WEBVTT
00:00:08.000 --> 00:00:14.000
Well, today is the last day on
Laplace transform and the first
00:00:12.000 --> 00:00:18.000
day before we start the rest of
the term, which will be spent on
00:00:16.000 --> 00:00:22.000
the study of systems.
I would like to spend it on one
00:00:20.000 --> 00:00:26.000
more type of input function
which, in general,
00:00:23.000 --> 00:00:29.000
your teachers in other courses
will expect you to have had some
00:00:28.000 --> 00:00:34.000
acquaintance with.
It is the kind associated with
00:00:32.000 --> 00:00:38.000
an impulse, so an input
consisted of what is sometimes
00:00:36.000 --> 00:00:42.000
called a unit impulse.
Now, what's an impulse?
00:00:40.000 --> 00:00:46.000
It covers actually a lot of
things.
00:00:43.000 --> 00:00:49.000
It covers a situation where you
withdraw from a bank account.
00:00:48.000 --> 00:00:54.000
For example,
take half your money out of a
00:00:52.000 --> 00:00:58.000
bank account one day.
It also would be modeled the
00:00:56.000 --> 00:01:02.000
same way.
But the simplest way to
00:00:59.000 --> 00:01:05.000
understand it the first time
through is as an impulse,
00:01:03.000 --> 00:01:09.000
if you know what an impulse is.
If you have a variable force
00:01:08.000 --> 00:01:14.000
acting over time,
and we will assume it is acting
00:01:12.000 --> 00:01:18.000
along a straight line so I don't
have to worry about it being a
00:01:17.000 --> 00:01:23.000
vector, then the impulse,
according to physicists,
00:01:20.000 --> 00:01:26.000
the physical definition,
the impulse of f of t
00:01:25.000 --> 00:01:31.000
over some time interval.
Let's say the time interval
00:01:30.000 --> 00:01:36.000
running from a to b is,
by definition,
00:01:33.000 --> 00:01:39.000
the integral from a to b of f
of t dt.
00:01:38.000 --> 00:01:44.000
Actually, I am going to do the
00:01:42.000 --> 00:01:48.000
most horrible thing this period.
I will assume the force is
00:01:46.000 --> 00:01:52.000
actually a constant force.
So, in that case,
00:01:50.000 --> 00:01:56.000
I wouldn't even have to bother
with the integral at all.
00:01:55.000 --> 00:02:01.000
If f of t is a constant,
let's say capital F,
00:01:58.000 --> 00:02:04.000
then the impulse is --
Well, that integral is simply
00:02:04.000 --> 00:02:10.000
the product of the two,
the impulse over that time
00:02:08.000 --> 00:02:14.000
interval is simply F times b
minus a.
00:02:12.000 --> 00:02:18.000
Just the product of those two.
The force times the length of
00:02:17.000 --> 00:02:23.000
time for which it acts.
Now, that is what I want to
00:02:21.000 --> 00:02:27.000
calculate, want to consider in
connection with our little mass
00:02:26.000 --> 00:02:32.000
system.
So, once again,
00:02:29.000 --> 00:02:35.000
I think this is probably the
last time you'll see the little
00:02:34.000 --> 00:02:40.000
spring.
Let's bid a tearful farewell to
00:02:37.000 --> 00:02:43.000
it.
There is our little mass on
00:02:39.000 --> 00:02:45.000
wheels.
And let's make it an undamped
00:02:42.000 --> 00:02:48.000
mass.
It has an equilibrium point and
00:02:45.000 --> 00:02:51.000
all the other little things that
go with the picture.
00:02:49.000 --> 00:02:55.000
And when I apply an impulse,
what I mean is applying a
00:02:54.000 --> 00:03:00.000
constant force to this over a
definite time interval.
00:03:00.000 --> 00:03:06.000
And that is what I mean by
applying an impulse over that
00:03:03.000 --> 00:03:09.000
time interval.
Now, what is the picture of
00:03:06.000 --> 00:03:12.000
such a thing?
Well, the force is only going
00:03:09.000 --> 00:03:15.000
to be applied,
in other words,
00:03:11.000 --> 00:03:17.000
I am going to push on the mass
or pull on the mass with a
00:03:15.000 --> 00:03:21.000
constant force.
With a little electromagnet
00:03:18.000 --> 00:03:24.000
here, we will assume,
there is a pile of iron filings
00:03:22.000 --> 00:03:28.000
or something inside there.
I turn on the electromagnet.
00:03:26.000 --> 00:03:32.000
It pulls with a constant force
just between time zero and time
00:03:30.000 --> 00:03:36.000
two seconds.
And then I stop.
00:03:34.000 --> 00:03:40.000
That is going to change the
motion of the thing.
00:03:37.000 --> 00:03:43.000
First it is going to start
pulling it toward the thing.
00:03:40.000 --> 00:03:46.000
And then, when it lets go,
it will zoom back and there
00:03:44.000 --> 00:03:50.000
will be a certain motion after
that.
00:03:46.000 --> 00:03:52.000
What the question is,
if I want to solve that problem
00:03:49.000 --> 00:03:55.000
of the motion of that in terms
of the Laplace transform,
00:03:53.000 --> 00:03:59.000
how am I going to model this
force?
00:03:55.000 --> 00:04:01.000
Well, let's draw a picture of
it first.
00:03:59.000 --> 00:04:05.000
It starts here.
It is zero for t,
00:04:02.000 --> 00:04:08.000
let's say the force is applied
between time zero to time h.
00:04:08.000 --> 00:04:14.000
And then its force is turned
on, it stays constant and then
00:04:14.000 --> 00:04:20.000
it is turned off.
And those vertical lines
00:04:18.000 --> 00:04:24.000
shouldn't be there.
But, since in practice,
00:04:23.000 --> 00:04:29.000
it takes a tiny bit of time to
turn a force on and off.
00:04:30.000 --> 00:04:36.000
It is, in practice,
not unrealistic to suppose that
00:04:34.000 --> 00:04:40.000
there are approximately vertical
lines there.
00:04:38.000 --> 00:04:44.000
They are slightly slanted but
not too much.
00:04:41.000 --> 00:04:47.000
Now, I want it to be unit
impulse.
00:04:44.000 --> 00:04:50.000
This is the force access and
this is the time access.
00:04:49.000 --> 00:04:55.000
Since the impulse is the area
under this curve,
00:04:53.000 --> 00:04:59.000
if I want that to be one,
then if this is h,
00:04:56.000 --> 00:05:02.000
the height to which I --
In other words,
00:05:01.000 --> 00:05:07.000
the magnitude of the force must
be one over h in order
00:05:05.000 --> 00:05:11.000
that the area be one,
in order, in other words,
00:05:09.000 --> 00:05:15.000
that this integral be one,
the area under that curve be
00:05:13.000 --> 00:05:19.000
one.
So the unit impulse looks like
00:05:15.000 --> 00:05:21.000
that.
The narrower it is here,
00:05:17.000 --> 00:05:23.000
the higher it has to be that
way.
00:05:20.000 --> 00:05:26.000
The bigger the force must be if
you want the end result to be a
00:05:24.000 --> 00:05:30.000
unit impulse.
Now, to solve a problem,
00:05:27.000 --> 00:05:33.000
a typical problem,
then, would be a spring.
00:05:32.000 --> 00:05:38.000
The mass is traveling on the
track.
