So in this video, we're going to continue 
our discussion of the frequency domain. 
We're thinking about signal structure, in 
the frequency domain. 
In addition to the time domain. 
We're first going to prove something 
called Parseval's Theorem, which shows us 
how to plot, compute the tower and a 
signal, entirely in the frequency domain. 
In many cases, it's easier to compute it 
that way than directly in the time limit. 
We'll then talk about constructing 
signals explicitly in the frequency 
domain. 
that's a digital communication example. 
It's a lot of fun to see a little bit, 
get a little glimpse of what digital 
communication is all about. 
And finally, we're going to filter a 
periodic signal. 
it's a fairly simple extension of what 
we've already done, already, when we 
talked about analog filters. 
Alright, so the crucial thing we're 
going to prove is something called 
Parseval's Theorem. 
And this is a very important result in 
the consideration of signals and of 
power. 
So just to review, 
this is the Fourier series of a periodic 
signal, S. 
It's got a period of capital T. 
We know these are the 48 coefficients, 
and it's expressed as a superposition of 
complex exponential signals whose 
frequencies are harmonics of 1/t. 
To find the Fourier coefficient we 
calculate this integral, we have to plug 
in what the signal is and just do the 
integral and find it for [INAUDIBLE] 
Okay, so here's Parseval's Theorem stated 
in words. 
The power in time-domain equals power 
calculated in the frequency domain. 
So let's see what that means. 
First of all power in a signal is given 
by this expression. 
Now, we have talked about power in terms 
of the circuits faultaging current. 
The idea here is that the average power 
of a periodic signal, Is going to be 
related to the integral of the square 
root of the signal, as if it were being 
passed through a one ohm resistor. 
So that's the idea, so, we look back at 
the power formulas. 
No matter if this is a voltage, or a 
current, this would be the average, 
power, expression, assuming it was, going 
through 1 ohm resistor. 
I hip, want to really point out, that 
this is a very general expression that 
the power over here is not in watts. It's 
proportional to the power expressed in 
watts. 
Of course it depends, if it's a voltage 
or current and what the value of the 
resistor it goes through as to whether 
this expression is numerically correct. 
It usually isn't but for theory we just 
we'll figure that out later, but this is 
the key element. 
The power in the signal of course depends 
on the signal wave form, how big it is, 
all kinds of things. 
So, how are we going to prove Parseval's 
Theorem? And 
the idea is, that I'm going to take this 
expression, for the, for s, as a Fourier 
series, and plug it into there. 
Now we're going to do a little bit of 
mathematics, because we have the square 
of the sum. 
So let's talk about, just in general. 
Suppose I have a sum, a k's, and I square 
them. 
What's the expression for that? Well it's 
not the sum, of ak squared. 
That's not the answer. 
It's the, it's a double sum, which I'm 
going to write as a sum over k and l, 
with a sub k, 
a sub l. 
And to see that, look at the very example 
of x+y^2. 
Well, that equal to x times x plus x 
times y is equal to all possible pairwise 
products. 
Right? So that's why this is equal to 
that. 
And, so, when I plug the, Fourier series 
into this expression, I'm going to make 
use of this property here. 
Okay. 
So, what is 1 divided by T, the integral, 
from 0 to t, of, the double sum, k and l, 
of c sub k, c sub l, e to the plus, and 
I'm going to combine the exponential 
parts. And this ought to look very 
familiar from the previous video, because 
this has to do with the orthogonality 
relationship I showed previously. 
What I'm going to do is push the interval 
inside, so it becomes clear, and now we 
know what the value of this interval This 
integral is equal to 0, if K does not 
equal -L. 
And with that situation you got 1, 
occurred when this term right here was 0. 
So, the integral is 0 when K does not 
equal -L. 
You see where the 1 from k=l. 
Well that's great. 
So this becomes, equal to, and by the way 
when, the 1/t makes this a 1 up here so, 
we have the sum over k. 
Of c sub k, c sub minus k. 
Well, remember our properties? What is c 
of minus k? That's equal to c k 
conjugate. 
Conjugate symmetry here. 
So, c sub k times c sub k conjugate, 
that's the magnitude squared of c sub k. 
So that's Parseval's Theorem, 
is that the power of a signal can be 
expressed, e, can, can be calculated, 
expressed, either in the time domain, or 
in the frequency domain. 
The expressions are very similar in that 
you square something and add them up. 
Integration is like adding in, in many 
ways, so, this is quite useful. 
You may decide that this is an easy way 
to calculate our, and this, and I'm about 
to give you some examples of when that's 
true. 
When that comes up, abd you think about 
using the, for a series, as an 
approximation when you have a finite 
number of terms. 
So, for all kinds of reasons you may 
decide I only want capital K terms in my, 
for a series Going from -K to +K is what 
I mean by capital K terms, rather than 
the full series. 
