0:00:00.630,0:00:04.380 Hi, and welcome to module 9.5 of digital [br]signal processing. 0:00:04.380,0:00:08.870 In this module, we will touch briefly on [br]some topics in receiver design. 0:00:08.870,0:00:12.251 A lot of things, unfortunately, happen to [br]the signal while it's traveling through 0:00:12.251,0:00:15.244 the channel. [br]The signal picks up noise, we have seen 0:00:15.244,0:00:18.417 that already. [br]It also gets distorted because the 0:00:18.417,0:00:21.703 channel will act as some sort of filter [br]that is not necessarily all pass and 0:00:21.703,0:00:25.646 linear phase. [br]Interference happens too, that might be 0:00:25.646,0:00:29.862 parts of the channel that we thought were [br]usable, and they're actually not, so the 0:00:29.862,0:00:33.892 receiver really has to deal with a copy [br]of the transmitted signal that is very, 0:00:33.892,0:00:40.460 very far from the idealized version we [br]have seen so far. 0:00:40.460,0:00:44.744 The way receivers, especially digital [br]receivers, can cope with the distortions 0:00:44.744,0:00:48.713 and noise introduced by the channel, is [br]by implemented adaptive filtering 0:00:48.713,0:00:52.625 techniques. [br]Now we will not have the time to go into 0:00:52.625,0:00:56.365 very many details about adaptive signal [br]processing, again these are topics that 0:00:56.365,0:01:01.380 you will be able to study in more [br]advanced signal processing classes. 0:01:01.380,0:01:05.540 But I think it is important to give you [br]an overview of things that have to happen 0:01:05.540,0:01:10.390 inside a receiver, inside your ADSL [br]receiver for instance. 0:01:10.390,0:01:13.840 So you can enjoy this high data rates [br]that are available today. 0:01:13.840,0:01:18.329 And the first technique that we will look [br]at is adaptive equalization and then we 0:01:18.329,0:01:22.349 will look at some very simple timing [br]recovery that it used in practice in 0:01:22.349,0:01:27.942 receivers. [br]Let's begin with a blast from the past. 0:01:27.942,0:01:33.222 [NOISE] Those of you that are a little [br]bit older will certainly have recognized 0:01:33.222,0:01:38.342 this sound as the obligatory soundtrack [br]every time you used to connect to the 0:01:38.342,0:01:43.847 internet. [br]And indeed this is the sound made by a 0:01:43.847,0:01:49.470 V34 modem that was the standard dial up [br]connection device seen in the 90s until 0:01:49.470,0:01:54.889 the early 2000. [br]Now if you have ever used a modem, you've 0:01:54.889,0:02:00.900 heard the sound and you probably wondered [br]what was going on, so we're going to 0:02:00.900,0:02:06.840 analyze what we just heard from the [br]graphical point of view. 0:02:06.840,0:02:11.130 If we look at the block diagram for the [br]receiver once again, what we're going to 0:02:11.130,0:02:15.420 do is we're going to plot the base band [br]complex samples as points on the complex 0:02:15.420,0:02:20.270 blank. [br]So we're going to take B, r of n has the 0:02:20.270,0:02:26.230 as the horizontal coordinate and b, i of [br]n as the vertical coordinate. 0:02:26.230,0:02:30.430 And before we do so let's just look for a [br]second at what happens inside the 0:02:30.430,0:02:35.120 receiver when the signal at the input is [br]a simple sinusoid, like cosine of omega c 0:02:35.120,0:02:41.300 plus omega zero n. [br]We are demodulating this very simple 0:02:41.300,0:02:45.960 signal with the two carriers, the cosine [br]of omega c n and sine of omega c n. 0:02:45.960,0:02:51.100 And then we're filtering the result with [br]a low pass field. 