00:05:34.000 --> 00:05:40.000
Let's suppose the spring
constant is one,
00:05:38.000 --> 00:05:44.000
so there would be a
differential equation.
00:05:41.000 --> 00:05:47.000
And the right-hand side would
be this f of t.
00:05:45.000 --> 00:05:51.000
Well, let's give it its name,
the name I gave it before.
00:05:49.000 --> 00:05:55.000
Remember, I called the unit box
function the thing which was one
00:05:55.000 --> 00:06:01.000
between zero and h and zero
everywhere else.
00:06:00.000 --> 00:06:06.000
The notation we used for that
was u, and then it had a double
00:06:04.000 --> 00:06:10.000
subscript from the starting
point and the finishing point.
00:06:07.000 --> 00:06:13.000
So oh-- u(oh) of t.
00:06:10.000 --> 00:06:16.000
This much represents the thing
if it only rose to the high one.
00:06:14.000 --> 00:06:20.000
But if it, instead,
rises to the height one over h
00:06:17.000 --> 00:06:23.000
in order to make that
area one, I have to multiply it
00:06:21.000 --> 00:06:27.000
by the factor one over h.
Now, if you want to solve this
00:06:25.000 --> 00:06:31.000
by the Laplace transform.
In other words,
00:06:28.000 --> 00:06:34.000
see what the motion of that
mass is as I apply this unit
00:06:32.000 --> 00:06:38.000
impulse to it over that time
interval.
00:06:34.000 --> 00:06:40.000
You have to take the Laplace
transform, if that is the way we
00:06:38.000 --> 00:06:44.000
are doing it.
Now, the left-hand side is just
00:06:41.000 --> 00:06:47.000
routine and would involve the
initial conditions.
00:06:44.000 --> 00:06:50.000
The whole interest is taking
the Laplace transform of the
00:06:48.000 --> 00:06:54.000
right-hand side.
And that is what I want to do
00:06:51.000 --> 00:06:57.000
now.
The problem is what is the
00:06:52.000 --> 00:06:58.000
Laplace transform of this guy?
00:07:02.000 --> 00:07:08.000
Well, remember,
to do everything else,
00:07:04.000 --> 00:07:10.000
you do everything by writing in
terms of the unit step function?
00:07:08.000 --> 00:07:14.000
This function that we are
talking about is one over h
00:07:12.000 --> 00:07:18.000
times what you get by
first stepping up to one.
00:07:16.000 --> 00:07:22.000
That is the unit step function,
which goes up by one and tries
00:07:20.000 --> 00:07:26.000
to stay at one ever after.
And then, when it gets to h,
00:07:24.000 --> 00:07:30.000
it has got to step down.
Well, the way you make it step
00:07:27.000 --> 00:07:33.000
down is by subtracting off the
function, which is the unit step
00:07:31.000 --> 00:07:37.000
function but where the step
takes place, not at time zero
00:07:35.000 --> 00:07:41.000
but at time h.
In other words,
00:07:38.000 --> 00:07:44.000
I translate the unit step
function of course with,
00:07:42.000 --> 00:07:48.000
I don't think I have to draw
that picture again.
00:07:45.000 --> 00:07:51.000
The unit step function looks
like zing.
00:07:47.000 --> 00:07:53.000
And if you translate it to the
right by h it looks like zing.
00:07:51.000 --> 00:07:57.000
And then make it negative to
subtract it off.
00:07:54.000 --> 00:08:00.000
And what you will get is this
box function.
00:07:56.000 --> 00:08:02.000
So we want to take the Laplace
transform of this thing.
00:08:01.000 --> 00:08:07.000
Well, let's assume,
for the sake of argument that
00:08:05.000 --> 00:08:11.000
you didn't remember.
Well, you had to use the
00:08:08.000 --> 00:08:14.000
formula at 2:00 AM this morning
and, therefore,
00:08:12.000 --> 00:08:18.000
you do remember it.
[LAUGHTER] So I don't have to
00:08:16.000 --> 00:08:22.000
recopy the formula onto the
board.
00:08:19.000 --> 00:08:25.000
Maybe if there is room there.
All right, let's put it up
00:08:24.000 --> 00:08:30.000
there.
It says that u of t minus a
00:08:26.000 --> 00:08:32.000
times f,
any f, so let's call it g so
00:08:31.000 --> 00:08:37.000
you won't confuse it with this
particular one,
00:08:34.000 --> 00:08:40.000
times g translated.
If you translate a function
00:08:39.000 --> 00:08:45.000
from t, if you translate it to
the right by a then its Laplace
00:08:44.000 --> 00:08:50.000
transform is e to the minus a s
times whatever the
00:08:48.000 --> 00:08:54.000
old Laplace transform was,
g of s.
00:08:51.000 --> 00:08:57.000
Multiply by an exponential on
the right.
00:08:54.000 --> 00:09:00.000
On the left that corresponds to
translation.
00:08:58.000 --> 00:09:04.000
Except you must remember to put
in that factor u for a secret
00:09:02.000 --> 00:09:08.000
reason which I spent half of
Wednesday explaining.
00:09:05.000 --> 00:09:11.000
What do we have here? The
Laplace transform of u of t,
00:09:09.000 --> 00:09:15.000
that is easy.
That is simply one over s.
00:09:13.000 --> 00:09:19.000
The Laplace transform of this
00:09:15.000 --> 00:09:21.000
other guy we get from the
formula.
00:09:17.000 --> 00:09:23.000
It is basically one over s.
No, the Laplace transform of u
00:09:21.000 --> 00:09:27.000
of t.
But because it has been
00:09:24.000 --> 00:09:30.000
translated to the right by h,
I have to multiply it by that
00:09:28.000 --> 00:09:34.000
factor e to the minus
h times s.
00:09:38.000 --> 00:09:44.000
That is the answer.
And, if you want to solve
00:09:40.000 --> 00:09:46.000
problems, this is what you would
feed into the equation.
00:09:44.000 --> 00:09:50.000
And you would calculate and
calculate and calculate it.
00:09:47.000 --> 00:09:53.000
But that is not what I want to
do now because that was
00:09:51.000 --> 00:09:57.000
Wednesday and this is Friday.
You have the right to expect
00:09:55.000 --> 00:10:01.000
something new.
Here is what I am going to do
00:09:57.000 --> 00:10:03.000
new.
I am going to let h go to zero.
00:10:01.000 --> 00:10:07.000
As h goes to zero,
this function gets narrower and
00:10:06.000 --> 00:10:12.000
narrower, but it also has to get
higher and higher because its
00:10:12.000 --> 00:10:18.000
area has to stay one.
What I am interested in,
00:10:16.000 --> 00:10:22.000
first of all,
is what happens to the Laplace
00:10:20.000 --> 00:10:26.000
transform as h goes to zero.
In other words,
00:10:25.000 --> 00:10:31.000
what is the limit,
as h goes to zero of --
00:10:30.000 --> 00:10:36.000
Well, what is that function?
One minus e to the negative hs
00:10:34.000 --> 00:10:40.000
divided by hs.
00:10:37.000 --> 00:10:43.000
Well, this is an 18.01 problem,
an ordinary calculus problem,
00:10:42.000 --> 00:10:48.000
but let's do it nicely.
You see, the nice way to do it
00:10:46.000 --> 00:10:52.000
is to make a substitution.
