0:00:00.440,0:00:03.600
Hi and welcome to Module 9.6 of Digital [br]Signal Processing.

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This is our last module in our Data [br]Communications section, and with

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everything that we've learned so far we [br]will be able to look inside an ADSL modem

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and see how that works. [br]We will start by examining the nature of

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the channel. [br]That ABSL works on.

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Namely the copper wire links your home to [br]the nearest central office.

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We will look at the signaling strategy [br]that is best put on place on such a

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channel. [br]And finally we will look at a very

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efficient implementation of that [br]signalling strategy that goes under the

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name of discrete multi-tone modulation. [br]If we just take an abstract view of the

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telephone network today, we see that we [br]have a link that goes.

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From the home to the central office, and [br]in the central office, a fundamental

0:00:43.400,0:00:47.895
split takes place. [br]The voice communication, when you talk on

0:00:47.895,0:00:52.823
the phone, is sent to the voice network [br]and then relayed to the, what is called

0:00:52.823,0:01:01.182
the plain old telephone system, POTS. [br]The data part of your communication when

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you use the ADSL is separated from the [br]voice content and sent to a DSLAM.

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DSLAM stands for digital subscriber line [br]access multiplier.

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And it's fundamentally a bank of modems [br]that manage to handle multiple

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communications at the same time. [br]And the data here then goes on to the

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internet in digital format. [br]So, if you want, what we're really

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interested in is how to send the data [br]from your home to the central office,

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because what happens afterwards is [br]already entirely in the digital domain.

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But here, we have what is called a last [br]mile, a piece of copper wire, namely an

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analog channel. [br]That connects your home to the central

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office. [br]Now a copper wire has, naturally, a very

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large bandwidth in excess of one 1MHz. [br]But because of the width of the bandwidth

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and because the wire is not shielded, it [br]is actually likely to pick up a lot of

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interference and noise. [br]If we look at how the ADSL channel is

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organized and we're showing here just the [br]positive frequencies.

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We can see three distinct regions. [br]The first one is the part reserved to the

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telephone conversation, this is the base [br]band part of the channel up to about four

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kilohertz. [br]Then we have a region that is devoted to

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the upstream part of the data [br]communication, the data that you send up

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to the internet, and then a downstream [br]part that is much larger that is used for

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data download. [br]This asymmetry between upstream and

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downstream is actually the reason why the [br]communication protocol is called ADSL.

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ADSL stands for Asymmetric Digital [br]Subscriber Line.

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If we now look at all the nasty things [br]that can happen on the channel when we

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send data. [br]We can identify three fundamental sources

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of worry. [br]The first one is an attenuation curve for

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the channel that is completely uneven. [br]This could be due to imperfection in the

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wire, parasitic capacitance, and so on, [br]so forth.

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Then we might have very large noise or [br]interference in certain regions of the

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spectrum. [br]For instance, you turn on your vacuum

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cleaner, and that raises the noise floor [br]in the certain frequency band.

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And thirdly we have very localized [br]interference from radio communications

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now the radio band starts well within the [br]bandwidth of the ADSL channel for

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instance from 15 to a hundred kilohertz [br]here you have ship to shore

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communication. [br]Up to 500 kilohertz, you have airplane

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communication, and over 500 kilohertz you [br]have the AM radio band.

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So if you live near a radio station, for [br]instance, tough luck.

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You have a lot of interference in the [br]upper regions of the ADSL band.

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Since the channel is so wide, and the [br]type of disturbances is so diverse, it

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would be extraordinarily difficult to try [br]to equalize and compensate for this

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problems on a global scale. [br]So the ideal, instead, is to divide the

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channel into independent sub channels. [br]And create each sub channel separately.

