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# Laplace Transform of the Unit Step Function : Introduction to the unit step function and its Laplace Transform

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- The whole point in learning differential equations is that
- eventually we want to model real physical systems. I know
- everything we've done so far has really just been a toolkit
- of being able to solve them, but the whole reason is that
- because differential equations can describe a lot of systems,
- and then we can actually model them that way.
- And we know that in the real world, everything isn't these
- nice continuous functions, so over the next couple of videos
- we're going to talk about functions that are a little
- bit more discontinuous than what you might be used to even
- in kind of a traditional calculus or traditional
- Precalculus class.
- And the first one is the unit step function.
- Let's write it as u, and then I'll put a little
- subscript c here of t.
- And it's defined as when t is-- let me put it this way.
- It's defined as 0 when t is less than-- whatever subscript
- I put here-- when t is less than c.
- And it's defined as 1-- that's why we call it the unit step
- function-- when t is greater than or equal to c.
- And if I had to graph this, and you could graph it as well
- but it's not too difficult to graph.
- Let me draw my x-axis right here.
- And I'll do a little thicker line.
- That's my x-axis right there.
- This is my y-axis right there.
- And when we talk about Laplace transforms, which we'll talk
- about shortly, we only care about t is greater than 0.
- Because we saw, in our definition of the Laplace
- transform, we're always taking the integral from 0 to
- infinity, so we're only dealing with
- the positive x-axis.
- But anyway, by this definition, it would be zero
- all the way until you get to some value c, so you'd be zero
- until you get to c.
- And then at c, you jump, and the point c is included x is
- equal to c here.
- So it's included, so I'll put a dot there, because it's
- greater than or equal to c.
- You're at 1, so this is 1 right here.
- And then you go forward for all of time.
- And you're like, Sal, you just said that differential
- equations, we're learning to model things, why is this type
- of a function useful?
- Well, in the real world, sometimes you do have
- something that essentially jolts something, that moves it
- from this position to that position.
- And obviously, nothing can move it immediately like this,
- but you might have some system, it could be an
- electrical system or mechanical system, where maybe
- the behavior looks something like this, where maybe it
- moves it like that or something.
- And this function is a pretty good analytic approximation
- for some type of real world behavior like this when
- something just gets moved.
- Whenever we solve these differential equations
- analytically, we're really just trying to get a pure
- model of something.
- Eventually, we'll see that it doesn't perfectly describe
- things, but it helps describe it enough for us to get a
- sense of what's going to happen.
- Sometimes it will completely describe things, but anyway,
- we can ignore that for now, so let me get rid of these things
- right there.
- So the first question is, well, you know, what if
- something doesn't jar just like that?
- What if I want to construct more fancy unit functions or
- more fancy step functions?
- Let's say I wanted to construct something that
- looked like this.
- Let me say this is my y-axis.
- This is my x-axis.
- And let's say I wanted to construct something that is
- at-- and let me do it in a different color.
- Let's say it's at 2 until I get to pi.
- And then from pi until forever it just stays at zero.
- So how could I construct this function right here using my
- unit step function?
- So what if I had written it as-- so my unit step
- function's at zero initially, so what if I make it 2 minus a
- unit step function that starts at pi?
- So if I define my function here, will this work?
- Well, this unit step function, when we pass pi, is only going
- to be equal to 1, but we want this thing
- to be equal to zero.
- So it has to be 2 minus 2, so I'll have to put at 2 here,
- and this should work.
- When we're at any value below pi, when t is less than pi
- here, this becomes a zero, so our function will just
- evaluate to 2, which is right there.
- But as soon as we hit t is equal to pi, that pi is the c
- in this example, as soon as we hit that, the unit step
- function becomes 1.
- We multiply that by 2, and we have 2 minus 2, and then we
- end up here with zero,
- Now, that might be nice and everything, but let's say you
- wanted for it to go back up.
- Let's say that instead of it going like this-- let me kind
- of erase that by overdrawing the x-axis again-- we want the
- function to jump up again.
- We want it to jump up again.
- And lets say at some value, let's say it's at 2pi, we want
- the function to jump up again.
- How could we construct this?
- And we could make it jump to anything, but let's say we
- want it to jump back to 2.
- Well, we could just add another unit step function
- here, something that would have been zero all along, all
- the way up until this point.
- But then at 2pi, it jumps, so in this case,
- our c would be 2pi.
- That's our unit step function, and we want it to jump to 2.
- This would just jump to 1 by itself.
- So let's multiply it by 2.