We will change h s to u
00:10:50.000 --> 00:10:56.000
because it is occurring
as a unit in both cases.
00:10:55.000 --> 00:11:01.000
This is going to be the same as
the limit as u goes to zero.
00:11:00.000 --> 00:11:06.000
I think there are too many u's
00:11:04.000 --> 00:11:10.000
here already.
I cannot use u,
00:11:07.000 --> 00:11:13.000
you cannot use t,
v is velocity,
00:11:10.000 --> 00:11:16.000
w is wavefunction.
There is no letter.
00:11:13.000 --> 00:11:19.000
All right, u.
It is one minus e to the
00:11:17.000 --> 00:11:23.000
negative u over u.
00:11:20.000 --> 00:11:26.000
So what is the answer?
Well, either you know the
00:11:25.000 --> 00:11:31.000
answer or you replace this by,
say, the first couple of terms
00:11:30.000 --> 00:11:36.000
of the Taylor series.
But I think most of you would
00:11:35.000 --> 00:11:41.000
use L'Hopital's rule,
so let's do that.
00:11:38.000 --> 00:11:44.000
The derivative of the top is
zero here.
00:11:40.000 --> 00:11:46.000
The derivative by the chain
rule of e to the negative u is e
00:11:44.000 --> 00:11:50.000
to the negative u times minus
one.
00:11:47.000 --> 00:11:53.000
And that minus one cancels that
minus.
00:11:49.000 --> 00:11:55.000
So the derivative of the top is
simply e to the negative u and
00:11:53.000 --> 00:11:59.000
the derivative of the bottom is
one.
00:11:55.000 --> 00:12:01.000
So, as u goes to zero,
that limit is one.
00:12:00.000 --> 00:12:06.000
Interesting.
Let's draw a picture this way.
00:12:03.000 --> 00:12:09.000
I will draw it schematically.
Up here is the function one
00:12:08.000 --> 00:12:14.000
over h times u zero h of t,
our box function,
00:12:14.000 --> 00:12:20.000
except it has the height one
over h instead of the
00:12:19.000 --> 00:12:25.000
height one.
We have just calculated that
00:12:23.000 --> 00:12:29.000
its Laplace transform is that
funny thing, one minus e to the
00:12:28.000 --> 00:12:34.000
minus hs divided by hs.
00:12:34.000 --> 00:12:40.000
That is the top line.
All this is completely kosher,
00:12:39.000 --> 00:12:45.000
but now I am going to let h go
to zero.
00:12:43.000 --> 00:12:49.000
And the question is what do we
get now?
00:12:47.000 --> 00:12:53.000
Well, I just calculated for you
that this thing approaches one,
00:12:53.000 --> 00:12:59.000
has the limit one.
And now, let's fill in the
00:12:58.000 --> 00:13:04.000
picture.
What does this thing approach?
00:13:03.000 --> 00:13:09.000
Well, it approaches a function
which is zero everywhere.
00:13:10.000 --> 00:13:16.000
As h approaches zero,
this green box turns into a box
00:13:17.000 --> 00:13:23.000
which is zero everywhere except
at zero.
00:13:22.000 --> 00:13:28.000
And there, it is infinitely
high.
00:13:26.000 --> 00:13:32.000
So, keep going up.
00:13:35.000 --> 00:13:41.000
Now, of course,
that is not a function.
00:13:38.000 --> 00:13:44.000
People call it a function but
it isn't.
00:13:41.000 --> 00:13:47.000
Mathematicians call it a
generalized function,
00:13:45.000 --> 00:13:51.000
but that is not a function
either.
00:13:48.000 --> 00:13:54.000
It is just a way of making you
feel comfortable by talking
00:13:53.000 --> 00:13:59.000
about something which isn't
really a function.
00:13:56.000 --> 00:14:02.000
It was given the name,
introduced formally into
00:14:00.000 --> 00:14:06.000
mathematics by a physicist,
Dirac.
00:14:05.000 --> 00:14:11.000
And he, looking ahead to the
future, did what many people do
00:14:09.000 --> 00:14:15.000
who introduce something into the
literature, a formula or a
00:14:13.000 --> 00:14:19.000
function or something which they
think is going to be important.
00:14:17.000 --> 00:14:23.000
They never name it directly
after themselves,
00:14:20.000 --> 00:14:26.000
but they always use as the
symbol for it the first letter
00:14:24.000 --> 00:14:30.000
of their name.
I cannot tell you how often
00:14:26.000 --> 00:14:32.000
that has happened.
Maybe even Euler called e for
00:14:31.000 --> 00:14:37.000
that reason, although he claims
it was in Latin because it has
00:14:37.000 --> 00:14:43.000
to do with exponentials.
Well, luckily his name began
00:14:42.000 --> 00:14:48.000
with an E, too.
That is Paul Dirac's delta
00:14:45.000 --> 00:14:51.000
function.
I won't dignify it by the name
00:14:49.000 --> 00:14:55.000
function by writing that out,
by putting the world function
00:14:54.000 --> 00:15:00.000
here, too, but it is called the
delta function.
00:15:00.000 --> 00:15:06.000
From this point on,
the entire rest of the lecture
00:15:03.000 --> 00:15:09.000
has a slight fictional element.
The entire rest of the lecture
00:15:08.000 --> 00:15:14.000
is in figurative quotation
marks, so you are not entirely
00:15:12.000 --> 00:15:18.000
responsible for anything I say.
This is a non-function,
00:15:16.000 --> 00:15:22.000
but you put it in there and
call it a function.
00:15:19.000 --> 00:15:25.000
And you naturally want to
complete, if it's a function
00:15:23.000 --> 00:15:29.000
then it must have a Laplace
transform, even though it
00:15:27.000 --> 00:15:33.000
doesn't, so the diagram is
completed that way.
00:15:32.000 --> 00:15:38.000
And its Laplace transform is
declared to be one.
00:15:35.000 --> 00:15:41.000
So let's start listing the
properties of this weird thing.
00:15:54.000 --> 00:16:00.000
The delta function,
its Laplace transform is one.
00:16:05.000 --> 00:16:11.000
Now, one of the things is we
have not yet expressed the fact
00:16:09.000 --> 00:16:15.000
that it is a unit impulse.
In other words,
00:16:13.000 --> 00:16:19.000
since the areas of all of these
boxes, they all have areas one
00:16:17.000 --> 00:16:23.000
as they are shrunk this way they
get higher that way.
00:16:22.000 --> 00:16:28.000
By convention,
one says that the area under
00:16:25.000 --> 00:16:31.000
the orange curve also remains
one in the limit.
00:16:30.000 --> 00:16:36.000
Now, how am I going to express
that?
00:16:32.000 --> 00:16:38.000
Well, it is done by the
following formula that the
00:16:35.000 --> 00:16:41.000
integral, the total impulse of
the delta function should be
00:16:39.000 --> 00:16:45.000
one.
Now, where do I integrate?
00:16:41.000 --> 00:16:47.000
Well, from any place that it is
zero to any place that it is
00:16:45.000 --> 00:16:51.000
zero on the other side of that
vertical line.
00:16:48.000 --> 00:16:54.000
But, in order to avoid
controversy, people integrate
00:16:52.000 --> 00:16:58.000
all the way from negative
infinity to infinity since it
00:16:55.000 --> 00:17:01.000
doesn't hurt.
