第 7 讲：连续时间傅里叶级数
本讲内容：连续时间线性时不变系统对复指数信号的响应及其特征函数性质；将周期信号表示为成谐波关系的复指数信号的线性组合；连续时间傅里叶级数的分析式与综合式；对称和反对称周期方波示例；周期信号的逼近；傅里叶级数的收敛性。
授课教师：艾伦·V. 奥本海姆教授
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PROFESSOR: Over the last several lectures, we've dealt
with the representation of linear time-invariant systems
through convolution.
And just to remind you of our basic strategy, essentially,
the idea was to exploit the notion of linearity by
decomposing the input into a sum of basic inputs and then
using linearity to tell us that the output can be
represented as the corresponding linear
combination of the associated outputs.
So, if we have a linear system, either continuous-time
or discrete-time, for example, with continuous time, if the
input is decomposed as a linear combination of basic
inputs, with each of these basic inputs generating an
associated output, and if the system is linear, then the
output of the system is the same linear combination of the
associated outputs.
And the same statement is identical both for continuous
time and discrete time.
So the strategy is to decompose the input into these
basic inputs.
And the inputs were chosen also with some particular
strategy in mind.
In particular, for both continuous time or discrete
time, in this representation, the basic inputs used in the
decomposition are chosen, first of all, so that a broad
class of signals could be represented in terms of these
basic inputs, and second of all, so that the response to
these basic inputs is, in some sense, easy to compute.
Now, in the representation which led us to convolution,
the particular choice that we made in the discrete-time case
for our basic inputs was a decomposition of the input in
terms of delayed impulses.
And the associated outputs that that generated were
delayed versions of the impulse response.
Decomposing the input into a linear combination of these,
the output into the corresponding linear
combination of these, then led to the convolution sum in the
discrete time case.
And in the continuous-time case, a similar kind of
decomposition, in terms of impulses, and associated
representation of the output, in terms of the impulse
response, led to the convolution integral.
Now, in this lecture, and for a number of the succeeding
lectures, we'll want to turn our attention to a very
different set of basic building blocks.
And in particular, the signals that we'll be using as the
building blocks for our more general signals, rather than
impulses, as we've dealt with before, will be, in general,
complex exponentials.
So, in a general sense, in the continuous-time case, we'll be
thinking in terms of a decomposition of our signals
as a linear combination of complex exponentials,
continuous-time, or, in the discrete-time case, complex
exponentials, where z_k is complex here in discrete time
and s sub k is complex here in continuous time.
Now, the basic strategy, of course, requires that we
choose a set of inputs, basic building blocks, which have
two properties.
One is that the system response be straightforward to
compute, or in some sense, easy to compute.
And second is that it be a fairly general set of building
blocks so that we can build lots of signals out of them.
What we'll find with complex exponentials, either
continuous-time or discrete-time, is that they
very nicely have those two properties.
In particular, the notion that the output of a linear
time-invariant system is easy to compute is tied to what's
referred to as the Eigenfunction function
property of complex exponentials, which we'll
focus on shortly in a little more detail.
And second of all, the fact that we can, in fact,
represent very broad classes of signals as linear
combinations of these will be a topic and an issue that
we'll develop in detail over, in fact, the next set of
lectures, this lecture, and the next set of lectures.
Now, in doing this, although we could, in fact, begin with
our attention focused on, in general, complex exponentials,
what we'll choose to do is first focus on the case in
which the exponent in the continuous-time case is purely
imaginary, as I indicate here, and in the discrete-time case,
where the magnitude of the complex number
z_k is equal to 1.
So what that corresponds to in the continuous-time case is a
set of building blocks of the form e^(j omega_k t), and in
the discrete-time case, a set of building blocks of the form
e^(j Omega_k n).
What we'll see is a representation in these terms
leads to what's referred to as Fourier analysis.
And that's what will be dealing with over the next set
of lectures.
We'll then be exploiting this representation actually
through most of the course.
