 lecture what 34 okay so most of this class we're gonna discuss about the exam today okay so we'll go through the solutions and yeah we'll see if there's some time left towards the end I'll do some brief review of what we've been doing so far okay so first of all how did the exam go everybody's happy you know smiling okay so I saw briefly I saw a few papers as glance through them I think the first question most people have done something attempted something basically because probably because it involved simple geometry people able to do it but even the second part of the first question which involved a simple q function calculations several people have not done okay so that was that was quite surprising to me okay so it was actually is just quickly just looking at the picture and writing down q to q functions some several people have not done it was really surprising even if you didn't do the decision regions you can quickly do an approximate computation of better right so anyway let me see tell you what I expected from the first question then you can see how it works okay so the okay so so the constellation was minus 2 plus 2 okay so this is 1 comma 5 and minus 1 comma minus 5 okay so so so of course the perpendicular bisectors are going to be involved one of the perpendicular bisectors has already been given which is the y-axis right so one perpendicular bisectors the y-axis so you don't have to do any work to do it and and you can see that that perpendicular bisectors will be very useful right so it's separating two things that are very close by so of course that perpendicular bisectors will be very useful so you can't throw that out so I think most people have done that the next thing is when you draw perpendicular bisectors for this line and this line they'll have to intersect somewhere right so they'll have to intersect at one common point that's what the triangles perpendicular bisectors do you might have learnt it at some point in your life right if you draw three perpendicular bisectors they intersect at some point what's so special about that point what is that point called circum center so if you draw a circle using that from that point with the radius is the length of the something it will pass through all that so you must have I'm sure you must have learned that some point in your life right so that's so I think most people did that roughly but people are confused about that also I mean they're quite surprised to see that the three lines were actually intersecting I mean anyway they intersected you will get something like this okay so maybe my drawing is not to scale so that'll be the perpendicular bisectors there so this picture is exactly symmetric so you don't have to worry so much about the perpendicular bisectors on the other side they'll have to intersect so once you do this finish that's it the decision region is done okay so you don't have to worry about the other perpendicular bisector because it won't add to the decision region in any way this is the only perpendicular bisectors that matter I think I don't know I think it might have taken some time for you to figure out but I didn't expect it to take 15 minutes okay so maybe maybe you took 15 minutes some of you okay so how do you find this point alright we'll go to 11 by 5 a quick way of finding it is to see that this vector and this vector have to be this vector this vector and this vector will have to be what will have to be perpendicular right so the dot products have to vanish so first thing is to find this point can you find that point very easily that midpoint yeah so that point will work out to 3 by 2 comma 5 by 2 okay so then you can find this vector you can find both the vectors in terms of that point why it's a question of figuring out simple equation that you'll get 11 by 5 okay so it's an easy easy thing to do so this point will be minus 11 by 5 and your decision region is very clearly set okay so that's the first problem okay so that's part a for part b I'd given the bit assignments as 0 0 0 1 am I right 1 0 here and then 1 1 if you look at the second bit right if you so so quick way of writing it is to fix one transmitted point okay so you fix the transmitted point doing pairwise right so you fix one transmitted point for instance minus 2 is a good point to fix you see fix that point then you ask the question what's the most likely way in which I'll make an error my second bit minus 2 will become plus 2 so you have to only worry about that pairwise error probability okay if minus 2 becomes 1 plus 5 5 j or minus 1 minus 5 j what happens right if it becomes 1 plus 5 j there's no error in bit 2 if it becomes minus 1 minus 5 j there can be an error in bit 2 but that is a much larger distance pairwise so you can ignore that error in relation with this error so when you do approximation that's the approximation you do okay so you can quickly write probability for bit 2 as q 2 by sigma right so this this will work roughly it's a rough estimate of the probability of error so this is fine I think most people found this thing difficult to write which is very surprising you don't need to do the decision regions to write this answer right for bit 1 what is the answer same once again fix minus 2 so you see bit 1 to be an error you should go to either 1 plus 5 j or minus 1 minus 5 j so this distance is important what is this distance root 34 okay but there are two possibilities so we'll get 2 times q root 34 