Does it?
00:16:57.000 --> 00:17:03.000
It is zero practically all the
time.
00:17:01.000 --> 00:17:07.000
This is the function whose
Laplace transform is one.
00:17:06.000 --> 00:17:12.000
Its integral from minus
infinity to infinity is one.
00:17:11.000 --> 00:17:17.000
How else can we calculate for
it?
00:17:15.000 --> 00:17:21.000
Well, I would like to calculate
its convolution.
00:17:20.000 --> 00:17:26.000
Here is f of t.
What happens if I convolute it
00:17:26.000 --> 00:17:32.000
with the delta function?
Well, if you go back to the
00:17:31.000 --> 00:17:37.000
definition of the convolution,
you know, it is that funny
00:17:35.000 --> 00:17:41.000
integral, you are going to do a
lot of head scratching because
00:17:40.000 --> 00:17:46.000
it is not really all that clear
how to integrate with the delta
00:17:44.000 --> 00:17:50.000
function.
Instead of doing that let's
00:17:46.000 --> 00:17:52.000
assume that it follows the laws
of the Laplace transform.
00:17:50.000 --> 00:17:56.000
In that case,
its Laplace transform would be
00:17:53.000 --> 00:17:59.000
what?
Well, the whole thing of a
00:17:55.000 --> 00:18:01.000
convolution is that the Laplace
transform of the convolution is
00:18:00.000 --> 00:18:06.000
the product of the two separate
Laplace transforms.
00:18:05.000 --> 00:18:11.000
So that is going to be F of s
times the Laplace
00:18:09.000 --> 00:18:15.000
transform of the delta function,
which is one.
00:18:13.000 --> 00:18:19.000
Now, what must this thing be?
Well, there is some ambiguity
00:18:18.000 --> 00:18:24.000
as to what it is for negative
values of t.
00:18:21.000 --> 00:18:27.000
But if we, by brute force,
decide for negative values of t
00:18:26.000 --> 00:18:32.000
it is going to have the value
zero, that is the way we make
00:18:31.000 --> 00:18:37.000
things unique.
In fact, why don't we make f of
00:18:35.000 --> 00:18:41.000
unique that way to start with?
00:18:38.000 --> 00:18:44.000
This is a function now that is
allowed to do anything it wants
00:18:41.000 --> 00:18:47.000
on the right-hand side of zero
starting at zero,
00:18:44.000 --> 00:18:50.000
but on the left-hand side of
zero it is wiped away and must
00:18:48.000 --> 00:18:54.000
be zero.
This is a definite thing now.
00:18:50.000 --> 00:18:56.000
Its convolution is this.
And the inverse Laplace
00:18:53.000 --> 00:18:59.000
transform is --
The answer, in other words,
00:18:57.000 --> 00:19:03.000
is the same thing as what u of
t f of t would be.
00:19:02.000 --> 00:19:08.000
It's the same thing,
F of s.
00:19:04.000 --> 00:19:10.000
And so, the conclusion is that
these are equal,
00:19:08.000 --> 00:19:14.000
since they must be unique.
They have been made unique by
00:19:12.000 --> 00:19:18.000
making them zero for t negative.
In other words,
00:19:16.000 --> 00:19:22.000
apply to a function,
well, I won't recopy it.
00:19:19.000 --> 00:19:25.000
But the point is that delta t,
for the convolution operation,
00:19:24.000 --> 00:19:30.000
is acting like an identity.
If I multiply,
00:19:29.000 --> 00:19:35.000
in the sense of convolution,
it is a peculiar operation.
00:19:33.000 --> 00:19:39.000
But algebraically,
it has a lot of the properties
00:19:36.000 --> 00:19:42.000
of multiplication.
It is communitive.
00:19:39.000 --> 00:19:45.000
It is linear in both factors.
In other words,
00:19:42.000 --> 00:19:48.000
it is almost anything you would
want with multiplication.
00:19:46.000 --> 00:19:52.000
It has an identity element,
identity function.
00:19:49.000 --> 00:19:55.000
And the identity function is
the Dirac delta function.
00:19:53.000 --> 00:19:59.000
Anything else here?
Yeah, I will throw in one more
00:19:57.000 --> 00:20:03.000
thing.
It would just require one more
00:20:01.000 --> 00:20:07.000
phony argument,
which I won't bother giving
00:20:04.000 --> 00:20:10.000
you, but it is not totally
implausible.
00:20:06.000 --> 00:20:12.000
After all, u of t,
the unit step function is not
00:20:11.000 --> 00:20:17.000
differentiable,
is not a differentiable
00:20:13.000 --> 00:20:19.000
function.
It looks like this.
00:20:15.000 --> 00:20:21.000
Here its derivative is zero,
here its derivative is zero,
00:20:19.000 --> 00:20:25.000
and in this class it is not
even defined in between.
00:20:23.000 --> 00:20:29.000
But, I don't care,
I will make it go straight up.
00:20:27.000 --> 00:20:33.000
The question is what's its
derivative?
00:20:31.000 --> 00:20:37.000
Well, zero here,
zero there and infinity at
00:20:34.000 --> 00:20:40.000
zero, so it must be the delta
function.
00:20:37.000 --> 00:20:43.000
That has exactly the right
properties.
00:20:40.000 --> 00:20:46.000
So the same people who will
tell you this will tell you that
00:20:44.000 --> 00:20:50.000
also.
And, in fact,
00:20:45.000 --> 00:20:51.000
when you use it to solve
differential equations it acts
00:20:50.000 --> 00:20:56.000
as if that is true.
I think I have given you an
00:20:53.000 --> 00:20:59.000
example on your homework.
Let me now show you a typical
00:20:57.000 --> 00:21:03.000
example of the way the Dirac
delta function would be used to
00:21:02.000 --> 00:21:08.000
solve a problem.
Let's go back to our little
00:21:06.000 --> 00:21:12.000
spring, since it is the easiest
thing.
00:21:09.000 --> 00:21:15.000
You are familiar with it from a
physical point of view,
00:21:13.000 --> 00:21:19.000
and it is the easiest thing to
illustrate on.
00:21:16.000 --> 00:21:22.000
We have our spring mass system.
Where is it?
00:21:20.000 --> 00:21:26.000
Is it on the board?
Up there.
00:21:22.000 --> 00:21:28.000
That one.
And the differential equation
00:21:25.000 --> 00:21:31.000
we are going to solve is y
double prime plus y
00:21:29.000 --> 00:21:35.000
equals --
And now, I am going to assume
00:21:33.000 --> 00:21:39.000
that the spring is kicked with
impulse a.
00:21:37.000 --> 00:21:43.000
I am not going to kick it at
time t equals zero,
00:21:42.000 --> 00:21:48.000
since that would get us into
slight technical difficulties.
00:21:47.000 --> 00:21:53.000
Anyway, it is more fun to kick
it at time pi over two.
00:21:52.000 --> 00:21:58.000
The thing is,
00:21:54.000 --> 00:22:00.000
what is happening?
Well, we have got to have
00:21:58.000 --> 00:22:04.000
initial conditions.
The initial conditions are
00:22:02.000 --> 00:22:08.000
going to be, let's start at time
zero.
00:22:05.000 --> 00:22:11.000
We will start it at the
position one.