And then toward the end of the course, we'll return to
generalizing the Fourier representation to a discussion
Laplace transforms and Z-transforms.
So for now, we want to restrict ourselves to complex
exponentials of a particular form, and in fact, also
initially to continuous-time signals and systems.
So let's begin with the continuous-time case and the
complex exponentials that we want to deal with and focus,
first of all, on what I refer to as the Eigenfunction
property of this particular set of building blocks.
We're talking about basic signals of the form e^(j
omega_k t).
And the statement is that for a linear time-invariant
system, the response to one of these is of exactly the same
form, just simply multiplied by a complex factor, that
complex factor depending on what the
frequency, omega_k, is.
Now more or less, the justification for this, or the
proof, follows by simply looking at the response to a
complex exponential, using the convolution integral.
So if we put a complex exponentials into a linear
time-invariant system with impulse response h(t), then we
can express the response as I've indicated here.
We can then recognize that this complex exponentials can
be factored into two terms.
And so we can rewrite this complex
exponential as this product.
Second, recognize that this term can be taken outside the
integral, over here, because of the fact that it depends
only on t and not on Tau.
And so what we're left with, when we track this through, is
that, with a complex exponential input, we get an
output which is the same complex exponential, namely
this factor, times this integral.
And this integral is what I refer to above as H(omega_k).
And so, in fact, we put in a complex exponential, we get
out a complex exponentials of the same frequency, multiplied
by a complex constant.
And that is what's referred to as the Eigenfunction property,
Eigenfunction meaning that an Eigenfunction of a system, or
mathematical expression, is a function which, when you put
it through the system, comes out looking exactly the same
except for a change in amplitude, the change in
amplitude being the Eigenvalue.
So in fact, this function is the Eigenfunction.
And this value is the Eigenvalue.
OK, now it's because of the Eigenfunction property that
complex exponentials are particularly convenient as
building blocks.
Namely you put it through the system, they come out with the
same form and simply scale.
The other part to the question, related to the
strategy that we've been pursuing, is to hope that
these signals can be used as building blocks to represent a
very broad class of signals through a linear combination.
And in fact, that turns out to be the case with complex
exponentials.
As we work our way through that, we'll first consider the
case of periodic signals.
And what that leads to is a representation of periodic
signals through what's called the Fourier series.
Following that, we'll turn our attention to non-periodic, or
as I refer to it, aperiodic signals.
And the representation that's developed in terms of linear
combinations of complex exponentials is what's
referred to as the Fourier transform.
So the first thing we want to deal with are periodic signals
and the Fourier series.
So what we're talking about then is the continuous-time
Fourier series.
And the Fourier series is a representation for periodic
continuous-time signals.
We have a signal, then, which is periodic.
And we're choosing T_0 to denote the period.
So it's T_0 that corresponds to the period of
our periodic signal.
omega_0 is 2 pi / T_0, as you recall from our discussion pf
periodic signals and sinusoids before.
And that's 2 pi f_0.
So this is the fundamental frequency.
Now let's examine, first of all, complex exponentials, and
recognize, first of all, that there is a complex exponential
that has exactly the same period and fundamental
frequency as our more general periodic signal, namely the
complex exponential e^(j omega_0 t), where omega_0 is 2
pi / T_0, or equivalently, T_0 is 2 pi / omega_0.
Now that's the complex exponential which has T_0 as
the fundamental period.
But there are harmonically related complex exponentials
that also have T_0 as a period, although in fact,
their fundamental period is shorter.
So we can also look at complex exponentials of the form e^(j
k omega_0 t).
These likewise are periodic with a period of T_0.
Although, in fact, their fundamental period is T_0 / k,
or equivalently, 2 pi divided by their fundamental
frequency, k omega_0.
So as k, an integer, varies, these correspond to
harmonically related complex exponentials.
Now what the Fourier series says, and we'll justify this
bit by bit as the discussion goes on, what the Fourier
series says, and in fact, what Fourier said, which was
essentially his brilliant insight, is that, if I have a
very general periodic signal, I can represent it as a linear
combination of these harmonically-related complex
exponentials.
