So, in this video, we are going to talk about signals again but from a very
different view point. From all the previous videos when we
talked about circuits, we discovered that if we could express a signal as a
superposition of complex exponentials different thinking as a sinusoid, Then
solving the circuit was easy. the complex exponential was a natural
input to study the, the, circuit, circuit's behavior with.
however, sinusoids by themselves are pretty limiting.
you need to talk about more general, signals.
It still would like to express a signal as a super position, a complex
exponentials can we do that? The answer is yes.
This, so we're going to begin by talking about periodic signals, something called
the Fourier series, and what we're going to discover Which is a very interesting
point, is that you can define a signal, as a function in time, or as a function
of frequency, and we can get between the 2 representations of exactly the same
quantity, using the Fourier series. And so we're going to start talking about
signals and what we call domains, we can talk about them as functions of time or
we can talk about them as functions of frequency, whichever seems to be most
convenient. But to get started we need to find the
Fourier series and understand it a little bit.
So, this is the Fourier series for a periodic signal.
This is very important. It's periodic with period T.
And, Fourier claimed that at the beginning of the 19th century, that you
could express a periodic signal as a superposition of complex exponentials.
The frequency of which was K over T, so this is the Kth harmonic of the
fundamental frequency 1 over T, so the fundamental is the first harmonic, then
there's the second. The third and the fourth.
You notice, K here goes negative, so we actually talk about the -1 harmonic, and
-2 harmonic, etcetera, and we'll see how to handle that in just a second.
The, coefficients of this expansion, are called the Fourier coefficients.
So what Fourier claimed is that, if you give me a, time signal.
There are some coefficients here, C sub k.
That allow me to, express. This signal is a weighted superposition
of complex exponentials. Now the question is what are those Cks?
How do you find them? Well we're going to use properties of the complex exponential
here, of what are called harmonically related complex exponentials, is
extremely important. And, it goes by the technical term,
orthogonality. So, I want to consider this interval
Between, a complex exponential having a harmonic number of k.
And one that has a harmonic number of -l. Well, let's do this interval.
So, we have 1 over t, and we have the interval.
I'm going to merge the two exponentials, and the exponents, okay, and do this
intervals very easy we just have to bring down the That the, quantity multiplying
t. Put that in the denominator.
And the interval gives us back the exponential, of course.
So we get 1 over t. 1 over j 2 pi, k.
-l T * times the same thing. So now, it's available, that would be 0
to 2. Well, what do we get when we evaluated at
the upper limit here? Well, the T is capital T so these cancel.
And what you're left with is e ^ +j 2 pi * k - l.
And, evaluating at 0 gives us, 1 of course.
Well, what is e ^ j 2 pi * an integer? For any integer.
Well, I hope you said that the answer is 1.
So, this term turns out to always be 1 for any value of K and N.
The difference between 2 integers is another integer.
So. You get the J2 pi and integer is gotta be
1, so it looks like we get 0 for an answer.
1-1 here, that seems to be kind of odd until we realize over here in the
denominator that we have a k - l. So, when k = l The numerator here, given
by this term, is 1, well the denominator is also zero, so we have a problem.
Numerator is zero, denominator is zero, what's, how we figure this out.
Let's go back and look at the original interval.
When k=l, well, rewriting it just like I did before, k=l, this is 0, which makes
this 1. And so the integral gets very easy.
We get the integral of 1, from 0 to t, that's capital T.
Divide by capital T, you get 1. So that's why this is true.
As long as k does not equal l, then you get, this interval of 0.
But when k and l agree, you get 1. Okay, well that's a very important
property, it's very useful So, suppose I, take my expression for s(t).
Alright, and i'm going to multiply both sides now by that, and integrate.
Well, we know over here what happens. Over here the only time you get anything
is when k=l, so what happens, all the terms in this sum disappear except for
the l, whatever l is, and that leaves us with the following result, that The c sub
l term is = to the original signal * a complex exponential, having a frequency
of -l over t, integrate, and / capital T. So the idea is, in order for this.
To be true, the c sub ks, the c sub ls, have to be given by this.
Or else it cannot be true. This is the way, that given the, periodic
signal, as a function of time, I can find, it's kth Fourier coefficient.
So there's an equivalence here. Which is an important thing to realize.