- And now we have this function.
- So you could imagine, you can make an arbitrarily
- complicated function of things jumping up and down to
- different levels based on different essentially linear
- combinations of these unit step functions.
- Now, what if I wanted to do something
- a little bit fancier?
- What if I wanted to do something that-- let's say I
- have some function that looks like this.
- Let me draw some function.
- I should draw straighter than that.
- I should have some standards.
- So let's say that just my regular f of t--
- let me, this is x.
- Actually, why am I doing x?
- This would be the t-axis.
- We're doing the time domain.
- It could have been x.
- And then we'll call this f of t.
- So let me draw some arbitrary f of t.
- Let's say my function looks something crazy like that.
- So this is my f of t.
- What if I'm modeling a physical system
- that doesn't do this?
- That actually at some point-- well, actually, let's say it
- stays at zero.
- It stays at zero until some value.
- Let's say it goes to zero until-- I don't know, I'll
- call that c again.
- And then at c, f of t kind of starts up.
- So right at c, f of t should start up, so it just kind of
- goes like this.
- So essentially what we have here is a combination of zero
- all the way, and then we have a shifted f of t.
- So at c, we have a shifted f of t, so it shifts that way.
- So how can we construct this yellow function, where it's
- essentially a shifted version of this green function, but
- it's zero below c?
- This green function might have continued.
- It might have gone something like this.
- It might have, continued and done something crazy, but what
- we did is we shifted it from here to there, and then we
- zeroed out everything before c.
- So how could we do that?
- Well, just shifting this function, you've learned in
- your Algebra II or your precalculus classes, to shift
- a function by c to the right, you just to replace your t
- with a t minus c.
- So this function right here is f of t minus c.
- And to make sure I get it right, what I always do is I
- imagine, OK, what's going to happen when t is equal to c?
- When t is equal to c, you're going to have a c minus a c,
- and you're going to have f of 0.
- So f of 0, it should be the same.
- So when t is equal to c, this value, the value of the
- function should be equivalent to the value of the original
- green function at zero, so it's equivalent to that value,
- which makes sense.
- If we go up one more above c, so let's say this is one more
- above c, so we get to this point, if t is c plus 1, then
- when you put c plus 1 minus c, you just have f of 1, and f of
- 1 is really just this point right here.
- And so it'll be that f of 1, so it makes sense.
- So as we move one forward here, we're essentially at the
- same function value as we were there, so the shift works.
- But I said that we have to also-- if I just shifted this
- function, you would have all this other stuff, because you
- would have had all this other stuff when the function was
- back here still going on.
- The function-- I'll draw it
- lightly-- would still continue.
- But I said I wanted to zero out this function
- before we reach c.
- So how can I zero out that function?
- Well, I think it's pretty obvious to you.
- I started this video talking about the unit step function.
- So what if I multiply the unit step
- function times this thing?
- What's going to happen?
- So what if I-- my new function, I call it the unit
- step function up until c of t times f of t minus c?
- So what's going to happen?
- Until we get to c, the unit step function is zero when
- it's less than c.
- So you're going to have zero times I don't care what this
- is Zero times anything is zero, so this function is
- going to be zero.
- Once you hit c, the unit step function becomes 1.
- So once you pass c, this thing becomes a 1, and you're just
- left with 1 times your function.
- So then your function can behave as it would like to
- behave, and you actually shifted it.
- This t minus c is what actually shifted this green
- function over to the right.
- And this is actually going to be a very
- useful constructed function.
- And in a second, wer'e going to figure out the Laplace
- transform of this, and you're going to appreciate, I think,
- why this is a useful function to look at.
- But now you understand at least what it is and why it
- essentially shifts a function and zeroes out everything
- before that point.
- Well, I told you that this is a useful function, so we
- should add its Laplace transform to our library of
- Laplace transforms. So let's do that.
- Let's take a the Laplace transform of this, of the unit
- step function up to c.
- I'm doing it in fairly general terms. In the next video,
- we'll do a bunch of examples where we can apply this, but
- we should at least prove to ourselves what the Laplace
- transform of this thing is.
- Well, the Laplace transform of anything, or our definition of
- it so far, is the integral from 0 to infinity of e to the
- minus st times our function.
- So our function in this case is the unit step function, u
- sub c of t times f of t minus c dt.
- And this seems very general.
- It seems very hard to evaluate this integral at first, but
- maybe we can make some form of a substitution to get it into
- a term that we can appreciate.
- So let's make a substitution here.
- Let me pick a nice variable to work with.
