution, if so how can it be equal to the area under it?

Jack
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1 Answers1

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Since $\delta(x) = \infty $ when $x = 0$ , otherwise $\delta(x)= 0$ , there is no finite peak value. However $\int \delta(x) dx = 1$ which is the area under $\delta(x)$ and so we might consider $1$ to be the strength. Then if we increase the 'strength' by c, that is, $c\int \delta(x) dx = c $ , we can see the new strength is equal to the new area.

user45664
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    Er, what do you mean "$\delta(x) \to \infty$ as $x \to \infty$"? In any reasonable sense that is wrong... – Ian Jun 06 '16 at 00:22
  • Sorry, I meant as x tends to zero – user45664 Jun 06 '16 at 00:38
  • That's still misleading, "as" it tends to zero it stays zero, then right at zero it "is infinite" heuristically speaking. – Ian Jun 06 '16 at 00:39
  • $\delta(x) \to \infty$ as $x \to 0$ is not acceptable. all you need to say is that if $\delta(x)$ was bounded $< C$ everywhere then $\int_{-a}^a \delta(x)$ would be $< 2a C$, and since $\delta(x)$ is zero on $]\epsilon,a]$ this is $=\int_{-\epsilon}^\epsilon \delta(x) < 2 \epsilon C$ – reuns Jun 06 '16 at 00:41
  • @Ian; Got it--thanks. – user45664 Jun 08 '16 at 00:11
  • @user1952009; Hope this addresses your concern. – user45664 Jun 08 '16 at 00:13