So. Given the signal, I can find the c sub
Ks, by using this expression. And, given the c sub Ks, I can find the
signal. I'm using this expression.
So I can go back and forth as the arrow suggests here.
And between the time domain, expression of the signal is that function of time,
and the frequency domain The expression of the signal as a function here after
harmonic index, where as a function of K, you have to give me the C sub K's, and
then the complex exponentials are understood because we know the period
T's. This is going to be a very important
concept as we go through. An important result for us.
Right now, because of the. Your position property.
Well, let's do, let's figure out the Fourier series from, for some simple,
wave forms. So here we have our, sine wave.
And I've given it a phase this time to make it a bit more general.
So the period is 1 over the frequency. So, this is the fundamental frequency.
It's just f 0. And, I could do the integral, but I want
to illustrate something by using Euler's formula.
So, as we all know, Euler's formula says, that a cosine can be written as e ^
Minus, of a complex exponential remaining e to the plus, for a complex exponential.
And, the a/2 is the scaling factors you need to make, you need to get it to,
work. Well, lets look at the Fourier series,
and consider what I call the syemetric terms.
So here's the Fourier series, and I'm just writing it out, and I'm going to
pick up the -kth and the kth terms here, and not worry about the other ones.
So, this is what I mean by symmetry, I'm picking up the terms where the in, the,
harmonic numbers are negatives to each other.
Well, notice that this looks a lot like that.
And that looks a lot like that. So I think, if we pick k= to 1.
I think we have found our Fourier series coefficients.
All the rest of these are going to be zero.
Because we found how to express the, sine wave by this sum.
And that's exactly what the answer is. Is that the, Fourier coefficients are all
zero, except when k=1. And we get these, quantities.
For the, harmonic fundamental frequency rather and it's negative counterpart.
This is another interesting fact, in that, to, we've seen this already but to
express a signal, as would now you'd be needing negative frequency term, in it's
pi's a good frequency. And we're going to see, that their
Fourier coefficients are related to each other in a very simple way, which we'll
get to in just a second. Now I want to go on to a square wave
example now. where were going to have to do the
interval to see what we get. So here's our square wave and, you note
that it's a discontinuous function. in fact, I didn't even define what it
was, at the point of discontinuity. We'll see that's an interesting point in
just a second. And, it turns out in order to do the
integral we're going to need to express it in, mathematically.
So, this is a mathematical expression for it.
Notice I'm leaving open what it is at t over 2.
Okay, so to do the integral, what we have to do is split the range of integration
up so that within each range we have an m loop formula for what the signal is.
So from 0 to t/2. It was just A from T/2 to t, it's minus
A. So now we can do the integrals, and these
integrals are pretty easy to do. I'm going to do the one on the left, so
we get 1/T. And, and again we get a, and then we're
going to get 1 over that exponent is j 2 pi, k/t * e to the -j2pi kt over t.
And I evaluate with that from zero to T/2.
Okay at T/2, the let's see the t, t's cancel and that 2 cancels and so what
we're left with is e^-j pi k. Minus 1.
The, exponential of 0 is, is 1. What is e^j*pi*k? What is that equal to?
It's equal to something very simple. In particular, I would point out, that
that's e, to the minus j pi, all to the k.
That's the same thing. Well, in some quarters, this is
considered one of the most interesting and important formulas in mathematics.
E to the - j pi. Even e with the + j pi is -1.
So, what we get here is that this is -1 to the k.
And then you bring on the -1. Turns out this other integral is the same
thing except with a - sign, and because of the - sign out here the two terms
merge and what we get is the following expression, so.
The Fourier coefficient for a square wave is given by this.
And when K is an odd number like 1, 3, etc.
this is mine, this is minus 1. You get minus 1 minus 1.
Gives you a minus 2 and some things cancel leaving you with this.
Oh when k is even, well that's one, and 1-1=0 so you always get zero here.
So, the result is that. The spectrum.
The coeffi-, Fourier coefficients of a square wave only consist of odd harmonic
terms. And furthermore, their harmonics are
decaying. As one over k.
As the harmonic number gets bigger and bigger.
The coefficient amplitude gets bigger, gets smaller and smaller, I'm sorry,
smaller and smaller. So, lets plot it, lets see what it looks
like. So, here's our, 4a coefficient formula,
and, here I'm plotting, the spectrum Of a square root.
