Images of graphs of dirac delta show an arrow head pointing upwards at t=0, what does that means?
Is it referring that at t=0, amplitude is infinity?
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2$\begingroup$ The delta is a distribution and not a function so it doesn't really have a value at $t=0$. The delta is defined through it's integral property, i.e. $\int_a^b \delta(t)dt$. This is 1 if the integration interval includes $t=0$ . Otherwise it's 0. $\endgroup$– HilmarCommented Mar 2, 2022 at 9:44
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1$\begingroup$ In case you are unfamiliar, there are multiple delta "functions". I recommend you familiarize yourself with the differences between the definitions and properties of the Dirac delta (a distribution) and the Kronecker delta (a piecewise function). $\endgroup$– AshCommented Mar 2, 2022 at 16:54
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$\begingroup$ You are mixing two things: the concept of the Dirac impulse -- which is really an abomination with zero values everywhere and infinite at one point, only, which gives perfrectly flat spectrum everywhere -- with the mathematical representation of it -- a distribution, or a limit. $\endgroup$– a concerned citizenCommented Mar 2, 2022 at 22:56
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4 Answers
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The arrow head is a symbol. It symbolizes "there's a Dirac delta at this position". That's all its meaning.
Is it referring that at t=0, amplitude is infinity?
Ahhh, no. You cannot say "amplitude is infinity", because "infinity" is not a value. The Dirac delta is not a function with a defined value at $t=0$. But that's a discussion for a different post (and certainly has been discussed here before) - the arrow just signifies "Dirac Delta here", and all properties follow from that, and not from the symbol - which is just some ink on paper, not the math behind that ink.
answered Mar 2, 2022 at 9:48
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1$\begingroup$ Plus, without the arrow (or any other symbol) the Dirac at $t=0$ would be invisible, at it is in line with the y-axis. $\endgroup$– MaxCommented Mar 2, 2022 at 14:59
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$\begingroup$ I'm not at all convinced. Surely the arrow refers to fact that (all?) functions that approach a Dirac delta in limit $\epsilon \to 0$ show $f(0) \to \infty$ as $\epsilon \to 0$? It is surely not coincidence that it's an arrow pointing up (instead of any other direction or symbol) $\endgroup$ Commented Mar 4, 2022 at 7:28
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On the wiki page for the Dirac delta function, you can find one meaning of the arrow. It somehow means that is not "defined" as a constant defined value, but more as a factor applied to evaluations, related to to so-called "unit-area" under the symbol.
answered Mar 2, 2022 at 19:39
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what you wanna imagine are a single-peak function where the peak has a simple height parameter and width parameter. then imagine adjusting the height parameter higher, while at the same time, adjusting the width parameter in inverse proportion to the height parameter.
as the width parameter is going to zero, the height parameter is growing without bound (that's sorta what we mean by "infinity"). that's sorta what the arrow is about.
but it gets skinny.
answered Mar 3, 2022 at 4:07
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3$\begingroup$ I like this illustration – especially because it iterates through (what looks like) Gaussians, because people tend to know that Gaussians Fourier-transform to Gaussians of (proportional to) inverse variance - and in the limit, the time-domain needle-Gaussian approaches zero variance, which means the frequency-domain Gaussian becomes "infinitely wide", i.e., flat. $\endgroup$ Commented Mar 3, 2022 at 13:49
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$\begingroup$ Here's an animated gif: that illustrates it better. $\endgroup$ Commented Mar 4, 2022 at 1:48
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To reiterate other good comments and answers: first, no, although it might be thought almost to be correct, it is not that $\delta$ has value "infinity" at $0$... and the arrow notation should not be construed as that.
Rather, the arrow notation is a sort of extension of the simpler ideas of "graphing functions", beyond just a curve that can be hand-drawn. Again, it cannot mean (though this is not utterly wrong...) that there is a value of $\delta$ at $0$, and it's "infinity". No, the arrow notation can properly be interpreted exactly as saying "it's $\delta$", which "does something strong at $0$"... and is described precisely by its interaction (however we want to put it) with other (nicer) functions.
So it's an extension of usual graphing conventions, specifically to include $\delta$, without telling lies, etc.
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$\begingroup$ From Richard Hamming: "Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane." From the POV of signal processing, I am not concerned about the difference between the dirac delta function is from a nascent delta function that is sufficiently skinny. The plane better fly in either case. $\endgroup$ Commented Mar 3, 2022 at 4:50
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$\begingroup$ @robertbristow-johnson well, the rigorous definition of the Dirac distribution encapsulates exactly that: that, as long as $\varepsilon$ is small enough, it doesn't matter anymore what exact nascent delta is used. But there's an important caveat: how small is small enough can (and does) depend on the application. It's not enough to pick just one "small-looking" constant value and use it for everything. ... $\endgroup$ Commented Mar 3, 2022 at 11:45
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$\begingroup$ ...You may be able to pick one value that works for the design of the plane, but if you then try using that same value for a medical apparatus you're in trouble. Vice versa, if you try to use the smaller value when designing the plane, it may theoretically speaking be ok, but your CFD simulations would end up taking 2000 years to complete, so that's not the path to a safe aircraft either. The correct thing is to derive all the theory using Dirac as a distribution as it it, not nascent, and thereby making it implicitly clear that selecting an appropriate $\varepsilon$ is up to the user. $\endgroup$ Commented Mar 3, 2022 at 11:45
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$\begingroup$ // how small is small enough can (and does) depend on the application. It's not enough to pick just one "small-looking" constant value and use it for everything.// ---- well, how 'bout a nascent delta function that has width of one Planck time? May I use that as a sufficiently small value to replace the Dirac in a physical context? $\endgroup$ Commented Mar 4, 2022 at 1:58