1 00:00:00,440 --> 00:00:03,600 Hi and welcome to Module 9.6 of Digital Signal Processing. 2 00:00:03,600 --> 00:00:07,20 This is our last module in our Data Communications section, and with 3 00:00:07,20 --> 00:00:10,980 everything that we've learned so far we will be able to look inside an ADSL modem 4 00:00:10,980 --> 00:00:15,595 and see how that works. We will start by examining the nature of 5 00:00:15,595 --> 00:00:18,260 the channel. That ABSL works on. 6 00:00:18,260 --> 00:00:22,460 Namely the copper wire links your home to the nearest central office. 7 00:00:22,460 --> 00:00:25,587 We will look at the signaling strategy that is best put on place on such a 8 00:00:25,587 --> 00:00:28,219 channel. And finally we will look at a very 9 00:00:28,219 --> 00:00:31,180 efficient implementation of that signalling strategy that goes under the 10 00:00:31,180 --> 00:00:35,635 name of discrete multi-tone modulation. If we just take an abstract view of the 11 00:00:35,635 --> 00:00:39,880 telephone network today, we see that we have a link that goes. 12 00:00:39,880 --> 00:00:43,400 From the home to the central office, and in the central office, a fundamental 13 00:00:43,400 --> 00:00:47,895 split takes place. The voice communication, when you talk on 14 00:00:47,895 --> 00:00:52,823 the phone, is sent to the voice network and then relayed to the, what is called 15 00:00:52,823 --> 00:01:01,182 the plain old telephone system, POTS. The data part of your communication when 16 00:01:01,182 --> 00:01:06,235 you use the ADSL is separated from the voice content and sent to a DSLAM. 17 00:01:06,235 --> 00:01:10,30 DSLAM stands for digital subscriber line access multiplier. 18 00:01:10,30 --> 00:01:13,157 And it's fundamentally a bank of modems that manage to handle multiple 19 00:01:13,157 --> 00:01:17,784 communications at the same time. And the data here then goes on to the 20 00:01:17,784 --> 00:01:22,94 internet in digital format. So, if you want, what we're really 21 00:01:22,94 --> 00:01:25,388 interested in is how to send the data from your home to the central office, 22 00:01:25,388 --> 00:01:30,650 because what happens afterwards is already entirely in the digital domain. 23 00:01:30,650 --> 00:01:34,682 But here, we have what is called a last mile, a piece of copper wire, namely an 24 00:01:34,682 --> 00:01:38,356 analog channel. That connects your home to the central 25 00:01:38,356 --> 00:01:41,329 office. Now a copper wire has, naturally, a very 26 00:01:41,329 --> 00:01:46,508 large bandwidth in excess of one 1MHz. But because of the width of the bandwidth 27 00:01:46,508 --> 00:01:49,784 and because the wire is not shielded, it is actually likely to pick up a lot of 28 00:01:49,784 --> 00:01:54,244 interference and noise. If we look at how the ADSL channel is 29 00:01:54,244 --> 00:01:58,300 organized and we're showing here just the positive frequencies. 30 00:01:58,300 --> 00:02:02,579 We can see three distinct regions. The first one is the part reserved to the 31 00:02:02,579 --> 00:02:07,543 telephone conversation, this is the base band part of the channel up to about four 32 00:02:07,543 --> 00:02:12,154 kilohertz. Then we have a region that is devoted to 33 00:02:12,154 --> 00:02:15,750 the upstream part of the data communication, the data that you send up 34 00:02:15,750 --> 00:02:19,780 to the internet, and then a downstream part that is much larger that is used for 35 00:02:19,780 --> 00:02:25,177 data download. This asymmetry between upstream and 36 00:02:25,177 --> 00:02:30,956 downstream is actually the reason why the communication protocol is called ADSL. 37 00:02:30,956 --> 00:02:34,440 ADSL stands for Asymmetric Digital Subscriber Line. 38 00:02:34,440 --> 00:02:37,490 If we now look at all the nasty things that can happen on the channel when we 39 00:02:37,490 --> 00:02:41,60 send data. We can identify three fundamental sources 40 00:02:41,60 --> 00:02:44,602 of worry. The first one is an attenuation curve for 41 00:02:44,602 --> 00:02:49,379 the channel that is completely uneven. This could be due to imperfection in the 42 00:02:49,379 --> 00:02:53,40 wire, parasitic capacitance, and so on, so forth. 