Hi and welcome to Module 9.6 of Digital 
Signal Processing. 
This is our last module in our Data 
Communications section, and with 
everything that we've learned so far we 
will be able to look inside an ADSL modem 
and see how that works. 
We will start by examining the nature of 
the channel. 
That ABSL works on. 
Namely the copper wire links your home to 
the nearest central office. 
We will look at the signaling strategy 
that is best put on place on such a 
channel. 
And finally we will look at a very 
efficient implementation of that 
signalling strategy that goes under the 
name of discrete multi-tone modulation. 
If we just take an abstract view of the 
telephone network today, we see that we 
have a link that goes. 
From the home to the central office, and 
in the central office, a fundamental 
split takes place. 
The voice communication, when you talk on 
the phone, is sent to the voice network 
and then relayed to the, what is called 
the plain old telephone system, POTS. 
The data part of your communication when 
you use the ADSL is separated from the 
voice content and sent to a DSLAM. 
DSLAM stands for digital subscriber line 
access multiplier. 
And it's fundamentally a bank of modems 
that manage to handle multiple 
communications at the same time. 
And the data here then goes on to the 
internet in digital format. 
So, if you want, what we're really 
interested in is how to send the data 
from your home to the central office, 
because what happens afterwards is 
already entirely in the digital domain. 
But here, we have what is called a last 
mile, a piece of copper wire, namely an 
analog channel. 
That connects your home to the central 
office. 
Now a copper wire has, naturally, a very 
large bandwidth in excess of one 1MHz. 
But because of the width of the bandwidth 
and because the wire is not shielded, it 
is actually likely to pick up a lot of 
interference and noise. 
If we look at how the ADSL channel is 
organized and we're showing here just the 
positive frequencies. 
We can see three distinct regions. 
The first one is the part reserved to the 
telephone conversation, this is the base 
band part of the channel up to about four 
kilohertz. 
Then we have a region that is devoted to 
the upstream part of the data 
communication, the data that you send up 
to the internet, and then a downstream 
part that is much larger that is used for 
data download. 
This asymmetry between upstream and 
downstream is actually the reason why the 
communication protocol is called ADSL. 
ADSL stands for Asymmetric Digital 
Subscriber Line. 
If we now look at all the nasty things 
that can happen on the channel when we 
send data. 
We can identify three fundamental sources 
of worry. 
The first one is an attenuation curve for 
the channel that is completely uneven. 
This could be due to imperfection in the 
wire, parasitic capacitance, and so on, 
so forth. 
Then we might have very large noise or 
interference in certain regions of the 
spectrum. 
For instance, you turn on your vacuum 
cleaner, and that raises the noise floor 
in the certain frequency band. 
And thirdly we have very localized 
interference from radio communications 
now the radio band starts well within the 
bandwidth of the ADSL channel for 
instance from 15 to a hundred kilohertz 
here you have ship to shore 
communication. 
Up to 500 kilohertz, you have airplane 
communication, and over 500 kilohertz you 
have the AM radio band. 
So if you live near a radio station, for 
instance, tough luck. 
You have a lot of interference in the 
upper regions of the ADSL band. 
Since the channel is so wide, and the 
type of disturbances is so diverse, it 
would be extraordinarily difficult to try 
to equalize and compensate for this 
problems on a global scale. 
So the ideal, instead, is to divide the 
channel into independent sub channels. 
And create each sub channel separately. 
So here for instance, this channel 
contains, highly localized, radio 
frequency interference would not probably 
be used because it would be to difficult 
to compensate for that. 
Here on these channels the noise level is 
different and so for instance we can use 
different signalling strategy according 
to the local signal to noise ratio. 
And similarly here for channels that have 
a very large attenuation, probably 
wouldn't be worth while to try and send 
data over these thins. 
But on the other hand that we will try to 
exploit the cleanest sub channels to send 
maximum amount of data. 
Now to formalize the sub channel 
structure. 
Suppose that we want to allocate N 
subchannels over the total positive 
bandwidth. 
We want the subchannels to have equal 
bandwidth, so their bandwidth will be F 
max over N, where F max is the maximum 
frequency allowed for by the channel. 
And we equally space the subchannels, by 
sensoring them over k F max over N, with 
k that goes from 0 to big N minus 1. 
This means that the first channel K equal 
to 0, will be bass-band. 
And then the subsequent channels will be 
bass-band, with center frequencies given 
by this formula. 
Now we want to translate this design, to 
the digital domain, so we pick a sampling 
frequency that is at least twice the 
maximum frequency in the channel. 
But careful now, because Fmax is quite 
high. 
The center frequency for each sub channel 
will be omega K equal to 2 pi kFmax over 
N divided by the sampling frequency. 
And if we sample at the[UNKNOWN] 
frequency, so Fs is equal twice Fmax, 
then omegak becomes simple 2 pi over 2N 
times k. 
We will not simplify the 2's in the 
fraction, because they will be useful 
later. 