Well how good is that approximation? 
Well, we can talk about the error, which 
is this, the original signal minus the 
approximation. 
Well, when you subtract out essentially 
the middle terms out of this sum, 
what you get left are the terms from 
minus infinity to minus K, minus 1, and 
from K plus 1 to infinity. 
Well, you think about this another way, 
this is just this, these terms over here 
are just the Fourier series for, for the 
error signal, here. 
So how much power, is in the error? the 
RMS value, square root power, is given by 
this expression. 
So, I did a few things here to point out 
some We know that for negative frequency, 
negative index. 
It's C sub minus K equals C K conjugate. 
That means the magnitude of C minus K 
squared is equal to the magnitude of C 
sub K squared. 
So when I sum up That is Parseval's 
Theorme says, from minus, minus infinity 
to infinity. 
The negative index terms, negative 
frequency terms are going to be the same 
as the positive frequency terms. 
So you can write the whole thing and this 
is a very good example of why you want to 
do that is only being over the positive 
index parts and you put in a 2. 
Because you can, that's, your concluding 
the negative frequency terms. 
Okay, 
so now we have an idea of how well the 
Fourier series with K terms approximates 
some signal. 
We can look at the RMS value of the error 
signal. 
Great example of using the frequency 
domain to calculate power. 
So we go back to our approximation of the 
square wave. Here we have a plot of the 
magnitude squared of K, and you can see 
it decreases like 1 over k squared, like 
we would expect. 
Now this the squared, 
and now if you look at the RMS error 
that's calculated from the expression 
given above, you see it's going down but 
it doesn't go down very quickly. 
that's just, so this tells you that if 
you want to approximate the Fourier, want 
to approximate a square way with a 
Fourier series, 
you're going to need lots and lots of 
terms. 
49 is a pretty high number and that gets 
the error down to right about there. 
I calculated it. 
So if you really wanted something with a 
very, very small area. 
you're going to need lots of terms in the 
Fourier series. 
It's a very interesting way of looking at 
it, I think. 
So there's an alternate view of using the 
Fourier series to approximate a signal, 
and that is to measure its distortion. So 
the idea is I have some periodic 
waveform. 
In this example it's just the square 
wave. 
And the question is, how close to a 
sinusoid is it? And this is called 
harmonic distortion, because this is used 
in audio to access how linear amplifiers 
really are. 
And the measure that's used is called the 
Total Harmonic Distortion, a very long 
term. So THD stands for Total Harmonic 
Distortion, and it has the following 
definition. 
The numerator is the power in the higher 
harmonics. 
Notice it starts at 2, 
and of course, this 2 has to do with the 
fact that we are including the negative, 
frequency terms as well. 
And, the denominator is the power in the 
signal. 
Now the, a little detail here is that you 
subtract off c0. And the reason for that 
is that c0 is known as the average value 
of the signal, and if you look at the 
formula for computing c0, you'll see it's 
just the integral of the signal over a 
period divided by the period. 
Well, that's called the average value. 
In computing harmonic distortion, I'm 
interested in how it departs, from a sine 
wave. 
I'm not really interested in whether 
there's a, a, com, a average value that 
isn't zero. 
It's shifted up or down. 
That's really not the issue. 
The issue is, how much does it depart, 
from a sine wave. 
So that's why this little term in here is 
kind of a detail and serving for my 
square wave, for example of c zero is 
zero, 
which you will notice the numerator 
ignores c zero as well. 
And so using our theory series 
expressions, through Parseval's Theorem, 
we see that the power in the signal, 
forgetting about the other value, is 
that, and, the numerator is that. 
Now, to compute the harmonic distortion, 
I would have to, plug in what the 4a 
series is, and do the sum. 
And, gotta tell you, that's really not a 
terribly good way of doing it, because 
those sums, by enlarge, don't have, do 
not have closed forms expressions. 
So let me point out something, that this 
is equal to the power in the original 
signal, which is all the terms summed up 
from 1 to infinity. That's the de, 
denominator minus the power in c1. 
The first harmonic is what that's 
supposed to mean. 
So I can use that to compute my 
numerator, only thing we have left is 
compute the power and the signal. 
Well, that's easily done, that's, you 
just square the signal. 
Which in the case of the square wave, is 
going to give us a value of 1. 
Integrate that over a period, divide by 
the length of the period, we get 1. 
So, we can find the power in the original 
signal, usually quite easily. 
So, c 1, if you remember what, the, 
Fourier series is, for a square wave, is 
just 2, over j pi. 
Which makes that numerator equal to one, 
minus two, times 2 over pi squared. 
And, this 1 Is the power in the signal 
which we calculated just by looking at 
what the waveform was and acting 
accordingly. 