0:02:51.100,0:02:54.480 So if we work out this formula with [br]standard trigonometric identities, we can 0:02:54.480,0:02:59.540 always express for instance the product [br]of two cosine functions as the sum. 0:02:59.540,0:03:03.680 Of the cosine of the sum of the angles [br]plus the cosine of the difference of the 0:03:03.680,0:03:06.435 angles. [br]And same for the product of cosine and 0:03:06.435,0:03:09.614 sine. [br]So if we do that, we get four terms, two 0:03:09.614,0:03:16.520 of which have a frequency that will fall [br]outside of the pass band of the filter h. 0:03:16.520,0:03:20.360 So when we apply the filter to these [br]terms, we're left only with cosine of 0:03:20.360,0:03:26.700 omega 0 n plus j sine of omega 0 n. [br]Which is of course e to the j omega 0 n. 0:03:26.700,0:03:30.934 So when the input to the receiver is a [br]cosine, the points in the complex 0:03:30.934,0:03:35.825 baseband sequence will be points around [br]the circle and a difference between two 0:03:35.825,0:03:42.714 successive points is the angle omega 0. [br]The reason why we might be called to 0:03:42.714,0:03:46.410 demodulate a simple sinusoid is because [br]the receiver will send what are called 0:03:46.410,0:03:49.826 pile tones, simple sinusoids that are [br]used to probe the line and gauge the 0:03:49.826,0:03:54.730 response of the channel at particular [br]frequencies. 0:03:54.730,0:03:58.132 So with this in mind, let's look at the [br]slow-motion analysis of the base band 0:03:58.132,0:04:01.744 signals. [br]Samples, when the input is the audio file 0:04:01.744,0:04:06.440 we just heard before. [br]So lets start with a part that goes like 0:04:06.440,0:04:09.706 this[SOUND]. [br]This signal contains several sinusoids, 0:04:09.706,0:04:14.780 that we can see here in the plot. [br]And the sinusoids also contain abrupt 0:04:14.780,0:04:18.460 phase reversal, meaning that, at some [br]given points in time The phase of the 0:04:18.460,0:04:23.324 sinusoid is augmented by pi. [br]You can see this as this small explosions 0:04:23.324,0:04:27.810 in the circular pattern in the plot. [br]These phase reversals, are used by the 0:04:27.810,0:04:31.330 transmitter and the receiver, as time [br]markers, to estimate the propagation 0:04:31.330,0:04:35.450 delay of the signal, from source to [br]destination. 0:04:35.450,0:04:38.844 The next part goes like this. [br][NOISE]. 0:04:38.844,0:04:42.610 And this is a training sequence. [br]The transmitter sends a sequence of known 0:04:42.610,0:04:46.670 symbols, namely the receiver knows the [br]symbol that are transmitted. 0:04:46.670,0:04:49.797 And so the receiver can use this [br]knowledge to train an equalizer to undo 0:04:49.797,0:04:53.876 the affects of the channel. [br]The last part is the data transmission 0:04:53.876,0:04:57.452 proper, the noisy part if you want, of [br]the audio file. 0:04:57.452,0:05:00.990 And the interesting thing is that the [br]transmitter and receiver perform a 0:05:00.990,0:05:04.876 handshake procedure, using a very low bit [br]rate QAM transmission using only four 0:05:04.876,0:05:10.632 points, therefore two bits per symbol. [br]To exchange the parameters of the real 0:05:10.632,0:05:13.780 data transmission that is going to [br]follow. 0:05:13.780,0:05:16.390 The speed, the constellation size, and so [br]on. 0:05:16.390,0:05:20.248 Using the four point QAM constellation in [br]the beginning ensures that, even in very 0:05:20.248,0:05:23.847 noisy conditions, transmitter and [br]receiver can exchange their vital 0:05:23.847,0:05:29.498 information. [br]So even from this simple qualitative 0:05:29.498,0:05:33.268 description of what happens in a real [br]communication scenario, we can see that 0:05:33.268,0:05:38.130 the task that the receiver is saddled [br]with is very complicated. 