 So, this is lecture 18, so in the last class we began to look at the case when there is a channel and you cannot make your received signal orthogonal when your symbols are going through the channel. So, that is the situation I saw, how did I write it down? So, you have some bits, I want to remind you how it works, some bits you are converting it using some mapper through some constellation to a symbol sequence S k and maybe k between 0 and L minus 1, so L symbol sequence is drawn from the alphabet, so there are quite a few of them now. And then what am I doing? I am putting it through a transmit filter gt and then that is going to go through a channel ct and then noise gets added to it and then you are trying to figure out what to do here. So, this R of t I wrote as summation k equals 0 to L minus 1 S k H of t minus kt plus nt. So, like I said for some reason this H of t minus kt do not form a orthogonal set of signals, so maybe you cannot control your g of t suitably whatever you do not know c of t and this H of t works out to what? H of t is g of t convolved with c of t which is probably an unknown channel. So, remember c of t is the complex baseband equivalent of a passband frequency it can be that also. In general H of t is going to be complex, so I am not going to ask require that to be real and like I said it is not orthonormal, so whether it is orthonormal or not the first thing to do in our receiver is always to find an orthonormal basis for your set of signals and project that is the first and nice thing to do, but it turns out when you deal with this in practice and receivers the algorithms get very ugly and you usually do not do very clean neat projections and all that. Obviously it is not done it is always approximate it is always adaptive you do a lot of things and it seems complicated, so that is why I do not want to do that first, so there is a way to do that also in this situation and we will do it later. First I want to write down some expressions and bring up some heuristics get you thinking about those things and deal with the ugly looking expressions in the receivers because that is what is usually done in practice. The receiver algorithms are seldom very clean and neat, so we will do the clean and neat stuff also, but beginning to begin with we will see something slightly ugly so that we get used to the way the whole thing is going to work. So, the way I wrote it down my heuristic was to minimize the distance between R of t and and sigma S of k a of k h of t minus k for all vectors a, so in that respect I defined a metric j a for some a belonging to what, x l right, so you draw l symbols from your alphabet and define the corresponding signal that you would have obtained which would be what, so basically this I defined as the distance between R of t and the signal corresponding to a, so what is the signal corresponding to a summation a k h of t minus k t, to be able to compute such things in the receiver clearly, so a k is square brackets, so clearly you should know h of t, so you might know g of t because that is your transmitters response, but c of t you may not know and there are easy techniques to find it, you sound the channel and you find c of t maybe, so that is one way of visualizing it, so you see already there is some ugliness built into the receiver, you have to do some, well I do not want to call it maybe it is not that ugly anyway, but still you have to do some such things to figure out what the channel is and you have to do some real engineering, it is not just sitting down in a signal space and projecting, so it is different from what you have to do, so you have to find h of t, so you do you find that signal and then you do it, so the first question is for a given a can we at least compute this fast, so that is the first question I asked and I am trying to reduce this to a form, but it is all discrete and it is based on filters and samples and symbol rate calculations, see if we can do that fast, even that does not really solve a problem, why? Suppose I can compute j a fast, does that really solve my problem? Not really, why? I have to do this for each a, because the reason is my s hat is what? It is not just based on j a, it is argument of minimum over all a in excel j a, so I have to not only compute j a for a particular a, I have to compute it for every single a and that is going to be really really large, so I might also, is there a question? So, if you have any comments also you are free to stop me and make any comments, it is okay. So, as L increases you see this becomes terribly difficult, you will do too many of these computations, maybe you reduce the computation with j a, but still there are more important problems that need to be solved before you can implement this receiver, so anyway for the first step we will try to see how to simplify the computations in j a and then we will do proceed further, to do that simplification I did the following, so I wrote down after a while this expression for j a, I wrote it down as e r minus 2 times real part of summation k equals 0 to L minus 1 a star k y k plus, so I have enough room I think I will write it down below, k plus double summation k equals 0 to L minus 1, j equals 0 to L minus 1 a k, well should be careful here, so I am trying to be as consistent as possible with notation, but usually it is difficult to be consistent with