If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Laplace transform of the dirac delta function

## Video transcript

in the last video I introduced you to what is probably the most bizarro function that you've encountered so far and that was the Dirac the Dirac Delta function Dirac Delta function Dirac Delta and I defined it to be and I'll do the shifted version of it you're already hopefully reasonably familiar with it or if I shift it so Dirac Delta of t minus C we can say that it equals is equals 0 when T does not equal C so it equals zero everywhere but it essentially pops up to infinity and we have to be careful with this infinity I'm going to write it in quotes it pops up to infinity because well well so and we even saw in the previous video it's kind of different degrees of infinity because you can still multiply this by other numbers to get larger Dirac Delta functions when T is equal to C but more important than this and this is kind of a pseudo definition here is the idea that when we take the integral when we take the area under the curve over the entire X or the entire T axis I guess we could say when we take the area under this curve and obviously it equals zero everywhere except at write equals at T is equal to C when we take this area this is the important point that the area is equal to 1 that the area is equal to 1 and so this is what I meant by kind of pseudo infinity because if I have 2 times the Dirac Delta function if I have 2 times the Dirac Delta function and if I'm taking the area under the curve of that of 2 times the Dirac Delta function t minus C DT this should be equal to 2 times the area of just under the Dirac Delta function two times from minus infinity to infinity of the Delta function shifted by C DT which is just 2 times we already showed you I just said by definition this is 1 so this will be equal to 2 so if if if I put a 2 out here this infinity will have to be twice as high so that the area is now 2 so that's why I put that infinity in parentheses but it's an interesting function I talked about it at the end of the last video it can help model things that kind of jar things all of a sudden but they impart a fixed amount of impulse on something you know a fixed amount of change in momentum and we'll understand that a little bit more in the future but let's kind of get the mathematical tools completely understood and let's try to figure out what the Dirac Delta function does when we multiply it what it does to the Laplace transform when we multiply it times some function so let's say I have my Dirac Delta function and I'm going to shift it that's a little bit more interesting and it you know if you want to unshipped it you just say okay well C equals zero what happens when C equals zero and I'm going to shift it and multiply it times some arbitrary function f of T if I wanted to figure out the Laplace transform of just the the Delta function by itself I could set f of T is equal to one so let's take the let's take our Laplace transform of this and we can just use the definition of the Laplace transform so this is equal to the area from 0 to infinity under neath or we call it the integral from 0 to infinity of e to the minus st that's just part of the Laplace transform definition times this thing and I'll just write it in this order times f of T times our Dirac Delta function Delta t minus C and times DT DT now here I'm going to make a little bit of an intuitive argument a lot of the math we do is kind of especially when you're if you want to be very rigorous and formal the Dirac Delta function starts to break down a lot of tools that you might have not realize it break down but I think intuitively we can still work with it so I'm going to I'm going to solve this integral for you intuitively and I think it'll make some sense so let's draw this let me draw this what we're trying to do so let me draw what we're trying to take the integral of and we only care from 0 to infinity so I'll only do it from 0 to infinity and I'll assume that C is greater than 0 that the Delta function pops up someplace and the positive T axis so what is this what is this first part going to look like what is that going to look like e to the minus s T times f of T I don't know it's going to be some function either minus s T starts at one and drops down but we're multiplying it some times some arbitrary function so I'll just I'll just draw it like this maybe it looks something like this this right here is e to the minus s T times F of T and the F of T is what kind of gives us its arbitrary shape fair enough now let's graph our Dirac Delta function it's 0 everywhere except right at C right at C right there it pops up infinitely high but we only draw an arrow that is of height 1 to show that it's area is 1 I mean normally when you graph things you don't draw arrows but this arrow shows that the area under this infinitely high thing is 1 so we do a 1 there so if we multiply this right we care about the area under this whole thing when we multiply these two functions when we multiply this times this times the Delta function this is let me write this this is the Delta function shifted to see if I multiply that times that what do I get this is this is kind of the key intuition here let me redraw my axes and I'll try to let me see if I can do it a little bit straighter okay don't judge me by the straightness of my axes so that's T so