by 2 sigma this is fine I mean that's all this is what I expected for part b think several people have not written this I don't know I was very surprised it's the simplest pairwise computation straightforward approximation should take only two minutes for any not even two minutes I think it takes 10 seconds for two people to write okay so that's the first problem so so model of the story is parallelogram is difficult to slice it's not as simple as the rectangle rectangle will work out very nicely parallelogram is difficult to slice and so in some wireless diversity applications your transmit constellation will be intentionally made a parallelogram or it will become a parallelogram because of the way the channel works the different fading on different dimensions okay and so in those cases you have to carefully slice a parallelogram at the receiver okay a better alternative would be to simply rotate it undo the rotation make it rectangular and do your previous slicing just implementing the slicer and in the receiver is more painful you have to compute so many straight lines you might be just better off computing the distance to each of the point and picking the minimum okay so better way to do it is to rotate it and then do a rectangular slice okay any questions on this okay all right so these expressions are very very critical okay so you should you should be able to look at a constellation look at the mapping and very quickly write the dominant bit error rate expression there's nothing in it just find this nearest neighbor which will give you a bit error right q of d by 2 sigma there's nothing more to it right when you compare these two which will be smaller the two bit error rates which will be smaller which bit has a lower probability of error bit one right be smaller so you quickly say that because a q argument is larger for bit one the outside constant doesn't matter that's that much q goes much much lower okay so you see bit one is more protected than bit all right so that's the first question and I think more or less people did it okay for the decision region but for some reason writing these q expressions was difficult for many people it's just quite surprising should be able to quickly write it there's nothing in it is just a approximate term it's no big no problem I think the question also mentioned approximate right 26 which one so this distance is important is what you're saying this is to 26 it's okay I mean if you did so instead of 24 if you did two times root 26 also is fine all these two distances are different is it oh yeah you're right these two distances are different so this is what code is 26 doesn't matter I mean one of these things it's good enough okay so maybe root 26 is not too bad a number think of them 26 34 whatever you want to write it's fine just pick one distance and work it out okay all right so that's the first question second question I should say also surprised me a little bit as I thought it was a direct formula application for the standard form so H of Z is given as one so any term involving H will be one don't have to worry about it and but SN of Z was given as what one by three minus two Z plus Z inverse by two I think many people wrote this and then surprisingly there was amazing discomfort in factoring this case like I couldn't believe that people think out of 40 people five people wrote one root of this I was quite surprised it's it's a quadratic equation I think if you can't factor a quadratic polynomial I don't know I don't know where to start okay so it's a serious problem I think it's it's more a crisis of confidence as opposed to anything else I think you're some for some reason you thought this would be a ultimately difficult spectral factorization it's just solving a quadratic equation there's nothing in it if you write it down what will this work out to 4 by 6 minus no 4 my is it for no to write sorry 2 by 4 minus Z square minus Z power minus 2 okay so I'll show you how to do this I think I should show that once because this may be maybe one step which is slightly counterintuitive so one way I like of doing it is I don't like factoring polynomials like this I always like it in the standard form so a good idea is to simply take minus Z power minus 2 out so then you get Z power 4 minus 4 Z square plus 1 okay so many of you know how to factor this polynomial okay so it seems like many people didn't know I think there's some pages after pages were written before they wrote down the final answer okay so very surprising and I didn't I didn't expect it when I wrote when I and I said the question paper I didn't expect people will take more than two minutes for this okay so if you took more than two minutes I don't know what to do okay so this will factor as 2 minus root 3 and 2 plus am I right 2 plus root 3 okay so how do you do this once again for people who are not very clear how do you do this factoring to use the age old formula minus B plus or minus square root of V square minus 4 AC by 2A okay so apparently the BBC did a survey of people who learned the quadratic formula and found that only 1% of the people who ever learned it actually use it so now I know why it's not because they don't have any use for it it's because forgotten maybe I don't know okay so it's as simple as this so now there is a slightly confusing thing to get it into the form that we want okay so what's the form that we want we want positive constant times a minimum