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PROFESSOR 1: OK, let's launch right into it.
Jacob and Uri were inspired by the echo channel
to try out a simulation of what I'd put on the board last time
or on my slides.
So what you're going to hear is Jacob's message
going into the echo channel.
Remember, that was something with a unit sample response
of the type delta n plus.
And actually, I think in their case,
now it's 0.999 delta n minus 1, so it is something like this.
Sorry?
Oh, n minus 4,000, OK.
And so you'll hear the original message, I think.
You'll hear the message going through the echo channel.
And then you'll hear the message cleaned up
with the receiver filter.
That's just the inverse filter.
And then you'll hear what noise does to it and two
flavors of noise, I think.
ECHO VOICE: This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
[BELL DINGING]
PROFESSOR 2: I'm just going to increase the delay,
so we can hear the echo more clearly.
ECHO VOICE: This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
PROFESSOR 2: So you can all hear the echo in that.
And the next one will be it cleaned up--
ECHO VOICE: This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
PROFESSOR 2: So this was just one
in deconvolution, assuming the channel had no noise.
The next one is what happens if there's even
a small amount of noise in the channel--
so little that you couldn't hear it in the echoed signal.
ECHO VOICE: This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
PROFESSOR 2: So you can hear the noise building up because
of the deconvolver.
If you actually had noise at a particular frequency,
you end up with an even--
ECHO VOICE: This is a test of the 602 deconvolution system.
If this is a real deconvolution, you
are instructed to go as quickly as possible
to the frequency domain to make an assessment
of the effectiveness of this approach.
PROFESSOR 2: So one of the difficulties with deconvolution
is it can be perfect if you have no noise.
But even small amounts of noise can really
mess up the deconvolution.
It magnifies the small noise sources.
PROFESSOR 1: OK, great.
Thank you.
And that was a few lines of MATLAB, right?
Same thing can be done in Python.
All right, we continue.
So we're going to talk today about spectral content
of signals after having spent some time on the frequency
domain--
the frequency response of LTI systems.
Let's get this up here.
So we've talked about frequency response.
You've seen the definition.
If I give you the unit output response of a system over here,
you know how to compute the frequency response.
And then we went through how you can go the other way
with the DTFT to compute the frequency
response and the inverse DTFT to compute the time domain signal
from the frequency response.
And what we said last time was that
what you're doing with a unit sample response,
you could actually do with any signal.
So you can take any signal x sub n,
compute from it the DTFT, the Discrete Time Fourier
Transform, with the same formula.
What you get is this object that can be used then
to reconstruct the signal--
again, the same formula, all right?
So there's just a change of perspective.
There's nothing different here.
The key observation now, though, is
that the formula on the left, the inverse DTFT,
is actually allowing us to represent x sub n, the time
domain signal, as a weighted combination of exponentials
of this type.
And the reason that's important is
that we know how to deal with signals of that type
very easily.
We already know that if you have e
to the j omega n going into a system with frequency
response--
h omega, so an LTI system with that frequency response.
Let me make this a specific frequency omega 0.
What comes out is the frequency response evaluated at omega 0,
multiplying what went in.
All right?
Nothing more complicated than that.
So what now if you had an x sub n
going in that was a weighted combination of terms
of this type?
And I'm going to take a weighted combination that's
actually a continuum.
It's not just a finite number of terms.
I'm going to actually take this particular weighted
combination.
So over some interval of length 2 pi,
I take a weighted combination like this.
So this is going over all frequencies in our interval.
Let's take minus pi to pi.
I might as well write in minus pi to pi--
keep it explicit.
You can think of this as being approximated
by a sum in the usual fashion.
I'm not going to write this out.
But we know how to approximate integrals by sums.
What the sum will have is, for instance,
a typical term of the type e to the j omega 0n.
And the weight that multiplies it will be x omega 0 d omega,
right?
So we think of x omega 0 d omega as being
the amount of the exponential at frequency omega 0
that's in x sub n.
So here is a representation of how the signal is made up.
So then what would you say is the output of the system?
If that's the input, and this is an LTI system,
what's the output going to be?
Any ideas?
Somebody?
I thought I saw a hand.
No ideas?
What if instead of this, I had a1 e to the j omega 1n plus a2
e to the j omega 2n going in?
Suppose that had gone in?
What would be coming out?
Folks, we're just two weeks from a quiz, here.
