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GEORGE VERGHESE: I want to actually spend
a little time, not too much, talking
about discrete-time Fourier transforms
versus discrete-time Fourier series.
You don't have any basis for comparison,
but I think the way we've told the spectral content
story this term is quite a bit simpler than in previous terms,
but that leaves you puzzling over chunks of the notes
and practice problems that refer to the discrete-time Fourier
series and to periodic signals and so on.
So I just wanted to give you a little insight into that,
and then we'll go on to talk about modulation
and demodulation some more.
OK, so we've seen that our interest is generally
in signals of finite duration because practical computation
has to deal with that.
And so we've got signals of this form--
0 outside of some window, and really without loss
of generality, I can take it to be some window from 0
to L minus 1.
If the signal shifts in time, we know
what to do with the Fourier transform.
Can you hear me all right, by the way, at the back?
Yeah, OK.
All right so if we've got non-zero values only
over a finite range, then the computation
of the discrete-time Fourier transform boils
down to a simple finite computation.
Now, what we'll typically do is give ourselves
a little more flexibility.
Since the signal is 0 outside of this interval anyway,
we might sometimes allow ourselves
to think of the signal as being longer, but still
with zeros out here.
So you might come all the way up to some P minus 1.
And what we're saying is, this is the window of interest.
Everywhere outside of this window, the signal is 0.
Now, the signal can be 0 at various points inside here,
as well, but what we're saying is, outside of this interval,
the signal is 0.
Therefore, I only need to compute this from 0
to P minus 1, all right?
And the nice thing is, it turns out
that you can actually recover the time-domain signals
from the samples of the DTFT through the formula
on the right side.
So what we're doing is we're actually
computing the DTFT just at isolated
points on the axis between minus pi and pi,
just P, capital P, points.
Or you can think of them as points on the unit circle that
correspond to each of those exponentials that
appear in the Fourier transform definition.
And we then recover the time-domain signals just
from those samples, OK?
And really what's driving this is the fact that the signal is
0 outside of a finite window.
OK.
We'll also typically-- and if you look in the books,
you'll see this, as well--
this notation often gets simplified, so x of omega sub k
gets simplified to just x sub k.
It's the k-th spectral coefficient.
All right, so all that is good.
And we have this nice algorithm for computing things, which
is the fast Fourier transform.
So we talked about how that significantly
reduces computation.
Now, there are properties of these formulas
that you can explore, and I have some listed here.
I'm not going to go through them.
They're essentially the same properties
we've seen for the DTFT.
I want to focus more on this formula for reconstruction
of the time signal from the spectral coefficients.
By the way, in a previous writing of this formula,
I had written the upper limit as P over 2.
It's actually P over 2 minus 1, so I'll
fix that in the earlier slides.
OK, so what you're guaranteed is that if you apply that formula,
you will recover every signal value in this window of length
capital P. But what happens outside of that window?
Well, if you look at this expression,
is the right-hand side here periodic?
You should suspect that it is because of the e to the j
something n is there, right?
If you look at the definition of omega sub k
and look at each of these terms, it turns out that each of these
will repeat periodically with period 2 pi.
Sorry, with period-- let's see.
I've said it badly.
This whole term will repeat when n increases by capital P,
all right?
So let me write it down.
And why is that?
Well, omega sub k is 2 pi over P.
So if you increase time by capital P,
you're going to increase the exponent by 2 pi,
and you've got the same exponential back again.
And you can do this for any integer multiple of capital P.
So what that tells you is that the expression
on the right-hand side is actually
going to repeat periodically outside of this interval.
So it's fine to use this formula to recover
the values in this window, but if you start
to evaluate this formula outside of that window,
you're going to start getting this whole thing repeated
periodically, so you're going to get--
at this point you'll get--
and so on, OK?
So the formula doesn't know what to do except
to replicate periodically.
It's up to you to know that this formula is
no good outside of this window.
All right?
There's another way to think of it, though,
which is that this formula gives you
a nice, compact representation for a periodic signal.
So if you started off with a periodic signal,
here's a way to represent it as just a sum of capital P
exponentials, and that's what a Fourier series is.
So you've seen in 1803 or other places,
in continuous time I imagine, that if you
had a periodic signal you could represent it
with a Fourier series.
This is actually a Fourier series
for this periodic signal.
But if you know that your action of interest
is all in this finite interval within one period,
then you can actually use the Fourier series just
to study what goes on in that one interval
without worrying about what's outside.