I'm plotting the magnitude in phase of the Fourier coefficient.
This should remind you of what we did for transfer functions where we plotted the
magnitude and phase of the transfer function.
And so I really am thinking about this Fourier coefficient though, as a function
of frequency Just like I thought about the transfer function as a function of
frequency. And, re-express this axis in frequency in
just a second. So, the magnitude.
You can see how the, coefficients go down, like 1 over k, when they're odd.
And when they're even, you get zero, which is what the expression says.
So this is the odd harmonic structure, if you will, of the superposition of, of
complex exponentials that express a square wave.
The phase is interesting. The Again, we have, something in Cartisan
form, so what's the phase of that? We need to know the phase of 1/J, and there
are lots of ways of thinking about that. That's equal to 1/e^j on over to And so
when that goes upstairs you get even minus j pi/2 so the phase is minus pi/2
okay? And since that term is the same for all of them the phase is a constant at
the odd harmonic terms and at the zero everywhere else because the phase is not
You can define when your magnitude is 0, you might as well call it zero.
So, that's the spectrum, and a spectrum is a function of frequency.
All I've done here, is relabel the axis, to ex, express exactly, what is, the
spectrum is a function of F, this is in Hertz.
Okay? So, it consists of something, a, complex exponential at 1/t, 3/t, 5/t,
etc. So with the period was 1 millisecond.
That'd make the fundamental frequency 1 kilohertz, so a square wave that has a
period of a millisecond, as Forier series coefficients, has a spectrum in it,
kilohertz, 3 kilohertz, 5 kilohertz, etcetera.
And the amplitude is going down, at 1 over the harmonic Now it expresses, the
superposition of complex exponentials, are what we need, to produce a square
wave. Very interesting.
And now let's do another example, that's a little bit easier to do.
but it's a, this is a very important example it turns just a little bit later
in this course. So, it's the periodic pulse sequence.
So here's our well-known pulse. It has a duration of delta, and I'm going
to repeat it every capital T, to make a, what's called a periodic pulse train.
So again, I need a formula for it over The period, and here is our formula for
the pulse. which I think is pretty straightforward.
And doing the interval is very straightforward.
You just plug it in, and you get the following answer.
Well, I'd like to plot this like I did for the square wave example.
But it isn't obvious to me at all what this looks like.
I could just type the expression in to the computer and have it plotted.
But I'd rather we get some more understanding here.
And I want to show you a cute little trick.
This is real important thing to appreciate.
And it has to do with this little part of it right here.
So, a little math. So, I want you to note, that I have a
term that looks like 1 minus e to the minus j theta.
Suppose I factor out, what we call half the phase.
So, I am dividing this phase over here by 2.
And I pull out. That factor.
hope you agree that this and this are the same thing.
Okay? That cancels that, and now I get my phase back to where it was.
Well, that's very exciting because if I divide this by 2j Multiplied by 2 j.
What's that? Well that turns out to be Euhler's formula for a sine wave.
And so, after all is said and done, this term here is equal to that.
And so there's a sine, buried in there, times a,uh,a complex exponential, a phase
term. So.
That means, using that result, I can plug that in, simplify the expression.
And here's what I get for a final answer. For the Fourier coefficient of a periodic
pulse train. It has a sined, what we call a sine x /x
behavior, known as the function of k here.
We get sine over the argument times the same thing.
So if I take this and multiply it by delta and divide it by T, and I'll do the
same thing up here, delta over T, then I have something that looks like sine X
over X. And this has got a name, this is called
the sinc function. And I need to know what a sinc function
looks like. And, here's the plot of a sinc function.
very important, for us, it turns it's going to reoccur in this course a lot.
So, sinc function, the reasons come through as 0 of course.
Because of the sinc, this amplitude is dying like 1 over X.
At the origin, that's the interesting part, notice you get 0 for 0.
And you have to use L'Hopital's rule to figure it out, and you'll quickly
discover that it's equal to 1, so this is a plot of our, sinc function, 1 at the
origin and then, it looks like a sinusoid with the amplitude going down at 1 over
x, so that means we can plot. Now, our spectrum, for a, periodic pulse
sequence. Again I'm plotting, this in terms of
harmonic number, and frequency is k/t. So the t is understood.