Yeah?
STUDENT: [INAUDIBLE]
PROFESSOR 1: So can you tell me explicitly
what I would get in this case?
If this was x sub n, then y of n would be?
STUDENT: [INAUDIBLE]
PROFESSOR 1: a1h.
STUDENT: [INAUDIBLE]
PROFESSOR 1: Is it just omega, or?
STUDENT: It's omega 1.
PROFESSOR 1: Omega 1, right?
It's the frequency response evaluated at the frequency
that you're interested in, and then
e to the j omega 1n, and then the response to the other term.
This is what superposition is about.
So that wasn't so hard.
What if instead, x sub n is given
by a continuum of such exponentials?
Integrals are essentially linear combinations,
but taken to the limit, where it's not just a finite number.
So I'm not asking for a proof of anything.
I'm asking for your conjecture as to what the answer might be.
This is how math is done, by the way.
You conjecture what the result might be based on gut instinct,
based on well-educated intuition.
And then you go back and construct a proof
and hide all your tracks.
But engineers like to work with intuition,
and often will stick with that.
Yeah?
STUDENT: [INAUDIBLE]
PROFESSOR 1: OK, so what would--
can you give me the explicit expression?
What's your guess?
This is going in.
It's a weighted combination of exponentials with weights that
are given by x omega d omega.
So--
STUDENT: So you would get x omega d omega [INAUDIBLE]..
PROFESSOR 1: Times what?
I didn't hear the last piece.
STUDENT: Times [INAUDIBLE].
PROFESSOR 1: Where's the j omega coming?
What's the j?
I'm missing-- oh, e to the j omega?
All right.
So start again.
Oh, you don't want to?
STUDENT: No.
PROFESSOR 1: OK.
You're on the right track.
You got us started.
Anybody else?
I could show it to you on the next slide,
but that will take all the fun out of it, right?
STUDENT: [INAUDIBLE]
PROFESSOR 1: OK, it's probably that I didn't hear.
Can you tell it to me?
STUDENT: x omega [INAUDIBLE].
PROFESSOR 1: X omega times h omega or e to the j omega n?
What did you say?
STUDENT: [INAUDIBLE]
PROFESSOR 1: I'm willing to--
STUDENT: [INAUDIBLE] omega n times [INAUDIBLE]..
PROFESSOR 1: That's still the part that went in, right?
Now it's got to get mapped by a frequency response.
STUDENT: Yeah, that's what I said.
PROFESSOR 1: OK, so maybe you'd said that,
and I didn't hear it.
But now we've got to assemble this
over all possible frequencies.
Well, we have the 1 over 2 pi.
Maybe this was said already, and I just couldn't hear it.
All we're doing is we're saying here's
a weighted combination of exponentials going in.
The weights are given by the x omegas or x omega d
omega, if you want to think of it that way.
So what comes out is the combination
of responses to each of those.
These are the things that go in.
For each frequency, you multiply by the corresponding value
of the frequency response and do that for all the frequencies
of the input.
So this is just applying linearity and superposition.
So if I compare that with this expression, which just tells me
how the time domain signal yn relates to its DTFT, right?
This is just writing the same thing for y
that I wrote for x over there.
If you compare the two, what you discover
about the DTFT of the output?
Where's the weighted combination?
It's whatever multiplies e to the j omega
n in this expression, right?
So it's just going to be h omega x omega.
So we've done a complete analysis
of the input-output response of the system
for essentially an arbitrary input with just
a simple multiplication.
So look what we've done.
We've taken the input that we were given, computed
the DTFT of it, which gives us the spectral content,
and I'll spend some time giving you intuition for that.
That's the spectral content of the input signal.
What we've discovered is that the spectral content
of the output is the spectral content of the input scale
by the frequency response.
And once you have the spectral content of the output,
you can reconstruct the time domain signal.
So the big difference here is there's no convolution.
You're just doing a multiplication.
Instead of doing y of n equals h convolved with xn,
we're just doing a multiplication.
So once again, you see the convolution of the time domain
maps to multiplication in the frequency domain.
All right, so we'll build up to the story again.
Let's get some intuition for spectral content.
And let's take a particular example.
So suppose I have an x of n that's a one-sided exponential.
You've probably done things like this in recitation already.
So this is a signal that starts at time 0
and then starts at the value 1 and halves it each time.
So it's a discrete time exponential.
