So today we're going to discuss Fourier analysis.
You'll see a lot of math, but I'll try to keep an eye on
the physics. Imagine that I pluck a string
which is fixed at both ends. So, here is X and here is Y.
And, this is the string at T equals zero.
And, imagine I let it go. And we know from what we have
learned that it will start to oscillate in the superposition
of its normal modes, which are standing waves.
And so, the $64 million question today is now,
in which modes will oscillate, and what are the [NOISE
OBSCURES] of those modes? If you write down Y as a
function of X and T, then we know from what we have
learned to date that you should be able to write this,
N equals one to infinity of amplitudes which I indicate with
a B now rather than with an A. But B is, of course,
an amplitude, has a dimension length,
times the sine of KNX times the cosine of omega NT.
And, because of the boundary conditions, K of N,
because the string is fixed at both ends, K of N is N pi
divided by L, and omega N is the velocity
propagation times K of N. And N, then,
is one, two, three, four,
five. So, we know this.
There is nothing new. We've had this recently.
So, I can also write down, then, that at T equals zero,
I can write down this shape in terms of a series of signs.
I will leave the time off now. So, at T equals zero,
then we have YX at time T equals zero would be B1 times
the sine of pi X over L plus B2 times the sine of 2 pi X over L
plus B3, and so on. And our task now today is what
values of N are necessary to make that pulse,
and what are, then, the amplitudes,
B, that we have here? And, that's what Fourier
analysis is all about. Fourier was born in 1768.
And, he died in 1830. So, this is work that was the
long time ago, a brilliant mathematician.
Now, I will give you the general approach first.
And then, I will come back to the special case of a string,
which is fixed at both ends. But I really think I owe you
the general format, the general procedure first.
So the idea behind Fourier, then, is that any periodic,
single valued function with period of 2 pi radians can be
represented by a Fourier series. And, I will write down this
Fourier series here. So, this is now a function of
X. X is now in radians.
It's not that X. I will come back to that.
So, a function of X can now be written as a constant.
We call it A zero divided by two, plus the sum,
one to infinity of AM cosine MX plus the sum of one to infinity
of BM times the sine of MX. And this X that you see here is
also called X in Beckafee [SP?] and Barrett is in radians.
I will call this first term one.
I will call this one two, and I will call this one three.
So, our task, now, is to find the values for
A and B if you know the function, FX.
And I will build up with you the recipe that allows you to
calculate the values for A and the values for B.
What you do, the first thing you do to find
A zero, you take the integral from minus pi to plus pi DX on
both sides of the aisle. This one side of the aisle,
and this is the other side of the aisle.
And when you do that, you will see that for all
values of M, Mary, all terms here will be zero.
And, the same is true for three.
For any value of M that you take, if you do an integral over
a sine over a complete period from minus pi to pi,
there is a complete period. You get zero,
of course. So, all values for three will
be zero. And so, what you're left with
is, then, that the integral from minus pi to plus pi of XXDX is,
then, the integral from minus pi to plus pi of A zero divided
by two DX. And, that is pi times A zero.
That's immediately obvious. So, we now already have our
first way of calculating A zero. A zero, if I use this result,
is nothing but one over pi times the integral from minus pi
to plus pi times the function, X, of DX.
So, you tell me what the function of X is,
and I can calculate A zero for you.
Keep in mind that A zero divided by two is nothing but
the average value of the function over the period two pi.
And, I will come back to that later during this lecture.
How are we going to calculate, now, the other values for A?
Well, what you do now is you take the integral for minus pi
to plus pi times the cosine of Nancy X DX.
And, you do that on both sides of the aisle.
You do it on the left side, and you do it on the right
side. And Nancy, now,
is one, two, three, and so on.
When you do that, you'll see that this term is
going to be zero. That's immediately obvious.
You do an integral over a cosine function times a
constant. You get zero.
So, the first one you don't worry about.
The third one, for any value of Mary and for
any value of Nancy, the third one is also zero.
And, if you don't believe that, check that on your own.
So then, you'd get a cosine of Nancy X times the sine of Mary X
for any value of N, any integer of N,
any integer of M. If you integrate it over one
whole period, you get zero.
So, this is also zero. If we now go to term number
two, it is also always zero except when Mary is the same as
Nancy. So, number two is also zero
except when M equals N. When M equals N,
you get here a cosine squared. And when you integrate a cosine
squared. And, when you integrate a
cosine squared, you do not get zero.
In other words, what you have to do if Mary
equals Nancy, you're going to get the cosine
squared of MX DX because Mary is Nancy.
And, I must integrate that between minus pi and pi.
And that equals pi. So now, we have a recipe for A
of M because A of M now is one divided by pi.
That is this pi that comes from the integral of the cosine
squared integrate with minus pi and plus pi, a function X times
the cosine of Mary XDX. So, we know now how to
calculate A zero, and we know how to calculate
the values for A of M. And, we will,
of course, do an example together.
