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So today we're going to discuss
Fourier analysis.
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You'll see a lot of math,
but I'll try to keep an eye on
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the physics.
Imagine that I pluck a string
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which is fixed at both ends.
So, here is X and here is Y.
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And, this is the string at T
equals zero.
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And, imagine I let it go.
And we know from what we have
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learned that it will start to
oscillate in the superposition
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of its normal modes,
which are standing waves.
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And so, the $64 million
question today is now,
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in which modes will oscillate,
and what are the [NOISE
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OBSCURES] of those modes?
If you write down Y as a
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function of X and T,
then we know from what we have
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learned to date that you should
be able to write this,
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N equals one to infinity of
amplitudes which I indicate with
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a B now rather than with an A.
But B is, of course,
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an amplitude,
has a dimension length,
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times the sine of KNX times the
cosine of omega NT.
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And, because of the boundary
conditions, K of N,
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because the string is fixed at
both ends, K of N is N pi
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divided by L,
and omega N is the velocity
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propagation times K of N.
And N, then,
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is one, two,
three, four,
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five.
So, we know this.
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There is nothing new.
We've had this recently.
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So, I can also write down,
then, that at T equals zero,
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I can write down this shape in
terms of a series of signs.
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I will leave the time off now.
So, at T equals zero,
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then we have YX at time T
equals zero would be B1 times
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the sine of pi X over L plus B2
times the sine of 2 pi X over L
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plus B3, and so on.
And our task now today is what
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values of N are necessary to
make that pulse,
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and what are,
then, the amplitudes,
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B, that we have here?
And, that's what Fourier
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analysis is all about.
Fourier was born in 1768.
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And, he died in 1830.
So, this is work that was the
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long time ago,
a brilliant mathematician.
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Now, I will give you the
general approach first.
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And then, I will come back to
the special case of a string,
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which is fixed at both ends.
But I really think I owe you
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the general format,
the general procedure first.
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So the idea behind Fourier,
then, is that any periodic,
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single valued function with
period of 2 pi radians can be
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represented by a Fourier series.
And, I will write down this
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Fourier series here.
So, this is now a function of
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X.
X is now in radians.
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It's not that X.
I will come back to that.
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So, a function of X can now be
written as a constant.
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We call it A zero divided by
two, plus the sum,
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one to infinity of AM cosine MX
plus the sum of one to infinity
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of BM times the sine of MX.
And this X that you see here is
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also called X in Beckafee [SP?]
and Barrett is in radians.
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I will call this first term
one.
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I will call this one two,
and I will call this one three.
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So, our task,
now, is to find the values for
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A and B if you know the
function, FX.
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And I will build up with you
the recipe that allows you to
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calculate the values for A and
the values for B.
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What you do,
the first thing you do to find
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A zero, you take the integral
from minus pi to plus pi DX on
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both sides of the aisle.
This one side of the aisle,
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and this is the other side of
the aisle.
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And when you do that,
you will see that for all
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values of M, Mary,
all terms here will be zero.
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And, the same is true for
three.
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For any value of M that you
take, if you do an integral over
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a sine over a complete period
from minus pi to pi,
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there is a complete period.
You get zero,
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of course.
So, all values for three will
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be zero.
And so, what you're left with
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is, then, that the integral from
minus pi to plus pi of XXDX is,
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then, the integral from minus
pi to plus pi of A zero divided
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by two DX.
And, that is pi times A zero.
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That's immediately obvious.
So, we now already have our
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first way of calculating A zero.
A zero, if I use this result,
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is nothing but one over pi
times the integral from minus pi
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to plus pi times the function,
X, of DX.
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So, you tell me what the
function of X is,
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and I can calculate A zero for
you.
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Keep in mind that A zero
divided by two is nothing but
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the average value of the
function over the period two pi.
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And, I will come back to that
later during this lecture.
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How are we going to calculate,
now, the other values for A?
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Well, what you do now is you
take the integral for minus pi
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to plus pi times the cosine of
Nancy X DX.
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And, you do that on both sides
of the aisle.
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You do it on the left side,
and you do it on the right
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side.
And Nancy, now,
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is one, two,
three, and so on.
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When you do that,
you'll see that this term is
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going to be zero.
That's immediately obvious.
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You do an integral over a
cosine function times a
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constant.
You get zero.
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So, the first one you don't
worry about.
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The third one,
for any value of Mary and for
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any value of Nancy,
the third one is also zero.
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And, if you don't believe that,
check that on your own.
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So then, you'd get a cosine of
Nancy X times the sine of Mary X
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for any value of N,
any integer of N,
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any integer of M.
If you integrate it over one
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whole period,
you get zero.
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So, this is also zero.
If we now go to term number
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two, it is also always zero
except when Mary is the same as
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Nancy.
So, number two is also zero
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except when M equals N.
When M equals N,
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you get here a cosine squared.
And when you integrate a cosine
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squared.
And, when you integrate a
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cosine squared,
you do not get zero.
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In other words,
what you have to do if Mary
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equals Nancy,
you're going to get the cosine
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squared of MX DX because Mary is
Nancy.
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And, I must integrate that
between minus pi and pi.
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And that equals pi.
So now, we have a recipe for A
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of M because A of M now is one
divided by pi.
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That is this pi that comes from
the integral of the cosine
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squared integrate with minus pi
and plus pi, a function X times
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the cosine of Mary XDX.
So, we know now how to
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calculate A zero,
and we know how to calculate
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the values for A of M.
And, we will,
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of course, do an example
together.
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It's now clear what you're
going to do to find [two
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values?] for B.
You're now going to integrate
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both sides of the aisle between
minus pi and plus pi times the
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sine of Nancy X DX.
And when you do that,
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you'll see exactly the same
that you had here.
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This will always be zero.
This will now always be zero,
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and this will always be zero
except when Mary is Nancy.
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Then you get here the sine
square, and the integral of the
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sine square will be pi,
just like the integral here of
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the cosine square was pi.
And so, you see now that we
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also have now the recipe to find
all the values for B of M.
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That's going to be one divided
by pi times the integral of
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minus pi to plus pi.
That is the period of the
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function times F of X times the
sine of Mary X DX.
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And here you see in its most
general form,
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the formalism of Fourier
analysis.
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Keep in mind that whenever you
have an integral for minus pi to
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pi, if you prefer zero to 2 pi,
that's fine because the
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function is periodic.
So, you can always replace this
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zero to 2 pi if that suits your
purpose.
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If you look at these three
recipes, you can do away with
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this one because if you make M
equals zero here,
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originally M was one,
two, three, four,
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five.
But if you also include M
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equals zero here,
you get exactly the same that
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you have here because when M
equals zero, the cosine is one.
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So this is then identical.
That is the only reason why we
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called this constant term A zero
divided by two.
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We could have called that a C
because, after all,
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a constant is a constant.
Why would you want to call it A
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zero divided by two?
If you have called this C,
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which is the average value of
the function,
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then you'd need this C here,
and you would have here one
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over 2 pi.
And therefore,
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you need three recipes to do
Fourier analysis,
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whereas now,
if we define this constant as A
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zero over two,
you only need two.
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So, that's the only reason.
It's just for practical
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purposes.
But A zero over two is a
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constant, and it is the average
value of the function over the
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period.
Now, all this may look a little
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bit opaque to you now,
and it will become clear,
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I hope, during this lecture
when I put this Fourier analysis
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to work.
What you see here,
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and that's the way I want you
to look at it,
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is that is simply a recipe.
And you and I,
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when we used this recipe,
we execute it.
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And we do not always ask
ourselves, why is the recipe the
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way it is?
If I apply Kramer's rule,
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I do not, every time,
say to myself,
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why is it really the way it is?
I have seen once why it is the
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way it is, and I apply it.
Of course, you have to know
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when you can use it and when you
cannot use it.
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So, now I would like to return
to the plucked string.
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But I first want to take a
close look at it at time T
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equals zero.
And, what I'm going to do now
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is to pluck it in a very unusual
way.
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And there is a reason why I do
that so unusual,
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because that is doable in one
lecture to do the Fourier
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00:13:47,261 --> 00:13:50,923
analysis on it.
So, I have a string,
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now, which is fixed between
zero and L.
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And, I'm going to pluck it in
an extremely obnoxious way,
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namely, like this.
So, I'm forcing it like a
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00:14:06,360 --> 00:14:11,363
square, very painful for the
string, but that's not my
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00:14:11,363 --> 00:14:14,951
problem.
And, let this be an amplitude,
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A, if I want to use Fourier
analysis, and you'll see that I
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do.
And you will also see how I'm
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going to do that.
I need a periodic function,
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and this is not periodic.
So I'm going to make it
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00:14:34,097 --> 00:14:39,216
periodic.
And the way I'm going to make
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it periodic is the following.
I'm going to pretend that the
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function is really this.
So, I'm adding this.
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00:14:52,955 --> 00:14:58,477
I make it 2L,
but I go much farther along,
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and go on, and go on,
and go on.
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So, I'm going to define it
between zero and 2L.
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00:15:08,233 --> 00:15:12,743
And that's periodic.
Notice that this pattern now,
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00:15:12,743 --> 00:15:15,781
up square down square is
periodic.
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00:15:15,781 --> 00:15:21,395
So, that means my period is 2L.
And also, notice that by doing
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that my A zero divided by two is
zero because the average value
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00:15:27,101 --> 00:15:31,243
of this function,
which is now defined between
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zero and 2L and much beyond 2L
and below zero that that
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00:15:36,213 --> 00:15:41,000
function has an average value of
zero.
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00:15:41,000 --> 00:15:46,619
And so, my function now,
my function of X,
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00:15:46,619 --> 00:15:54,157
you can write down Y of X if
you want to is plus A for X
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being between L and zero.
And, it is minus A for X
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between 2L and L because this
here is minus A.
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00:16:08,000 --> 00:16:14,421
Now, I want to express the
shape into a Fourier series.
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00:16:14,421 --> 00:16:20,962
But before I can do that,
I have to make some changes in
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the recipe because in the recipe
we had radians.
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And radians are apples.
Here we have X.
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00:16:31,518 --> 00:16:35,367
But that's not radians.
That is meters.
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And meters are coconuts.
Do you have a question?
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00:16:40,126 --> 00:16:44,582
Between 2L and L it is minus A.
Between zero,
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00:16:44,582 --> 00:16:48,835
is that what you want?
Thank you very much.