00:22:08.000 --> 00:22:14.000
So I take my spring,
I drag it to the position one,
00:22:11.000 --> 00:22:17.000
I take the little mass there
and then let it go.
00:22:15.000 --> 00:22:21.000
And so it starts going birr.
But right when it gets to the
00:22:19.000 --> 00:22:25.000
equilibrium point I give it a,
"cha!" with unit impulse.
00:22:23.000 --> 00:22:29.000
I started it from rest.
Those will be the initial
00:22:27.000 --> 00:22:33.000
conditions.
And I want to say that I kicked
00:22:31.000 --> 00:22:37.000
it, not with unit impulse,
but with the impulse a.
00:22:35.000 --> 00:22:41.000
Bigger.
And I did that at time pi over
00:22:38.000 --> 00:22:44.000
two.
So how are we going to say
00:22:41.000 --> 00:22:47.000
that?
Well, kick it means delivered
00:22:43.000 --> 00:22:49.000
that impulse over an extremely
short time interval,
00:22:47.000 --> 00:22:53.000
but in such a way kicked it
sufficiently hard that the total
00:22:52.000 --> 00:22:58.000
impulse was a.
The way to say that is kick it
00:22:56.000 --> 00:23:02.000
with the Dirac delta function.
Translate it to the point time
00:23:02.000 --> 00:23:08.000
pi over two.
Not at zero any longer.
00:23:06.000 --> 00:23:12.000
t minus pi over two.
00:23:09.000 --> 00:23:15.000
But that would kick it with a
unit impulse.
00:23:12.000 --> 00:23:18.000
I want it to kick it with the
impulse a, so I will just
00:23:16.000 --> 00:23:22.000
multiply that by the constant
factor a.
00:23:20.000 --> 00:23:26.000
Let's put this over here.
y of zero equals one,
00:23:24.000 --> 00:23:30.000
that's the starting value.
Now we have a problem.
00:23:30.000 --> 00:23:36.000
The only thing new in solving
this with the Laplace transform
00:23:33.000 --> 00:23:39.000
is I have this funny right-hand
side.
00:23:36.000 --> 00:23:42.000
But it corresponds to a
physical situation.
00:23:38.000 --> 00:23:44.000
Let's do it.
You take the Laplace transform
00:23:41.000 --> 00:23:47.000
of both sides of the equation.
Remember how to do that?
00:23:44.000 --> 00:23:50.000
You have to take account of the
initial conditions.
00:23:48.000 --> 00:23:54.000
The Laplace transform of the
second derivative is you
00:23:51.000 --> 00:23:57.000
multiply by s squared,
and then you have to subtract.
00:23:55.000 --> 00:24:01.000
You have to use these initial
conditions.
00:23:59.000 --> 00:24:05.000
This one won't give you
anything, but the first one
00:24:02.000 --> 00:24:08.000
means I have to subtract one
times s.
00:24:05.000 --> 00:24:11.000
That is the Laplace transform
of y double prime.
00:24:09.000 --> 00:24:15.000
The Laplace transform of y,
of course, is just capital Y.
00:24:13.000 --> 00:24:19.000
And how about the Laplace
00:24:16.000 --> 00:24:22.000
transform of the right-hand
side.
00:24:18.000 --> 00:24:24.000
Well, we will have the constant
factor a because the Laplace
00:24:22.000 --> 00:24:28.000
transform is linear.
And now, the delta function
00:24:25.000 --> 00:24:31.000
would have the transform one.
But when I translate it,
00:24:30.000 --> 00:24:36.000
pi over two,
that means I have to use that
00:24:33.000 --> 00:24:39.000
formula.
Translate it by pi over two
00:24:35.000 --> 00:24:41.000
means take the one that it would
have been otherwise and multiply
00:24:40.000 --> 00:24:46.000
it by e, that exponential
factor.
00:24:42.000 --> 00:24:48.000
It would be e to the minus pi
over two,
00:24:46.000 --> 00:24:52.000
that is the A times s times
one, which would be the g of s,
00:24:49.000 --> 00:24:55.000
the Laplace transform
or the delta function before it
00:24:54.000 --> 00:25:00.000
had been translated.
But I don't have to put that in
00:24:57.000 --> 00:25:03.000
because it's one.
I am multiplying by one.
00:25:01.000 --> 00:25:07.000
And to do everything now is
routine.
00:25:03.000 --> 00:25:09.000
Solve for the Laplace
transform.
00:25:05.000 --> 00:25:11.000
Well, what is it?
It is y is equal to.
00:25:08.000 --> 00:25:14.000
I put the s on the other side.
That makes the right-hand side
00:25:12.000 --> 00:25:18.000
the sum of two terms.
And I divide by the coefficient
00:25:15.000 --> 00:25:21.000
of y, which is s squared plus
one.
00:25:18.000 --> 00:25:24.000
The s is over on the right-hand
side and it is divided by s
00:25:22.000 --> 00:25:28.000
squared plus one.
And the other factor is there,
00:25:25.000 --> 00:25:31.000
too.
And it, too,
00:25:26.000 --> 00:25:32.000
is divided by s squared plus
one.
00:25:35.000 --> 00:25:41.000
Now, we take the inverse
Laplace transform of those two
00:25:38.000 --> 00:25:44.000
terms and add them up.
00:26:00.000 --> 00:26:06.000
What will we get?
Well, y is equal to,
00:26:02.000 --> 00:26:08.000
the inverse Laplace transform
of s over s squared plus one is
00:26:07.000 --> 00:26:13.000
cosine t.
00:26:11.000 --> 00:26:17.000
Now, for this thing we will
have to use our formula.
00:26:15.000 --> 00:26:21.000
If this weren't here,
the inverse Laplace transform
00:26:19.000 --> 00:26:25.000
of a over s squared plus one
would
00:26:24.000 --> 00:26:30.000
be what?
Well, it would be a times the
00:26:27.000 --> 00:26:33.000
sine of t.
00:26:35.000 --> 00:26:41.000
In other words,
if this is the g of s
00:26:37.000 --> 00:26:43.000
then the function on the left
would be basically A sine t.
00:26:41.000 --> 00:26:47.000
But because it has been
00:26:43.000 --> 00:26:49.000
multiplied by that exponential
factor, e to the minus as
00:26:47.000 --> 00:26:53.000
where a is pi over two,
00:26:50.000 --> 00:26:56.000
the left-hand side has to be
changed from A sine t
00:26:53.000 --> 00:26:59.000
to what it would be
with the translated form.
00:26:58.000 --> 00:27:04.000
So the rest of it is u of t
minus pi over two,
00:27:01.000 --> 00:27:07.000
because a is pi over
two, times what it would have
00:27:05.000 --> 00:27:11.000
been just from the factor g of s
itself.
00:27:09.000 --> 00:27:15.000
In other words,
A times the sine of,
00:27:11.000 --> 00:27:17.000
again, t minus pi over two.
00:27:15.000 --> 00:27:21.000
I am applying that formula,
but I am applying it in that
00:27:19.000 --> 00:27:25.000
direction.
I started with this,
00:27:21.000 --> 00:27:27.000
and I want to recover the
left-hand side.
00:27:24.000 --> 00:27:30.000
And that is what it must look
like.
00:27:26.000 --> 00:27:32.000
The A, of course,
just gets dragged along for the
00:27:29.000 --> 00:27:35.000
free ride.
