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Hi, and welcome to module 9.5 of digital 
signal processing.

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In this module, we will touch briefly on 
some topics in receiver design.

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A lot of things, unfortunately, happen to 
the signal while it's traveling through

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the channel. 
The signal picks up noise, we have seen

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that already. 
It also gets distorted because the

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channel will act as some sort of filter 
that is not necessarily all pass and

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linear phase. 
Interference happens too, that might be

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parts of the channel that we thought were 
usable, and they're actually not, so the

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receiver really has to deal with a copy 
of the transmitted signal that is very,

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very far from the idealized version we 
have seen so far.

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The way receivers, especially digital 
receivers, can cope with the distortions

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and noise introduced by the channel, is 
by implemented adaptive filtering

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techniques. 
Now we will not have the time to go into

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very many details about adaptive signal 
processing, again these are topics that

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you will be able to study in more 
advanced signal processing classes.

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But I think it is important to give you 
an overview of things that have to happen

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inside a receiver, inside your ADSL 
receiver for instance.

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So you can enjoy this high data rates 
that are available today.

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And the first technique that we will look 
at is adaptive equalization and then we

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will look at some very simple timing 
recovery that it used in practice in

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receivers. 
Let's begin with a blast from the past.

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[NOISE] Those of you that are a little 
bit older will certainly have recognized

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this sound as the obligatory soundtrack 
every time you used to connect to the

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internet. 
And indeed this is the sound made by a

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V34 modem that was the standard dial up 
connection device seen in the 90s until

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the early 2000. 
Now if you have ever used a modem, you've

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heard the sound and you probably wondered 
what was going on, so we're going to

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analyze what we just heard from the 
graphical point of view.

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If we look at the block diagram for the 
receiver once again, what we're going to

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do is we're going to plot the base band 
complex samples as points on the complex

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blank. 
So we're going to take B, r of n has the

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as the horizontal coordinate and b, i of 
n as the vertical coordinate.

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And before we do so let's just look for a 
second at what happens inside the

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receiver when the signal at the input is 
a simple sinusoid, like cosine of omega c

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plus omega zero n. 
We are demodulating this very simple

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signal with the two carriers, the cosine 
of omega c n and sine of omega c n.

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And then we're filtering the result with 
a low pass field.

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So if we work out this formula with 
standard trigonometric identities, we can

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always express for instance the product 
of two cosine functions as the sum.

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Of the cosine of the sum of the angles 
plus the cosine of the difference of the

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angles. 
And same for the product of cosine and

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sine. 
So if we do that, we get four terms, two

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of which have a frequency that will fall 
outside of the pass band of the filter h.

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So when we apply the filter to these 
terms, we're left only with cosine of

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omega 0 n plus j sine of omega 0 n. 
Which is of course e to the j omega 0 n.

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So when the input to the receiver is a 
cosine, the points in the complex

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baseband sequence will be points around 
the circle and a difference between two

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successive points is the angle omega 0. 
The reason why we might be called to

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demodulate a simple sinusoid is because 
the receiver will send what are called

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pile tones, simple sinusoids that are 
used to probe the line and gauge the

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response of the channel at particular 
frequencies.

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So with this in mind, let's look at the 
slow-motion analysis of the base band

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signals. 
Samples, when the input is the audio file

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we just heard before. 
So lets start with a part that goes like

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this[SOUND]. 
This signal contains several sinusoids,

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that we can see here in the plot. 
And the sinusoids also contain abrupt

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phase reversal, meaning that, at some 
given points in time The phase of the

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sinusoid is augmented by pi. 
You can see this as this small explosions

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in the circular pattern in the plot. 
These phase reversals, are used by the

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transmitter and the receiver, as time 
markers, to estimate the propagation

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delay of the signal, from source to 
destination.

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The next part goes like this. 
[NOISE].

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And this is a training sequence. 
The transmitter sends a sequence of known

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symbols, namely the receiver knows the 
symbol that are transmitted.

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And so the receiver can use this 
knowledge to train an equalizer to undo

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the affects of the channel. 
The last part is the data transmission

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proper, the noisy part if you want, of 
the audio file.