And so what's the DTFT?
Well, you're going to some from m equals
minus infinity to infinity.
But actually, this only exists for positive--
for non-negative time.
So you're going to get 0.5 to the n.
Or let's keep it at m.
That's just the definition.
Isn't it the definition?
I'll just use the definition.
So you can now sum an infinite series here.
And what do you get?
You get 1 over 1 minus 0.5e to the minus j omega.
This is just summing a geometric series because each term here
follows from the previous one by multiplying by the factor 0.5e
to the minus j omega.
So it's a geometric series with that ratio.
And so this is what the sum works out to be.
So if you wanted to figure out the spectral content,
you first compute the DTFT.
And then the most helpful way to get a feel for the signal
is to look at the magnitude of the DTFT.
And that's what's actually plotted in this case.
This is taken from somewhere that
used slightly different notation,
but we'll talk through it.
What happened to the top of my slide, here?
OK, it doesn't matter.
What I've plotted here is the magnitude of x
and the phase of x.
To get that, you'll actually have to convert this
to magnitude in angle form.
I'm not doing that for you.
I'm assuming you've had practice or will get practice
in recitation.
But the result of that is a magnitude
that looks like this and a phase that looks like this.
The horizontal scale, just to make the point that this
is a periodic object, just like frequency response,
it actually goes from minus 4 pi to plus 4 pi.
But the interval of interest is really just minus pi
to plus pi.
All right?
So it's just that central portion
that's of interest-- similarly here, minus pi to plus pi.
And then it replicates periodically outside of that.
So we didn't really need to show it to you outside of that.
This is just to make the point.
Another thing to observe is that the magnitude
is an even function of frequency,
and the phase is an odd function of frequency.
So these are elementary checks that you should make.
If you get an answer that doesn't
satisfy those properties, you've gone wrong somewhere
along the way.
If you look at the top plot in this set,
the top plot is exactly a signal of that type,
except I've chosen a different number.
Instead of 0.5 to the n times u of n, it's something else.
But here's a geometric--
sorry, a discrete time exponential or geometric
series.
Now I've just plotted the DTFT from minus pi to pi
to show you what the spectral content of the signal is.
And I'm just plotting the magnitude.
Ignore these labels.
These are the same figures you saw earlier
for little h and big H. I'm using them again,
except now I'm thinking of this as a signal,
and this is its DTFT.
This is a signal, x sub n, and this is its spectral content.
The relationship is exactly what we had with frequency response.
So where is the spectral content concentrated?
Is it at low frequencies or high frequencies
or intermediate frequencies for this first example?
Concentrated around 0, so you'd say
it's concentrated at low frequencies.
There is content on all frequencies,
though, so this doesn't dip down to 0 anywhere else.
You have to assemble a combination of sinusoids
at all frequencies to construct this signal here.
Here is a signal.
I've had to change the horizontal scale
because this is a signal that evolves much more slowly.
Again, I can ask what's the spectral content of it?
I get something which has a peak near 0 frequency.
So it's only got low frequencies in it
and has very little high frequency content.
You can also start to develop ideas for how fast--
how high a frequency you ought to expect to see here.
So for instance, what's the fastest wiggle
that you see in this signal?
About how long does it take to--
if you were thinking of underlying sinusoids,
what's the fastest wiggling you're seeing over here?
Well, to my eye, this kind of rises and curves within about
18 or 20 samples, right?
So this might be a half period of an underlying sinusoid.
So if I thought of the period as being--
these are just rough calculations.
But it helps you understand what we mean by spectral content
and helps you as a way of checking answers.
But let's see.
If I said 18 was approximately a half period of an underlying
sinusoid, and I don't see anything faster than that,
period is 2pi over omega 0.
So I'm saying that's approximately 18.
So the frequency that I expect to see,
the fastest frequency there, is 2pi over 18 or pi over 9.
Is that roughly consistent with what we're seeing there?
Does the spectral content drop off?
Well, here is pi over 4.
Here's pi over 8.
Somewhere on pi over 9, we've run out
of underlying components.
It's because the frequency content
is limited to that range that the associated signal doesn't
wiggle any faster than this.
Here's another example-- a signal
that actually-- well, in this case,
it seems to have some fairly regular periodicity to it,
and then it damps out.
By the way, in all these cases, I'm assuming--
actually, in all these cases, I'm
assuming the signal's identically 0 outside--
outside of what I've shown you here.