phase causal times that minimum phase is maximum phase so so it's very simple the next identification is to figure out which of these roots are inside the unit circle okay so once again I think many of you have calculators it might be easy to calculate the 2 minus root 3 is less than 1 okay when you take its square root also it will be less than 1 okay so you can quickly figure out that this part has to correspond to the first term here has to correspond to the minimum phase so definitely the other term has to correspond to a maximum phase right so it has to happen because this is a valid power spectral density it has to factor like that so it's actually enough pretty much if you find the minimum phase component the maximum phase component will work work itself out and then you can figure the constant out you can do it in various ways but I'll show you how I how I did it so it's not too difficult so I want this to be 1 minus some constant times z power minus 2 so it's good to multiply the z power minus 2 out that way so you get 2 minus root 3 z power minus 2 okay then you have another term which has to definitely work out to what 1 minus 2 minus root 3 z squared it has to work out to that you can either write it that way and then figure out the constant or you can directly also figure out what the constant has to be so you'll see this plus sign has to go to the other side it'll become 2 plus root 3 minus z squared and then what you should do what should you do just a question of making this monic so pull the 2 plus root 3 outside you get 2 plus root 3 times 1 minus 2 minus root 3 z power minus 2 so maybe this is a cost for contribution so 1 by 2 plus root 3 but this is 2 minus root 3 what is the solution for that what do you do you have to make the denominator rational right so you multiply and divide by 2 minus root 3 you'll see amazingly this becomes also 1 minus 2 minus root 3 z squared okay so some reason I think many people were not confident enough to do this that's very surprising to me so maybe your DSP has been learnt and forgotten for a long time but but you should be able to factor polynomials so it's very simple polynomial I didn't I didn't really expect people to be stumped on this but I saw a lot of people are working very hard at it writing complicated equations to do this so I don't know okay so that's the first thing so you're able to quickly identify gamma n squared mnz and then this becomes oh well these are all 1 by right because it's coming okay so let me let me write the proper thing out and then write the denominator so this is just the denominator so I should remember when I factor sn of z these things are going to go to the denominator so I'll get 2 by 2 plus root 3 times 1 by 1 minus 2 minus root 3 z power minus 2 1 by 1 minus 2 minus root 3 z square is this okay so you see this becomes gamma n squared this becomes mnz this becomes mn star of 1 by z star okay so there was a problem involving cos square omega factoring in the one of the tutorials okay so hoping people will remember that also when they did this but seems like it's been forgotten okay so once you do this the other factoring is also very easy so when you find sz of z what do you get okay one more thing was what what do you take ES what do you take for ES one just take one don't keep ES as a constant somewhere so it's okay doesn't matter doesn't make a big difference and some people are taking ES I know I didn't mention it explicitly in the paper but I was hoping somebody will take it as one many people took it as one but some people have taken it as ES itself and worried about it but anyway we'll see so you'll have mod h square which is just plus 1 plus sn right so sn will work out as 2 by 4 minus z square minus z power minus 2 so when you do this you see what you get for sz is 6 minus z square minus z power minus 2 by 4 minus z square minus z power minus 2 so one factoring you've already done so it's just a question of repeating the same thing for 6 minus z square minus z power minus 8 so if you do that I believe if I remember correctly you get something like 3 plus root 8 am I right okay so you get 3 plus root 8 times 1 minus 3 minus root 8 z power minus 2 1 minus 3 minus root 8 z square divided by 2 plus root 3 divided by 1 minus 2 minus root 3 z power minus 2 divided by 1 minus 2 minus root 3 z square okay so once you do this it's very easy to identify this as gamma z square this as mzz this as mz star 1 by z star that's it so it's it's after this it is just straight forward formula application so you do zero forcing DFE MMSE DFE you know exactly what the precursor is what the pros cursor is just plug it in you'll get the answer structure for the zero forcing DFE and the MMSE DFE and I've also asked DFE so even the mean square error is trivial to compute once you do the factorization right so the mean square error for ZF DFE will work out as what gamma n squared which is 2 by 2 plus root 3 okay and for the MMSE DFE you will get 2 by 3 plus root 8 okay if you do that calculation so you see clearly this is less than this okay so you can quickly see that but if you see the precursor for MMSE DFE it will be a much more complicated filter than the precursor for zero forcing DFE which will be a simple well three tap filter okay the post cursor is okay for both of them but the precursor for MMSE DFE will be a complicated poll zero thing in fact it might even be a IAR anti-causal stable version