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I appreciate corrections
because it's awkward to do it
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00:16:54,202 --> 00:16:55,822
later.
Thank you.
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00:16:55,822 --> 00:17:02,000
So, radians are apples,
and meters are coconuts.
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00:17:02,000 --> 00:17:09,062
And so, what should I do not
make the coconuts into apples?
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00:17:09,062 --> 00:17:15,151
I now have to take the old X,
which was in radians,
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00:17:15,151 --> 00:17:22,335
which is the one that I had
here, and is the one that I have
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00:17:22,335 --> 00:17:26,476
here.
And I'm going to replace that
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00:17:26,476 --> 00:17:35,000
in that formalism by pi X over
L, and this X now is in meters.
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00:17:35,000 --> 00:17:39,503
And if I have done that,
you can see that if my X,
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00:17:39,503 --> 00:17:43,639
now, becomes 2L,
then this has moved over 2 pi
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00:17:43,639 --> 00:17:46,856
radians.
So, that's exactly what you
226
00:17:46,856 --> 00:17:50,625
have to do.
It is unfortunate that we call
227
00:17:50,625 --> 00:17:53,566
this X.
We could have called this
228
00:17:53,566 --> 00:17:56,507
formalism Z or some other
symbol.
229
00:17:56,507 --> 00:18:02,596
But, in general we give X.
We could have called this Z.
230
00:18:02,596 --> 00:18:08,253
Well, you would have to write
down that the X there is then pi
231
00:18:08,253 --> 00:18:12,426
Z divided by L.
Now we have apples here and we
232
00:18:12,426 --> 00:18:16,506
have apples there.
So, that means that now my
233
00:18:16,506 --> 00:18:22,441
Fourier series are going to look
very much like what I have here.
234
00:18:22,441 --> 00:18:25,408
But I prefer to write them
again.
235
00:18:25,408 --> 00:18:28,932
So, my function,
Y, as a function of X,
236
00:18:28,932 --> 00:18:34,681
which of course is my function
of X, I know that it is in the Y
237
00:18:34,681 --> 00:18:40,187
direction.
That function should now be
238
00:18:40,187 --> 00:18:47,152
written somewhat differently:
A zero divided by two plus the
239
00:18:47,152 --> 00:18:54,236
sum from one to infinity of A of
M times the cosine of M pi X
240
00:18:54,236 --> 00:18:57,777
over L.
This is now [R/our?] X.
241
00:18:57,777 --> 00:19:04,861
This is X in meters plus the
sum of one to infinity of B of M
242
00:19:04,861 --> 00:19:11,000
times the sine of MX times pi
divided by L.
243
00:19:11,000 --> 00:19:16,712
And so, now we have made a
modification to the general
244
00:19:16,712 --> 00:19:22,317
idea, which is in radians,
to be applicable to a case
245
00:19:22,317 --> 00:19:29,000
whereby we don't have radians
but why we have X in meters.
246
00:19:29,000 --> 00:19:34,514
Before we can execute the
analysis, I also had to change
247
00:19:34,514 --> 00:19:38,524
the recipe in terms of A of M
and B of M.
248
00:19:38,524 --> 00:19:42,935
First of all,
I don't have an integral over a
249
00:19:42,935 --> 00:19:48,550
period 2 pi, but I have to do an
integral from zero to L,
250
00:19:48,550 --> 00:19:53,863
or from minus L to plus L.
So, where earlier I had one
251
00:19:53,863 --> 00:19:59,979
over pi in the role minus pi two
plus pi, I had to change this
252
00:19:59,979 --> 00:20:06,095
boundary either from zero to 2L,
or if you prefer from minus L
253
00:20:06,095 --> 00:20:08,908
to L.
I don't care.
254
00:20:08,908 --> 00:20:12,544
It's the same thing because the
function is periodic.
255
00:20:12,544 --> 00:20:16,668
But I also have to change this
one over pi, which we have in
256
00:20:16,668 --> 00:20:20,513
front here because that was the
result of this integral.
257
00:20:20,513 --> 00:20:24,778
And, if you do the integral in
our space, you would have found
258
00:20:24,778 --> 00:20:27,364
L and not pi.
This is half the period,
259
00:20:27,364 --> 00:20:31,000
and half the period of our
function is L.
260
00:20:31,000 --> 00:20:36,741
So, this one over pi from minus
pi to plus now must be changed
261
00:20:36,741 --> 00:20:40,694
into one divided by L integrals
zero to 2L.
262
00:20:40,694 --> 00:20:45,211
And so, to make sure that
you're not going to get
263
00:20:45,211 --> 00:20:49,823
confused, I'm going to rewrite
the recipe for you.
264
00:20:49,823 --> 00:20:53,588
And I might as will do that
again in red.
265
00:20:53,588 --> 00:20:59,047
So, now that RA of N is one
divided by L times an integral,
266
00:20:59,047 --> 00:21:07,325
let's just go from zero to 2L.
And then we get our function of
267
00:21:07,325 --> 00:21:11,930
X.
And then we're going to get the
268
00:21:11,930 --> 00:21:20,581
cosine of M pi X divided by L DX
and B of M is then one divided
269
00:21:20,581 --> 00:21:28,116
by L times the integral from
zero to 2L of my function,
270
00:21:28,116 --> 00:21:36,488
X, which is my plucked string
times the sine of M pi X over L
271
00:21:36,488 --> 00:21:42,161
times DX.
And so, now we have not changed
272
00:21:42,161 --> 00:21:47,667
anything in terms of the
formalism of Fourier analysis.
273
00:21:47,667 --> 00:21:53,684
But we have adapted the recipe.
So, we now have a new series
274
00:21:53,684 --> 00:21:59,802
written in terms of our new X.
And, we have the new recipe to
275
00:21:59,802 --> 00:22:04,289
calculate the A values in terms
of our new X,
276
00:22:04,289 --> 00:22:10,000
and in terms of the new period
which is now 2L.
277
00:22:10,000 --> 00:22:15,479
So, now you may think that I am
ready to plunge into the math,
278
00:22:15,479 --> 00:22:20,419
and to start executing it.
But, I'm not going to do that
279
00:22:20,419 --> 00:22:23,203
yet.
I would like to take a look
280
00:22:23,203 --> 00:22:28,053
first at the values of A.
We already agreed that A zero
281
00:22:28,053 --> 00:22:30,838
is zero.
So, that's not an issue
282
00:22:30,838 --> 00:22:33,083
anymore.
But, let us look,
283
00:22:33,083 --> 00:22:36,766
for instance,
at the function A1 times the
284
00:22:36,766 --> 00:22:43,133
cosine of pi X over L.
That would be the first term of
285
00:22:43,133 --> 00:22:48,086
the cosine series.
And the second term would be A2
286
00:22:48,086 --> 00:22:51,422
times the cosine of 2 pi X over
L.
287
00:22:51,422 --> 00:22:56,274
Well, let us assume that we
found a value for A1.
288
00:22:56,274 --> 00:23:02,137
So, I'm going to plot now into
this function of ours the A1
289
00:23:02,137 --> 00:23:04,158
value.
So, it's here,
290
00:23:04,158 --> 00:23:08,000
here, here, here,
and here.
291
00:23:08,000 --> 00:23:14,601
And so, here is that function.
And this amplitude,
292
00:23:14,601 --> 00:23:18,911
then, is A1.
That's unacceptable.
293
00:23:18,911 --> 00:23:24,704
Why is this acceptable?
Look at my function,
294
00:23:24,704 --> 00:23:34,000
and then look at the red curve,
which is my A1 cosine function.
295
00:23:34,000 --> 00:23:39,347
That red curve is supposed to
help construct this function.
296
00:23:39,347 --> 00:23:43,680
Here it is positive,
so I demand that it must be
297
00:23:43,680 --> 00:23:47,460
negative here.
Otherwise, I can never make
298
00:23:47,460 --> 00:23:51,425
this [trough?],
whereas it's going to add up
299
00:23:51,425 --> 00:23:53,822
positive here.
So, this A1:
300
00:23:53,822 --> 00:23:58,063
out of the question,
can never contribute to my
301
00:23:58,063 --> 00:24:02,274
function.
But, not only can A1 not
302
00:24:02,274 --> 00:24:07,296
contribute, but any cosine
function that you draw will
303
00:24:07,296 --> 00:24:12,793
always look here like this.
It will always be positive here
304
00:24:12,793 --> 00:24:17,341
and positive there,
whereas I will demand that it
305
00:24:17,341 --> 00:24:21,511
has to be positive here and make
it up there.
306
00:24:21,511 --> 00:24:25,586
Otherwise I can never build up
his function.
307
00:24:25,586 --> 00:24:28,997
To put it in a more
intellectual way,
308
00:24:28,997 --> 00:24:32,977
the way that mathematicians
would use them,
309
00:24:32,977 --> 00:24:38,000
a cosine function is an even
function.
310
00:24:38,000 --> 00:24:43,876
That means that the function of
X is the same as the function of
311
00:24:43,876 --> 00:24:47,421
minus X.
We call that an even function.
312
00:24:47,421 --> 00:24:51,339
Here, X is zero.
A cosine function of X has
313
00:24:51,339 --> 00:24:54,697
exactly the same value as at
minus X.
314
00:24:54,697 --> 00:24:58,242
But, our function is odd.
Our function,
315
00:24:58,242 --> 00:25:03,000
the way we defined it,
zero here, is odd.
316
00:25:03,000 --> 00:25:08,986
And, an odd function,
the function of X equals minus
317
00:25:08,986 --> 00:25:15,208
the function of minus X.
And, to put it in a nutshell,
318
00:25:15,208 --> 00:25:21,665
you can never fit an odd
function with even functions in
319
00:25:21,665 --> 00:25:25,773
Fourier.
Neither can you fit an even
320
00:25:25,773 --> 00:25:33,194
function with odd functions.
So, that's perhaps a better way
321
00:25:33,194 --> 00:25:37,729
of looking at it.
So, we can already conclude
322
00:25:37,729 --> 00:25:43,501
that all cosine functions are
out for this specific case,
323
00:25:43,501 --> 00:25:47,417
of course.
Let's now take a look at the
324
00:25:47,417 --> 00:25:51,952
values for B.
So, keep in mind B are the sign
325
00:25:51,952 --> 00:25:54,941
functions.
So, A zero is zero.
326
00:25:54,941 --> 00:26:00,094
All values for A are zero.
