[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.44,0:00:03.60,Default,,0000,0000,0000,,Hi and welcome to Module 9.6 of Digital \NSignal Processing.
Dialogue: 0,0:00:03.60,0:00:07.20,Default,,0000,0000,0000,,This is our last module in our Data \NCommunications section, and with
Dialogue: 0,0:00:07.20,0:00:10.98,Default,,0000,0000,0000,,everything that we've learned so far we \Nwill be able to look inside an ADSL modem
Dialogue: 0,0:00:10.98,0:00:15.60,Default,,0000,0000,0000,,and see how that works. \NWe will start by examining the nature of
Dialogue: 0,0:00:15.60,0:00:18.26,Default,,0000,0000,0000,,the channel. \NThat ABSL works on.
Dialogue: 0,0:00:18.26,0:00:22.46,Default,,0000,0000,0000,,Namely the copper wire links your home to \Nthe nearest central office.
Dialogue: 0,0:00:22.46,0:00:25.59,Default,,0000,0000,0000,,We will look at the signaling strategy \Nthat is best put on place on such a
Dialogue: 0,0:00:25.59,0:00:28.22,Default,,0000,0000,0000,,channel. \NAnd finally we will look at a very
Dialogue: 0,0:00:28.22,0:00:31.18,Default,,0000,0000,0000,,efficient implementation of that \Nsignalling strategy that goes under the
Dialogue: 0,0:00:31.18,0:00:35.64,Default,,0000,0000,0000,,name of discrete multi-tone modulation. \NIf we just take an abstract view of the
Dialogue: 0,0:00:35.64,0:00:39.88,Default,,0000,0000,0000,,telephone network today, we see that we \Nhave a link that goes.
Dialogue: 0,0:00:39.88,0:00:43.40,Default,,0000,0000,0000,,From the home to the central office, and \Nin the central office, a fundamental
Dialogue: 0,0:00:43.40,0:00:47.90,Default,,0000,0000,0000,,split takes place. \NThe voice communication, when you talk on
Dialogue: 0,0:00:47.90,0:00:52.82,Default,,0000,0000,0000,,the phone, is sent to the voice network \Nand then relayed to the, what is called
Dialogue: 0,0:00:52.82,0:01:01.18,Default,,0000,0000,0000,,the plain old telephone system, POTS. \NThe data part of your communication when
Dialogue: 0,0:01:01.18,0:01:06.24,Default,,0000,0000,0000,,you use the ADSL is separated from the \Nvoice content and sent to a DSLAM.
Dialogue: 0,0:01:06.24,0:01:10.30,Default,,0000,0000,0000,,DSLAM stands for digital subscriber line \Naccess multiplier.
Dialogue: 0,0:01:10.30,0:01:13.16,Default,,0000,0000,0000,,And it's fundamentally a bank of modems \Nthat manage to handle multiple
Dialogue: 0,0:01:13.16,0:01:17.78,Default,,0000,0000,0000,,communications at the same time. \NAnd the data here then goes on to the
Dialogue: 0,0:01:17.78,0:01:22.94,Default,,0000,0000,0000,,internet in digital format. \NSo, if you want, what we're really
Dialogue: 0,0:01:22.94,0:01:25.39,Default,,0000,0000,0000,,interested in is how to send the data \Nfrom your home to the central office,
Dialogue: 0,0:01:25.39,0:01:30.65,Default,,0000,0000,0000,,because what happens afterwards is \Nalready entirely in the digital domain.
Dialogue: 0,0:01:30.65,0:01:34.68,Default,,0000,0000,0000,,But here, we have what is called a last \Nmile, a piece of copper wire, namely an
Dialogue: 0,0:01:34.68,0:01:38.36,Default,,0000,0000,0000,,analog channel. \NThat connects your home to the central
Dialogue: 0,0:01:38.36,0:01:41.33,Default,,0000,0000,0000,,office. \NNow a copper wire has, naturally, a very
Dialogue: 0,0:01:41.33,0:01:46.51,Default,,0000,0000,0000,,large bandwidth in excess of one 1MHz. \NBut because of the width of the bandwidth
Dialogue: 0,0:01:46.51,0:01:49.78,Default,,0000,0000,0000,,and because the wire is not shielded, it \Nis actually likely to pick up a lot of
Dialogue: 0,0:01:49.78,0:01:54.24,Default,,0000,0000,0000,,interference and noise. \NIf we look at how the ADSL channel is
Dialogue: 0,0:01:54.24,0:01:58.30,Default,,0000,0000,0000,,organized and we're showing here just the \Npositive frequencies.
