So, this video represents the combination 
of our exploration of signals in the 
frequency domain. 
Today, we're going to talk about 
non-periodic signals. 
These are the ones where we finally need 
to build up to. 
After today, we could handle both 
periodic and periodic signals with equal 
facility. 
The mathematical tool we're going to use 
is known as the Fourier transform. 
And so, we need to explore those 
properties and I think there will be no 
surprises there, we'll, we have seen a 
lot of them already. 
And finally, we will show you how you can 
figure out what happens when you filter a 
signal, 
be it periodic or non-periodic. 
Okay. 
So, first of all, we have to derive what 
the Fourier transform is. 
And it turns out what I'm going to do is 
I'm going to play with the period T. 
So, I'm going to be explicit here about 
what the period of the signal is. 
And as we have had and seen, the Fourier 
series coefficient is given by a formula 
that looks like this. 
And the one difference here are the range 
of integration. 
Turns out you can find the Fourier 
coefficient by integrating over any 
period, 
as long as you integrate over one period. 
So, for this derivation, I'm integrating 
from -T/2 to T/2 instead of zero to T. 
It doesn't really make any difference. 
I'm also throwing in over here the fact 
that the C sub k is are a function of 
what the period is. 
That doesn't really change anything. 
Now, I'm going to make a little simple 
definition. 
And that is that ST of f is just T * Ck 
of t. 
So, what I'm doing is just bringing that 
to the other side. 
And, of course, that's what you get. 
The other thing here is I'm defining f to 
be k/T, so that appears in the formula 
there and it appears over there. 
So, so far I'm just manipulating into 
finding things I haven't really done 
anything. 
Now, we know what the Fourier series 
expression is for the signal in terms of 
the ST(f). 
Since the ST(f) is just this in order to 
have the Fourier series formula, I have 
to divide by T and f is k/T. 
So it's, nothing has really changed. 
So, here's the fun. 
Here's where we start the mathematics. 
What happens if we make the period go to 
infinity? So, if you think about, for 
example, the periodic pulse string. 
So, it looks like that. 
Drawing way out there. 
T, there's even one over here at -T. 
I'm sorry, -T over here, goes on and on. 
What happens if T goes to infinity? Well, 
what's going to happen is all of the 
signals, 
all the pulses here are going to go out 
there because they have to be shifted 
out. 
And this one, these are going to go to 
minus infinity. 
So, the only thing that's left is a 
single pulse around the origin. 
It doesn't shift. 
And the reason for that, just to recall, 
is that if this is p(t), this is p(t) - 
T. 
So, when we let T go to infinity it 
shifts out. 
Okay. So, we, that's the mathematics. 
We need to look and, at what happens when 
t goes to infinity. So just repeating 
what we just looked at, and now, what 
happens if t goes to infinity. 
Well, what happens is that this, you can 
consider as a remind sum. 
So, when T goes to infinity, 
1/T turns out to be the separation 
between any two frequency values. 
And so, that becomes df. 
What's happening here is that since f in 
here is k/T as you're letting t go to 
infinity, 
you have to change the harmonic numbers 
so to keep you at that same frequency. 
And, that's essentially a detail, but 
we're now done. And the sum which is 
represented by the Fourier series 
expression, becomes an integral as we let 
the period go to infinity. 
So, to show you that that's kind of thing 
that's happening, 
I show two examples here of the period 
pulse sequence, periodic pulse train, 
with two different periods. 
Here, the T is 1. 
Here, I made it 5 times bigger, and you 
can see how the spectral lines fill in 
here. 
They get very dense. So, the sum, 
which is, corresponds to this term 
becomes an integral because the lines get 
closer and closer and closer until it get 
into a continuum. 
Okay. So now, we have what's known as the 
Fourier Transform Pair. 
So, if you're given a signal, you can 
find its spectrum by just doing, doing 
this calculation. 
If you have the spectrum, you can find 
the signal in the time domain. 
Here's the signal in the frequency 
domain. 
Here's the signal in the time domain. 
So, just like before with the Fourier 
series, 
you can express a signal either in a 
frequency domain or in a time domain. 
And the Fourier transform is your way 
going between the two. 
The definition is, this slide is what's 
called the Fourier transform, and you've 
heard that terminology. 
That's what they mean going from time to 
frequency. 
And the inverse Fourier transform is 
going from frequency to time. 