so it may be one of those things so the precursor in the MMSE DFE is more difficult to implement okay so that's the only thing you have to make okay so this is what I was expecting I think I think mostly it was some confusion on people's part maybe you didn't believe that this question is as simple as it is okay so I don't know I think for some reason the speckle factorization defeated many people and I was quite I was quite surprised at this okay any questions on this okay hopefully it's okay so one more thing is basically the some recent digital communication is considered to be a tough course and people go into it with lots of confusion it's actually very simple it involves high school geometry and high school algebra there's nothing more to it okay so anything that's more complicated than that nobody does it's only the factoring polynomials and drawing decision reasons no you don't need any background all the terminology around it is completely confusing but ultimately you're doing only that okay so I have a lot of confidence when you go into the exam all I don't I don't ask complicated questions I'll never ask questions for which we have to do 10 pages to find the root it'll never happen so if you find that you're writing pages after pages then something is wrong go back and check see if there is or maybe the question is wrong ask me what the right question was okay so don't go on and on and do for 30 minutes and do that so and unfortunately the last question pretty much nobody attempted was there anybody here who really attempted the last question I think nobody really even looked at it so I'll have to do some serious thinking about how to grade that but anyway so let me let me do that the last question also there is a major red herring in the last question as in this is big description of g of f and all that but I just gave it for completion you don't need the g of f okay so this you have to look at it carefully and see what the meaning of that question was okay for instance look at the first part of that last question what is the white end match filter right the match filter you can write down without any thinking there is no reason to worry about the match filter what is the match filter g of f times c of f conjugate that's the match filter there's nothing more to it right that people might have been able to write but I saw very few people even who wrote down that okay I think for some reason they were totally confused about the folded spectrum right away you don't have to worry about the folded spectrum when you write the match filter match filter is just h star of f just in this case it's just h of f everything is real so that's all you just write h of f is g of f times c of f which is what 1 plus e power minus j so if you're worried about implementing it how do you implement it you might want to do it in with some delay lines in analog or you can even do it digitally by interchanging etc but that all that is different but match filter writing it down is very simple well there is a I should be careful let me take back whatever saying c of f is actually complex you'll have to do c star carefully there okay so h of f is what is g of f times which is real times c of f which will work out as 1 plus I can't remember the constant I think 5 by 2 was it 5 by 2 e power minus j 2 pi f t and then the next one was 1 is it next one was 1 e power minus j 4 pi f t okay this is h of f so the match filter is simply h star of f this would have given you like some three marks you know very few people have written it but I'll see I'll see what to do about it so it says g of f maybe g star of f is for comfort 1 plus 5 by 2 e power j 2 pi f t plus e power j 4 pi f t yeah yeah this is the match filter zone no this is part of the white and match filter right so the first part is the match filter the next step comes the whitening right so do it step by step first step everybody should be able to do right you know even if you can't figure out what the folded spectrum is you can do the match filter that's what I'm trying to say match filter you can do very easily just close your eyes and write the formula for the match filters nobody will question you for that the next thing is to find the folded spectrum and factor it okay yeah maybe there is something non-trivial there it's also not very non-trivial but it's something something there find the folded spectrum so that division you should know the match filter does not require any spectral factorization the whitening requires the optimal whitening requires spectral factorization right so this people must have been able to write so think everybody started with the the response after the match filter automatically assuming the h star of thinking that it's not very trivial okay so anyway so how do you find the folded spectrum after the sampler the match filter okay so you know already what the formula is right so it's going to be if i write say so v part j the folded spectrum v part j 2 pi f t is going to work out as 1 by t what summation m equals minus infinity to infinity i'll do it in one way and then i'll give you an intuition for why that answer makes sense okay so h of f minus m by t squared okay so keep it in this form don't simplify any further than that so if you do this you'll see you get 1 by t summation m equals minus infinity to infinity mod g of f minus m by t squared times what times simply mod cf squared why will m by t go away if you