I'd like to talk.
Thank you. One of the things I'd like to
give a little insight into today is the mathematical basis for
hearing. For example,
if a musical tone, a pure musical tone would
consist of a pure oscillation in terms of the vibration of the
air. It would be a pure oscillation.
So, [SINGS], and if you superimpose upon
that, suppose you sing a triad, [SINGS], those are three tones.
Each has its own period of oscillation, and then another
one, which is the top one, which is even faster.
The higher it is, the faster the thing.
Anyway, what you hear, then, is the sum of those
things. So, C plus E plus G,
let's say, what you hear is the wave form.
It's periodics, still, but it's a mess.
I don't know, I can't draw it.
So, this is periodic, but a mess, some sort of mess.
Now, of course, if you hear the three tones
together, most people, if they are not tone deaf,
anyway, can hear the three tones that make up that.
So, in other words, if this is the function which
is the sum of those three, some sort of messy function,
f of t, you're able to do Fourier
analysis on it, and break it up.
You're able to take that f of t, and somehow mentally express
it as the sum of three pure oscillations.
That's Fourier analysis. We've been doing it with an
infinite series, but it's okay.
It's still Fourier analysis if you do it with just three.
So, in other words, the f of t is going to
be the sum of, let's say, sine,
I don't know, it's going to be the sign of
one frequency plus the sine of another frequency plus the sine
of a third, maybe with coefficients here.
So, somehow, since you were born,
you have been able to take the f of t,
and express it as the sum of the three signs.
And, here, therefore, the three tones that make up
the triad. Now, the question is,
how did you do that Fourier analysis?
In other words, does your brain have a little
integrator in it, which calculates the
coefficients of that series? Of course, the answer is no.
It has to do something else. So, one of the things I'd like
to aim at in this lecture is just briefly explaining what,
in fact, actually happens to do that.
Now, to do that, we'll have to make some little
detours, as always. So, first I'm going to,
throughout the lecture, in fact, I gave you last time a
couple of shortcuts for calculating Fourier series based
on evenness and oddness, and also some expansion of the
idea of Fourier series where we use the different,
but things didn't have to be periodic or period two pi,
but it can have an arbitrary period, 2L, and we could still
get a Fourier expansion for it. Let me, therefore,
begin just as a problem, another type of shortcut
exercise, to do a Fourier calculation, which we are going
to be later in the period to explain the music problem.
So, let's suppose we're starting with the function,
f of t, which is a real square wave,
and I'll make its period different from the one,
not two pi. So, suppose we had a function
like this. So, this is one,
and this is one. So, the height is one,
and this point is one as well. And then, it's periodic ever
after that. I'll tell you what,
let's do like the electrical engineers do and put these
vertical lines there even though they don't exist.
Okay, so the height is one and it goes over.
The half period is one. This really is a square wave.
I mean, it's really a square, not what they usually call a
square wave. So, my question is,
what's its Fourier series? Well, it's neither even nor
odd. That's a little dismaying.
It sounds like we're going to have to calculate an's and bn's.
So, the shortcuts I gave you last time don't seem to be
applicable. Now, of course,
nor is the period two pi, but that shouldn't be too bad.
In fact, you ought to look for an expansion in terms of things
that look like sine of n, well, what should it be? Since L is equal to one,
the half period is equal to one.
Remember, the period is 2L, not L.
It's n pi over L, but if L is one,
we should be looking for an expansion in terms of functions
that look like this. Now, since we've already done
the work for the official square way, which looks something like
this, what you always try to do is reduce these things to
problems that you've already solved.
This is a legitimate one, since I solved it in lecture
for you. So, we can consider it as
something we know. So, I observed that since I am
very lazy, that if I lower this function by one half,
it will become an odd function. Now it's an odd function.
Okay, I just cut the work in half.
So, let's call this function, let's call this,
I don't know, S of t.
The green one is the one we wanted to start with.
So, f of t is a green function.
But, I can improve things even more because the function that
we calculated in the lecture is a lot like this salmon function.
That's why I called it S. But, the difference is that the
function we calculated with this one.
In the first place, it went down further.
It went not to negative one half, which is where that one
goes. But, it went down to negative
one, and then went up here to plus one.
And, it went over to pi. So, it came down again,
but not, but at the point, pi.