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And the interesting thing is that the 
transmitter and receiver perform a

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handshake procedure, using a very low bit 
rate QAM transmission using only four

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points, therefore two bits per symbol. 
To exchange the parameters of the real

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data transmission that is going to 
follow.

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The speed, the constellation size, and so 
on.

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Using the four point QAM constellation in 
the beginning ensures that, even in very

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noisy conditions, transmitter and 
receiver can exchange their vital

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information. 
So even from this simple qualitative

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description of what happens in a real 
communication scenario, we can see that

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the task that the receiver is saddled 
with is very complicated.

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So it's a dirty job, but a receiver has 
to do it.

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And a receiver has to cope with four 
potential sources of problem.

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Interference. 
The propagation delay, so the delay

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introduced by the channel. 
The linear distortion introduced by the

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channel. 
And drifts in internal clocks between the

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digital system inside the transmitter and 
the digital system inside the receiver.

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So when it comes to interference the 
handshake procedure, and the line probing

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pilot tones are used in clever ways to 
circumvent the major sources of

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interference. 
We will see some example later on when we

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discuss ADSL. 
The propagation delay, is tackled by a

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delay estimation procedure, that we will 
look at in just a second.

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The distortion to this by the channel is 
compensated using adaptive equalization

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techniques, and we will see some examples 
of that as well.

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And clock drifts are tackled by timing 
recovery techniques that in and of

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themselves are quite sophisticated and 
therefore we leave them to more advanced

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classes. 
Graphically, if we sum up the chain of

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events that occur between the 
transmissions of the original digital

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signal and the beginning of the 
demodulation of the received signal.

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We have a digital to analog converter and 
a transmitter, this is transmitter part

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of the chain that operates with a given 
sample period Ts.

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This generates an analog signal which is 
sent over a channel.

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We can represent the channel for the time 
being as a linear filter in the

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continuous time domain. 
With frequency response d of j omega.

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At the input of the receiver, we have a 
continuous time signal, s hat of t.

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Which is a distorted and delayed version 
of the original analog signal.

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We will neglect noise for the time being. 
This signal is sampled by an a to d

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converter that operates at a period t 
prime of s.

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And we obtain the sequence of samples 
that will be input to the modulator.

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So this is the receiver part of the 
chain.

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We have to take into account the 
distortion introduced by the channel, and

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we have to take into account the 
potentially time varying discrepancies in

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the clocks between the transmitter and 
the receiver.

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These two systems are geographically 
remote and there is no guarantee that the

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two internal clocks that're used in the A 
to D and D to A converters are

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synchronized or run exactly at the same 
frequency.

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Let's start with problem of delay 
compensation.

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To simplify the analysis we'll assume 
that the clocks at transmitter and

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receiver are synchronized and 
synchronous.

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So T prime of s is equal to T s, and the 
channel acts as a simple delay.

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So the received signal is simply a 
delayed version of the transmitted

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signal. 
Which implies that the frequency response

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of the channel is simply e to the minus j 
omega d.

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So, the channel introduces a delay of d 
seconds.

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We can express this in samples in the 
following way.

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We write d as the product of the sampling 
period, times b plus tau.

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Where b is an integer. 
And tau is strictly less than one-half in

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magnitude. 
So b is called the bulk delay because it

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gives us an integer number of samples of 
delay at the receiver and tau is the

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fractional delay. 
So the fraction of samples introduced by

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the continuous time delay of d. 
So, how do we compensate for this delay?

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Well, the bulk delay is rather easy to 
tackle.

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Imagine the transmitter begins 
transmission by sending just an impulse

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over the channel. 
So, the discreet time signal is this one,

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it's just a delta and a 0. 
It gets sent to D-to-A converter.

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And the converter will output a 
continuous time signal that looks like an

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interpolation function like the sinc. 
And like all interpolation functions, it

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will have a maximum peak at zero that 
corresponds to the known zero sample.

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This signal gets transmitted over the 
channel.

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And it gets to the receiver after a delay 
d that we can estimate for instance by

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looking at the displacement of the peak 
of the interpolation function.