notation, so we will try that, rho h j minus k, so I wrote this down and what was y k, you obtain y k by sending r of t, first what is e r, e r is the modulus of r t square, so what is modulus of r t, it is the L2 norm, integral mod square r t, remember all these k's are complex, so the modulus means something, you cannot just throw it out, all these k's are complex and y k I defined as r of t filtered by a matched filter, matched to h t, so h t not g t, it is a star of minus t, then you sample at symbol rate to get y k, remember in practice there will be additional ugliness here in the form of delay and synchronization, so there will be some delay due to various things, your g of t might have been non-causal, you might have delayed it purposefully for that, h of t might be causal, there might be other things, your receiver might have to turn on before the transmitter got started, so there will be all kinds of delays, so you have to adjust for all those things and there are ways to adjust for it in practice, so what people do is people send some tones initially and synchronize themselves first, known signals, you synchronize first and then after a while you send the actual thing, so you have to do all these things, all that is part and parcel of how the whole system will work, but here in this neat theory I am going to just write k t, but you should know that built into it there is some approximation, so y k, so y k if you want to write down it will become integral minus infinity to infinity r of t h star of t minus k t, so that is the symbol rate calculation, maybe one can do it, so the number of computations in this term is only around L computations, do you agree, maybe L minus 1 multiple, L multiplications and L minus 1 additions etc., so it is not too bad, so it is complex, so each multiplication might be more and all these things, but all those things do not change the order, it is order L, you only order L. And what about the next term, the next term rho h is a little bit more involved, so you take h of t, send it through the same matched filter h star of minus t to get and then sample at k t to get rho h k, so if the Fourier transform for h of t is h f, what is the Fourier transform for h star of minus t, h star of f, so that you know, then what is the Fourier transform for rho h k, is it just mod h of square, what does the sampling do, it has to make it periodic and all that, so it is alias, so this spectrum will actually be 1 by t summation m equals minus infinity to infinity mod h of f minus m by t square, remember what is my 1 by t, it is my symbol rate, so it is coming from that, how many that the impulse train S k, the time interval between that, that is my 1 by t. So, this is to be expected, so you see the Fourier transform for rho h of k is real and non-negative, so that is to be expected because rho h of k is a sampling of the autocorrelation of h of t, so autocorrelation has that property and when you do a Fourier transform on it, it is going to be real and non-negative. So, I would sometimes refer to this as S h of f, remember the better notation for this is what, S h e power j 2 pi f t, so that is the better notation for this, remember that is the proper way of writing it, but just quickly writing it as S h f, hopefully it is clear. So, the reason why I use S as opposed to capital rho or something is, I know it is going to be a, it is the power spectral density for h, so that is the way, well the sampled version, so it is a discrete time power spectral density for h, so that is why I use this, so this is also called the folded spectrum, the alias spectrum, I will use all these things interchangeably, so I will say folded spectrum for h, aliased spectrum for h, all these things, so it happens because of the sampling. So, that is rho h of k and you notice rho h of k I can compute ahead of time at the receiver if I know h of t, I do not have to wait and do that computation every time, I can compute it for enough k and put it in some memory and keep it there, so I can use it, even if I do that the computation of the second term needs how many order, what is the rough order of the number of computations in the second term, l, can you say l, l squared, so it is going to be l squared, so about l squared computations, it requires l square and that is not nice, so the reason why that is not nice is linear with l is okay, as you transmit more and more symbols and your computations is linear with l, anyway you have to wait for linearly more time to decode everything, that seems reasonable, but once it goes to l squared, it becomes really ugly because then you have to wait for for every increase you have to wait for much longer before you put out everything, so l squared maybe is not very nice in computation, so that is what I meant by saying even though I have come from discrete time, continuous time to discrete time which makes my computations look better, I still haven't reduced it to a very nice form where I can at least do linear in l computations for each a, so that is something I want to do and there is a wonderful way of simplifying this computation to make it linear in l, so that is the way I am going to introduce this next idea, so you have to basically do what is called a spectral factorization on SH of f and that will end up simplifying this computation from l squared to l, so that is basically because of