So that representation is what I've indicated here.
And this summation is what will be referred to as the
Fourier series.
And as we proceed with the discussion, there are two
issues that will develop.
One is, assuming that our periodic signal can be
represented this way, how do we determine the Fourier
series coefficients, as they're referred to, a_k.
That's one question.
And the second question will be how broad a class of
signals, in fact, can be represented this way.
And that's another question that we'll deal with
separately.
Now just focusing on this representation for a minute,
this representation of the Fourier series, which I've
repeated again here, is what's referred to as the complex
exponential form of the Fourier series.
And it's important to note, incidentally, that the
summation involves frequencies, k omega_0, that
are both positive and negative.
In other words, this index k runs over limits that include
both negative values and positive values.
Now that complex exponential form is one representation for
the Fourier series.
And in fact, it's the one that we will be principally relying
on in this course.
There is another representation that perhaps
you've come across previously and that in a variety of other
contexts is typically used, which is called the
trigonometric form for the Fourier series.
Without really tracking through the algebra,
essentially we can get to the trigonometric form from the
complex exponential form by recognizing that if we express
the complex coefficient in polar form or in rectangular
form and expand the complex exponential term out in terms
of cosine plus j sine, using just simply Euler's relation,
then we will end up with a representation for the
periodic signal, or a re-expression of the Fourier
series expression that we had previously, either in the form
that I indicate here, where now the periodic signal is
expressed in terms of a summation of cosines with
appropriate amplitude and phase.
Or another equivalent trigonometric form involves
rearranging this in terms of a combination
of cosines and sines.
Now in this representation, the frequencies of the
sinusoids vary only over positive frequencies.
And typically one thinks of periodic signals as having
positive frequencies associated with them.
However, let's look back and the complex exponential form
for the Fourier series at the top of the board.
And in that representation, when we use this
representation, we'll find it convenient to refer to both
positive frequencies and negative frequencies.
So the representation that we will most typically be using
is the complex exponential form.
And in that form, what we'll find as we think of
decomposing a periodic signal into its components at
different frequencies, it will involve both positive
frequencies and negative frequencies.
Okay, now we have the Fourier series representation, as I've
indicated here.
Again, so far I've sidestepped the issue as to whether this
in fact represents all the signals that
we'd like to represent.
Let's first address the issue of how we determine these
coefficients a_k, assuming that, in fact, this
representation is valid.
And again, I'll kind of move through the
algebra fairly quickly.
The algebraic steps are ones that you can pursue more
leisurely just to kind of verify them and
step through them.
But essentially, the algebra develops out of the
recognition that if we integrate a complex
exponential over one period, T_0--
and I mean by this notation that this is an integral over
a period, where I don't particularly care where the
period starts and where the period stops, in other words,
exactly what period I picked--
that this integral is equal to T_0 when m is equal to 0.
And it's equal to 0 if m is not equal to 0.
That follows simply from the fact that if we substitute in
for using or Euler's relation, so that we have the integral
of a cosine plus j times the sine, if m is not equal to 0,
then both of these integrals over a period are 0.
The integral of a of a periodic of a sinusoid, cosine
or sine, over an integral number of periods is 0.
Whereas, if m is equal to 0, this integral will be equal to
T_0, the integral of the cosine.
And the integral of the sine is equal to 0.
Okay, well, the next step in developing the expression for
the coefficient a_k is to refer back to the Fourier
series expression, which was that x(t) is equal to the sum
of a_k e^(j k omega_0 t).
If we multiply both sides of that by e^(-j n omega_0 t),
and integrate that over a period--
both sides of the equation integrated over a period, so
these two equations are equal--
and then in essence, interchange the summation and
the integration so that this part of the expression comes
outside the sum, and then we combine these two complex
exponentials together, where we come out is the expression
that I've indicated here.
