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Well, today is the last day on
Laplace transform and the first
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day before we start the rest of
the term, which will be spent on
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the study of systems.
I would like to spend it on one
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more type of input function
which, in general,
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your teachers in other courses
will expect you to have had some
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acquaintance with.
It is the kind associated with
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an impulse, so an input
consisted of what is sometimes
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called a unit impulse.
Now, what's an impulse?
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It covers actually a lot of
things.
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It covers a situation where you
withdraw from a bank account.
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For example,
take half your money out of a
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bank account one day.
It also would be modeled the
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same way.
But the simplest way to
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understand it the first time
through is as an impulse,
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if you know what an impulse is.
If you have a variable force
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acting over time,
and we will assume it is acting
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along a straight line so I don't
have to worry about it being a
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vector, then the impulse,
according to physicists,
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the physical definition,
the impulse of f of t
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over some time interval.
Let's say the time interval
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running from a to b is,
by definition,
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the integral from a to b of f
of t dt.
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Actually, I am going to do the
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most horrible thing this period.
I will assume the force is
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actually a constant force.
So, in that case,
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I wouldn't even have to bother
with the integral at all.
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If f of t is a constant,
let's say capital F,
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then the impulse is --
Well, that integral is simply
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the product of the two,
the impulse over that time
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interval is simply F times b
minus a.
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Just the product of those two.
The force times the length of
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time for which it acts.
Now, that is what I want to
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calculate, want to consider in
connection with our little mass
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system.
So, once again,
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I think this is probably the
last time you'll see the little
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spring.
Let's bid a tearful farewell to
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it.
There is our little mass on
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wheels.
And let's make it an undamped
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mass.
It has an equilibrium point and
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all the other little things that
go with the picture.
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And when I apply an impulse,
what I mean is applying a
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constant force to this over a
definite time interval.
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And that is what I mean by
applying an impulse over that
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time interval.
Now, what is the picture of
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such a thing?
Well, the force is only going
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to be applied,
in other words,
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I am going to push on the mass
or pull on the mass with a
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constant force.
With a little electromagnet
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here, we will assume,
there is a pile of iron filings
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or something inside there.
I turn on the electromagnet.
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It pulls with a constant force
just between time zero and time
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two seconds.
And then I stop.
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That is going to change the
motion of the thing.
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First it is going to start
pulling it toward the thing.
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And then, when it lets go,
it will zoom back and there
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will be a certain motion after
that.
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What the question is,
if I want to solve that problem
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of the motion of that in terms
of the Laplace transform,
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how am I going to model this
force?
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Well, let's draw a picture of
it first.
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It starts here.
It is zero for t,
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let's say the force is applied
between time zero to time h.
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And then its force is turned
on, it stays constant and then
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it is turned off.
And those vertical lines
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shouldn't be there.
But, since in practice,
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it takes a tiny bit of time to
turn a force on and off.
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It is, in practice,
not unrealistic to suppose that
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there are approximately vertical
lines there.
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They are slightly slanted but
not too much.
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Now, I want it to be unit
impulse.
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This is the force access and
this is the time access.
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Since the impulse is the area
under this curve,
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if I want that to be one,
then if this is h,
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the height to which I --
In other words,
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the magnitude of the force must
be one over h in order
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that the area be one,
in order, in other words,
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that this integral be one,
the area under that curve be
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one.
So the unit impulse looks like
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that.
The narrower it is here,
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the higher it has to be that
way.
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The bigger the force must be if
you want the end result to be a
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unit impulse.
Now, to solve a problem,
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a typical problem,
then, would be a spring.
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The mass is traveling on the
track.
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Let's suppose the spring
constant is one,
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so there would be a
differential equation.
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And the right-hand side would
be this f of t.
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Well, let's give it its name,
the name I gave it before.
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Remember, I called the unit box
function the thing which was one
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between zero and h and zero
everywhere else.
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The notation we used for that
was u, and then it had a double
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subscript from the starting
point and the finishing point.
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So oh-- u(oh) of t.
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This much represents the thing
if it only rose to the high one.
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But if it, instead,
rises to the height one over h
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in order to make that
area one, I have to multiply it
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by the factor one over h.
Now, if you want to solve this
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by the Laplace transform.
In other words,
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see what the motion of that
mass is as I apply this unit
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impulse to it over that time
interval.
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You have to take the Laplace
transform, if that is the way we
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are doing it.
Now, the left-hand side is just
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routine and would involve the
initial conditions.
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The whole interest is taking
the Laplace transform of the
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right-hand side.
And that is what I want to do
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now.
The problem is what is the
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Laplace transform of this guy?
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Well, remember,
to do everything else,
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you do everything by writing in
terms of the unit step function?
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This function that we are
talking about is one over h
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times what you get by
first stepping up to one.
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That is the unit step function,
which goes up by one and tries
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to stay at one ever after.
And then, when it gets to h,
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it has got to step down.
Well, the way you make it step
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down is by subtracting off the
function, which is the unit step
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function but where the step
takes place, not at time zero
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but at time h.
In other words,
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I translate the unit step
function of course with,
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I don't think I have to draw
that picture again.
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The unit step function looks
like zing.
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And if you translate it to the
right by h it looks like zing.
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And then make it negative to
subtract it off.
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And what you will get is this
box function.
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So we want to take the Laplace
transform of this thing.
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Well, let's assume,
for the sake of argument that
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you didn't remember.
Well, you had to use the
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formula at 2:00 AM this morning
and, therefore,
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you do remember it.
[LAUGHTER] So I don't have to
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recopy the formula onto the
board.
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Maybe if there is room there.
All right, let's put it up
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there.
It says that u of t minus a
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times f,
any f, so let's call it g so
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you won't confuse it with this
particular one,
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times g translated.
If you translate a function
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from t, if you translate it to
the right by a then its Laplace
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transform is e to the minus a s
times whatever the
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old Laplace transform was,
g of s.
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Multiply by an exponential on
the right.
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On the left that corresponds to
translation.