the symmetry in rho h and all that, you really do not need l squared, it is enough if you have l in your computation, so that is the logic behind it, but later on when we study the proper correlation and projection receivers in the signal space, the optimal receivers, the spectral factorization will show up in a wonderfully fundamental way, so the spectral factorization is a much more fundamental and deep idea than just an idea to reduce this computation from l squared to l, but anyway this also is equally important and we will go through it in that fashion, so that is the next few things we will be seeing, so how to do this computation ja in order of l computations and for that you will see we need to do some fancy nice interesting spectral factorization, so I am going to give you the result first and then maybe go through a couple of simple ideas for proving it, so once again this is not a DSP course and if you have not seen spectral factorization, this will be new to you and I will urge you to go back and read through books and understand what it is at a more leisurely pace, I am not going to do it in great detail, so spectral factorization, suppose I have, so I will do it with this notation itself, I will not change the notation sh of e power j theta which is what, this is the Fourier transform for rho hk, discrete time Fourier transform, this I know is real and non-negative, so anytime I have a discrete time signal whose Fourier transform is, discrete time Fourier transform is real and non-negative, it turns out rho h of k can be written as gamma squared which is a real constant times some signal mk convolved with some signal m star of minus k, so this might be, one can easily motivate this from several things, what is greatest you can say this decomposition will be unique and you can force m of k to be monic, causal and loosely minimum phase, so this will become unique when you say m of k has to be monic, causal and loosely minimal phase, so it is possible to write down any discrete time signal whose discrete time Fourier transform is real and non-negative in this form, gamma squared which is a real positive constant times some signal which is monic, causal, loosely minimum phase convolved with m star of minus k, so once you say monic, causal, loosely minimum phase that m of k becomes unique, so if you do not say that there are various other possibilities of writing this down, so this is the, what is called the spectral factorization theorem, is there a question, loosely, so minimum phase is poles and zeros inside the unit circle, loosely is, zeros can be on the unit circle, so basically zeros on unit circle is okay, I have to allow it, I am only saying real and non-negative for sh of e power j theta, so obviously there can be zeros on the unit circle, there is no problem, so what is monic, m of zero is one, what is causal, m k is zero for k less than zero and loosely minimum phase, is because loosely minimum phase means all poles and zeros for m k have to be inside the unit circle and zeros can be on the unit circle also, so poles have to be inside, zeros on or inside the unit circle, so minimum phase signals have a lot of nice properties, so maybe in a good course on DSP you might see all those properties, one property that you can easily guess is what, what will be the inverse of m of k, it will be stable, causal, everything, so all these things you can also say, so it is nicely invertible, that is one property, the other property which is more crucial is most of the energy you can show will be contained in the first few tabs of m k, so the actual result is among all signals which have the same magnitude response, magnitude frequency spectrum, this minimum phase one will have the most energy in every beginning, first few tabs, so I am not stating it formally, so basically what happens is in a minimum phase sequence, all the energy will be in the first two tabs, when I say, so when you do energy of m k, all of them will be in the first few tabs, so eventually it will die down pretty fast, so of course there is some careful riders there about, you are only comparing among sequences with the same magnitude response, among those the minimum phase one will have the, will have most energy in the any first few tabs, so those are all very nice properties and we will exploit those things maybe later on as we go along, but for now how does this help us in simplifying the computation, that is what I am going to do and so I am thinking if I should do some brief diversion to spectral factorization now or no, I think I am going to show how it simplifies the computation first and then do a small diversion into spectral factorization and show how to do this in practice or why it makes sense, so that is something I will quickly do as we go along, so the same result can also be specified in the z domain, so in the z domain you can write sh of z as gamma square mz times what, m star of 1 by z star, so notice this confirms a lot of things you know about real non-negative Fourier transforms, so if some sequence has real non-negative Fourier transform, real is to fall, forget about non-negative, real Fourier transform then you know if you have a 0 what else should be a 0, the conjugate reciprocal should be a 0, if you have a pole then the conjugate reciprocal should be a pole, that just comes about by fact that on the unit circle the