And that's really what we've done this term,
is we've kind of ignored periodic signals.
We've said all the attention is in a finite interval.
Within that interval, we have this Fourier representation.
It's easily computed by the FFT, and everything works nicely.
So just to give a concrete illustration of how we end up
applying this in a particular situation that
should be familiar, if I had an input going into an LTI system
producing an output, and if the input was non-zero only from 0
to, let's say some n sub x, and if the unit sample
response of the system was zero only from 0 to n sub h,
is there a particular interval of time
that you can guarantee for me will
contain all the non-zero values of the output?
I want you to find for me an interval outside
of which the output is guaranteed to be 0.
Anybody?
Yeah?
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: Yeah, good.
There are many ways to think of this.
One is to say, well, the input value at time 0
fires off a unit sample of duration,
goes from 0 to n sub h, and then the input value at time 1
fires off a unit sample response that starts one time later,
and so on.
So each of these fires off a unit sample response.
Well, you've got inputs extending from 0
to n x, and so you're going to have
an output that extends from 0 to n x plus n h, all right?
So you're guaranteed that all the action of interest
happens in this finite interval.
And given that that's the case, you can actually--
whoops, what happened?
You can actually do this kind of spectral representation,
use the FFT, and all of that.
You're going to just work on a finite interval, 0
to P minus 1, defined by that or greater than that, OK?
So this is actually one of the most frequent uses of the FFT.
It's to study systems where all the action happens
in a finite window and you know a priori what
the length of that window is, and you can then
do all your computations there.
And you never look outside that window
because you've already guaranteed that everything
of interest happens there.
But when you read the notes, you'll
find it's essentially the same story,
but when you talk about Fourier series
you're actually talking about the whole signal,
the periodic signal, all right?
One bit of notation, also, as you're reading the notes,
just to go back a second here.
We've been working entirely in terms
of these samples of the DTFT.
When you're thinking of Fourier series, when you're
thinking of this as a Fourier series,
it's typical to write X omega k over P
as just the Fourier coefficient, A sub k.
So you'll see in the notes an A sub k.
That's just a normalized version of the Fourier transform
sample, OK?
All right.
That's as much as I wanted to say on this,
so let's get back to talking about modulation
and demodulation.
If you have questions on what I talked about,
you can bring them up in recitation.
All right, so just to review where we are.
We've got some signal, x of n at baseband.
Baseband just means that its frequency content
is centered around zero.
You've not done any modulation or shifting yet.
You've been allotted some part of the frequency axis
to do your transmission in because someone's told you,
perhaps, that the medium that you're going to use
can only transmit in that range, or the FCC
has decreed that you're only going to use that region.
So you want to send that signal somehow
in another frequency band.
So modulation was a process by which
we converted up to some carrier frequency,
and then demodulation was what you did with the receiver
to get back down.
So just to look at that in a little more detail.
This is the modulation process we talked about last time.
You've got a time-domain signal, your information signal.
You multiply it by the cosine to get
an amplitude-modulated transmitted signal.
So t of n is the signal that you transmit, OK?
There are other names for this.
This process of multiplying a signal
by a cosine at a particular frequency
is referred to as heterodyning.
That's a term from the earliest days of Amplitude Modulation,
I think invented by Fessenden, who also invented AM,
and of course it's specifically amplitude modulation for us.
All right, so just to think spectrally,
we had a simplified version of this picture last time,
but let's first assume that this signal has some spectrum, which
is shown by a cartoon here.
I'm assuming a real signal.
So we know that the spectrum has a real part that's even
and an imaginary part that's odd,
and that's what's shown for you here on this figure, OK?
So we're going to track the spectrum of the signal
by tracking the real and imaginary parts separately
because the spectrum is in general a complex function
of frequency.
We've seen last time what happens when
you multiply by the cosine.
You take the spectrum, and you replicate it
at the locations of the carrier.
So if your carrier frequency is omega c,
here's your frequency band, going from minus pi
somewhere there to plus pi somewhere here.
You've got a plus and minus omega c, the carrier frequency.
So what happens when you modulate
is you take the spectrum and you plunk it down
on plus omega c and minus omega c,
and you scale by 1/2, all right?
So if the real part had amplitude a before,
it now has amplitude a over 2, and the imaginary part,
similarly.
I haven't drawn these to scale, but hopefully the labels
are clear enough.
OK, so the modulation is not simple
when you're thinking of what it does in the frequency domain.