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The receiver converts this into a 
discrete time sequence.

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Now in the figure here it looks as if the 
sample incidence at the transmitter and

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receiver are perfectly aligned. 
Now this is not necessarily the case

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because the starting time for the 
interpolator at the transmitter and the

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sampler at the receiver are not 
necessarily synchronous.

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But any difference in starting time can 
be integrated into the propagation delay

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as long as the sampling periods are the 
same.

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So with this, all we need to do at the 
receiver is to look for the maximum value

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in the sequence of samples. 
Because of the shape of the interpolating

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function, we know that the real maximum 
will be at most half a sample in either

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direction of the location, of the maximum 
sample value.

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So, add the receiver to offset the bulk 
delay, we will just set the nominal time

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n equal to 0, to coincide with the 
location of the maximum value of the

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sample sequence. 
Now of course we need to compensate for

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the fractional delay, so we need to 
estimate tau.

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And to do that we'll use a different 
technique.

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Let me add in passing, that in real 
communication devices, of course we're

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not using impulses to offset the bulk 
delay.

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Because impulses are full-band signals 
and so they would be filtered out by the

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passband characteristic of the channel. 
The trick is to embed discontinuities in

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pilot tones and to recognize this 
discontinuities at the receiver.

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As we have seen in the animation at the 
beginning of this module, we use phase

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reversals, which are abrupt 
discontinuities in sinusoids, to provide

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a recognizable instant in time for the 
receiver to latch on.

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Okay, so what about the fractional delay? 
Well, for the fractional delay, we use a

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sinusoid instead of a delta, so we build 
a bass band signal, which is simply a

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complex exponential at a known frequency, 
omega 0.

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This will be converted to a real signal 
before being sent to the D-to-A

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converter, and so what we transmit 
actually is cosine of omega c, the

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carrier frequency, plus the pilot 
frequency omega 0, times n.

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The receiver will receive a delayed 
version of this, which contains the delay

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now in sample, and fraction of sample, b 
plus tau.

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After we demodulate this cosine you 
remember we got a complex exponential and

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we can also compensate already for the 
bulk delay which we know.

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So for an integer number of sample b we 
obtain a base band signal at b of n which

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is e to the j omega and minus tau. 
Since we know the frequency of omega 0 we

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can just multiply this quantity by e to 
the minus j omega 0 n and obtain e to the

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minus j omega 0 tau, which is a constant. 
And which we can invert given that we

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know the frequency omega 0. 
And so now we have an estimate for both

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the bulk delay and the fractional delay. 
Now we have to bring back the signal to

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the original timing. 
The bulk delay is really no problem.

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It's just an integer number of samples. 
What creates a problem is the fractional

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delay because that will shift the peaks 
with respect to the sampling intervals.

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So if we want to compensate for the bulk 
delay we need to compute subsample values

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and in theory to do that we should use a 
sinc fractional delay namely a filter

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with impulse response sinc of n plus tau. 
In practice however, we will use a local

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interpolation and this is a very 
practical application of the Lagrange

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interpolation technique that we saw in 
module 6.2.

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So graphically the situation is like so, 
we have a stream of samples coming in.

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And for each sample, we want to compute 
the subsample value with a distance of

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tau from the nearest sample interval. 
And we want to only use a local

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neighborhood of samples to estimate this. 
Now, you remember from module 6.2.

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The Lagrange approximation works by 
building a linear combination of Lagrange

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polynomials weighed by the samples of the 
function.

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So, as per usual, we choose the sampling 
interval equal to 1, so that we lighten

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the notation. 
We have a continuous time function x of

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t, and we want to compute x of n plus 
tau, with tau less than one half in

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magnitude. 
So we have samples of this function at

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integers, n, and the local Lagrange 
approximation around n, is given by this

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linear combination of Lagrange 
polynomials.

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Weighted by the samples of the functions 
around the approximation point.

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So we use the notation x L of n and t. 
N is the center point and t is the value

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from the center point at which we want to 
compute the approximation.