It's now clear what you're going to do to find [two
values?] for B. You're now going to integrate
both sides of the aisle between minus pi and plus pi times the
sine of Nancy X DX. And when you do that,
you'll see exactly the same that you had here.
This will always be zero. This will now always be zero,
and this will always be zero except when Mary is Nancy.
Then you get here the sine square, and the integral of the
sine square will be pi, just like the integral here of
the cosine square was pi. And so, you see now that we
also have now the recipe to find all the values for B of M.
That's going to be one divided by pi times the integral of
minus pi to plus pi. That is the period of the
function times F of X times the sine of Mary X DX.
And here you see in its most general form,
the formalism of Fourier analysis.
Keep in mind that whenever you have an integral for minus pi to
pi, if you prefer zero to 2 pi, that's fine because the
function is periodic. So, you can always replace this
zero to 2 pi if that suits your purpose.
If you look at these three recipes, you can do away with
this one because if you make M equals zero here,
originally M was one, two, three, four,
five. But if you also include M
equals zero here, you get exactly the same that
you have here because when M equals zero, the cosine is one.
So this is then identical. That is the only reason why we
called this constant term A zero divided by two.
We could have called that a C because, after all,
a constant is a constant. Why would you want to call it A
zero divided by two? If you have called this C,
which is the average value of the function,
then you'd need this C here, and you would have here one
over 2 pi. And therefore,
you need three recipes to do Fourier analysis,
whereas now, if we define this constant as A
zero over two, you only need two.
So, that's the only reason. It's just for practical
purposes. But A zero over two is a
constant, and it is the average value of the function over the
period. Now, all this may look a little
bit opaque to you now, and it will become clear,
I hope, during this lecture when I put this Fourier analysis
to work. What you see here,
and that's the way I want you to look at it,
is that is simply a recipe. And you and I,
when we used this recipe, we execute it.
And we do not always ask ourselves, why is the recipe the
way it is? If I apply Kramer's rule,
I do not, every time, say to myself,
why is it really the way it is? I have seen once why it is the
way it is, and I apply it. Of course, you have to know
when you can use it and when you cannot use it.
So, now I would like to return to the plucked string.
But I first want to take a close look at it at time T
equals zero. And, what I'm going to do now
is to pluck it in a very unusual way.
And there is a reason why I do that so unusual,
because that is doable in one lecture to do the Fourier
analysis on it. So, I have a string,
now, which is fixed between zero and L.
And, I'm going to pluck it in an extremely obnoxious way,
namely, like this. So, I'm forcing it like a
square, very painful for the string, but that's not my
problem. And, let this be an amplitude,
A, if I want to use Fourier analysis, and you'll see that I
do. And you will also see how I'm
going to do that. I need a periodic function,
and this is not periodic. So I'm going to make it
periodic. And the way I'm going to make
it periodic is the following. I'm going to pretend that the
function is really this. So, I'm adding this.
I make it 2L, but I go much farther along,
and go on, and go on, and go on.
So, I'm going to define it between zero and 2L.
And that's periodic. Notice that this pattern now,
up square down square is periodic.
So, that means my period is 2L. And also, notice that by doing
that my A zero divided by two is zero because the average value
of this function, which is now defined between
zero and 2L and much beyond 2L and below zero that that
function has an average value of zero.
And so, my function now, my function of X,
you can write down Y of X if you want to is plus A for X
being between L and zero. And, it is minus A for X
between 2L and L because this here is minus A.
Now, I want to express the shape into a Fourier series.
But before I can do that, I have to make some changes in
the recipe because in the recipe we had radians.
And radians are apples. Here we have X.
But that's not radians. That is meters.
And meters are coconuts. Do you have a question?
Between 2L and L it is minus A. Between zero,
is that what you want? Thank you very much.
I appreciate corrections because it's awkward to do it
later. Thank you.
So, radians are apples, and meters are coconuts.
And so, what should I do not make the coconuts into apples?
I now have to take the old X, which was in radians,
which is the one that I had here, and is the one that I have
here. And I'm going to replace that
in that formalism by pi X over L, and this X now is in meters.
And if I have done that, you can see that if my X,
now, becomes 2L, then this has moved over 2 pi
radians. So, that's exactly what you
have to do. It is unfortunate that we call
this X. We could have called this
formalism Z or some other symbol.
But, in general we give X. We could have called this Z.
Well, you would have to write down that the X there is then pi
Z divided by L. Now we have apples here and we
have apples there. So, that means that now my
Fourier series are going to look very much like what I have here.
But I prefer to write them again.
So, my function, Y, as a function of X,
which of course is my function of X, I know that it is in the Y
direction. That function should now be
written somewhat differently: A zero divided by two plus the
sum from one to infinity of A of M times the cosine of M pi X
over L. This is now [R/our?] X.
This is X in meters plus the sum of one to infinity of B of M
times the sine of MX times pi divided by L.
And so, now we have made a modification to the general
idea, which is in radians, to be applicable to a case
whereby we don't have radians but why we have X in meters.