Now, it is not simple, but this picture is a little deceptive,
perhaps, because I made an implicit assumption here.
Otherwise, the picture would be a bit messier.
What am I relying on here to get this simple picture?
Yeah?
Sorry?
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: Centered?
What's centered at zero?
Oh, the spectrum here?
OK, yeah, the spectrum centered at here
gives me a simple picture, yeah.
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: OK, exactly.
You see, when I drew this here, you
can still recognize the triangles that
came from over there.
But if the baseband signal had a frequency content that
extended way over, then the replication that I have
here would actually leak into the replication
that I have there, and I get a more complicated picture,
all right?
So if you want the simple picture,
you actually have to limit the frequency content
of your baseband signal, OK?
You can see here, if the signal only
extends omega c on either side, then I'm OK.
The two replications will not smear into each other,
all right?
So we need a limit on the frequency content.
Now, the specific limit that you have
depends on the application.
We'll see later when you do frequency-division
multiplexing, where you're trying
to put many different signals in the same general frequency
band, that the restrictions might be different.
But the basic idea is this, that you
want any replication of your signal,
if you're going to extract it later on downstream somewhere,
you want the replication to not be corrupted by images of it
somewhere else, or images of some other signal.
So actually, the example that I showed you last time
wasn't perfect in that regard, right?
Remember, this was the spectrum of our typical baseband.
We had 256 samples, like this, and then 0's.
And we looked at the spectral content.
It was given by a sinc-like function,
and this is the spectral content magnitude after modulation,
and therefore it's the two replicas.
I'd modulated this onto a 1,000 Hertz carrier.
So this is what we saw.
And you can see here that there's funny stuff going on
in here because the tails of the two replications
are merging with each other, OK?
So it's not perfectly symmetrical around here.
And actually, these sinc-like functions decay very slowly,
so even though it won't be visible to your eye,
there's a considerable amount of this
that's actually due to the replica out here, OK?
So this case doesn't quite satisfy
that band-limited condition.
If you shape your pulse a little bit more
carefully, for instance, if you had more rounded edges,
then you can pull in the frequency content,
and you might do a better approximation
to keeping the replica separate.
Or you might use a higher carrier frequency.
That pull them apart and have less interference,
but it's certainly an issue that you need to think about.
OK.
So what happens at the receiver?
We already saw this briefly at the end of lecture.
If what you receive is what you transmit-- in other words,
if it's the signal, then extracting the x of n is easy.
We said what you do is you basically
do the same heterodyning again, right?
You take the signal that comes in,
multiply it by cosine of the carrier frequency.
That's your signal after demodulation,
and a little bit of algebra shows that you actually
have your original signal of interest,
and then something that's your original signal
modulated by a cosine at twice the carrier frequency.
So now there's some hope that you can actually
pull these things apart.
All right.
So one question, of course, is what
does the spectrum of this look like, and we'll look at that.
And then the other question is, again,
what constraint on the bandwidth of the signal
that you originally sent from the transmitter--
what constraint is needed to recover?
So let's look at the spectrum of the received signal first.
We're assuming the channel is not distorting
and that we don't have noise, so what's transmitted
is also what's received.
So here is the spectrum of what's received.
It's exactly the spectrum I showed you earlier, right?
It was the baseband spectrum, but replicated at plus omega c
and minus omega c.
So this is what comes in off the channel,
assuming no distortion.
And I'm going to multiply it again
by a cosine at the carrier frequency.
So what is it that I have to do?
I take this entire spectrum, plunk down
a copy centered at plus omega c, and another copy
at minus omega c.
Because my demodulation, just to remind you,
my demodulation is multiplication
by cosine omega c again.
It's a multiplication by cosine omega c.
Well, we know what that does in the transform domain, so
here is the picture.
And the piece that we want is the center piece.
So what we need to do is filter it out
of what's resulted from the heterodyning.
So what kind of cutoff frequency would you-- what kind of filter
and what kind of cutoff frequency would you want?
Any suggestions?
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: Sorry, I didn't hear where that came from.
Yeah?
AUDIENCE: Lowpass filter?
GEORGE VERGHESE: Lowpass filter.
So, for instance, an ideal lowpass filter
would be great, right?
If you had a filter with a frequency-domain characteristic
that was perfectly flat in some region and then cut off,
let me say, at some frequency omega 0,
so something like that-- well, actually, we want a factor of 2
to compensate for the demodulation process,
if we want to get exactly the same thing back.
