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### 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 dirac delta function

Figuring out the Laplace Transform of the Dirac Delta Function. Created by Sal Khan.

## Want to join the conversation?

- At the end of the video (10:50-11:50) it is stated that:

L{d(t)} = 1

L{d(t-c)} = e^(-cs)

Both d(t) and d(t-c) represent a line of the same height, one line is on 0 and the other is on c. Since both d(t) and d(t-c) are lines of the same height, howcome their area (and therefore their result to the laplace-transform) are different?

Thanks a lot for your help and another great thanks to Khan. Very helpful indeed.(14 votes)- if c=0, e^(-cs)=1, so those two equations you wrote are the same

Its not that the delta functions have different areas, just that the laplace transforms look different(18 votes)

- At10:55, Sal finds the Laplace transform of the delta function where c is 0. He concludes that the result is 1, but surely it should be 1/2, as only half of the spike is within the range from 0 to infinity.(4 votes)
- Through the magic of calculus, the entire area of 1 is exactly at zero. No part of it is "hanging off" either side of zero.(7 votes)

- Can someone explain me intuitively the difference between phase and time shifting in Laplace , I don't mean the formula , rather meaning behind it ?(6 votes)
- at11:20why is f(c)=1? shouldn't it be just f(0) since c=0? have we defined what f(x) is?(4 votes)
- integrate f(t-c) from infinity to zero, which includes the value of ZERO. If you recall the last video (Dirac delta function), it did mention the value of τ could be infinite small to make infinite high in y-axis. In this case, the integration range is mainly dependent on -τ and τ. Assumed τ infinitely close to 0, then the range can be assumed as zero.

Back to your question. when c=0, 0 is still in the integration domain and therefore the calculated result is still the unit.(2 votes)

- What is the inverse laplace of "s"(4 votes)
- The inverse Laplace transform of F(s)=s is actually 𝛿'(t), the derivative of the Dirac delta function.

In general the inverse Laplace transform of F(s)=s^n is 𝛿^(n), the nth derivative of the Dirac delta function.

This can be verified by examining the Laplace transform of the Dirac delta function (i.e. the 0th derivative of the Dirac delta function) which we know to be 1 =s^0.(2 votes)

- So the inverse Laplace of 1 is the Dirac delta function?(4 votes)
- Why do the limits for the dirac delta function not matter? Sal said doesn't matter 0 to infinity or -infinity to infinity. I don't understand.(3 votes)
- it's because of the definition of the dirac delta function. the function, which is actually a type of general function called a distribution, has the properties that the function has a value of zero everywhere except at zero, where it's "infinite", and any definite integral of this function which includes zero in its bounds will have a value of 1, and zero otherwise.(1 vote)

- Seems like you are assuming c>0(1 vote)
- yes, he writes it at4:14(4 votes)

- At8:23, I don't agree with your explanation of bringing the constant outside the integral bc u have to evaluate the integral at all points. From 0 to infinity, e^(-st)*f(t) does have values. It is misleading and I hope you or some1 else can come up with a better explanation.(0 votes)
- He pulls e^(-sc) * f(c) outside of the integral not e^(-st)*f(t). Because we have a c in place of t the entire expression is a constant and can be pulled out of the integral(8 votes)

- Is L^-1{1} = d(t), or can it be something else? (Is the inverse Laplace transform of 1 unique?)(2 votes)

