Dirac Delta Function And Its Properties Pdf

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In this section, we will use the Dirac delta function to analyze mixed random variables. Technically speaking, the Dirac delta function is not actually a function.

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Dirac delta function

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Laplace transform of cos t and polynomials. Laplace transform of the unit step function. Inverse Laplace examples. Dirac delta function. Laplace transform of the dirac delta function. Next lesson. Current timeTotal duration Google Classroom Facebook Twitter. Video transcript When I introduced you to the unit step function, I said, you know, this type of function, it's more exotic and a little unusual relative to what you've seen in just a traditional Calculus course, what you've seen in maybe your Algebra courses.

But the reason why this was introduced is because a lot of physical systems kind of behave this way. That all of a sudden nothing happens for a long period of time and then bam!

Something happens. And you go like that. And it doesn't happen exactly like this, but it can be approximated by the unit step function. Similarly, sometimes you have nothing happening for a long period of time. Nothing happens for a long period of time, and then whack! Something hits you really hard and then goes away, and then nothing happens for a very long period of time.

And you'll learn this in the future, you can kind of view this is an impulse. And we'll talk about unit impulse functions and all of that. So wouldn't it be neat if we had some type of function that could model this type of behavior?

And in our ideal function, what would happen is that nothing happens until we get to some point and then bam! It would get infinitely strong, but maybe it has a finite area. And then it would go back to zero and then go like that. So it'd be infinitely high right at 0 right there, and then it continues there.

And let's say that the area under this, it becomes very-- to call this a function is actually kind of pushing it, and this is beyond the math of this video, but we'll call it a function in this video. But you say, well, what good is this function for? How can you even manipulate it? And I'm going to make one more definition of this function. Let's say we call this function represented by the delta, and that's what we do represent this function by.

It's called the Dirac delta function. And we'll just informally say, look, when it's in infinity, it pops up to infinity when x equal to 0. And it's zero everywhere else when x is not equal to 0.

And you say, how do I deal with that? How do I take the integral of that? And to help you with that, I'm going to make a definition. I'm going to tell you what the integral of this is. This is part of the definition of the function.

I'm going to tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it to be equal to 1. I'm defining this. Now, you might say, Sal, you didn't prove it to me. No, I'm defining it. I'm telling you that this delta of x is a function such that its integral is 1.

So it has this infinitely narrow base that goes infinitely high, and the area under this-- I'm telling you-- is of area 1. And you're like, hey, Sal, that's a crazy function. I want a little bit better understanding of how someone can construct a function like this. So let's see if we can satisfy that a little bit more. But then once that's satisfied, then we're going to start taking the Laplace transform of this, and then we'll start manipulating it and whatnot.

Let's see, let me complete this delta right here. Let's say that I constructed another function. Let's call it d sub tau And this is all just to satisfy this craving for maybe a better intuition for how this Dirac delta function can be constructed. And let's say my d sub tau of-- well, let me put it as a function of t because everything we're doing in the Laplace transform world, everything's been a function of t.

So let's say that it equals 1 over 2 tau, and you'll see why I'm picking these numbers the way I am. And let's say it's 0 everywhere else. So this type of equation, this is more reasonable. This will actually look like a combination of unit step functions, and we can actually define it as a combination of unit step functions. So if I draw, that's my x-axis. And then if I put my y-axis right here. That's my y-axis. Sorry, this is a t-axis. I have to get out of that habit.

This is the t-axis, and, I mean, we could call it the y-axis or the f of t-axis, or whatever we want to call it. That's the dependent variable. So what's going to happen here? It's going to be zero everywhere until we get to minus t, and then at minus t, we're going to jump up to some level. Just let me put that point here. So this is minus tau, and this is plus tau. So it's going to be zero everywhere, and then at minus tau, we jump to this level, and then we stay constant at that level until we get to plus tau.

And that level, I'm saying is 1 over 2 tau. So this point right here on the dependent axis, this is 1 over 2 tau. So why did I construct this function this way? Well, let's think about it. What happens if I take the integral? Let me write a nicer integral sign. If I took the integral from minus infinity to infinity of d sub tau of t dt, what is this going to be equal to?

Well, if the integral is just the area under this curve, this is a pretty straightforward thing to calculate. You just look at this, and you say, well, first of all, it's zero everywhere else. It's zero everywhere else, and it's only the area right here. I mean, I could rewrite this integral as the integral from minus tau to tau-- and we don't care if infinity and minus infinity or positive infinity, because there's no area under any of those points-- of 1 over 2 tau d tau.

