Now this picture should ring a bell, and 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 we said that DFT reconstruction formula could be interpreted as a bank of n oscillators. Each oscillator would operate at a frequency that was 2 pi over N times k, and we would scale each oscillator with an amplitude A of k and with phase offset 5k, we would run this machine for big N samples, and we would 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 5 k are kept constant for the whole duration of the generation process. So while n goes from 0 to begin minus 1, A of k and 5 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 of the up-sampling period. So we will show how to do that and we will show that if we manage to do that the ADSL transmission can be efficiently implemented with simply an inverse FFT. The name of this technique is Discrete Multitone Modulation. So the great ADSL trick is very simple. Instead of using a sign in the up-sampler, let's us a bad filter. Simply the indicator function for the interval, 0 to 2N -1, and see what happens. So the impulse response of the up-sampling filter is now this one and please notice that this is just an unnormalized 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 off, 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 won't 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 up-sampling, samples out of the modulator come out at a rate of 2NB samples per second. So, this part works much faster than this part. Now, the carrier is periodic with period 2 n. 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 sound. On the other hand, If we use the indicator function as the input's response, the net result is that the values of Bk of n will be constant over chunks of 2 N 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 2 N. This frequency is periodic with period 2 N, so, and for each chunk of 2 N samples, we go look for the symbol that corresponds to this interval. So if say n is equal to 0 here, n is equal to 2 N here. n = 4N here and n = 6N here. For this interval we will go look for a[0] and multiply this portion of the carrier by this value. Here we will look for a[1], here a[2]. And so on and so forth. So with this simplification the whole transmitter can be sketched like so. We have the symbols from different subchannels that get multiplied by the carrier and kept constant over intervals of 2n output samples. The whole thing gets summed together, we get the aggregate band path signal. We take the real part and we're ready for the D to A converter. We can now write explicitly the formula for the aggregate band pass 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 2 pi over 2N nk. Now, because of the way the index to the a, k 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 zero to 2N- 1, these values will stay constant. And again for values of the output index that go from 2N to 4N-1. So we could compute two big n values for the sequence c[n] 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 2 big N 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-1. But that's not a problem, we can supplement those elements. And so we can compute a chunk of two big N output samples in one go as an inverse DFT over 2N points of a vector that is given by N channel symbols and has another big N zeros added to the end of it. To the index for the sub channel symbols is given by he value of the output index divided by 2N 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 [n]. And we can write that real part as c [n] plus the conjugate of c [n] divided by 2. Now it is easy to prove, and it's left as an exercise, that the conjugate or 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 the finite length vector, it's useful to think of the periodic extension. With this result, and knowing that c [n] is equal to 2 N times the inverse DFT of vector that is zeros in it's latter part, we can sum c n with its conjugate to obtain that the real part of c n is equal to N times the the inverse DFT of a vector that is given by twice the symbol for the base band subchannel. The reason why we can write this is because the base band signal will always have real valued symbols, because it is base band, followed by the complex symbols for the m -1 remaining subchannels, followed by the conjugate of the symbols for the n -1 remaining subchannels. But going from channel n- 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 base band symbol. Followed by the symbols for the subchannels from 1 to N-1. 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 2N output samples in one go. We use a parallel to serial device to output the samples one at a time. And here we have our D to A 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 mappers of the subchannels. And we repeat the process. An actual ADSL modem uses a maximum frequency for the channel of 1104 kilohertz. Divides this channel into 256 subchannels. Each QAM modem for the subchannel can independently choose between 0 and 15 bits per symbol. Now, the first seven channels are left off because that is the band used by the voice communication over the telephone channel. Channels 7 to 31 are used for data upstream. And the rest is left for data downstream. For a maximal 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 online modem that you most probably used to watch this online class.