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So here for instance, this channel [br]contains, highly localized, radio

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frequency interference would not probably [br]be used because it would be to difficult

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to compensate for that. [br]Here on these channels the noise level is

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different and so for instance we can use [br]different signalling strategy according

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to the local signal to noise ratio. [br]And similarly here for channels that have

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a very large attenuation, probably [br]wouldn't be worth while to try and send

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data over these thins. [br]But on the other hand that we will try to

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exploit the cleanest sub channels to send [br]maximum amount of data.

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Now to formalize the sub channel [br]structure.

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Suppose that we want to allocate N [br]subchannels over the total positive

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bandwidth. [br]We want the subchannels to have equal

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bandwidth, so their bandwidth will be F [br]max over N, where F max is the maximum

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frequency allowed for by the channel. [br]And we equally space the subchannels, by

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sensoring them over k F max over N, with [br]k that goes from 0 to big N minus 1.

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This means that the first channel K equal [br]to 0, will be bass-band.

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And then the subsequent channels will be [br]bass-band, with center frequencies given

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by this formula. [br]Now we want to translate this design, to

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the digital domain, so we pick a sampling [br]frequency that is at least twice the

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maximum frequency in the channel. [br]But careful now, because Fmax is quite

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high. [br]The center frequency for each sub channel

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will be omega K equal to 2 pi kFmax over [br]N divided by the sampling frequency.

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And if we sample at the[UNKNOWN] [br]frequency, so Fs is equal twice Fmax,

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then omegak becomes simple 2 pi over 2N [br]times k.

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We will not simplify the 2's in the [br]fraction, because they will be useful

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later. [br]The bandwidth of each subchannel, is also

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2 pi over 2N. [br]And so, if we want to send symbols over

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any of these subchannels, remember the [br]modulation scheme that we've seen in the

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previous modules, then we will have to [br]use an up-sampling factor, k, that is at

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least 2N. [br]If we plot the result in visual domain.

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Let's suppose that we just want to have [br]three sub-channels, we have something

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that looks like this. [br]The center frequencies will be multiples

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of 2 pi over 6, so we'll have 0. [br]2 pi over 6 and 4 pi over 6 and we will

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center channels over these frequencies [br]and the band width of each channel will

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be 2 pi over 6. [br]So the 1st channel is the base band

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channel and then we have two pass band [br]channels with of course their negative

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frequency counterpart. [br]The next step in ADSL communication is to

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put a QAM modem on each subchannel [br]independently and we will decide on the

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data rate for each modem. [br]Based on the signal to noise ratio of

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each sub channel. [br]So if the noise floor is low, then we

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will have a large constellation for that [br]sub channel and vice versa.

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On channels that are unusable because of [br]noise or interference, we will just send

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zeroes and we will not care about that. [br]The structure of the SQL scheme is of

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course going to be communicated From the [br]transmitter to the receiver so that the

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receiver knows where to expect data. [br]This is part of the handshaking procedure

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between transmitter and receiver. [br]Now let's look more in detail at the

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structure of the modem that we use on [br]each sub-channel.

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This is a classic modulation scheme where [br]we start with a sequence of symbols as

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produced by the mapper. [br]Then we have an up-sampling by a factor

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of 2 N. [br]So inserting two n minus one zero's every

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other sample and then filtering the [br]sequence with a low pass, usually a

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raised cosine, with a cutoff frequency [br]two pi over two n, in this case.

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This produces the complex base band [br]signal bk of n and this.

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Complex baseband signal gets modulated [br]with a complex exponential whose

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frequency is indexed by the channel [br]number.

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And this is the center frequency of each [br]channel.

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Omega ck is equal to 2pi over 2n times k. [br]And here we have, finally, the pass band

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signal. [br]That fits the prescribed bandwidth of the

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kth channel. [br]And normally, if we just had one channel,

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we would put here a block that computes [br]the real part and then our d2a converter.

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But here we have several modems in [br]parallel.

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And so we have a structure that looks [br]like this.