Before we can execute the analysis, I also had to change
the recipe in terms of A of M and B of M.
First of all, I don't have an integral over a
period 2 pi, but I have to do an integral from zero to L,
or from minus L to plus L. So, where earlier I had one
over pi in the role minus pi two plus pi, I had to change this
boundary either from zero to 2L, or if you prefer from minus L
to L. I don't care.
It's the same thing because the function is periodic.
But I also have to change this one over pi, which we have in
front here because that was the result of this integral.
And, if you do the integral in our space, you would have found
L and not pi. This is half the period,
and half the period of our function is L.
So, this one over pi from minus pi to plus now must be changed
into one divided by L integrals zero to 2L.
And so, to make sure that you're not going to get
confused, I'm going to rewrite the recipe for you.
And I might as will do that again in red.
So, now that RA of N is one divided by L times an integral,
let's just go from zero to 2L. And then we get our function of
X. And then we're going to get the
cosine of M pi X divided by L DX and B of M is then one divided
by L times the integral from zero to 2L of my function,
X, which is my plucked string times the sine of M pi X over L
times DX. And so, now we have not changed
anything in terms of the formalism of Fourier analysis.
But we have adapted the recipe. So, we now have a new series
written in terms of our new X. And, we have the new recipe to
calculate the A values in terms of our new X,
and in terms of the new period which is now 2L.
So, now you may think that I am ready to plunge into the math,
and to start executing it. But, I'm not going to do that
yet. I would like to take a look
first at the values of A. We already agreed that A zero
is zero. So, that's not an issue
anymore. But, let us look,
for instance, at the function A1 times the
cosine of pi X over L. That would be the first term of
the cosine series. And the second term would be A2
times the cosine of 2 pi X over L.
Well, let us assume that we found a value for A1.
So, I'm going to plot now into this function of ours the A1
value. So, it's here,
here, here, here, and here.
And so, here is that function. And this amplitude,
then, is A1. That's unacceptable.
Why is this acceptable? Look at my function,
and then look at the red curve, which is my A1 cosine function.
That red curve is supposed to help construct this function.
Here it is positive, so I demand that it must be
negative here. Otherwise, I can never make
this [trough?], whereas it's going to add up
positive here. So, this A1:
out of the question, can never contribute to my
function. But, not only can A1 not
contribute, but any cosine function that you draw will
always look here like this. It will always be positive here
and positive there, whereas I will demand that it
has to be positive here and make it up there.
Otherwise I can never build up his function.
To put it in a more intellectual way,
the way that mathematicians would use them,
a cosine function is an even function.
That means that the function of X is the same as the function of
minus X. We call that an even function.
Here, X is zero. A cosine function of X has
exactly the same value as at minus X.
But, our function is odd. Our function,
the way we defined it, zero here, is odd.
And, an odd function, the function of X equals minus
the function of minus X. And, to put it in a nutshell,
you can never fit an odd function with even functions in
Fourier. Neither can you fit an even
function with odd functions. So, that's perhaps a better way
of looking at it. So, we can already conclude
that all cosine functions are out for this specific case,
of course. Let's now take a look at the
values for B. So, keep in mind B are the sign
functions. So, A zero is zero.
All values for A are zero. Now, let's turn to the B
values. So, I'm going to draw the
function again here. We realize, of course,
it's only defined from zero to L.
The rest is only a mathematical way of doing the Fourier
analysis. So we are going to define it
all the way up to 2L. And then, we make it periodic.
So let us put in, there, the function B1 sine pi
B1 times the sine of pi X over L.
That is the very first one in the series of the B's.
That one is wonderful. That is exactly what we want.
This is sinusoid, which would have an amplitude
B1. It couldn't be better.
Here it is a mountain, and here it is a valley.
Well, we need a mountain here and we need a value there.
No surprise because a sine function is an odd function.
And, our function is odd. So clearly, odd functions will
do very well. Let's now take a look at B2.
B2 times the sine of 2 pi X over L, let's put it in there.
I'm full of expectations that B2 will do a great job.
So, we have here zero crossings.
So, here is my B2. And, this amplitude is B2,
out of the question. There is no way that B2 can do
me any good. Why?
Yeah? Exactly.
There are various ways of looking at it.
Here, you see a mountain that helps building this up.
But, look what you see here. You want a valley.
And, it's not adding to the valley.
That's one way of looking at it.
There's another way of looking at it to say,
look, everything has to be symmetric about the line one
half L. If this builds up,
it's got to build up there. And it's not doing that.
So, we must conclude that B2 must become zero.
And by analogy, all even values of M,
B will be zero. So, this is a prediction that I
make. We haven't done any fancy
Fourier analysis. We have already concluded that
for our specific function, for our specific string,
which is fixed at both ends, that all values for A will be
zero, and that only odd values for B for which the end value is
odd will we get non-zero answers.
So, now we're ready to actually execute the Fourier recipe.