0:05:38.130,0:05:40.760 So it's a dirty job, but a receiver has [br]to do it. 0:05:40.760,0:05:45.452 And a receiver has to cope with four [br]potential sources of problem. 0:05:45.452,0:05:49.900 Interference. [br]The propagation delay, so the delay 0:05:49.900,0:05:52.818 introduced by the channel. [br]The linear distortion introduced by the 0:05:52.818,0:05:55.602 channel. [br]And drifts in internal clocks between the 0:05:55.602,0:06:00.940 digital system inside the transmitter and [br]the digital system inside the receiver. 0:06:00.940,0:06:04.834 So when it comes to interference the [br]handshake procedure, and the line probing 0:06:04.834,0:06:08.256 pilot tones are used in clever ways to [br]circumvent the major sources of 0:06:08.256,0:06:13.000 interference. [br]We will see some example later on when we 0:06:13.000,0:06:16.414 discuss ADSL. [br]The propagation delay, is tackled by a 0:06:16.414,0:06:20.990 delay estimation procedure, that we will [br]look at in just a second. 0:06:20.990,0:06:24.992 The distortion to this by the channel is [br]compensated using adaptive equalization 0:06:24.992,0:06:29.130 techniques, and we will see some examples [br]of that as well. 0:06:29.130,0:06:33.156 And clock drifts are tackled by timing [br]recovery techniques that in and of 0:06:33.156,0:06:37.644 themselves are quite sophisticated and [br]therefore we leave them to more advanced 0:06:37.644,0:06:41.916 classes. [br]Graphically, if we sum up the chain of 0:06:41.916,0:06:45.744 events that occur between the [br]transmissions of the original digital 0:06:45.744,0:06:51.442 signal and the beginning of the [br]demodulation of the received signal. 0:06:51.442,0:06:56.660 We have a digital to analog converter and [br]a transmitter, this is transmitter part 0:06:56.660,0:07:01.300 of the chain that operates with a given [br]sample period Ts. 0:07:01.300,0:07:05.500 This generates an analog signal which is [br]sent over a channel. 0:07:05.500,0:07:08.472 We can represent the channel for the time [br]being as a linear filter in the 0:07:08.472,0:07:13.190 continuous time domain. [br]With frequency response d of j omega. 0:07:13.190,0:07:17.280 At the input of the receiver, we have a [br]continuous time signal, s hat of t. 0:07:17.280,0:07:22.690 Which is a distorted and delayed version [br]of the original analog signal. 0:07:22.690,0:07:27.600 We will neglect noise for the time being. [br]This signal is sampled by an a to d 0:07:27.600,0:07:32.110 converter that operates at a period t [br]prime of s. 0:07:32.110,0:07:36.100 And we obtain the sequence of samples [br]that will be input to the modulator. 0:07:36.100,0:07:38.210 So this is the receiver part of the [br]chain. 0:07:38.210,0:07:41.620 We have to take into account the [br]distortion introduced by the channel, and 0:07:41.620,0:07:45.850 we have to take into account the [br]potentially time varying discrepancies in 0:07:45.850,0:07:49.483 the clocks between the transmitter and [br]the receiver. 0:07:49.483,0:07:52.981 These two systems are geographically [br]remote and there is no guarantee that the 0:07:52.981,0:07:56.200 two internal clocks that're used in the A [br]to D and D to A converters are 0:07:56.200,0:08:00.310 synchronized or run exactly at the same [br]frequency. 0:08:00.310,0:08:03.280 Let's start with problem of delay [br]compensation. 0:08:03.280,0:08:07.100 To simplify the analysis we'll assume [br]that the clocks at transmitter and 0:08:07.100,0:08:10.346 receiver are synchronized and [br]synchronous. 0:08:10.346,0:08:14.470 So T prime of s is equal to T s, and the [br]channel acts as a simple delay. 