Now, let's turn to the B
327
00:26:00,094 --> 00:26:03,392
values.
So, I'm going to draw the
328
00:26:03,392 --> 00:26:09,062
function again here.
We realize, of course,
329
00:26:09,062 --> 00:26:12,062
it's only defined from zero to
L.
330
00:26:12,062 --> 00:26:17,218
The rest is only a mathematical
way of doing the Fourier
331
00:26:17,218 --> 00:26:20,687
analysis.
So we are going to define it
332
00:26:20,687 --> 00:26:25,468
all the way up to 2L.
And then, we make it periodic.
333
00:26:25,468 --> 00:26:29,875
So let us put in,
there, the function B1 sine pi
334
00:26:29,875 --> 00:26:34,000
B1 times the sine of pi X over
L.
335
00:26:34,000 --> 00:26:38,585
That is the very first one in
the series of the B's.
336
00:26:38,585 --> 00:26:43,171
That one is wonderful.
That is exactly what we want.
337
00:26:43,171 --> 00:26:47,307
This is sinusoid,
which would have an amplitude
338
00:26:47,307 --> 00:26:49,555
B1.
It couldn't be better.
339
00:26:49,555 --> 00:26:53,692
Here it is a mountain,
and here it is a valley.
340
00:26:53,692 --> 00:26:58,637
Well, we need a mountain here
and we need a value there.
341
00:26:58,637 --> 00:27:03,493
No surprise because a sine
function is an odd function.
342
00:27:03,493 --> 00:27:11,193
And, our function is odd.
So clearly, odd functions will
343
00:27:11,193 --> 00:27:16,924
do very well.
Let's now take a look at B2.
344
00:27:16,924 --> 00:27:24,892
B2 times the sine of 2 pi X
over L, let's put it in there.
345
00:27:24,892 --> 00:27:33,000
I'm full of expectations that
B2 will do a great job.
346
00:27:33,000 --> 00:27:38,837
So, we have here zero
crossings.
347
00:27:38,837 --> 00:27:47,123
So, here is my B2.
And, this amplitude is B2,
348
00:27:47,123 --> 00:27:56,538
out of the question.
There is no way that B2 can do
349
00:27:56,538 --> 00:27:59,551
me any good.
Why?
350
00:27:59,551 --> 00:28:02,574
Yeah?
Exactly.
351
00:28:02,574 --> 00:28:05,375
There are various ways of
looking at it.
352
00:28:05,375 --> 00:28:09,109
Here, you see a mountain that
helps building this up.
353
00:28:09,109 --> 00:28:12,412
But, look what you see here.
You want a valley.
354
00:28:12,412 --> 00:28:14,853
And, it's not adding to the
valley.
355
00:28:14,853 --> 00:28:17,079
That's one way of looking at
it.
356
00:28:17,079 --> 00:28:20,167
There's another way of looking
at it to say,
357
00:28:20,167 --> 00:28:24,045
look, everything has to be
symmetric about the line one
358
00:28:24,045 --> 00:28:25,840
half L.
If this builds up,
359
00:28:25,840 --> 00:28:31,147
it's got to build up there.
And it's not doing that.
360
00:28:31,147 --> 00:28:35,174
So, we must conclude that B2
must become zero.
361
00:28:35,174 --> 00:28:38,395
And by analogy,
all even values of M,
362
00:28:38,395 --> 00:28:42,512
B will be zero.
So, this is a prediction that I
363
00:28:42,512 --> 00:28:45,196
make.
We haven't done any fancy
364
00:28:45,196 --> 00:28:49,402
Fourier analysis.
We have already concluded that
365
00:28:49,402 --> 00:28:53,876
for our specific function,
for our specific string,
366
00:28:53,876 --> 00:28:58,977
which is fixed at both ends,
that all values for A will be
367
00:28:58,977 --> 00:29:04,525
zero, and that only odd values
for B for which the end value is
368
00:29:04,525 --> 00:29:09,000
odd will we get non-zero
answers.
369
00:29:09,000 --> 00:29:13,410
So, now we're ready to actually
execute the Fourier recipe.
370
00:29:13,410 --> 00:29:16,984
And, if you insist t hat you
want to do the A's,
371
00:29:16,984 --> 00:29:19,646
be my guest.
You can do all the A's,
372
00:29:19,646 --> 00:29:22,840
and you will see that all of
them are zero.
373
00:29:22,840 --> 00:29:25,730
So, I will not waste your time
on that.
374
00:29:25,730 --> 00:29:29,000
But, I will certainly do the
B's.
375
00:29:29,000 --> 00:29:35,000
376
00:29:35,000 --> 00:29:42,880
So, we now execute our recipe.
And, here it is adjusted to our
377
00:29:42,880 --> 00:29:48,047
X, and so we're going to get
that B of M.
378
00:29:48,047 --> 00:29:55,282
Now the function is defined
between zero and L is plus A.
379
00:29:55,282 --> 00:30:02,000
So, I can bring the A outside
and put it here.
380
00:30:02,000 --> 00:30:07,406
And then, I get the integral
from zero to 2L.
381
00:30:07,406 --> 00:30:13,427
And then, I get the sine of N
pi X divided by LDX.
382
00:30:13,427 --> 00:30:17,237
That is not between zero and
2L.
383
00:30:17,237 --> 00:30:23,872
You should have screamed.
That is only between zero and
384
00:30:23,872 --> 00:30:27,559
L.
It's only between zero and L
385
00:30:27,559 --> 00:30:33,456
that it is plus A.
But, when we integrate from L
386
00:30:33,456 --> 00:30:38,114
to 2L, then it becomes minus A.
So, the next part of the
387
00:30:38,114 --> 00:30:42,010
integral is minus A.
So, I bring the A outside.
388
00:30:42,010 --> 00:30:47,177
So, I get minus A divided by L
times the integral from L to 2L
389
00:30:47,177 --> 00:30:50,057
times the sine of N pi X over L
DX.
390
00:30:50,057 --> 00:30:54,715
So, the function comes in two
parts [UNINTELLIGIBLE] our
391
00:30:54,715 --> 00:30:59,374
Fourier analysis why the
function comes in more than two
392
00:30:59,374 --> 00:31:04,107
parts, believe me.
This is an easy case.
393
00:31:04,107 --> 00:31:08,822
That's why I chose it.
This integral is trivial,
394
00:31:08,822 --> 00:31:11,832
of course.
If I do an integral,
395
00:31:11,832 --> 00:31:17,050
I get minus the cosine.
So, I get A divided by L out,
396
00:31:17,050 --> 00:31:21,866
and I get M pi downstairs.
And, I get L upstairs.
397
00:31:21,866 --> 00:31:27,585
And then, I get the cosine of M
pi X over L, and I have to
398
00:31:27,585 --> 00:31:32,000
evaluate that between zero and
L.
399
00:31:32,000 --> 00:31:35,595
And here, I get a minus sign
now.
400
00:31:35,595 --> 00:31:41,438
So, I get plus A divided by L.
I get M pi downstairs.
401
00:31:41,438 --> 00:31:46,494
I get L upstairs,
and then I get the cosine of
402
00:31:46,494 --> 00:31:52,224
Mary pi X divided by L.
And, I have to evaluate that
403
00:31:52,224 --> 00:31:56,831
between L and 2L.
That's all I have to do.
404
00:31:56,831 --> 00:32:01,213
That is the integral,
a very simple one.
405
00:32:01,213 --> 00:32:05,033
Now, when M is odd,
this one alone,
406
00:32:05,033 --> 00:32:12,000
don't look at this,
this one alone is minus two.
407
00:32:12,000 --> 00:32:14,484
You have enough knowledge to
confirm that.
408
00:32:14,484 --> 00:32:17,757
You can see if you substitute X
equals L, for instance,
409
00:32:17,757 --> 00:32:19,878
you take M equals one,
which is odd,
410
00:32:19,878 --> 00:32:22,424
you get the cosine of pi.
That's minus one.
411
00:32:22,424 --> 00:32:24,545
And then, the cosine of zero is
one.
412
00:32:24,545 --> 00:32:27,939
But you have to subtract that.
You get another minus one.
413
00:32:27,939 --> 00:32:30,000
So, you get minus two.
414
00:32:30,000 --> 00:32:36,000
415
00:32:36,000 --> 00:32:38,818
And so this one,
if you do that,
416
00:32:38,818 --> 00:32:43,181
you get plus two.
Check it and you'll see that it
417
00:32:43,181 --> 00:32:48,454
is plus two because the borders,
the boundaries L to 2L are
418
00:32:48,454 --> 00:32:52,181
different.
It's not the same as zero to L.
419
00:32:52,181 --> 00:32:56,636
But, if M is even,
you get a zero here and you get
420
00:32:56,636 --> 00:33:00,090
a zero there.
And so, what comes out is
421
00:33:00,090 --> 00:33:05,000
exactly what we predicted,
that all even values of Mary
422
00:33:05,000 --> 00:33:10,090
will give zero values for V,
and that only the odd values
423
00:33:10,090 --> 00:33:15,000
will give me values that are not
zero.
424
00:33:15,000 --> 00:33:20,045
Before I write this out in a
complete Fourier form,
425
00:33:20,045 --> 00:33:26,100
you should appreciate the fact
that this entire integral here
426
00:33:26,100 --> 00:33:32,155
gives me exactly the same answer
as this entire integral here
427
00:33:32,155 --> 00:33:37,000
because minus times minus two is
plus two.
428
00:33:37,000 --> 00:33:40,962
And, this plus times plus two
is also plus two.
429
00:33:40,962 --> 00:33:46,044
And, that is always the case
when you have a string which is
430
00:33:46,044 --> 00:33:50,007
fixed at both ends,
that the zero to L integral
431
00:33:50,007 --> 00:33:55,434
always gives you the same answer
as this imaginary L to 2L which
432
00:33:55,434 --> 00:34:00,000
you introduced to make the
function periodic.
433
00:34:00,000 --> 00:34:05,109
It is for that and only for
that reason that Tony French
434
00:34:05,109 --> 00:34:10,775
gives you a much easier way to
calculate Fourier components in
435
00:34:10,775 --> 00:34:14,306
strings.
And he says all you have to do
436
00:34:14,306 --> 00:34:17,836
is say that B of M is two
divided by L.
437
00:34:17,836 --> 00:34:20,715
He multiplies the recipe by
two.