And, if you insist t hat you want to do the A's,
be my guest. You can do all the A's,
and you will see that all of them are zero.
So, I will not waste your time on that.
But, I will certainly do the B's.
So, we now execute our recipe. And, here it is adjusted to our
X, and so we're going to get that B of M.
Now the function is defined between zero and L is plus A.
So, I can bring the A outside and put it here.
And then, I get the integral from zero to 2L.
And then, I get the sine of N pi X divided by LDX.
That is not between zero and 2L.
You should have screamed. That is only between zero and
L. It's only between zero and L
that it is plus A. But, when we integrate from L
to 2L, then it becomes minus A. So, the next part of the
integral is minus A. So, I bring the A outside.
So, I get minus A divided by L times the integral from L to 2L
times the sine of N pi X over L DX.
So, the function comes in two parts [UNINTELLIGIBLE] our
Fourier analysis why the function comes in more than two
parts, believe me. This is an easy case.
That's why I chose it. This integral is trivial,
of course. If I do an integral,
I get minus the cosine. So, I get A divided by L out,
and I get M pi downstairs. And, I get L upstairs.
And then, I get the cosine of M pi X over L, and I have to
evaluate that between zero and L.
And here, I get a minus sign now.
So, I get plus A divided by L. I get M pi downstairs.
I get L upstairs, and then I get the cosine of
Mary pi X divided by L. And, I have to evaluate that
between L and 2L. That's all I have to do.
That is the integral, a very simple one.
Now, when M is odd, this one alone,
don't look at this, this one alone is minus two.
You have enough knowledge to confirm that.
You can see if you substitute X equals L, for instance,
you take M equals one, which is odd,
you get the cosine of pi. That's minus one.
And then, the cosine of zero is one.
But you have to subtract that. You get another minus one.
So, you get minus two.
And so this one, if you do that,
you get plus two. Check it and you'll see that it
is plus two because the borders, the boundaries L to 2L are
different. It's not the same as zero to L.
But, if M is even, you get a zero here and you get
a zero there. And so, what comes out is
exactly what we predicted, that all even values of Mary
will give zero values for V, and that only the odd values
will give me values that are not zero.
Before I write this out in a complete Fourier form,
you should appreciate the fact that this entire integral here
gives me exactly the same answer as this entire integral here
because minus times minus two is plus two.
And, this plus times plus two is also plus two.
And, that is always the case when you have a string which is
fixed at both ends, that the zero to L integral
always gives you the same answer as this imaginary L to 2L which
you introduced to make the function periodic.
It is for that and only for that reason that Tony French
gives you a much easier way to calculate Fourier components in
strings. And he says all you have to do
is say that B of M is two divided by L.
He multiplies the recipe by two.
So, instead of having, where is my recipe,
instead of having one divided by L, he says,
no, you should really [be?] two divided by L.
But then he says all you have to do is now integrate between
zero and L of the function of X times the sine times Mary pi X
divided by L times DX. And, this is equation 632 in
French. If that's all we knew about
Fourier analysis, we would have a very narrow
picture because it is an extremely special case.
But it's true that whenever you have a string which is fixed at
both ends, this will do. I felt an obligation to you to
show you the Fourier formalism, which is a beautiful formalism
in more general terms. So, we are now ready to write
down the complete Fourier series.
Notice that the sum of these two become 4A divided by M pi.
So, B of M is going to be 4A divided by M pi.
But, M is only odd. And so, with that in mind,
I can write down now here our function, which was one of our
big steps that Y, I put here zero here because
this is a time equals zero when I have this string in this crazy
shape can now be written as 4A divided by M pi.
I could write it down as simply pi.
And then I get here the sine pi X divided by L plus one third.
That is that M equals three, which is three times lower
because you have an M three here times the sine of three pi X
divided by L plus one fifth. And then you get the sine of
five pi X divided by L, and so on, and so on.
So now I would like to put in my function.
I would like to put in the first two terms.
And I'll do that here on the blackboard.
So I'm going to concentrate now only on my function zero to L.
In reality, the string is only here.
And now I would like to put in there, this is A.
And now I would like to know, what is B1?
Well, B1 is going to be 4A divided by pi,
which is approximately 1.278. Well, let's put it in there.
This is A, so I have to put in one third more so it comes up to
this point roughly. So, there it is.
I love it. It's not quite a square yet,
but we're getting there. We are on our way to building
up the square. The next one is B3.
B3 is 4A divided by 3 pi. So it is one third of 1.27,
which is about 0.42 A. And, B5 is one fifth times B1.
Let's put in the B3. So, B3 has an amplitude which
is about 0.4, which is about this much.
It's about the height, 0.4, a little lower.
And, so here we have the maximum.
And then, we have a zero here. We have a zero here,
and so we have here. And then, we go through this
point. So, the curve is something like
this.
So this, now, is B3.
And, this is B1. Now, look at B3.
Look at the beauty of Fourier analysis.
This is higher than it should be.
B3 says I'll take care of that. I will subtract something
there. I will take that off there.