43 00:02:53,40 --> 00:02:57,392 Then we might have very large noise or interference in certain regions of the 44 00:02:57,392 --> 00:03:00,292 spectrum. For instance, you turn on your vacuum 45 00:03:00,292 --> 00:03:04,460 cleaner, and that raises the noise floor in the certain frequency band. 46 00:03:04,460 --> 00:03:08,940 And thirdly we have very localized interference from radio communications 47 00:03:08,940 --> 00:03:13,280 now the radio band starts well within the bandwidth of the ADSL channel for 48 00:03:13,280 --> 00:03:17,130 instance from 15 to a hundred kilohertz here you have ship to shore 49 00:03:17,130 --> 00:03:23,335 communication. Up to 500 kilohertz, you have airplane 50 00:03:23,335 --> 00:03:29,210 communication, and over 500 kilohertz you have the AM radio band. 51 00:03:29,210 --> 00:03:31,730 So if you live near a radio station, for instance, tough luck. 52 00:03:31,730 --> 00:03:35,440 You have a lot of interference in the upper regions of the ADSL band. 53 00:03:35,440 --> 00:03:38,726 Since the channel is so wide, and the type of disturbances is so diverse, it 54 00:03:38,726 --> 00:03:42,224 would be extraordinarily difficult to try to equalize and compensate for this 55 00:03:42,224 --> 00:03:47,418 problems on a global scale. So the ideal, instead, is to divide the 56 00:03:47,418 --> 00:03:53,480 channel into independent sub channels. And create each sub channel separately. 57 00:03:53,480 --> 00:03:57,575 So here for instance, this channel contains, highly localized, radio 58 00:03:57,575 --> 00:04:01,985 frequency interference would not probably be used because it would be to difficult 59 00:04:01,985 --> 00:04:06,718 to compensate for that. Here on these channels the noise level is 60 00:04:06,718 --> 00:04:09,961 different and so for instance we can use different signalling strategy according 61 00:04:09,961 --> 00:04:14,865 to the local signal to noise ratio. And similarly here for channels that have 62 00:04:14,865 --> 00:04:17,903 a very large attenuation, probably wouldn't be worth while to try and send 63 00:04:17,903 --> 00:04:21,910 data over these thins. But on the other hand that we will try to 64 00:04:21,910 --> 00:04:26,470 exploit the cleanest sub channels to send maximum amount of data. 65 00:04:26,470 --> 00:04:28,900 Now to formalize the sub channel structure. 66 00:04:28,900 --> 00:04:32,322 Suppose that we want to allocate N subchannels over the total positive 67 00:04:32,322 --> 00:04:35,398 bandwidth. We want the subchannels to have equal 68 00:04:35,398 --> 00:04:39,52 bandwidth, so their bandwidth will be F max over N, where F max is the maximum 69 00:04:39,52 --> 00:04:45,466 frequency allowed for by the channel. And we equally space the subchannels, by 70 00:04:45,466 --> 00:04:52,590 sensoring them over k F max over N, with k that goes from 0 to big N minus 1. 71 00:04:52,590 --> 00:04:58,346 This means that the first channel K equal to 0, will be bass-band. 72 00:04:58,346 --> 00:05:03,456 And then the subsequent channels will be bass-band, with center frequencies given 73 00:05:03,456 --> 00:05:07,405 by this formula. Now we want to translate this design, to 74 00:05:07,405 --> 00:05:10,925 the digital domain, so we pick a sampling frequency that is at least twice the 75 00:05:10,925 --> 00:05:16,712 maximum frequency in the channel. But careful now, because Fmax is quite 76 00:05:16,712 --> 00:05:19,890 high. The center frequency for each sub channel 77 00:05:19,890 --> 00:05:25,290 will be omega K equal to 2 pi kFmax over N divided by the sampling frequency. 