0:08:14.470,0:08:17.370 So the received signal is simply a [br]delayed version of the transmitted 0:08:17.370,0:08:20.822 signal. [br]Which implies that the frequency response 0:08:20.822,0:08:24.112 of the channel is simply e to the minus j [br]omega d. 0:08:24.112,0:08:27.000 So, the channel introduces a delay of d [br]seconds. 0:08:27.000,0:08:30.800 We can express this in samples in the [br]following way. 0:08:30.800,0:08:34.790 We write d as the product of the sampling [br]period, times b plus tau. 0:08:34.790,0:08:38.414 Where b is an integer. [br]And tau is strictly less than one-half in 0:08:38.414,0:08:41.376 magnitude. [br]So b is called the bulk delay because it 0:08:41.376,0:08:45.282 gives us an integer number of samples of [br]delay at the receiver and tau is the 0:08:45.282,0:08:50.846 fractional delay. [br]So the fraction of samples introduced by 0:08:50.846,0:08:56.480 the continuous time delay of d. [br]So, how do we compensate for this delay? 0:08:56.480,0:08:59.560 Well, the bulk delay is rather easy to [br]tackle. 0:08:59.560,0:09:02.793 Imagine the transmitter begins [br]transmission by sending just an impulse 0:09:02.793,0:09:06.206 over the channel. [br]So, the discreet time signal is this one, 0:09:06.206,0:09:10.290 it's just a delta and a 0. [br]It gets sent to D-to-A converter. 0:09:10.290,0:09:13.157 And the converter will output a [br]continuous time signal that looks like an 0:09:13.157,0:09:17.370 interpolation function like the sinc. [br]And like all interpolation functions, it 0:09:17.370,0:09:21.420 will have a maximum peak at zero that [br]corresponds to the known zero sample. 0:09:21.420,0:09:23.545 This signal gets transmitted over the [br]channel. 0:09:23.545,0:09:27.100 And it gets to the receiver after a delay [br]d that we can estimate for instance by 0:09:27.100,0:09:31.870 looking at the displacement of the peak [br]of the interpolation function. 0:09:31.870,0:09:34.670 The receiver converts this into a [br]discrete time sequence. 0:09:34.670,0:09:38.360 Now in the figure here it looks as if the [br]sample incidence at the transmitter and 0:09:38.360,0:09:41.920 receiver are perfectly aligned. [br]Now this is not necessarily the case 0:09:41.920,0:09:44.958 because the starting time for the [br]interpolator at the transmitter and the 0:09:44.958,0:09:48.709 sampler at the receiver are not [br]necessarily synchronous. 0:09:48.709,0:09:52.789 But any difference in starting time can [br]be integrated into the propagation delay 0:09:52.789,0:09:56.310 as long as the sampling periods are the [br]same. 0:09:56.310,0:10:00.531 So with this, all we need to do at the [br]receiver is to look for the maximum value 0:10:00.531,0:10:05.565 in the sequence of samples. [br]Because of the shape of the interpolating 0:10:05.565,0:10:09.850 function, we know that the real maximum [br]will be at most half a sample in either 0:10:09.850,0:10:13.554 direction of the location, of the maximum [br]sample value. 0:10:13.554,0:10:17.389 So, add the receiver to offset the bulk [br]delay, we will just set the nominal time 0:10:17.389,0:10:20.752 n equal to 0, to coincide with the [br]location of the maximum value of the 0:10:20.752,0:10:25.114 sample sequence. [br]Now of course we need to compensate for 0:10:25.114,0:10:28.720 the fractional delay, so we need to [br]estimate tau. 0:10:28.720,0:10:31.230 And to do that we'll use a different [br]technique. 0:10:31.230,0:10:34.578 Let me add in passing, that in real [br]communication devices, of course we're 0:10:34.578,0:10:37.728 not using impulses to offset the bulk [br]delay. 