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Except you must remember to put
in that factor u for a secret
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reason which I spent half of
Wednesday explaining.
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What do we have here? The
Laplace transform of u of t,
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that is easy.
That is simply one over s.
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The Laplace transform of this
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other guy we get from the
formula.
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It is basically one over s.
No, the Laplace transform of u
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of t.
But because it has been
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translated to the right by h,
I have to multiply it by that
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factor e to the minus
h times s.
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That is the answer.
And, if you want to solve
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problems, this is what you would
feed into the equation.
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And you would calculate and
calculate and calculate it.
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But that is not what I want to
do now because that was
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Wednesday and this is Friday.
You have the right to expect
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something new.
Here is what I am going to do
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new.
I am going to let h go to zero.
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As h goes to zero,
this function gets narrower and
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narrower, but it also has to get
higher and higher because its
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area has to stay one.
What I am interested in,
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first of all,
is what happens to the Laplace
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transform as h goes to zero.
In other words,
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what is the limit,
as h goes to zero of --
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Well, what is that function?
One minus e to the negative hs
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divided by hs.
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Well, this is an 18.01 problem,
an ordinary calculus problem,
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but let's do it nicely.
You see, the nice way to do it
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is to make a substitution.
We will change h s to u
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because it is occurring
as a unit in both cases.
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This is going to be the same as
the limit as u goes to zero.
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I think there are too many u's
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here already.
I cannot use u,
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you cannot use t,
v is velocity,
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w is wavefunction.
There is no letter.
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All right, u.
It is one minus e to the
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negative u over u.
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So what is the answer?
Well, either you know the
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answer or you replace this by,
say, the first couple of terms
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of the Taylor series.
But I think most of you would
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use L'Hopital's rule,
so let's do that.
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The derivative of the top is
zero here.
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The derivative by the chain
rule of e to the negative u is e
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to the negative u times minus
one.
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And that minus one cancels that
minus.
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So the derivative of the top is
simply e to the negative u and
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the derivative of the bottom is
one.
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So, as u goes to zero,
that limit is one.
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Interesting.
Let's draw a picture this way.
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I will draw it schematically.
Up here is the function one
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over h times u zero h of t,
our box function,
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except it has the height one
over h instead of the
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height one.
We have just calculated that
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its Laplace transform is that
funny thing, one minus e to the
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minus hs divided by hs.
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That is the top line.
All this is completely kosher,
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but now I am going to let h go
to zero.
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And the question is what do we
get now?
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Well, I just calculated for you
that this thing approaches one,
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has the limit one.
And now, let's fill in the
201
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picture.
What does this thing approach?
202
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Well, it approaches a function
which is zero everywhere.
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As h approaches zero,
this green box turns into a box
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which is zero everywhere except
at zero.
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And there, it is infinitely
high.
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So, keep going up.
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208
00:13:35 --> 00:13:41
Now, of course,
that is not a function.
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People call it a function but
it isn't.
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00:13:41 --> 00:13:47
Mathematicians call it a
generalized function,
211
00:13:45 --> 00:13:51
but that is not a function
either.
212
00:13:48 --> 00:13:54
It is just a way of making you
feel comfortable by talking
213
00:13:53 --> 00:13:59
about something which isn't
really a function.
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It was given the name,
introduced formally into
215
00:14:00 --> 00:14:06
mathematics by a physicist,
Dirac.
216
00:14:05 --> 00:14:11
And he, looking ahead to the
future, did what many people do
217
00:14:09 --> 00:14:15
who introduce something into the
literature, a formula or a
218
00:14:13 --> 00:14:19
function or something which they
think is going to be important.
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00:14:17 --> 00:14:23
They never name it directly
after themselves,
220
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but they always use as the
symbol for it the first letter
221
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of their name.
I cannot tell you how often
222
00:14:26 --> 00:14:32
that has happened.
Maybe even Euler called e for
223
00:14:31 --> 00:14:37
that reason, although he claims
it was in Latin because it has
224
00:14:37 --> 00:14:43
to do with exponentials.
Well, luckily his name began
225
00:14:42 --> 00:14:48
with an E, too.
That is Paul Dirac's delta
226
00:14:45 --> 00:14:51
function.
I won't dignify it by the name
227
00:14:49 --> 00:14:55
function by writing that out,
by putting the world function
228
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here, too, but it is called the
delta function.
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00:15:00 --> 00:15:06
From this point on,
the entire rest of the lecture
230
00:15:03 --> 00:15:09
has a slight fictional element.
The entire rest of the lecture
231
00:15:08 --> 00:15:14
is in figurative quotation
marks, so you are not entirely
232
00:15:12 --> 00:15:18
responsible for anything I say.
This is a non-function,
233
00:15:16 --> 00:15:22
but you put it in there and
call it a function.
234
00:15:19 --> 00:15:25
And you naturally want to
complete, if it's a function
235
00:15:23 --> 00:15:29
then it must have a Laplace
transform, even though it
236
00:15:27 --> 00:15:33
doesn't, so the diagram is
completed that way.
237
00:15:32 --> 00:15:38
And its Laplace transform is
declared to be one.
238
00:15:35 --> 00:15:41
So let's start listing the
properties of this weird thing.
239
00:15:41 --> 00:15:47
240
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The delta function,
its Laplace transform is one.
241
00:16:00 --> 00:16:06
242
00:16:05 --> 00:16:11
Now, one of the things is we
have not yet expressed the fact
243
00:16:09 --> 00:16:15
that it is a unit impulse.
In other words,
244
00:16:13 --> 00:16:19
since the areas of all of these
boxes, they all have areas one
245
00:16:17 --> 00:16:23
as they are shrunk this way they
get higher that way.
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00:16:22 --> 00:16:28
By convention,
one says that the area under
247
00:16:25 --> 00:16:31
the orange curve also remains
one in the limit.
248
00:16:30 --> 00:16:36
Now, how am I going to express
that?