Isn't that beautiful? So, you're going to flatten
this out already, already with two terms,
B1 and B3. You are already beginning to
see the building up of this crazy square.
And look here. B1 says, sorry,
I cannot fill this up for you. This is the best I can do.
B3 says I'm going to help you. I'm going to help stop there,
and it's going to do the same there.
And so, you see that if you now keep adding odd values of M,
that you will gradually approach that square,
which is an amazing concept. And, what I will demonstrate to
you is making a square. And we will make it like this
just like we did here. We can do that on an
oscilloscope. We can have a signal that makes
this trace. Admittedly, this is time.
But the way you'll see it, it is space.
So you can think of it as being space.
It's no different from the function that we had.
And then, we will show you the B1 value.
In this case, since it is in time,
it actually has a frequency. It's 440 Hz.
We can make you listen to it. And then, we will show you the
B3 value, which has three times higher frequency.
And then, we will show you the B5, and the B7,
and the B9. And then, we will add them all
up, and you will see, it begins to look like a
square. And so, let us set that up,
supposed to see that on the central there.
What you see here is simply the B1.
So you don't see very much, do you?
You just see a sine curve. Well, you're looking at this
sine curve. Now I'm going to show,
can we here it to Marcos? Here's the 440 Hz.
Now, I'm going to show you the B3.
Notice, it is three times smaller in amplitude.
That's the way we set it as a three times higher frequency.
And now, I'm going to show you B5, which is five times smaller
than the, hold it, no, no, no, no,
no. OK, it's five times smaller
than B1. And, the frequency,
of course, is five times higher.
You can hear it. There it is.
And now, I'm going to show you seven, seven times smaller than
B1. And, here is B9.
This is nine times smaller than B1.
And, now I'm going to show them all five to you.
And, it's really beginning to look like a square pulse.
It is clear that B1 is very, very important.
So, if I take the one out, which I will do now,
it doesn't even look like a square.
That shows you how important that first term is.
If I take B3 out, that is less disastrous,
but it is still pretty disastrous.
But you already begin to see something that looks like a
square slowly. And it should be clear to you
now that in order to get the real sharp edges here,
you need very high harmonics. You have to go to M values of
49, 101, 201. They're all odd.
You have to go to very high values, and the higher you go,
the closer you will get to the square.
So, this is a nice moment to have a break.
And I know you're dying to do your fifth mini quiz.
I will hand it out now, and then we will all start at
the same time with the mini quiz so that each one of you has the
same amount of time. [SOUND OFF/THEN ON]
I owed you the histogram of the exam.
You see that here? I think it's a wonderful
histogram as far as I'm concerned.
It is too early to talk passing and failing, and A,
B, C's, and D's. But, if I knew nothing else,
no other grades, then I would have to put you in
the danger zone if you score less than 45.
That's all I can say. I can add nothing more than
what I already wrote you by e-mail.
So, let us now return to our string is still at T equals
zero. So, our string is still in this
position, dying to be released. That was the goal.
Remember, we would pluck it, and we would let it go.
At time T equals zero, this is the Fourier analysis.
These are the Fourier components of that string.
But now, I say OK, go.
And what happens now is that this one is going to oscillate
with omega 1T. This one is going to oscillate
omega 2T, omega 3T. And this one is going to
oscillate with cosine omega 5T. Each one of these are standing
waves. And so, each one of these are
going to do this with their own amplitude and with their own
frequency. And, omega M is going to be V
times K of M. And, the speed of propagation
is the square root of T divided by mu.
And, that's what's going to happen.
So, now, you may wonder what you're going to see if you let
this spring go. And, chances are that if I ask
you, what do you think will happen, that you may think that
this will happen with the string as a whole.
But that's not true. And the best way,
actually, to answer the question of what you're going to
see is to go back to a simple case of a triangular pulse on a
string. I'll do that here.
If I have a triangular pulse on a string, I exaggerate,
of course, the amplitude highly fixed here at zero and fixed
here at L. And I let it go.
Yes, I can decompose that in terms of the Fourier components.
And each of these Fourier components are going [to?]
standing waves up and down. It doesn't give me much
insight. But, it is intriguing that
that's what happens. I can also think of it that
nature is seeing a disturbance and it says, well,
we are going to propagate this disturbance.
Nature doesn't know the difference between left and
right. And so, clearly will happen if
this has in amplitude, A, there will be a pulse was in
amplitude of one half, which goes in this direction,
and one with the same one half, which goes into that direction.
And that is exactly what you're going to see when you let it go.
So, a little later in time, you'll see this going in this
direction, going in this direction.
At the moment that the top of the mountain reaches L,
and the top of this mountain reaches zero,
you're going to see nothing, absolutely nothing because the
reflected one and the incident one exactly cancel each other.
And then, a little later in time, they will be on their way
back. And then a little later in
time, you'll see this one but completely flipped over when
these two triangles meet each other.
And so, that's when you'll see the thing upside down.