z transform has to evaluate to a real quantity, so that is something very easily you can show just based on very simple properties, the non-negativity will require something more, so it will require that if you have 0's on the unit circle they have to be a 0's of even multiplicity, so that is an additional thing, so in general if you look at the pole 0 plot for this, if you have every 0 here will also correspond to a 0 somewhere outside, every pole here will also correspond to a pole somewhere outside, every 0 on the unit circle will have even multiplicity, so you see immediately why you get a nice spectral factorization like this, you take all the poles and zeros inside the unit circle and assign it to m of z and take half the zeros on the unit circle and assign it to m of z, m star of 1 by z star will automatically give you all the other things and there should be a nice minimum phase, so I guess that is the only diversion into spectral factorization I wanted to go into and I have already done it, so it is very easy to motivate these things, so the question is what if SH of z is not rational, what if there are, what if its poles and zeros are too many, I do not know how many of them, then there are more complicated ways of doing spectral factorization and this result is true whether or not SH of z is rational, so this is true always, if it is rational the spectral factorization is easy to do, if it is not rational it is slightly more confusing but it can be done, so anyway, so that is the only background I wanted to go to, so by the way there is a good book on signal analysis, title is signal analysis, it is available in Tata McGraw-Hill publication, so it is cheap also by Popolus, so if at all you think your signals and systems background is not as strong as you want it to be, you think there are a lot of weak links, this is a very good book, it is not a descriptive book, so it is not a beginner's book but it has got everything there and it is a good book to read, it connects up all the loosens, okay, so let us proceed, so I have done that, so what does it mean, what happens to me, so once I do that, so once again the next step is to show some interesting things that happen, so what we are going to do now to simplify the calculation further, once more it will look a little bit ad hoc but I will come back and justify it later and we will make sense of all of this as we go along, so right now it will look a little bit ad hoc, so what we are going to do to simplify this computation is the following, so we are going to take R of T as before and we are going to filter with head star of minus T and sample at KT to get YK, so what did I know my YK will be, what is my YK now, so the way I wrote it down, it seems not too clear what YK is going to be but anyway YK is going to be related to rho H, so do you see that, so R of T, so if you want to ask the question from S, what did I do to S here, I did this and then noise got added here, this is a complete picture, so I had S, I am interested in S, so maybe that is SK, so I should draw it as SK, so can we write YK in terms of SK, how will I write that down, do you see that, do you agree that will happen, can I say this will be SK convolved with rho HK plus some NK, can I say that, so this is okay, most people are with me now, no problem, so actually I should say A, I am trying to compute JA, so I am going to say AK, so if I do this, I am going to get here, AK convolved with rho H of K plus NK, I am sorry, no I am going to just look at AK, so hold on for a minute, so you want to have SK instead of AK, maybe we can have SK instead of AK, it is okay, so it is okay, I mean fine, I will have SK, if that is what you want we will have that, so that is fine, so a couple of questions I want to ask before we proceed, now that we know something about this H star of minus T and H of T and all that, so if you want maybe you can write it differently, but what about NK, what can you say about NK here, so you see that, so the answer came very quickly, maybe enough people did not have time to think about it, N of T I know is a white Gaussian process, but H star of minus T when filtered and sampled is not the same as a series of orthogonal functions, when I do that I am not doing correlations with the series of orthonormal functions, so NK I cannot expect to be white Gaussian with mean 0 and variance N0 by 2, I cannot expect that because I know H star of minus T is not doing an orthonormal projection, so H of T, H of T minus KT and all that are not orthogonal anymore, I do not know whether they are orthogonal or not, so the only thing you can say is NK will have statistics, the power spectral density will be something like SHF, so PSD will be this, it will still be Gaussian, can I say it will be Gaussian, yeah it is Gaussian, Gaussian filtered is still Gaussian, but the power spectral density will be the power spectral density of H star of minus T which is the same as power spectral density of H of T which will be SH of F, so that is the first difference you notice here when S of K comes in, previously when we had orthogonal projections, my noise I was sure was uncorrelated between for different K, so now noise is becoming correlated, so maybe I should put N of K here instead of NK, so just to be consistent it is okay to have NK as well, so that is the first observation which is slightly disturbing, but