 So, this is lecture 20, right, ok. So, right now we are looking at the ISI case when you cannot assume that your channel is ideal anymore and Nyquist criteria is not satisfied by the received pulse spectrum. So, you will have ISI necessarily, ok. So, I am going to once again draw this picture, ok. So, you have a symbol sequence SK. So, you usually think of this as a finite sequence say length L symbol sequence, ok. So, the transmitter first thing you do is send it through a transmit filter, ok. So, you send it through a transmit filter and then in reality if you are doing passband transmission what should you do next? You should do upconversion, right. So, imagine G of t is I mean S of k can be complex. You usually think of G of t as real, but S of k is complex and you do an upconversion, you go to some frequency and then your channel actually is there in that frequency and then you do a downconversion back. So, instead of doing all that we do a passband equivalent picture where we simply say we have a complex baseband channel C of t, ok. So, this is complex baseband, ok. And then noise gets added to this, ok. Once again this is actually the complex baseband equivalent of noise, ok. So, all these things are there, one can do that, alright. So, here this is your received signal R of t, ok. So, we initially started off by saying what do you do with R of t? We thought of doing a minimum distance receiver from there we were able to see that convolving with h star of minus t seems like a very natural thing to do and sampling at sample rate. And then we also later on saw that if you want to do an optimal projection onto an orthonormal basis also you can convolve with h star of minus t, then sample at symbol rate and then do a discrete time filter, ok. So, that is the those are all equivalent ways of thinking about it and I am going to do that next, ok. So, the first operation that you would want to do is to filter with h star of minus t, what is h of t? And sample at signal rate, symbol rate, ok. So, this immediately means c of t should be known at the receiver, ok. Otherwise you cannot hope to implement h star of minus t at the receiver, ok. So, typically the way I have written it down it looks like h star of minus t if at all it is implemented in the receiver it is going to be in analog, ok. So, you never do things like that, how do you do it? Oversample R of t, sample it to a very large amount and then you do even the filtering by h star of minus t in digital, ok. So, you never do these things in analog because you cannot control the filter shape and all that very easily, right. So, you do k t. Of course, once again the sampling here there might be a delay to accommodate for various things, ok. So, you do that you get y k, ok. And this y k has a very simple description the way I wrote it down, ok. So, y k will actually be sk convolved with the autocorrelation of h, ok. So, it has is i on both sides, ok. y k will depend on sk sk minus 1 sk plus 1 sk minus 2 sk plus 2 all these things in a symmetric way rho h of k, ok. So, it seemed like one thing to do to get rid of the anti-causal is i was to filter with 1 by gamma square m star of 1 by z star. How did I get that? So, you look at the spectrum of rho h of k, ok. So, you see rho h of k is actually what it is, it is h of t convolved with h star of minus t sampled at k t, ok. So, that is rho h of k. You do a Fourier transform discrete time Fourier transform of course, you get sh of e bar j 2 pi f t, ok, which would be a valid power spectral density for which you can do spectral factorization, ok. You would get spectral factorization to get a minimum phase m, right, ok. So, this actually corresponds to m k which is monic causal and minimum phase, loosely minimum phase. I am going to simply say minimum phase. Remember that the zeros can be on the unit circle, ok. So, what was gamma square? Gamma square if you write it down will work out to exp 1 by 2 pi integral minus pi to pi log sh of e power j theta d theta, ok. So, this is the formula for gamma square, ok. So, it will work out like that, ok. So, if your sh of z is rational then you can do the minimum phase factorization in a simple way, ok. So, you simply take all the poles and zeros in m of z inside the unit circle and then half the zeros on the unit circle into m of z and the remaining will automatically come in m star of 1 by z star because sh of e power j 2 pi f t is real and non-negative on the unit circle, ok. So, those things will work out. And gamma square you can choose suitably. Here the gamma square will work out to this formula, all right. So, if you see all this then you notice that you can do a filtering here with 1 by gamma square m star of e power j 2 pi f t, ok. So, however small I write it it ends up over shooting. So, I am going to simply say m star of 1 by z star, ok. So, hopefully hopefully it is clear to people m star of 1 by z star works out to this, this filter and out of this filter you get z k, ok. So, we saw that this entire operation filtering by h star of minus t sampling and then filtering with 1 by gamma square m star of 1 by z star was nothing but doing an orthonormal projection, right on the signal space corresponding to the signal received at r of t, ok. The signal component r of t defines the signal space on that signal space if you do an orthonormal projection you would be doing the exact same thing. But the difference from what we did before was z of k will still contain i s i you cannot get rid of i s i because your pulse does not satisfy the Nyquist data in general. So, it cannot get rid of the i s i in total. So, z of k will contain i s i now but only causal i s i because I picked my m suitably, ok. So, z