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# Laplace Transform of the Dirac Delta Function : Figuring out the Laplace Transform of the Dirac Delta Function

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- 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 delta function.
- And I defined it to be-- and I'll do the
- shifted version of it.
- You're already hopefully reasonably familiar with it.
- So Dirac delta of t minus c.
- We can say that it equals 0, when t does not equal c, so it
- equals 0 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.
- 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
- pseudodefinition 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 t is equal to c, when we
- take this area, this is the important point, that the area
- is equal to 1.
- And so this is what I meant by pseudoinfinity, because 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 2
- 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 I put a 2 out here, this infinity will have to be twice
- as high, so that the area is now 2.
- 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 that it can
- help model things that kind of jar things all of a sudden,
- but they impart a fixed amount of impulse on something and 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 if you want to unshift it, you just say, OK,
- well, c equals 0.
- What happens when c equals 0?
- 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
- delta function by itself, I could say f of
- t is equal to 1.
- So 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,
- or we could call it the integral from 0 to infinity of
- e to the minus -- 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.
- 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 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 realized it would break down, but I think
- intuitively, we can still work with it.
- So 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 zero to infinity, so I'll only do it
- from zero to infinity.
- And I'll assume that c is greater than zero, that the
- delta function pops up someplace in
- the positive t-axis.
- So what is this first part going to look like?
- What is that going to look like? e to the minus
- st times f of t?
- I don't know.
- It's going to be some function. e to the minus st
- starts at 1 and drops down, but we're multiplying it times
- some arbitrary function, so I'll just draw it like this.
- Maybe it looks something like this.
- This right here is e to the minus st times f of t.
- And the f of t is what kind of gives it its arbitrary shape.
- Fair enough.
- Now, let's graph our Dirac delta function.
- With zero 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 its 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, 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 c.
- If I multiply that times that, what do I get?
- This is kind of the key intuition here.
- Let me redraw my axes.
- Let me see if I can do it a little bit straighter.
- 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 zero.
- So it's zero times anything.
- I don't care what this function is going to do, it's
- going to be zero.
- So it's going to be zero everywhere, except something
- 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.
- 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.
- I'll mark it right here on the y-axis, or on the f of t,
- whatever you want to call it.
- This is going to be e to the minus sc 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 5, 5 times this, you're just getting 5 times
- the Dirac delta function.
- Or in this case, it's not 5.
- It's this little more abstract thing.
- I could just draw it like this.
- 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 it in terms of t.
- s, it becomes something when we go into the Laplace world,
- but from t's point of view, it's just a constant.
- All of these are just constants, so this might as
- well just be 5.
- 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 this height is still going to be infinitely high, but
- it's infinitely high scaled in such a way that its area is
- going to be not 1.
- And I'll show it to you.
- So what's the integral of this thing?
- Taking the integral of this thing from minus infinity to
- infinity, since this thing is this thing, it should be the
- same thing as taking the integral of this thing from
- minus infinity to infinity.
- So let's do that.
- Actually, we don't have to do it from minus infinity.
- I said from zero to infinity.
- So if we take from zero to infinity, 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 this 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 st 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 give you these
- things to look at.
- If we take out the constants 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.
- This is not an f.
- I have 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 a
- delta of t minus c dt.
- Let me get the right color.
- Now, what is this thing by definition?
- This thing is 1.
- I mean, we could put it from minus infinity to infinity, it
- doesn't matter.
- The only time where it has any area is 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 shifted 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 shifted delta function t
- minus c times some function f of t, it equals e
- to the minus c.
- Essentially, we're just evaluating e to the minus st
- evaluated at c.
- So e to the minus cs 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 the Laplace transform-- actually, what is
- the Laplace transform of just the plain
- vanilla delta function?
- Well, in this case, we have c is equal to 0, and f of t is
- equal to 1.
- 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 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.
- 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 cs times f
- of c, but f is just a constant, f is just 1 here.
- So it's times 1, or it's just e to the mine cs.
- So just like that, using a kind of visual evaluation of
- the integral, we were able to figure out the Laplace
- transforms for a bunch of different situations involving
- the Dirac delta function.
- Anyway, hopefully, you found that reasonably useful.

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