put f minus m by t what will happen to these things it's all symbol rate events okay so they will not change when you fold and alias okay so anything periodic like that will not change it has to make sense right and the same thing is going to happen after t after 2 t and i'm doing everything at symbol rate of the receiver so if i take care there's no isi with my transmit and receive filters so only isi should be caused by the channel's delay okay if the channel is not at symbol rate then you have trouble okay like you do in the wireless multipart situation okay then you have to do all kinds of equalization but if your channel tabs channel is doing just something at symbol rate you can clearly do very simple equalize no reason to do anything else okay it's not doing anything else so that's the so this is a very mathematical way of showing it but intuitively it should be cleared you it's enough if you kill the isi for one one path from the channel the other parts will also get cleared simultaneously the only thing that will be left is what the channel introduced at symbol rate okay so you don't have to deal with that so you pull the mod c of f square outside what will happen to 1 by t summation that it'll become 1 so you simply get the folded spectrum as mod c of f squared okay so now what is s of z okay so it's c of z times c star of 1 by z star so you'll get 1 plus 5 by 2 z inverse plus z power minus 2 times 1 plus 5 by 2 z plus z square okay all right so so this step is a little bit tricky okay so I'm not saying it's very trivial you should have understood some very very many things about digital communication to make this jump but if you blindly apply the formula you get the answer still there's no problem and if you know how to manipulate it you still get the answer okay so that level this question is also not too complicated okay if you had a if you were confident about the problem and if you see the problem you would have done it there's no problem I think if you're already worried and upset and thinking of so many things it's tough to do okay so that's the problem so next time you write this exam first of all believe that you can do the problem and I don't ask too many difficult things it's very simple things that I'll ask just write the formula down the answer will follow okay and have some trust and faith in the way it is the question okay so you write this so now what should you do what's the next step factor so what do you need to factor this okay once again you need the quadratic thing okay but you have to be careful here okay so here also the terms will come out very carefully I believe this will factor as 4 times 1 minus half z inverse squared 1 minus half z squared okay so you might want to check this okay so I might have made some mistake but it'll factor something something like this okay so you identify this as gamma z squared gamma squared this becomes your m star of 1 by z star and your y-turner is 1 by 4 1 minus half z squared that's your y-turner and you see your y-turner is what it's it's not a good filter it's a AR anti-costal right so but if you do it then you get what you get the minimum phase channel response to be exactly this 1 minus half z inverse square so the whole channel the equivalent channel is simply sk going through oh one plus it's plus is it it's the same as this no no so he's pointing out he's factoring this okay so be very careful lot of people gave this answer it's really disturbed me the minimum phase part of the factorization has z and z inverse does that make sense okay minimum phase the minimum phase part of the spectral factorization has to be monic causal and minimum phase moment you have z inverse and z what does it mean it's got something else it can't be minimum phase lot of people factored the 4 minus z square minus z power minus 1 with the z in the answer for the minimum phase okay so that doesn't make sense so what he's doing is also something like that he's got a zero inside the unit circle and he's got a zero outside the unit circle the zero outside the unit circle where should it go it should go to the maximum phase part it cannot be in the minimum phase part okay so that's how you do it so am i right i think this is fine the sign was wrong of course but i think this is fine okay so you will get 1 minus half z inverse square and then noise this is your z k all right so now you do the match filter bound and the and i think i asked zero forcing dfe the it's very easy all those things are great so you can do those things in just two steps for this problem okay so this was the third question and okay so i'll tell you what's disturbing me about this whole exam though so it's fine i mean the actual specific answers might be okay but overall i think people have close to zero confidence that they can do something in digital communication okay so i think i'm sure that a few people who are not like that but most people seem to be very unsure of themselves when they look at a problem which is really which is very scary for me because i think i think the way i'm doing this course i'm just telling you simple ways of doing the problems right so ultimately that's what i'm doing very simple ideas for suppose this is a channel the post is a transmit filter this is a channel this is a receipt what do you do what do you do what i've been doing suddenly for some reason i think maybe the problem was spectral factorization some of you think you're