## Video transcript

In the last video, I introduced
you to what is probably the most bizarro
function that you've encountered so far. And that was the Dirac
delta function. And I defined it to be--
and I'll do the shifted version of it. You're already hopefully
reasonably familiar with it. So Dirac delta of t minus c. We can say that it equals 0,
when t does not equal c, so it equals 0 everywhere, but it essentially pops up to infinity. And we have to be careful
with this infinity. I'm going to write
it in quotes. It pops up to infinity. And we even saw in the previous
video, it's kind of different degrees of infinity,
because you can still multiply this by other numbers to get
larger Dirac delta functions when t is equal to c. But more important than this,
and this is kind of a pseudodefinition here, is the
idea that when we take the integral, when we take the area
under the curve over the entire x- or the entire t-axis,
I guess we could say, when we take the area under this
curve, and obviously, it equals zero everywhere except
at t is equal to c, when we take this area, this is the
important point, that the area is equal to 1. And so this is what I meant by
pseudoinfinity, because if I have 2 times the Dirac delta
function, and if I'm taking the area under the curve of
that, of 2 times the Dirac delta function t minus c dt,
this should be equal to 2 times-- the area of just under
the Dirac delta function 2 times from minus infinity to
infinity of the delta function shifted by c dt, which is just
2 times-- we already showed you, I just said, by definition,
this is 1, so this will be equal to 2. So if I put a 2 out here, this
infinity will have to be twice as high, so that the
area is now 2. That's why I put that infinity
in parentheses. But it's an interesting
function. I talked about it at the end of
the last video that it can help model things that kind of
jar things all of a sudden, but they impart a fixed amount
of impulse on something and a fixed amount of change
in momentum. And we'll understand that a
little bit more in the future. But let's kind of get the
mathematical tools completely understood. And let's try to figure out what
the Dirac delta function does when we multiply it, what
it does to the Laplace transform when we multiply
it times some function. So let's say I have my Dirac
delta function and I'm going to shift it. That's a little bit
more interesting. And if you want to unshift
it, you just say, OK, well, c equals 0. What happens when c equals 0? And I'm going to shift it and
multiply it times some arbitrary function f of t. If I wanted to figure out the
Laplace transform of just the delta function by itself,
I could say f of t is equal to 1. So let's take our Laplace
transform of this. And we can just use the
definition of the Laplace transform, so this is equal to
the area from 0 to infinity, or we could call it the integral
from 0 to infinity of e to the minus -- that's just
part of the Laplace transform definition-- times this thing--
and I'll just write it in this order-- times
f of t times our Dirac delta function. Delta t minus c and times dt. Now, here I'm going to
make a little bit of an intuitive argument. A lot of the math we do is kind
of-- especially if you want to be very rigorous and
formal, the Dirac delta function starts to break down a
lot of tools that you might have not realized it would
break down, but I think intuitively, we can still
work with it. So I'm going to solve this
integral for you intuitively, and I think it'll
make some sense. So let's draw this. Let me draw this, what
we're trying to do. So let me draw what we're
trying to take the integral of. And we only care from zero to
infinity, so I'll only do it from zero to infinity. And I'll assume that c is
greater than zero, that the delta function pops
up someplace in the positive t-axis. So what is this first part
going to look like? What is that going to look
like? e to the minus st times f of t? I don't know. It's going to be some function.
e to the minus st starts at 1 and drops down, but
we're multiplying it times some arbitrary function, so I'll
just draw it like this. Maybe it looks something
like this. This right here is e to the
minus st times f of t. And the f of t is what kind of
gives it its arbitrary shape. Fair enough. Now, let's graph our Dirac
delta function. With zero everywhere except
right at c, right at c right there, it pops up infinitely
high, but we only draw an arrow that is of height 1 to
show that its area is 1. I mean, normally when you graph
things you don't draw arrows, but this arrow shows
that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care
about the area under this whole thing. When we multiply these two
functions, when we multiply this times this times the delta
function, this is-- let me write this. This is the delta function
shifted to c. If I multiply that times
that, what do I get? This is kind of the key
intuition here. Let me redraw my axes. Let me see if I can do it
a little bit straighter. Don't judge me by the
straightness of my axes. So that's t. So what happens when I
multiply these two? Everywhere, when t equals
anything other than c, the Dirac delta function is zero. So it's zero times anything. I don't care what this function