Well you plug in the numbers, you do the 
calculations. 
You wind up with about 20% harmonic 
distortion. 
What that means is that the square wave 
has about 20% of its power in higher 
harmonics and another way of saying it is 
the sine wave approximates the square 
wave in such a way that it captures 80% 
of the power. 
Well it turns out that that is a 
spectacularly large harmonic distortion. 
the idea is in audio that you want to 
assess how linear an amplifier is. 
If the amplifier were perfectly linear if 
you put in a sign wave into your 
amplifier, you put in a sign wave, you 
should get out a sign wave. 
Perfect sine wave, which would make, the 
harmonic distortion 0, because all of 
these are 0. 
Well, amplifiers have some moderate 
amount of distortion and so you want one 
number that quantifies how linear it is 
and they use a total harmonic distortion 
for that through a series a test 
equipment. 
harmonic distortion of even 1% is 
considered pretty large. 
In fact, all audio amplifiers are 
specified through their harmonic 
distortion. 
Let's explore the frequency domain in 
another case of digital communications. 
Let me tell you How digital communication 
works. 
We're going to explore this in much more 
detail later. 
So, here's the idea. 
I want to send a bit, either a 0 or a 1 
to, to the sin, to the receiver. 
And one way of doing it, is either to 
send nothing. 
Over an, of really/g, interval of, T, to 
represent a 0, or you send a sine wave to 
represent a 1. 
And, to send a sequence of bits, you 
alternate between a 0 and a sine wave, 
depending on what the bit is you want to 
send at any one time. 
Okay, 
this is a one bit at a time scheme, so 
I'm sending 1 bit within the, interval 0 
to T. 
Now you can think of a fancier scheme, 
that sends two bits at a time. 
So, in, instead of just thinking, you 
know, same T second interval, I can, send 
2 bit 0 0. 
I'll just send nothing. 
If I see zero one, I'll send a zero plus 
the sine wave of a given frequency, 
but if want to send 1 0, I'll send a sine 
wave of a different frequency. 
And finally if I have both bits of 1, 
I'll send to sum. 
Essentially, what I'm doing is 
constructing the transmitted signal in 
the frequency domain by essentially using 
super position and indicating which are 
the two bits are by choosing which 
frequencies are turned on and turned off. 
So that is a total construction of a 
signal in the frequency domain. 
Here's an example, where I got, here's 
the frequency f 1, it's over here and the 
frequency f 2 is over there, and here I 
have the sequence 1 0. 
Here's the sequence one one, here's the 
bit sequence, zero one, and, here's the, 
waveforms that are produced. 
Entirely constructed by thinking about 
encoding information, encoding bits, as 
freq, in the frequency domain, by 
representing them by frequencies. 
So. 
Again, you want to think about signals 
both in the time domain, or in the 
frequency domain depending on what the 
application is. 
Very useful, to be flexible and be able 
to go back and forth, depending on, what 
the application is. 
Okay, now we need to filter, a signal. 
So we know how this works for what we've 
done with circuits. 
We know that if the signal is a complex 
exponential and the filter is linear in 
timing variant, that the output is given 
by the transfer function times a complex 
exponential of the same frequency as the 
input, So, given a transfer function, 
you'll know what the filter is going to 
do. 
And, of course if the frequency is just 
harmonic of some one over t, we just plug 
in into the transferred function what 
that frequency is, and of course we carry 
along the frequency of the input. 
And as I showed you when we talked about 
circuits, if you have a superposition of 
two complex exponentials, the output is a 
superposition. 
This is the definition of a linear, 
system, so guess what happens when you 
have a Fourier series? It's just a 
superposition of lots and lots of complex 
exponentials, so the output is given by 
the transfer, sum of the transfer 
functions evaluated at those harmonic 
frequencies, 
times, the c sub k's, for the original 
signal. 
So, in essence, this is the Fourier 
coefficient, for the output. 
So what do you do? To figure out what a 
filter does to a signal, you start with 
the signal, usually in the time domain. 
You find it's Fourier series 
representation. 
You find those C sub K's. 
You plug that into this expression along 
with the transfer function, evaluate it 
for those harmonic frequencies, and 
reconstruct what y is. 
So, start in the time domain, go into the 
frequency domain, 
figure out what's going on there and then 
reconstruct in the time domain. 
This may seem tedious, but it turns out 
to really be a easy way of doing things 
and very efficient, as we'll see a little 
bit later. 
Okay. 
So let's do an example. 
So, here's a filter. 
And, the input I'm going to use, is a 
periodic pulse train. 
And we know that spectrum is, I'm not 
going to, write the expression again, 
it's pretty long-winded. 
But of course it has a Fourier series. 
And the filter I'm going to put it 
through, is our old friend the low-pass 
filter. 