Just a terminology, 
nothing else. 
But, we can now calculate either the, the 
spectrum from the time domain's function, 
functional version of the signal or the 
time domain version from a frequency 
domain function or version of this. 
we point out this as a very interesting 
symmetry to it or they look very similar. 
The only difference between these two 
formulas is the sign of that exponent. 
So, the Fourier transform gets the minus 
sign in the exponential and the inverse 
Fourier transform gets the plus sign. 
That's going to be a very handy thing to 
note when we explore it's properties. 
Let's do a little example first. 
Let's take the Forier transform of a 
pulse. 
So, if you remember, a pulse is a 
standard definition, we've seen it many 
times. 
It's a pulse of type one, 
and with delta. 
Since it is zero out here, that makes our 
life very simple. 
When we plug it into this formula, we 
don't need to integrate from minus 
infinity to infinity, we know the 
integral over these areas is going to be 
zero. 
So, we just integrate from zero to delta. 
So, 
we plug it in and we just integrate one 
from zero to delta times the complex 
exponential. 
And that integral is easily calculated. 
And this should look very familiar to 
you. 
We're going to pull our same math trick. 
And we're going to write this. 
Pull out half the phase and simplifying, 
you get our very familiar sinc function. 
So, this is the spectrum of a pulse. 
And now, there are no spectral lines. 
It's a continuous function of f, the 
magnitude of p of f. 
Looks like a sync. 
Okay? And then, the first zero which is 
always an interesting place, is going to 
be the frequency at which this is equal 
to one because the sign is zero at 
integer multiples of pi. 
So, the first one out here is when f 
delta equals one, and so this frequency 
is one over delta. 
So, if you will, most of the energy and 
power in the spectrum is, is between -1 
over delta and one over delta, and then 
it decays by one/f. 
Okay. 
Let me do another example for you. 
And, this one is a decaying exponential 
times a unit step. 
Okay. 
So, what does this thing look like? So 
since the unit step is zero, for t less 
than zero, how about here that product is 
zero? For t greater than zero, the unit 
step is one. 
So, out here, we just get the decaying 
exponential. 
Okay. So, the a is positive in this case. 
So, to plug this into our Fourier 
transform expression, again we're just 
going to ingrate the exponential from 
zero to infinity, alright? And here's 
what we get for an answer. 
And I'll show you how to get that. 
So, instead you get the integral from 
zero to infinity, with e^-at, e^-j2πft 
dt. 
And you combine the exponentials [SOUND] 
to get a very easy integral to do. And I 
think it's pretty easy to see is that 
this is what you're going to get for the 
result. 
there's an important little detail in 
here and that a is positive so that when 
you evaluate the exponent and the 
infinity you get zero because this will 
dominate. 
Okay. So, this should look like a very 
familiar quantity here. 
and we'll get ot that in just a second. 
Okay. 
So, here's some interesting properties of 
spectra. 
None of these should be a surprise. 
We have a Parseval's Theorem. 
So, signal power is equal, has roughly 
the same expression except for the 
magnitude in either the time domain or 
frequency domain, will look very, very 
simple. So, you can calculate power in 
the frequency domain or time domain, 
whichever is easier. 
And if you decide one integral looks very 
hard, you can use the Fourier transform 
to go into the other domain and see if 
the integral looks simpler over there. 
We play that kind of game all the time 
because of the Fourier transform's 
properties. 
And, we have conjugate symmetry for real 
signals. And again, if it's even, the 
spectrum is even. 
And if it's odd, the signal in the time 
domain is odd, it's odd in the frequency 
domain. 
And furthermore, because of the conjugate 
symmetry properties, it's either real or 
imaginary depending if the time domain 
version is even or odd. 
No real surprises there, you should be 
able to prove these yourself quite 
easily. 
Let's go into some more interesting 
Fourier Transform Properties, 
okay? 
First one is, one that you've seen 
before. 
And then, if I delay in the time domain, 
that corresponds to the multiplying in 
the frequency domain by e^-j2πf times the 
delay, 
okay? 
We've seen this with the with the Foutier 
series and I think it's pretty easy to 
see when you plug that same formula, use 
the same mathematics in this case, too. 
However, what's really interesting is 
what happens if you delay in frequency. 
You take this spectrum and shift it over 
by f0. 
