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## Differential equations

### Course: Differential equations > Unit 3

Lesson 2: Properties of the Laplace transform- Laplace as linear operator and Laplace of derivatives
- Laplace transform of cos t and polynomials
- "Shifting" transform by multiplying function by exponential
- Laplace transform of t: L{t}
- Laplace transform of t^n: L{t^n}
- Laplace transform of the unit step function
- Inverse Laplace examples
- Dirac delta function
- Laplace transform of the dirac delta function

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# Laplace transform of the unit step function

Introduction to the unit step function and its Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

- I keep getting stuck around 20 min where Sal explains that x is just a letter. I can't get my head around this since I thought that initially we made the substitution x = t - c

i.e. that f(t - c) = f(x)

Doesn't this mean that at the end we have to re-substitute t - c into the function such that we have the Laplace transform of the function f(t - c) factored by exp(-sc) as opposed to just the Laplace transform of the function f(t) factored by exp(-sc).

Thank you for any help :)(22 votes)- Took me a while to understand this but I finally figured it out.

The issue here is in the normal way we write laplace transforms, the variable of integration is implied. This is because it doesn't normally matter, but it's biting us here.

The result Sal got was`e^(-sc) * laplace f(x)`

`integrated with respect to x`

This is exactly the same as`e^(-sc) * laplace f(t) integrated with respect to t`

which is a property of integrals.

If you*did*substitute`x = t - c`

, and wrote`e^(-sc) * laplace f(t-c)`

you are actually writing`e^(-sc) * laplace f(t-c) integrated with respect to t`

because we would assume the variable of integration is`t`

.**However**, to correctly substitute`x = t - c`

you would need to write`e^(-sc) * laplace f(t-c) integrated with respect to t - c`

**This**integral is the same as`e^(-sc) * laplace f(t) integrated with respect to t`

which is what Sal concludes.

I hope this helps someone.(8 votes)

- Don't we have to assume that c > 0 for this? The bounds of integration in the original definition of the Laplace transform were from 0 to infinity. I understand that t values below c for the unit step function will cancel out the entire integral, but if we were to have a c value which was below 0 then wouldn't this in a sense violate our original definition of a Laplace transformation as the integral will give even more area under the curve? Or am I looking at this incorrectly?(7 votes)
- I think your point is what Sal mentioned in the beginning of the video. It's not that the unit step function requires the assumption c>0 in and of itself. Rather that for use with the Laplace transform, we are only interested in c values greater than zero since, as you mentioned, the transform is only concerned with t [0,inf).(5 votes)

- Defining the second unit step function, around 4min: when you say that you make the function jump up again to a y value of 2 at x=2π, the function also continues (and therefore is not a function) at y=0, unless you've somehow told it not to. The question then is, did you tell it not to, and if you did, how?(3 votes)
- I think you're talking about this function: f(t)=2-2u[pi](t)+2u[2pi](t)...

The unit step function (the u-something function of t) basically tells you where it is taking the step, in this case the first is at pi and the second at two-pi. The sign tells you which direction it is going, i.e. if it is negative it is going down and positive it is moving up. the coefficient in front, in this case they are both two's, tells you how far of a jump it is making. So in this function we start at a constant function of two, then when we have -2u[pi](t) for the next part we know when we get to pi on the t-axis it should decrease by 2, here equalling zero, and then when we have +2u[2pi](t) we know that after continuing at the zero from the previous part, when we get to two-pi we should increase by 2. I hope this clears it up a bit for you, let me know if you have any more questions. These guys can be a little tricksy sometimes.(8 votes)

- we assume that the term u_c(t) is 1 but this value seems to change for this function as it is peaks and troughs? or is he only referring to the y value of the function at the unit step?(2 votes)
- The graph is u_c(t) times f(t-c). It is the function f that is varying. u_c(t) merely multiplies f by 0 or 1.(5 votes)

- I have a question abt the video around 20.31 where SAL said that the laplace transformation of f(x) is almost equivalent to f(t). If x and t were independent variables, then i had no problem assuming that but here x and t are not independent variables because SAL makes a substitution earlier in the video that x = t-c , then wont the laplace transformation of f(x) would be equivalent to laplace transform of f(t-c). Please correct me where i m wrong?(2 votes)
- I had trouble wrapping my head around it too and so I substituted t back in and saw that it leads to the same conclusion. From what I understand, it's the presence of the unit step function (and that the entire function is 0 until t = c) that makes the Laplace transforms of f(x) and f(t) basically the same.

As a reminder,

t = x + c or x = t - c

Laplace{u_c(t) f(t-c)} = e^(-sc) * integral from x=0 to infinity of e^(-sx) f(x) dx

^Those equations were from around19:30if that wasn't clear. Substituting back in t,

= e^(-sc) * integral from t=c to infinity of (e^(-s(t-c)) * u_c(t) * f(t-c) dt

If we were to subtract the bounds of the integral by c and replace t with t + c in the integrand, you get an equivalent integral. (Test this out yourself with a simpler integral if you find it difficult to swallow.) Also, infinity minus an arbitrary constant c is still infinity so the upper bound remains the same. So,

= e^(-sc) * integral from t=0 to infinity of e^(-s((t + c) - c)) * u_c(t + c) * f(t + c - c) dt

u_c (t + c) is always 1 within the interval [0, infinity) so we don't have to worry about it. Simplifying,

= e^(-sc) * integral from t=0 to infinity of e^(-st) * f(t) dt

That integral should look familiar enough to us as the Laplace transform of f(t). So,

Laplace{u_c(t) f(t-c)} = e^(-sc) * Laplace{f(t)}

I hope this made sense.(4 votes)

- In this video Sal takes x as t and write f(x)=f(t) and say that both are just letters but what about that substitution x=t-c?(3 votes)
- wait I'm confused.. isn't this supposed to be the laplace transform of ONLY the step function?

The video is showing the laplace transform of step function u(t)*f(t-c)...(2 votes)- Yes, the title of the video is slightly misleading, but if you want the laplace transform of just the unit step function, then you can just say f(t-c) = 1.(4 votes)

- i think it's also called the heavyside function, isn't it?(2 votes)
- Is the laplace transform of unit step function L{ua(t)} = e^(-as)/s (s>0, a>0)??

That seems different from the result in this video.(2 votes)- In the video, he never takes the Laplace transform of just the unit step function. If f(t)=1, then, as you wrote, L{ua(t)*f(t-a)]=e^(-as)*L{f(t)}=e^(-as)*L{1} = e^(-as)*(1/s) = e^(-as)/s.(3 votes)

- How is the unit step function an actual function? Doesn't it violate the vertical line test by passing through infinite times? Thanks in advance.(1 vote)
- No. it only passes through a vertical line once. If the vertical line is at t < c, then the function only intersects it at 0. If the vertical line is at t ≥ c, then the function only intersects it at 1.(5 votes)

## Video transcript

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.