So, there are two very different ways of looking at
what happens when you release a plucked string.
One way is to say they're oscillations or standing waves a
la Fourier, and the other way is to say, well,
we're going to always cut it in two pulses, each of half the
amplitude, and let them go back and forth, and let them reflect
at the ends. Now, keep in mind that standing
waves are the result of the superposition of traveling
waves. So, the two different ways of
looking are, of course, connected.
They have the same underlying physics.
And, now I want to demonstrate this to you using a program that
was developed by Professor Wyslouch when he was lecturing
8.03, and also Nergis. I think she's in the audience.
Nergis, are you hiding? Yeah, you're hiding.
Nergis has also worked on this program.
It's a wonderful program. I'm going to put a triangle
first on a string. The width of the triangle is
about one third of the length. And then, I will release it.
And I will let it go. And we will see the Fourier
components. And at the end,
when the whole show is over, after one half the time for the
fundamental period, we will wait one half period of
the fundamental. We will inspect very closely
the Fourier components. Because of the symmetry of this
problem, all even values of M will have zero values of B.
And, you will be able to see that.
Now, were we going to change the light setting for this or
were we not? No?
I thought we were. Why don't you turn that off,
also? That's too difficult perhaps.
No, the bar, the bar, the bar.
All right, so what you're going to see first is the triangle.
The number of Fourier components that we have is 25.
Did I do something wrong? Excuse me?
We have no what? You broke the line?
Oh, that's nice. Thank you.
The number of terms we have is 25.
We have M equals one, M equals three,
M equals five, all the way up to 49.
That's when we cut it off. All even values for M have no B
value. And, if you are ready,
I'm ready. So, let's first run this one.
There, you see the triangle and the blue lines are the Fourier
components. They are each doing this thing,
their own thing, standing waves up and down.
But, the net result is something that you are quite
familiar with. You see two pulses with half
the amplitude of the original one.
They move to each their own side.
They reflect, and they come back.
And then, we'll stop when we have half the period of the
fundamental. Now, look at the individual
Fourier components. This is B1.
This one is B3. You see, it helps building up
that triangle. This is B5.
It helps building up the triangle.
This is B7 and this is B9. And here, they all conspire.
It's amazing. There are 25 sinusoids that are
conspiring there. And they show you nothing,
and the same here. And it's hard to believe that
all these sinusoids can add up to zero here,
and how beautifully they connect here.
And, the only reason why this tip is not very sharp is we only
have 25 terms. If I would run it with 400
terms, then you'd see this sharper.
But it would take a lot more time.
Let's look at the Fourier components which you'll see
here. So, this is B1,
which we have arbitrarily called one.
You see, all the even values, this is B2, B4,
B6, B8. All the even values are zero.
The B3 is negative in this case.
That was not the case for our square.
Remember, in the case of the square, B1, B3,
B5, B7 were all positive. They all had the same sign.
That's not the case with the triangle.
They alternate. The next one that I want to run
for you is a square which has a width half the length of the
string. And it is centered at the
middle of the string.
And again, two pulses are each going their own direction with
half the amplitudes. They are being reflected.
And strange things happen at the ends.
And now there comes a time one quarter of the period of the
fundamental that you see nothing.
And now they are coming back from the reflection.
Very different from what you expect, isn't it?
Now, look at B1. That's our real biggie.
Look at B1, boy, here, is that a surprise?
No, because clearly B1 is not quite as square.
But it's almost a square, right?
You just have to subtract something here.
Well, B3 says, that's OK, I will subtract
there, and B3 will subtract here.
And, B1 is a little shallow here.
It's a little bit too, not generous enough.
And then B3 says, OK, I'll help you building it
up. And so, if we look at the
Fourier components, you see now that,
again, all the even values are zero, B1, the definition one,
and now you see this one is negative.
And, B5 is also negative, not so obvious.
And B7 is positive, and B9 is positive.
And now, I'll show you one whereby I'm going to offset the
square. The square is only one quarter
of the length, and I'm going to offset it.
And, when I offset it from the center, you need both odd and
even values. And you're going to see that.
So, now you cannot just get away with only even values of
Mary. So, here you see the wave
spinning up in two, just like you expected.
It has a width a quarter of the length of the string.
This one has already reflected. This one is now on its way to
being reflected. Now this one is coming back.
There it is. This one is already on its
return. And of course they are going to
meet again. And now, we are one half period
of the fundamental later. And now look.
This now is the first harmonic. But now, there is also a second
harmonic. M equals two has a B value.
Here it is. This is the B2 value,
large amplitude. You noticed that?
And, look how important B4 is. I think this is the B4 value.
Yes indeed, here is the B3. This is the B4.
It has a huge amplitude. And look how important it is,
because right here where the pulse is, you need this bulge to
push this further out because the B1 alone cannot reach that
point. Let's take a look at the
Fourier spectrum here. Here you see B1,
which we, by definition, call one.
You see B2 is there. It's large.