So, if you recall The, transfer function 
is given by that expression, and if you 
plot the magnitude as a function, if you 
can see it rolls off and has a cut off 
frequency which in this case, is equal to 
1 over 2 pi rc. 
We showed this in previous videos. 
I want to figure out what happens when I 
put this periodic pulse train as the 
source signal so that's going to be x(t). 
The output, vout. 
That's going to be y of t. 
So, how do I find it? I use the, 
superposition principle. 
And this is what we get. 
here's the transfer function. 
And here it is evaluated at k over t. 
The kth harmonic. 
And multiply it times c. 
And then reconstruct the signal. 
Well, this reconstruction, to say the 
least, 
We can't do it, analytically. 
And so this is where you just, call upon 
a computer that can evaluate these things 
and add them up. 
And here's what you get, 
really kind of interesting. 
So, first of all, 
I had to tell you, or I had to pick what 
t was, the period of our periodic pulse 
train. 
I picked 1 millisecond. 
What does that make the fundamental 
frequency? The first harmonic of that 
signal? What is that frequency? And of 
course the answer is, it's 1/t so it's 1 
kilohertz. 
Let's keep that in mind. 
And along with the previous example that 
I gave for the periodic pulse train the 
width of the pulse is about 20% of the, 
period. 
Okay which corresponds to the examples 
I'll show you. 
So, I'm now am looking at what happens to 
the output and the outputs are shown down 
here. 
This is the spectrum of the output. 
So these are the Fourier coefficients. 
For filter of various frequencies. 
so I'm changing r times c to produce 
these various cut off frequencies. 
Okay, don't forget that the fundamental 
is at 1 kilohertz. So, you recall the 
Fourier series for periodic pulse train 
has power at all permonics of the 
fundamental. 
So it has power of 1 kilohertz, 2 
kilohertz, 3 kilohertz, 4 kilohertz, etc. 
Okay. 
So, picking a call frequency of a 100 
hertz, that's really lowpass filtering. 
The the, first harmonic is well above, 10 
times the cutoff frequency of the filter. 
So it attinuates the the spectrum of the 
input a lot, as you can see by this 
figure. 
When you get produced, a somewhat lumpy 
looking output, it's not even clear that 
their were pulses in there originally, 
it's really low past filtering. 
So low past filtering tends to produce, 
tends to produce signals that don't bury 
very much in time. 
They can't wiggle very quickly because 
the higher frequencies have been 
surpressed. 
By the filter. 
They're low pass, remember they tend to 
attenuate high frequencies. 
When the cut off frequency is now one 
kilohertz, which is right at the first 
harmonic. 
The first harmonic won't be suppressed 
very much, be suppressed by 1 of the 
square root of 2. 
You can sort of see the pulses coming 
out, kind of, it is not clear the origin 
of the pulses which tend to get which i 
call the short fins, it goes up the comes 
down, goes up and comes down, this is 
periodic, now this is periodic of course. 
And then when the cutoff frequency of the 
filter is well above the first harmonic, 
well above the, eighth harmonic, ninth 
harmonic, it's way up there. 
What you will see, is that the output 
very closely resembles the original, 
pulse, and there you can see, if you look 
very carefully, of course you can see it, 
that it's rounded, the edges here, both, 
at the tops and at the bottoms. 
So, it doesn't filter very much, but, if 
it's, the cutoff frequencies really low 
compared to the first harmonic. 
You get a lot of change of the output. 
So, and this is a common use for lowpass 
filters, is to change the wave form to be 
something that you want given some 
various simple wave form that you could 
probably produce in the lab. 
Okay. 
So, signals can be defined either in time 
or frequency. 
What I concentrated on in this video, 
frequency domain, characterizations of 
signals. 
So the idea was I started with s sub ks. 
And I use that to construct signals. 
In going the other way you can take your 
signal of course and figure out what its 
frequency domain representation is. 
And that really means you can go either 
way, you can start in time, go to 
frequency, start in frequency and then go 
to time. 
all, all choices are available to you. 
So, you can define them in either domain 
depending on the application. 
You can also study, the structure of the 
signal in either domain. 
You may be very interested in what the 
pulse width is for example, but you're 
also maybe interested in how high up, 
what's the highest harmonic that's needed 
to have a 90% approximation of the 
original pulse train. 
Either way of thinking about it it may be 
important, depending on the application. 
And, finally, for linear, time-invariant 
filters, we can determine their outputs 
for periodic inputs because we can use 
the Fourier series results that we've got 
in superposition to get the answer. 
So we're really becoming more 
sophisticated. 
Next video, what we're going to do is 
drop periodic. 
We're going to talk about how to 
generalize these results so we can talk 
in general about general signals, whether 
they be periodic or not, and that means 
we'll really know a lot about signals and 
systems.