Now, as I emphasized to you
00:27:34.000 --> 00:27:40.000
last time, and I hope you did on
your homework that you handed
00:27:38.000 --> 00:27:44.000
in, you mustn't leave it in that
form.
00:27:41.000 --> 00:27:47.000
You have to make cases because
people will expect you to tell
00:27:46.000 --> 00:27:52.000
them what the meaning of this
is.
00:27:49.000 --> 00:27:55.000
Now, if t is less than pi over
two, this is zero.
00:27:52.000 --> 00:27:58.000
And, therefore,
that term does not exist.
00:27:56.000 --> 00:28:02.000
So the first part of it is just
the cosine t term if
00:28:00.000 --> 00:28:06.000
t lies between zero and pi over
two.
00:28:05.000 --> 00:28:11.000
If t is bigger than pi over two
then this factor is
00:28:10.000 --> 00:28:16.000
one.
It's the unit step function.
00:28:12.000 --> 00:28:18.000
And I, therefore,
must add in this term.
00:28:16.000 --> 00:28:22.000
Now, what is that term?
What is the sine of t minus pi
00:28:20.000 --> 00:28:26.000
over two?
The sine of t looks
00:28:25.000 --> 00:28:31.000
like that.
The sine of t,
00:28:27.000 --> 00:28:33.000
if I translate it,
looks like this.
00:28:31.000 --> 00:28:37.000
If I translate it by pi over
two.
00:28:33.000 --> 00:28:39.000
And let's finish it up,
the pi that was over here moved
00:28:38.000 --> 00:28:44.000
into position.
That curve is the curve
00:28:41.000 --> 00:28:47.000
negative cosine t.
00:28:51.000 --> 00:28:57.000
And so the answer is if t is
bigger than pi over two,
00:28:55.000 --> 00:29:01.000
it is cosine t minus A
times cosine t.
00:29:00.000 --> 00:29:06.000
Or, in other words,
00:29:02.000 --> 00:29:08.000
it is one minus A times cosine
t.
00:29:11.000 --> 00:29:17.000
Now, do those match up?
They have always got to match
00:29:13.000 --> 00:29:19.000
up, or you have made a mistake.
You always have to get a
00:29:17.000 --> 00:29:23.000
continuous function when you
have just discontinuities.
00:29:20.000 --> 00:29:26.000
Do we get a continuous
function?
00:29:22.000 --> 00:29:28.000
Yeah, when t is pi over two
the value here is
00:29:25.000 --> 00:29:31.000
zero.
The value of this is also zero
00:29:27.000 --> 00:29:33.000
at pi over two.
There is no conflict in the
00:29:31.000 --> 00:29:37.000
values.
Values doesn't suddenly jump.
00:29:33.000 --> 00:29:39.000
The function is continuous.
It is not differential but it
00:29:38.000 --> 00:29:44.000
is continuous.
Well, what function does that
00:29:41.000 --> 00:29:47.000
look like?
There are cases.
00:29:43.000 --> 00:29:49.000
It starts out life as the
function cosine t.
00:29:47.000 --> 00:29:53.000
So it gets to here.
And at t equals pi over two,
00:29:51.000 --> 00:29:57.000
the mass gets kicked
and that changes the function.
00:29:55.000 --> 00:30:01.000
Now, what are the values?
Well, if A is bigger than one
00:30:01.000 --> 00:30:07.000
this is a negative
number and it therefore becomes
00:30:07.000 --> 00:30:13.000
the function negative
cosine t.
00:30:11.000 --> 00:30:17.000
Now, negative cosine t looks
like this, the blue guy.
00:30:16.000 --> 00:30:22.000
Negative cosine t is a function
that looks like this.
00:30:21.000 --> 00:30:27.000
So it goes from here,
it reverses direction,
00:30:26.000 --> 00:30:32.000
the mass reverses direction
from what you thought it was
00:30:31.000 --> 00:30:37.000
going to do.
And it does that because A is
00:30:36.000 --> 00:30:42.000
so large that that impulse was
enough to make it reverse
00:30:40.000 --> 00:30:46.000
direction.
Of course it might only do
00:30:42.000 --> 00:30:48.000
this, but this is what will
happen if A is bigger than one.
00:30:47.000 --> 00:30:53.000
This will be A,
which is a lot bigger than one.
00:30:50.000 --> 00:30:56.000
If it's not so much bigger than
one it might look like that.
00:30:55.000 --> 00:31:01.000
So A is just bigger than one.
How's that?
00:30:59.000 --> 00:31:05.000
Well, what if A is less than
one?
00:31:01.000 --> 00:31:07.000
Well, in that case it stays
positive.
00:31:04.000 --> 00:31:10.000
If A is less than one,
this is now still a positive
00:31:07.000 --> 00:31:13.000
number.
And, therefore,
00:31:09.000 --> 00:31:15.000
the cosine continues on its
merry way.
00:31:12.000 --> 00:31:18.000
The only thing is it might be a
little more sluggish or it might
00:31:17.000 --> 00:31:23.000
be very peppy and do that.
Let's just go that far.
00:31:20.000 --> 00:31:26.000
This will be the case A less
than one.
00:31:23.000 --> 00:31:29.000
Well, of course,
the most interesting case is
00:31:26.000 --> 00:31:32.000
what happens if A is exactly
equal to one?
00:31:32.000 --> 00:31:38.000
The porridge is exactly just
right, I think that's the
00:31:37.000 --> 00:31:43.000
phrase.
Too hot.
00:31:38.000 --> 00:31:44.000
Too cold.
Just right.
00:31:40.000 --> 00:31:46.000
When A is equal to one,
it is zero.
00:31:44.000 --> 00:31:50.000
It starts out as cosine t.
00:31:48.000 --> 00:31:54.000
When it gets to t,
it continues on ever after as
00:31:52.000 --> 00:31:58.000
the function zero.
I have a visual aid for the
00:31:57.000 --> 00:32:03.000
only time this term.
It didn't work at all.
00:32:02.000 --> 00:32:08.000
I mean, on the other hand,
the last hour,
00:32:06.000 --> 00:32:12.000
the people who worked it were
not intrinsically baseball
00:32:11.000 --> 00:32:17.000
players, so we will use the
equation of the pendulum
00:32:15.000 --> 00:32:21.000
instead.
That is a lot easier than mass
00:32:18.000 --> 00:32:24.000
spring.
This is a pendulum.
00:32:20.000 --> 00:32:26.000
It is undamped because I
declare it to be and it swings
00:32:25.000 --> 00:32:31.000
back and forth.
And here I am releasing it.
00:32:30.000 --> 00:32:36.000
The variable is not x or y but
theta, the angle through.
00:32:34.000 --> 00:32:40.000
Here theta is one,
let's say.
00:32:37.000 --> 00:32:43.000
That's about one radian.
It starts there and swings back
00:32:41.000 --> 00:32:47.000
and forth.
It is not damped,
00:32:44.000 --> 00:32:50.000
so it never loses amplitude,
particularly if I swish it,
00:32:48.000 --> 00:32:54.000
if I move my hand a little bit.
I want someone who knows how to
00:32:53.000 --> 00:32:59.000
bat a baseball.
That was the problem last hour.
00:32:57.000 --> 00:33:03.000
Two people.
One to release it.
00:33:01.000 --> 00:33:07.000
I will stand up and try to hold
it here.