So this would be in the frequency domain.
Ideal lowpass filter.
And we know how to get approximations to this, right?
Because this is not really implementable.
If you wanted to implement this, what kind of unit sample
response would you need?
A sinc function, right, but extending infinitely
in both directions.
But we could truncate that sinc, and we could shift it
forwards in time to get a causal approximation to this filter.
And the resulting frequency response
will, if you plot it out, if you compute it and plot it,
won't look too different from this.
If I plotted the magnitude, you'll
get-- it's something that's a plausible approximation
to this lowpass filter.
But what cutoff frequency would you want?
What omega 0 should you pick?
Any suggestions?
Anybody?
We're trying to extract this piece.
So omega c would be a pretty safe choice, right?
Omega c would be one that passed everything here, and would
basically extract any signal that
satisfied that initial constraint that we mentioned.
So if your baseband signal originally
extended from minus omega c to plus omega c,
then a lowpass filter that extracted
that would do fine without pulling
in any of the replication here.
So omega c is certainly fine.
But if the signal that you transmitted at baseband
actually had a narrower bandwidth than that,
then you might just want to get away
with a lowpass filter with a lower cutoff.
Can you think of why you might want to do that?
Is there anything that motivates you
to use as small a bandwidth as possible?
Yeah?
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: To limit the amount of noise, right?
We suppress noise in this whole story.
So if you're going to build a filter like this
but all the interesting action is over here,
well, all of the rest of the filter
is doing is letting other signals get in, especially
noise, and then that's going to add to the output
and make things more difficult. So you'd really
like to get the smallest bandwidth that
suffices to pass the signal part of what you're interested in
but keep out the noise, all right?
But if you didn't know anything about the signal
and its spread, or you believe that the spectrum extended
really from minus omega c to plus omega c,
then you would want to make omega not equal
to the carrier frequency, right?
But you've got to look at your particular situation
and see what it is you're going to do.
OK, so this is the picture that we have at demodulation.
You're going to take the received-- well, no, sorry.
This is the modulation part.
Ah, no, it's not.
Sorry, this is not well-drawn.
That shouldn't be x sub n.
That should be the received signal, OK?
So the received signal comes in, gets multiplied locally
by cosine, gives you the demodulated signal,
and then you have the lowpass filter,
so I'll change that before I post it.
Simple enough?
OK.
Now, there are some problems that you can run into,
and doing all of this in the lab you actually see that
very quickly.
So let me actually put this on the board here.
What we said is that our demodulated signal is
going to be our received signal times cosine omega c n, right?
And if we assume no distortion in the channel this,
is x n cosine omega c n.
But there's a bit of a problem here,
which is that, even if you've been
told what carrier frequency your center is going to use,
you might not know exactly what phase.
It's typically the case that you don't know what
the phase is on this cosine.
So you know omega c, but you don't know exactly the phase,
which means that your local carrier, your local oscillator
or your local carrier multiplication here,
will end up having some offset relative to the carrier used
at the transmitting end, OK?
And so the question is, if we track this through,
what happens through the demodulation process,
so that's really what this is trying to do.
So we're saying d of n is your received signal,
but the local oscillator or the local cosine that you're
heterodyning with at the receiver
doesn't know exactly what phase was used as the transmitter,
so you've got to assume that there
is going to be some offset.
So this is actually what the multiplication is.
And now you use a simple trig identity.
It's the cosine of something times the cosine of something.
That splits into this.
And so what we're actually going to get from the heterodyning
at the receiver is x of n times all of this, OK?
So what we're going to get is--
I should write it down here.
We're going to get 0.5 x of n.
And then there's two pieces here.
There is the cosine phi, and then there
is the cosine 2 omega c n minus phi.
OK?
When you don't have any phase error,
the cosine phi term is 1, but now it's reduced from that.
The rest of the process is the same.
You're going to do some filtering to get rid
of this piece, the double frequency piece,
and you're going to pull out just what you're interested in.
Except now it's no longer x of n itself.
It's x of n multiplied by this cosine.
So can you see that this could lead you into trouble?
What's the worst case here?
Sorry, worst case-- yeah?
AUDIENCE: [INAUDIBLE]
GEORGE VERGHESE: Yeah, if phi is pi over 2,
then cosine phi is 0, and you get nothing, all right?
So if you're unlucky in the offset
between your local sinusoid and the sinusoid that
was used at the transmitter, you could end up with nothing, OK?