of k, z k you can show this is very important, z k you can show is s k convolved with m k plus n k. Remember all these guys are complex now. See s k is complex, m k is complex, n k is also complex, ok. So, each component of n k will be i i d normal it means 0 and variance n naught by 2 gamma square, ok, right. Each in each dimension n of k will be like this, ok. So, that is the orthonormal projection you project into project with an orthonormal basis every dimension will be normal with 0 mean and variance equal to the variance that have that was there before, ok. Initially you might think the variance is different, but after you filter with h star of minus t you have a gamma square entering the no actually it becomes colored and then you have a 1 by gamma square. So, that will multiply it out and you will get it. So, but those constants are irrelevant do not worry about this n naught by 2 gamma square, ok. Only in an exam problem it those things will become relevant, ok. In reality it is some noise with some variance it is the only thing you worry about, ok. You measure it in practice and then you know what your SNR is, ok. So, nothing it is not too significant there, ok. So, this model is very very crucial. So, what people do usually is they say they will denote, ok. So, you remember what we did for the non ISI case. We went from a received transmit constellation directly to a received constellation from SK to SK plus noise, ok. So, we went there directly. So, now people usually bunch up this whole thing together and get what is called a discrete time model for ISI, ok. You go from SK to ZK directly and the only thing that happens from SK to ZK is what? Filtering by MK and addition of noise which is IID 0 mean some variance, ok. So, that is the discrete time ISI AWGN model and this is very very popular in practice, ok. So, you will see most people when they do simulation studies and in fact, even in practice you always work with this model, ok. You do not look at the whole system in complete detail. If your channel is known at the receiver you can equivalently work with this model, ok. So, where S of K goes through MZ, MZ is what? Monic, Causal, loosely minimum phase, ok. So, MK is that and what you get out is going to be SK convolved, ok. You have to add noise, sorry, followed by addition of noise which is NK to get ZK which is SK convolved with MK plus MK, ok. So, this guy is IID normal in each dimension, ok. So, with may 0 mean and variance sigma square and this is the BK, ok, which is received value, received symbol value, signal value. In the non-ISI case, we simply had S of K plus noise. Now, you cannot do that. You will have S of K convolved with MK, ok, right. MK is Causal and all that. So, this BK can be written in a very nice form, ok. So, you can write it as K equals 0 to infinity, M, ok, L I think, L equals 0 to infinity. I want to write it so that I get, sorry, let me write it this way, ML, S, K minus L, ok. I can do this because I know MK is Causal. So, only the positive things will come, ok. So, you notice B of 0 contains components from S of 0 and something before and we did not transmit anything before S of 0. So, nothing else will come, ok. B of 1 will contain S of 1 and then S 0 as well. B of 2 will contain some component from S 2, S 1 as well as S 0, ok. So, of course, you cannot observe B of K directly. You can observe only B of K plus noise, ok. So, that is the only thing you observe, but B of K is the received signal value, ok. So, now it is possible to draw things like received constellations, ok. So, a transmit constellation was nice S of K and say some BPSK, QPSK or 16 QAM. Your received constellation, if you plot the BK, will no longer be the same thing. So, it will be something else. We will try and we will try that exercise soon enough, ok. We will see some examples where you plot the received constellation and it will be different, ok. But one thing I know definitely is the noise that gets added to BK is IID Gaussian 0 mean, ok. So, that those are all nice things to do which we can use in our detector, ok. So, that is the overall idea. So, most simulations that people do for ISI is AWGN. It will always be in this model. Nobody is going to simulate that whole thing with G of t and C of t and H star of my machine. It is not necessary because the whole thing is equivalent to a discrete time model and a discrete time model is much easier to simulate in software but, right. So, you just and it is all symbol rate, ok. So, you do not need to do too much of, do not need to have too much memory and all. So, this is how people usually simulate, ok. Any questions on how this model was derived, ok. And the crucial assumption of course is you know H of t, yeah, right. If you do not know H of t, you cannot really do it with the receipt. So, if you do not do that, then your model will be different. We will see those cases later. For now, we will assume it is known at the receipt, ok. Any questions, any comments, it is ok. Everything is fine. So, the next thing I want to do is to look at, ok. So, I want to look at the detector, but before that, let me just quickly do a few more things that is common in practice when people think of simulating such things. So, what is usually done is when you are thinking of a channel M of z, you want to normalize M of z so that the energy is 1 in M of z, ok. So, there is a good reason why you want to do it. So, typically normalize such that summation over k mod mk squared equals 1, ok. So, the reason why you want to normalize this is now energy in sk will be equal to energy in what? Bk, ok. So, once you normalize this Mk to have unit energy, there is no energy artificially being