convinced the spectral factorization is completely complicated you don't have to use any root finding methods okay so that's the that seems to have stumped most people okay it's also i mean people were saying it's difficult to do it as part of a dsp course so maybe the thing to do would have been for me to spend some time i should have probably spent some time on spectral factorization in the beginning i thought i did but i maybe i didn't do it in this specific way to show you how it can be done in a very simple way without too many complications okay so so so so i think that's that pretty much covers it probably the only thing that that requires some care is this third problem part a but other than that the other things must must be okay okay so any questions any questions so people are okay everybody's smiling and happy and i don't know i saw a record number of people show up for the exam was 40 people everybody everybody was there class usually there's hardly at least the week before the exam they were hardly like 10 15 people so suddenly surprise us all a lot of people have never seen before why does this keep going from as usual they're not here in today's class also so let me do a rough count i'm 21 yeah 29 25 28 27 people in class this is the usual number anyway okay so so so so so yeah so one question to ponder about this what happens what happens if these guys are not are not at symbol time okay suppose this this is some tau 1 and this is some tau 2 okay so this is exactly the problem that people deal with in wireless communication so when you're moving for instance in mobile communication whatever you transmit you pretty much get this channel you get a channel which does what's called multi path the same signal will come to you at different times you don't see anything else so but but the multi path is not guaranteed to be at symbol symbol time okay so it'll be some other time so then look at what happens to your folding spectrum okay so then see what you should do and that'll give you a lot of ideas for how to work with the work with the multi paths but the current popular way of working with it is slightly different what's the very popular way of working with these multi path things so something called wife dm which is used all the time so it'll take care of all these things in a very nice way okay so that's that's one more thing maybe we'll see it if we have some time but but that's a different problem it changes the problem all together once this is not multiple of symbol time these factors won't go away they'll stay for every term and this spectrum changes folded spectrum changes completely so what you have to do is very different when the multi path is not at symbol time okay all right so that's pretty much the only comment I wanted to make so I don't know about this question so is this is this okay this is okay right so so in future you might get questions where you have to do is that level of spectral factorization so this will be the only thing that will be required nothing more so don't worry about if there are some z terms here all things become more complicated but it's very unlikely that I'll ever ask such things in an exam okay so maybe in an assignment when you have matlab or something I might ask it but if you don't have matlab very unlikely that I'll ask okay don't worry about it okay and uh yeah so this problem requires some care so these are all approximate expressions so so I mean I don't want to draw the actual structure for the zero forcing DFE and the MMSE DFA I think you can figure it out it's just a formula you know where the what the filters are should be fine here also it's okay I think the match filter bound is okay so I think that brings us conclusion of quiz two quiz two discussion so let's move on to so what I want to do for the rest of this class is briefly review what we've been doing in equalization okay so the last thing we were doing was what what was the last thing we were doing we were doing adaptive equalization right so so let me see so I should should try to remind you what the okay so so overall the way the way most equalizers work okay so this is the the adaptive equalizer is how most equalizers are actually implemented of course in OFDM you do something else but in if you want to implement a time domain equalizer mostly it's done like this and the principle hopefully you remember so if you have a if you have what so what is the model I think did I did I pick any nice model for the whole thing let's just pick the model okay so suppose you have so the way the whole model works is you have an SK at the transmitter you put it through a transmit filter right this is a symbol rate then it goes through the channel okay of course there's an upconversion which I never show okay and then there's a channel which is the baseband equivalent okay and then the question is what do you do for processing at the receiver and usually like I said in most cases most people would do g star minus t okay so you match it only to the transmit filter because that's the only thing that is known you don't know the channel so usually you match it to the transmit filter or you might even not want to match it to the transmit filter you simply want to do a low low pass filtering that also might be okay so you could do that also some ways of doing it but you can also match it to the transmit filter that's one possibility there's no no no thing wrong with that