Because these two formulas were so 
similar, you know that the answer has to 
be it's multiplied by complex 
exponential. However, because of this 
sign difference in this formula, it's 
either the plus here. So, delaying 
infrequency corresponds to multiplying in 
the time domain but either the +j2πft0t. 
Whereas, if your delay in time 
corresponds in multiplying by e^-j2πf 
tap. 
Okay? So, this is where the, you can 
really use to your advantage these 
symmetry properties, the fact that these 
look so similar. 
You can mathematically, 
if you can drive one property, one 
domain, you can use it to see what 
happens in kind of the reverse sense, 
okay? 
Let's explore another property here. 
And this one is really easy. And that the 
signal at the origin in the time domain 
and it's value at the origin has to be 
the integral of the spectrum over all 
frequency. 
And the way to see that is to take this 
formula and just replace this by zero. 
Well, the complex exponential is of zero 
is just one, and so you just left with 
the integral. 
And that's pretty easy to see. 
And, of course, use the same thing over 
here and just just f at zero. 
So, the value of the spectrum at the 
origin is equal to the integral of the 
signal. 
And here's one, and it's a bit more 
exciting. 
The spectrum of the derivative of a 
signal is j2πf times the spectrum of the 
original signal S, 
okay? And how do you derive that? And the 
answer is, I'm simply going to take the 
derivative of this formula. 
So, ds(t)/dt is equal to, 
[SOUND] 
okay. 
So, what I'm going to do is I'm going to 
take this derivative and move it inside 
the integral. 
The only place that t occurs is right 
there in that complex exponential. 
Well, the derivative of complex 
exponential is just j2πf times the 
complex exponential times e^+j2πf t/dt. 
Well, this is the inverse Fourier 
transform formula. 
However, what it has in it, what you're 
taking inverse transform of is that. So 
that means these are related to each 
other. 
So, really pretty cool and that if you 
take a spectrum and you multiply it by 
j2πf, that corresponds in the time domain 
to taking a derivative. 
Very interesting. 
Well, if multiplying by j2πf corresponds 
to taking the derivative, 
guess what? 
Dividing by j2πf corresponds, too. Well, 
that's got to correspond to the equation. 
And again, you can show that by plugging 
this into the formula over here. 
I'll let you do that. 
It's pretty easy. 
and you can easily see that this is the 
way it works. 
I have a little question for you. 
What happens, what is the spectrum of 
this second derivative of a signal? 
What's the spectrum the second derivative 
of a signal? So, d^2s(t)/dt, well that's 
just a derivative of the derivative, 
right? So, the spectrum of this thing is 
j2 path. 
The spectrum corresponding to this is to 
multiply by a minus, another, another 
j2πf. 
And so, what you get is -4π^2f^2 *. 
s(f), 
okay? The minus sign of course is j^2. 
So, you can also talk about the, the 
double integral, the signal if you will, 
integrating it twice and all that. 
That, of course, when it's dividing twice 
by j2πf. 
Really interesting point. 
So, here's the summary of all those 
examples and properties that have done in 
the previous slides. 
And, they are very important for us. 
And we will use these properties over and 
over and over again with an example 
that's coming up is going to be 
particularly important to remember the 
integral, and to remember this. You'll 
see that in just a second. 
And I guess, that one's going to be 
important. 
First, I want to talk about a very 
interesting case that's very important. 
What is the spectrum of one plus the 
signal times the cosine? Okay. 
And s here is assumed to be a 
non-periodic signal, it has no special 
properties other than it's some signal. 
In fact, it could be something like a 
speech signal or something. 
So first of all, we note that if you just 
multiply everything through that we have 
a periodic part and we have a 
non-periodic part. 
The reason that this is non-periodic, the 
cosine might be periodic, but once you 
multiply by something that's not 
periodic, you get something that is 
clearly not going to be periodic. 
Okay. 
So, how do you find the spectrum of this? 
The way you do it, is you use the Fourier 
series for the periodic part. 
And now we know what to do. 
That gives you terms only at plus and 
minus fc, 
you'll see that in just a second. 
And we're going to use the Fourier 
transform on the non-periodic part. And 
then, because the Fourier transform is 
linear, the integral of a sum is the sum 
of the integral so it's easily seen to be 
linear. We're going to add the results 
together. 
Our only work is going to be on this 
second term. 
Okay. 
So, I don't want to, I don't like doing 
integral. 
