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Dirac delta function

Introduction to the Dirac Delta Function. Created by Sal Khan.

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  • blobby green style avatar for user E.E.Busby
    At , I guess you could say he's building a "tau-er". Haha.
    (74 votes)
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  • leaf orange style avatar for user nebbionegiuseppe
    Hi everyone, i wanted to know, why at Sal modelled the force applied with a Dirac's Delta ? Can't understand the motivation, and why this is possible...

    (7 votes)
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    • aqualine ultimate style avatar for user McWilliams, Cameron
      Consider a physical system, in which a object is at rest, and an external force is quickly applied, accelerating it to a velocity of 2 m/s. We know that this force could not have accelerated the object instantly, but for our purposes, let's assume it did. To model this, we would need a function the represents an infinite acceleration (to accelerate the object in an infinitely small time) but has a finite area (the area under the acceleration function is velocity). Sounds like the Dirac delta function, huh? In this scenario, the force applied to the object could be modeled as 2*m*delta(x), where m is the mass of the accelerated object. Physicists and engineers make the assumption that some things happen instantly, because they are so fast that trying to model them using actually equations would over complicate the problem, with no gain. So they use the Dirac delta function to make these "instantaneous" models. It may also help to think of the Dirac delta function as the derivative of the step function. The Dirac delta function usually occurs as the derivative of the step function in physics. In the above example I gave, and also in the video, the velocity could be modeled as a step function.
      (22 votes)
  • blobby green style avatar for user Yulia Borukhina
    I was wondering if there's a typo in the video, or if I'm misunderstanding something... if we say function equals 2*tau, while tau varies between - and +, doesn't it mean that the function graph goes under X axis (as the function value is negative when tau is negative)?
    (3 votes)
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    • sneak peak green style avatar for user Alex Knauth
      Here, he's kind of assuming that Tau will always be positive (for the purpose of this video). Tau doesn't vary between - and + here, because of that implied assumption. Also, Tau isn't the independent variable here. The function is dependent on the variable t, or time. Because of that, Tau can be whatever Sal says it is no matter what t is.
      (3 votes)
  • leaf green style avatar for user Gazi Kothiasin
    so does this mean that infinity * 0 = 1 ?
    (1 vote)
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  • leaf red style avatar for user Diego Song
    So from to , you talk about shifting the delta function over the t axis.
    But what happens if you, say, multiply t inside the delta function by a number? e.g. int (\delta (5t-3))? So in this example, I shift 3 units to the right, but then, when I evaluate integrals based on the definition that int (\delta(t) dt) = 1, I should divide by 1/5, right?

    Also, what happens when we are talking about volume integrals on a delta function with a similar situation like the one postulated above? So int(\delta^3(5t-3))? I am not sure how I should "take the 5 out" to easily solve the integral. Should I also cube what I divide? ([1/5]^3)
    (1 vote)
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    • leaf green style avatar for user Gordon N. Fleming
      The integral of delta(x) is = 1 IF the variable of integration is x, i.e.,
      Int delta(x) dx = 1. Similarly, Int delta(x - 2) dx = Int delta(x - 2) d(x - 2) = 1.
      But Int delta(2x) dx = Int delta(2x) d(2x/2) = (1/2) Int delta(2x) d(2x) = (1/2) 1 = (1/2). Similarly, for a > 0,
      Int delta(ax + b) dx = (1/a) Int delta(ax + b) d(ax + b) = (1/a),
      while for a < 0 we get,
      Int delta(ax + b) dx = - (1/a) because the negative nature of a inverts the order of the integration.
      (1 vote)
  • male robot hal style avatar for user William Ortez
    Can you use he dirac delta function to model the collapse of a wave function for a particle? Please help
    (2 votes)
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  • blobby green style avatar for user fatonhaliti.89
    is the same δ(t+1) = δ(1-t)
    (1 vote)
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  • mr pink red style avatar for user Enya Hsiao
    Sal discussed the integral of the dirac delta function by giving a example of dtau(t) at , which fully explained the answer, but what if Dtau(t) isn't 1/2Pi at tau<t<tau, does it produce the same results?
    (1 vote)
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  • blobby green style avatar for user Mark Hayes
    Where were you in the '80s when I was forced to take this stuff. Those profs were as clear as mud - you make it look simple.
    (1 vote)
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    • male robot hal style avatar for user Ham
      We didn't have internet, computers were pretty new, and professors were (relatively) rare. Not to mention, few people actually understood high level stuff like this. Even now, this is not common knowledge (not everyone majors in science), but this should be (perhaps even taught in high school?)
      (1 vote)
  • purple pi purple style avatar for user k3edge
    You shouldn't be able to use the regular Riemann integral on this particular function because the dirac delta function takes on certain values at discrete x values, so under these circumstances shouldn't we use a lebesgue integral? Since the number of all x's such that f(x) = infinity is countable (in this case 1 since the only x that makes f(x) = infinity is 0), that means its measure is 0, therefore making there area under which x makes f(x) = infinity 0. All the other x values make the function 0, so therefore it makes the area under the entire graph 0. Am I missing something?
    (1 vote)
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    • ohnoes default style avatar for user Tejas
      You are missing that we can make the infinity as large as we want in order to make the integral of f(x) be 1. Thus, when using a Lebesgue integral, we would get A{0} + A{0} + A{0} + .... Since {0} is countable, its area is infinitesimal, but we also add it infinitely many times. To make the integral finite, we are allowed to say that we add the A{0} uncountably infinitely many times.
      (1 vote)