You can also get the negative of what
was sent, and so on, so you can go
through the whole set of possibilities there.
So the case of a phase error of pi over 2
corresponds to looking at a signal that
was transmitted on a cosine and multiplying it by a sine, OK?
And you can think through that in the spectral setting,
as well.
Maybe you'll do some of this in recitation,
or maybe you already have.
But when you multiply by a sine in the spectral domain--
So if you've got--
you've got your received signal, r of n,
and now you're multiplying it by sine omega c n, right?
Sine omega c n, well, that's 1 over 2j e to the j omega
c n minus e to the minus j omega c n, right?
So in the spectral domain, what happens?
Well, you've got r of n multiplied by 1 over 2j times
this first exponential.
In the spectral domain, that does a shifting
and a multiplying by 1 over 2j, and then you've
got this term doing the same kind of thing.
So you're going to have a shift of the spectrum of r of n
in the frequency domain and a scaling by 1 over 2j.
So if you think through what the shifting and scaling does,
you see that it's a little bit more
of a complicated picture of what you had over here.
Well, the real part gets replicated around the 2 omega c
region, but flipped over.
The imaginary part gets carried over intact.
And then the replications around minus 2-- sorry,
around minus omega c, that is--
the imaginary part gets flipped over,
and the real part gets carried over directly.
Except what was real before becomes imaginary now.
What was imaginary before becomes real now.
So you can track through all of that.
And it just comes from applying the standard DTFT results
to what the spectrum of the product of an r of n and this
is, OK?
But the interesting thing is now,
the two replications, when you sum them up,
will leave you with nothing at 0 because this piece here
will cancel out exactly with that piece there.
So if you think through in the spectral domain what's
going on, you'll understand exactly
that if you've put your signal on the cosine
and you demodulate with a sign, you're
going to get nothing in that lowpass region, OK?
So that's just the same result but seen spectrally.
All right, so that's uncertainty between the phase
of the transmitter and the phase of the receiver.
Here's another thing that has a similar effect, which
is an unknown delay on the channel, OK?
So at the transmitting end, you've got your baseband signal
multiplying the carrier.
This is what's transmitted.
But then you have a time delay, let's say
D samples, capital D samples.
So then what's received is actually t of n minus D,
and that's what's going to get multiplied
by the local carrier.
And I'm assuming for now that you have the phase, so we can
bring them both together later.
But I'm assuming now there's no phase out locally,
but there's an unknown delay on the channel.
And you can see it's going to be the same kind of thing.
You've got a cosine times a cosine,
and the arguments are slightly different from each other.
And you use the same trig identity,
and what you find is the output of this process
is not the input delayed, which is what
you would like to get ideally.
You aren't going to compensate for the delay
with a causal filter, but it's also going to be scaled.
And it's going to be scaled by an unknown amount that depends
on that delay, all right?
So it's the same kind of thing that happens.
So the question is, how do we get around this?
And here's one idea that works well,
and which you're actually exploring in the lab.
Which is to use both the sine and the cosine, OK?
So use both the sign on the cosine to demodulate.
If you go completely bad on one channel
because you've got the phase completely wrong
with the cosine, you're going to do
all right on the sine channel.
If you do completely bad on the sine channel
because you've got the phase wrong,
then you're going to do all right on the cosine channel.
So at least one of them will work,
and more typically both of them will work a little bit,
and what you'll then do is combine the two outputs.
OK?
So you're going to have the signal coming in.
There's a cosine multiplication and the sine multiplication,
and then the lowpass filtering.
We refer to this as the in-phase component,
assuming that you were modulating on a cosine,
and this is referred to as the quadrature component.
So there's in-phase and quadrature.
"Quadrature" just means at right angles.
So this is the I and the Q components.
And if you work out what these are, assuming now
that there's both a time delay and a phase offset,
you can see that the in-phase component
will be the signal that you want, but multiplied
by cosine phi.
The quadrature component will be the signal
that you want, but multiplied by sine phi.
And from there, it's not so hard to imagine
that you could actually get back to the signal of interest.
And here's one way to do it that works fine
if you've got on-off signaling.
So what you would do is, here's the I. Here's the Q.
And I've just represented it graphically here.
This is typical to do.
So here is the I component.
Here is the Q component.
And you could take the root sum of squares
to basically get rid of that sine theta and cosine theta
term, right?
