Dirac Delta Function

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Dirac Delta Function : Introduction to the Dirac Delta Function

相關課程

0 / 750

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.

留言:

載入中...

新增一則留言(尚未登入)

dirac delta function 討論區:

載入中...