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. 1 over 2 tau when t is less then tau and greater than minus tau. 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. Maybe I'm beating a dead horse. I think you're satisfied that the area under this is going to be 1, regardless of what tau was. I kept this abstract. Now, if I take smaller and smaller values of tau, what's going to happen? If my new tau is going to be here, let's say my new tau is going to be there, I'm just going to pick up my new tau there, then my 1 over 2 tau, the tau is now a smaller number. So when it's in the denominator, my 1 over 2 tau is going to be something like this, right? I mean, I'm just saying, if I pick smaller and smaller taus. So then if I pick an even smaller tau than that, then my height is going to be have to be higher. My 1 over 2 tau is going to have to even be higher than that. And so I think you see where I'm going with this. What happens as the limit as tau approaches zero? So what is the limit as tau approaches zero of my little d sub tau function? What's the limit of this? Well, these things are going to go infinitely close to zero, but this is the limit. They're never going to be quite at zero. And your height here is going to go infinitely high, but the whole time, I said no matter what my tau is, because it was defined very arbitrarily, was my area is always going to be 1. So you're going to end up with your Dirac delta function. Let me write it now. I was going to write an x again. Your Dirac delta function is a function of t, and because of this, if you ask what's the limit as tau approaches zero of the integral from minus infinity to infinity of d sub tau of t dt, well, this should still be 1, right? Because this thing right here, this evaluates to 1. So as you take the limit as tau approaches zero-- and I'm being very generous with my definitions of limits and whatnot. I'm not being very rigorous. But I think you can kind of understand the intuition of where I'm going. This is going to be equal to 1. And so by the same intuitive argument, you could say that the limit from minus infinity to infinity of our Dirac delta function of t dt is also going to be 1. And likewise, the Dirac delta function-- I mean, this thing pops up to infinity at t is equal to 0. This thing, if I were to draw my x-axis like that, and then right at t equals 0, my Dirac delta function pops up like that. And you normally draw it like that. And you normally draw it so it goes up to 1 to kind of depict its area. But you actually put an arrow there, and so this is your Dirac delta function. But what happens if you want to shift it? How would I represent my-- let's say I want to do t minus 3? What would the graph of this be? Well, this would just be shifting it to the right by 3. For example, when t equals 3, this will become the Dirac delta of 0. So this graph will just look like this. This will be my x-axis. And let's say that this is my y-axis. Let me just make that 1. And let me just draw some points here, so it's 1, 2, 3 That's t is equal to 3. Did I say that was the x-axis? That's my t-axis. This is t equal to 3. And what I'm going to do here is the Dirac delta function is going to be zero everywhere. But then right at 3, it goes infinitely high. And obviously, we don't have enough paper to draw an infinitely high spike right there. So what we do is we draw an arrow. We draw an arrow there. And the arrow, we usually draw the magnitude of the area under that spike. So we do it like this. And let me be clear. This is not saying that the function just goes to 1 and then spikes back down. This tells me that the area under the function is equal to 1. This spike would have to be infinitely high to have any area, considering it has an infinitely small base, so the area under this impulse function or under this Dirac delta function. Now, this one right here is t minus 3, but your area under this is still going to be 1. And that's why I made the arrow go to 1. Let's say I wanted to graph-- let me do it in another color. Let's say I wanted to graph 2 times the Dirac delta of t minus 2. How would I graph this? Well, I would go to t minus 2. When t is equal to 2, you get the Dirac delta of zero, so that's where you would have your spike. And we're multiplying it by 2, so you would do a spike twice as high like this. Now, both of these go to infinity, but this goes twice as high to infinity. And I know this is all being a little ridiculous now. But the idea here is that the area under this