And I'm, this plot applies, for this special case, where the pulse width is
1/5 of the period. Just, a special case to plot it.
And now, we get, we see that all the, the Fourier coefficients are non 0, except in
the case where they, are harmonically related to 1 over, delta over T.
And that's the way, because of this sync behavior.
Right? It goes to 0 at some places. Okay.
So there's our sync function. Except we're plotting the magnitude, so
those negative parts in the previous slide go above the axis, and we have to
change the phase. But, the point is, again, the spectrum of
periodic pulses. Consist of a lot of, of harmonic terms
whose amplitudes are generally decaying but now they, they ripple up and down
depending on what the values are but generally they are going down.
The phase is interesting. it turns out it looks like the phase
should. A linear function, where the phase is
just that argument. But, but is multiplying J.
And it looks like it's linearly decreasing.
It's got a negative slope. It looks like it should do that forever.
And why do I have this funny plot where it pops.
It looks like it's doing what it's supposed to, then it pops up and does it
again. And that has to do with a complication of
the sync function. In this area where it goes negative, that
adds a phase of pi to the in the phase term, right, because you
can, a magnitude cannot go negative, so that pops into the phase.
Well, the next value on this straight line is going to be minus pi.
I add a phase of pi because it's going negative, and it pops it back So that's
what all these have to do with these are returning to their original values
because of the phase. and insert the phase of the sinc coming
into play, as a kind of of a subtle phase.
It may be real value but it still has a phase.
Negative numbers have a phase of minus of not of pi.
And -pi depending on how you want to think of it.
Okay. So, lets think about this now.
We have a signal, that's in both the time domain and the frequency domain.
I can, express it, as a super position of Of sine itself of rather complex
exponentials. Can i really do that, are they really
equal to each other. So, i need to explore them in more
details and what I am going to do is, look at Get what's called a finite
approximation of this term, and see what happens as I let K get B.
So, if we start with our square wave and look at it for some value K, well, this
finite sum shouldn't look like a square wave.
But as I take more and more and more and more turns in, it had better become the
square way or I cant write, this equality is just wrong.
Well let's see what happens, and this get interesting.
So if you only go out to the. K=1 term, which is the first harmonic,
you see that you get the sine wave like you should, you should.
And it kind of looks like a square wave, but it's not certainly not equal to it.
As k gets big, I go down to the 49th term here.
You see that it's coming closer and closer.
To the square waves. Really very interesting.
You should plot this for yourself sometime.
It's really quite exciting how it works. However, Josiah Gibbs, an American
mathematical physicist pointed out something.
He noticed. See these little bumps up up here? And it
may look like, they're diminishing. The ripples inside here definitely are
getting smaller. But those little bumps, which he called
ears, they're known as Gibbs' ears, don't go away.
And it turns out they will always be there.
And they get closer and closer to the, point of discontinuity.
Turns out the discontinuity's our problem.
Well, let's think about this. if we start with a discontinuous signal,
like a square wave. Well complex exponentials are all
continuous. How can a sum of continuous things add up
to be a discontinuous thing? How can this be true? And so we've got to call in to
question the fact that this may not Work right.
Does the gray seies really work? Well, it turns out, when Fourier presented his
work in the theory of heat, to the French Academy of Sciences, he, his work was
examined by, 3 mathematicians Lapluass. Legendre and Lagrange, three
mathematicians you may have heard of, pretty, and they awarded him the prize
for the best piece of work, but they really called into question what happens
when s is discontinuous. It seemed like it'd work, like I showed
you, but there's these little points And these, the skid here that things don't
quite work right. How do you justify that? Well, we have to
get, go off and do a little, do a little bit of mathematics.
And it's pretty easy, so, it's been shown, that if you have an s(t) that's
continuous, like a sine wave Or something else that's a, periodic, and has to be
periodic. If it's continuous, the Fourier series,
the technical word is converges for each t.
In other words, this side and this side are equal, for every value of t, as long
as s is continuous. The issue comes when it's discontinuous.
And, if s of t has discontinuities, like the square wave, or the periodic pulses
we looked at, the Fourier series does not converge, at the points of discontinuity.