I like using the properties. 
And, the way to get at those properties 
is through Euler's formula. 
Euler's formula comes up a lot and 
there's a reason for that. 
It's very very useful. 
So, cosine is easy plus, plus e to the 
minus. 
And put the s of t to the side. 
But from our properties we know that when 
you multiply in the time domain by a 
complex exponential, that corresponds to 
a delay or an advance in the frequency. 
And that's what each of these terms is. 
So, you write down the answer. 
It's easy to show that the Fourier 
transform of some signal times the cosine 
is going to be the spectrum of that 
signal, delayed, and advanced in 
frequency. 
And the amount of the frequency delayer 
advance is f sub c, 
okay? And the half is there because you 
get the half of Euler's form. 
So, and now we know what the Fourier 
series is and we now can add the 
spectrum. 
And we get the following plot. 
So, I like to use this example spectrum. 
it's called a rooftop spectrum because it 
sort of looks like a rooftop. 
And notice it has to be this, has to be 
symmetric, right? What happens in 
positive frequency has to occur in 
negative frequency and it has to be a 
conjugate symmetric which means for what 
I'm showing here, it's got to be an even 
function. 
We're going to use this a lot. 
So, the part that's due to the periodic 
part, that was of those lines, like we 
always do stem plots for things relative 
Fourier series. 
And the frequency at which they occur is 
-fc and +fc. 
For the non-periodic part, that's this, 
that's why you get this thing delayed and 
advanced in frequency. 
Let me add up. 
So you now, once you multiply by the 
cosine, you shift it up so the spectrum 
of the original signal now gets centered 
to f sub, f sub c. 
Okay. 
This is a very important example because 
this is amplitude modulation, and we're 
going to be very interested in how that, 
how radio signals work. 
but we have to do that a little bit 
later. 
Right now, it's just and example for us 
exploring Fourier transforms. 
Okay. Let's start filtering. 
I think it's pretty easy to see that the 
integral of a signal is just a sum. 
So, we're summing up complex exponentials 
to create our signal X(t). 
So, in terms of the spectra in the 
frequency domain, the Fourier transform 
of the output is just equal to the 
transfer function times the Fourier 
transform of the input. 
So, 
to figure out how to filter a signal, 
like on our friend the pulse, with our 
friend the RC low-pass, 
what we're going to do is all we have to 
do is find the Fourier transform of our 
pulse, which we've already done. 
Transfer functions are already in the 
frequency domain, 
in most cases. So, we're going to figure 
out what X(f) is going to be p, P(f), 
going to be that, multiply them together. 
And then, we have to find the inverse 
transform of of Y(f) to figure out what 
Y(t) is. 
That's the program. 
Okay. So, we know what the spectrum is. 
I'm not going to write it in terms of the 
sync function because there's a reason 
that you'll see in just a second, 
okay? So, we've already calculated the 
spectrum little bit earlier. 
And now, we're going to multiply them 
together and I think, and this is the 
answer for the H * X is this complicated 
thing. 
So, all we need to do is find it's 
inverse for a transform. 
Okay. And so, we just need to plug all of 
that, [LAUGH] into there. 
I don't know about you but I don't think 
that's going to be in an integral table 
anywhere. 
It's a little too complicated. 
And it's got lots of working, moving 
pieces. 
And again, I hate doing integrals, I'd 
rather use the properties if I can. 
This is what I normally do. 
I try to use the properties first. And if 
those fail, then I result to doing 
integral. 
You get a lot more incite of what's going 
on if you use the properties. 
In, I'm going to use this example to show 
you what I mean by that. 
So, I'm just going to look at this 
expression of the Y(f) in some detail. 
Well, some things ought to stand out for 
you. 
First of all, what's this do? So here we 
have some spectrum over here. 
We multiply it by one, we just get it 
back. 
What happens when we multiply it by 
e^-j2πf delta? 
Multiply the spectrum e^-j2πf delta, 
what happens in the time domain? So, if 
we call this S(t), that's this inverse 
Fourier transform, this means time delay. 
And so, this term in the time domain is 
going to be S(t) minus delta. 
So, all we have to figure out is what's 
the inverse transform of this term. 
Okay. 
To do that, I'm going to reorder things 
just a little change to make it a little 
bit clearer. 
And what this looks like to me is I have 
some spectrum multiplied by 1/j2πf. 