And here, negative pi went up again.
Okay, let me remind you what this one was.
Suppose we call it, O doesn't look good,
I don't know, how about g of u?
Let's, for a secret reason, call the variable u this time,
okay? So, the previous knowledge that
I'm relying on was that I derived the Fourier series for
you by an orthodox calculation. And, it's not too hard to do
because this is an odd function. And therefore,
you only have to calculate the bn's.
And, half of them turn out to be zero, although you don't know
that in advance. But anyway, the answer was four
over pi times the sum of just the odd ones,
the sine of n u, and that you had to
divide by n. So, this is the expansion of g,
this function, g of u, the Fourier expansion
of this function. Since it's an odd function,
it only involves the signs. There's no funny stuff here
because the period is now two pi.
And, this came from the first lecture on Fourier series,
or from the book, wherever you want it,
or solutions to the notes. There are lots of sources for
that. The solution's in the notes.
Okay, now, that looks so much like the salmon function,
--- -- I ought to be able to
convert one into the other. Now, I will do that by
shrinking the axis. But, since this can get rather
confusing, what I'll do is overlay this.
What I prefer to do is I think u, okay, I'm changing,
I'm keeping the thing the same. But, I'm going to change the
name of the variable, the t, in such a way that on
the t-axis, this becomes the point, one.
If I do that, then this function will turn
exactly into that one, except it will go not from
minus a half to a half, but it will go from negative
one to one, since I haven't done anything to the vertical axis.
So, how I do that? What's the relation between u
and t? Well, u is equal to pi times t,
or the other way around. You know that it's going to be
approximately this. Try one, and then check that it
works. When t is equal to one,
u is pi, which is what it's supposed to be.
So, this is the relation between the two.
And therefore, without further ado,
I can say that, let's write the relation
between them. f of t is what I want.
Well, what's f of t if I subtract one half of that?
So, that's going to be equal to the salmon function plus one
half, right, or the salmon function is f of t lowered by
one half. One thing is the same as the
other. And, what's the relation
between this salmon function and the orange function?
Well, the salmon function is, so, let's convert,
so, S of t -- it's more convenient, as I wrote the
formula g of u. Let's start it from that end.
If I start from g of u, what do I have to do to convert
it into S of-- into the salmon function?
Well, take one half of it. So, if I put them all together,
the conclusion is that f of t is equal to one half
plus S of t, which is one half of g of u, but u is pi t.
So, it's four pi, four over pi times the sum of
the sine of n. And, for u, I will write pi t divided by n.
And, sorry, I forgot to say that sum is only over the odd
values of n, not all values of n.
So, the sum over n odd of that, and, of course,
the two will cancel that. So, here we have,
in other words, just by this business of
shrinking or just stretching or shrinking the axis,
lowering it and squishing it that way a little bit.
We get from this Fourier series, we get that one just by
this geometric procedure. I'd like you to be able to do
that because it saves a lot of time.
Okay, so let's put this answer up in, I'm going to need it in a
minute, but I don't really want to recopy it.
So, let me handle it by erasing.
So, let's call that plus two over pi,
and there is our formula for that green function that we
wrote before. So, I'll put that in green.
So, we'll have a color-coded lecture again.
Now, what we're going to be doing ultimately,
to getting at the music problem that I posed at the beginning of
the lecture, is we want to solve, and this is what a study
of Fourier series has been aiming at, to solve second-order
linear equations with constant coefficients were the right-hand
side was a more general function than the kind we've been
handling. So, now, in order to simplify,
and we don't have a lot of time in the course,
I'd have to take another day to make more complicated
calculations, which I don't want to do since
you will learn a lot from them, anyway.
I think you will find you've had enough calculation by the
time Friday morning rolls around.
So, let's look at the undamped case, which is simpler,
or undamped spring, or undamped anything because it
doesn't have that extra term, which requires extra
calculations. So, I'll follow the book now
and some of the notes and the visuals, and called the
independent variable-- the dependent variable I'm going to
call x now. And, the independent variable
is, as usual, time.
So, this is going to be, in general, f of t,
and I'm going to use it by calculating example,
this is the actual f of t I'm going to be using.
But, the general problem for a general f of t is to solve this,
or at least to find a particular solution.