- I don't know, we're not using an x anywhere.
- We might as well use an x.
- That's the most fun variable to work with.
- Sometimes, you'll see in a lot of math classes, they
- introduce these crazy Latin alphabets, and that by itself
- makes it hard to understand.
- So I like to stay away from those crazy Latin alphabets,
- so we'll just use a regular x.
- Let's make a substitute.
- Let's say that x is equal to t minus c.
- Or you could, if we added t to both sides, we could say that
- t is equal to x plus c.
- Let's see what happens to our subsitution.
- And also, if we took the derivative of both sides of
- this, or I guess the differential, you would get dx
- is equal to dt.
- Or I mean, if you took dx with respect to dt, you would get
- that to equal to 1. c is just a constant.
- Then if you multiply both sides by dt, you get dx is
- equal to dt, and that's a nice substitution.
- So what is our integral going to become with this
- substitution?
- So our integral this was t equals 0 to
- t is equal to infinity.
- When t is equal to 0, what is x going to be equal to?
- Well, x is going to be equal to minus c.
- Actually, before I go there, let me actually take a step
- back, because we could progress.
- We could go in this direction.
- But we could actually simplify it more before we do that.
- Let's go back to out original integral before we even made
- our substitution.
- If we're taking the integral from 0 to infinity of this
- thing, we already said what does this integral look like
- or what does this function look like?
- It's zero.
- We have this unit step function sitting right here.
- We have the unit step function sitting right there.
- So this whole expression is going to be zero
- until we get to c.
- This whole thing, by definition, this unit step
- function is zero until we get to c.
- So this everything's going to be zeroed out
- until we get to c.
- So we could essentially say, you know, we don't have to
- take the integral from t equals 0 to t equals infinity.
- We could take the integral-- let me write it here.
- I'll just use that old integral sign.
- We could just take the integral from t is equal to c
- to t is equal to infinity of e to the minus st, the unit step
- function, uc of t times f of t minus c dt.
- In fact, at this point, this unit step function, it has no
- use anymore.
- Because before t is equal to c, it's 0, and now that we're
- only worried about values above c, it's equal to 1, so
- it equals 1 in this context.
- I want to make that very clear to you.
- What did I do just here?
- I changed our bottom boundary from 0 to c.
- And I think you might realize why I did it when I was
- working with the substitution, because this will simplify
- things if we do this ahead of time.
- So if we have this unit step function, this thing is going
- to zero out this entire integral before we get to c.
- Remember, this definite integral is really just the
- area under this curve of this whole function, of the unit
- step function times all of this stuff.
- All of this stuff, when we multiply it, is going to be
- zero until we get to some value c.
- And then above c, it's going to be e to the minus st times
- f of t minus c.
- So it's going to start doing all this crazy stuff.
- So if we want to essentially find the area under this
- curve, we can ignore all the stuff that happens before c.
- So instead of going from t equals 0 to infinity, we can
- go from t is equal to c to infinity because there was no
- area before t was equal to c.
- So that's all I did here.
- And then the other thing I said is that the unit step
- function, it's going to be 1 over this entire range of
- potential t-values, so we can just kind of ignore it.
- It's just going to be 1 this entire time, so our integral
- simplifies to the definite integral from t is equal to c
- to t is equal to infinity of e to the minus st times f of t
- minus is c dt.
- And this will simplify it a good bit.
- I was going down the other road when I did the
- substitution first, which would have worked, but I think
- the argument as to why I could have changed the boundaries
- would've been a harder argument to make.
- So now that we had this, let's go back and make that
- substitution that x is equal to t minus c.
- So our integral becomes-- I'll do it in green-- when t is
- equal to c, what is x?
- Then x is 0, right?
- c minus c is 0.
- When t is equal to infinity, what is x?
- Well x is, you know, infinity minus any constant is still
- going to be infinity, or if the limit is t approaches
- infinity, x is still going to be infinity here.
- And it's the integral of e to the minus s, but now instead
- of a t, we have the substitution.
- If we said x is equal to t minus c, then we can just add
- c to both sides and get t is equal to x plus c.
- So you get x plus c there, and then times the function f of t
- minus c, but we said t minus c is the same thing as x.
- And dt is the same thing is dx.
- I showed you that right there, so we can write this as dx.
- Now this is starting to look a little bit interesting.
- So what is this equal to?
- This is equal to the integral from 0 to infinity-- let me
- expand this out-- of e to the minus sx minus sc
- times f of x dx.
- Now, what is the equal to?
