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PROFESSOR: OK.
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This is one more thing to tell
you about Laplace transforms,
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and introducing a new
word, convolution.
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And so we're going to
find our old formula
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in new language, a new way.
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But the formula is familiar.
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And the problem is our
basic problem, second order,
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linear, constant coefficient
with a forcing term.
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And we know that the
Laplace-- and I'll
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take zero boundary conditions.
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So that the Laplace
transform is just
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s squared y, sy, and that's
the transform of our equation.
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No problem.
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OK, now I'll divide by that.
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So I move that as 1
over, and I call it G.
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So this G is 1 over s squared,
plus Bs plus C. And that
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has the name transfer function.
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And then this is the
transform of the forcing term.
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OK.
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So here we have a nice
formula for y of s,
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after I do that division.
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It's a product.
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The transform of the
solution that we want
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is that transform
times that transform.
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This is the transform
of the impulse response.
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This is the transform
of the right-hand side.
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Now I just have a Laplace
transform question.
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Suppose my transform
is one function of s
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times another function of s,
what is the inverse transform?
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What is the inverse transform?
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What function y of t
gives me G times F?
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And I'm just going
to answer that.
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The answer is the
g and the f, those
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are the ones that give that.
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But I do not just
multiply those.
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The new operation that
gives the right answer
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is called convolution.
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And I'll use a star.
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So right now I'm
going to say what
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does that convolution mean.
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So this is a general question.
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If I have two functions
multiplied together,
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then I want the
inverse transform,
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then I take the separate inverse
transforms, little g and little
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f, and I convolve
them, I do convolution.
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And let me tell you
what convolution is.
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So convolution is-- here is
the formula for convolution.
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It's an integral from 0
to t of one function--
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maybe I better use
capital T, better-- times
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the other function, integrated.
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That's what convolution is.
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So what have I achieved here?
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The same old formula.
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The formula which we
described way back
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at the beginning as
inputs f, growth factors
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over the remaining time, g.
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Put all those together
by integration.
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Put all the inputs with
their growth factors.
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Integrate to put
them all together.
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And that is y.
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So it's a familiar formula,
with only a new word.
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But you see that I could
jump to the answer,
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once I knew about the
convolution formula,
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and I knew that this is the
function whose transform its--
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let me say again.
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Its transform is GF.
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So if I multiply transforms,
I convolve functions.
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And looking at it the other
way, if I multiply functions I
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would convolve their transforms.
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So convolution grows
the number of functions
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that we can deal with
on Laplace transform.
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Because it tells us what to
do with products, capital G
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capital F. Or it tells us what
to do with little g little f.
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So I'm almost through,
because I don't plan to check.
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I could.
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But this isn't the right place.
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The book does it accurately.
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I don't plan to check that
this statement is true
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that the transform of
that one is that one.
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But it is true.
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But I do plan to do an example.
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Now second degree
gets a little messy.
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So let me do a first
degree example.
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Example, I'll take the
equation dy dt minus ay.
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That's our usual first
degree differential equation.
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And I'll take e to the ct
on the right-hand side.
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OK.
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I'm doing those, because I can
take the transforms and check
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everything.
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So let me transform both
of those starting from 0.
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So the transform of that is
s y of s, minus a y of s,
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equals, well I know
the transform f of s.
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I know the transform of
that is 1 over s minus c.
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So this is just, s
minus a factors out.
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So well y of s is 1 over
s minus a, and s minus c.
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Again, this is the simplest
differential equation
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with a forcing term that
I could use as an example.
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So now I'm looking for what is
y of T. I'm looking for y of T.
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And I'm now going to use
the language of convolution.
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This is the transform
of e to the at.
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This is the
transform of-- so you
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see I'm thinking of that as
the transform of e to at,
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and the transform of e to the
st. So there is one factor.
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And there's the other factor.
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So according to the
convolution formula,
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I can write down the
inverse transform, the y
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of t I want as the integral.
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I'm just going to
copy the convolution.
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But I know the
functions for that.
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So it's an integral from 0 to t.
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What do I have? g of t minus t.
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What is the inverse transform
of 1 over s minus a?
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It's e to the a t minus t.
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And what is the inverse
transform of 1 over s minus c?
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e to the cT dT.
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So I have used the-- I've
just put in what I know
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in the convolution formula.
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And this should be
the correct answer.
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And I can do this integral.
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And what do I get?
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Well, I'm pretty sure
that I get e to the-- down
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below there will be
a-- you see I'm going
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to combine those exponentials.
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So I'll have a c minus a.
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It comes out perfectly.
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e to the ct, minus e to the at.
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That's the right answer.
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It's not only what the
convolution formula tells me,
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it's what I know.
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So that example is a good
one to show that when--
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so I didn't use
partial fractions.
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Normally I would separate
this into partial fractions,
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and then I would recognize
those two pieces of the answer.
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I didn't do that this time.
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Instead of using partial
fractions, the algebra,
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I used the convolution formula,
and did the integral or almost
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did it.
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We can do it.
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And we get that answer.
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OK.
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So the point of
this video is simply
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to introduce the idea
of a convolution, which
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is the quantity we need,
the function we need,
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when the transform is a
product of two transforms.
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OK.
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Thank you.