That's what most of the work is, because we already know how
from that to get the general solution by adding the solution
to the reduced equation, the associated homogeneous
equation. So, all our work has been,
this past couple of weeks, in how you find a particular
solution. Now, the case in which we know
what to do is, so we can find our particular
solution. Let's call that x sub p. We could find x sub p if the
right hand side is cosine omega, well, in general,
an exponential, but since we are not going to
use complex exponentials today, all these things are real.
And I'd like to keep them real. If it's either cosine omega t
or sine omega t, or some multiple of that by linearity, it's just as good.
We already know how to find the thing, and to find a particular
solution. So, the procedure is use
complex exponentials, and that magic formula I gave
you. But, right now,
just to save a little time, since I already did that on the
lecture on resonance, I solved it explicitly for
that, and you've had adequate practice I think in the problem
sets. Let's simply write down the
answer that comes out of that. The answer for the particular
solution is cosine omega t or sine omega t. That's the top.
And, it's over a constant. And, the constant is omega
naught squared. That's the natural frequency
which comes from the system, minus the imposed frequency,
the driving frequency that the system, the spring or whatever
it is, undamped spring, is being driven with.
Okay, understand the notation. Cosine this over that,
or sine, depending on whether you started driving it with
cosine or sine. So, this is from the lecture,
if you like, from the lecture on resonance,
but again it's, I hope by now,
a familiar fact. Let me remind you what this had
to do with resonance. Then, the observation was that
if omega, the driving frequency is very close to the natural
frequency, then this is close to that.
The denominator is almost zero, and that makes the amplitude of
the response very, very large.
And, that was the phenomenon of resonance.
Okay, now what I'd like to do is apply those formulas to
finding out what happens for a general f(t),
or in particular this one. So, in general,
I'll keep using the notation, f of t,
even though I've sorted used it for that.
But in general, what's the situation?
If f of t is a sine series, cosine series,
all right, let's do everything. Suppose it's,
in other words, the procedure is,
take your f of t, expand it in a Fourier series.
Well, doesn't that assume it's periodic?
Yes, sort of. So, suppose it's a Fourier
series. I'll make a very general
Fourier series, write it this way:
cosine (omega)n t, and then the sine terms, sine (omega)n t
from one to infinity where the omegas are,
omega n is short for that. Well, it's going to have the n
in it, of course, but I want, now,
to make the general period to be 2L.
So, it would be n pi over L. Of course, if L is equal to one, then it's n pi.
Or, if L equals pi, those are the two most popular
cases, by far. Then, it's simply n itself,
the driving frequency. But, this would be the general
case, n pi over L if the period is the period of f
of t is 2L. So, that's what the Fourier
series looks like. Okay, then the particular
solution will be what? Well, I got these formulas.
In other words, what I'm using is superposition
principle. If it's just this,
then I know what the answer is for the particular solution,
the response. So, if you make a sum of these
things, a sum of these inputs, you are going to get a sum of
the responses by superposition. So, let's write out the ones we
are absolutely certain of. What's the response to here?
Well, it's (a)n cosine omega n t. The only thing is, now it's divided by omega
naught squared. This constant has changed,
and the same thing here. Of course, by linearity,
if this is multiplied by a, then the answer is multiplied
by, the response is also multiplied by a.
So, the same thing happens here.
Here, it's (b)n and over, again, omega naught squared
minus omega times the sine of omega t. So, in other words,
as soon as you have the Fourier expansion, the Fourier series
for the input, you automatically get this by
just writing it down the Fourier series for the response.
That's the fundamental idea of Fourier series,
at least applied in this context.
They have many other contexts, approximations,
so on and so forth. But, that's the idea here.
All right, what about that constant term?
Well, this formula still works if omega equals zero.
If omega equals zero, then this is the constant,
one. The formula is still correct.
Omega is zero here. The only thing you have to
remember is that the original thing is written in this form.
So, the response will be, what will it be?
Well, it's one divided by omega naught squared,
if I'm in the case omega zero is equal to zero.
So, it's a zero divided by two omega naught squared.
And, as you will see, it looks just like the others.
You're just taking omega, and making it equal to zero for
that particular case. Sorry, this should be omega n's
all the way through here. All right, well,
let's apply this to the green function.