438
00:34:20,715 --> 00:34:24,524
So, instead of having,
where is my recipe,
439
00:34:24,524 --> 00:34:28,519
instead of having one divided
by L, he says,
440
00:34:28,519 --> 00:34:34,000
no, you should really [be?] two
divided by L.
441
00:34:34,000 --> 00:34:39,977
But then he says all you have
to do is now integrate between
442
00:34:39,977 --> 00:34:46,156
zero and L of the function of X
times the sine times Mary pi X
443
00:34:46,156 --> 00:34:51,222
divided by L times DX.
And, this is equation 632 in
444
00:34:51,222 --> 00:34:54,666
French.
If that's all we knew about
445
00:34:54,666 --> 00:34:59,124
Fourier analysis,
we would have a very narrow
446
00:34:59,124 --> 00:35:05,000
picture because it is an
extremely special case.
447
00:35:05,000 --> 00:35:10,992
But it's true that whenever you
have a string which is fixed at
448
00:35:10,992 --> 00:35:16,211
both ends, this will do.
I felt an obligation to you to
449
00:35:16,211 --> 00:35:22,107
show you the Fourier formalism,
which is a beautiful formalism
450
00:35:22,107 --> 00:35:27,037
in more general terms.
So, we are now ready to write
451
00:35:27,037 --> 00:35:31,000
down the complete Fourier
series.
452
00:35:31,000 --> 00:35:37,530
Notice that the sum of these
two become 4A divided by M pi.
453
00:35:37,530 --> 00:35:42,485
So, B of M is going to be 4A
divided by M pi.
454
00:35:42,485 --> 00:35:47,552
But, M is only odd.
And so, with that in mind,
455
00:35:47,552 --> 00:35:54,195
I can write down now here our
function, which was one of our
456
00:35:54,195 --> 00:35:59,262
big steps that Y,
I put here zero here because
457
00:35:59,262 --> 00:36:06,356
this is a time equals zero when
I have this string in this crazy
458
00:36:06,356 --> 00:36:13,000
shape can now be written as 4A
divided by M pi.
459
00:36:13,000 --> 00:36:16,469
I could write it down as simply
pi.
460
00:36:16,469 --> 00:36:22,693
And then I get here the sine pi
X divided by L plus one third.
461
00:36:22,693 --> 00:36:28,204
That is that M equals three,
which is three times lower
462
00:36:28,204 --> 00:36:34,326
because you have an M three here
times the sine of three pi X
463
00:36:34,326 --> 00:36:40,648
divided by L plus one fifth.
And then you get the sine of
464
00:36:40,648 --> 00:36:45,000
five pi X divided by L,
and so on, and so on.
465
00:36:45,000 --> 00:36:51,000
466
00:36:51,000 --> 00:36:55,941
So now I would like to put in
my function.
467
00:36:55,941 --> 00:37:01,002
I would like to put in the
first two terms.
468
00:37:01,002 --> 00:37:05,702
And I'll do that here on the
blackboard.
469
00:37:05,702 --> 00:37:13,053
So I'm going to concentrate now
only on my function zero to L.
470
00:37:13,053 --> 00:37:17,271
In reality, the string is only
here.
471
00:37:17,271 --> 00:37:22,935
And now I would like to put in
there, this is A.
472
00:37:22,935 --> 00:37:27,756
And now I would like to know,
what is B1?
473
00:37:27,756 --> 00:37:32,576
Well, B1 is going to be 4A
divided by pi,
474
00:37:32,576 --> 00:37:40,652
which is approximately 1.278.
Well, let's put it in there.
475
00:37:40,652 --> 00:37:46,526
This is A, so I have to put in
one third more so it comes up to
476
00:37:46,526 --> 00:37:49,842
this point roughly.
So, there it is.
477
00:37:49,842 --> 00:37:53,442
I love it.
It's not quite a square yet,
478
00:37:53,442 --> 00:37:58,463
but we're getting there.
We are on our way to building
479
00:37:58,463 --> 00:38:01,589
up the square.
The next one is B3.
480
00:38:01,589 --> 00:38:08,924
B3 is 4A divided by 3 pi.
So it is one third of 1.27,
481
00:38:08,924 --> 00:38:16,482
which is about 0.42 A.
And, B5 is one fifth times B1.
482
00:38:16,482 --> 00:38:23,604
Let's put in the B3.
So, B3 has an amplitude which
483
00:38:23,604 --> 00:38:30,000
is about 0.4,
which is about this much.
484
00:38:30,000 --> 00:38:33,375
It's about the height,
0.4, a little lower.
485
00:38:33,375 --> 00:38:35,946
And, so here we have the
maximum.
486
00:38:35,946 --> 00:38:39,964
And then, we have a zero here.
We have a zero here,
487
00:38:39,964 --> 00:38:43,821
and so we have here.
And then, we go through this
488
00:38:43,821 --> 00:38:46,794
point.
So, the curve is something like
489
00:38:46,794 --> 00:38:48,000
this.
490
00:38:48,000 --> 00:38:53,000
491
00:38:53,000 --> 00:38:54,875
So this, now,
is B3.
492
00:38:54,875 --> 00:38:58,035
And, this is B1.
Now, look at B3.
493
00:38:58,035 --> 00:39:01,787
Look at the beauty of Fourier
analysis.
494
00:39:01,787 --> 00:39:04,946
This is higher than it should
be.
495
00:39:04,946 --> 00:39:10,475
B3 says I'll take care of that.
I will subtract something
496
00:39:10,475 --> 00:39:13,734
there.
I will take that off there.
497
00:39:13,734 --> 00:39:18,473
Isn't that beautiful?
So, you're going to flatten
498
00:39:18,473 --> 00:39:22,422
this out already,
already with two terms,
499
00:39:22,422 --> 00:39:26,174
B1 and B3.
You are already beginning to
500
00:39:26,174 --> 00:39:32,000
see the building up of this
crazy square.
501
00:39:32,000 --> 00:39:34,685
And look here.
B1 says, sorry,
502
00:39:34,685 --> 00:39:39,870
I cannot fill this up for you.
This is the best I can do.
503
00:39:39,870 --> 00:39:45,333
B3 says I'm going to help you.
I'm going to help stop there,
504
00:39:45,333 --> 00:39:48,574
and it's going to do the same
there.
505
00:39:48,574 --> 00:39:54,037
And so, you see that if you now
keep adding odd values of M,
506
00:39:54,037 --> 00:39:58,111
that you will gradually
approach that square,
507
00:39:58,111 --> 00:40:03,574
which is an amazing concept.
And, what I will demonstrate to
508
00:40:03,574 --> 00:40:09,747
you is making a square.
And we will make it like this
509
00:40:09,747 --> 00:40:13,726
just like we did here.
We can do that on an
510
00:40:13,726 --> 00:40:17,894
oscilloscope.
We can have a signal that makes
511
00:40:17,894 --> 00:40:21,305
this trace.
Admittedly, this is time.
512
00:40:21,305 --> 00:40:24,905
But the way you'll see it,
it is space.
513
00:40:24,905 --> 00:40:28,410
So you can think of it as being
space.
514
00:40:28,410 --> 00:40:34,000
It's no different from the
function that we had.
515
00:40:34,000 --> 00:40:37,345
And then, we will show you the
B1 value.
516
00:40:37,345 --> 00:40:40,176
In this case,
since it is in time,
517
00:40:40,176 --> 00:40:43,608
it actually has a frequency.
It's 440 Hz.
518
00:40:43,608 --> 00:40:48,670
We can make you listen to it.
And then, we will show you the
519
00:40:48,670 --> 00:40:52,788
B3 value, which has three times
higher frequency.
520
00:40:52,788 --> 00:40:56,648
And then, we will show you the
B5, and the B7,
521
00:40:56,648 --> 00:41:00,166
and the B9.
And then, we will add them all
522
00:41:00,166 --> 00:41:04,026
up, and you will see,
it begins to look like a
523
00:41:04,026 --> 00:41:09,471
square.
And so, let us set that up,
524
00:41:09,471 --> 00:41:14,742
supposed to see that on the
central there.
525
00:41:14,742 --> 00:41:19,114
What you see here is simply the
B1.
526
00:41:19,114 --> 00:41:23,485
So you don't see very much,
do you?
527
00:41:23,485 --> 00:41:30,428
You just see a sine curve.
Well, you're looking at this
528
00:41:30,428 --> 00:41:35,185
sine curve.
Now I'm going to show,
529
00:41:35,185 --> 00:41:39,458
can we here it to Marcos?
Here's the 440 Hz.
530
00:41:39,458 --> 00:41:42,737
Now, I'm going to show you the
B3.
531
00:41:42,737 --> 00:41:47,307
Notice, it is three times
smaller in amplitude.
532
00:41:47,307 --> 00:41:53,070
That's the way we set it as a
three times higher frequency.
533
00:41:53,070 --> 00:41:59,131
And now, I'm going to show you
B5, which is five times smaller
534
00:41:59,131 --> 00:42:02,410
than the, hold it,
no, no, no, no,
535
00:42:02,410 --> 00:42:07,458
no.
OK, it's five times smaller
536
00:42:07,458 --> 00:42:10,916
than B1.
And, the frequency,
537
00:42:10,916 --> 00:42:14,886
of course, is five times
higher.
538
00:42:14,886 --> 00:42:18,472
You can hear it.
There it is.
539
00:42:18,472 --> 00:42:26,285
And now, I'm going to show you
seven, seven times smaller than
540
00:42:26,285 --> 00:42:30,000
B1.
And, here is B9.
541
00:42:30,000 --> 00:42:33,849
This is nine times smaller than
B1.
542
00:42:33,849 --> 00:42:39,169
And, now I'm going to show them
all five to you.
543
00:42:39,169 --> 00:42:45,283
And, it's really beginning to
look like a square pulse.
544
00:42:45,283 --> 00:42:50,150
It is clear that B1 is very,
very important.
545
00:42:50,150 --> 00:42:55,358
So, if I take the one out,
which I will do now,
546
00:42:55,358 --> 00:42:59,207
it doesn't even look like a
square.
547
00:42:59,207 --> 00:43:06,000
That shows you how important
that first term is.
548
00:43:06,000 --> 00:43:09,770
If I take B3 out,
that is less disastrous,
549
00:43:09,770 --> 00:43:12,804
but it is still pretty
disastrous.