78 00:05:25,290 --> 00:05:30,220 And if we sample at the[UNKNOWN] frequency, so Fs is equal twice Fmax, 79 00:05:30,220 --> 00:05:35,810 then omegak becomes simple 2 pi over 2N times k. 80 00:05:35,810 --> 00:05:38,930 We will not simplify the 2's in the fraction, because they will be useful 81 00:05:38,930 --> 00:05:43,205 later. The bandwidth of each subchannel, is also 82 00:05:43,205 --> 00:05:47,2 2 pi over 2N. And so, if we want to send symbols over 83 00:05:47,2 --> 00:05:51,156 any of these subchannels, remember the modulation scheme that we've seen in the 84 00:05:51,156 --> 00:05:55,186 previous modules, then we will have to use an up-sampling factor, k, that is at 85 00:05:55,186 --> 00:06:00,930 least 2N. If we plot the result in visual domain. 86 00:06:00,930 --> 00:06:04,830 Let's suppose that we just want to have three sub-channels, we have something 87 00:06:04,830 --> 00:06:09,635 that looks like this. The center frequencies will be multiples 88 00:06:09,635 --> 00:06:14,755 of 2 pi over 6, so we'll have 0. 2 pi over 6 and 4 pi over 6 and we will 89 00:06:14,755 --> 00:06:19,705 center channels over these frequencies and the band width of each channel will 90 00:06:19,705 --> 00:06:24,338 be 2 pi over 6. So the 1st channel is the base band 91 00:06:24,338 --> 00:06:27,922 channel and then we have two pass band channels with of course their negative 92 00:06:27,922 --> 00:06:32,748 frequency counterpart. The next step in ADSL communication is to 93 00:06:32,748 --> 00:06:36,835 put a QAM modem on each subchannel independently and we will decide on the 94 00:06:36,835 --> 00:06:42,160 data rate for each modem. Based on the signal to noise ratio of 95 00:06:42,160 --> 00:06:45,840 each sub channel. So if the noise floor is low, then we 96 00:06:45,840 --> 00:06:51,280 will have a large constellation for that sub channel and vice versa. 97 00:06:51,280 --> 00:06:55,88 On channels that are unusable because of noise or interference, we will just send 98 00:06:55,88 --> 00:06:59,910 zeroes and we will not care about that. The structure of the SQL scheme is of 99 00:06:59,910 --> 00:07:03,796 course going to be communicated From the transmitter to the receiver so that the 100 00:07:03,796 --> 00:07:08,985 receiver knows where to expect data. This is part of the handshaking procedure 101 00:07:08,985 --> 00:07:12,384 between transmitter and receiver. Now let's look more in detail at the 102 00:07:12,384 --> 00:07:15,340 structure of the modem that we use on each sub-channel. 103 00:07:15,340 --> 00:07:19,102 This is a classic modulation scheme where we start with a sequence of symbols as 104 00:07:19,102 --> 00:07:23,718 produced by the mapper. Then we have an up-sampling by a factor 105 00:07:23,718 --> 00:07:27,454 of 2 N. So inserting two n minus one zero's every 106 00:07:27,454 --> 00:07:31,643 other sample and then filtering the sequence with a low pass, usually a 107 00:07:31,643 --> 00:07:38,310 raised cosine, with a cutoff frequency two pi over two n, in this case. 108 00:07:38,310 --> 00:07:43,570 This produces the complex base band signal bk of n and this. 109 00:07:43,570 --> 00:07:47,476 Complex baseband signal gets modulated with a complex exponential whose 110 00:07:47,476 --> 00:07:50,870 frequency is indexed by the channel number. 111 00:07:50,870 --> 00:07:53,446 And this is the center frequency of each channel. 112 00:07:53,446 --> 00:08:00,487 Omega ck is equal to 2pi over 2n times k. And here we have, finally, the pass band 113 00:08:00,487 --> 00:08:03,970 signal. That fits the prescribed bandwidth of the 114 00:08:03,970 --> 00:08:07,550 kth channel. And normally, if we just had one channel, 115 00:08:07,550 --> 00:08:13,798 we would put here a block that computes the real part and then our d2a converter. 116 00:08:13,798 --> 00:08:18,640 But here we have several modems in parallel. 