So what that's going to give you is the absolute value of x
of n minus D. So you can certainly get back
the absolute value of what was used to modulate the carrier,
and that may be all you need.
If you have on-off signaling, that's all you need.
If your modulating signal never goes negative,
its absolute value is the same as the signal, so this is fine.
So what you will discover is that you get some signal out
there, and you're looking for its length.
When the length is non-zero, you say you have a 1 cent.
When the length of 0, you say you have a 0 cent.
In the presence of noise, of course,
it won't be exactly at the origin.
There might be some cloud of points there.
And similarly for the 1 level.
What if you were interested in the polarity, though?
So suppose it mattered to you whether the signal was
positive or negative.
Well, you could then just plot the point
and don't take the magnitude.
So you'll get something that looks a bit more like this.
OK?
So what you'll have is, when a 1 is sent,
perhaps you'll get that value in the absence of noise.
When a minus 1 is sent-- sorry, when
a 0 is sent, corresponding to minus 1,
this is what you'd get.
This is-- sorry, I should have said that.
This is assuming bipolar signaling, right?
Bipolar signaling is the case where
you're interested in the sign of the signal.
You use plus 1 to send a 1.
You use minus 1 to send a 0, OK?
So you get some diagram like this.
The only problem here is, if you've got uncertain phase
and delay, you actually don't know which of these two points
corresponds to the plus 1 and which
corresponds to the minus 1.
So there's that additional ambiguity
that needs to somehow be resolved,
and there are different procedures you might use.
You could, for instance, have some preamble
with a sign that's agreed on and use that as a basis
for figuring out which is a plus and which is a minus.
And there are other ways of doing it, as well, where what's
called differential coding, where basically it's
not where that is, but whether it flips over
to the other side or not that signals a bit.
And so what you could do is, to transmit a 1,
you'll step the phase by pi, and that can be detected,
and to transmit a 0, you don't change
the phase in the next bit slot.
So if from one bit slot to the next the dot stays there,
you know you've just received a 0.
If from one bit slot to the next it
flips over to the other side, you
know you've just received a 1, OK?
So even with the ambiguity, if you change the way you
code at the sending end, you can actually compensate for this.
OK, now playing this game with sines and cosines
can actually also be done at the transmitting end,
and we haven't explored that in class,
but it's something that you could think about.
So we've been talking about taking the samples
and multiplying them onto a cosine carrier.
You could have another bitstream whose samples you
multiply onto a sine carrier.
And you can just add them together and send them
over the channel.
At the receiving end, you multiply by cosine.
Well, that'll only pull out-- you multiply that cosine
and then filter, lowpass filter--
that'll only pull out the first stream in the ideal case.
Multiply by a sine and filter, you'll
get exactly the second stream.
So you can simultaneously send two streams
on a given carrier using this scheme, this method, OK?
So depending on how you make out in the lab in problem set 6--
I don't know how many simultaneous carriers
you're getting, but whatever you end up with you
can actually try now to transmit twice as much on each carrier
by using this kind of a scheme.
Could be fun.
All right, so this kind of bipolar signaling,
what's called phase-shift keying--
I didn't explain that really, did I?
I said-- we've said it before, but--
we've talked about bipolar or phase-shift keying.
All that we mean is, if you get a signal with voltage plus 1
and minus 1 for your bit 0 on bit 1,
by the time you modulate what you're going to end up doing
is sending a burst of carrier here with the plus 1.
And then when you come to the minus 1 region,
you're going to multiply that carrier by minus 1,
so you're going to suddenly step the phase, right?
So amplitude modulation with an amplifier
that switches between these levels
can be also thought of as a phase shifting.
So you're keying between a phase of 0 degrees
and a phase of 180.
So this kind of scheme is used all over the place.
And I actually have a slide that lists a whole bunch of schemes
that you're familiar with, you see every day
in all sorts of literature.
You know, 802, and Bluetooth, and Zigbee and so on.
In all of these standards, there's
some piece of it or some domain or some regime
in which what's going on is some variant of what
we've learned here.
And they get fancier and more sophisticated,
but you really have the key ideas here.
OK, let's now talk about putting multiple signals on a given
piece of the spectrum, OK?
This is exactly the situation you have in your lab now.
You've got the speaker that can transmit in a certain band,
and you're trying to put multiple simultaneous signals
on it by using different carriers.
So this is what's called Frequency-Division
Multiplexing, or FDM.
And the idea is very simple.