550
00:43:12,804 --> 00:43:17,862
But you already begin to see
something that looks like a
551
00:43:17,862 --> 00:43:21,816
square slowly.
And it should be clear to you
552
00:43:21,816 --> 00:43:26,413
now that in order to get the
real sharp edges here,
553
00:43:26,413 --> 00:43:33,438
you need very high harmonics.
You have to go to M values of
554
00:43:33,438 --> 00:43:36,877
49, 101, 201.
They're all odd.
555
00:43:36,877 --> 00:43:43,636
You have to go to very high
values, and the higher you go,
556
00:43:43,636 --> 00:43:48,023
the closer you will get to the
square.
557
00:43:48,023 --> 00:43:52,885
So, this is a nice moment to
have a break.
558
00:43:52,885 --> 00:44:00,000
And I know you're dying to do
your fifth mini quiz.
559
00:44:00,000 --> 00:44:03,270
I will hand it out now,
and then we will all start at
560
00:44:03,270 --> 00:44:07,232
the same time with the mini quiz
so that each one of you has the
561
00:44:07,232 --> 00:44:10,000
same amount of time.
[SOUND OFF/THEN ON]
562
00:44:10,000 --> 00:44:16,000
563
00:44:16,000 --> 00:44:19,502
I owed you the histogram of the
exam.
564
00:44:19,502 --> 00:44:23,589
You see that here?
I think it's a wonderful
565
00:44:23,589 --> 00:44:26,800
histogram as far as I'm
concerned.
566
00:44:26,800 --> 00:44:31,664
It is too early to talk passing
and failing, and A,
567
00:44:31,664 --> 00:44:36,438
B, C's, and D's.
But, if I knew nothing else,
568
00:44:36,438 --> 00:44:40,531
no other grades,
then I would have to put you in
569
00:44:40,531 --> 00:44:44,102
the danger zone if you score
less than 45.
570
00:44:44,102 --> 00:44:48,282
That's all I can say.
I can add nothing more than
571
00:44:48,282 --> 00:44:51,243
what I already wrote you by
e-mail.
572
00:44:51,243 --> 00:44:56,033
So, let us now return to our
string is still at T equals
573
00:44:56,033 --> 00:44:59,168
zero.
So, our string is still in this
574
00:44:59,168 --> 00:45:05,012
position, dying to be released.
That was the goal.
575
00:45:05,012 --> 00:45:10,714
Remember, we would pluck it,
and we would let it go.
576
00:45:10,714 --> 00:45:16,416
At time T equals zero,
this is the Fourier analysis.
577
00:45:16,416 --> 00:45:21,670
These are the Fourier
components of that string.
578
00:45:21,670 --> 00:45:24,018
But now, I say OK,
go.
579
00:45:24,018 --> 00:45:30,503
And what happens now is that
this one is going to oscillate
580
00:45:30,503 --> 00:45:35,422
with omega 1T.
This one is going to oscillate
581
00:45:35,422 --> 00:45:41,571
omega 2T, omega 3T.
And this one is going to
582
00:45:41,571 --> 00:45:48,107
oscillate with cosine omega 5T.
Each one of these are standing
583
00:45:48,107 --> 00:45:51,857
waves.
And so, each one of these are
584
00:45:51,857 --> 00:45:58,178
going to do this with their own
amplitude and with their own
585
00:45:58,178 --> 00:46:02,664
frequency.
And, omega M is going to be V
586
00:46:02,664 --> 00:46:06,522
times K of M.
And, the speed of propagation
587
00:46:06,522 --> 00:46:09,921
is the square root of T divided
by mu.
588
00:46:09,921 --> 00:46:13,044
And, that's what's going to
happen.
589
00:46:13,044 --> 00:46:18,372
So, now, you may wonder what
you're going to see if you let
590
00:46:18,372 --> 00:46:22,506
this spring go.
And, chances are that if I ask
591
00:46:22,506 --> 00:46:27,834
you, what do you think will
happen, that you may think that
592
00:46:27,834 --> 00:46:31,784
this will happen with the string
as a whole.
593
00:46:31,784 --> 00:46:36,635
But that's not true.
And the best way,
594
00:46:36,635 --> 00:46:41,927
actually, to answer the
question of what you're going to
595
00:46:41,927 --> 00:46:47,701
see is to go back to a simple
case of a triangular pulse on a
596
00:46:47,701 --> 00:46:50,106
string.
I'll do that here.
597
00:46:50,106 --> 00:46:55,302
If I have a triangular pulse on
a string, I exaggerate,
598
00:46:55,302 --> 00:47:00,979
of course, the amplitude highly
fixed here at zero and fixed
599
00:47:00,979 --> 00:47:03,988
here at L.
And I let it go.
600
00:47:03,988 --> 00:47:07,693
Yes, I can decompose that in
terms of the Fourier components.
601
00:47:07,693 --> 00:47:10,843
And each of these Fourier
components are going [to?]
602
00:47:10,843 --> 00:47:13,931
standing waves up and down.
It doesn't give me much
603
00:47:13,931 --> 00:47:16,030
insight.
But, it is intriguing that
604
00:47:16,030 --> 00:47:18,933
that's what happens.
I can also think of it that
605
00:47:18,933 --> 00:47:21,897
nature is seeing a disturbance
and it says, well,
606
00:47:21,897 --> 00:47:24,491
we are going to propagate this
disturbance.
607
00:47:24,491 --> 00:47:27,579
Nature doesn't know the
difference between left and
608
00:47:27,579 --> 00:47:31,275
right.
And so, clearly will happen if
609
00:47:31,275 --> 00:47:35,296
this has in amplitude,
A, there will be a pulse was in
610
00:47:35,296 --> 00:47:39,165
amplitude of one half,
which goes in this direction,
611
00:47:39,165 --> 00:47:43,868
and one with the same one half,
which goes into that direction.
612
00:47:43,868 --> 00:47:48,648
And that is exactly what you're
going to see when you let it go.
613
00:47:48,648 --> 00:47:52,896
So, a little later in time,
you'll see this going in this
614
00:47:52,896 --> 00:47:55,475
direction, going in this
direction.
615
00:47:55,475 --> 00:47:59,420
At the moment that the top of
the mountain reaches L,
616
00:47:59,420 --> 00:48:02,531
and the top of this mountain
reaches zero,
617
00:48:02,531 --> 00:48:06,931
you're going to see nothing,
absolutely nothing because the
618
00:48:06,931 --> 00:48:13,000
reflected one and the incident
one exactly cancel each other.
619
00:48:13,000 --> 00:48:17,918
And then, a little later in
time, they will be on their way
620
00:48:17,918 --> 00:48:20,546
back.
And then a little later in
621
00:48:20,546 --> 00:48:25,380
time, you'll see this one but
completely flipped over when
622
00:48:25,380 --> 00:48:28,347
these two triangles meet each
other.
623
00:48:28,347 --> 00:48:32,757
And so, that's when you'll see
the thing upside down.
624
00:48:32,757 --> 00:48:36,997
So, there are two very
different ways of looking at
625
00:48:36,997 --> 00:48:42,000
what happens when you release a
plucked string.
626
00:48:42,000 --> 00:48:45,970
One way is to say they're
oscillations or standing waves a
627
00:48:45,970 --> 00:48:49,104
la Fourier, and the other way is
to say, well,
628
00:48:49,104 --> 00:48:53,213
we're going to always cut it in
two pulses, each of half the
629
00:48:53,213 --> 00:48:57,532
amplitude, and let them go back
and forth, and let them reflect
630
00:48:57,532 --> 00:49:00,527
at the ends.
Now, keep in mind that standing
631
00:49:00,527 --> 00:49:04,218
waves are the result of the
superposition of traveling
632
00:49:04,218 --> 00:49:06,656
waves.
So, the two different ways of
633
00:49:06,656 --> 00:49:10,000
looking are, of course,
connected.
634
00:49:10,000 --> 00:49:13,000
They have the same underlying
physics.
635
00:49:13,000 --> 00:49:18,027
And, now I want to demonstrate
this to you using a program that
636
00:49:18,027 --> 00:49:22,567
was developed by Professor
Wyslouch when he was lecturing
637
00:49:22,567 --> 00:49:26,783
8.03, and also Nergis.
I think she's in the audience.
638
00:49:26,783 --> 00:49:31,000
Nergis, are you hiding?
Yeah, you're hiding.
639
00:49:31,000 --> 00:49:34,251
Nergis has also worked on this
program.
640
00:49:34,251 --> 00:49:38,701
It's a wonderful program.
I'm going to put a triangle
641
00:49:38,701 --> 00:49:42,638
first on a string.
The width of the triangle is
642
00:49:42,638 --> 00:49:47,601
about one third of the length.
And then, I will release it.
643
00:49:47,601 --> 00:49:51,709
And I will let it go.
And we will see the Fourier
644
00:49:51,709 --> 00:49:53,933
components.
And at the end,
645
00:49:53,933 --> 00:49:58,982
when the whole show is over,
after one half the time for the
646
00:49:58,982 --> 00:50:03,261
fundamental period,
we will wait one half period of
647
00:50:03,261 --> 00:50:08,779
the fundamental.
We will inspect very closely
648
00:50:08,779 --> 00:50:14,139
the Fourier components.
Because of the symmetry of this
649
00:50:14,139 --> 00:50:19,698
problem, all even values of M
will have zero values of B.
650
00:50:19,698 --> 00:50:22,974
And, you will be able to see
that.
651
00:50:22,974 --> 00:50:28,632
Now, were we going to change
the light setting for this or
652
00:50:28,632 --> 00:50:30,121
were we not?
No?
653
00:50:30,121 --> 00:50:36,400
I thought we were.
Why don't you turn that off,
654
00:50:36,400 --> 00:50:40,530
also?
That's too difficult perhaps.
655
00:50:40,530 --> 00:50:44,052
No, the bar,
the bar, the bar.
656
00:50:44,052 --> 00:50:51,340
All right, so what you're going
to see first is the triangle.
657
00:50:51,340 --> 00:50:57,534
The number of Fourier
components that we have is 25.
658
00:50:57,534 --> 00:51:03,000
Did I do something wrong?
Excuse me?
659
00:51:03,000 --> 00:51:06,704
We have no what?
You broke the line?
660
00:51:06,704 --> 00:51:09,456
Oh, that's nice.
Thank you.
661
00:51:09,456 --> 00:51:12,948
The number of terms we have is
25.
