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.