If you take the DTFT of this, you
find that the spectral content is what-- low frequency,
mid frequency, high frequency?
STUDENT: Mid?
PROFESSOR 1: Mid frequency right?
Because this is low frequency.
Here is 0.
This is high frequency.
This is just a reflection on the left side.
So at some intermediate frequency,
there's a peak in the spectral content.
And again, you can go through the rough calculation
I just made.
So we see some oscillation here.
Let's see.
Let's estimate the period.
That's 1, 2, 3, 4, 5.
Let's say it's about a period of 5 for those oscillations.
So 2 pi over omega 0 is approximately 5.
So omega 0 is approximately 2 over 5 pi.
So we expect to see a spectral peak somewhere around 0.4 pi.
Here's 0.25 pi.
Here's 0.5 pi.
We're about right.
So make these sorts of checks.
Yeah, question?
STUDENT: So for that first calculation
with the [INAUDIBLE].
PROFESSOR 1: Oh, sorry.
Yeah, I should have done twice this, right?
What I did was estimate-- thanks for catching that.
I estimated at about 18 is the half period.
And so the period I should have had--
36 here.
Good.
This is all ballpark, of course, but no reason
to make it worse than it has to be.
Vibrating, right?
And that's not exactly where it sits down there,
but it's in the right region.
Now here's the part that we've already seen on the board.
Once you know the spectral content of the input,
you can assemble-- or you can think of the time domain signal
as being made up of those components.
Correspondingly, that's what the output is.
And all we're doing is invoking the fact
that this is an LTI system for which we
know the frequency response gives us
the output for exponential inputs.
And then we compare that with what
we expect to be seeing for the DTFT of y,
and we make this conclusion.
So this is exactly what we had earlier.
One thing to keep in mind, the DTFT, the frequency response,
the DTFT--
these are all complex functions of omega in general.
Each of them will have a magnitude and an angle.
So make sure you understand why the magnitude of y
is the product of the magnitudes of h and x
and why the angle of y is the sum of the individual angles,
all right?
It's basically the fact that for a complex number c,
you can write it as the magnitude times e
to the j angle.
So that's really what's being invoked there.
So really, what the story is about-- we've done
a lot of math along the way.
But this is really the story.
And I've only exposed a little part of it
for you because I've only dealt with DT signals.
But the same thing holds for continuous time signals.
A huge class of such signals can be
written as linear combinations of sinusoids.
And when I say "linear combination,"
it could be a combination of a discrete set, finite
or infinite.
Or it could be a continuous combination of exponentials
or sinusoids under an integral sign, but the idea is the same.
If you've done 1803, you've seen this kind of thing happening,
at least for periodic signals.
And then the other piece of what we rely on
is that LTI systems are very easy to understand in terms
of their action on sinusoids.
So once you put these two pieces together,
you've got a very powerful way to analyze LTI systems.
So just to go back to the kind of example
I had last time, in, which you'll be--
or you're already dealing with in the lab--
we're talking about an audio channel, for instance.
The frequency response, in this case, is the magnitude.
Some characteristic here-- this is a bit of a cartoon.
But let me show you more typical experimental plots
of frequency response.
This is frequency response magnitude.
In most of these plots, people don't show you the phase.
Part of the reason is, or maybe the major reason
is, that for audio, the ear is not
all that sensitive to phase.
If we were doing the analogous thing for video,
then you'd be very concerned about phase.
But in audio characteristics, people
will typically only show you the magnitude
because the phase distortions aren't
picked up quite that readily.
So here is three speakers.
If you look on the site there, you'll see many more tested.
This is the frequency range.
Now, I should make some comments about that.
We're talking about doing minus pi to pi.
We've been talking about filters with frequency responses
that we show from minus pi to pi.
So for instance, if I had a band pass filter,
it would be something with--
in the ideal case, something like this, right?
This is because we're writing things
in terms of big omega for a discrete time filter.
These are actually written--
the scale here is hertz.
And they're talking about the action
on an underlying continuous time signal.
So you actually need a way to go from an underlying continuous
time signal that sampled at a particular sampling rate--
let's say f of s samples per second--
to a corresponding omega for the underlying discrete time
sequence.
So the question is, how does fs map to omega?
And I had a slide last time.
I haven't gone through it in detail.
Maybe we'll have you work through it
on a homework problem.