Sorry, 1 over 2 tau dt. So we could write it this way too, right? Because we can just take the boundaries from here to here, because we get nothing whether t goes to positive infinity or minus infinity. And then over that boundary, the function is a constant, 1 over 2 tau, so we could just take this integral.

And either way we evaluate it. We don't even have to know calculus to know what this integral's going to evaluate to. This is just the area under this, which is just the base. What's the base? The base is 2 tau. You have one tau here and then another tau there. So it's equal to 2 tau times your height. And your height, I just said, is 1 over 2 tau. So your area for this function, or for this integral, is going to be 1. You could evaluate this. You could get this is going to be equal to-- you take the antiderivative of 1 over 2 tau, you get-- I'll do this just to satiate your curiosity-- t over 2 tau, and you have to evaluate this from minus tau to tau.

And when you would put tau in there, you get tau over 2 tau, and then minus minus tau over 2 tau, and then you get tau plus tau over 2 tau, that's 2 tau over 2 tau, which is equal to 1.

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It is used to model the density of an idealized point mass or point charge as a function equal to zero everywhere except for zero and whose integral over the entire real line is equal to one. As a distribution, the Dirac delta function is a linear functional that maps every function to its value at zero. In engineering and signal processing , the delta function, also known as the unit impulse symbol, [6] may be regarded through its Laplace transform , as coming from the boundary values of a complex analytic function of a complex variable. The formal rules obeyed by this function are part of the operational calculus , a standard tool kit of physics and engineering. In many applications, the Dirac delta is regarded as a kind of limit a weak limit of a sequence of functions having a tall spike at the origin in theory of distributions, this is a true limit.

It is worth noting that the Dirac δ function is not strictly speaking a valid function. We define the delta function δ(x) as an object with the following properties: δ(x)={∞x=00otherwise. δ(x)=ddxu(x), where u(x) is the unit step function (Equation );.


Examples of this kind of forcing function would be a hammer striking an object or a short in an electrical system. In both of these cases a large force or voltage would be exerted on the system over a very short time frame. The Dirac Delta function is used to deal with these kinds of forcing functions. There are many ways to actually define the Dirac Delta function. To see some of these definitions visit Wolframs MathWorld.

On the Fractional Derivative of Dirac Delta Function and Its Application

On the other hand, the fractional-order system gets more and more attention. This paper investigates the fractional derivative of the Dirac delta function and its Laplace transform to explore the solution for fractional-order system. The paper presents the Riemann-Liouville and the Caputo fractional derivative of the Dirac delta function, and their analytic expression.

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Выли сирены. Как весенний лед на реке, потрескивал корпус ТРАНСТЕКСТА. - Я спущусь вниз и отключу электропитание, - сказал Стратмор, положив руку на плечо Сьюзан и стараясь ее успокоить.  - И сразу же вернусь. Сьюзан безучастно смотрела, как он направился в шифровалку. Это был уже не тот раздавленный отчаянием человек, каким она видела его десять минут .

 Что. Этого не может. Он заперт внизу. - Нет. Он вырвался оттуда. Нужно немедленно вызвать службу безопасности.

 - Я не. Я сейчас же отправлю ее домой. - Боюсь, вы опоздали, - внушительно заявил Беккер и прошелся по номеру.  - У меня к вам предложение. - Ein Vorschlag? - У немца перехватило дыхание.

Он вылетел из-за поворота на уровне лодыжек подобно рапире фехтовальщика.

 - Что скажешь. А потом мы могли бы… - Выкинь это из головы. - Сколько в тебе снобизма.  - Хейл вздохнул и повернулся к своему компьютеру. В этом вся ее сущность.

 Какого чер… В распечатке был список последних тридцати шести файлов, введенных в ТРАНСТЕКСТ. За названием каждого файла следовали четыре цифры - код команды добро, данной программой Сквозь строй.

Не сомневаюсь, - подумала. Сьюзан никогда еще не видела шефа столь подавленным. Его редеющие седые волосы спутались, и даже несмотря на прохладу, создаваемую мощным кондиционером, на лбу у него выступили капельки пота.

Все было бесполезно. До поворота оставалось еще триста метров, а такси от него отделяло всего несколько машин. Беккер понимал, что через несколько секунд его застрелят или собьют, и смотрел вперед, пытаясь найти какую-нибудь лазейку, но шоссе с обеих сторон обрамляли крутые, покрытые гравием склоны.

4 Response
  1. Amy H.

    Dirac clearly had precisely such ideas in mind when, in §15 of his Quantum. Mechanics,1 that the “delta function”—which he presumes to satisfy the conditions. ∫ +∞ properties of δ(•) which are important to the theory of Green's functions.

  2. Leon K.

    Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.

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