The bandwidth of each subchannel, is also 
2 pi over 2N. 
And so, if we want to send symbols over 
any of these subchannels, remember the 
modulation scheme that we've seen in the 
previous modules, then we will have to 
use an up-sampling factor, k, that is at 
least 2N. 
If we plot the result in visual domain. 
Let's suppose that we just want to have 
three sub-channels, we have something 
that looks like this. 
The center frequencies will be multiples 
of 2 pi over 6, so we'll have 0. 
2 pi over 6 and 4 pi over 6 and we will 
center channels over these frequencies 
and the band width of each channel will 
be 2 pi over 6. 
So the 1st channel is the base band 
channel and then we have two pass band 
channels with of course their negative 
frequency counterpart. 
The next step in ADSL communication is to 
put a QAM modem on each subchannel 
independently and we will decide on the 
data rate for each modem. 
Based on the signal to noise ratio of 
each sub channel. 
So if the noise floor is low, then we 
will have a large constellation for that 
sub channel and vice versa. 
On channels that are unusable because of 
noise or interference, we will just send 
zeroes and we will not care about that. 
The structure of the SQL scheme is of 
course going to be communicated From the 
transmitter to the receiver so that the 
receiver knows where to expect data. 
This is part of the handshaking procedure 
between transmitter and receiver. 
Now let's look more in detail at the 
structure of the modem that we use on 
each sub-channel. 
This is a classic modulation scheme where 
we start with a sequence of symbols as 
produced by the mapper. 
Then we have an up-sampling by a factor 
of 2 N. 
So inserting two n minus one zero's every 
other sample and then filtering the 
sequence with a low pass, usually a 
raised cosine, with a cutoff frequency 
two pi over two n, in this case. 
This produces the complex base band 
signal bk of n and this. 
Complex baseband signal gets modulated 
with a complex exponential whose 
frequency is indexed by the channel 
number. 
And this is the center frequency of each 
channel. 
Omega ck is equal to 2pi over 2n times k. 
And here we have, finally, the pass band 
signal. 
That fits the prescribed bandwidth of the 
kth channel. 
And normally, if we just had one channel, 
we would put here a block that computes 
the real part and then our d2a converter. 
But here we have several modems in 
parallel. 
And so we have a structure that looks 
like this. 
Each channel will have two things that 
vary with respect to the others: the 
frequency of the modulation and of 
course, the series of symbols produced by 
the mapper. 
We sum all the complex base band signals 
together before taking the real part, and 
then send in the signal To the D to A 
converter. 
Now this picture should ring a bell. 
An indeed we have seen something that was 
very very close to this back in module 
4.3. 
So here's the picture to jog your memory 
and remember that DFT reconstruction 
formula could be interpreted as. 
A bank of n oscillators. 
Each oscillator would operate at a 
frequency that was two pi over n times k. 
And we would scale each oscillator with 
an amplitude, a of k, and with a phase 
offset, five k. 
We would run this machine for big n 
samples. 
And we will get our signal out. 
Now, the difference between this scheme 
and what we just saw is fundamentally 
that, in this scheme, a of k and phi of k 
are kept constant for the whole duration 
of the generation process. 
So while n goes from zero to big n minus 
one, a of k and phi of k stay the same. 
Whereas in the modem scheme that we seen 
before. 
The symbol sequence which is a complex 
symbol sequence so embeds both magnitude 
and phase will change at each new value 
of n. 
So is there a way to map the modem 
structure to the inverse DFT structure? 
We can do that if we manage to find a way 
to keep the symbols constant over the 
whole duration. 
So we will show how to do that and we 
will show that if we manage to do that, 
then ADSL transmission can be efficiently 
implemented with simple an inverse FFT. 
The name of this technique is discrete 
multitone modulation. 
So the great ADSL trick is very simple. 
Instead of using a[INAUDIBLE] sign in the 
up sampler, let's use a bad filter. 
Simply the indicator function for the 
interval 0 to 2n-1 and see what happens. 
So the impulse response of the upsampling 
filter is now this one. 
And please notice that this is just an 
un-normalized moving average filter and 
the frequency response of course is this 
one, we have seen it many times before. 
With the first 0 here in pi over N. 
If we compare the frequency response of 
the moving average if you want, or the 
indicator function with that of the 
filter that we should be using, namely a 
low pass filter with cut of pi over 2N. 
Then we see that the performance of the 
filter is not very good. 
Nonetheless, the thing will work, 
especially thanks to some clever little 
tricks in the way we choose the 
transmission symbols. 
But we will not have time to go into 
that. 
So let's go back to the subchannel modem. 
The thing to remark here is that the 
symbols from the mapper. 
Come in at a rate of B symbols per 
second. 
And because of the upsampling, samples 
out of the modulator, come out at a rate 
of 2NB samples per second. 
So this part works much faster than this 
part here. 
Now the carrier, is periodic with period 
2N. 
And so, each symbol Will influence a full 
period of the carrier. 