But this is actually the mapping.
If you have a sequence that comes
from an underlying continuous time sinusoid
by sampling at fs samples per second,
and you're doing all your calculations
in the discrete time domain, if you
want to think about what that means for the underlying
continuous time domain, you want to map pi to fs over 2.
It's not the omega that maps to that.
It's the pi.
So for instance, in the lab, I think
you're using 48 kilohertz, for instance,
at least for some part of it, as a sampling rate.
You get a discrete time sequence out of that.
You do various DTFT-type computations--
spectral content or frequency response.
Then you plot them on this kind of a scale.
If you want to think about what the underlying continuous line
frequency is, well, that's 24 kilohertz in this case,
and minus 24 kilohertz, it's at this end.
So when you're trying to visualize
what this characteristic is telling you about what you're
seeing with a discrete time sequence,
that's really the mapping.
The other thing about these characteristics
is that people only plot the positive frequency part.
So they ignore the negative frequency because
of the symmetries there.
In applications, when they give you a frequency response,
they will typically just give you the positive frequency
part of that.
All right, so this is what we're seeing here.
This is the characteristic of the LTI system
that you're going to be sending signals into.
And then you've got to characterize the signals
that you're going to send through it--
voice or music or whatever.
And this is a figure I showed you last time.
But basically, you're looking at the spectral content
of the signal of interest and seeing how it matches up
with the channel that you have.
And if you compare with--
well, let's actually look at the previous case
and get a few landmarks, here.
So let's take the Sony speaker, for instance, down here.
OK, so it's got a fairly flat frequency response
for a range of frequencies.
But you've got to get fairly high up before you get there.
For frequencies lower than about 100 hertz,
this is not doing a very good job of propagating the sound.
The frequency response is measured
by having a microphone at a fixed distance from the speaker
in an anechoic chamber.
And you can see that this one is actually
perhaps the poorest of the-- it is
the poorest of the speakers in that it does it very poorly
with the low-frequency sound.
So for this particular one, you would
hope that the spectral content of what you're
trying to send through the channel
lives somewhere in maybe 300 to--
300 hertz to 10 kilohertz-- if you want to get it across
the channel--
from the speaker with high fidelity.
But if you've got low-frequency signal
that you're trying to send, and you use the speaker,
well, you're going to be out of luck.
It's not going to propagate it very well.
So thinking in terms of frequency response
and spectral content is really key to making
sense of a lot of this.
All right, let's get a little more practice with this.
And I just want to show you that once you've
learned how to deal with frequency response,
there's not new stuff that you're going to do
to deal with spectral content.
It's just a change of perspective.
So let's see.
If I asked you for a signal that had its frequency
content uniformly distributed in some finite range,
can someone tell me what that signal is going to look like?
I'm asking you for a signal whose spectral
content is uniformly in some range minus omega
c to plus omega c.
Have you met such a signal before?
Anyone?
STUDENT: [INAUDIBLE]
PROFESSOR 1: Sorry?
STUDENT: [INAUDIBLE]
PROFESSOR 1: It was the unit--
we've seen the same kind of thing with the unit sample
response to the low pass filter, right?
So if this was a frequency response,
then the associated unit sample response
would be the signal we're talking about.
So remember what that was called?
We called it a sinc function--
sinc function in time.
So if the spectral content is this in frequency,
then the signal that you're talking about
is going to be a sinc function.
Now you can actually work that out.
You don't have to take my word for it.
You want flat spectral content in the range minus omega
c to plus omega c.
So the signal that you're going to get as a result
you can extract from this computation.
And this is exactly the same function that we saw last time.
So the DTFT does what your eye may not do very well.
If I had just given you the signal
and asked you to take a guess as to what
the spectral content of that is, you're
not very likely to have ended up deducing
that the spectral content is flat in some range and 0
outside of that.
So the DTFT is valuable in actually
doing this analysis for you.
So there is a signal that has flat spectral content.
More examples of a similar type--
and again, we first encountered these
in the context of unit sample responses and frequency
responses.
But now, I just want to change perspective and think
in terms of time domain signals and their associated DTFTs.
So if we look at the top one there,
this is the case we just saw, except I've
truncated the sinc function.
And so what I get is not the perfectly uniform distribution
of frequencies in some interval.
There's a little bit of a wiggle to it.
But this is essentially the sinc function
and its spectral content.