It turns out, that for a square wave, I've plotted a square wave here.
for second, and you can evaluate the complex exponentials, all at T over 2,
and add them up, and it turns out, it's going to give you a value, even though
for the square wave, I didn't assign a value, and the value it's going to assign
It turns out, is the, is the, average of the values on either side of the
discontinuity, so it's going to assign a value of zero to what happens at
precisely T over 2, even though I didn't define it, of course, so it doesn't
converge. It gives a value, but it may not be the
value that you'd want. So, how do we think about this.
Well, do want to point out something. So, if you look at the difference between
the signal you have and this finite gray series approximation You look at the
power in that error. And you take the limit as k goes to
infinity. It turns out you can prove that you
always get zero. So, this is the error in your, your
approximation. Case, k term approximation.
And it turns out that error, as the number of terms gets big, has no power,
has no energy, there's no, there's no difference between the 2 in terms of
power. So, mathemeticians said, it does really
Work, except you have to change your definition of, equality to mean this,
okay? Two things are equal if their, the difference between the two has no power,
and this is called, convergence in mean-square, or mean-square equality.
And it really is the power in the error that becomes the important thing.
So, for all intents and purposes, the Fourier series really does work.
It's just that it doesn't quite work at points of discontinuity.
But we don't care about points of discontinuity anyway.
What do I care? But it is equal. So the Fourier series really does work.
A far as were concerned. Alright.
So, let's look at the properties of the forier coefficients.
So, again, here's my time domain, if you will.
That, representation of the signal as a function of t.
Here's my frequency domain, representation as the function of k,
which governs the, the Fourier coefficients, and here's how you get
between the 2. And lets look at the properties.
So, first of all, if S of T is real valued, you can show that the harmonics
that are neglies of each other are complex conjugates of each other.
We have seen this before. This says that the real part Of c minus
k, is equal to the real part, of c k. And that the imaginary part, of c of
minus k, is equal to the minus, the negative dimension part of c k.
And so this was conjugate symmetry. That we talked about before.
The real part is even. The imaginary part is odd.
And so you can show that the Fourier coefficients evade is.
And I think, if you check all the previous examples we've worked on,
they've had this property. Furthermore, if s(t), itself, is real and
even. It's an even, signal about the origin.
Then the, c, c sub -k-=c sub k. Which means the c sub k's are always
real, and they're also even, that's what even means.
So, you have really an even, for the signal, the 4a coefficients are also real
and even. And if you have an odd, signal, that's, a
real value, and odd, and that's what odd means, it turns out the coefficients are
also odd, but they're pure imaginary. Okay? That's what it takes to be
consistent with this, this result. Up here, okay? So very important
symmetries that you can use to simplify your life, and essentially, what I'm
doing is figuring out what the negative frequency axis is from the positive
frequency axis. So I can fill in what happens At negative
harmonic numbers, negative frequency from what's having, happening at the positive
indices. And one little cute property here, that's
very important, suppose I delay the signal, what is its Fourier coefficient
going to be? So, the idea is that I have let's say a square wave Okay.
And, it turns out it, to make it periodic, we've gotta do that.
And now, when I delay it, let's say by a little bit, so it does this.
You would think that the Fourier coefficients would be related somehow,
and it turns out they are, through this linear phase right here.
The, phase turns out to be depending on the delay.
And then the frequency of the Fourier coefficient enters in but it looks like a
linear delay. I mean linear phase.
Delay in time signal corresponds to multiplying the spectral Version of the
signal by a linear phase. It's linear because, this argument up
here, which is 2 pi, I'm going to write it a different way, tao*k, is a linear
function of k. Okay, well, through the Fourier series,
we can study a signal in both time and frequency.
They're equivalent to each other. We can go back and forth.
And this is going to be a very convenient thing for us to use to explore how
signals ought, what their structures are and how systems operate on them.
So the periodic signals, the Fourier series is the way we obtain the signal's
spectrum. The signals can exist in either time or
frequency domains. This is really important thing.
You may start, we have started with signals as being defined as functions of
time and we used the interval here to figure out what.
What it's spectrum is. I'm going to give you an example in the
next video, where we start with the super position, and that's the natural thing to
do. Now we're start with a frequency domain
specification to signal, and see how it's built, what the result of the signal is.
So, either one, it's perfectly valid. And this is very exciting, and will help
us figure out how signals and systems work in modern information systems.