Well, that means integrate. 
What's the spectrum? Okay. 
I'm going to point out that one of the 
examples we did was this decaying 
exponential formula, and we found out 
that the Fourier transform of this was 
that. 
This is almost as the same form as that. 
And to give it in that form, I'm going to 
multiply and divide this by RC, and I 
think we can see now that a in the 
formula corresponds to R1/C. We have this 
one up top. 
I'm going to divide by multiply by a, and 
that multiplies this side by a. 
Again, because the Fourier transform 
formulas are linear. 
If you multiply one by a constant, the 
other one gets multiplied by a constant. 
And so, I know that the inverse Fourier 
transform of this term is this. 
And now, we're almost done. 
Now, I just have to, I know that to j, 
1/j2πf means integrate, so I just have to 
integrate this term again. 
This is an integral from minus infinity 
to T, 
it's always a definite integral, that's 
what integrate means. 
And so, that's what you get When you, 
perform the integral. 
And note that since this signal is zero 
for t less than zero, so is its integral. 
And everything in infinity is zero. 
Okay. 
And so, we're almost done. 
We know that this turnout here means 
delay. 
So, the final answer is that. 
It looks a little complicated, but 
actually we've seen this already. 
Remember, what happened when we took the 
periodic pulse train and sent it to our 
RC low-pass filter? I showed you some 
plots. 
I want to bring back one of those plots. 
So, the input here, if you recall, was 
the periodic pulse train. 
So, we have pulse there and the period 
was one millisecond. 
And it was like that. 
And you got that for an answer. 
Well, 
I can think about this periodic pulse 
train as being a superposition of pulses 
delayed, 
okay? 
So, that means I can think of the output 
as being the superposition of these 
outputs. 
And so, this formula should apply to that 
term. 
So, let's just not ignore this part for a 
second because that corresponds to this. 
And what we've done is put in a single 
pulse to figure out the answer. 
So now, I have a much clearer 
understanding of how this signal is 
constructed, but its structure is. 
Okay. 
So, let's go through it. 
The, this starts at zero, and what its 
going to do is go up and keep on going. 
And it has an asymptotic value of one, 
okay? The second term here, u(t) minus 
delta. 
Notice everything here is delay, delayed 
by delta, 
so this doesn't start until delta and 
it's going to be the negative of what we 
have over here, okay? So, at delta, you 
get the same thing headed toward -1. 
And when you add them up, you get this 
adds up and then it turns down, it goes 
down and it's going to go to zero since 
+1 and -1 will cancel when you add them 
up. 
And that's what is going on in this wave 
form. 
So, the shark fins that I referred to 
when we talked about the Fourier series, 
they actually consist of a rising 
exponential even, when minus, even -t of 
RC and then a decaying exponential, they 
are properties of this circuit because RC 
only comes from the circuit. 
I should point out that the formula for 
decaying exponential is frequently 
written like this, and it has a tail. 
And this is called the time constant of 
the exponential, just terminology. 
And so, the time constant of these 
exponentials here is RC. 
just a little fact, and R, so units of R 
* C by the way, are time, which is kind 
of interesting. 
Okay. 
So, this shows you how at least I 
calculate the output of a linear filter 
to any signal coming in. 
I find the Fourier transform of the 
signal, and then try to use the 
properties of the integrals, properties 
rather of the Fourier transform, to 
figure out what's going on. 
So we have finally finished our story of 
singnals in the frequency domain. 
So, if you have a periodic signal its 
spectrum is given by a Fourier series. 
If we have a non-periodic signal, it's 
the Fourier transform and its inverse 
Fourier transform that describe how you 
go back and forth. 
We now know how to filter anything in the 
frequency domain. 
we, if it's periodic, we use the Fourier 
series version of things. 
And if it's not periodic, we use the 
Fourier transform version of things. 
we either always decompose a signal into 
it's periodic, non-periodic parts, and 
then filter each separately and then add 
the results back in. 
Superposition, we use over, and over, and 
over again, it's very important. 
Well, we now know a lot, about how to 
think about signals. 
We can think about their structure in the 
time domain, we can think about their 
structure in the frequency domain. 
Our next video is going to talk about 
speech. 
I'm going to talk about the structure of 
speech and how it has both time domain 
and frequency domain parts. 
Really kind of interesting, in fact.