0:10:37.728,0:10:41.292 Because impulses are full-band signals [br]and so they would be filtered out by the 0:10:41.292,0:10:46.410 passband characteristic of the channel. [br]The trick is to embed discontinuities in 0:10:46.410,0:10:51.310 pilot tones and to recognize this [br]discontinuities at the receiver. 0:10:51.310,0:10:54.775 As we have seen in the animation at the [br]beginning of this module, we use phase 0:10:54.775,0:10:58.750 reversals, which are abrupt [br]discontinuities in sinusoids, to provide 0:10:58.750,0:11:02.948 a recognizable instant in time for the [br]receiver to latch on. 0:11:02.948,0:11:07.702 Okay, so what about the fractional delay? [br]Well, for the fractional delay, we use a 0:11:07.702,0:11:11.860 sinusoid instead of a delta, so we build [br]a bass band signal, which is simply a 0:11:11.860,0:11:16.894 complex exponential at a known frequency, [br]omega 0. 0:11:16.894,0:11:20.664 This will be converted to a real signal [br]before being sent to the D-to-A 0:11:20.664,0:11:24.369 converter, and so what we transmit [br]actually is cosine of omega c, the 0:11:24.369,0:11:30.270 carrier frequency, plus the pilot [br]frequency omega 0, times n. 0:11:30.270,0:11:34.560 The receiver will receive a delayed [br]version of this, which contains the delay 0:11:34.560,0:11:39.190 now in sample, and fraction of sample, b [br]plus tau. 0:11:39.190,0:11:43.155 After we demodulate this cosine you [br]remember we got a complex exponential and 0:11:43.155,0:11:47.900 we can also compensate already for the [br]bulk delay which we know. 0:11:47.900,0:11:52.373 So for an integer number of sample b we [br]obtain a base band signal at b of n which 0:11:52.373,0:11:58.325 is e to the j omega and minus tau. [br]Since we know the frequency of omega 0 we 0:11:58.325,0:12:02.420 can just multiply this quantity by e to [br]the minus j omega 0 n and obtain e to the 0:12:02.420,0:12:09.200 minus j omega 0 tau, which is a constant. [br]And which we can invert given that we 0:12:09.200,0:12:12.469 know the frequency omega 0. [br]And so now we have an estimate for both 0:12:12.469,0:12:16.527 the bulk delay and the fractional delay. [br]Now we have to bring back the signal to 0:12:16.527,0:12:20.140 the original timing. [br]The bulk delay is really no problem. 0:12:20.140,0:12:23.550 It's just an integer number of samples. [br]What creates a problem is the fractional 0:12:23.550,0:12:28.375 delay because that will shift the peaks [br]with respect to the sampling intervals. 0:12:28.375,0:12:33.266 So if we want to compensate for the bulk [br]delay we need to compute subsample values 0:12:33.266,0:12:37.865 and in theory to do that we should use a [br]sinc fractional delay namely a filter 0:12:37.865,0:12:45.198 with impulse response sinc of n plus tau. [br]In practice however, we will use a local 0:12:45.198,0:12:48.558 interpolation and this is a very [br]practical application of the Lagrange 0:12:48.558,0:12:52.560 interpolation technique that we saw in [br]module 6.2. 0:12:52.560,0:12:56.220 So graphically the situation is like so, [br]we have a stream of samples coming in. 0:12:56.220,0:13:00.828 And for each sample, we want to compute [br]the subsample value with a distance of 0:13:00.828,0:13:06.112 tau from the nearest sample interval. [br]And we want to only use a local 0:13:06.112,0:13:11.510 neighborhood of samples to estimate this. [br]Now, you remember from module 6.2. 0:13:11.510,0:13:14.994 The Lagrange approximation works by [br]building a linear combination of Lagrange 0:13:14.994,0:13:18.312 polynomials weighed by the samples of the [br]function. 