00:33:04.000 --> 00:33:10.000
Somebody releases it.
And then somebody who has to be
00:33:08.000 --> 00:33:14.000
very skillful should apply a
unit impulse of exactly one when
00:33:13.000 --> 00:33:19.000
it gets to the equilibrium
point.
00:33:16.000 --> 00:33:22.000
So who can do that?
Who can play baseball here?
00:33:20.000 --> 00:33:26.000
Come on.
Somebody elected?
00:33:30.000 --> 00:33:36.000
All right. Come on. [APPLAUSE]
00:33:41.000 --> 00:33:47.000
Somebody release it,
too.
00:33:44.000 --> 00:33:50.000
Somebody tall to handle it all.
I think that will be me.
00:33:52.000 --> 00:33:58.000
Just hold it at what you would
take to be one radian.
00:34:00.000 --> 00:34:06.000
He releases it.
When it gets to the bottom,
00:34:04.000 --> 00:34:10.000
you will have to get way down,
and maybe on this side.
00:34:11.000 --> 00:34:17.000
Are you a lefty or a righty?
Rightly.
00:34:15.000 --> 00:34:21.000
Okay.
Bat it what part.
00:34:17.000 --> 00:34:23.000
Give it a good swat.
I will stand up higher.
00:34:22.000 --> 00:34:28.000
Help.
I'm not very stable.
00:34:25.000 --> 00:34:31.000
[APPLAUSE] A trial run.
Again.
00:34:30.000 --> 00:34:36.000
Okay.
A little further out.
00:34:32.000 --> 00:34:38.000
First of all,
you have to see where it's
00:34:36.000 --> 00:34:42.000
going.
Why don't you stand,
00:34:38.000 --> 00:34:44.000
oh, you bat rightly.
That's right.
00:34:41.000 --> 00:34:47.000
Okay.
Let's try it again.
00:34:59.000 --> 00:35:05.000
Strike one.
It's okay.
It's the beginning of the
baseball season.
One more.
The Red Sox are having trouble,
too.
Not bad. [APPLAUSE]
00:35:13.000 --> 00:35:19.000
If he had hit even harder it
would have reversed direction
00:35:16.000 --> 00:35:22.000
and gone that way.
If you hadn't hit it quite as
00:35:20.000 --> 00:35:26.000
hard it would have continued on,
still at cosine t,
00:35:24.000 --> 00:35:30.000
but with less amplitude.
But if you hit it exactly right
00:35:28.000 --> 00:35:34.000
--
It is fun to try to do.
00:35:31.000 --> 00:35:37.000
Toomre in our department is a
master at this,
00:35:36.000 --> 00:35:42.000
but he has been practicing for
years.
00:35:40.000 --> 00:35:46.000
He can take a little mallet and
go blunk, and it stops
00:35:45.000 --> 00:35:51.000
absolutely dead.
It is unbelievable.
00:35:49.000 --> 00:35:55.000
I should have had him give the
lecture.
00:35:53.000 --> 00:35:59.000
Now, I would like to do
something truly serious.
00:36:00.000 --> 00:36:06.000
Here, I guess.
Because there is a certain
00:36:03.000 --> 00:36:09.000
amount of engineering lingo you
have to learn.
00:36:07.000 --> 00:36:13.000
It is used by almost everybody.
Not architects and biologists
00:36:12.000 --> 00:36:18.000
probably quite yet,
but anybody that uses the
00:36:16.000 --> 00:36:22.000
Laplace transform will use these
words in connection with it.
00:36:21.000 --> 00:36:27.000
I really think,
since it is such a widespread
00:36:25.000 --> 00:36:31.000
technique, that these are things
you should know.
00:36:31.000 --> 00:36:37.000
Anyway, it will be easy.
It is just the enrichment of
00:36:34.000 --> 00:36:40.000
your vocabulary.
It is always fun to learn new
00:36:37.000 --> 00:36:43.000
vocabulary words.
So, let's just consider a
00:36:40.000 --> 00:36:46.000
general second order equation.
By the way, all this applies to
00:36:45.000 --> 00:36:51.000
higher order equations,
too.
00:36:47.000 --> 00:36:53.000
It applies to systems.
The same words are used,
00:36:50.000 --> 00:36:56.000
but let's use something that
you know.
00:36:52.000 --> 00:36:58.000
Here is a system.
It could be a spring mass
00:36:55.000 --> 00:37:01.000
dashpot system.
It could be an RLC circuit.
00:37:00.000 --> 00:37:06.000
Or that pendulum,
a damped pendulum,
00:37:02.000 --> 00:37:08.000
anything that is modeled by
that differential equation with
00:37:06.000 --> 00:37:12.000
constant coefficients,
second-order.
00:37:08.000 --> 00:37:14.000
This is the input.
The input can be any kind of a
00:37:11.000 --> 00:37:17.000
function.
Exponential functions,
00:37:13.000 --> 00:37:19.000
sine, cosine.
It could be a Dirac delta
00:37:16.000 --> 00:37:22.000
function.
It could be a sum of these
00:37:18.000 --> 00:37:24.000
things.
It could be a Fourier series.
00:37:21.000 --> 00:37:27.000
Anything of the sort of stuff
we have been talking about
00:37:24.000 --> 00:37:30.000
throughout the last few weeks.
And let's have simple initial
00:37:30.000 --> 00:37:36.000
conditions so that doesn't louse
things up, the simplest possible
00:37:34.000 --> 00:37:40.000
ones.
The mass starts at the
00:37:36.000 --> 00:37:42.000
equilibrium point from rest.
Of course, it doesn't stay that
00:37:40.000 --> 00:37:46.000
way because there is an input
that is asking it to move along.
00:37:45.000 --> 00:37:51.000
Now all I want to do is solve
this in general with a Laplace
00:37:49.000 --> 00:37:55.000
transform.
If I do it in general,
00:37:51.000 --> 00:37:57.000
that is always easier than
doing it in particular since you
00:37:56.000 --> 00:38:02.000
don't ever have to do any
calculations.
00:38:00.000 --> 00:38:06.000
It is s squared Y.
There are no other terms here
00:38:05.000 --> 00:38:11.000
because the initial conditions
are zero.
00:38:08.000 --> 00:38:14.000
This part will be a times s Y.
00:38:12.000 --> 00:38:18.000
Again, no other terms because
the initial conditions are zero.
00:38:17.000 --> 00:38:23.000
Plus b times Y.
And all that is equal to
00:38:21.000 --> 00:38:27.000
whatever the Laplace transform
is of the right-hand side.
00:38:26.000 --> 00:38:32.000
So it is F of s.
Next step.
00:38:31.000 --> 00:38:37.000
Boy, this is an easy problem.
You solve for Y.
00:38:35.000 --> 00:38:41.000
Well, Y is F of s times one
over s squared plus as plus b.
00:38:45.000 --> 00:38:51.000
Now, what is that?
The next step now is to figure
00:38:50.000 --> 00:38:56.000
out what the answer to the
problem is, what's the Y of t?
00:38:55.000 --> 00:39:01.000
Well, you do that by taking the
00:39:00.000 --> 00:39:06.000
inverse Laplace transform.
But because these are general
00:39:04.000 --> 00:39:10.000
functions, I don't have to write
down any specific answer.