And then essentially what happens at this point,
algebraically, is that we use the result that we just
developed to evaluate this integral.
So multiplying both sides of the Fourier series and then
doing the integration leads us, after the appropriate
manipulation, to the expression
that I have up here.
And this integral is equal to T_0 if k is equal to n,
corresponding to 0 up here.
And it's 0 otherwise, which is what we had demonstrated or
argued previously.
And the upshot of all that, then, is that the right hand
side of this expression disappears except for the term
when k is equal to n.
And so finally, we have what I indicate here, taking T_0 and
moving it over to the other side of the equation, that
then tells us how we determine the Fourier series
coefficients a_n, or a_k.
So that, in effect, then is what we refer to as the
analysis equation, the equation that begins with x(t)
and tells us how to get the Fourier series coefficients.
What I'll refer to as the Fourier series synthesis
equation is the equation that tells us how to build x(t) out
of these complex exponentials.
So we have the synthesis equation, which is the one we
started from.
We have the analysis equation, which is the equation that we
just developed.
So we in effect have gone through the issue of, assuming
that a Fourier series representation is in fact
valid, how we get the coefficients.
We'll want to address somewhat the question of how broad a
class of signals are we talking about.
And what's in fact amazing, and was Fourier's amazing
insight, was that it's a very broad class of signals.
But let's first look at just some examples in which we take
a signal, assume that it has the Fourier series
representation, and see what the Fourier series
coefficients look like.
So we'll begin with what I refer to as an antisymmetric
periodic square wave--
periodic of course, because we're talking about periodic
signals; square wave referring to its shape; and
antisymmetric referring to the fact that it
is an odd time function.
In other words, it is
antisymmetric about the origin.
Now the expression for the Fourier series coefficients
tells us that we determine a_k by taking 1 / T0 times the
integral over a period of x(t), e^(-j k omega_0 t) dt.
The most convenient thing in this case is to choose a
period, which let's say goes from -T_0 / 2 to +T_0 / 2.
So here x(t) is -1.
Here x(t) is +1.
And so I've expressed the Fourier series coefficients as
this integral, that's from -T_0 / 2 to 0.
And then added to that is the positive part of the cycle.
And so we have these two integrals.
Now, I don't want to track through the details of the
algebra again.
I guess I've decided that that's much more fun for you
to do on your own.
But the way it comes out when you go through it is the
expression that I finally indicate after suggesting that
there are few more steps to follow.
And what develops is that those two integrals together,
for k not equal to 0, come out to this expression.
And that expression is not valid for k = 0.
For k equal to 0, we can go back to the basic expression
for the Fourier series, which is 1 / T_0, the integral over
a period, x(t) e^(-j k omega_0 t) dt.
For k = 0, of course this term just simply becomes 1.
And so the zeroth coefficient is 1 / T_0 times the integral
of x(t) over a period.
Now, going back to the original function that we
have, what we're saying then is that the zeroth coefficient
is 1 / T_0 times the integral over one period, which is, in
effect, the average value.
And it's straightforward to verify for this case that
average value is equal to 0.
Now let's look at these Fourier series coefficients on
a bar graph.
And I've indicated that here.
The expression for the Fourier series
coefficients we just developed.
And it involves--
it's 0 for k = 0, it's a factor of this
form for k =/= 0.
Plotted on a bar graph, then we see values like this, 0 at
k = 0 and then associated values.
And there are a number of things to focus on when you
look at this.
One is the fact that the Fourier series coefficients
for this example are purely imaginary.
A second is that the Fourier series coefficients for this
example are an odd sequence.
In other words, if you look at this sequence, what you see
are these values for -k flipped over.
So they're imaginary and odd.
And what that results in, when you look at the trigonometric
form of the Fourier series, is that in fact, those
conditions, if you put the terms all together, lead you
to a trigonometric representation, which involves
only sine terms--
in other words, no cosine terms.
Let me just draw your attention to the fact that,
since a_k's are imaginary, this j takes care of that fact
so that these coefficients are in fact real.