introduced in your model, ok. And that will cause some confusion because if you for instance scale by 10 to get from sk to Bk, it is not right, ok. So, it is those noises not getting scaled. It is just scaling this and that does not cause all kinds of confusion, ok. So, in your model. So, it is always good to normalize your channel in the ISI model to unit energy and when you do not have to worry about differences in transmit constellation and receive constellation energy. It is all normalized and it is nice. So, you do that, ok. So, this will make this will make the. So, if I do this one property of M of z might be lost, what is that property? By the normalization. Monic. Monic is the only thing that will be lost. Other things will not be lost, ok. So, one by scaling nothing will happen, but the monic property might be lost, ok. So, that is the only thing to keep in mind. So, this is done to make energy of sk equal to energy of Bk, ok. So, one thing you can do for instance if you are worried about SNR, how do you define SNR now, ok. Energy in s divided by 2 times sigma square. The reason why I am doing 2 is what? Because I want to account for both dimensions. When I do energy in s, I am going to find energy in 2 dimensions. So, I want to divide by noise energy in 2 dimensions. But even if you do not put that 2, the only thing that will happen is your entire curve will shift by 3 dB left or right. But if you plot uncoded versus coded or something, comparisons will still be valid as long as you stick to the same model, ok. So, it is not a big deal, but it is usually there is a 2 factor, ok. So, SNR is defined as in your model E s divided by 2 sigma square, ok. So, if you want to think of it as N0 by 2 gamma square, then it will become N0 by gamma square in the denominator. So, gamma also enters the picture if you want to go back to the continuous time model. But like I said, it is not too important to worry about those kind of these things. All right. So, now let us see a few examples of this discrete time model and how things work out. The reason why I am doing it is I have to solve the next problem, ok. So, I have not solved the receiver completely. What is the next step? Detection, right. So, if I have got the z case, I have to now detect my s case, and it is not the same as before. I cannot do independent detection for each k because I have got ISI, ok. So, I have to detect for the whole thing together. So, how do I do the detection is the next question. For that we have to understand the discrete time model very well, ok. So, for that we are going to see a few examples and then I will give you a nice view of the discrete time model which will help us move to a detect, ok. So, that is the kind of road map roughly. So, let us see. So, here are a few examples. First example I am going to take for mz is 1 by root 2 minus 1 by root 2 z inverse, ok. So, in fact I might even violate the minimum phase criteria, ok. So, when I pick mz, ok. So, that is because I do not want to think of minimum phase when I come up with examples. I want to be able to just quickly come up with examples. In reality, you might have a minimum phase thing, ok. So, I will just quickly come up with examples. Do not shout at me if it is not minimum phase, ok. So, now what is, ok. So, couple of other things I need to define, ok. So, if m of z is this, z k becomes what? z k becomes 1 by root 2 s k minus 1 by root 2 s k minus 1 plus n k, right. This guy is my what I have called as b k, which is the received signal value, ok, after the just the ISI, ok. So, this is nice. So, if I want to now get a hold of the received constellation and all that, I have to fix the transmit constellation, ok. So, my simple example, first time I will take x to b b ps k, ok. So, which is the simplest example one can take, ok. So, if it is b ps k, what will be the received constellation, ok. I am drawing two axes, I actually do not need two axes, it is only one dimension, it is all real. But what will be the values when the received constellation? Yeah. So, there is a 1 by root 2 floating around. So, just be careful about that. So, s k and s k minus 1 can take 4 different possibilities. So, technically there should be 4 different points in your received constellation, but some of them might overlap with each other, ok. So, less than or equal to 4 points that I know for sure, ok. So, after that is just a question of substituting the different possibilities and finding different b k's. If I substitute both to be equal to 1 1, what will I get? 0, ok. So, I get 0 as 1 point. If I substitute both to be minus 1, what do I get? 0 again. So, that is an overlap, but that need not necessarily happen in general, for a constellation it can happen. What are the other two values? root 2 and minus root 2, right. Am I right? So, that will be your kind of received constellation for each k, but remember each k is not independent, right. Do you agree with me? If I give you the constellation or if I give you the actual occurrence of b k at k, the next occurrence is closely controlled. It cannot be anything else that you want, ok. So, keep that in mind. So, it is all each k it is not independent and the transmit side it is independent, right. I do a constellation and for each k I pick a point independently from the constellation. Here the receiver I do not do that, I only pick the transmit and the receiver comes forward, ok. So, that is important when I draw the constellation remember that it is not independent for each k, ok. All right. So, let us do a slightly more complicated case. We will do the constellation, the resistance, transmit constellation being QPSK, ok. So, I want you to think for a while, consider all possibilities and draw the received constellation. It is a good exercise. I will give you some clarity about what is going on. First of all, how many points do you expect? 