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And the Lagrange polynomials are given by 
this formula here, which is the same as

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in module 6.2. 
So the delayed compensated input signal

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will be set equal to the Lagrange 
approximation at tau.

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So let's look at an example. 
Assume that we want a second order

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approximation. 
So we pick N equal to 1 and we will have

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three Lagrange polynomials. 
And so, we will need to use three samples

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of the sequence to compute interpolation. 
These three polynomials will be centered

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in n minus 1 and in n plus 1 and scaled 
by the values of the samples at these

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locations. 
And finally, we will sum the poll numbers

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together and compute their value in n 
plus tau.

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So, we start with the first one, which is 
centered in n minus 1.

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And, like all interpolation polynomials, 
its value is 1 in n minus 1, and 0, at

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other integer values of the argument. 
The second polynomial will be centered in

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n, and the third polynomial will be 
centered in n plus 1.

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When we sum them together, we obtain a 
second order curve that goes through the

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points, that interpolates the three 
points, and then we can compute the

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approximation as the value of this curve 
in n plus tau.

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Now the nice thing about this approach, 
is that if we look at the approximation,

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if we take the Lagrange approximation 
around n.

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We can define a set of coefficients, d 
tau of k, which are the values of each

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Lagrange polynomial in tau. 
So d tau of k, are 2 N plus 1 values, the

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form, the coefficients, of an FIR filter. 
And we can compute the value of the

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Lagrange approximation simply as the 
convolution of the incoming sequence with

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this interpolation filter. 
So for example, if these are the three

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Lagrange polynomials for n equal to 1, we 
can compute these polynomials for t equal

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to tau, where tau is the fractional delay 
that we estimated before.

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And we will obtain three coefficients, 
like here, for instance, is an example

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for tau equal to 0.2. 
Three coefficients that give us an FIR

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filter, and then we can just simply 
filter the samples coming into the

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receiver with this filter, to compensate 
for the fractional delay.

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So again, the algorithm is, estimate the 
fractional delay.

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The bulk delay is no problem, again. 
Compute the 2 N plus 1 Lagrangian

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coefficients and filter it with the 
resulting FIR.

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The added advantage of this strategy is 
that if the delay changes over time for

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any reason, all we need to do is to keep 
the estimation running and update the FIR

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coefficients as the estimation changes 
over time.

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Okay, now that we know how to compensate 
for the propagation delay introduced by

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the channel. 
Let's go see the rechannel it with an

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arbitrary frequency response D j of 
omega.

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And the transmission chain goes from the 
pass band signal s of n, discreet time,

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into a D-to-A converter, analog signal s 
of t, it gets filtered by the channel,

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gives us hat s of t, which is sampled at 
the receiver to give us.

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A received fast band signal, hat s of n. 
But now we have seen in the previous

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module that this block diagram can be 
converted into an all digital scheme

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where our band pass signal s of n gets 
filtered by the discrete time equivalent

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of the channel. 
And, gives us a filtered version of the

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bandpass signal, as would appear inside 
the receiver.

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So the problem now, is that we would like 
to undo the effects of the channel, on

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the transmitted signal. 
And the classic way to do that, is to

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filter the received signal hat s of n, by 
a filter E, that compensates for the

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distortion or the filtering introduced by 
the channel.

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So the target is that the output of the 
filtering operation gives us a signal hat

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s e of n, which is equal to the 
transmitted signal.

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How do we do that? 
In theory, it would be enough to pick a

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transfer function for the filter E, which 
is just the reciprocal of the equivalent

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transfer function of the channel. 
But the problem is that we don't know the

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transfer function of the channel in 
advance because each time we transmit

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data over the channel, this transfer 
function may change.

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And also, even while we're transmitting 
data, the transfer function might change

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because it is a physical system that 
might be subject to.

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Drifts and modifications. 
So what do we do?

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We need to use adaptive equalization. 
So the filter that compensates for the

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distortion introduced by the channel is 
called an equalizer.

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And what we want to do Is to change the 
filter in time, so change the filter

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coefficients in a DPS realization as a 
function of the error that we obtain when

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we compare the output of the filter with 
the signal that we would like to obtain.