So, what have we got? We have its Fourier series.
So, if the green function is, if the input in other words is
this square wave, the green square wave,
so in your notes, this guy, this particular f of
t is the input. And, the equation is x double
prime plus omega naught squared x equals f of t. Then, the response is,
well, I can't draw you a picture of the response because
I don't know what the Fourier series actually looks like.
But, let's at least write down what the Fourier series is.
The Fourier series will be, well, what is it?
It's one half. The constant out front is one
half, except it's one over two omega naught squared. So, this is my function,
f of t. That's the general formula for
how the input is related to the response.
And, I'm applying it to this particular function,
f of t. And, the answer is plus.
Well, my Fourier series involves only odd sums,
only the summation over odd, and only of the sign.
So, it is going to be two over pi,
sorry, so it's going to be two over pi out front.
That constant will carry along by linearity.
And, I'm going to sum over odd, n odd values only.
The basic thing in the upstairs is going to be the sine of omega
n t. But, what is (omega)n?
Well, (omega)n is n pi. So, it's n pi t.
And, how about the bottom? The bottom is going to be omega
naught squared minus omega n squared. And, this is my (omega)n,
minus n pi squared. What's that? Well, I don't know.
All I could do would be to calculate it.
You could put it on MATLAB and ask MATLAB to calculate and plot
for you the first few terms, and get some vague idea of what
it looks like. That's nice,
but it's not what's interesting to do.
What's interesting to do is to look at the size of the
coefficients. And, again, rather than do it
in the abstract, let's take a specific value.
Let's suppose that the natural frequency of the system,
in other words, the frequency at which that
little spring wants to go vibrate back and forth,
whatever you got vibrating. Let's suppose the natural
frequency that's omega naught is ten for the sake of
definiteness, as they say.
Okay, if that's ten, all I want to do is calculate
in the crudest possible way what a few of these terms are.
So, the response is, so let's see,
we've got to give that a name. The response is (x)p of t. What's (x)p of t?
I'm just going to calculate it very approximately.
This means, you know, throwing caution to the winds
because I don't have a calculator with me.
And, I want you to look at this thing without a calculator.
The first term is one over 200. Okay, that's the only term I can get exactly right.
[LAUGHTER] Or, I could if I could calculate.
I suppose it's 0.005, right?
That's the constant term. Okay, so the next term,
let's see, two over pi is two thirds.
I'll keep that in mind, right?
Plus two thirds, 0.6, let's say,
that's an indication of the accuracy with which these things
are going to be performed. I think in Texas for a long
while, the legislature declared pi to be three,
anyways. One of those states did it to
save calculation time. I'm not kidding,
by the way. All right, so what's the first
term? If n equals one,
I have the sine of pi t. That's the n equals one term. What's the denominator like?
That's about 100 minus 9 squared. Let's say it's 91, sine t over 91.
What's the next term? Sine of three pi t, remember, I am omitting,
I'm only using the odd values of n because those are the only
ones that enter into the Fourier expansion for this function,
which is at the bottom of everything.
All right, what's the sine three pi t?
Well, now, I've got 100 minus three pi, --
-- that's 9 squared is 81. So, no, what am I doing?
So, we have 100 minus three times pi is 9,
squared. Well, let's say a little more.
Let's say 85. So, that's 15.
How bout the next one? Well, it's sine 5 pi t. I think I'll stop here as soon
as we do this one because at this point it's clear what's
happening. This is 100 squared minus,
that's 15 squared is 225, so that's about 125 with a
negative sign. So, minus this divided by 125.
And, after this they are going to get really quite small
because the next one will be seven pi squared.
That's 400, and this is becoming negligible.
So, what's happening? So, it's approximately,
in other words, 0.005 plus the next coefficient
is, let's see, 6/10, let's say 100,
sine pi t. And, what comes next?
Well, it's now 1/20th. It's about a 20th.
Let's call that 0.005 sine three pi t,
and now so small, minus 0.01, let's say times
this last one, sine 5 pi t.
What you find, in other words,
is that the frequencies which make up the response do not
occur with the same amplitude. What happens is that this
amplitude is roughly five times larger than any of the
neighboring ones. And after that,
it's a lot larger than the ones that come later.
In other words, the main frequency which occurs
in the response is the frequency three pi.