curve should be twice the area under this curve. And that's why we make the arrow go to 2 to say that the area under this arrow is 2. The spike would have to go infinitely high. So this is all a little abstract, but this is a useful way to model things that are kind of very jarring. Obviously, nothing actually behaves like this, but there are a lot of phenomena in physics or the real world that have this spiky behavior. Instead of trying to say, oh, what does that spike exactly look like? We say, hey, that's a Dirac delta function. And we'll dictate its impulse by something like this. And just to give you a little bit of motivation behind this, and I was going to go here in the last video, but then I kind of decided not to. But I'm just going to show it, because I've been doing a lot of differential equations and I've been giving you no motivation for how this applies in the real world. But you can imagine, if I have just a wall, and then I have a spring attached to some mass right there, and let's say that this is a natural state of the spring, so that the spring would want to be here, so it's been stretched a distance y from its kind of natural where it wants to go. And let's say I have some external force right here. Let's say I have some external force right here on the spring, and, of course, let's say it's ice on ice. There's no friction in all of this. And I just want to show you that I can represent the behavior of this system with the differential equation. And actually things like the unit step functions, the Dirac delta function, actually start to become useful in this type of environment. So we know that F is equal to mass times acceleration. That's basic physics right there. Now, what are all of the forces on this mass right here? Well, you have this force right here. And I'll say this is a positive rightward direction, so it's that force, and then you have a minus force from the spring. The force from the spring is Hooke's Law. It's proportional to how far it's been stretched from its kind of natural point, so its force in that direction is going to be ky, or you could call it minus ky, because it's going in the opposite direction of what we've already said is a positive direction. So the net forces on this is F minus ky, and that's equal to the mass of our object times its acceleration. Now, what's its acceleration? If its position is y, so if y is equal to position, if we take the derivative of y with respect to t, y prime, which we could also say dy dt, this is going to be its velocity. And then if we take the derivative of that, y prime prime, which is equal to d squared y with respect to dt squared, this is equal to acceleration. So instead of writing a, we could right y prime prime. And so, if we just put this on the other side of the equation, what do we get? We get the force-- this force, not just this force; this was just F equals ma-- but this force is equal to the mass of our object, times the acceleration of the object plus whatever the spring constant is for the spring plus k times our position, times y. So if you had no outside force, if this was zero you'd have a homogeneous differential equation. And in that case, the spring would just start moving on its own. But now this F, all of a sudden, it's kind of a non-homogeneous term, it's what the outside force you're applying to this mass. So if this outside force was some type of Dirac delta function-- so let's say it's t minus 2 is equal to our mass times y prime prime plus our spring constant times y, this is saying that at time is equal to 2 seconds, we're just going to jar this thing to the right. And it's going to have an-- and I'll talk more about it-- it's going to have an impulse of 2. It's force times time is going to be-- or its impulse is going to have 1. And I don't want to get too much into the physics here, but its impulse or its change in momentum, is going to be of magnitude 1, depending on what our units are. But anyway, I just wanted to take a slight diversion, because you might think Sal is introducing me to these weird, exotic functions. What are they ever going to be good for? But this is good for the idea that sometimes you just jar this thing by some magnitude and then let go. And you do it kind of infinitely fast, but you do it enough to change the momentum of this in a well-defined way. Anyway, in the next video, we'll continue with the Dirac delta function. We'll figure out its Laplace transform and see what it does to the Laplace transforms of other functions.