- Well, we could factor out an e to the minus sc and bring it
- outside of the integral, because this has nothing to do
- with what we're taking the integral with respect to.
- So let's do that.
- Let me take this guy out, and maybe just to not confuse you,
- let me rewrite the whole thing.
- 0 to infinity.
- I could rewrite this e term as e-- actually, let me
- write it this way.
- I'll do what was already in green as e to the minus sx
- times e to the minus sc.
- Common base.
- So if I were to multiply these two, I could just add the
- exponents, which you would get that up there, times
- f of x, d of x.
- This is a constant term with respect to x, so we can just
- factor it out.
- We can just factor this thing out right there, so then you
- get e to the minus sc times the integral from 0 to
- infinity of e to the minus sx times f of x dx.
- Now, what were we doing here the whole time?
- We were taking the Laplace transform of the unit step
- function that goes up to c, and then it's 0 up to c, and
- it's 1 after that, of t times some shifted function
- f of t minus c.
- And now we got that as being equal to this thing, and we
- made a substitution.
- We simplified it a little bit.
- e to the minus sc times the integral from 0 to infinity of
- e to the minus sx f of x dx.
- Something about the tablet doesn't work properly right
- around this period.
- But this should look interesting to you.
- What is this?
- This is the Laplace transform of f of x.
- Let me write that down.
- What's the Laplace transform of-- well, I could write it as
- f of t or f of x.
- The Laplace transform of f of t is equal to the integral
- from 0 to infinity of e to the minus st times f of t dt.
- This and this are the exact same thing.
- We're just using a t here.
- We're using an x here.
- No difference.
- They're just letters.
- This is f of t.
- e to the minus st times f of t dt.
- I could have also rewritten it as the Laplace
- transform of f of t.
- I could write this as the integral from 0 to infinity of
- e to the minus sy times f of y dy.
- I could do it by anything because this
- is a definite integral.
- The y's are going to disappear,
- and we've seen that.
- All you're left with is a function of s.
- This ends up being some capital, well, you know, we
- could write some capital function of s.
- So this is interesting.
- This is the Laplace transform of f of t times some scaling
- factor, and that's what we set out to show.
- So we can now show that the Laplace transform of the unit
- step function times some function t minus c is equal to
- this function right here, e to the minus sc, where this c is
- the same as this c right here, times the Laplace
- transform of f of t.
- Times the Laplace transform-- I don't know what's going on
- with the tablet right there-- of f of t.
- Let me write that.
- This is equal to-- because it's looking funny there-- e
- to the minus sc times the Laplace transform of f of t.
- So this is our result.
- Now, what does this mean?
- Oh, look it back-filled it somehow.
- What does this mean?
- What can we do with this?
- Well, let's say we wanted to figure out the Laplace
- transform of the unit step function that
- starts up at pi of t.
- And let's say we're taking something that we know well:
- sine of t minus pi.
- So we shifted it, right?
- This thing is really malfunctioning at this point
- right here.
- Let me pause it.
- I just paused.
- Sorry if that was a little disconcerting.
- I just paused the video because it was having trouble
- recording at some point on my little board.
- So let me rewrite the result that we proved just now.
- We showed that the Laplace transform of the unit step
- function t, and it goes to 1 at some value c times some
- function that's shifted by c to the right.
- It's equal to e to the minus cs times the Laplace transform
- of just the unshifted function.
- That was our result.
- That was the big takeaway from this video.
- And if this seems like some Byzantine, hard-to-understand
- result, we can apply it.
- So let's say the Laplace transform, this is what I was
- doing right before the actual pen tablet started
- malfunctioning.
- If we want to take the Laplace transform of the unit step
- function that goes to 1 at pi, t times the sine function
- shifted by pi to the right, we know that this is going to be
- equal to e to the minus cs.
- c is pi in this case, so minus pi s times the Laplace
- transform of the unshifted function.
- So in this case, it's the Laplace
- transform of sine of t.
- And we know what the Laplace transform of sine of t is.
- It's just 1 over s squared plus 1.
- So the Laplace transform of this thing here, which before
- this video seemed like something crazy, we now know
- is this times this.
- So it's e to the minus pi s times this, or we could just
- write it as e to the minus pi s over s squared plus 1.
- We'll do a couple more examples of this in the next
- video, where we go back and forth between the Laplace
- world and the t and between the s domain
- and the time domain.
- And I'll show you how this is a very useful result to take a
- lot of Laplace transforms and to invert a lot of Laplace
- transforms.

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