00:39:09.000 --> 00:39:15.000
The only thing is to use the
convolution because this is the
00:39:13.000 --> 00:39:19.000
product of two functions of s.
The inverse transform will be
00:39:18.000 --> 00:39:24.000
the convolution of their
respective things.
00:39:21.000 --> 00:39:27.000
The answer is going to be the
convolution of F of t,
00:39:26.000 --> 00:39:32.000
the input function in other
words, convoluted with the
00:39:30.000 --> 00:39:36.000
inverse Laplace transform of
that thing.
00:39:35.000 --> 00:39:41.000
Now, we have to have a name for
that, and those are the two
00:39:39.000 --> 00:39:45.000
words I want to introduce you to
because they are used
00:39:43.000 --> 00:39:49.000
everywhere.
The function,
00:39:44.000 --> 00:39:50.000
on the right-hand side,
this function one over s
00:39:48.000 --> 00:39:54.000
squared plus as plus b,
00:39:51.000 --> 00:39:57.000
notice it only depends upon the
left-hand side of the
00:39:55.000 --> 00:40:01.000
differential equation,
on the damping constant.
00:40:00.000 --> 00:40:06.000
The spring constant if you are
thinking of a mass spring
00:40:03.000 --> 00:40:09.000
dashpot system.
So this depends only on the
00:40:06.000 --> 00:40:12.000
system, not on what input is
going into it.
00:40:08.000 --> 00:40:14.000
And it is called the transfer
function.
00:40:11.000 --> 00:40:17.000
Is usually called capital W of,
sometimes it is
00:40:15.000 --> 00:40:21.000
capital H of s,
there are different things,
00:40:18.000 --> 00:40:24.000
but it is always called the
transfer function.
00:40:27.000 --> 00:40:33.000
What we are interested in
putting here its inverse Laplace
00:40:31.000 --> 00:40:37.000
transform.
Well, I will call that W of t
00:40:34.000 --> 00:40:40.000
to go with the capital
W of s by the usual
00:40:39.000 --> 00:40:45.000
notation.
Its inverse Laplace transform,
00:40:42.000 --> 00:40:48.000
well, I cannot calculate that.
I will just give it a name,
00:40:46.000 --> 00:40:52.000
W of t.
And that is called the weight
00:40:49.000 --> 00:40:55.000
function of the system.
This is the transfer function
00:40:53.000 --> 00:40:59.000
of the system,
so put in "of the system" if
00:40:57.000 --> 00:41:03.000
you are taking notes.
And so the answer is that
00:41:02.000 --> 00:41:08.000
always the solution is the
convolution to this differential
00:41:06.000 --> 00:41:12.000
equation that we have been
solving for the last three or
00:41:11.000 --> 00:41:17.000
four weeks.
It is the convolution of that.
00:41:14.000 --> 00:41:20.000
And, therefore,
the solution is expressed as a
00:41:18.000 --> 00:41:24.000
definite integral of the
function of the input on the
00:41:22.000 --> 00:41:28.000
right-hand side,
what is forcing the equation,
00:41:26.000 --> 00:41:32.000
times this magic function but
flipped and translated by t.
00:41:32.000 --> 00:41:38.000
That says du for you guys over
there.
00:41:34.000 --> 00:41:40.000
In other words,
the solution to the
00:41:37.000 --> 00:41:43.000
differential equation is
presented as a definite
00:41:41.000 --> 00:41:47.000
integral.
Marvelous.
00:41:42.000 --> 00:41:48.000
And the only thing is the
definite integral involves this
00:41:47.000 --> 00:41:53.000
funny function W of t.
To understand why that is the
00:41:52.000 --> 00:41:58.000
solution, you have to understand
what W of t is.
00:41:55.000 --> 00:42:01.000
Well, formally,
of course, it's that.
00:42:00.000 --> 00:42:06.000
But what does it really mean?
The problem is what is W of t
00:42:05.000 --> 00:42:11.000
really? Not just formally,
00:42:08.000 --> 00:42:14.000
but what does it really mean?
I mean, is it real?
00:42:12.000 --> 00:42:18.000
I think the simplest way of
thinking of it,
00:42:16.000 --> 00:42:22.000
once you know about the delta
function is just to think of
00:42:21.000 --> 00:42:27.000
this differential equation y
double prime plus a y prime plus
00:42:27.000 --> 00:42:33.000
b. Except use as the input the
00:42:32.000 --> 00:42:38.000
Dirac delta function.
In other words,
00:42:35.000 --> 00:42:41.000
we are kicking the mass.
The mass starts at rest,
00:42:39.000 --> 00:42:45.000
so the initial conditions are
going to be what they were
00:42:43.000 --> 00:42:49.000
before. y of zero,
00:42:46.000 --> 00:42:52.000
y prime of zero. Both zero.
00:42:49.000 --> 00:42:55.000
The mass starts at rest from
the equilibrium position,
00:42:53.000 --> 00:42:59.000
and it is kicked in the
positive direction,
00:42:57.000 --> 00:43:03.000
I guess that's this way,
with unit impulse.
00:43:02.000 --> 00:43:08.000
At time zero with unit impulse.
In other words,
00:43:05.000 --> 00:43:11.000
kick it just hard enough so you
impart a unit impulse.
00:43:10.000 --> 00:43:16.000
So that situation is modeled by
this differential equation.
00:43:15.000 --> 00:43:21.000
The kick at time zero is
modeled by this input,
00:43:19.000 --> 00:43:25.000
the Dirac delta function.
And now, what happens if I
00:43:23.000 --> 00:43:29.000
solve it?
Well, you see,
00:43:25.000 --> 00:43:31.000
everything in the solution is
the same.
00:43:30.000 --> 00:43:36.000
The left stays the same,
but on the right-hand side I
00:43:34.000 --> 00:43:40.000
should have not f of s here.
00:43:37.000 --> 00:43:43.000
Since this is the delta
function, I should have one.
00:43:42.000 --> 00:43:48.000
What I get is,
on the left-hand side,
00:43:45.000 --> 00:43:51.000
s squared Y plus as Y plus bY
equals,
00:43:50.000 --> 00:43:56.000
for the Laplace transform of
the right-hand side is simply
00:43:56.000 --> 00:44:02.000
one.
And, therefore,
00:43:57.000 --> 00:44:03.000
Y is what?
Y is one over exactly the
00:44:02.000 --> 00:44:08.000
transform function.
And therefore its inverse
00:44:05.000 --> 00:44:11.000
Laplace transform is that weight
function.
00:44:09.000 --> 00:44:15.000
That is the simplest
interpretation I know of what
00:44:13.000 --> 00:44:19.000
this magic weight function is,
which gives the solution to all
00:44:18.000 --> 00:44:24.000
the differential equations,
no matter what the input is.
00:44:23.000 --> 00:44:29.000
The weight function is the
response of the system at rest
00:44:27.000 --> 00:44:33.000
to a sharp kick at time zero
with unit impulse.
00:44:33.000 --> 00:44:39.000
And read the notes because they
will explain to you why this
00:44:37.000 --> 00:44:43.000
could be thought of as the
superposition of a lot of sharp
00:44:42.000 --> 00:44:48.000
kicks times zero a little later.
Kick, kick, kick,
00:44:46.000 --> 00:44:52.000
kick.
And that's what makes the
00:44:49.000 --> 00:44:55.000
solution.
Next time we start systems.