So what this says is that for the antisymmetric square wave,
in effect, the Fourier series is a sine series.
The antisymmetric square wave is an odd function.
Sinusoids are odd functions.
And so this is all kind of reasonable, that we're
building an odd function out of odd functions.
As an additional aside, which I won't exploit or refer to
any further here, but just draw your attention to, is
that another aspect of this periodic square wave, the
particular one that we chose, is that it is what's referred
to as an odd harmonic function.
In other words, for even values of k, the Fourier
series coefficients are 0.
They're are only non-zero for odd values of k.
Now let's look at another example.
Another example is the symmetric
periodic square wave.
And this is in fact example 4.5, worked out in more detail
in the text.
Then I won't bother to work this out in detail here,
except to draw your attention to several points.
Here is the symmetric periodic square wave.
And what I mean by symmetric is that it's
an even time function.
Now just kind of extrapolating your intuition, what you
should expect is that if it's only an even time function, it
should be built up or buildable, if it's buildable
at all, out of only even sinusoids.
And in fact, that's the case.
So if we look at the Fourier series coefficients for this,
is zeroth coefficient, again, is the average value, which in
this case, is 1/2.
Here I've plotted pi times the Fourier series coefficients.
So the zeroth value is pi / 2.
The coefficients are now an even sequence, in other words,
symmetric about k = 0.
And the consequence of that is that when you take these
coefficients and put together the equivalent trigonometric
form, the trigonometric form involves only cosines and no
sine terms.
Now you'll see this in other examples, not that we'll do in
the lecture, but examples in the text and in the video
manual, if in fact the square wave was neither symmetric or
antisymmetric, then the trigonometric form would
involve both sines and cosines.
And that is, of course, the more general case.
Furthermore, in the two examples I've shown here, in
both cases, the signal is odd harmonic.
In other words, for even values of k, the coefficients
are equal to 0.
Although I won't justify that here, that's a consequence of
the fact that this symmetry is exactly about half a period.
And if you made the on time of the square wave different in
relation to the off time, then that
property would also disappear.
Now what's kind of amazing, actually, is that if we take a
square wave, like I have here or as I had in the
antisymmetric case, the implication is that I can
build that square wave by adding up
enough sines or cosines.
And it really seems kind of amazing because the square
wave, after all, is a very discontinuous function.
Sinusoids are very continuous.
And it seems puzzling that in fact you can do that.
Well let's look in a little bit of detail how the
sinusoidal terms add up to build a square wave.
And to do that, let's first define what I refer to as a
partial sum.
So here we have the expression which is the synthesis
equation, telling us how x(t) could be represented as
complex exponentials if it can be.
And let's consider just a finite number of
terms in this sum.
And so x_n(t), of course, as n goes to infinity, approaches
the infinite sum that we're talking about.
And although we could do this more generally, let's not.
Let's focus on the symmetric square wave case, where
because of the symmetry of these coefficients, namely
that a_k is equal to a_(-k), we can rewrite these terms as
cosine terms.
And so this partial sum can be expressed the way I'm
expressing it here.
Well let's look at a few of these terms.
On the graph, I have, first of all, x(t), which is our
original square wave.
The term that I indicate here is the factor of 1/2,
which is this term.
With n = 1, that would correspond to adding one
cosine term to that.
And so the sum of those two would be this, which looks a
little closer to the square wave, but certainly not very
close to it at all.
And in fact, it's somewhat hard to imagine without seeing
the terms build up how in fact, by adding more and more
terms, we can generate something that is essentially
flat, except at the discontinuities.
So let's look at this example.
And what I'd like to show is this example, but now as we
add many more terms to it.
And let's see in fact how these individual terms add up
to build up the square wave.
So this is the square wave that we want to build up
through the Fourier series as a sum of sinusoids.
And the term for k = 0 will be a constant which represents
the DC value of this.
And so in the partial sum, as we develop it, the first thing
that we'll show is just the term for k = 0.