is going to do, it's going to be zero. So it's going to be zero
everywhere, except something interesting happens at
t is equal to c. At t equals c, what's the
value of the function? Well, it's going to
be the value of the Dirac delta function. It's going to be the Dirac delta
function times whatever height this is. This is going to be this point
right here or this right there, that point. This is going to be this
function evaluated at c. I'll mark it right here on the
y-axis, or on the f of t, whatever you want to call it. This is going to be e to the
minus sc times f of c. All I'm doing is I'm just
evaluating this function at c, so that's the point
right there. So if you take this point, which
is just some number, it could be 5, 5 times this, you're
just getting 5 times the Dirac delta function. Or in this case, it's not 5. It's this little more
abstract thing. I could just draw
it like this. When I multiply this thing times
my little delta function there, I get this. The height, it's a delta
function, but it's scaled now. It's scaled, so now my
new thing is going to look like this. If I just multiply that times
that, I essentially get e to the minus sc times f of c. This might look like some fancy
function, but it's just a number when we consider
it in terms of t. s, it becomes something when we
go into the Laplace world, but from t's point of view,
it's just a constant. All of these are just constants,
so this might as well just be 5. So it's this constant times my
Dirac delta function, times delta of t minus c. When I multiply that thing times
that thing, all I'm left with is this thing. And this height is still going
to be infinitely high, but it's infinitely high scaled in
such a way that its area is going to be not 1. And I'll show it to you. So what's the integral
of this thing? Taking the integral of this
thing from minus infinity to infinity, since this thing is
this thing, it should be the same thing as taking the
integral of this thing from minus infinity to infinity. So let's do that. Actually, we don't have to do
it from minus infinity. I said from zero to infinity. So if we take from zero to
infinity, what I'm saying is taking this integral
is equivalent to taking this integral. So e to the minus sc f of c
times my delta function t minus c dt. Now, this thing right here, let
me make this very clear, I'm claiming that this is
equivalent to this. Because everywhere else, the
delta function zeroes out this function, so we only care about
this function, or e to the minus st f of t when
t is equal to c. And so that's why we were able
to turn it into a constant. But since this is a constant,
we can bring it out of the integral, and so this is equal
to-- I'm going to go backwards here just to kind of save space
and still give you these things to look at. If we take out the constants
from inside of the integral, we get e to the minus sc times f
of c times the integral from 0 to infinity of f
of t minus c dt. Oh sorry, not f of t minus c. This is not an f. I have to be very careful. This is a delta. Let me do that in a
different color. I took out the constant terms
there, and it's going to be a delta of t minus c dt. Let me get the right color. Now, what is this thing
by definition? This thing is 1. I mean, we could put it from
minus infinity to infinity, it doesn't matter. The only time where it has any
area is right under c. So this thing is equal to 1. So this whole integral right
there has been reduced to this right there, because this
is just equal to 1. So the Laplace transform of
our shifted delta function times some other function is
equal to e to the minus sc times f of c. Let me write that
again down here. Let me write it all at once. So the Laplace transform of our
shifted delta function t minus c times some function
f of t, it equals e to the minus c. Essentially, we're just
evaluating e to the minus st evaluated at c. So e to the minus
cs times f of c. We're essentially just
evaluating these things at c. This is what it equals. So from this we can get a lot
of interesting things. What is the Laplace transform--
actually, what is the Laplace transform
of just the plain vanilla delta function? Well, in this case, we have c
is equal to 0, and f of t is equal to 1. It's just a constant term. So if we do that, then the
Laplace transform of this thing is just going to be e to
the minus 0 times s times 1, which is just equal to 1. So the Laplace transform of our
delta function is 1, which is a nice clean thing
to find out. And then if we wanted to just
figure out the Laplace transform of our shifted
function, the Laplace transform of our shifted delta
function, this is just a special case where f
of t is equal to 1. We could write it times 1, where
f of t is equal to 1. So this is going to be equal to
e to the minus cs times f of c, but f is just a constant,
f is just 1 here. So it's times 1, or it's
just e to the mine cs. So just like that, using a kind
of visual evaluation of the integral, we were able
to figure out the Laplace transforms for a bunch of
different situations involving the Dirac delta function. Anyway, hopefully, you found
that reasonably useful.