It's a little larger than 0.7. And then, you have a B3,
which is negative, and a B4, which is negative,
and a B5, which is negative. B4 is off scale.
And now, you get B6 positive, B7 positive.
B8 is zero, and B16 is zero. Well, that's probably the
result of the way we set up the square, which has a width of one
quarter of a length. And we offset it from the
middle by one quarter of the length, remarkable example of
Fourier analysis now, which is way more complicated
than if you nicely center it. All values of B,
now, all values of M, are important.
You will now understand, when I said if we pluck the
string of a harp or the string of a violin, and you pluck it
this way, fixed here and fixed here, the tone that you hear is
different than when you pluck it this way.
You now understand that because the Fourier components of this
one and this one are very different than the Fourier
components from this one. And so, when he let the string
go, the frequencies that will be produced are at the frequencies
of these normal modes of these B values.
But the various B values are very different here from there.
And so, the sound that you hear is distinctly different when you
pluck a string on the side than when you pluck it in the middle.
You may remember that during that lecture,
I told you that the hammer of a piano hits the string about 1/7
of the length of the string from one side.
That is done to suppress the seventh harmonic.
For some reason that beats me, people don't like the seventh
harmonic. So that's the way they kill it.
So, it does depend on where you pluck and where you hit.
And, now you can put that in context because now you
understand that you analyze the thing here in terms of Fourier
components. And, each one will then
oscillate at their own frequency with their own amplitude.
So, what we have discussed now at length is a Fourier analysis
in space in X in meters. And we decomposed a function in
the sum of many harmonics, which together make up,
then, the shape of the string. Now, if we look at sound,
sound, of course, is something that is a function
in time. Suppose I look at the sound
signal made by a tuning fork of 440 Hz.
And I have here that signal in time.
Suppose I have one second of data, 440 Hz.
So this is now time. But if you want to think of
this as being in space, be my guest.
Now, there are special programs, special algorithms
that take this one second of data, perform a Fourier
analysis. They have a name.
We call them fast Fourier transforms.
And, it's going to tell me what the amplitude of B1 is and A1 of
B2 and A2 of B3 and A3, and so on.
If I now think in terms of hertz rather than in terms of
omega, there is nothing at one hertz.
So, though A's and B's are zero, there is nothing at two
hertz. So, those A's and B's are zero.
But by the time I reach 440 Hz, the system says,
yippee, there's a lot of power there at 440 Hz.
And what we now do, we make a plot of the square of
A plus the square of B. And, the reason why we square
them is that energy is always proportional to the amplitude
squared. And so, you take the A squares
and the B squares, and you add them.
We call those a power density spectrum, a Fourier spectrum.
And, here you have your omega or you have your F value in
hertz. The difference is only 2 pi.
And, if you now do a Fourier analysis of that signal,
you'll see at 440 Hz a huge value.
And you'll see almost nothing anywhere else.
So, you have now done a Fourier analysis of a time signal.
And you have decomposed it into its Fourier components,
which in this case is only 440 Hz.
If you did the same with the middle A of the piano,
which is 440 Hz, you would see a big value here.
But, you would see also something at 880.
And, you may even see something at the third harmonic.
And, the same is true for a piano.
And it is also for a violin. If you take a violin,
and you ask the violinist to produce the 440,
you always get a little bit more of the 880.
And, you get some of three times the fundamental.
And, this is something that we can demonstrate.
We have here an admittedly somewhat poor man's version of
what's called the Fourier analyzer.
We take in sound for about 1/5 of the second period and then we
asked the computer to Fourier analyze that for us,
and show us the Fourier spectrum in terms of where the
frequencies were. And, we will see that here,
for which I think we also invented the special light
condition, didn't we?
The scale that you see from left to right is 2 kHz.
I'll first make you see the 440 Hz.
[SOUND PLAYS] You see that huge spike?
That's the 440 Hz. So, the end of the scale is 2
kHz. [SOUND PLAYS] 256 Hz.
[SOUND PLAYS] It's lower. It's lower.
I have here a flute. And remember during the lecture
on musical instruments. We believed,
and I demonstrated that you can really think this one resonate
only in its fundamental. I will now show you with this
Fourier analyzer that there is also a little bit of the second
harmonic, maybe even the third harmonic.
Look at it now. [SOUND PLAYS]
You see that? That was the fundamental second
and third harmonic even. [SOUND PLAYS] Ah.
So this is a beautiful way of analyzing the frequencies and
sound. I will whistle for you.
[SOUND PLAYS] Amazing. Would you have guessed it?
Very nicely separated. What is the frequency?
[SOUND PLAYS] It's about here.
That's about 1 kHz. It's about 900 Hz.
[SOUND PLAYS] And, the second harmonic,
not so obvious, my whistle.
Very high frequency. [SOUND PLAYS] It is so high
that it's off scale. And I did that purposely
because I know you hate the whistle.
So, I didn't want you to see what it really looks like.
We can listen to the radio. And then we can do a Fourier
analysis on what we hear. [RADIO PLAYS] Oh,
I don't have a question for you.