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Each channel will have two things that [br]vary with respect to the others: the

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frequency of the modulation and of [br]course, the series of symbols produced by

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the mapper. [br]We sum all the complex base band signals

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together before taking the real part, and [br]then send in the signal To the D to A

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converter. [br]Now this picture should ring a bell.

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An indeed we have seen something that was [br]very very close to this back in module

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4.3. [br]So here's the picture to jog your memory

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and remember that DFT reconstruction [br]formula could be interpreted as.

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A bank of n oscillators. [br]Each oscillator would operate at a

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frequency that was two pi over n times k. [br]And we would scale each oscillator with

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an amplitude, a of k, and with a phase [br]offset, five k.

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We would run this machine for big n [br]samples.

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And we will get our signal out. [br]Now, the difference between this scheme

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and what we just saw is fundamentally [br]that, in this scheme, a of k and phi of k

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are kept constant for the whole duration [br]of the generation process.

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So while n goes from zero to big n minus [br]one, a of k and phi of k stay the same.

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Whereas in the modem scheme that we seen [br]before.

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The symbol sequence which is a complex [br]symbol sequence so embeds both magnitude

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and phase will change at each new value [br]of n.

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So is there a way to map the modem [br]structure to the inverse DFT structure?

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We can do that if we manage to find a way [br]to keep the symbols constant over the

0:10:00.298,0:10:05.654
whole duration. [br]So we will show how to do that and we

0:10:05.654,0:10:09.875
will show that if we manage to do that, [br]then ADSL transmission can be efficiently

0:10:09.875,0:10:16.372
implemented with simple an inverse FFT. [br]The name of this technique is discrete

0:10:16.372,0:10:21.230
multitone modulation. [br]So the great ADSL trick is very simple.

0:10:21.230,0:10:26.530
Instead of using a[INAUDIBLE] sign in the [br]up sampler, let's use a bad filter.

0:10:26.530,0:10:33.540
Simply the indicator function for the [br]interval 0 to 2n-1 and see what happens.

0:10:33.540,0:10:37.100
So the impulse response of the upsampling [br]filter is now this one.

0:10:37.100,0:10:41.390
And please notice that this is just an [br]un-normalized moving average filter and

0:10:41.390,0:10:47.980
the frequency response of course is this [br]one, we have seen it many times before.

0:10:47.980,0:10:52.752
With the first 0 here in pi over N. [br]If we compare the frequency response of

0:10:52.752,0:10:57.205
the moving average if you want, or the [br]indicator function with that of the

0:10:57.205,0:11:05.250
filter that we should be using, namely a [br]low pass filter with cut of pi over 2N.

0:11:05.250,0:11:08.406
Then we see that the performance of the [br]filter is not very good.

0:11:08.406,0:11:11.556
Nonetheless, the thing will work, [br]especially thanks to some clever little

0:11:11.556,0:11:14.810
tricks in the way we choose the [br]transmission symbols.

0:11:14.810,0:11:17.700
But we will not have time to go into [br]that.

0:11:17.700,0:11:20.652
So let's go back to the subchannel modem. [br]The thing to remark here is that the

0:11:20.652,0:11:24.658
symbols from the mapper. [br]Come in at a rate of B symbols per

0:11:24.658,0:11:28.541
second. [br]And because of the upsampling, samples

0:11:28.541,0:11:34.190
out of the modulator, come out at a rate [br]of 2NB samples per second.

0:11:34.190,0:11:37.150
So this part works much faster than this [br]part here.

0:11:37.150,0:11:41.545
Now the carrier, is periodic with period [br]2N.

0:11:41.545,0:11:48.580
And so, each symbol Will influence a full [br]period of the carrier.