Dialogue: 0,0:01:58.30,0:02:02.58,Default,,0000,0000,0000,,We can see three distinct regions. \NThe first one is the part reserved to the
Dialogue: 0,0:02:02.58,0:02:07.54,Default,,0000,0000,0000,,telephone conversation, this is the base \Nband part of the channel up to about four
Dialogue: 0,0:02:07.54,0:02:12.15,Default,,0000,0000,0000,,kilohertz. \NThen we have a region that is devoted to
Dialogue: 0,0:02:12.15,0:02:15.75,Default,,0000,0000,0000,,the upstream part of the data \Ncommunication, the data that you send up
Dialogue: 0,0:02:15.75,0:02:19.78,Default,,0000,0000,0000,,to the internet, and then a downstream \Npart that is much larger that is used for
Dialogue: 0,0:02:19.78,0:02:25.18,Default,,0000,0000,0000,,data download. \NThis asymmetry between upstream and
Dialogue: 0,0:02:25.18,0:02:30.96,Default,,0000,0000,0000,,downstream is actually the reason why the \Ncommunication protocol is called ADSL.
Dialogue: 0,0:02:30.96,0:02:34.44,Default,,0000,0000,0000,,ADSL stands for Asymmetric Digital \NSubscriber Line.
Dialogue: 0,0:02:34.44,0:02:37.49,Default,,0000,0000,0000,,If we now look at all the nasty things \Nthat can happen on the channel when we
Dialogue: 0,0:02:37.49,0:02:41.60,Default,,0000,0000,0000,,send data. \NWe can identify three fundamental sources
Dialogue: 0,0:02:41.60,0:02:44.60,Default,,0000,0000,0000,,of worry. \NThe first one is an attenuation curve for
Dialogue: 0,0:02:44.60,0:02:49.38,Default,,0000,0000,0000,,the channel that is completely uneven. \NThis could be due to imperfection in the
Dialogue: 0,0:02:49.38,0:02:53.40,Default,,0000,0000,0000,,wire, parasitic capacitance, and so on, \Nso forth.
Dialogue: 0,0:02:53.40,0:02:57.39,Default,,0000,0000,0000,,Then we might have very large noise or \Ninterference in certain regions of the
Dialogue: 0,0:02:57.39,0:03:00.29,Default,,0000,0000,0000,,spectrum. \NFor instance, you turn on your vacuum
Dialogue: 0,0:03:00.29,0:03:04.46,Default,,0000,0000,0000,,cleaner, and that raises the noise floor \Nin the certain frequency band.
Dialogue: 0,0:03:04.46,0:03:08.94,Default,,0000,0000,0000,,And thirdly we have very localized \Ninterference from radio communications
Dialogue: 0,0:03:08.94,0:03:13.28,Default,,0000,0000,0000,,now the radio band starts well within the \Nbandwidth of the ADSL channel for
Dialogue: 0,0:03:13.28,0:03:17.13,Default,,0000,0000,0000,,instance from 15 to a hundred kilohertz \Nhere you have ship to shore
Dialogue: 0,0:03:17.13,0:03:23.34,Default,,0000,0000,0000,,communication. \NUp to 500 kilohertz, you have airplane
Dialogue: 0,0:03:23.34,0:03:29.21,Default,,0000,0000,0000,,communication, and over 500 kilohertz you \Nhave the AM radio band.
Dialogue: 0,0:03:29.21,0:03:31.73,Default,,0000,0000,0000,,So if you live near a radio station, for \Ninstance, tough luck.
Dialogue: 0,0:03:31.73,0:03:35.44,Default,,0000,0000,0000,,You have a lot of interference in the \Nupper regions of the ADSL band.
Dialogue: 0,0:03:35.44,0:03:38.73,Default,,0000,0000,0000,,Since the channel is so wide, and the \Ntype of disturbances is so diverse, it
Dialogue: 0,0:03:38.73,0:03:42.22,Default,,0000,0000,0000,,would be extraordinarily difficult to try \Nto equalize and compensate for this
Dialogue: 0,0:03:42.22,0:03:47.42,Default,,0000,0000,0000,,problems on a global scale. \NSo the ideal, instead, is to divide the
Dialogue: 0,0:03:47.42,0:03:53.48,Default,,0000,0000,0000,,channel into independent sub channels. \NAnd create each sub channel separately.
Dialogue: 0,0:03:53.48,0:03:57.58,Default,,0000,0000,0000,,So here for instance, this channel \Ncontains, highly localized, radio
Dialogue: 0,0:03:57.58,0:04:01.98,Default,,0000,0000,0000,,frequency interference would not probably \Nbe used because it would be to difficult
Dialogue: 0,0:04:01.98,0:04:06.72,Default,,0000,0000,0000,,to compensate for that. \NHere on these channels the noise level is
Dialogue: 0,0:04:06.72,0:04:09.96,Default,,0000,0000,0000,,different and so for instance we can use \Ndifferent signalling strategy according
Dialogue: 0,0:04:09.96,0:04:14.86,Default,,0000,0000,0000,,to the local signal to noise ratio. \NAnd similarly here for channels that have
Dialogue: 0,0:04:14.86,0:04:17.90,Default,,0000,0000,0000,,a very large attenuation, probably \Nwouldn't be worth while to try and send
Dialogue: 0,0:04:17.90,0:04:21.91,Default,,0000,0000,0000,,data over these thins. \NBut on the other hand that we will try to
Dialogue: 0,0:04:21.91,0:04:26.47,Default,,0000,0000,0000,,exploit the cleanest sub channels to send \Nmaximum amount of data.