0:13:18.312,0:13:21.822 So, as per usual, we choose the sampling [br]interval equal to 1, so that we lighten 0:13:21.822,0:13:25.882 the notation. [br]We have a continuous time function x of 0:13:25.882,0:13:29.858 t, and we want to compute x of n plus [br]tau, with tau less than one half in 0:13:29.858,0:13:34.764 magnitude. [br]So we have samples of this function at 0:13:34.764,0:13:39.500 integers, n, and the local Lagrange [br]approximation around n, is given by this 0:13:39.500,0:13:44.670 linear combination of Lagrange [br]polynomials. 0:13:44.670,0:13:48.560 Weighted by the samples of the functions [br]around the approximation point. 0:13:48.560,0:13:54.102 So we use the notation x L of n and t. [br]N is the center point and t is the value 0:13:54.102,0:13:59.760 from the center point at which we want to [br]compute the approximation. 0:13:59.760,0:14:03.244 And the Lagrange polynomials are given by [br]this formula here, which is the same as 0:14:03.244,0:14:07.524 in module 6.2. [br]So the delayed compensated input signal 0:14:07.524,0:14:11.300 will be set equal to the Lagrange [br]approximation at tau. 0:14:11.300,0:14:14.120 So let's look at an example. [br]Assume that we want a second order 0:14:14.120,0:14:17.470 approximation. [br]So we pick N equal to 1 and we will have 0:14:17.470,0:14:22.850 three Lagrange polynomials. [br]And so, we will need to use three samples 0:14:22.850,0:14:28.135 of the sequence to compute interpolation. [br]These three polynomials will be centered 0:14:28.135,0:14:31.852 in n minus 1 and in n plus 1 and scaled [br]by the values of the samples at these 0:14:31.852,0:14:35.832 locations. [br]And finally, we will sum the poll numbers 0:14:35.832,0:14:38.613 together and compute their value in n [br]plus tau. 0:14:38.613,0:14:42.500 So, we start with the first one, which is [br]centered in n minus 1. 0:14:42.500,0:14:47.108 And, like all interpolation polynomials, [br]its value is 1 in n minus 1, and 0, at 0:14:47.108,0:14:52.990 other integer values of the argument. [br]The second polynomial will be centered in 0:14:52.990,0:14:56.760 n, and the third polynomial will be [br]centered in n plus 1. 0:14:56.760,0:14:59.880 When we sum them together, we obtain a [br]second order curve that goes through the 0:14:59.880,0:15:02.760 points, that interpolates the three [br]points, and then we can compute the 0:15:02.760,0:15:07.680 approximation as the value of this curve [br]in n plus tau. 0:15:07.680,0:15:12.300 Now the nice thing about this approach, [br]is that if we look at the approximation, 0:15:12.300,0:15:16.686 if we take the Lagrange approximation [br]around n. 0:15:16.686,0:15:20.956 We can define a set of coefficients, d [br]tau of k, which are the values of each 0:15:20.956,0:15:26.610 Lagrange polynomial in tau. [br]So d tau of k, are 2 N plus 1 values, the 0:15:26.610,0:15:32.872 form, the coefficients, of an FIR filter. [br]And we can compute the value of the 0:15:32.872,0:15:37.224 Lagrange approximation simply as the [br]convolution of the incoming sequence with 0:15:37.224,0:15:42.482 this interpolation filter. [br]So for example, if these are the three 0:15:42.482,0:15:47.450 Lagrange polynomials for n equal to 1, we [br]can compute these polynomials for t equal 0:15:47.450,0:15:54.440 to tau, where tau is the fractional delay [br]that we estimated before. 0:15:54.440,0:15:58.730 And we will obtain three coefficients, [br]like here, for instance, is an example 0:15:58.730,0:16:02.736 for tau equal to 0.2. [br]Three coefficients that give us an FIR 0:16:02.736,0:16:05.810 filter, and then we can just simply [br]filter the samples coming into the 0:16:05.810,0:16:10.530 receiver with this filter, to compensate [br]for the fractional delay. 0:16:10.530,0:16:13.130 So again, the algorithm is, estimate the [br]fractional delay. 