16 points, right. So, those are my points. I have how many points do I have? Put 9 points, am I right? And that is where they occur, ok. So, each of these guys, the squares will be 1 by root 2 plus or minus 1 plus or minus j, ok. So, those are the 4 points, all right. So, one interesting exercise which is a standard quiz question once again is to find energy of a particular received constellation at 1 k, ok. So, how do you find energy now? So, remember all these points will not occur with equal probability, ok. So, you have to be careful. SK occurs with equal probability. For each of these points, you have to compute probability. For instance, in the BPSK case, it is easy to figure out what the probabilities are. What are the probabilities? Yeah, the probability half here, 1 by 4 here, 1 by 4 here. So, you do that and compute, you will get your energy back again to be 1, same as before, ok. So, but you have to do this computation. So, you have to compute the probability at which, with which each point might occur. It is not uniform now, ok, because of the way in which my SK is worked out, all right. So, this is an interesting exercise and you can see this exercise is going to get complicated when, what happens? When will this exercise begin to get complicated? There are 2 things. When there are more taps, ok. When you have more taps in your M of Z, this is going to get complicated and then the second possibility is when there are more symbols in your transmit constellation, ok. So, in fact, you can get an exact relationship for the maximum number of points in your received constellation in terms of these 2 things, ok. So, you can show if you have a mu tap M of Z. So, what do I mean by a mu tap M of Z? M0, M1 Z inverse, M mu Z power minus mu, ok. So, this is what I mean by saying mu tap. So, when I say one tap, it is not a scalar multiple, ok. It is actually one ISA. So, if you want to say multiplication by a constant, I will say 0 tap, ok. So, it is a constant thing, ok. So, remember this notation, this can get confusing because different people use different convention here. My convention here when I say mu tap is mu plus 1 order, ok. So, you have got M0 through M mu, ok. So, remember that, ok. So, if you have this and if your transmit constellation is X, then number of points in received constellation is less than or equal to what? Yeah, size of X power mu, ok. So, it grows exponentially in mu, ok. So, that is a bit of a problem, ok. You have to be worried about this problem because if you want to do detection on the ZK and if M of Z has too many taps, you have to worry about too many points in your constellation, mu plus 1, ok. Is that a mu plus 1? Ok, yeah, that is a mu plus 1, ok. So, I have to worry about too many points in my received constellation and it is going to get quickly complicated, ok. So, that is something that you have to keep in mind and today everybody wants very high spectral efficiency. So, people are using 64 QAM, 32 QAM without any problem. So, at least 16 QAM is being used. So, when you have 16 QAM and then you have and you want a signal very fast, which means you have to face a lot of ISI, ok. And you have 10 taps, you are pretty much done. You cannot even hope for doing any such a keeping track of any such received constellation, ok. So, those are problems and we have to address those problems as we go along. For now, I will assume it is manageable and talk about the ideal detector, ok. So, I will talk about the ideal detector first and we will quickly see the ideal detector is not that implementable as it becomes more complicated and then we will worry about other things, ok. So, I want to do one more example, just to show you that this can get a little bit more painful and it is also an example which I am going to have around for the other things as we go along. So, this is the second example. Once again, I have no idea if this is a minimum face thing or not, but I know it is monic and it is not. I mean it is FIR as in it is causal and it is normalized, ok. So, that is the only thing I care about. I do not know if it is minimum face or not, ok. So, that is my M of Z, ok. So, you can see it is normalized, ok. Suppose I say X is BPSK, ok. So, I want you to plot the received constellation, ok. So, the same exercise if you try with BPSK, it will get even more complicated and ugly, but it is possible to do it, ok. So, you will get many more points, but BPSK is not too bad, ok. One can do it, what would you get, ok. So, there are 8 possibilities 1 plus 1 by root 2, ok. And then minus 1 minus 1 by root 2, ok. And then 1 by root 2 minus 1 by root 2, any other possibility, ok. So, minus 1 plus 1 by root 2, 1 minus 1 by root 2, any other possibility, ok. That is it, ok. The only 6 points, ok, right. So, one can plot these things, ok. So, you can do 1 by root 2 is what, 0.7, right. So, it is closer to 1 than 1 minus 1 by root 2. So, you will have, you will have points like this, ok. So, I do that, 6 points. And probabilities you can compute, you can compute the received energy and all that, ok. So, you try for QPSK, it becomes a bit more of a headache, ok. So, it is difficult to do these things and keep track of these things, ok. But simulations you can do very fast and quickly generate this, ok. So, all right. So, I am thinking I mean I want to return your answer scripts and discuss some things there as well. So, I think I am going to stop here for this class. I will return your quiz answer sheets and then we will pick up from here in the next class.