Dialogue: 0,0:04:26.47,0:04:28.90,Default,,0000,0000,0000,,Now to formalize the sub channel \Nstructure.
Dialogue: 0,0:04:28.90,0:04:32.32,Default,,0000,0000,0000,,Suppose that we want to allocate N \Nsubchannels over the total positive
Dialogue: 0,0:04:32.32,0:04:35.40,Default,,0000,0000,0000,,bandwidth. \NWe want the subchannels to have equal
Dialogue: 0,0:04:35.40,0:04:39.52,Default,,0000,0000,0000,,bandwidth, so their bandwidth will be F \Nmax over N, where F max is the maximum
Dialogue: 0,0:04:39.52,0:04:45.47,Default,,0000,0000,0000,,frequency allowed for by the channel. \NAnd we equally space the subchannels, by
Dialogue: 0,0:04:45.47,0:04:52.59,Default,,0000,0000,0000,,sensoring them over k F max over N, with \Nk that goes from 0 to big N minus 1.
Dialogue: 0,0:04:52.59,0:04:58.35,Default,,0000,0000,0000,,This means that the first channel K equal \Nto 0, will be bass-band.
Dialogue: 0,0:04:58.35,0:05:03.46,Default,,0000,0000,0000,,And then the subsequent channels will be \Nbass-band, with center frequencies given
Dialogue: 0,0:05:03.46,0:05:07.40,Default,,0000,0000,0000,,by this formula. \NNow we want to translate this design, to
Dialogue: 0,0:05:07.40,0:05:10.92,Default,,0000,0000,0000,,the digital domain, so we pick a sampling \Nfrequency that is at least twice the
Dialogue: 0,0:05:10.92,0:05:16.71,Default,,0000,0000,0000,,maximum frequency in the channel. \NBut careful now, because Fmax is quite
Dialogue: 0,0:05:16.71,0:05:19.89,Default,,0000,0000,0000,,high. \NThe center frequency for each sub channel
Dialogue: 0,0:05:19.89,0:05:25.29,Default,,0000,0000,0000,,will be omega K equal to 2 pi kFmax over \NN divided by the sampling frequency.
Dialogue: 0,0:05:25.29,0:05:30.22,Default,,0000,0000,0000,,And if we sample at the[UNKNOWN] \Nfrequency, so Fs is equal twice Fmax,
Dialogue: 0,0:05:30.22,0:05:35.81,Default,,0000,0000,0000,,then omegak becomes simple 2 pi over 2N \Ntimes k.
Dialogue: 0,0:05:35.81,0:05:38.93,Default,,0000,0000,0000,,We will not simplify the 2's in the \Nfraction, because they will be useful
Dialogue: 0,0:05:38.93,0:05:43.20,Default,,0000,0000,0000,,later. \NThe bandwidth of each subchannel, is also
Dialogue: 0,0:05:43.20,0:05:47.20,Default,,0000,0000,0000,,2 pi over 2N. \NAnd so, if we want to send symbols over
Dialogue: 0,0:05:47.20,0:05:51.16,Default,,0000,0000,0000,,any of these subchannels, remember the \Nmodulation scheme that we've seen in the
Dialogue: 0,0:05:51.16,0:05:55.19,Default,,0000,0000,0000,,previous modules, then we will have to \Nuse an up-sampling factor, k, that is at
Dialogue: 0,0:05:55.19,0:06:00.93,Default,,0000,0000,0000,,least 2N. \NIf we plot the result in visual domain.