249
00:16:32 --> 00:16:38
Well, it is done by the
following formula that the
250
00:16:35 --> 00:16:41
integral, the total impulse of
the delta function should be
251
00:16:39 --> 00:16:45
one.
Now, where do I integrate?
252
00:16:41 --> 00:16:47
Well, from any place that it is
zero to any place that it is
253
00:16:45 --> 00:16:51
zero on the other side of that
vertical line.
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But, in order to avoid
controversy, people integrate
255
00:16:52 --> 00:16:58
all the way from negative
infinity to infinity since it
256
00:16:55 --> 00:17:01
doesn't hurt.
Does it?
257
00:16:57 --> 00:17:03
It is zero practically all the
time.
258
00:17:01 --> 00:17:07
This is the function whose
Laplace transform is one.
259
00:17:06 --> 00:17:12
Its integral from minus
infinity to infinity is one.
260
00:17:11 --> 00:17:17
How else can we calculate for
it?
261
00:17:15 --> 00:17:21
Well, I would like to calculate
its convolution.
262
00:17:20 --> 00:17:26
Here is f of t.
What happens if I convolute it
263
00:17:26 --> 00:17:32
with the delta function?
Well, if you go back to the
264
00:17:31 --> 00:17:37
definition of the convolution,
you know, it is that funny
265
00:17:35 --> 00:17:41
integral, you are going to do a
lot of head scratching because
266
00:17:40 --> 00:17:46
it is not really all that clear
how to integrate with the delta
267
00:17:44 --> 00:17:50
function.
Instead of doing that let's
268
00:17:46 --> 00:17:52
assume that it follows the laws
of the Laplace transform.
269
00:17:50 --> 00:17:56
In that case,
its Laplace transform would be
270
00:17:53 --> 00:17:59
what?
Well, the whole thing of a
271
00:17:55 --> 00:18:01
convolution is that the Laplace
transform of the convolution is
272
00:18:00 --> 00:18:06
the product of the two separate
Laplace transforms.
273
00:18:05 --> 00:18:11
So that is going to be F of s
times the Laplace
274
00:18:09 --> 00:18:15
transform of the delta function,
which is one.
275
00:18:13 --> 00:18:19
Now, what must this thing be?
Well, there is some ambiguity
276
00:18:18 --> 00:18:24
as to what it is for negative
values of t.
277
00:18:21 --> 00:18:27
But if we, by brute force,
decide for negative values of t
278
00:18:26 --> 00:18:32
it is going to have the value
zero, that is the way we make
279
00:18:31 --> 00:18:37
things unique.
In fact, why don't we make f of
280
00:18:35 --> 00:18:41
unique that way to start with?
281
00:18:38 --> 00:18:44
This is a function now that is
allowed to do anything it wants
282
00:18:41 --> 00:18:47
on the right-hand side of zero
starting at zero,
283
00:18:44 --> 00:18:50
but on the left-hand side of
zero it is wiped away and must
284
00:18:48 --> 00:18:54
be zero.
This is a definite thing now.
285
00:18:50 --> 00:18:56
Its convolution is this.
And the inverse Laplace
286
00:18:53 --> 00:18:59
transform is --
The answer, in other words,
287
00:18:57 --> 00:19:03
is the same thing as what u of
t f of t would be.
288
00:19:02 --> 00:19:08
It's the same thing,
F of s.
289
00:19:04 --> 00:19:10
And so, the conclusion is that
these are equal,
290
00:19:08 --> 00:19:14
since they must be unique.
They have been made unique by
291
00:19:12 --> 00:19:18
making them zero for t negative.
In other words,
292
00:19:16 --> 00:19:22
apply to a function,
well, I won't recopy it.
293
00:19:19 --> 00:19:25
But the point is that delta t,
for the convolution operation,
294
00:19:24 --> 00:19:30
is acting like an identity.
If I multiply,
295
00:19:29 --> 00:19:35
in the sense of convolution,
it is a peculiar operation.
296
00:19:33 --> 00:19:39
But algebraically,
it has a lot of the properties
297
00:19:36 --> 00:19:42
of multiplication.
It is communitive.
298
00:19:39 --> 00:19:45
It is linear in both factors.
In other words,
299
00:19:42 --> 00:19:48
it is almost anything you would
want with multiplication.
300
00:19:46 --> 00:19:52
It has an identity element,
identity function.
301
00:19:49 --> 00:19:55
And the identity function is
the Dirac delta function.
302
00:19:53 --> 00:19:59
Anything else here?
Yeah, I will throw in one more
303
00:19:57 --> 00:20:03
thing.
It would just require one more
304
00:20:01 --> 00:20:07
phony argument,
which I won't bother giving
305
00:20:04 --> 00:20:10
you, but it is not totally
implausible.
306
00:20:06 --> 00:20:12
After all, u of t,
the unit step function is not
307
00:20:11 --> 00:20:17
differentiable,
is not a differentiable
308
00:20:13 --> 00:20:19
function.
It looks like this.
309
00:20:15 --> 00:20:21
Here its derivative is zero,
here its derivative is zero,
310
00:20:19 --> 00:20:25
and in this class it is not
even defined in between.
311
00:20:23 --> 00:20:29
But, I don't care,
I will make it go straight up.
312
00:20:27 --> 00:20:33
The question is what's its
derivative?
313
00:20:31 --> 00:20:37
Well, zero here,
zero there and infinity at
314
00:20:34 --> 00:20:40
zero, so it must be the delta
function.
315
00:20:37 --> 00:20:43
That has exactly the right
properties.
316
00:20:40 --> 00:20:46
So the same people who will
tell you this will tell you that
317
00:20:44 --> 00:20:50
also.
And, in fact,
318
00:20:45 --> 00:20:51
when you use it to solve
differential equations it acts
319
00:20:50 --> 00:20:56
as if that is true.
I think I have given you an
320
00:20:53 --> 00:20:59
example on your homework.
Let me now show you a typical
321
00:20:57 --> 00:21:03
example of the way the Dirac
delta function would be used to
322
00:21:02 --> 00:21:08
solve a problem.