What's happened is, in other words,
near resonance has occurred. So, if omega is ten,
very near resonance, that is, it's not too close,
but it's not too far away either, occurs for the frequency
three pi in the input. Now, where's the frequency
three pi in the input? It isn't there.
It's just that green thing. Where in that is the frequency
three pi? I can't answer that for you,
but that's the function of Fourier series,
to say that you can decompose that green function into a sum
of frequencies, as it were, and the Fourier
coefficients tell you how much frequency goes into each of
those f of t's. Now, so, f of t is decomposed
into the sum of frequencies by the Fourier analysis.
But, the system isn't going to respond equally to all those
frequencies. It's going to pick out and
favor the one which is closest to its natural frequency.
So, what's happened, these frequencies,
the frequencies and their relative importance in f of t
are hidden, as it were.
They're hidden because we can't see them unless you do the
Fourier analysis, and look at the size of the
coefficients. But, the system can pick out.
The system picks out and favors, picks out for resonance,
or resonates with, resonates with the frequencies
closest to its natural frequency.
Well, suppose the system had natural frequency,
not ten. This is a put up job.
Suppose it had natural frequency five.
Well, in that case, none of them are close to the
hidden frequencies in f of t, and there would be no
resonance. But, because of the particular
value I gave here, I gave the value ten,
it's able to pick out n equals three as the most important,
the corresponding three pi as the most important frequency in
the input, and respond to that. Okay, so this is the way we
hear, give or take a few thousand pages.
So, what does the ear do? How does the ear,
so, it's got that thing, messy curve,
which I erased, which has a secret,
which just has three hidden frequencies.
Okay, from now on I hand wave, right, like they do in other
subjects. So, we got our frequency.
So, it's got a [SINGS]. That's one frequency.
[SINGS] And, what goes in there is the sum
of those three, and the ear has to do something
to say out of all the frequencies in the world,
I'm going to respond to that one, that one,
and that one, and send a signal to the brain,
which the brain, then, will interpret as a
beautiful triad. Okay, so what happens is that
the ear, I don't talk physiology, and I never will
again. I know nothing about it,
but anyway, the ear, when you get far enough in
there, there are little three bones, bang, bang,
bang; this is the eardrum, and then there's the part which
has wax. Then, there's the eardrum which
vibrates, at least if there is not too much wax in your ear.
And then, the vibrations go through three little bones which
send the vibrations to the inner ear, which nobody ever sees.
And, the inner ear, then, is filled with thick
fluid and a membrane, and the last bone hits up
against the membrane, and the membrane vibrates.
And, that makes the fluid vibrate.
Okay, good. So, it's vibrating according to
the function f of t. Well, what then?
Well, that's the marvelous part.
It's almost impossible to believe, but there is this,
sort of like a snail thing inside.
I've forgotten the name. It's cochlea.
And, it has these hairs. They are not hairs really.
I don't know what else to call them. They're not hairs. But, there are things so long,
you know, they stick up. And, there are 20,000 of them.
And, they are of different lengths.
And, each one is tuned to a certain frequency.
Each one has a certain natural frequency, and they are all
different, and they are all graded, just like a bunch of
organ pipes. And, when that complicated wave
hits, the complicated wave hits, each one resonates to a hidden
frequency in the wave, which is closest to its natural
frequency. Now, most of them won't be
resonating at all. Only the ones close to the
frequency [SINGS], they'll resonate,
and the nearby guys will resonate, too,
because they will be nearby, almost have the same natural
frequency. And, over here,
there will be a few which resonate to [SINGS],
and finally over here a few which go [SINGS],
and each of those little hairs, little groups of hairs will
signal, send that signal to the auditory nerve somehow or other,
which will then carry these three inputs to the brain,
and the brain, then, will interpret that as
you are hearing [SINGS]. So, the Fourier analysis is
done by resonance. You here resonance because each
of these things has a certain natural frequency which is able,
then, to pick out a resonant frequency in the input.
I'd like to finish our work on Fourier series.
So, for homework I'm asking you to do something similar.
Taken an input. I gave you a frequency here,
a different omega naught, a different input,
as you by means of this Fourier analysis to find out which it
will resonate, which of the hidden frequencies
in the input the system will resonate to, just so you can
work it out yourself and do it. Now, I'd like to first try to
match up what I just did by this formula with what's in your
book, since your book handles the identical problem but a
little differently, and it's essentially the same.