Dialogue: 0,0:06:00.93,0:06:04.83,Default,,0000,0000,0000,,Let's suppose that we just want to have \Nthree sub-channels, we have something
Dialogue: 0,0:06:04.83,0:06:09.64,Default,,0000,0000,0000,,that looks like this. \NThe center frequencies will be multiples
Dialogue: 0,0:06:09.64,0:06:14.76,Default,,0000,0000,0000,,of 2 pi over 6, so we'll have 0. \N2 pi over 6 and 4 pi over 6 and we will
Dialogue: 0,0:06:14.76,0:06:19.70,Default,,0000,0000,0000,,center channels over these frequencies \Nand the band width of each channel will
Dialogue: 0,0:06:19.70,0:06:24.34,Default,,0000,0000,0000,,be 2 pi over 6. \NSo the 1st channel is the base band
Dialogue: 0,0:06:24.34,0:06:27.92,Default,,0000,0000,0000,,channel and then we have two pass band \Nchannels with of course their negative
Dialogue: 0,0:06:27.92,0:06:32.75,Default,,0000,0000,0000,,frequency counterpart. \NThe next step in ADSL communication is to
Dialogue: 0,0:06:32.75,0:06:36.84,Default,,0000,0000,0000,,put a QAM modem on each subchannel \Nindependently and we will decide on the
Dialogue: 0,0:06:36.84,0:06:42.16,Default,,0000,0000,0000,,data rate for each modem. \NBased on the signal to noise ratio of
Dialogue: 0,0:06:42.16,0:06:45.84,Default,,0000,0000,0000,,each sub channel. \NSo if the noise floor is low, then we
Dialogue: 0,0:06:45.84,0:06:51.28,Default,,0000,0000,0000,,will have a large constellation for that \Nsub channel and vice versa.
Dialogue: 0,0:06:51.28,0:06:55.88,Default,,0000,0000,0000,,On channels that are unusable because of \Nnoise or interference, we will just send
Dialogue: 0,0:06:55.88,0:06:59.91,Default,,0000,0000,0000,,zeroes and we will not care about that. \NThe structure of the SQL scheme is of
Dialogue: 0,0:06:59.91,0:07:03.80,Default,,0000,0000,0000,,course going to be communicated From the \Ntransmitter to the receiver so that the
Dialogue: 0,0:07:03.80,0:07:08.98,Default,,0000,0000,0000,,receiver knows where to expect data. \NThis is part of the handshaking procedure
Dialogue: 0,0:07:08.98,0:07:12.38,Default,,0000,0000,0000,,between transmitter and receiver. \NNow let's look more in detail at the
Dialogue: 0,0:07:12.38,0:07:15.34,Default,,0000,0000,0000,,structure of the modem that we use on \Neach sub-channel.
Dialogue: 0,0:07:15.34,0:07:19.10,Default,,0000,0000,0000,,This is a classic modulation scheme where \Nwe start with a sequence of symbols as
Dialogue: 0,0:07:19.10,0:07:23.72,Default,,0000,0000,0000,,produced by the mapper. \NThen we have an up-sampling by a factor
Dialogue: 0,0:07:23.72,0:07:27.45,Default,,0000,0000,0000,,of 2 N. \NSo inserting two n minus one zero's every
Dialogue: 0,0:07:27.45,0:07:31.64,Default,,0000,0000,0000,,other sample and then filtering the \Nsequence with a low pass, usually a
Dialogue: 0,0:07:31.64,0:07:38.31,Default,,0000,0000,0000,,raised cosine, with a cutoff frequency \Ntwo pi over two n, in this case.
Dialogue: 0,0:07:38.31,0:07:43.57,Default,,0000,0000,0000,,This produces the complex base band \Nsignal bk of n and this.
Dialogue: 0,0:07:43.57,0:07:47.48,Default,,0000,0000,0000,,Complex baseband signal gets modulated \Nwith a complex exponential whose
Dialogue: 0,0:07:47.48,0:07:50.87,Default,,0000,0000,0000,,frequency is indexed by the channel \Nnumber.
Dialogue: 0,0:07:50.87,0:07:53.45,Default,,0000,0000,0000,,And this is the center frequency of each \Nchannel.
Dialogue: 0,0:07:53.45,0:08:00.49,Default,,0000,0000,0000,,Omega ck is equal to 2pi over 2n times k. \NAnd here we have, finally, the pass band
Dialogue: 0,0:08:00.49,0:08:03.97,Default,,0000,0000,0000,,signal. \NThat fits the prescribed bandwidth of the
Dialogue: 0,0:08:03.97,0:08:07.55,Default,,0000,0000,0000,,kth channel. \NAnd normally, if we just had one channel,
Dialogue: 0,0:08:07.55,0:08:13.80,Default,,0000,0000,0000,,we would put here a block that computes \Nthe real part and then our d2a converter.
Dialogue: 0,0:08:13.80,0:08:18.64,Default,,0000,0000,0000,,But here we have several modems in \Nparallel.
Dialogue: 0,0:08:18.64,0:08:20.54,Default,,0000,0000,0000,,And so we have a structure that looks \Nlike this.
Dialogue: 0,0:08:20.54,0:08:24.69,Default,,0000,0000,0000,,Each channel will have two things that \Nvary with respect to the others: the
Dialogue: 0,0:08:24.69,0:08:28.98,Default,,0000,0000,0000,,frequency of the modulation and of \Ncourse, the series of symbols produced by
Dialogue: 0,0:08:28.98,0:08:33.91,Default,,0000,0000,0000,,the mapper. \NWe sum all the complex base band signals
Dialogue: 0,0:08:33.91,0:08:38.13,Default,,0000,0000,0000,,together before taking the real part, and \Nthen send in the signal To the D to A
Dialogue: 0,0:08:38.13,0:08:41.95,Default,,0000,0000,0000,,converter. \NNow this picture should ring a bell.