Let's go back to our little
323
00:21:06 --> 00:21:12
spring, since it is the easiest
thing.
324
00:21:09 --> 00:21:15
You are familiar with it from a
physical point of view,
325
00:21:13 --> 00:21:19
and it is the easiest thing to
illustrate on.
326
00:21:16 --> 00:21:22
We have our spring mass system.
Where is it?
327
00:21:20 --> 00:21:26
Is it on the board?
Up there.
328
00:21:22 --> 00:21:28
That one.
And the differential equation
329
00:21:25 --> 00:21:31
we are going to solve is y
double prime plus y
330
00:21:29 --> 00:21:35
equals --
And now, I am going to assume
331
00:21:33 --> 00:21:39
that the spring is kicked with
impulse a.
332
00:21:37 --> 00:21:43
I am not going to kick it at
time t equals zero,
333
00:21:42 --> 00:21:48
since that would get us into
slight technical difficulties.
334
00:21:47 --> 00:21:53
Anyway, it is more fun to kick
it at time pi over two.
335
00:21:52 --> 00:21:58
The thing is,
336
00:21:54 --> 00:22:00
what is happening?
Well, we have got to have
337
00:21:58 --> 00:22:04
initial conditions.
The initial conditions are
338
00:22:02 --> 00:22:08
going to be, let's start at time
zero.
339
00:22:05 --> 00:22:11
We will start it at the
position one.
340
00:22:08 --> 00:22:14
So I take my spring,
I drag it to the position one,
341
00:22:11 --> 00:22:17
I take the little mass there
and then let it go.
342
00:22:15 --> 00:22:21
And so it starts going birr.
But right when it gets to the
343
00:22:19 --> 00:22:25
equilibrium point I give it a,
"cha!" with unit impulse.
344
00:22:23 --> 00:22:29
I started it from rest.
Those will be the initial
345
00:22:27 --> 00:22:33
conditions.
And I want to say that I kicked
346
00:22:31 --> 00:22:37
it, not with unit impulse,
but with the impulse a.
347
00:22:35 --> 00:22:41
Bigger.
And I did that at time pi over
348
00:22:38 --> 00:22:44
two.
So how are we going to say
349
00:22:41 --> 00:22:47
that?
Well, kick it means delivered
350
00:22:43 --> 00:22:49
that impulse over an extremely
short time interval,
351
00:22:47 --> 00:22:53
but in such a way kicked it
sufficiently hard that the total
352
00:22:52 --> 00:22:58
impulse was a.
The way to say that is kick it
353
00:22:56 --> 00:23:02
with the Dirac delta function.
Translate it to the point time
354
00:23:02 --> 00:23:08
pi over two.
Not at zero any longer.
355
00:23:06 --> 00:23:12
t minus pi over two.
356
00:23:09 --> 00:23:15
But that would kick it with a
unit impulse.
357
00:23:12 --> 00:23:18
I want it to kick it with the
impulse a, so I will just
358
00:23:16 --> 00:23:22
multiply that by the constant
factor a.
359
00:23:20 --> 00:23:26
Let's put this over here.
y of zero equals one,
360
00:23:24 --> 00:23:30
that's the starting value.
Now we have a problem.
361
00:23:30 --> 00:23:36
The only thing new in solving
this with the Laplace transform
362
00:23:33 --> 00:23:39
is I have this funny right-hand
side.
363
00:23:36 --> 00:23:42
But it corresponds to a
physical situation.
364
00:23:38 --> 00:23:44
Let's do it.
You take the Laplace transform
365
00:23:41 --> 00:23:47
of both sides of the equation.
Remember how to do that?
366
00:23:44 --> 00:23:50
You have to take account of the
initial conditions.
367
00:23:48 --> 00:23:54
The Laplace transform of the
second derivative is you
368
00:23:51 --> 00:23:57
multiply by s squared,
and then you have to subtract.
369
00:23:55 --> 00:24:01
You have to use these initial
conditions.
370
00:23:59 --> 00:24:05
This one won't give you
anything, but the first one
371
00:24:02 --> 00:24:08
means I have to subtract one
times s.
372
00:24:05 --> 00:24:11
That is the Laplace transform
of y double prime.
373
00:24:09 --> 00:24:15
The Laplace transform of y,
of course, is just capital Y.
374
00:24:13 --> 00:24:19
And how about the Laplace
375
00:24:16 --> 00:24:22
transform of the right-hand
side.
376
00:24:18 --> 00:24:24
Well, we will have the constant
factor a because the Laplace
377
00:24:22 --> 00:24:28
transform is linear.
And now, the delta function
378
00:24:25 --> 00:24:31
would have the transform one.
But when I translate it,
379
00:24:30 --> 00:24:36
pi over two,
that means I have to use that
380
00:24:33 --> 00:24:39
formula.
Translate it by pi over two
381
00:24:35 --> 00:24:41
means take the one that it would
have been otherwise and multiply
382
00:24:40 --> 00:24:46
it by e, that exponential
factor.
383
00:24:42 --> 00:24:48
It would be e to the minus pi
over two,
384
00:24:46 --> 00:24:52
that is the A times s times
one, which would be the g of s,
385
00:24:49 --> 00:24:55
the Laplace transform
or the delta function before it
386
00:24:54 --> 00:25:00
had been translated.
But I don't have to put that in
387
00:24:57 --> 00:25:03
because it's one.
I am multiplying by one.
388
00:25:01 --> 00:25:07
And to do everything now is
routine.
389
00:25:03 --> 00:25:09
Solve for the Laplace
transform.
390
00:25:05 --> 00:25:11
Well, what is it?
It is y is equal to.
391
00:25:08 --> 00:25:14
I put the s on the other side.
That makes the right-hand side
392
00:25:12 --> 00:25:18
the sum of two terms.
And I divide by the coefficient
393
00:25:15 --> 00:25:21
of y, which is s squared plus
one.