662
00:51:12,948 --> 00:51:16,759
We have M equals one,
M equals three,
663
00:51:16,759 --> 00:51:20,463
M equals five,
all the way up to 49.
664
00:51:20,463 --> 00:51:26,496
That's when we cut it off.
All even values for M have no B
665
00:51:26,496 --> 00:51:29,459
value.
And, if you are ready,
666
00:51:29,459 --> 00:51:34,602
I'm ready.
So, let's first run this one.
667
00:51:34,602 --> 00:51:40,076
There, you see the triangle and
the blue lines are the Fourier
668
00:51:40,076 --> 00:51:43,846
components.
They are each doing this thing,
669
00:51:43,846 --> 00:51:47,705
their own thing,
standing waves up and down.
670
00:51:47,705 --> 00:51:52,192
But, the net result is
something that you are quite
671
00:51:52,192 --> 00:51:55,961
familiar with.
You see two pulses with half
672
00:51:55,961 --> 00:52:00,000
the amplitude of the original
one.
673
00:52:00,000 --> 00:52:03,147
They move to each their own
side.
674
00:52:03,147 --> 00:52:06,295
They reflect,
and they come back.
675
00:52:06,295 --> 00:52:11,704
And then, we'll stop when we
have half the period of the
676
00:52:11,704 --> 00:52:15,540
fundamental.
Now, look at the individual
677
00:52:15,540 --> 00:52:18,491
Fourier components.
This is B1.
678
00:52:18,491 --> 00:52:22,819
This one is B3.
You see, it helps building up
679
00:52:22,819 --> 00:52:25,278
that triangle.
This is B5.
680
00:52:25,278 --> 00:52:30,000
It helps building up the
triangle.
681
00:52:30,000 --> 00:52:34,537
This is B7 and this is B9.
And here, they all conspire.
682
00:52:34,537 --> 00:52:38,235
It's amazing.
There are 25 sinusoids that are
683
00:52:38,235 --> 00:52:41,848
conspiring there.
And they show you nothing,
684
00:52:41,848 --> 00:52:45,798
and the same here.
And it's hard to believe that
685
00:52:45,798 --> 00:52:49,411
all these sinusoids can add up
to zero here,
686
00:52:49,411 --> 00:52:52,521
and how beautifully they
connect here.
687
00:52:52,521 --> 00:52:57,647
And, the only reason why this
tip is not very sharp is we only
688
00:52:57,647 --> 00:53:02,022
have 25 terms.
If I would run it with 400
689
00:53:02,022 --> 00:53:04,668
terms, then you'd see this
sharper.
690
00:53:04,668 --> 00:53:07,235
But it would take a lot more
time.
691
00:53:07,235 --> 00:53:11,281
Let's look at the Fourier
components which you'll see
692
00:53:11,281 --> 00:53:12,837
here.
So, this is B1,
693
00:53:12,837 --> 00:53:15,638
which we have arbitrarily
called one.
694
00:53:15,638 --> 00:53:19,061
You see, all the even values,
this is B2, B4,
695
00:53:19,061 --> 00:53:21,862
B6, B8.
All the even values are zero.
696
00:53:21,862 --> 00:53:24,274
The B3 is negative in this
case.
697
00:53:24,274 --> 00:53:27,075
That was not the case for our
square.
698
00:53:27,075 --> 00:53:30,421
Remember, in the case of the
square, B1, B3,
699
00:53:30,421 --> 00:53:36,688
B5, B7 were all positive.
They all had the same sign.
700
00:53:36,688 --> 00:53:40,372
That's not the case with the
triangle.
701
00:53:40,372 --> 00:53:44,952
They alternate.
The next one that I want to run
702
00:53:44,952 --> 00:53:50,826
for you is a square which has a
width half the length of the
703
00:53:50,826 --> 00:53:54,012
string.
And it is centered at the
704
00:53:54,012 --> 00:53:57,000
middle of the string.
705
00:53:57,000 --> 00:54:01,000
706
00:54:01,000 --> 00:54:05,970
And again, two pulses are each
going their own direction with
707
00:54:05,970 --> 00:54:09,698
half the amplitudes.
They are being reflected.
708
00:54:09,698 --> 00:54:12,763
And strange things happen at
the ends.
709
00:54:12,763 --> 00:54:17,568
And now there comes a time one
quarter of the period of the
710
00:54:17,568 --> 00:54:20,218
fundamental that you see
nothing.
711
00:54:20,218 --> 00:54:24,195
And now they are coming back
from the reflection.
712
00:54:24,195 --> 00:54:29,000
Very different from what you
expect, isn't it?
713
00:54:29,000 --> 00:54:32,367
Now, look at B1.
That's our real biggie.
714
00:54:32,367 --> 00:54:35,908
Look at B1, boy,
here, is that a surprise?
715
00:54:35,908 --> 00:54:39,793
No, because clearly B1 is not
quite as square.
716
00:54:39,793 --> 00:54:42,470
But it's almost a square,
right?
717
00:54:42,470 --> 00:54:45,924
You just have to subtract
something here.
718
00:54:45,924 --> 00:54:49,378
Well, B3 says,
that's OK, I will subtract
719
00:54:49,378 --> 00:54:52,142
there, and B3 will subtract
here.
720
00:54:52,142 --> 00:54:54,905
And, B1 is a little shallow
here.
721
00:54:54,905 --> 00:55:00,000
It's a little bit too,
not generous enough.
722
00:55:00,000 --> 00:55:03,510
And then B3 says,
OK, I'll help you building it
723
00:55:03,510 --> 00:55:05,647
up.
And so, if we look at the
724
00:55:05,647 --> 00:55:08,395
Fourier components,
you see now that,
725
00:55:08,395 --> 00:55:12,898
again, all the even values are
zero, B1, the definition one,
726
00:55:12,898 --> 00:55:15,645
and now you see this one is
negative.
727
00:55:15,645 --> 00:55:18,698
And, B5 is also negative,
not so obvious.
728
00:55:18,698 --> 00:55:21,598
And B7 is positive,
and B9 is positive.
729
00:55:21,598 --> 00:55:25,949
And now, I'll show you one
whereby I'm going to offset the
730
00:55:25,949 --> 00:55:28,772
square.
The square is only one quarter
731
00:55:28,772 --> 00:55:31,902
of the length,
and I'm going to offset it.
732
00:55:31,902 --> 00:55:36,405
And, when I offset it from the
center, you need both odd and
733
00:55:36,405 --> 00:55:41,449
even values.
And you're going to see that.
734
00:55:41,449 --> 00:55:46,179
So, now you cannot just get
away with only even values of
735
00:55:46,179 --> 00:55:48,712
Mary.
So, here you see the wave
736
00:55:48,712 --> 00:55:52,260
spinning up in two,
just like you expected.
737
00:55:52,260 --> 00:55:56,652
It has a width a quarter of the
length of the string.
738
00:55:56,652 --> 00:56:01,719
This one has already reflected.
This one is now on its way to
739
00:56:01,719 --> 00:56:06,353
being reflected.
Now this one is coming back.
740
00:56:06,353 --> 00:56:09,547
There it is.
This one is already on its
741
00:56:09,547 --> 00:56:12,742
return.
And of course they are going to
742
00:56:12,742 --> 00:56:16,272
meet again.
And now, we are one half period
743
00:56:16,272 --> 00:56:19,466
of the fundamental later.
And now look.
744
00:56:19,466 --> 00:56:24,678
This now is the first harmonic.
But now, there is also a second
745
00:56:24,678 --> 00:56:27,704
harmonic.
M equals two has a B value.
746
00:56:27,704 --> 00:56:30,394
Here it is.
This is the B2 value,
747
00:56:30,394 --> 00:56:34,388
large amplitude.
You noticed that?
748
00:56:34,388 --> 00:56:39,208
And, look how important B4 is.
I think this is the B4 value.
749
00:56:39,208 --> 00:56:42,639
Yes indeed, here is the B3.
This is the B4.
750
00:56:42,639 --> 00:56:46,969
It has a huge amplitude.
And look how important it is,
751
00:56:46,969 --> 00:56:51,870
because right here where the
pulse is, you need this bulge to
752
00:56:51,870 --> 00:56:56,690
push this further out because
the B1 alone cannot reach that
753
00:56:56,690 --> 00:56:59,140
point.
Let's take a look at the
754
00:56:59,140 --> 00:57:03,224
Fourier spectrum here.
Here you see B1,
755
00:57:03,224 --> 00:57:05,751
which we, by definition,
call one.
756
00:57:05,751 --> 00:57:08,124
You see B2 is there.
It's large.
757
00:57:08,124 --> 00:57:12,258
It's a little larger than 0.7.
And then, you have a B3,
758
00:57:12,258 --> 00:57:15,779
which is negative,
and a B4, which is negative,
759
00:57:15,779 --> 00:57:19,148
and a B5, which is negative.
B4 is off scale.
760
00:57:19,148 --> 00:57:22,287
And now, you get B6 positive,
B7 positive.
761
00:57:22,287 --> 00:57:26,344
B8 is zero, and B16 is zero.
Well, that's probably the
762
00:57:26,344 --> 00:57:31,167
result of the way we set up the
square, which has a width of one
763
00:57:31,167 --> 00:57:36,329
quarter of a length.
And we offset it from the
764
00:57:36,329 --> 00:57:41,640
middle by one quarter of the
length, remarkable example of
765
00:57:41,640 --> 00:57:46,300
Fourier analysis now,
which is way more complicated
766
00:57:46,300 --> 00:57:50,493
than if you nicely center it.
All values of B,
767
00:57:50,493 --> 00:57:53,754
now, all values of M,
are important.
768
00:57:53,754 --> 00:57:58,506
You will now understand,
when I said if we pluck the
769
00:57:58,506 --> 00:58:04,004
string of a harp or the string
of a violin, and you pluck it
770
00:58:04,004 --> 00:58:09,688
this way, fixed here and fixed
here, the tone that you hear is
771
00:58:09,688 --> 00:58:15,000
different than when you pluck it
this way.
772
00:58:15,000 --> 00:58:19,388
You now understand that because
the Fourier components of this
773
00:58:19,388 --> 00:58:23,057
one and this one are very
different than the Fourier
774
00:58:23,057 --> 00:58:27,014
components from this one.
And so, when he let the string
775
00:58:27,014 --> 00:58:31,546
go, the frequencies that will be
produced are at the frequencies
776
00:58:31,546 --> 00:58:35,000
of these normal modes of these B
values.