If we use a standard low pass filter 
here, for every value of n here, so for 
every value of the carrier there will be 
a different value in this Base band 
sequence that comes out of the sample. 
On the other hand if we use the indicator 
function as the input response the net 
result is that the values of bk of n will 
be constant over chunks of 2N samples. 
In that case we can simplify this whole 
scheme like so where now the only clock 
in the system is the output clock N. 
So now the oscillator in the modulator. 
Runs freely, at a frequency which is a 
multiple of 2 pi over 2N. 
This frequency is periodic, with period 
2N, so. 
And for each chunk of 2N samples, we go 
look for the symbol. 
The corresponds to this interval so if 
say n is equal to 0 here, n is equal to 
2n here, n is equal to 4n here and n is 
equal to 6n here for this interval we 
will go look for a of zero and multiply 
this portion of the carrier by this 
value. 
Able to look for a one, here a two, and 
so on and so on. 
So with this simplification, the whole 
transmitter can be sketched like so. 
We have the symbols from different sub 
channels that get multiplied by the 
carrier and kept constant over intervals 
of two and outward samples. 
The whole thing gets summed together. 
We get the aggregate bandpass signal, we 
take the real part, and we're ready for 
the D2A converter. 
We can now write explicitly, the formula 
for the aggregate bandpass signal c n. 
And this is the sum over all subchannels, 
of the symbol For that subchannel for 
that interval, multiplied by e to the j 
two pi over two n, nk. 
Now because of the way the index to the 
ak sequence is computed, these symbols 
will stay constant over intervals for the 
output index. 
That are 2n long. 
So for instance, for small n that goes 
from 0, to 2n minus 1, these values will 
stay constant. 
And again, for values of the output index 
that g from 2n. 
2, 4 and minus 1. 
So we could compute two big N values for 
the sequence CN in one fell swoop if we 
exploit the fact that this guy looks 
remarkably like an inverse DFT. 
As a matter of fact, by looking at the 
argument here, we can say that this is 
almost an inverse DFT over two big end 
points. 
The two things that are missing are the 
normalizing factor in front, 1 over 2N 
And the terms in the sum for the index k 
that goes from n to 2n minus 1. 
But that's not a problem. 
We can supplement this elements. 
And so we can compute a chunk of two big 
n output samples in one go as an inverse 
dft over two n points of a vector that is 
given by n channel symbols. 
And has another big n zeros appended to 
the end of it. 
The index for the subchannel symbols is 
given by the value of the output index 
divided by 2 n and we take the integer 
part. 
But we can do even better because in the 
end, remember, we're interested in the 
real part of the vector c of n. 
And we can write that real part as c of n 
plus the conjugate of c of n divided by 
2. 
Now it is easy to prove, and it's left as 
an exercise, that the conjugate of the 
inverse DFT of a vector, is equal to the 
inverse DFT of the conjugate of the time 
reversed vector. 
And, when we time reverse a finite length 
vector It's useful to think of the 
periodic extension. 
With this result and knowing that c of n 
is equal to 2n times the inverse DFT of a 
vector that is zeros in its latter part. 
We can sum cn with it's conjugate to 
obtain that the real part of cn is equal 
to N times the inverse DFT of a vector 
that is given by twice the symbol for the 
base band sub channel. 
The reason why we can write this is 
because the baseband signal will always 
have real value symbols, because it's a 
baseband, followed by the complex symbols 
for the n minus one remaining 
subchannels, followed by The conjugate of 
the symbols, for the N minus 1 remaining 
subchannels, but going from channel N 
minus 1, to channel 1. 
Schematically, we can draw up the ADSL 
transmitter, as one big inverse FFT, and 
the inputs to this FFT are twice the 
baseband symbol. 
Followed by the symbols for the 
subchannels from one to n minus one, and 
then we take these values, we conjugate 
them, and we flip their order, and we put 
those in the remaining inputs of the FFT. 
Now we run the inverse FFT and we get two 
n output samples in one go. 
We use a parrel to serial device to 
output the samples one at a time and here 
we have our d to i converter to put them 
on the channel. 
Once the n samples have been put out we 
go back, we fetch another set of n 
symbols from the n mamppers of the sub 
channels. 
Then we repeat the process. 
An actual ADSL modem uses a maximum 
frequency for the channel of 1004 
kilohertz divides this channel in to 256 
sub-channels. 
Each QAM modem for the sub-channel. 
Can independently choose between zero and 
fifteen bits per symbol. 
Now the first seven channels are left off 
because that is the band used by the 
voice communication over a telephone 
channel. 
Channels seven through 31 are used for 
data upstream. 
And the rest is left for data downstream 
for a maximum theoretical throughput of 
14.9 megabits per second. 
This would happen if all the downstream 
sub channels could use their maximum 
theoretical rate Which is a rare 
occurrence. 
And these are the specs of the on-line 
modem that you most probably used to 
watch this on-line class.