Now for k = 1, we would add to that one sinusoidal term.
And so the sum of the term for k = 1 and k = 0
is represented here.
Now when we go to k = 2, because of the fact that this
is an odd harmonic function, in fact, the term for k = 2
will have zero amplitude and so this won't change.
Here we show the Fourier series with k = 2
and there's no change.
And then we will go to k = 3.
And we will be adding, then, one
additional sinusoidal term.
Here is k = 3.
When we go to k = 4, again, there won't be any change.
But there will be another term that's added at k = 5 here.
Then k = 6, again, because it's odd harmonic, no change.
And finally k = 7 is shown here.
And we can begin to see that this starts to look somewhat
like the square wave.
But now to really emphasize how this builds up, let's more
rapidly add many more terms, and in fact increase the
number of terms up to about 100, recognizing that the
shape will only change on the inclusion of the odd-numbered
terms, not the even-numbered terms, because it's an odd
harmonic function.
So now we're increasing and we're building up toward k =
100, 100 terms.
And notice that it is the higher-order terms that tend
to build up the discontinuity corresponding to the notion
that the discontinuity, or sharp edges in a signal, in
fact, are represented through the higher frequencies in the
Fourier series.
And here we have a not-too-unreasonable
approximation to the original square wave.
There is the artifact of the ripples at the discontinuity.
And in fact, that rippling behavior at the discontinuity
is referred to the Gibbs phenomenon.
And it's an inherent part of the Fourier series
representation at discontinuities.
Now to emphasize this, let's decrease the number
of terms back down.
And we will carry this down to k = 1, again to emphasize how
the sinusoids are building up the square wave.
Here we are back at k = 1.
And then finally, we will add back in the sinusoids
that we took out.
And let's build this back up to 100 terms, showing the
approximation that we generated with 100 terms to
the square wave.
Okay, so what you saw is that, in fact, we got awfully close
to a square wave.
And the other thing that was kind of interesting about it
as it went along was the fact that, with the low
frequencies, what we were tending to build was the
general behavior.
And as the higher frequencies came in, that tended
contribute to the discontinuity.
And in fact, something that will stand out more and more
as we go through our discussion of Fourier series
and Fourier transforms, is that general statement, that
it's the low-frequency terms that represent the broad time
behavior, and it's the high-frequency terms that are
used to build up the sharp
transitions in the time domain.
Now we need to get a little more precise about the
question of how, in fact, the Fourier series, or when the
Fourier series represents the functions that we're talking
about and in what sense they represent them.
And so if we look again at the synthesis equation, what we
really want to ask is, if we add up enough of these terms,
in what sense does this sum represent this time function?
Well, let's again use the notion of our partial sum.
So we have the partial sum down here.
And we can think of the difference between this
partial sum and the original time function as the error.
And I've defined the error here.
And what we would like to know is does this error decrease as
we add more and more terms?
And in fact, in what sense, if the error does decrease, in
what sense does it decrease?
Now in detail this is a fairly complicated
and elaborate topic.
I don't mean to make that sound frightening.
It's mainly a statement that I don't want to
explore in a lot of detail.
But it relates to what it's referred to as the issue of
convergence of the Fourier series.
And the convergence of the Fourier series, the bottom
line on it, the kind of end statement, can be made in
several ways.
One statement related to the convergence of the Fourier
series is the following.
If I have a time function, which is what is referred to
as square integrable, namely its integral, over
a period, is finite.
Then what you can show, kind of amazingly, is that the
energy in that error, in other words, the energy and the
difference between the original function and the
partial sum, the energy in that, goes to 0
as n goes to infinity.
A somewhat tighter condition is a condition referred to as
it Dirichlet conditions, which says that if the time function
is absolutely integrable, not square integrable, but
absolutely integrable--
and I've kind of hedged the issue by just simply referring
to x(t) as being well behaved--
then the statement is that the error in fact goes to 0 as n
increases, except at the discontinuities.