117 00:08:18,640 --> 00:08:20,540 And so we have a structure that looks like this. 118 00:08:20,540 --> 00:08:24,694 Each channel will have two things that vary with respect to the others: the 119 00:08:24,694 --> 00:08:28,982 frequency of the modulation and of course, the series of symbols produced by 120 00:08:28,982 --> 00:08:33,911 the mapper. We sum all the complex base band signals 121 00:08:33,911 --> 00:08:38,132 together before taking the real part, and then send in the signal To the D to A 122 00:08:38,132 --> 00:08:41,950 converter. Now this picture should ring a bell. 123 00:08:41,950 --> 00:08:46,305 An indeed we have seen something that was very very close to this back in module 124 00:08:46,305 --> 00:08:50,53 4.3. So here's the picture to jog your memory 125 00:08:50,53 --> 00:08:55,564 and remember that DFT reconstruction formula could be interpreted as. 126 00:08:55,564 --> 00:09:00,426 A bank of n oscillators. Each oscillator would operate at a 127 00:09:00,426 --> 00:09:07,65 frequency that was two pi over n times k. And we would scale each oscillator with 128 00:09:07,65 --> 00:09:12,910 an amplitude, a of k, and with a phase offset, five k. 129 00:09:12,910 --> 00:09:16,860 We would run this machine for big n samples. 130 00:09:16,860 --> 00:09:21,142 And we will get our signal out. Now, the difference between this scheme 131 00:09:21,142 --> 00:09:25,234 and what we just saw is fundamentally that, in this scheme, a of k and phi of k 132 00:09:25,234 --> 00:09:31,290 are kept constant for the whole duration of the generation process. 133 00:09:31,290 --> 00:09:37,280 So while n goes from zero to big n minus one, a of k and phi of k stay the same. 134 00:09:37,280 --> 00:09:40,750 Whereas in the modem scheme that we seen before. 135 00:09:40,750 --> 00:09:45,708 The symbol sequence which is a complex symbol sequence so embeds both magnitude 136 00:09:45,708 --> 00:09:50,50 and phase will change at each new value of n. 137 00:09:50,50 --> 00:09:55,90 So is there a way to map the modem structure to the inverse DFT structure? 138 00:09:55,90 --> 00:10:00,298 We can do that if we manage to find a way to keep the symbols constant over the 139 00:10:00,298 --> 00:10:05,654 whole duration. So we will show how to do that and we 140 00:10:05,654 --> 00:10:09,875 will show that if we manage to do that, then ADSL transmission can be efficiently 141 00:10:09,875 --> 00:10:16,372 implemented with simple an inverse FFT. The name of this technique is discrete 142 00:10:16,372 --> 00:10:21,230 multitone modulation. So the great ADSL trick is very simple. 143 00:10:21,230 --> 00:10:26,530 Instead of using a[INAUDIBLE] sign in the up sampler, let's use a bad filter. 144 00:10:26,530 --> 00:10:33,540 Simply the indicator function for the interval 0 to 2n-1 and see what happens. 145 00:10:33,540 --> 00:10:37,100 So the impulse response of the upsampling filter is now this one. 146 00:10:37,100 --> 00:10:41,390 And please notice that this is just an un-normalized moving average filter and 147 00:10:41,390 --> 00:10:47,980 the frequency response of course is this one, we have seen it many times before. 148 00:10:47,980 --> 00:10:52,752 With the first 0 here in pi over N. If we compare the frequency response of 149 00:10:52,752 --> 00:10:57,205 the moving average if you want, or the indicator function with that of the 150 00:10:57,205 --> 00:11:05,250 filter that we should be using, namely a low pass filter with cut of pi over 2N. 151 00:11:05,250 --> 00:11:08,406 Then we see that the performance of the filter is not very good. 152 00:11:08,406 --> 00:11:11,556 Nonetheless, the thing will work, especially thanks to some clever little 153 00:11:11,556 --> 00:11:14,810 tricks in the way we choose the transmission symbols. 154 00:11:14,810 --> 00:11:17,70 But we will not have time to go into that. 155 00:11:17,70 --> 00:11:20,652 So let's go back to the subchannel modem. The thing to remark here is that the 156 00:11:20,652 --> 00:11:24,658 symbols from the mapper. Come in at a rate of B symbols per 157 00:11:24,658 --> 00:11:28,541 second. And because of the upsampling, samples 158 00:11:28,541 --> 00:11:34,190 out of the modulator, come out at a rate of 2NB samples per second. 