394
00:25:18 --> 00:25:24
The s is over on the right-hand
side and it is divided by s
395
00:25:22 --> 00:25:28
squared plus one.
And the other factor is there,
396
00:25:25 --> 00:25:31
too.
And it, too,
397
00:25:26 --> 00:25:32
is divided by s squared plus
one.
398
00:25:30 --> 00:25:36
399
00:25:35 --> 00:25:41
Now, we take the inverse
Laplace transform of those two
400
00:25:38 --> 00:25:44
terms and add them up.
401
00:25:40 --> 00:25:46
402
00:26:00 --> 00:26:06
What will we get?
Well, y is equal to,
403
00:26:02 --> 00:26:08
the inverse Laplace transform
of s over s squared plus one is
404
00:26:07 --> 00:26:13
cosine t.
405
00:26:11 --> 00:26:17
Now, for this thing we will
have to use our formula.
406
00:26:15 --> 00:26:21
If this weren't here,
the inverse Laplace transform
407
00:26:19 --> 00:26:25
of a over s squared plus one
would
408
00:26:24 --> 00:26:30
be what?
Well, it would be a times the
409
00:26:27 --> 00:26:33
sine of t.
410
00:26:30 --> 00:26:36
411
00:26:35 --> 00:26:41
In other words,
if this is the g of s
412
00:26:37 --> 00:26:43
then the function on the left
would be basically A sine t.
413
00:26:41 --> 00:26:47
But because it has been
414
00:26:43 --> 00:26:49
multiplied by that exponential
factor, e to the minus as
415
00:26:47 --> 00:26:53
where a is pi over two,
416
00:26:50 --> 00:26:56
the left-hand side has to be
changed from A sine t
417
00:26:53 --> 00:26:59
to what it would be
with the translated form.
418
00:26:58 --> 00:27:04
So the rest of it is u of t
minus pi over two,
419
00:27:01 --> 00:27:07
because a is pi over
two, times what it would have
420
00:27:05 --> 00:27:11
been just from the factor g of s
itself.
421
00:27:09 --> 00:27:15
In other words,
A times the sine of,
422
00:27:11 --> 00:27:17
again, t minus pi over two.
423
00:27:15 --> 00:27:21
I am applying that formula,
but I am applying it in that
424
00:27:19 --> 00:27:25
direction.
I started with this,
425
00:27:21 --> 00:27:27
and I want to recover the
left-hand side.
426
00:27:24 --> 00:27:30
And that is what it must look
like.
427
00:27:26 --> 00:27:32
The A, of course,
just gets dragged along for the
428
00:27:29 --> 00:27:35
free ride.
Now, as I emphasized to you
429
00:27:34 --> 00:27:40
last time, and I hope you did on
your homework that you handed
430
00:27:38 --> 00:27:44
in, you mustn't leave it in that
form.
431
00:27:41 --> 00:27:47
You have to make cases because
people will expect you to tell
432
00:27:46 --> 00:27:52
them what the meaning of this
is.
433
00:27:49 --> 00:27:55
Now, if t is less than pi over
two, this is zero.
434
00:27:52 --> 00:27:58
And, therefore,
that term does not exist.
435
00:27:56 --> 00:28:02
So the first part of it is just
the cosine t term if
436
00:28:00 --> 00:28:06
t lies between zero and pi over
two.
437
00:28:05 --> 00:28:11
If t is bigger than pi over two
then this factor is
438
00:28:10 --> 00:28:16
one.
It's the unit step function.
439
00:28:12 --> 00:28:18
And I, therefore,
must add in this term.
440
00:28:16 --> 00:28:22
Now, what is that term?
What is the sine of t minus pi
441
00:28:20 --> 00:28:26
over two?
The sine of t looks
442
00:28:25 --> 00:28:31
like that.
The sine of t,
443
00:28:27 --> 00:28:33
if I translate it,
looks like this.
444
00:28:31 --> 00:28:37
If I translate it by pi over
two.
445
00:28:33 --> 00:28:39
And let's finish it up,
the pi that was over here moved
446
00:28:38 --> 00:28:44
into position.
That curve is the curve
447
00:28:41 --> 00:28:47
negative cosine t.
448
00:28:44 --> 00:28:50
449
00:28:51 --> 00:28:57
And so the answer is if t is
bigger than pi over two,
450
00:28:55 --> 00:29:01
it is cosine t minus A
times cosine t.
451
00:29:00 --> 00:29:06
Or, in other words,
452
00:29:02 --> 00:29:08
it is one minus A times cosine
t.
453
00:29:07 --> 00:29:13
454
00:29:11 --> 00:29:17
Now, do those match up?
They have always got to match
455
00:29:13 --> 00:29:19
up, or you have made a mistake.
You always have to get a
456
00:29:17 --> 00:29:23
continuous function when you
have just discontinuities.
457
00:29:20 --> 00:29:26
Do we get a continuous
function?
458
00:29:22 --> 00:29:28
Yeah, when t is pi over two
the value here is
459
00:29:25 --> 00:29:31
zero.
The value of this is also zero
460
00:29:27 --> 00:29:33
at pi over two.
There is no conflict in the
461
00:29:31 --> 00:29:37
values.
Values doesn't suddenly jump.
462
00:29:33 --> 00:29:39
The function is continuous.
It is not differential but it
463
00:29:38 --> 00:29:44
is continuous.
Well, what function does that
464
00:29:41 --> 00:29:47
look like?
There are cases.
465
00:29:43 --> 00:29:49
It starts out life as the
function cosine t.
466
00:29:47 --> 00:29:53
So it gets to here.
And at t equals pi over two,
467
00:29:51 --> 00:29:57
the mass gets kicked
and that changes the function.
468
00:29:55 --> 00:30:01
Now, what are the values?
Well, if A is bigger than one
469
00:30:01 --> 00:30:07
this is a negative
number and it therefore becomes
470
00:30:07 --> 00:30:13
the function negative
cosine t.