You've got three signals here and this illustration,
the blue signal, red, and green.
Pick a carrier frequency for each of them.
Do the modulation, and then just add them on the channel.
If you've got a linear medium, then the signals
will superpose, so what's received
is just the sum of these, and now
you can do the same kind of thing.
And what we're relying on here is, again,
the heterodyning principle.
Whoops, sorry.
OK, so if you've got frequencies omega red, omega blue,
omega green in the signal that you're receiving,
and you multiply this with some local sinusoidal frequency,
omega 0, where will your various spectra
be centered in the result?
So the way to think of it is, all the sums
at different frequencies here will now appear.
So you get omega 0 plus omega r.
You get omega 0 minus omega r, and similarly for all of these.
OK, it's the same thing that you saw with a single transmitted
the signal, except now it's a more elaborate constellation.
It's actually this that's being transmitted.
So the receiving end, you'll pick a particular frequency
to multiply the incoming signal by.
The result will have pieces of the spectrum centered
at each of these.
So if you want to center one of these in your lowpass filter,
how should you pick the local oscillator?
If you want to tune in a channel,
what is it that you want to do?
You want to get one of these center frequencies
to sit right in the window of your lowpass filter.
So what you'll end up doing is pick your local oscillator
frequency to be the carrier frequency of the station you're
interested in, or the signal you're interested in, OK?
So it's the same idea.
You have a lowpass filter, and you're
using heterodyning to shift the piece
of the spectrum of interest into the passband
of the lowpass filter.
All right?
Now, what about the bandwidth of the lowpass filter?
What should it be?
So now it depends on how closely spaced your carriers are,
right?
So for instance, if I ended up heterodyning
such that my blue signal, my blue signal
came into the window of interest--
and I've got the red spectrum sitting somewhere here
and the other piece sitting somewhere here.
OK?
I've shifted things so that this is at 0.
What's this frequency now?
Omega r minus omega--
what was it?
Blue, right?
So I've basically shifted these frequencies.
So this is at 0, sitting in my lowpass filter.
Use a different color.
And I want to reject everything else.
So how should I pick that lowpass filter?
Well, presumably, you want the cutoff
to lie between these two frequencies.
So you want half the distance to the nearest carrier, right?
Half the distance to the nearest carrier frequency.
OK, so you can think through these.
And notice how all our thinking has
been in the spectral domain.
Thinking in the frequency domain clarifies this whole thing.
You really would not have been able to do
what you're doing thinking entirely in the time domain.
Now, all of this comes to us, really, from--
let's see.
I had that already?
Yeah.
All of this comes to us from a rich legacy in AM radio.
We're not using this for transmission of analog signals
by amplitude modulation, but it's the same principle.
So these principles actually were studied
from the early 1900s, actually.
And the AM radio that we see around us
now is actually set up exactly to do the kind of thing
we're talking about.
So you've got some frequency spectrum
that the FCC is allowing you to use.
Different stations are given different carrier frequencies
that they can operate on.
They're also instructed on what bandwidth they can occupy.
So basically, the carriers are 10 kilohertz apart,
the way that the stations are assigned.
So if you're transmitting from your station,
you'd better lowpass filter what you're
sending out to 5 kilohertz before you transmit it.
Because if you don't, you're going
to interfere with a nearby station.
Assuming there is a station in the same geographic area that's
been assigned to a neighboring carrier frequency, all right?
So all of these issues come in.
Another thing that actually is--
what did I do here?
I think I--
I mashed together two slides.
But the other thing that--
it turns out, for instance, AM radio,
at nighttime, because of the way radio transmits,
the signal can propagate much further.
So these stations are asked to reduce
their signal strength, the carrier strength, at nighttime
so that they're not interfering with nearby stations.
The stations that they would not interfere
with during the day they could interfere with at night
because propagation characteristics
turn out to change.
So all of this business of your signal
not interfering with your neighbor's carrier
or your neighbor's portion of the spectrum, all of that
ends up being important.
OK, I think we've probably said as
much as we want to say about the signals part of this class.
One of the things about 602 is that those of you
who master it come out knowing the subject better than any
of us that teach it because there's none of us that's
able to teach the course right through start to finish.
Well, that's not entirely true.
Harry knows how to do it Chris Terman knows how to do it.
The recitation instructors hang in there for the whole term,
but it's very hard for one person to do that.
So I'm done.
I'm going to be sitting there from lecture onwards,
so thank you all for your attention, and thank you.