that is okay, I mean we will deal with it as we go along, also look at this guy, SK convolved with rho H of K, so what is rho H of K, S of K goes from K equals 0 to L minus 1, what about rho H of K, it is an autocorrelation function, so what do I know about its properties, for one it will be, yeah non-causal and symmetric about 0, is that clear, so if I look at Y of 0, what will Y of 0 be, forget about the noise term for K equals 0, forget about the noise term, what will be this convolution for K equals 0, it is S and then you convolve with H with just a flip, there is no slipping, so it is solved, but still what happens, all the SSR involved in Y of 0, so the same thing will happen as you keep slipping the rho H of K through different steps, so this notation is bad because I mean you see this and you want to substitute K equals 0 here and the convolution you should not substitute K equals 0, what should you do, you should write down the whole, L equals minus infinity to infinity, SL H of rho H of K minus L, so that is what this means, so that is why the convolution notation is bad, here you can substitute K equals 0, not there, you substitute there then you will get totally meaningless results, so if you substitute K equals 0 you see the answer, rho H of minus L is not 0, see it is a symmetric thing, so you notice here at YK there is ISI both causal and anti-causal, so you have both causal as well as anti-causal ISI as far as YK is concerned, so all these observations are important because later on we will see receiver structures will be based on these observations, so what I want to point out is this has got both causal and anti-causal ISI terms, so this is important, so previously we never had this problem, when it was orthogonal what were we sure of, when H of T minus KT made an orthogonal set Y of K will only depend on SK, there is nothing more because I know my autocorrelation would have been, what would have been rho HK, I mean a delta, it is only one, nothing more is there, then my noise also is well behaved, it is Gaussian, I can do my detection happily, so you see when you do a non-orthogonal projection, lot of things change, so first of all you keep getting anti-causal, I mean causal as well as anti-causal ISI, but it is symmetric and there is some structure to it, so now I will use some of the spectral factorization structure that I know, what is the spectral factorization structure I know, rho H of K I can write as gamma square convolved with M of K convolved with M star of minus K, so once you see that you see S of K convolved with M of K is going to be only a nice causal contribution, but the convolution with M star of minus K is going to give me anti-causal thing, so one of the natural things to do with YK to get rid of the anti-causal thing is what, so how did I write this rho H of K, gamma square times convolved with M K convolved with M star of minus K, how do I get rid of the M star of minus K and YK, so you notice the L square complexity is coming because of this anti-causal nature, so if you had only L you would not have the L square complexity, it is coming because of the anti-causal nature, so maybe I want to get rid of the anti-causal part of my ISI and that is being caused by M star of minus K, so one thing I can do is what, convolved with, so I have S of K convolved with M of K convolved with M star of minus K, so maybe M star of minus K think of it as some some other Y of K, no not Y of K, some Z of K or whatever, how will I get rid of a convolution term, so if you have convolved with the inverse, it seems like a natural suggestion, so if I convolve with the inverse of M star of minus K, I can expect that in YK my M star of minus K will go away, so that is the way I am going to think about it. Is that clear? So that is the first suggestion that seems to suggest to get rid of the anti-causal ISI, maybe you should convolve with the inverse of M star of minus K, what is the inverse of M star of minus K? How do you go about finding it? What is the Z transform of M star of minus K? Just now gave me the answer, it is M star of 1 by Z star, so what will be the inverse of that? 1 by M star of 1 by Z star, will that be a reasonable filter? Yeah, it will be a reasonable filter, right, why? All the poles and zeros of M star of 1 by Z star are where? Yeah, somewhere outside, so if you do 1 by, they will all come in, so it is a reasonable filter, so maybe you can implement it, so it is not something too bad to think about, is that clear? So it seems like one of the things to do to YK to at least get rid of part of the ISI in a simple way is to filter with 1 by M star of 1 by Z star, so you will see this has a lot of deep meanings later on, you will see some orthogonality that happens because of this. So we will see that later as we go along, but for now my motivation is to simplify the computation in the YK, so for that I am going to do a filtering with that, so that is the thing that is going to happen. So let me write that down next, so what I am going to do in my JA, so let me remind you what this JA was, so maybe I should just go back and cut and paste, what was my JA, there it is, so that seems to be a good enough suggestion, so let us see what we can accomplish by doing this filtering, so I am going to take R of t, I am going to convolve with H star of minus t and sample at KT to get YK and