471
00:30:11 --> 00:30:17
Now, negative cosine t looks
like this, the blue guy.
472
00:30:16 --> 00:30:22
Negative cosine t is a function
that looks like this.
473
00:30:21 --> 00:30:27
So it goes from here,
it reverses direction,
474
00:30:26 --> 00:30:32
the mass reverses direction
from what you thought it was
475
00:30:31 --> 00:30:37
going to do.
And it does that because A is
476
00:30:36 --> 00:30:42
so large that that impulse was
enough to make it reverse
477
00:30:40 --> 00:30:46
direction.
Of course it might only do
478
00:30:42 --> 00:30:48
this, but this is what will
happen if A is bigger than one.
479
00:30:47 --> 00:30:53
This will be A,
which is a lot bigger than one.
480
00:30:50 --> 00:30:56
If it's not so much bigger than
one it might look like that.
481
00:30:55 --> 00:31:01
So A is just bigger than one.
How's that?
482
00:30:59 --> 00:31:05
Well, what if A is less than
one?
483
00:31:01 --> 00:31:07
Well, in that case it stays
positive.
484
00:31:04 --> 00:31:10
If A is less than one,
this is now still a positive
485
00:31:07 --> 00:31:13
number.
And, therefore,
486
00:31:09 --> 00:31:15
the cosine continues on its
merry way.
487
00:31:12 --> 00:31:18
The only thing is it might be a
little more sluggish or it might
488
00:31:17 --> 00:31:23
be very peppy and do that.
Let's just go that far.
489
00:31:20 --> 00:31:26
This will be the case A less
than one.
490
00:31:23 --> 00:31:29
Well, of course,
the most interesting case is
491
00:31:26 --> 00:31:32
what happens if A is exactly
equal to one?
492
00:31:32 --> 00:31:38
The porridge is exactly just
right, I think that's the
493
00:31:37 --> 00:31:43
phrase.
Too hot.
494
00:31:38 --> 00:31:44
Too cold.
Just right.
495
00:31:40 --> 00:31:46
When A is equal to one,
it is zero.
496
00:31:44 --> 00:31:50
It starts out as cosine t.
497
00:31:48 --> 00:31:54
When it gets to t,
it continues on ever after as
498
00:31:52 --> 00:31:58
the function zero.
I have a visual aid for the
499
00:31:57 --> 00:32:03
only time this term.
It didn't work at all.
500
00:32:02 --> 00:32:08
I mean, on the other hand,
the last hour,
501
00:32:06 --> 00:32:12
the people who worked it were
not intrinsically baseball
502
00:32:11 --> 00:32:17
players, so we will use the
equation of the pendulum
503
00:32:15 --> 00:32:21
instead.
That is a lot easier than mass
504
00:32:18 --> 00:32:24
spring.
This is a pendulum.
505
00:32:20 --> 00:32:26
It is undamped because I
declare it to be and it swings
506
00:32:25 --> 00:32:31
back and forth.
And here I am releasing it.
507
00:32:30 --> 00:32:36
The variable is not x or y but
theta, the angle through.
508
00:32:34 --> 00:32:40
Here theta is one,
let's say.
509
00:32:37 --> 00:32:43
That's about one radian.
It starts there and swings back
510
00:32:41 --> 00:32:47
and forth.
It is not damped,
511
00:32:44 --> 00:32:50
so it never loses amplitude,
particularly if I swish it,
512
00:32:48 --> 00:32:54
if I move my hand a little bit.
I want someone who knows how to
513
00:32:53 --> 00:32:59
bat a baseball.
That was the problem last hour.
514
00:32:57 --> 00:33:03
Two people.
One to release it.
515
00:33:01 --> 00:33:07
I will stand up and try to hold
it here.
516
00:33:04 --> 00:33:10
Somebody releases it.
And then somebody who has to be
517
00:33:08 --> 00:33:14
very skillful should apply a
unit impulse of exactly one when
518
00:33:13 --> 00:33:19
it gets to the equilibrium
point.
519
00:33:16 --> 00:33:22
So who can do that?
Who can play baseball here?
520
00:33:20 --> 00:33:26
Come on.
Somebody elected?
521
00:33:23 --> 00:33:29
522
00:33:30 --> 00:33:36
All right. Come on. [APPLAUSE]
523
00:33:33 --> 00:33:39
524
00:33:41 --> 00:33:47
Somebody release it,
too.
525
00:33:44 --> 00:33:50
Somebody tall to handle it all.
I think that will be me.
526
00:33:52 --> 00:33:58
Just hold it at what you would
take to be one radian.
527
00:34:00 --> 00:34:06
He releases it.
When it gets to the bottom,
528
00:34:04 --> 00:34:10
you will have to get way down,
and maybe on this side.
529
00:34:11 --> 00:34:17
Are you a lefty or a righty?
Rightly.
530
00:34:15 --> 00:34:21
Okay.
Bat it what part.
531
00:34:17 --> 00:34:23
Give it a good swat.
I will stand up higher.
532
00:34:22 --> 00:34:28
Help.
I'm not very stable.
533
00:34:25 --> 00:34:31
[APPLAUSE] A trial run.
Again.
534
00:34:30 --> 00:34:36
Okay.
A little further out.
535
00:34:32 --> 00:34:38
First of all,
you have to see where it's
536
00:34:36 --> 00:34:42
going.
Why don't you stand,
537
00:34:38 --> 00:34:44
oh, you bat rightly.
That's right.
538
00:34:41 --> 00:34:47
Okay.
Let's try it again.
539
00:34:44 --> 00:34:50
540
00:34:59 --> 00:35:05
Strike one.
It's okay.
541
00:34:59 --> 00:35:05
It's the beginning of the
baseball season.
542
00:34:59 --> 00:35:05
One more.
The Red Sox are having trouble,
543
00:34:59 --> 00:35:05
too.
Not bad. [APPLAUSE]
544
00:35:05 --> 00:35:11
545
00:35:13 --> 00:35:19
If he had hit even harder it
would have reversed direction
546
00:35:16 --> 00:35:22
and gone that way.