Dialogue: 0,0:08:41.95,0:08:46.30,Default,,0000,0000,0000,,An indeed we have seen something that was \Nvery very close to this back in module
Dialogue: 0,0:08:46.30,0:08:50.53,Default,,0000,0000,0000,,4.3. \NSo here's the picture to jog your memory
Dialogue: 0,0:08:50.53,0:08:55.56,Default,,0000,0000,0000,,and remember that DFT reconstruction \Nformula could be interpreted as.
Dialogue: 0,0:08:55.56,0:09:00.43,Default,,0000,0000,0000,,A bank of n oscillators. \NEach oscillator would operate at a
Dialogue: 0,0:09:00.43,0:09:07.65,Default,,0000,0000,0000,,frequency that was two pi over n times k. \NAnd we would scale each oscillator with
Dialogue: 0,0:09:07.65,0:09:12.91,Default,,0000,0000,0000,,an amplitude, a of k, and with a phase \Noffset, five k.
Dialogue: 0,0:09:12.91,0:09:16.86,Default,,0000,0000,0000,,We would run this machine for big n \Nsamples.
Dialogue: 0,0:09:16.86,0:09:21.14,Default,,0000,0000,0000,,And we will get our signal out. \NNow, the difference between this scheme
Dialogue: 0,0:09:21.14,0:09:25.23,Default,,0000,0000,0000,,and what we just saw is fundamentally \Nthat, in this scheme, a of k and phi of k
Dialogue: 0,0:09:25.23,0:09:31.29,Default,,0000,0000,0000,,are kept constant for the whole duration \Nof the generation process.
Dialogue: 0,0:09:31.29,0:09:37.28,Default,,0000,0000,0000,,So while n goes from zero to big n minus \None, a of k and phi of k stay the same.
Dialogue: 0,0:09:37.28,0:09:40.75,Default,,0000,0000,0000,,Whereas in the modem scheme that we seen \Nbefore.
Dialogue: 0,0:09:40.75,0:09:45.71,Default,,0000,0000,0000,,The symbol sequence which is a complex \Nsymbol sequence so embeds both magnitude
Dialogue: 0,0:09:45.71,0:09:50.50,Default,,0000,0000,0000,,and phase will change at each new value \Nof n.
Dialogue: 0,0:09:50.50,0:09:55.90,Default,,0000,0000,0000,,So is there a way to map the modem \Nstructure to the inverse DFT structure?
Dialogue: 0,0:09:55.90,0:10:00.30,Default,,0000,0000,0000,,We can do that if we manage to find a way \Nto keep the symbols constant over the
Dialogue: 0,0:10:00.30,0:10:05.65,Default,,0000,0000,0000,,whole duration. \NSo we will show how to do that and we
Dialogue: 0,0:10:05.65,0:10:09.88,Default,,0000,0000,0000,,will show that if we manage to do that, \Nthen ADSL transmission can be efficiently
Dialogue: 0,0:10:09.88,0:10:16.37,Default,,0000,0000,0000,,implemented with simple an inverse FFT. \NThe name of this technique is discrete
Dialogue: 0,0:10:16.37,0:10:21.23,Default,,0000,0000,0000,,multitone modulation. \NSo the great ADSL trick is very simple.
Dialogue: 0,0:10:21.23,0:10:26.53,Default,,0000,0000,0000,,Instead of using a[INAUDIBLE] sign in the \Nup sampler, let's use a bad filter.
Dialogue: 0,0:10:26.53,0:10:33.54,Default,,0000,0000,0000,,Simply the indicator function for the \Ninterval 0 to 2n-1 and see what happens.
Dialogue: 0,0:10:33.54,0:10:37.10,Default,,0000,0000,0000,,So the impulse response of the upsampling \Nfilter is now this one.
Dialogue: 0,0:10:37.10,0:10:41.39,Default,,0000,0000,0000,,And please notice that this is just an \Nun-normalized moving average filter and
Dialogue: 0,0:10:41.39,0:10:47.98,Default,,0000,0000,0000,,the frequency response of course is this \None, we have seen it many times before.
Dialogue: 0,0:10:47.98,0:10:52.75,Default,,0000,0000,0000,,With the first 0 here in pi over N. \NIf we compare the frequency response of
Dialogue: 0,0:10:52.75,0:10:57.20,Default,,0000,0000,0000,,the moving average if you want, or the \Nindicator function with that of the
Dialogue: 0,0:10:57.20,0:11:05.25,Default,,0000,0000,0000,,filter that we should be using, namely a \Nlow pass filter with cut of pi over 2N.