0:11:48.580,0:11:52.840
If we use a standard low pass filter [br]here, for every value of n here, so for

0:11:52.840,0:11:57.242
every value of the carrier there will be [br]a different value in this Base band

0:11:57.242,0:12:04.837
sequence that comes out of the sample. [br]On the other hand if we use the indicator

0:12:04.837,0:12:09.893
function as the input response the net [br]result is that the values of bk of n will

0:12:09.893,0:12:17.300
be constant over chunks of 2N samples. [br]In that case we can simplify this whole

0:12:17.300,0:12:22.220
scheme like so where now the only clock [br]in the system is the output clock N.

0:12:22.220,0:12:28.479
So now the oscillator in the modulator. [br]Runs freely, at a frequency which is a

0:12:28.479,0:12:35.462
multiple of 2 pi over 2N. [br]This frequency is periodic, with period

0:12:35.462,0:12:39.551
2N, so. [br]And for each chunk of 2N samples, we go

0:12:39.551,0:12:45.146
look for the symbol. [br]The corresponds to this interval so if

0:12:45.146,0:12:50.250
say n is equal to 0 here, n is equal to [br]2n here, n is equal to 4n here and n is

0:12:50.250,0:12:55.618
equal to 6n here for this interval we [br]will go look for a of zero and multiply

0:12:55.618,0:13:03.572
this portion of the carrier by this [br]value.

0:13:03.572,0:13:10.300
Able to look for a one, here a two, and [br]so on and so on.

0:13:10.300,0:13:14.110
So with this simplification, the whole [br]transmitter can be sketched like so.

0:13:14.110,0:13:18.698
We have the symbols from different sub [br]channels that get multiplied by the

0:13:18.698,0:13:25.150
carrier and kept constant over intervals [br]of two and outward samples.

0:13:25.150,0:13:29.280
The whole thing gets summed together. [br]We get the aggregate bandpass signal, we

0:13:29.280,0:13:33.600
take the real part, and we're ready for [br]the D2A converter.

0:13:33.600,0:13:38.364
We can now write explicitly, the formula [br]for the aggregate bandpass signal c n.

0:13:38.364,0:13:43.124
And this is the sum over all subchannels, [br]of the symbol For that subchannel for

0:13:43.124,0:13:48.620
that interval, multiplied by e to the j [br]two pi over two n, nk.

0:13:48.620,0:13:53.100
Now because of the way the index to the [br]ak sequence is computed, these symbols

0:13:53.100,0:13:58.150
will stay constant over intervals for the [br]output index.

0:13:58.150,0:14:03.413
That are 2n long. [br]So for instance, for small n that goes

0:14:03.413,0:14:08.750
from 0, to 2n minus 1, these values will [br]stay constant.

0:14:08.750,0:14:13.980
And again, for values of the output index [br]that g from 2n.

0:14:13.980,0:14:17.434
2, 4 and minus 1. [br]So we could compute two big N values for

0:14:17.434,0:14:21.826
the sequence CN in one fell swoop if we [br]exploit the fact that this guy looks

0:14:21.826,0:14:28.540
remarkably like an inverse DFT. [br]As a matter of fact, by looking at the

0:14:28.540,0:14:32.880
argument here, we can say that this is [br]almost an inverse DFT over two big end

0:14:32.880,0:14:37.784
points. [br]The two things that are missing are the

0:14:37.784,0:14:42.950
normalizing factor in front, 1 over 2N [br]And the terms in the sum for the index k

0:14:42.950,0:14:48.570
that goes from n to 2n minus 1. [br]But that's not a problem.

0:14:48.570,0:14:52.548
We can supplement this elements. [br]And so we can compute a chunk of two big

0:14:52.548,0:14:56.900
n output samples in one go as an inverse [br]dft over two n points of a vector that is

0:14:56.900,0:15:03.994
given by n channel symbols. [br]And has another big n zeros appended to

0:15:03.994,0:15:08.273
the end of it. [br]The index for the subchannel symbols is

0:15:08.273,0:15:12.604
given by the value of the output index [br]divided by 2 n and we take the integer

0:15:12.604,0:15:16.172
part. [br]But we can do even better because in the

0:15:16.172,0:15:19.832
end, remember, we're interested in the [br]real part of the vector c of n.