777
00:58:35,000 --> 00:58:39,116
But the various B values are
very different here from there.
778
00:58:39,116 --> 00:58:43,511
And so, the sound that you hear
is distinctly different when you
779
00:58:43,511 --> 00:58:47,906
pluck a string on the side than
when you pluck it in the middle.
780
00:58:47,906 --> 00:58:50,767
You may remember that during
that lecture,
781
00:58:50,767 --> 00:58:55,093
I told you that the hammer of a
piano hits the string about 1/7
782
00:58:55,093 --> 00:58:57,953
of the length of the string from
one side.
783
00:58:57,953 --> 00:59:02,000
That is done to suppress the
seventh harmonic.
784
00:59:02,000 --> 00:59:06,122
For some reason that beats me,
people don't like the seventh
785
00:59:06,122 --> 00:59:08,917
harmonic.
So that's the way they kill it.
786
00:59:08,917 --> 00:59:12,759
So, it does depend on where you
pluck and where you hit.
787
00:59:12,759 --> 00:59:16,323
And, now you can put that in
context because now you
788
00:59:16,323 --> 00:59:20,585
understand that you analyze the
thing here in terms of Fourier
789
00:59:20,585 --> 00:59:22,960
components.
And, each one will then
790
00:59:22,960 --> 00:59:26,943
oscillate at their own frequency
with their own amplitude.
791
00:59:26,943 --> 00:59:31,205
So, what we have discussed now
at length is a Fourier analysis
792
00:59:31,205 --> 00:59:37,263
in space in X in meters.
And we decomposed a function in
793
00:59:37,263 --> 00:59:42,421
the sum of many harmonics,
which together make up,
794
00:59:42,421 --> 00:59:48,210
then, the shape of the string.
Now, if we look at sound,
795
00:59:48,210 --> 00:59:53,263
sound, of course,
is something that is a function
796
00:59:53,263 --> 00:59:56,947
in time.
Suppose I look at the sound
797
00:59:56,947 --> 1:00:02,000
signal made by a tuning fork of
440 Hz.
798
1:00:02,000 --> 1:00:05,750
And I have here that signal in
time.
799
1:00:05,750 --> 1:00:10,142
Suppose I have one second of
data, 440 Hz.
800
1:00:10,142 --> 1:00:15,178
So this is now time.
But if you want to think of
801
1:00:15,178 --> 1:00:18,928
this as being in space,
be my guest.
802
1:00:18,928 --> 1:00:24,285
Now, there are special
programs, special algorithms
803
1:00:24,285 --> 1:00:29,750
that take this one second of
data, perform a Fourier
804
1:00:29,750 --> 1:00:33,375
analysis.
They have a name.
805
1:00:33,375 --> 1:00:36,289
We call them fast Fourier
transforms.
806
1:00:36,289 --> 1:00:41,387
And, it's going to tell me what
the amplitude of B1 is and A1 of
807
1:00:41,387 --> 1:00:44,057
B2 and A2 of B3 and A3,
and so on.
808
1:00:44,057 --> 1:00:48,508
If I now think in terms of
hertz rather than in terms of
809
1:00:48,508 --> 1:00:51,421
omega, there is nothing at one
hertz.
810
1:00:51,421 --> 1:00:55,872
So, though A's and B's are
zero, there is nothing at two
811
1:00:55,872 --> 1:01:00,000
hertz.
So, those A's and B's are zero.
812
1:01:00,000 --> 1:01:05,896
But by the time I reach 440 Hz,
the system says,
813
1:01:05,896 --> 1:01:11,667
yippee, there's a lot of power
there at 440 Hz.
814
1:01:11,667 --> 1:01:17,940
And what we now do,
we make a plot of the square of
815
1:01:17,940 --> 1:01:24,464
A plus the square of B.
And, the reason why we square
816
1:01:24,464 --> 1:01:31,741
them is that energy is always
proportional to the amplitude
817
1:01:31,741 --> 1:01:36,366
squared.
And so, you take the A squares
818
1:01:36,366 --> 1:01:39,126
and the B squares,
and you add them.
819
1:01:39,126 --> 1:01:43,701
We call those a power density
spectrum, a Fourier spectrum.
820
1:01:43,701 --> 1:01:48,118
And, here you have your omega
or you have your F value in
821
1:01:48,118 --> 1:01:50,800
hertz.
The difference is only 2 pi.
822
1:01:50,800 --> 1:01:54,901
And, if you now do a Fourier
analysis of that signal,
823
1:01:54,901 --> 1:01:57,504
you'll see at 440 Hz a huge
value.
824
1:01:57,504 --> 1:02:02,000
And you'll see almost nothing
anywhere else.
825
1:02:02,000 --> 1:02:07,073
So, you have now done a Fourier
analysis of a time signal.
826
1:02:07,073 --> 1:02:11,879
And you have decomposed it into
its Fourier components,
827
1:02:11,879 --> 1:02:14,816
which in this case is only 440
Hz.
828
1:02:14,816 --> 1:02:19,267
If you did the same with the
middle A of the piano,
829
1:02:19,267 --> 1:02:23,450
which is 440 Hz,
you would see a big value here.
830
1:02:23,450 --> 1:02:27,010
But, you would see also
something at 880.
831
1:02:27,010 --> 1:02:31,727
And, you may even see something
at the third harmonic.
832
1:02:31,727 --> 1:02:36,000
And, the same is true for a
piano.
833
1:02:36,000 --> 1:02:39,531
And it is also for a violin.
If you take a violin,
834
1:02:39,531 --> 1:02:42,702
and you ask the violinist to
produce the 440,
835
1:02:42,702 --> 1:02:45,801
you always get a little bit
more of the 880.
836
1:02:45,801 --> 1:02:49,261
And, you get some of three
times the fundamental.
837
1:02:49,261 --> 1:02:52,576
And, this is something that we
can demonstrate.
838
1:02:52,576 --> 1:02:56,612
We have here an admittedly
somewhat poor man's version of
839
1:02:56,612 --> 1:03:00,000
what's called the Fourier
analyzer.
840
1:03:00,000 --> 1:03:05,558
We take in sound for about 1/5
of the second period and then we
841
1:03:05,558 --> 1:03:09,951
asked the computer to Fourier
analyze that for us,
842
1:03:09,951 --> 1:03:14,703
and show us the Fourier
spectrum in terms of where the
843
1:03:14,703 --> 1:03:18,648
frequencies were.
And, we will see that here,
844
1:03:18,648 --> 1:03:23,220
for which I think we also
invented the special light
845
1:03:23,220 --> 1:03:26,000
condition, didn't we?
846
1:03:26,000 --> 1:03:30,000
847
1:03:30,000 --> 1:03:35,921
The scale that you see from
left to right is 2 kHz.
848
1:03:35,921 --> 1:03:39,947
I'll first make you see the 440
Hz.
849
1:03:39,947 --> 1:03:44,328
[SOUND PLAYS] You see that huge
spike?
850
1:03:44,328 --> 1:03:49,894
That's the 440 Hz.
So, the end of the scale is 2
851
1:03:49,894 --> 1:03:52,855
kHz.
[SOUND PLAYS] 256 Hz.
852
1:03:52,855 --> 1:04:05,000
[SOUND PLAYS] It's lower.
It's lower.
853
1:04:05,000 --> 1:04:08,888
I have here a flute.
And remember during the lecture
854
1:04:08,888 --> 1:04:11,557
on musical instruments.
We believed,
855
1:04:11,557 --> 1:04:16,208
and I demonstrated that you can
really think this one resonate
856
1:04:16,208 --> 1:04:20,249
only in its fundamental.
I will now show you with this
857
1:04:20,249 --> 1:04:24,900
Fourier analyzer that there is
also a little bit of the second
858
1:04:24,900 --> 1:04:27,873
harmonic, maybe even the third
harmonic.
859
1:04:27,873 --> 1:04:31,000
Look at it now.
[SOUND PLAYS]
860
1:04:31,000 --> 1:04:35,415
You see that?
That was the fundamental second
861
1:04:35,415 --> 1:04:39,529
and third harmonic even.
[SOUND PLAYS] Ah.
862
1:04:39,529 --> 1:04:45,349
So this is a beautiful way of
analyzing the frequencies and
863
1:04:45,349 --> 1:04:48,259
sound.
I will whistle for you.
864
1:04:48,259 --> 1:04:53,076
[SOUND PLAYS] Amazing.
Would you have guessed it?
865
1:04:53,076 --> 1:04:57,491
Very nicely separated.
What is the frequency?
866
1:04:57,491 --> 1:05:02,404
[SOUND PLAYS]
It's about here.
867
1:05:02,404 --> 1:05:07,963
That's about 1 kHz.
It's about 900 Hz.
868
1:05:07,963 --> 1:05:13,673
[SOUND PLAYS] And,
the second harmonic,
869
1:05:13,673 --> 1:05:17,580
not so obvious,
my whistle.
870
1:05:17,580 --> 1:05:24,642
Very high frequency.
[SOUND PLAYS] It is so high
871
1:05:24,642 --> 1:05:32,717
that it's off scale.
And I did that purposely
872
1:05:32,717 --> 1:05:38,138
because I know you hate the
whistle.
873
1:05:38,138 --> 1:05:46,501
So, I didn't want you to see
what it really looks like.
874
1:05:46,501 --> 1:05:55,020
We can listen to the radio.
And then we can do a Fourier
875
1:05:55,020 --> 1:06:01,525
analysis on what we hear.
[RADIO PLAYS] Oh,
876
1:06:01,525 --> 1:06:06,326
I don't have a question for
you.
877
1:06:06,326 --> 1:06:11,592
I'm sorry.
[LAUGHTER] [RADIO PLAYS]
878
1:06:11,592 --> 1:06:36,449
[SINGING]
Not the right type for me.
879
1:06:36,449 --> 1:06:51,393
[MUSIC PLAYS]
Too complicated for me.
880
1:06:51,393 --> 1:06:56,704
Any one of you want to sing?
Come on.
881
1:06:56,704 --> 1:07:00,983
Be brave.
Anyone want to sing?
882
1:07:00,983 --> 1:07:02,901
Yeah.
Come on.
883
1:07:02,901 --> 1:07:07,764
Give it a try.
Nice to be here.
884
1:07:07,764 --> 1:07:11,403
Believe me.