But I think I'd better say something about it.
So, the book's method, and to the extent which any of
these problems are worked out in the notes, the notes do this,
too. Use substitution.
Base uses differentiation of Fourier series term by term.
The work is almost exactly the same as here.
And, it has a slight advantage, that it allows you,
the book's method has a slight advantage that it allows you to
forget this formula. You don't have to know this
formula. It will come out in the wash.
Now, for some of you, that may be of colossal
importance, in which case, by all means,
use the book's method, term by term.
So, it requires no knowledge of this formula because after all,
I base this solution, I simply wrote down the
solution and I based it on the fact that I was able to write
down immediately the solution to this and put as being that
response. And for that,
I had to remember it, or be willing to use complex
exponentials quickly to remind myself.
There's very, very little difference between
the two. Even if you have to re-derive
that formula, the two take almost about the
same length of time. But anyway, the idea is simply
this. With the book,
you assume. In other words,
you take your function, f of t.
You expand it in a Fourier series.
Of course, which signs and cosines you use will depend upon
what the period is. So, you assume the solution of
the form-- Well, if I, for example,
carried out in this particular case, I don't know if I will do
all the work, but it would be natural to
assume a solution of the form, since the input looks like the
green guy. Assume a solution which looks
the same. In other words,
it will have a constant term because the input does.
But all the rest of the terms will be sines.
So, it will be something like (c)n times the sine of n pi t. The only question is,
what are the (c)n's? Well, I found one method up
there. But, the general method is just
plug-in. Substitute into the ODE.
Substitute into the ODE. You differentiate this twice to
do it. So, I'll do the double
differentiation and I won't stop the lecture there,
but I will stop the calculation there because it has nothing new
to offer. And, this is the way all the
calculations in the books and the solutions and the notes are
carried out. So, I don't think you'll have
any trouble. Well, this term vanishes.
This term becomes what? If I differentiate this twice,
I get summation, so, this is one to infinity
because I don't know which of these are actually going to
appear. Summation one to infinity,
(c)n times, well, if you differentiate the sine
twice, you get negative sine, right?
Do it once: you get cosine. Second time:
you get negative sine. But, each time you will get
this extra factor n pi from the chain rule.
And so, the answer will be negative (c)n times n pi squared
times the sine of n pi t. And so, the procedure is, very simply,
you substitute (x)p double prime into the differential
equation. In other words,
if you do it, we will multiply this by omega
naught squared. And, you add them.
And then, on the left-hand side, you are going to get a sum
of terms, sine n pi t times coefficients
involving the (c)n's. And, on the right,
so, you're going to get a sum involving the (c)n's,
and the sines n pi t, and on the right,
you're going to get the Fourier series for f of t,
which is exactly the same kind of expression.
The only difference is, now the sines have come with
definite coefficients. And then, you simply click the
coefficients on the left and the coefficients on the right,
and figure out what the (c)n's are.
So, by equating coefficients, you get the (c)n's.
Would you like me to carry it out?
Yeah, okay, I was going to do something else,
but I wouldn't have time to do it anyway.
So, why don't I take two minutes to complete the
calculation just so you can see you get the same answer?
All right, what do we get? If you add them up,
you get c naught, out front, plus (c)n is
multiplied by what? Well, from the top it's
multiplied by omega naught squared.
On the bottom, it's multiplied by n
pi squared. Ah-ha, where have I seen that
combination? The sum is equal to,
sorry, one half plus what is it, sum over n odd of sine n pi
t over n. So, the conclusion is that--
I'm sorry, it should be c naught times omega naught squared. So, what's the conclusion?
If c zero is one over two omega naught squared, and that (c)n,
only for n odd, the others will be even.
The others will be zero. The (c)n is going to be equal
to two over pi here. So, it's going to be two pi, two over pi times one over n
times one over omega naught squared minus n over
pi squared. This is terrible, which is the same answer we got
before, I hope. Did I cover it up?
Same answer. So, that answer at the
left-hand end of the board is the same one.
I've calculated, in other words,
what the c zeros are. And, I got the same answer as
before.