Here's another signal whose spectral content
is at high frequencies and essentially
0 in the low-frequency range.
What does it look like in the time domain?
Well, you can actually work it out.
And here's what you see-- that this has actually more wiggle
to it than the sinc does.
Alternate samples seem to take opposite signs, at least
of the dominant ones over here, reflecting
the high-frequency content of that signal.
Here's something that's intermediate.
This also has the oscillation in sign,
but it's not necessarily in alternate samples.
It's a little bit more leisurely.
Here's something that has low frequency and high frequency,
but not intermediate frequency.
So you see a component that's rapid wiggling,
but you also see this lower-frequency content
in there.
So this is what the DTFT does for us.
Now there's an issue of how you compute these,
because if you look at the formula for the DTFT,
you could certainly do analytical things
with that expression.
And that's the case that--
we've treated cases of that type, where you write down
an analytical formula for the DTFT.
And then you do things with that, like plotting.
But if I gave you some numerical sequence here,
there's certain simplifications.
For one thing, you really aren't going
to expect to compute this at a continuum of values of omega.
You're not really going to expect
to construct the values from minus pi to pi
at every real number omega in that interval, right?
That would take you a long time.
So what you're likely to be doing
is asking for what the DTFT is at some grid of points.
So you'll form a little grid.
And it's on that grid of points that you want the DTFT.
That's the only practical thing you can do.
You're not going to compute it at all omega outside
of toy examples like that.
So if you had a numerical sequence collected in the lab,
for instance, this is what you'd be aiming to do.
What's the other thing that's likely to be the case if you've
got a numerical sequence collected in the lab?
Any thoughts here?
It's unlikely that my summation is
going to go from minus infinity to infinity
because I'd be waiting a long time to collect
that signal, right?
So in practice, what we're dealing with
are signals of finite duration, typically assumed
to be 0 outside of that interval,
though you might have reasons in some context for assuming
otherwise.
We're always going to take finite length signals.
So the summation will be over a finite interval.
And we're going to want to compute the DTFT
on a finite group of points.
And that makes for some simplifications.
So let's see here.
You've probably heard people talk of FFT,
or the fast Fourier transform.
The fast Fourier transform is not a new kind of transform,
so the name is a little bit misleading.
It's a good way of computing samples of a DTFT.
So you don't have to learn a new transform.
We're still talking about this object, the DTFT.
The FFT is an efficient way of computing
the DTFT on a grid of points, given
a signal of finite duration.
Now, I've got a lot on this slide, and I hope all of it
is right.
But let me talk you through the basic idea, here.
So we're going to compute the DTFT
on a finite grid of points.
So that's the omega k's that I've shown you over there.
We've only got a finite duration signal.
Let me say that it exists only from 0 to p minus 1.
So the signal is 0 outside of that interval.
And therefore, all the other terms drop out of this.
So there's nothing new in this formula.
This is just acknowledging that I only
want to compute the DTFT at a grid point,
and I only have a finite duration signal.
Now the interesting thing is that if your signal is
0 outside of this interval, well,
that means your signal is completely
specified if xn is known to be non-zero only on the interval,
let's say, 0 to p minus 1.
So that's p values.
Then you would hope that just having p samples of the DTFT
will allow you to go the other way.
We know for sure that if I gave you
the entire DTFT, if I gave you the entire DTFT,
that you could go the other way, because we
have this expression.
If I gave you the entire DTFT, you
would just plug it into here.
And you'd get the time domain signal.
What's interesting, though, as it turns out--
and you might expect this.
Since your signal takes non-zero values only at p points,
you only need p samples of the DTFT
to get an exact reconstruction.
And here is the formula.
And the derivation is not hard.
I've omitted it here.
It's the same kind of idea that we used in the full case.
But you can actually--
using the values of the DTFT at these grid points,
you can reconstruct the signal x sub n.
So with these simplifications, you actually
have a simple pair that gets you through the numerics.
If you followed these formulas exactly as they're written,
you'd end up doing work on the order of p squared,
because you see each of these summations
involves taking p products.
And then you've got to sum them.
But you've got to do it at p different frequencies.
And the same thing on the other side--
you've got to do p products, but you've got to do it p times.
So it's order p squared computation.
The fast Fourier transform is actually
a clever way of using the symmetries associated
with these exponentials to group the computations
and make it much faster.
And you can actually reduce it to order p log p.