0:16:13.130,0:16:16.954 The bulk delay is no problem, again. [br]Compute the 2 N plus 1 Lagrangian 0:16:16.954,0:16:19.984 coefficients and filter it with the [br]resulting FIR. 0:16:19.984,0:16:23.689 The added advantage of this strategy is [br]that if the delay changes over time for 0:16:23.689,0:16:27.451 any reason, all we need to do is to keep [br]the estimation running and update the FIR 0:16:27.451,0:16:32.140 coefficients as the estimation changes [br]over time. 0:16:32.140,0:16:35.825 Okay, now that we know how to compensate [br]for the propagation delay introduced by 0:16:35.825,0:16:39.280 the channel. [br]Let's go see the rechannel it with an 0:16:39.280,0:16:41.935 arbitrary frequency response D j of [br]omega. 0:16:41.935,0:16:46.951 And the transmission chain goes from the [br]pass band signal s of n, discreet time, 0:16:46.951,0:16:51.891 into a D-to-A converter, analog signal s [br]of t, it gets filtered by the channel, 0:16:51.891,0:16:59.550 gives us hat s of t, which is sampled at [br]the receiver to give us. 0:16:59.550,0:17:04.600 A received fast band signal, hat s of n. [br]But now we have seen in the previous 0:17:04.600,0:17:07.780 module that this block diagram can be [br]converted into an all digital scheme 0:17:07.780,0:17:11.680 where our band pass signal s of n gets [br]filtered by the discrete time equivalent 0:17:11.680,0:17:17.450 of the channel. [br]And, gives us a filtered version of the 0:17:17.450,0:17:22.330 bandpass signal, as would appear inside [br]the receiver. 0:17:22.330,0:17:25.978 So the problem now, is that we would like [br]to undo the effects of the channel, on 0:17:25.978,0:17:30.361 the transmitted signal. [br]And the classic way to do that, is to 0:17:30.361,0:17:34.708 filter the received signal hat s of n, by [br]a filter E, that compensates for the 0:17:34.708,0:17:40.230 distortion or the filtering introduced by [br]the channel. 0:17:40.230,0:17:45.246 So the target is that the output of the [br]filtering operation gives us a signal hat 0:17:45.246,0:17:50.380 s e of n, which is equal to the [br]transmitted signal. 0:17:50.380,0:17:53.127 How do we do that? [br]In theory, it would be enough to pick a 0:17:53.127,0:17:56.922 transfer function for the filter E, which [br]is just the reciprocal of the equivalent 0:17:56.922,0:18:01.952 transfer function of the channel. [br]But the problem is that we don't know the 0:18:01.952,0:18:04.680 transfer function of the channel in [br]advance because each time we transmit 0:18:04.680,0:18:08.270 data over the channel, this transfer [br]function may change. 0:18:08.270,0:18:12.650 And also, even while we're transmitting [br]data, the transfer function might change 0:18:12.650,0:18:16.200 because it is a physical system that [br]might be subject to. 0:18:16.200,0:18:19.740 Drifts and modifications. [br]So what do we do? 0:18:19.740,0:18:25.562 We need to use adaptive equalization. [br]So the filter that compensates for the 0:18:25.562,0:18:30.409 distortion introduced by the channel is [br]called an equalizer. 0:18:30.409,0:18:34.669 And what we want to do Is to change the [br]filter in time, so change the filter 0:18:34.669,0:18:39.355 coefficients in a DPS realization as a [br]function of the error that we obtain when 0:18:39.355,0:18:47.610 we compare the output of the filter with [br]the signal that we would like to obtain. 0:18:47.610,0:18:52.260 In our case the signal that we would like [br]to obtain is the transmitted signal. 0:18:52.260,0:18:56.660 And so we take the received signal, we [br]filter it with the equalizer. 