If you hadn't hit it quite as
547
00:35:20 --> 00:35:26
hard it would have continued on,
still at cosine t,
548
00:35:24 --> 00:35:30
but with less amplitude.
But if you hit it exactly right
549
00:35:28 --> 00:35:34
--
It is fun to try to do.
550
00:35:31 --> 00:35:37
Toomre in our department is a
master at this,
551
00:35:36 --> 00:35:42
but he has been practicing for
years.
552
00:35:40 --> 00:35:46
He can take a little mallet and
go blunk, and it stops
553
00:35:45 --> 00:35:51
absolutely dead.
It is unbelievable.
554
00:35:49 --> 00:35:55
I should have had him give the
lecture.
555
00:35:53 --> 00:35:59
Now, I would like to do
something truly serious.
556
00:36:00 --> 00:36:06
Here, I guess.
Because there is a certain
557
00:36:03 --> 00:36:09
amount of engineering lingo you
have to learn.
558
00:36:07 --> 00:36:13
It is used by almost everybody.
Not architects and biologists
559
00:36:12 --> 00:36:18
probably quite yet,
but anybody that uses the
560
00:36:16 --> 00:36:22
Laplace transform will use these
words in connection with it.
561
00:36:21 --> 00:36:27
I really think,
since it is such a widespread
562
00:36:25 --> 00:36:31
technique, that these are things
you should know.
563
00:36:31 --> 00:36:37
Anyway, it will be easy.
It is just the enrichment of
564
00:36:34 --> 00:36:40
your vocabulary.
It is always fun to learn new
565
00:36:37 --> 00:36:43
vocabulary words.
So, let's just consider a
566
00:36:40 --> 00:36:46
general second order equation.
By the way, all this applies to
567
00:36:45 --> 00:36:51
higher order equations,
too.
568
00:36:47 --> 00:36:53
It applies to systems.
The same words are used,
569
00:36:50 --> 00:36:56
but let's use something that
you know.
570
00:36:52 --> 00:36:58
Here is a system.
It could be a spring mass
571
00:36:55 --> 00:37:01
dashpot system.
It could be an RLC circuit.
572
00:37:00 --> 00:37:06
Or that pendulum,
a damped pendulum,
573
00:37:02 --> 00:37:08
anything that is modeled by
that differential equation with
574
00:37:06 --> 00:37:12
constant coefficients,
second-order.
575
00:37:08 --> 00:37:14
This is the input.
The input can be any kind of a
576
00:37:11 --> 00:37:17
function.
Exponential functions,
577
00:37:13 --> 00:37:19
sine, cosine.
It could be a Dirac delta
578
00:37:16 --> 00:37:22
function.
It could be a sum of these
579
00:37:18 --> 00:37:24
things.
It could be a Fourier series.
580
00:37:21 --> 00:37:27
Anything of the sort of stuff
we have been talking about
581
00:37:24 --> 00:37:30
throughout the last few weeks.
And let's have simple initial
582
00:37:30 --> 00:37:36
conditions so that doesn't louse
things up, the simplest possible
583
00:37:34 --> 00:37:40
ones.
The mass starts at the
584
00:37:36 --> 00:37:42
equilibrium point from rest.
Of course, it doesn't stay that
585
00:37:40 --> 00:37:46
way because there is an input
that is asking it to move along.
586
00:37:45 --> 00:37:51
Now all I want to do is solve
this in general with a Laplace
587
00:37:49 --> 00:37:55
transform.
If I do it in general,
588
00:37:51 --> 00:37:57
that is always easier than
doing it in particular since you
589
00:37:56 --> 00:38:02
don't ever have to do any
calculations.
590
00:38:00 --> 00:38:06
It is s squared Y.
There are no other terms here
591
00:38:05 --> 00:38:11
because the initial conditions
are zero.
592
00:38:08 --> 00:38:14
This part will be a times s Y.
593
00:38:12 --> 00:38:18
Again, no other terms because
the initial conditions are zero.
594
00:38:17 --> 00:38:23
Plus b times Y.
And all that is equal to
595
00:38:21 --> 00:38:27
whatever the Laplace transform
is of the right-hand side.
596
00:38:26 --> 00:38:32
So it is F of s.
Next step.
597
00:38:31 --> 00:38:37
Boy, this is an easy problem.
You solve for Y.
598
00:38:35 --> 00:38:41
Well, Y is F of s times one
over s squared plus as plus b.
599
00:38:41 --> 00:38:47
600
00:38:45 --> 00:38:51
Now, what is that?
The next step now is to figure
601
00:38:50 --> 00:38:56
out what the answer to the
problem is, what's the Y of t?
602
00:38:55 --> 00:39:01
Well, you do that by taking the
603
00:39:00 --> 00:39:06
inverse Laplace transform.
But because these are general
604
00:39:04 --> 00:39:10
functions, I don't have to write
down any specific answer.
605
00:39:09 --> 00:39:15
The only thing is to use the
convolution because this is the
606
00:39:13 --> 00:39:19
product of two functions of s.
The inverse transform will be
607
00:39:18 --> 00:39:24
the convolution of their
respective things.
608
00:39:21 --> 00:39:27
The answer is going to be the
convolution of F of t,
609
00:39:26 --> 00:39:32
the input function in other
words, convoluted with the
610
00:39:30 --> 00:39:36
inverse Laplace transform of
that thing.
611
00:39:35 --> 00:39:41
Now, we have to have a name for
that, and those are the two
612
00:39:39 --> 00:39:45
words I want to introduce you to
because they are used
613
00:39:43 --> 00:39:49
everywhere.
The function,
614
00:39:44 --> 00:39:50
on the right-hand side,
this function one over s
615
00:39:48 --> 00:39:54
squared plus as plus b,
616
00:39:51 --> 00:39:57
notice it only depends upon the
left-hand side of the
617
00:39:55 --> 00:40:01
differential equation,
on the damping constant.