this YK plays a role in the computation and then there is this rho HK and it seems like to work with rho HK and the reasonable thing to do is to take this YK and convolve with and filter with 1 by, I will introduce the gamma square here, that is just for convenience you can, it is just a factor, so maybe I should write this down very clearly because it is an important step to get a ZK, so it seems like instead of worrying about YK and trying to compute JA in terms of YK, if you go to filter it further by 1 by gamma square M star of 1 by Z star, the ZK presumably has lesser ISI and maybe that one can come up with a computation which is simpler than before, so you will see this various ways of figuring it out and once you write the whole JA in terms of ZK, you will see it is simplified to order of L computations everywhere, so that is the trick, so you do a filtering to get rid of the anti-causal ISI, come to ZK and then try to write JA in terms of ZK as opposed to YK and that will give you a nice computation and I am not going to go through that step by step, I am going to do that fairly fast but I will write ZK in terms of YK first, just to write that, ZK is what? YK convolved with the inverse of M star of minus K and there is also a gamma square, so if I write it differently I will get that ZK convolved with what? M star of minus K has to be YK, I will add the gamma square just to make sure gamma square is taken care of, so this has to happen, YK convolved with the inverse of M star of minus K is ZK, so I can bring that inverse to this side, so it becomes this, so if you want you can write it down explicitly YK then becomes, no Z, YK then becomes gamma square summation L equals K to infinity ZL M star L minus K, so this requires some careful interpretation of convolution to write this down because it is M star of minus K, you should know what it means in terms of convolution, I have written it down carefully and I get this, I can use the fact that M of K is causal and monic, so I know M of minus K will be anti-causal, so I can use all that and simplify this computation, so I should write ZL, I am sorry just keep going back to this old notation because that is the notation I have used in my notes and for some reason it is all mixed up here, ZL M star L minus K, so that is YK and once you make this substitution here, so it is not clear how you quickly make that substitution, so I want you to think about it, you can make that substitution, work with it carefully, it requires some careful thought and the way it works out and if you do that JA will work out to the following expression, which is nice, JA becomes ER plus gamma square, okay so why am I putting two bars, it is just absolute value ZK minus AK convolved with, okay so I should put two, sorry I will write it carefully and I will explain what the notation is, square minus gamma square modulus ZK square, so what is my this notation when I say this for a signal XK, what I mean is summation K equals minus infinity to infinity modulus XK square, okay that is my notation for that double arrow, okay that is JA, okay so now you see there is nothing much that is too disturbing about complexity in terms of L now, okay the reason is what is this or what is this guy, ZK convolved with this, so L equals 0 to L minus 1 AL M K minus L, okay so that is the simple computation to do from ZK it will work out, okay so you notice JA now there is no double summation basically it is only single summations but there is a ZK involved that is all right and in addition you can notice few other things, okay the first term is independent of A, okay whatever A you have you have the same thing for the first term, likewise the last term is also independent of A, okay so if some of you are thinking about how I went from that to here it is not a simple one step simplification you have to do quite a few things before you get there but you can back work from here to there and can be done very easily, okay so do that do it later, okay some completion of squares and all involved it is a little bit not too trivial, okay but you can try it, okay so but nice thing is those two terms are independent of A, okay so what will happen to my decoder my decoder which decides S hat will have to do argument of minimization over A what only the thing in the middle which is ZK minus okay I will write it in this form summation L equals 0 to L minus 1 Al Mk minus L square, okay so if you want to see exactly what that thing is maybe we can do that, okay so I think you can restrict from k equals 0 to infinity you don't have to go through the whole k, okay so ZK minus summation 0 to L minus 1 Al Mk minus L so, okay so suddenly this looks like each term here requires only about L computations, okay so maybe the k equals 0 to infinity is disturbing you but that's okay I mean there will be only very few ks which are significant so we don't have to worry too much about it but anyway that thing was already there in the previous expression also so that's not going to simplify anything for you but this is the one which is interesting so you've got only L computations inside, okay so go back and try to work with this the J in terms of Z of k and work it back and show that it agrees with the same term in terms of Yk it's an interesting computation but anyway it is ugly and you have to work through it very carefully and show both of