I'm sorry. [LAUGHTER] [RADIO PLAYS]
[SINGING] Not the right type for me.
[MUSIC PLAYS] Too complicated for me.
Any one of you want to sing? Come on.
Be brave. Anyone want to sing?
Yeah. Come on.
Give it a try. Nice to be here.
Believe me. You can see all these,
OK, it's your decision. All right, so you see how you
can do a Fourier analysis of sound and then I decompose the
signal in terms of its Fourier components.
Neutron stars were discovered in 1967 by Jocelyn Bell.
Jocelyn was a graduate student in Cambridge,
England at the time. And her supervisor,
Anthony Hewish, had built a new type of radio
telescope. And she was in charge of
analyzing the data. And, she was sure that she had
received a periodic signal, which came almost every 1.3
seconds. They realized that that could
have been the discovery of the century because they thought
that they were receiving signals from the intelligent life
elsewhere in the universe. And so, they called in the
little green man. And then, a few months later,
Jocelyn discovered another one. And so, they called the first
one little green man one, and then little green man two,
LGM1 and LGM2. And then, when they discovered
a third one, they abandoned the idea of LGM's.
We now call these pulsars, and we know that the period is
the spin period of the neutron stars.
In 1974, Anthony Hewish received the Nobel Prize for
this discovery. It is a shame,
and it is scandalous that Jocelyn did not share in the
Nobel Prize. She actually made the
discovery. I've known her very well.
I've discussed it with her many times.
She was a graduate student. Maybe that was the reason why
the Nobel committee didn't think it was appropriate to give a
Nobel Prize to a graduate student.
It's ridiculous but perhaps true.
She was a woman, and there is,
perhaps, a very sad case of sex discrimination again.
We will never know, but she did not share in the
Nobel Prize that she should have.
Neutron stars have a mass about 1 1/2 times that of the sun.
They are 100,000 times smaller. There are only 20 km across.
So, they have a density which is ten to the 15th times higher
than the sun, ten to the 15th times higher
than water, which is higher than nuclear density.
And, I had some e-mail contact with Jocelyn does a few days
ago. She's now at Oxford.
And I said, I really would like to show the class you're
picture. And so, she sent me a picture,
which I'm going to show you. Are we still connected?
Where's Jocelyn? Is Jocelyn hiding?
No, there she is. There she is.
She sent me a picture at the time that she made the
discovery. You see the radio telescope
here. And you see her standing there
very modestly. She's a very modest woman.
And she made one of the most important discoveries of the
past century and did not share in the Nobel Prize.
We now know of hundreds of pulsars in the sky.
And so, the question, now, is how do you find these
pulsars? You can observe the sky with
radio telescopes. You can do it also with x-ray
observatories. And what you do,
now, if you take the data just like we took the data of our
sound signal, and you perform a Fourier
transform. You ask that data,
what are the Fourier components that I hid that I cannot see but
are hidden? And, in 1998,
Rudy [Winons?] in Amsterdam, an independently at Morgan here
at MIT, were analyzing data from a no x-ray source in our galaxy.
And they were performing a Fourier analysis,
which is standard nowadays in astronomy.
And they discovered that the neutral star was rotating with a
spin period of 2 1/2 milliseconds.
They noticed in the Fourier power spectrum a huge spike at
401 Hz. It means that at the equator of
the neutral star, the speed going around was
about 1/10 of the speed of light.
And I asked [UNINTELLIGIBLE] who would worked with me for
several years here at MIT to send me some of the data that he
obtained from which he could, then, finally derive that we
were dealing with a neutron star rotating around at two half
milliseconds. And so, he said,
well, Walter, why don't you show the class
only 1/5 of a second of my data, 1/5 of the second that you see
here? 2/10 of a second?
So, this is the timescale. And, when one x arrives,
you see a vertical bar there. And this time,
they were so close together that you see there two x-rays.
If you count the total number of x-rays in that 1/5 of a
second, you observe 33 x-rays. During that time,
the neutron star was rotating 80 times around already because
it rotates every two half milliseconds.
So, when you look at the data, you have no idea that here is
an underlying neutron star with a period of 2 1/2 milliseconds
with a frequency of 401 Hz. But now you take 3,000 seconds
of data. And if you have taken 8.03,
you know how to perform a fast Fourier transform.
Those programs, by the way, are readily
available on the market. And here you see a power
density spectrum. Vertically, it's the sum of A
squared plus B squared. And horizontally,
it's the frequency in hertz. And look what you see.
At 401 Hz, you see a huge spike.
And that is the underlying 2 1/2 milliseconds neutron star.
This is the frontier of astrophysics.
Nowadays, you cannot even think of astronomy or astrophysics
without Fourier analysis. It has an enormously important
impact on our research. My graduate students and my
post-docs perform Fourier transforms everyday.
And so, what we have discussed today is not just intellectually
interesting. It is at the forefront of
research. See you Thursday.