159 00:11:34,190 --> 00:11:37,150 So this part works much faster than this part here. 160 00:11:37,150 --> 00:11:41,545 Now the carrier, is periodic with period 2N. 161 00:11:41,545 --> 00:11:48,580 And so, each symbol Will influence a full period of the carrier. 162 00:11:48,580 --> 00:11:52,840 If we use a standard low pass filter here, for every value of n here, so for 163 00:11:52,840 --> 00:11:57,242 every value of the carrier there will be a different value in this Base band 164 00:11:57,242 --> 00:12:04,837 sequence that comes out of the sample. On the other hand if we use the indicator 165 00:12:04,837 --> 00:12:09,893 function as the input response the net result is that the values of bk of n will 166 00:12:09,893 --> 00:12:17,3 be constant over chunks of 2N samples. In that case we can simplify this whole 167 00:12:17,3 --> 00:12:22,220 scheme like so where now the only clock in the system is the output clock N. 168 00:12:22,220 --> 00:12:28,479 So now the oscillator in the modulator. Runs freely, at a frequency which is a 169 00:12:28,479 --> 00:12:35,462 multiple of 2 pi over 2N. This frequency is periodic, with period 170 00:12:35,462 --> 00:12:39,551 2N, so. And for each chunk of 2N samples, we go 171 00:12:39,551 --> 00:12:45,146 look for the symbol. The corresponds to this interval so if 172 00:12:45,146 --> 00:12:50,250 say n is equal to 0 here, n is equal to 2n here, n is equal to 4n here and n is 173 00:12:50,250 --> 00:12:55,618 equal to 6n here for this interval we will go look for a of zero and multiply 174 00:12:55,618 --> 00:13:03,572 this portion of the carrier by this value. 175 00:13:03,572 --> 00:13:10,300 Able to look for a one, here a two, and so on and so on. 176 00:13:10,300 --> 00:13:14,110 So with this simplification, the whole transmitter can be sketched like so. 177 00:13:14,110 --> 00:13:18,698 We have the symbols from different sub channels that get multiplied by the 178 00:13:18,698 --> 00:13:25,150 carrier and kept constant over intervals of two and outward samples. 179 00:13:25,150 --> 00:13:29,28 The whole thing gets summed together. We get the aggregate bandpass signal, we 180 00:13:29,28 --> 00:13:33,60 take the real part, and we're ready for the D2A converter. 181 00:13:33,60 --> 00:13:38,364 We can now write explicitly, the formula for the aggregate bandpass signal c n. 182 00:13:38,364 --> 00:13:43,124 And this is the sum over all subchannels, of the symbol For that subchannel for 183 00:13:43,124 --> 00:13:48,620 that interval, multiplied by e to the j two pi over two n, nk. 184 00:13:48,620 --> 00:13:53,100 Now because of the way the index to the ak sequence is computed, these symbols 185 00:13:53,100 --> 00:13:58,15 will stay constant over intervals for the output index. 186 00:13:58,15 --> 00:14:03,413 That are 2n long. So for instance, for small n that goes 187 00:14:03,413 --> 00:14:08,750 from 0, to 2n minus 1, these values will stay constant. 188 00:14:08,750 --> 00:14:13,98 And again, for values of the output index that g from 2n. 189 00:14:13,98 --> 00:14:17,434 2, 4 and minus 1. So we could compute two big N values for 190 00:14:17,434 --> 00:14:21,826 the sequence CN in one fell swoop if we exploit the fact that this guy looks 191 00:14:21,826 --> 00:14:28,540 remarkably like an inverse DFT. As a matter of fact, by looking at the 192 00:14:28,540 --> 00:14:32,880 argument here, we can say that this is almost an inverse DFT over two big end 193 00:14:32,880 --> 00:14:37,784 points. The two things that are missing are the 194 00:14:37,784 --> 00:14:42,950 normalizing factor in front, 1 over 2N And the terms in the sum for the index k 195 00:14:42,950 --> 00:14:48,570 that goes from n to 2n minus 1. But that's not a problem. 