them are the same, okay so it's a good exercise but finally we have a very nice expression at least for the minimization, okay so you're not doing any fancy double summation it's only a simple summation over k equals 0 to infinity, okay all right so that's the first simplification in computing J a, okay any questions any comments something that's disturbing you, okay so central to this whole derivation is the fact that rho h of k is going to factor spectrally if it did not factor spectrally I wouldn't have been able to do that filtering by 1 by gamma square m star of 1 by c star and get rid of the anti-causal part and Yk, okay then which gave me only a causal part and you see here what's happening here is what you can see what's happening here I'll draw I'll draw a picture of depicting what's exactly going on you'll see how it worked out, okay so it's a very interesting pictorial view of what's actually happening here and that's that's fairly important, okay so let's do that next, okay so I'll show you how this computation is going to work, okay so you actually have an r of t which came from s, okay so from s you did I'll simply write h of t, okay then noise got added to it you got r of t what was the first step we did h star of minus t which we knew was not a orthogonal projection, okay so you're not expecting all the ISI to be removed at that point clearly it's not an orthogonal projection but what happens because of that you get not only ISI on the causal side you also get ISI on the non-causal side because that's what you should expect right any if you do filtering and then sampling you will do aliasing, okay which means the autocorrelation of h will get involved and autocorrelation is symmetric about zero you can't expect non-ticosal part to go away, okay but then we took that autocorrelation did a spectral factorization and got rid of the anti-causal part by filtering with 1 by gamma square m star of 1 by z star, okay so once I did that I got a zk which I now know has only causal ISI, okay so I'm going to write down more expressions for zk we'll come to that slowly enough and what is my decoder doing it's taking ak which is a candidate sequence from my constellation then filtering with what filtering with what go back and look at this previous expression it's filtering with what mz, okay so mk or mz in the z domain and then what you're subtracting these two and then doing well sum square right or energy computation to get ja, okay so this is the pictorial view of what is happening is that clear, okay so it seems not too difficult for each a what is the real computation that you're doing you're evaluating the filtering with m of z which is just an order of l computation and then z of k stays the same for all k so you do the subtraction and do sum square you get ja so it's possible there's no l square computation for each a, okay so everything else works out quite reasonably, okay and now to go ahead and implement the entire receiver what should you do you should repeat this computation for each a and then pick that a which gave you minimum ja, okay so well this is not exactly ja this is this is I'll call it ja prime, okay so reason is it's not ja right the way I defined it this k is ja prime, okay it's that k but it's equivalent in terms of metric, okay so this is what we are doing so let's write down a few expressions for yk wrote down an expression it's what sk convolved with rho h of k plus nk, okay I know the statistics of nk here are not white anymore it is psd what sh of f right that's the psd for this, okay it's not white anymore okay so this is what I got what happens when I for zk can I write a similar expression for zk I'm convolving with the inverse of m star of minus k so basically I'm going through this filter so that will give me sk convolved with m of k, okay so you can see that that will cancel out and the gamma square will also go away and then the interesting question is what will happen to n, okay any guesses what do you think will happen to n psd square of, okay so think about what happens to n prime okay it's very easy to derive it it's not that difficult, okay it will work out to something like n naught by 2 gamma square or something and it will be white, okay so you can see that also, okay so you see why that happens so we'll see that as we go along later, okay so this 1 by gamma square m star of 1 by z star is doing wonderful things for us, okay seems to be okay do you see what will be the psd of n prime of k how do you compute it psd of n of k multiplied by 1 by gamma square m star of 1 by z star times what 1 by gamma square m of z, okay you'll see one of those things will cancel with s h of z you'll be only dealt with the gamma square you'll get n naught by 2 gamma square and it will be flat, okay right so psd is n naught by 2 gamma square and it's white, okay so maybe I did that filtering in a kind of a weird way to motivate the computations but what it's actually doing is it is getting rid of the anti-causal ISI that we saw not only that it's also making the white noise the colored noise white, okay so all these things are very nice and they'll play a crucial role in showing that such structures are optimal in some way, okay so we'll see all these things as we go along but this is the overall receiver structure that we have finally got and I want to stop here and pick up from here tomorrow.