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In our case the signal that we would like 
to obtain is the transmitted signal.

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And so we take the received signal, we 
filter it with the equalizer.

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We look at the result. 
We take the difference, with respect to

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the original signal, and we use the 
error, which should be zero in the ideal

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case, to drive the adaptation of the 
equalizer.

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But wait, how do we get the exact 
transmitted signal at the receiver?

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Well, we use two tricks. 
The first one is boot strapping.

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The transmitter will send a prearranged 
sequence of symbols to the receiver.

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So let's call the sequence of symbols a t 
of n.

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This gets modulated and generates a pass 
band signal s of n.

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Now at the receiver the sequence a t of n 
is known.

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And the receiver has an exact copy of the 
modulator, of the transmitter, inside of

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00:19:41,974 --> 00:19:46,485
itself. 
So the transmitter can generate locally,

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00:19:46,485 --> 00:19:50,138
an exact copy of the pass band signal s 
of n.

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00:19:50,138 --> 00:19:54,428
And so, for the bootstrapping part of the 
adaptation, we actually have an exact

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00:19:54,428 --> 00:19:58,718
copy of the transmitted pass band signal 
that we can use to drive the adaptation

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00:19:58,718 --> 00:20:03,890
of the coefficients. 
The training sequence is just long enough

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00:20:03,890 --> 00:20:06,620
to bring the equalizer to a workable 
state.

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00:20:06,620 --> 00:20:10,290
For the handshake procedure that we saw 
in the video before, for instance.

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00:20:10,290 --> 00:20:13,911
This would correspond to the moment where 
the receiver starts demodulating the four

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00:20:13,911 --> 00:20:17,550
point QIM. 
At that moment, the receiver will switch

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00:20:17,550 --> 00:20:20,890
strategy and implement a data driven 
adaptation.

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00:20:20,890 --> 00:20:25,770
The thing works like this. 
The received signal gets equalized, gets

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00:20:25,770 --> 00:20:30,606
demodulated and then the slicer will 
recover the sequence of transmitted

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00:20:30,606 --> 00:20:34,107
symbols. 
Since the receiver has a copy of the

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00:20:34,107 --> 00:20:38,880
transmitter inside of itself, it can use 
the sequence of transmitted symbol.

293
00:20:38,880 --> 00:20:42,430
To build a local copy of the transmitted 
signal.

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00:20:42,430 --> 00:20:47,120
Now of course, errors might happen in the 
slicing process, and so this local copy

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00:20:47,120 --> 00:20:52,496
is not completely error-free. 
But the assumption is that the equalizer

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00:20:52,496 --> 00:20:56,272
is doing already enough of a good job to 
keep the number of errors in this

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00:20:56,272 --> 00:21:01,548
sequence sufficiently low. 
So that the difference, with respect to

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00:21:01,548 --> 00:21:05,708
the received signal, is enough to refine 
the adaptation of the equalizer, and

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00:21:05,708 --> 00:21:11,800
especially to track the time varying 
conditions of the channel.

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00:21:11,800 --> 00:21:14,293
What we have seen, is just a qualitative 
overview of what happens inside of a

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00:21:14,293 --> 00:21:17,765
receiver. 
And there're still so many questions that

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00:21:17,765 --> 00:21:21,940
we would have to answer to be thorough. 
For instance, how do we carry out the

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00:21:21,940 --> 00:21:25,140
adaptation of the coefficients in the 
equalizer?

304
00:21:25,140 --> 00:21:30,394
How do we compensate for different clock 
rates in geographically diverse receivers

305
00:21:30,394 --> 00:21:34,544
and transmitters? 
How do we recover from the interference

306
00:21:34,544 --> 00:21:40,900
from other transmission devices, and how 
do we improve the resilience to noise?

307
00:21:40,900 --> 00:21:45,500
The answers to all those questions 
require a much deeper understanding of

308
00:21:45,500 --> 00:21:50,650
adaptive signal processing, and hopefully 
that'll be the topic of your next signal

309
00:21:50,650 --> 00:21:54,119
processing class.