618
00:40:00 --> 00:40:06
The spring constant if you are
thinking of a mass spring
619
00:40:03 --> 00:40:09
dashpot system.
So this depends only on the
620
00:40:06 --> 00:40:12
system, not on what input is
going into it.
621
00:40:08 --> 00:40:14
And it is called the transfer
function.
622
00:40:11 --> 00:40:17
Is usually called capital W of,
sometimes it is
623
00:40:15 --> 00:40:21
capital H of s,
there are different things,
624
00:40:18 --> 00:40:24
but it is always called the
transfer function.
625
00:40:22 --> 00:40:28
626
00:40:27 --> 00:40:33
What we are interested in
putting here its inverse Laplace
627
00:40:31 --> 00:40:37
transform.
Well, I will call that W of t
628
00:40:34 --> 00:40:40
to go with the capital
W of s by the usual
629
00:40:39 --> 00:40:45
notation.
Its inverse Laplace transform,
630
00:40:42 --> 00:40:48
well, I cannot calculate that.
I will just give it a name,
631
00:40:46 --> 00:40:52
W of t.
And that is called the weight
632
00:40:49 --> 00:40:55
function of the system.
This is the transfer function
633
00:40:53 --> 00:40:59
of the system,
so put in "of the system" if
634
00:40:57 --> 00:41:03
you are taking notes.
And so the answer is that
635
00:41:02 --> 00:41:08
always the solution is the
convolution to this differential
636
00:41:06 --> 00:41:12
equation that we have been
solving for the last three or
637
00:41:11 --> 00:41:17
four weeks.
It is the convolution of that.
638
00:41:14 --> 00:41:20
And, therefore,
the solution is expressed as a
639
00:41:18 --> 00:41:24
definite integral of the
function of the input on the
640
00:41:22 --> 00:41:28
right-hand side,
what is forcing the equation,
641
00:41:26 --> 00:41:32
times this magic function but
flipped and translated by t.
642
00:41:32 --> 00:41:38
That says du for you guys over
there.
643
00:41:34 --> 00:41:40
In other words,
the solution to the
644
00:41:37 --> 00:41:43
differential equation is
presented as a definite
645
00:41:41 --> 00:41:47
integral.
Marvelous.
646
00:41:42 --> 00:41:48
And the only thing is the
definite integral involves this
647
00:41:47 --> 00:41:53
funny function W of t.
To understand why that is the
648
00:41:52 --> 00:41:58
solution, you have to understand
what W of t is.
649
00:41:55 --> 00:42:01
Well, formally,
of course, it's that.
650
00:42:00 --> 00:42:06
But what does it really mean?
The problem is what is W of t
651
00:42:05 --> 00:42:11
really? Not just formally,
652
00:42:08 --> 00:42:14
but what does it really mean?
I mean, is it real?
653
00:42:12 --> 00:42:18
I think the simplest way of
thinking of it,
654
00:42:16 --> 00:42:22
once you know about the delta
function is just to think of
655
00:42:21 --> 00:42:27
this differential equation y
double prime plus a y prime plus
656
00:42:27 --> 00:42:33
b. Except use as the input the
657
00:42:32 --> 00:42:38
Dirac delta function.
In other words,
658
00:42:35 --> 00:42:41
we are kicking the mass.
The mass starts at rest,
659
00:42:39 --> 00:42:45
so the initial conditions are
going to be what they were
660
00:42:43 --> 00:42:49
before. y of zero,
661
00:42:46 --> 00:42:52
y prime of zero. Both zero.
662
00:42:49 --> 00:42:55
The mass starts at rest from
the equilibrium position,
663
00:42:53 --> 00:42:59
and it is kicked in the
positive direction,
664
00:42:57 --> 00:43:03
I guess that's this way,
with unit impulse.
665
00:43:02 --> 00:43:08
At time zero with unit impulse.
In other words,
666
00:43:05 --> 00:43:11
kick it just hard enough so you
impart a unit impulse.
667
00:43:10 --> 00:43:16
So that situation is modeled by
this differential equation.
668
00:43:15 --> 00:43:21
The kick at time zero is
modeled by this input,
669
00:43:19 --> 00:43:25
the Dirac delta function.
And now, what happens if I
670
00:43:23 --> 00:43:29
solve it?
Well, you see,
671
00:43:25 --> 00:43:31
everything in the solution is
the same.
672
00:43:30 --> 00:43:36
The left stays the same,
but on the right-hand side I
673
00:43:34 --> 00:43:40
should have not f of s here.
674
00:43:37 --> 00:43:43
Since this is the delta
function, I should have one.
675
00:43:42 --> 00:43:48
What I get is,
on the left-hand side,
676
00:43:45 --> 00:43:51
s squared Y plus as Y plus bY
equals,
677
00:43:50 --> 00:43:56
for the Laplace transform of
the right-hand side is simply
678
00:43:56 --> 00:44:02
one.
And, therefore,
679
00:43:57 --> 00:44:03
Y is what?
Y is one over exactly the
680
00:44:02 --> 00:44:08
transform function.
And therefore its inverse
681
00:44:05 --> 00:44:11
Laplace transform is that weight
function.
682
00:44:09 --> 00:44:15
That is the simplest
interpretation I know of what
683
00:44:13 --> 00:44:19
this magic weight function is,
which gives the solution to all
684
00:44:18 --> 00:44:24
the differential equations,
no matter what the input is.
685
00:44:23 --> 00:44:29
The weight function is the
response of the system at rest
686
00:44:27 --> 00:44:33
to a sharp kick at time zero
with unit impulse.
687
00:44:33 --> 00:44:39
And read the notes because they
will explain to you why this
688
00:44:37 --> 00:44:43
could be thought of as the
superposition of a lot of sharp
689
00:44:42 --> 00:44:48
kicks times zero a little later.
Kick, kick, kick,
690
00:44:46 --> 00:44:52
kick.
And that's what makes the
691
00:44:49 --> 00:44:55
solution.
Next time we start systems.