196 00:14:48,570 --> 00:14:52,548 We can supplement this elements. And so we can compute a chunk of two big 197 00:14:52,548 --> 00:14:56,900 n output samples in one go as an inverse dft over two n points of a vector that is 198 00:14:56,900 --> 00:15:03,994 given by n channel symbols. And has another big n zeros appended to 199 00:15:03,994 --> 00:15:08,273 the end of it. The index for the subchannel symbols is 200 00:15:08,273 --> 00:15:12,604 given by the value of the output index divided by 2 n and we take the integer 201 00:15:12,604 --> 00:15:16,172 part. But we can do even better because in the 202 00:15:16,172 --> 00:15:19,832 end, remember, we're interested in the real part of the vector c of n. 203 00:15:19,832 --> 00:15:23,986 And we can write that real part as c of n plus the conjugate of c of n divided by 204 00:15:23,986 --> 00:15:27,846 2. Now it is easy to prove, and it's left as 205 00:15:27,846 --> 00:15:32,582 an exercise, that the conjugate of the inverse DFT of a vector, is equal to the 206 00:15:32,582 --> 00:15:38,800 inverse DFT of the conjugate of the time reversed vector. 207 00:15:38,800 --> 00:15:42,768 And, when we time reverse a finite length vector It's useful to think of the 208 00:15:42,768 --> 00:15:47,658 periodic extension. With this result and knowing that c of n 209 00:15:47,658 --> 00:15:55,390 is equal to 2n times the inverse DFT of a vector that is zeros in its latter part. 210 00:15:55,390 --> 00:16:00,514 We can sum cn with it's conjugate to obtain that the real part of cn is equal 211 00:16:00,514 --> 00:16:05,806 to N times the inverse DFT of a vector that is given by twice the symbol for the 212 00:16:05,806 --> 00:16:12,960 base band sub channel. The reason why we can write this is 213 00:16:12,960 --> 00:16:17,985 because the baseband signal will always have real value symbols, because it's a 214 00:16:17,985 --> 00:16:22,485 baseband, followed by the complex symbols for the n minus one remaining 215 00:16:22,485 --> 00:16:27,660 subchannels, followed by The conjugate of the symbols, for the N minus 1 remaining 216 00:16:27,660 --> 00:16:35,854 subchannels, but going from channel N minus 1, to channel 1. 217 00:16:35,854 --> 00:16:41,464 Schematically, we can draw up the ADSL transmitter, as one big inverse FFT, and 218 00:16:41,464 --> 00:16:47,760 the inputs to this FFT are twice the baseband symbol. 219 00:16:47,760 --> 00:16:51,831 Followed by the symbols for the subchannels from one to n minus one, and 220 00:16:51,831 --> 00:16:56,385 then we take these values, we conjugate them, and we flip their order, and we put 221 00:16:56,385 --> 00:17:03,780 those in the remaining inputs of the FFT. Now we run the inverse FFT and we get two 222 00:17:03,780 --> 00:17:09,242 n output samples in one go. We use a parrel to serial device to 223 00:17:09,242 --> 00:17:13,338 output the samples one at a time and here we have our d to i converter to put them 224 00:17:13,338 --> 00:17:18,282 on the channel. Once the n samples have been put out we 225 00:17:18,282 --> 00:17:22,458 go back, we fetch another set of n symbols from the n mamppers of the sub 226 00:17:22,458 --> 00:17:26,624 channels. Then we repeat the process. 227 00:17:26,624 --> 00:17:32,153 An actual ADSL modem uses a maximum frequency for the channel of 1004 228 00:17:32,153 --> 00:17:38,766 kilohertz divides this channel in to 256 sub-channels. 229 00:17:38,766 --> 00:17:43,842 Each QAM modem for the sub-channel. Can independently choose between zero and 230 00:17:43,842 --> 00:17:48,66 fifteen bits per symbol. Now the first seven channels are left off 231 00:17:48,66 --> 00:17:51,665 because that is the band used by the voice communication over a telephone 232 00:17:51,665 --> 00:17:55,516 channel. Channels seven through 31 are used for 233 00:17:55,516 --> 00:17:59,329 data upstream. And the rest is left for data downstream 234 00:17:59,329 --> 00:18:04,690 for a maximum theoretical throughput of 14.9 megabits per second. 235 00:18:04,690 --> 00:18:08,274 This would happen if all the downstream sub channels could use their maximum 236 00:18:08,274 --> 00:18:11,510 theoretical rate Which is a rare occurrence. 237 00:18:11,510 --> 00:18:17,732 And these are the specs of the on-line modem that you most probably used to 238 00:18:17,732 --> 00:18:21,909 watch this on-line class.