Dialogue: 0,0:11:05.25,0:11:08.41,Default,,0000,0000,0000,,Then we see that the performance of the \Nfilter is not very good.
Dialogue: 0,0:11:08.41,0:11:11.56,Default,,0000,0000,0000,,Nonetheless, the thing will work, \Nespecially thanks to some clever little
Dialogue: 0,0:11:11.56,0:11:14.81,Default,,0000,0000,0000,,tricks in the way we choose the \Ntransmission symbols.
Dialogue: 0,0:11:14.81,0:11:17.70,Default,,0000,0000,0000,,But we will not have time to go into \Nthat.
Dialogue: 0,0:11:17.70,0:11:20.65,Default,,0000,0000,0000,,So let's go back to the subchannel modem. \NThe thing to remark here is that the
Dialogue: 0,0:11:20.65,0:11:24.66,Default,,0000,0000,0000,,symbols from the mapper. \NCome in at a rate of B symbols per
Dialogue: 0,0:11:24.66,0:11:28.54,Default,,0000,0000,0000,,second. \NAnd because of the upsampling, samples
Dialogue: 0,0:11:28.54,0:11:34.19,Default,,0000,0000,0000,,out of the modulator, come out at a rate \Nof 2NB samples per second.
Dialogue: 0,0:11:34.19,0:11:37.15,Default,,0000,0000,0000,,So this part works much faster than this \Npart here.
Dialogue: 0,0:11:37.15,0:11:41.54,Default,,0000,0000,0000,,Now the carrier, is periodic with period \N2N.
Dialogue: 0,0:11:41.54,0:11:48.58,Default,,0000,0000,0000,,And so, each symbol Will influence a full \Nperiod of the carrier.
Dialogue: 0,0:11:48.58,0:11:52.84,Default,,0000,0000,0000,,If we use a standard low pass filter \Nhere, for every value of n here, so for
Dialogue: 0,0:11:52.84,0:11:57.24,Default,,0000,0000,0000,,every value of the carrier there will be \Na different value in this Base band
Dialogue: 0,0:11:57.24,0:12:04.84,Default,,0000,0000,0000,,sequence that comes out of the sample. \NOn the other hand if we use the indicator
Dialogue: 0,0:12:04.84,0:12:09.89,Default,,0000,0000,0000,,function as the input response the net \Nresult is that the values of bk of n will
Dialogue: 0,0:12:09.89,0:12:17.30,Default,,0000,0000,0000,,be constant over chunks of 2N samples. \NIn that case we can simplify this whole
Dialogue: 0,0:12:17.30,0:12:22.22,Default,,0000,0000,0000,,scheme like so where now the only clock \Nin the system is the output clock N.
Dialogue: 0,0:12:22.22,0:12:28.48,Default,,0000,0000,0000,,So now the oscillator in the modulator. \NRuns freely, at a frequency which is a
Dialogue: 0,0:12:28.48,0:12:35.46,Default,,0000,0000,0000,,multiple of 2 pi over 2N. \NThis frequency is periodic, with period
Dialogue: 0,0:12:35.46,0:12:39.55,Default,,0000,0000,0000,,2N, so. \NAnd for each chunk of 2N samples, we go
Dialogue: 0,0:12:39.55,0:12:45.15,Default,,0000,0000,0000,,look for the symbol. \NThe corresponds to this interval so if
Dialogue: 0,0:12:45.15,0:12:50.25,Default,,0000,0000,0000,,say n is equal to 0 here, n is equal to \N2n here, n is equal to 4n here and n is
Dialogue: 0,0:12:50.25,0:12:55.62,Default,,0000,0000,0000,,equal to 6n here for this interval we \Nwill go look for a of zero and multiply
Dialogue: 0,0:12:55.62,0:13:03.57,Default,,0000,0000,0000,,this portion of the carrier by this \Nvalue.
Dialogue: 0,0:13:03.57,0:13:10.30,Default,,0000,0000,0000,,Able to look for a one, here a two, and \Nso on and so on.
Dialogue: 0,0:13:10.30,0:13:14.11,Default,,0000,0000,0000,,So with this simplification, the whole \Ntransmitter can be sketched like so.
Dialogue: 0,0:13:14.11,0:13:18.70,Default,,0000,0000,0000,,We have the symbols from different sub \Nchannels that get multiplied by the
Dialogue: 0,0:13:18.70,0:13:25.15,Default,,0000,0000,0000,,carrier and kept constant over intervals \Nof two and outward samples.