what happens when I multiply these two everywhere when T equals anything other than C the Dirac Delta function is 0 right so it's 0 times anything I don't care what this function is do it's going to be 0 so it's going to be 0 everywhere except something hat interesting happens at T is equal to C at T equals C what's the value of the function well it's going to be the value of the Dirac Delta function it's going to be the Dirac Delta function times whatever height this is right this is going to be this point right here or this right there that point this is going to be this function evaluated at C so it's going to be a market right here on the y axis or on the f of TI whatever you want to call it this is going to be e to the minus s C times F of C all I'm doing is I'm just evaluating this function at C so that's the point right there so if you take this point which is just some number it could be five five times this you're just getting five times the Dirac Delta function or in this case it's not five is this a little bit more abstract thing I could just draw it like this the grunt when I multiply this thing times my little Delta function there I get this the height it's a delta function but it's scaled now it's scaled so now my new thing is going to look like this if I just multiply that times that I essentially get e to the minus SC times f of C this might look like some fancy function but it's just a number when we consider in terms of T s you know it'll become something when we go into the plus world but and if from T's point of view it's just a constant all of these are just constants so this might as well just be five so it's this constant times my Dirac Delta function times delta of t minus c when i multiply that thing times that thing all I'm left with is this thing and it has and this height is still going to be infinitely high but it's infinitely high scaled in such a way that it's area it's area is going to be not one and I'll show it to you so what's the integral of this thing right taking the integral of this thing from minus infinity to infinity since this thing is this thing it should be the same thing taking the integral of this thing from minus infinity to infinity so let's do that actually we only have to do four minus infinity I said from zero to infinity so if we take from zero to infinity I'm what I'm saying is taking this integral is equivalent to taking this integral so e to the minus SC f of C times my delta function t minus C DT now this thing right here let me make it very clear I'm claiming that this is equivalent to this because everywhere else the Delta function zeroes out this function so we only care about this function or e to the minus s T F of T when T is equal to C and so that's why we were able to turn it into a constant but since this is a constant we can bring it out of the integral and so this is equal to I'm going to go backwards here just to kind of save space and still have you give you these things to look at if we take out the constants from outside of the from inside of the integral we get e to the minus SC times F of C times the integral from 0 to infinity of f of t minus C DT oh sorry not F of t minus C let me see how far back I can times this is not an F that had to be very careful this is a delta let me do that in a different color I took out the constant terms there and it's going to be of delta of t minus C DT and DT let me get the right color d T now what is this thing by definition this thing is 1 I mean we could put it from minus infinity to infinity doesn't matter the only time where it has any areas right under C so this thing is equal to 1 so this whole integral right there has been reduced to this right there because this is just equal to 1 so the Laplace transform of our ship did Delta function times some other function is equal to e to the minus SC times F of C let me write that again down here let me write it all at once so the Laplace transform of our ship did Delta function t minus C times some function f of T it equals it equals e to the minus C essentially we're just evaluating e to the minus s T evaluated at C so e to the minus C s times F of C we're essentially just evaluating these things at C this is what it equals so from this we can get a lot of interesting things what is what is the Laplace transform actually what is the Laplace transform of just our the plain vanilla Delta function well in this case we have C is equal to 0 and f of T is equal to 1 right it's just a constant term so if we do that then the Laplace transform of this thing is just going to be e to the minus 0 times s times 1 which is just equal to 1 so the Laplace transform or of our Delta function is 1 which is a nice clean thing to find out and then if we wanted to just figure out the Laplace transform of our shifted function the Laplace transform of our shifted Delta function this is just a special case where f of T is equal to 1 right this is just F of there's an we could write it times 1 where f of T is equal to 1 so this is going to be equal to e to the minus C s times F of C but F is just a constant you know there's f is just one here so times 1 or just e to the minus C s so just like that using a kind of a visual evaluation of the integral we're able to figure out the Laplace transform for for a bunch of different situations involving the Dirac Delta function anyway hopefully found that reasonably useful