and then you do symbol rate sampling okay suppose you have a scenario like this okay this is one way of doing it remember there are other options here for the match filtering you can do if you know cft you can do the optimal match filtering or you can do something other so many other possibilities there or low pass filtering is always something that can be done followed by a symbol rate sample okay so overall if you imagine this is zk okay so noise gets added here I'm sorry somebody said I forgot noise okay so overall from sk to zk you're going to have a channel which is which is similar to the generalized equalization channel that we had right so you can expect that this will be some h of z noise added uh nk zk okay so whatever you do here this could also be a low pass filter okay remember okay so you want to pick this uh match filter in some way so you do that once you pick it you actually have a expression like this so you can imagine you're actually trying to equalize a finite tab channel roughly most cases it'll work okay so that's something that's always done so you try to equalize a channel okay so what do you do first thing we saw was what do you do if there is no constraint on the complexity of the equalizers and all these things so you we derived a whole bunch of things zero forcing dfe zero forcing linear equalizer uh mmsc linear equalizer mmsc dfe all of them had some implementation problems they were all okay we were able to compare it and they worked okay so usually the mmsc dfe is a good option okay if you can implement it or the zero forcing dfe if you cannot implement the mmsc dfe you do the zero forcing dfe so basically if no if there's no constraint on complexity of your filtering you want to do mostly say zero forcing dfe or or mmsc dfe okay so that's pretty much a very good strategy to use and like I said the mmsc dfe maybe the unbiased mmsc dfe has some canonical structure which people have shown is capacity approaching also that's not that much suboptimality in doing that okay so one can do the mmsc dfe quite safely but in practice you're not going to have such things and you don't even know what the h of z is so when you don't know h of z okay no constraint on complexity you know h of z okay so the next case we considered was finite tap equalizers as in both the precursor as well as the post cursor is finite tap if you're doing only precursor no post cursor you pick your precursor to be both sided right so you pick minus l and plus l minus n plus n i forget i think minus p and plus p i took so you pick something like that and if you know h of z you can solve a linear equation to get the optimal taps i fully set it up for the linear equalizer case for the dfe also i showed you how to set it up but i well i didn't really show you how to set up the optimal thing for the dfe we only saw the adaptive version right but anyway for the dfe also you can set up a similar thing so maybe an assignment problem i'll give for setting up something like that so you know h of z means you know the statistics of zk once you know the statistics of zk it becomes a classic estimation problem finding the signals that were sent and it's very very simple and easy to see the second order things are important for the minimum mean square error criteria and you get this phi inverse alpha okay so you do that you get the linear equation solving okay the practical case the really practical case is when you have a finite tap equalizer when you want a finite tap equalizer and in addition don't know k don't know h i'm sorry in which case you won't even know the statistics of the zk that are coming in so the estimation problem becomes that much more complicated you'll have to also estimate several things in the way okay you don't want to do a very complicated big estimation so you simply do a iterative solution which is a very nice way of doing it what what i showed was this gradient gradient search or lms algorithm right so the lms is the is the best way of doing things the way it works the structure is roughly as follows so you have zk coming in so i'll show for the dfv type situation you have a precursor equalizer which i think what was my rotation for the precursor c of z c of z okay and then you have plus minus our slicer here and then you have say d of c okay i'm not sure my notation might have been complicated okay so you usually pick c of z in a dfe case to be what of the form m equals minus p to 0 cm c power m okay well i don't know i think 0 to p maybe okay so you see so this is this is uh precursor is chosen to be l one sided right in the anti-causal direction and d of z is going to be chosen as m equals say 0 to p if you want or say p prime okay dm z power minus so it's going to be picked as oh no i think this should be i'm getting confused should be minus m again okay so you pick d of z to be something like this okay and then you set up a set up one vector which includes all the coefficients of c and the coefficients of d okay and and you set up your iterative method for the lms okay so i might have written it down i forget probably i don't have it here okay so okay so maybe i'll write it down maybe i wrote it down somewhere else don't seem to have it here the last class i might have might have written it down okay so the basic thing to keep in mind is initially when you start out c of z and d of z will have to be randomly chosen values and there's no reason why you should get good