0:15:19.832,0:15:23.986
And we can write that real part as c of n [br]plus the conjugate of c of n divided by

0:15:23.986,0:15:27.846
2. [br]Now it is easy to prove, and it's left as

0:15:27.846,0:15:32.582
an exercise, that the conjugate of the [br]inverse DFT of a vector, is equal to the

0:15:32.582,0:15:38.800
inverse DFT of the conjugate of the time [br]reversed vector.

0:15:38.800,0:15:42.768
And, when we time reverse a finite length [br]vector It's useful to think of the

0:15:42.768,0:15:47.658
periodic extension. [br]With this result and knowing that c of n

0:15:47.658,0:15:55.390
is equal to 2n times the inverse DFT of a [br]vector that is zeros in its latter part.

0:15:55.390,0:16:00.514
We can sum cn with it's conjugate to [br]obtain that the real part of cn is equal

0:16:00.514,0:16:05.806
to N times the inverse DFT of a vector [br]that is given by twice the symbol for the

0:16:05.806,0:16:12.960
base band sub channel. [br]The reason why we can write this is

0:16:12.960,0:16:17.985
because the baseband signal will always [br]have real value symbols, because it's a

0:16:17.985,0:16:22.485
baseband, followed by the complex symbols [br]for the n minus one remaining

0:16:22.485,0:16:27.660
subchannels, followed by The conjugate of [br]the symbols, for the N minus 1 remaining

0:16:27.660,0:16:35.854
subchannels, but going from channel N [br]minus 1, to channel 1.

0:16:35.854,0:16:41.464
Schematically, we can draw up the ADSL [br]transmitter, as one big inverse FFT, and

0:16:41.464,0:16:47.760
the inputs to this FFT are twice the [br]baseband symbol.

0:16:47.760,0:16:51.831
Followed by the symbols for the [br]subchannels from one to n minus one, and

0:16:51.831,0:16:56.385
then we take these values, we conjugate [br]them, and we flip their order, and we put

0:16:56.385,0:17:03.780
those in the remaining inputs of the FFT. [br]Now we run the inverse FFT and we get two

0:17:03.780,0:17:09.242
n output samples in one go. [br]We use a parrel to serial device to

0:17:09.242,0:17:13.338
output the samples one at a time and here [br]we have our d to i converter to put them

0:17:13.338,0:17:18.282
on the channel. [br]Once the n samples have been put out we

0:17:18.282,0:17:22.458
go back, we fetch another set of n [br]symbols from the n mamppers of the sub

0:17:22.458,0:17:26.624
channels. [br]Then we repeat the process.

0:17:26.624,0:17:32.153
An actual ADSL modem uses a maximum [br]frequency for the channel of 1004

0:17:32.153,0:17:38.766
kilohertz divides this channel in to 256 [br]sub-channels.

0:17:38.766,0:17:43.842
Each QAM modem for the sub-channel. [br]Can independently choose between zero and

0:17:43.842,0:17:48.660
fifteen bits per symbol. [br]Now the first seven channels are left off

0:17:48.660,0:17:51.665
because that is the band used by the [br]voice communication over a telephone

0:17:51.665,0:17:55.516
channel. [br]Channels seven through 31 are used for

0:17:55.516,0:17:59.329
data upstream. [br]And the rest is left for data downstream

0:17:59.329,0:18:04.690
for a maximum theoretical throughput of [br]14.9 megabits per second.

0:18:04.690,0:18:08.274
This would happen if all the downstream [br]sub channels could use their maximum

0:18:08.274,0:18:11.510
theoretical rate Which is a rare [br]occurrence.

0:18:11.510,0:18:17.732
And these are the specs of the on-line [br]modem that you most probably used to

0:18:17.732,0:18:21.909
watch this on-line class.