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# Dirac Delta Function : Introduction to the Dirac Delta Function

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- 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.

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