performance so usually when you do when you don't know h of z and you have to adapt to an unknown h of t there will be two distinct phases in your receiver in one phase the first phase which is called the training phase you have to assume that you know what is sk okay so this output of the slicer which is s hat of k this is known during the training phase okay so depending on how complicated or how much you can afford you have a long enough training sequence okay so a sequence which is known both to the transmitter and the receiver and which is transmitted at first by the transmitter so you might want to pick say hundred symbols thousand symbols ten thousand symbols how many of you need depending on the number of tabs you have and how you how well you want to adapt you will first send that okay so the slicer pretty much will have to work but you won't use the s hat of k from the slicer during the training phase during the training phase you will use the symbols from the known sequence okay and compute the error okay the error term is crucial in your in your adapting right so if you look at the formula this this well the formula works roughly like this okay so let me see see if we have the formula properly written down okay ck plus 1 the ck plus beta times pk zk star right am i right this is how we wrote down and the ck will actually be will actually contain both c minus p all the way to c0 then it will also contain well this will be m equals 1 right so d1 to dp prime okay it will be a vector which contains all your coefficients and you'll have to compute the error also suitably okay so you'll have to adjust everything according to how this computation is done okay so this ek during the training phase will be computed only with the known sequence and you move to the next phase is called the decision directed phase when you believe that you've acquired the channel reasonably so it's called acquisition channel acquisition during the training phase you acquire the channel now you have acquired your c and d to a acceptable fashion and then you move to a decision directed phase where you actually use the output of the slicer for computing the error okay so you do that and you run your uh equalize okay so that's that's one way of doing a time domain equalizer in practice okay so this beta is also a thing that has to be chosen it's mostly a fudge factor so you you just pick something okay so something like 0.1 just pick that number then if you see that you're not converging or things are going all over the place you either decrease it or you increase it if you see that you're converging very slowly okay so you do one of those things adjusted according accordingly and you can do that okay so that's how you pick your beta in this way you can adapt your channel equalizer taps and get to the channel that you want and after that move to the decision directed phase okay in the decision directed phase your equalizer will vary if the channel varies slowly right if it varies too fast it may not be able to keep up with it but if it varies slowly your channel equalizer can still keep up with it okay how much the variation and all that we can quantify using this eigenvalue stuff okay so how fast it varies there are some bounds how quickly it converges right the convergence speed is decided by the eigenvalues of the of what the autocorrelation matrix for zk okay so you do zk's autocorrelation matrix it's eigenvalues and the spread of its eigenvalues will determine how quickly your equalizer converges so that will also in turn determine how quickly your channel can vary how much you can keep track of so all these things are I mean there's no strict way of properly analyzing them but some rough way of getting a feel for how that works and eigenvalue spread also depends on what the channel responds okay so I gave you some bounds so the channel is very widely varying then you can expect the eigenvalues also to be very largely spread okay so that might be an interesting thing to look at okay so this is this is where we are and so quickly running out so I'll give you a brief overview of what we are going to do for the next few classes the next few classes the first thing I'll consider is a very practical receiver which includes many of the blocks that you'll see in a practical receiver okay so things like career career synchronization timing synchronization all these things I've never talked about okay so all those things I'll try to introduce very briefly you're very high overview of what's done we won't go into great details here most of that is still considered an art okay so you just do it anyway you want you can do it nobody will question you too much there's no optimality as far as that's one thing we'll see and then the next thing I want to see is briefly look at some some more practical techniques techniques like things like DPSK I don't know if you've heard about a differential phase shift keying and a few other things which are which are useful in practice some things we have never seen but some nice and smart ideas which which help you avoid some problems in practice okay so that's one more thing we'll see and after that time permitting we'll see some OFDM and after that time permitting we might see some simple coding so with that we'll pretty much sign off with this course okay so hopefully I'll be able to return your answer sheets by tomorrow or day after and after that you can see how it goes