Dialogue: 0,0:13:25.15,0:13:29.28,Default,,0000,0000,0000,,The whole thing gets summed together. \NWe get the aggregate bandpass signal, we
Dialogue: 0,0:13:29.28,0:13:33.60,Default,,0000,0000,0000,,take the real part, and we're ready for \Nthe D2A converter.
Dialogue: 0,0:13:33.60,0:13:38.36,Default,,0000,0000,0000,,We can now write explicitly, the formula \Nfor the aggregate bandpass signal c n.
Dialogue: 0,0:13:38.36,0:13:43.12,Default,,0000,0000,0000,,And this is the sum over all subchannels, \Nof the symbol For that subchannel for
Dialogue: 0,0:13:43.12,0:13:48.62,Default,,0000,0000,0000,,that interval, multiplied by e to the j \Ntwo pi over two n, nk.
Dialogue: 0,0:13:48.62,0:13:53.10,Default,,0000,0000,0000,,Now because of the way the index to the \Nak sequence is computed, these symbols
Dialogue: 0,0:13:53.10,0:13:58.15,Default,,0000,0000,0000,,will stay constant over intervals for the \Noutput index.
Dialogue: 0,0:13:58.15,0:14:03.41,Default,,0000,0000,0000,,That are 2n long. \NSo for instance, for small n that goes
Dialogue: 0,0:14:03.41,0:14:08.75,Default,,0000,0000,0000,,from 0, to 2n minus 1, these values will \Nstay constant.
Dialogue: 0,0:14:08.75,0:14:13.98,Default,,0000,0000,0000,,And again, for values of the output index \Nthat g from 2n.
Dialogue: 0,0:14:13.98,0:14:17.43,Default,,0000,0000,0000,,2, 4 and minus 1. \NSo we could compute two big N values for
Dialogue: 0,0:14:17.43,0:14:21.83,Default,,0000,0000,0000,,the sequence CN in one fell swoop if we \Nexploit the fact that this guy looks
Dialogue: 0,0:14:21.83,0:14:28.54,Default,,0000,0000,0000,,remarkably like an inverse DFT. \NAs a matter of fact, by looking at the
Dialogue: 0,0:14:28.54,0:14:32.88,Default,,0000,0000,0000,,argument here, we can say that this is \Nalmost an inverse DFT over two big end
Dialogue: 0,0:14:32.88,0:14:37.78,Default,,0000,0000,0000,,points. \NThe two things that are missing are the
Dialogue: 0,0:14:37.78,0:14:42.95,Default,,0000,0000,0000,,normalizing factor in front, 1 over 2N \NAnd the terms in the sum for the index k
Dialogue: 0,0:14:42.95,0:14:48.57,Default,,0000,0000,0000,,that goes from n to 2n minus 1. \NBut that's not a problem.
Dialogue: 0,0:14:48.57,0:14:52.55,Default,,0000,0000,0000,,We can supplement this elements. \NAnd so we can compute a chunk of two big
Dialogue: 0,0:14:52.55,0:14:56.90,Default,,0000,0000,0000,,n output samples in one go as an inverse \Ndft over two n points of a vector that is
Dialogue: 0,0:14:56.90,0:15:03.99,Default,,0000,0000,0000,,given by n channel symbols. \NAnd has another big n zeros appended to
Dialogue: 0,0:15:03.99,0:15:08.27,Default,,0000,0000,0000,,the end of it. \NThe index for the subchannel symbols is
Dialogue: 0,0:15:08.27,0:15:12.60,Default,,0000,0000,0000,,given by the value of the output index \Ndivided by 2 n and we take the integer
Dialogue: 0,0:15:12.60,0:15:16.17,Default,,0000,0000,0000,,part. \NBut we can do even better because in the
Dialogue: 0,0:15:16.17,0:15:19.83,Default,,0000,0000,0000,,end, remember, we're interested in the \Nreal part of the vector c of n.
Dialogue: 0,0:15:19.83,0:15:23.99,Default,,0000,0000,0000,,And we can write that real part as c of n \Nplus the conjugate of c of n divided by
Dialogue: 0,0:15:23.99,0:15:27.85,Default,,0000,0000,0000,,2. \NNow it is easy to prove, and it's left as
Dialogue: 0,0:15:27.85,0:15:32.58,Default,,0000,0000,0000,,an exercise, that the conjugate of the \Ninverse DFT of a vector, is equal to the
Dialogue: 0,0:15:32.58,0:15:38.80,Default,,0000,0000,0000,,inverse DFT of the conjugate of the time \Nreversed vector.
Dialogue: 0,0:15:38.80,0:15:42.77,Default,,0000,0000,0000,,And, when we time reverse a finite length \Nvector It's useful to think of the
Dialogue: 0,0:15:42.77,0:15:47.66,Default,,0000,0000,0000,,periodic extension. \NWith this result and knowing that c of n
Dialogue: 0,0:15:47.66,0:15:55.39,Default,,0000,0000,0000,,is equal to 2n times the inverse DFT of a \Nvector that is zeros in its latter part.