And what well behaved means in that statement is that, as
discussed in the book, there are a finite number of maxima
and minima in any period and a finite number of finite
discontinuities, which is, essentially, always the case.
So under square integrability what we have is the statement
not that the partial sum goes to the right value at every
point, but that the energy in the error goes to 0.
Under the Dirichlet conditions, it says that, in
fact, the signal goes to the right value at every time
instant except at the discontinuities.
So going back to the square wave, the square wave
satisfies either one of those conditions.
And so what the consequence is is that, with the square wave,
if we looked at the error, then in fact what we would
find is that the energy in the error would go to zero as we
add more and more terms in the partial sum.
And in fact, since the square wave also satisfies the
Dirichlet conditions, the actual value of the error, the
difference between the partial sum and the true value, will
actually go to 0.
That difference will go to 0 except at the discontinuities.
And that, in fact, is kind of evident as we watch the
function build up by adding up these terms.
And so in fact, let's go back and see again the development
of the partial sums in relation to the
original time function.
Let's observe, this time again, basically what we saw
before, which is that it builds up to the right answer.
And furthermore what we'll plot this time, also as a
function of time, is the energy in the error.
And what we'll see is that the energy in the error will be
tending towards 0 as the number of terms increases.
So once again, we have the square wave.
And we want to again show the buildup of the Fourier series,
this time showing also how the energy in the error decreases
as we add more and more terms.
Well, once again, we'll begin with k = 0, corresponding to
the constant term.
And what's shown on the bottom trace is the energy in the
error between those two.
And we'll then add the term k = 1 to the DC term and we'll
see that the energy will decrease when we do that.
Here we have then the sum of k = 0 and k = 1.
Now with k = 2, the energy won't decrease any further
because it's an odd harmonic function.
That's what we've just added in.
When we add in the term for k = 3, again, we'll see the
energy in the error decrease as reflected
in the bottom curve.
So there we are at k = 3.
When we go to k = 4, there again is no
change in the error.
At k = 5, again the error decreases.
k = 6, there will be no change again.
And at k = 7, the energy decreases.
And now let's show how the error decreases by building up
the number of terms much more rapidly.
Already the error has gotten somewhat small on the scale in
which we're showing it, so let's expand out the error
scale, the vertical axis displaying the energy in the
error, so that we could watch how the energy decreases as we
add more and more terms.
So here we have the vertical scale expanded.
And now what we'll do is increase the number of terms
in the Fourier series and watch the energy in the error
decreasing, always decreasing, of course, on the inclusion of
the odd-numbered terms and not on the inclusion of the
even-numbered terms because of the fact that it's an odd
harmonic function.
Now the energy in the error asymptotically will approach
0, although point by point, the Fourier series will never
be equal to the square wave.
It will, at every instant of time, except at the
discontinuities, where there will always be some ripple
corresponding to what's referred to as the Gibbs
phenomenon.
So what we've seen, then, is a quick look at the Fourier
series representation of periodic signals.
We more broadly want to have a more general representation of
signals in terms of complex exponentials.
And so our next step will be to move toward a
representation of nonperiodic or aperiodic signals.
Now the details of this, I leave for the next lecture.
The only thought that I want to introduce at this point is
the basic strategy which is somewhat amazing and kind of
interesting to reflect on in the interim.
The basic strategy with an aperiodic signal is to think
of representing this aperiodic signal as a linear combination
of complex exponentials by the simple trick of periodically
replicating this signal, generating a periodic signal
using a Fourier series representation for that
periodic signal, and then simply letting the period go
to infinity.
As the period goes to infinity, that periodic signal
becomes the original aperiodic one that we had before.
And the Fourier series representation then becomes
what we'll refer to the Fourier transform.
So that's just a quick look at the basic idea and approach
that we'll take.
In the next lecture, we'll develop this a little more
carefully and more fully, moving from the Fourier
series, which we've used for periodic signals, to develop
the Fourier transform, which will then be representation
for aperiodic signals.
Thank you.