0:18:56.660,0:19:00.720 We look at the result. [br]We take the difference, with respect to 0:19:00.720,0:19:04.440 the original signal, and we use the [br]error, which should be zero in the ideal 0:19:04.440,0:19:08.580 case, to drive the adaptation of the [br]equalizer. 0:19:08.580,0:19:13.898 But wait, how do we get the exact [br]transmitted signal at the receiver? 0:19:13.898,0:19:17.340 Well, we use two tricks. [br]The first one is boot strapping. 0:19:17.340,0:19:22.730 The transmitter will send a prearranged [br]sequence of symbols to the receiver. 0:19:22.730,0:19:26.470 So let's call the sequence of symbols a t [br]of n. 0:19:26.470,0:19:31.720 This gets modulated and generates a pass [br]band signal s of n. 0:19:31.720,0:19:35.990 Now at the receiver the sequence a t of n [br]is known. 0:19:35.990,0:19:41.974 And the receiver has an exact copy of the [br]modulator, of the transmitter, inside of 0:19:41.974,0:19:46.485 itself. [br]So the transmitter can generate locally, 0:19:46.485,0:19:50.138 an exact copy of the pass band signal s [br]of n. 0:19:50.138,0:19:54.428 And so, for the bootstrapping part of the [br]adaptation, we actually have an exact 0:19:54.428,0:19:58.718 copy of the transmitted pass band signal [br]that we can use to drive the adaptation 0:19:58.718,0:20:03.890 of the coefficients. [br]The training sequence is just long enough 0:20:03.890,0:20:06.620 to bring the equalizer to a workable [br]state. 0:20:06.620,0:20:10.290 For the handshake procedure that we saw [br]in the video before, for instance. 0:20:10.290,0:20:13.911 This would correspond to the moment where [br]the receiver starts demodulating the four 0:20:13.911,0:20:17.550 point QIM. [br]At that moment, the receiver will switch 0:20:17.550,0:20:20.890 strategy and implement a data driven [br]adaptation. 0:20:20.890,0:20:25.770 The thing works like this. [br]The received signal gets equalized, gets 0:20:25.770,0:20:30.606 demodulated and then the slicer will [br]recover the sequence of transmitted 0:20:30.606,0:20:34.107 symbols. [br]Since the receiver has a copy of the 0:20:34.107,0:20:38.880 transmitter inside of itself, it can use [br]the sequence of transmitted symbol. 0:20:38.880,0:20:42.430 To build a local copy of the transmitted [br]signal. 0:20:42.430,0:20:47.120 Now of course, errors might happen in the [br]slicing process, and so this local copy 0:20:47.120,0:20:52.496 is not completely error-free. [br]But the assumption is that the equalizer 0:20:52.496,0:20:56.272 is doing already enough of a good job to [br]keep the number of errors in this 0:20:56.272,0:21:01.548 sequence sufficiently low. [br]So that the difference, with respect to 0:21:01.548,0:21:05.708 the received signal, is enough to refine [br]the adaptation of the equalizer, and 0:21:05.708,0:21:11.800 especially to track the time varying [br]conditions of the channel. 0:21:11.800,0:21:14.293 What we have seen, is just a qualitative [br]overview of what happens inside of a 0:21:14.293,0:21:17.765 receiver. [br]And there're still so many questions that 0:21:17.765,0:21:21.940 we would have to answer to be thorough. [br]For instance, how do we carry out the 0:21:21.940,0:21:25.140 adaptation of the coefficients in the [br]equalizer? 0:21:25.140,0:21:30.394 How do we compensate for different clock [br]rates in geographically diverse receivers 0:21:30.394,0:21:34.544 and transmitters? [br]How do we recover from the interference 0:21:34.544,0:21:40.900 from other transmission devices, and how [br]do we improve the resilience to noise? 0:21:40.900,0:21:45.500 The answers to all those questions [br]require a much deeper understanding of 0:21:45.500,0:21:50.650 adaptive signal processing, and hopefully [br]that'll be the topic of your next signal 0:21:50.650,0:21:54.119 processing class.