Dialogue: 0,0:15:55.39,0:16:00.51,Default,,0000,0000,0000,,We can sum cn with it's conjugate to \Nobtain that the real part of cn is equal
Dialogue: 0,0:16:00.51,0:16:05.81,Default,,0000,0000,0000,,to N times the inverse DFT of a vector \Nthat is given by twice the symbol for the
Dialogue: 0,0:16:05.81,0:16:12.96,Default,,0000,0000,0000,,base band sub channel. \NThe reason why we can write this is
Dialogue: 0,0:16:12.96,0:16:17.98,Default,,0000,0000,0000,,because the baseband signal will always \Nhave real value symbols, because it's a
Dialogue: 0,0:16:17.98,0:16:22.48,Default,,0000,0000,0000,,baseband, followed by the complex symbols \Nfor the n minus one remaining
Dialogue: 0,0:16:22.48,0:16:27.66,Default,,0000,0000,0000,,subchannels, followed by The conjugate of \Nthe symbols, for the N minus 1 remaining
Dialogue: 0,0:16:27.66,0:16:35.85,Default,,0000,0000,0000,,subchannels, but going from channel N \Nminus 1, to channel 1.
Dialogue: 0,0:16:35.85,0:16:41.46,Default,,0000,0000,0000,,Schematically, we can draw up the ADSL \Ntransmitter, as one big inverse FFT, and
Dialogue: 0,0:16:41.46,0:16:47.76,Default,,0000,0000,0000,,the inputs to this FFT are twice the \Nbaseband symbol.
Dialogue: 0,0:16:47.76,0:16:51.83,Default,,0000,0000,0000,,Followed by the symbols for the \Nsubchannels from one to n minus one, and
Dialogue: 0,0:16:51.83,0:16:56.38,Default,,0000,0000,0000,,then we take these values, we conjugate \Nthem, and we flip their order, and we put
Dialogue: 0,0:16:56.38,0:17:03.78,Default,,0000,0000,0000,,those in the remaining inputs of the FFT. \NNow we run the inverse FFT and we get two
Dialogue: 0,0:17:03.78,0:17:09.24,Default,,0000,0000,0000,,n output samples in one go. \NWe use a parrel to serial device to
Dialogue: 0,0:17:09.24,0:17:13.34,Default,,0000,0000,0000,,output the samples one at a time and here \Nwe have our d to i converter to put them
Dialogue: 0,0:17:13.34,0:17:18.28,Default,,0000,0000,0000,,on the channel. \NOnce the n samples have been put out we
Dialogue: 0,0:17:18.28,0:17:22.46,Default,,0000,0000,0000,,go back, we fetch another set of n \Nsymbols from the n mamppers of the sub
Dialogue: 0,0:17:22.46,0:17:26.62,Default,,0000,0000,0000,,channels. \NThen we repeat the process.
Dialogue: 0,0:17:26.62,0:17:32.15,Default,,0000,0000,0000,,An actual ADSL modem uses a maximum \Nfrequency for the channel of 1004
Dialogue: 0,0:17:32.15,0:17:38.77,Default,,0000,0000,0000,,kilohertz divides this channel in to 256 \Nsub-channels.
Dialogue: 0,0:17:38.77,0:17:43.84,Default,,0000,0000,0000,,Each QAM modem for the sub-channel. \NCan independently choose between zero and
Dialogue: 0,0:17:43.84,0:17:48.66,Default,,0000,0000,0000,,fifteen bits per symbol. \NNow the first seven channels are left off
Dialogue: 0,0:17:48.66,0:17:51.66,Default,,0000,0000,0000,,because that is the band used by the \Nvoice communication over a telephone
Dialogue: 0,0:17:51.66,0:17:55.52,Default,,0000,0000,0000,,channel. \NChannels seven through 31 are used for
Dialogue: 0,0:17:55.52,0:17:59.33,Default,,0000,0000,0000,,data upstream. \NAnd the rest is left for data downstream
Dialogue: 0,0:17:59.33,0:18:04.69,Default,,0000,0000,0000,,for a maximum theoretical throughput of \N14.9 megabits per second.
Dialogue: 0,0:18:04.69,0:18:08.27,Default,,0000,0000,0000,,This would happen if all the downstream \Nsub channels could use their maximum
Dialogue: 0,0:18:08.27,0:18:11.51,Default,,0000,0000,0000,,theoretical rate Which is a rare \Noccurrence.
Dialogue: 0,0:18:11.51,0:18:17.73,Default,,0000,0000,0000,,And these are the specs of the on-line \Nmodem that you most probably used to
Dialogue: 0,0:18:17.73,0:18:21.91,Default,,0000,0000,0000,,watch this on-line class.