You can see all these,
885
1:07:11,403 --> 1:07:17,136
OK, it's your decision.
All right, so you see how you
886
1:07:17,136 --> 1:07:23,532
can do a Fourier analysis of
sound and then I decompose the
887
1:07:23,532 --> 1:07:28,053
signal in terms of its Fourier
components.
888
1:07:28,053 --> 1:07:35,000
Neutron stars were discovered
in 1967 by Jocelyn Bell.
889
1:07:35,000 --> 1:07:38,950
Jocelyn was a graduate student
in Cambridge,
890
1:07:38,950 --> 1:07:42,533
England at the time.
And her supervisor,
891
1:07:42,533 --> 1:07:46,575
Anthony Hewish,
had built a new type of radio
892
1:07:46,575 --> 1:07:49,699
telescope.
And she was in charge of
893
1:07:49,699 --> 1:07:54,201
analyzing the data.
And, she was sure that she had
894
1:07:54,201 --> 1:07:59,162
received a periodic signal,
which came almost every 1.3
895
1:07:59,162 --> 1:08:03,078
seconds.
They realized that that could
896
1:08:03,078 --> 1:08:07,236
have been the discovery of the
century because they thought
897
1:08:07,236 --> 1:08:11,322
that they were receiving signals
from the intelligent life
898
1:08:11,322 --> 1:08:15,050
elsewhere in the universe.
And so, they called in the
899
1:08:15,050 --> 1:08:18,347
little green man.
And then, a few months later,
900
1:08:18,347 --> 1:08:22,648
Jocelyn discovered another one.
And so, they called the first
901
1:08:22,648 --> 1:08:26,591
one little green man one,
and then little green man two,
902
1:08:26,591 --> 1:08:29,745
LGM1 and LGM2.
And then, when they discovered
903
1:08:29,745 --> 1:08:32,971
a third one, they abandoned the
idea of LGM's.
904
1:08:32,971 --> 1:08:36,985
We now call these pulsars,
and we know that the period is
905
1:08:36,985 --> 1:08:41,000
the spin period of the neutron
stars.
906
1:08:41,000 --> 1:08:45,371
In 1974, Anthony Hewish
received the Nobel Prize for
907
1:08:45,371 --> 1:08:47,857
this discovery.
It is a shame,
908
1:08:47,857 --> 1:08:52,399
and it is scandalous that
Jocelyn did not share in the
909
1:08:52,399 --> 1:08:55,228
Nobel Prize.
She actually made the
910
1:08:55,228 --> 1:08:59,000
discovery.
I've known her very well.
911
1:08:59,000 --> 1:09:01,839
I've discussed it with her many
times.
912
1:09:01,839 --> 1:09:06,138
She was a graduate student.
Maybe that was the reason why
913
1:09:06,138 --> 1:09:10,743
the Nobel committee didn't think
it was appropriate to give a
914
1:09:10,743 --> 1:09:13,276
Nobel Prize to a graduate
student.
915
1:09:13,276 --> 1:09:15,732
It's ridiculous but perhaps
true.
916
1:09:15,732 --> 1:09:17,958
She was a woman,
and there is,
917
1:09:17,958 --> 1:09:21,949
perhaps, a very sad case of sex
discrimination again.
918
1:09:21,949 --> 1:09:25,557
We will never know,
but she did not share in the
919
1:09:25,557 --> 1:09:28,013
Nobel Prize that she should
have.
920
1:09:28,013 --> 1:09:34,000
Neutron stars have a mass about
1 1/2 times that of the sun.
921
1:09:34,000 --> 1:09:38,559
They are 100,000 times smaller.
There are only 20 km across.
922
1:09:38,559 --> 1:09:43,196
So, they have a density which
is ten to the 15th times higher
923
1:09:43,196 --> 1:09:46,365
than the sun,
ten to the 15th times higher
924
1:09:46,365 --> 1:09:50,074
than water, which is higher than
nuclear density.
925
1:09:50,074 --> 1:09:54,557
And, I had some e-mail contact
with Jocelyn does a few days
926
1:09:54,557 --> 1:09:56,412
ago.
She's now at Oxford.
927
1:09:56,412 --> 1:10:00,662
And I said, I really would like
to show the class you're
928
1:10:00,662 --> 1:10:03,599
picture.
And so, she sent me a picture,
929
1:10:03,599 --> 1:10:09,113
which I'm going to show you.
Are we still connected?
930
1:10:09,113 --> 1:10:12,237
Where's Jocelyn?
Is Jocelyn hiding?
931
1:10:12,237 --> 1:10:14,994
No, there she is.
There she is.
932
1:10:14,994 --> 1:10:19,589
She sent me a picture at the
time that she made the
933
1:10:19,589 --> 1:10:22,989
discovery.
You see the radio telescope
934
1:10:22,989 --> 1:10:26,205
here.
And you see her standing there
935
1:10:26,205 --> 1:10:29,881
very modestly.
She's a very modest woman.
936
1:10:29,881 --> 1:10:35,027
And she made one of the most
important discoveries of the
937
1:10:35,027 --> 1:10:41,000
past century and did not share
in the Nobel Prize.
938
1:10:41,000 --> 1:10:45,748
We now know of hundreds of
pulsars in the sky.
939
1:10:45,748 --> 1:10:51,025
And so, the question,
now, is how do you find these
940
1:10:51,025 --> 1:10:54,824
pulsars?
You can observe the sky with
941
1:10:54,824 --> 1:10:59,678
radio telescopes.
You can do it also with x-ray
942
1:10:59,678 --> 1:11:03,450
observatories.
And what you do,
943
1:11:03,450 --> 1:11:08,708
now, if you take the data just
like we took the data of our
944
1:11:08,708 --> 1:11:12,153
sound signal,
and you perform a Fourier
945
1:11:12,153 --> 1:11:14,692
transform.
You ask that data,
946
1:11:14,692 --> 1:11:20,403
what are the Fourier components
that I hid that I cannot see but
947
1:11:20,403 --> 1:11:22,579
are hidden?
And, in 1998,
948
1:11:22,579 --> 1:11:27,928
Rudy [Winons?] in Amsterdam,
an independently at Morgan here
949
1:11:27,928 --> 1:11:35,000
at MIT, were analyzing data from
a no x-ray source in our galaxy.
950
1:11:35,000 --> 1:11:38,702
And they were performing a
Fourier analysis,
951
1:11:38,702 --> 1:11:42,061
which is standard nowadays in
astronomy.
952
1:11:42,061 --> 1:11:47,227
And they discovered that the
neutral star was rotating with a
953
1:11:47,227 --> 1:11:50,069
spin period of 2 1/2
milliseconds.
954
1:11:50,069 --> 1:11:54,977
They noticed in the Fourier
power spectrum a huge spike at
955
1:11:54,977 --> 1:11:58,250
401 Hz.
It means that at the equator of
956
1:11:58,250 --> 1:12:01,952
the neutral star,
the speed going around was
957
1:12:01,952 --> 1:12:06,000
about 1/10 of the speed of
light.
958
1:12:06,000 --> 1:12:10,468
And I asked [UNINTELLIGIBLE]
who would worked with me for
959
1:12:10,468 --> 1:12:15,256
several years here at MIT to
send me some of the data that he
960
1:12:15,256 --> 1:12:19,805
obtained from which he could,
then, finally derive that we
961
1:12:19,805 --> 1:12:24,513
were dealing with a neutron star
rotating around at two half
962
1:12:24,513 --> 1:12:26,827
milliseconds.
And so, he said,
963
1:12:26,827 --> 1:12:30,099
well, Walter,
why don't you show the class
964
1:12:30,099 --> 1:12:35,047
only 1/5 of a second of my data,
1/5 of the second that you see
965
1:12:35,047 --> 1:12:38,000
here?
2/10 of a second?
966
1:12:38,000 --> 1:12:42,938
So, this is the timescale.
And, when one x arrives,
967
1:12:42,938 --> 1:12:47,185
you see a vertical bar there.
And this time,
968
1:12:47,185 --> 1:12:52,814
they were so close together
that you see there two x-rays.
969
1:12:52,814 --> 1:12:58,246
If you count the total number
of x-rays in that 1/5 of a
970
1:12:58,246 --> 1:13:03,311
second, you observe 33 x-rays.
During that time,
971
1:13:03,311 --> 1:13:07,938
the neutron star was rotating
80 times around already because
972
1:13:07,938 --> 1:13:10,868
it rotates every two half
milliseconds.
973
1:13:10,868 --> 1:13:15,419
So, when you look at the data,
you have no idea that here is
974
1:13:15,419 --> 1:13:20,123
an underlying neutron star with
a period of 2 1/2 milliseconds
975
1:13:20,123 --> 1:13:24,519
with a frequency of 401 Hz.
But now you take 3,000 seconds
976
1:13:24,519 --> 1:13:27,218
of data.
And if you have taken 8.03,
977
1:13:27,218 --> 1:13:32,000
you know how to perform a fast
Fourier transform.
978
1:13:32,000 --> 1:13:34,939
Those programs,
by the way, are readily
979
1:13:34,939 --> 1:13:38,651
available on the market.
And here you see a power
980
1:13:38,651 --> 1:13:42,209
density spectrum.
Vertically, it's the sum of A
981
1:13:42,209 --> 1:13:45,303
squared plus B squared.
And horizontally,
982
1:13:45,303 --> 1:13:49,171
it's the frequency in hertz.
And look what you see.
983
1:13:49,171 --> 1:13:51,569
At 401 Hz, you see a huge
spike.
984
1:13:51,569 --> 1:13:56,055
And that is the underlying 2
1/2 milliseconds neutron star.
985
1:13:56,055 --> 1:14:00,000
This is the frontier of
astrophysics.
986
1:14:00,000 --> 1:14:04,541
Nowadays, you cannot even think
of astronomy or astrophysics
987
1:14:04,541 --> 1:14:08,775
without Fourier analysis.
It has an enormously important
988
1:14:08,775 --> 1:14:12,624
impact on our research.
My graduate students and my
989
1:14:12,624 --> 1:14:16,088
post-docs perform Fourier
transforms everyday.
990
1:14:16,088 --> 1:14:20,861
And so, what we have discussed
today is not just intellectually
991
1:14:20,861 --> 1:14:23,709
interesting.
It is at the forefront of
992
1:14:23,709 --> 1:14:24,402
research.
See you Thursday.