So it's a huge simplification.
I've got some illustrative numbers down there.
So the FFT actually is a major reason
for advances in numerical computations,
including signal processing of various kinds--
the fact that you can get this reduction
from p squared to p log p.
All right, I don't think I need to say much
about the grid of points.
But let's move on to thinking about spectral content
of signals going through channels.
All right, so we'll get closer to signals of the type
that we're interested in, which are
these signals in this case for on-off keying
that are signaling 1's and 0's and that we're
trying to get across a channel-- for instance,
the audio channel.
So this might be a typical finite length sequence.
This particular case, we had chosen 7 samples per bit.
That's why the shortest interval you see
has 7 non-zero bits, there.
Here's the spectral content of the signal.
And actually, I've taken this figure
from an earlier version of the course, where
we talked about the discrete time Fourier series,
not the discrete time Fourier transform.
The discrete time Fourier series turns out
to be something very similar to the formulas
I showed you for the FFT.
So the discrete time Fourier series,
apart from a scale factor, is essentially a story
built around this relationship.
And that's developed in some detail in Section 13.2,
but we're actually bypassing about this term
to try and keep the story simpler.
So when you see these plots, you'll
see on a, at magnitude a of k, that's
a symbol associated with the discrete time Fourier series.
These are actually Fourier coefficients
associated with the periodic replication of this signal
outside.
So it's a discrete time version of the Fourier series
you may have seen in 1803.
All you have to do when you see a plot like this
is think of it as a scaled version of the DTFT samples.
So we're just talking about samples of the DTFT taken
at a grid of points.
The scale factor may be off by a factor of p--
the length of the signal.
But the shape is entirely told to you here.
So think of this as samples of a DTFT.
We're going from minus pi to plus pi.
What's down at the bottom here is
imagining that you've sent this signal over a channel that
could absorb the entire spectral content of the signal.
So suppose you had a channel whose bandwidth--
suppose I had a low pass channel whose bandwidth
has got some cutoff.
This is low pass.
Suppose this bandwidth could absorb
the entire spectral content of the signal.
In other words, what I mean is that all these DTFT numbers
that are significant actually fit in under this.
So suppose your channel was such that it didn't attenuate
the DTFT coefficients.
So the spectral content is unmodified when
it gets through the channel.
And so if you resynthesize the signal using this formula
at the receiving end, you'd get back the same thing again,
because the channel's not induced any distortion.
Let me go past this and actually show you
what happens when you start to distort the--
what goes across.
So here, what we're doing in this succession of experiments
is sending that same signal through a channel
with successively smaller bandwidth.
So in the first case, everything goes through.
There's no distortion.
You get the same thing back again.
In this case, the channel actually
has a cutoff that ends up zeroing out
all spectral content outside of some frequency range.
So it's a low pass channel whose bandwidth
is not enough to take the spectral content of what you're
feeding across.
So what do you expect to happen?
Well, the higher-frequency components of the signal
have been zeroed out.
So what should happen?
You expect the signal to be more rounded because it can't
make these sharp transitions.
It takes high-frequency content to make sharp transitions.
So what happens is you trim the spectral content by sending it
through a channel that's not wide enough
to contain all of the signal is that you
get a more rounded signal at the other end.
So you sent this.
This is what you're receiving.
This is the distortion that the channel
has imposed on your signal.
And you can imagine if you tried to find a place to sample this,
you might run into some trouble.
If you go even more extreme, here
is an even narrower channel.
What comes out is even more rounded than what we had there
because you've taken away more high-frequency components.
The signal just can't wiggle that fast,
so it takes its leisurely time going through its paces here.
And you can imagine that you can be thrown off
when you try and take samples.
This is actually even more evident on the eye diagram,
here.
So these are eye diagrams-- again, the same kind of thing.
As you successively transmit fewer and fewer--
let's say less and less of the high-frequency content
of the signal, what gets picked up--
what gets received is a more rounded version
of what was sent in.
And the corresponding eye diagrams
that you construct-- well, at a certain point,
I guess somewhere around here, you'd
be a little nervous about trying to find a place to threshold
and decide on what signal you have.
So this is not a noise issue.
This is a distortion issue.
It's distortion induced by the channel.
And it can all be understood in terms
of what the channel is doing to the spectral content
of the input.
I think we'll continue next time to get more insight into this
and start on the topic of modulation.