 So, this is lecture 13 and so what did we see last class, last class I think was talking about probability of error for MPAM, so we did probability of error and well this worked out approximately to be 2Q, is there a question, are you guys are okay, he is still asking you something, is there a question, you are okay, so guy is right behind Dinesh, did you have a question, you are okay, alright, so there was a, the way I wrote it down I said why this should be, root and not by 2, is that fine, right and from the way the energy of the signal, energy of the signal was related to D and energy of the noise was related to N0, so we could define a simple SNR, I defined that SNR in a specific way, once you define that SNR this works out to be Q what, square root of 3 times SNR divided by M squared minus 1, so that is how the SNR work, the probability of error worked out first and then we defined normalized SNR, it was just for convenience, but we saw this is a good enough approximation, it is not a very bad approximation at all, so it works out very well and I think I do not know if I completed the M squared QAM thing, maybe I did it, did I do it, I completed the M squared QAM, so this also worked out to something very similar except for some factors of 2, so I think this worked out as Q root 3 times SNR by 2 M squared minus 1, am I right, so this is how this worked out, so M squared QAM is nothing but 2 M PAMs and it simply worked out as 2 times roughly, so there is a by 2 there, it is a 3 dB penalty you pay for doing that, so that is roughly M squared QAM, so for Mre PSK for instance, for MPSK a similar derivation becomes complicated because it does not split into independent real and imaginary parts, you have to actually deal with the two dimensional Gaussian PDF and integrate, so it is not a very easy simplification to do and we will not do it, I will simply write down the final expression after a lot of approximations, you will see final expression will work out to 2 times Q root sine squared pi by M times SNR, so remember what is my MPSK constellation, it is on the circle and I am going to take the radius of the circle to be square root of E s, why would I do that, then my average energy is going to be E s which is what I wanted and then I can define once again my average noise power or energy is N naught by 2 and then my SNR would once again become E s by E m, which is the same which is a comparable quantity across different constellations, so this thing is on this circle, we have a lot of points, so this angle would be what, what would be this angle, 2 pi by m, so this is my constellation, for that if you work it out, if you work out all the expressions for probability of error, you will get this sine squared pi by M SNR, so now I want to compare these expressions and make some claims about how good is PSK when compared to QAM or PAM, so the first impulse looking at the expressions is to compare MPSK and MPAM, so I am going to argue first of all that it is not a fair comparison, why is it not a fair comparison, yes no maybe you can compare, nothing wrong in comparing it, but MPAM is real in baseband and MPSK is not real in baseband, so technically MPSK is already using double the bandwidth of PAM, so it is not a good comparison, but M squared PSK and M squared QAM I can very well compare, so in terms of what I am doing in baseband and what bandwidth I am using, M squared PSK and M squared QAM are going to give me a very good comparison, so no problem, so we can compare and see how that works, so let us try to do that, so if you look at, so the way we do it you will see it is a little bit interesting, so I will do it also on the graph, but I will first do it with the expressions then I will relate it into a graph, so let us look at M squared PSK versus M squared QAM, so let me write down the probability of other expressions, because I wrote it for MPSK, the M squared PSK is going to be slightly different, so the probability of error, so for M squared QAM works out to I will say equal to it is all approximate and up to certain 3 by 2 M squared minus 1 SNR and probability of error for M squared PSK is going to be what, 2 times Q square root sine squared pi by M squared SNR, so the first thing I will say is this 2 and 4 multiplying the Q outside will not even show up on my plot, if I plot log of PE versus SNR and DB, this 4 and 2 outside you cannot even see it on the plot, so maybe in the log scale 1 to 2 is a reasonable jump from 2 to 3 and 4 and all that there is really no significant thing, so you cannot see anything, so that is not a big deal, so we will ignore that part pretty much, and then we will ask the question, so we will assume, so we will ignore this differences, so we will say this we will ignore. Now I am going to ask the question, suppose these two probabilities of errors are equal, suppose these two have to be equal, what SNR do I need for M squared QAM and what SNR do I need for M squared PSK and I am going to compare those two things, so that is what I am going to do, so I am going to say suppose PE M squared QAM equals PE M squared PSK, what at SNR Q here and at SNR some PE here, so what am I doing here, if you plot both these PE's on the same graph versus SNR, I will get two different curves, I am fixing myself at one point on the y-axis, then I am asking the question, where do I get the intersections on the x-axis, so what are the two SNR's at which these things are achieved, so if you do that what should I do, I should simply equate what is inside the square root as an argument of Q, so it is enough if I do that, so this will happen if 3 by 2 M squared minus 1 SNR Q equals sine squared pi by M squared times SNR P, so we have been doing so many approximations, so we might as well approximate sine, for large M pi is just 3 close to 3, so large M sine pi by M squared is going to be what, pi by M squared, so you square that, you get pi squared by M power 4, so you do all that then you will see SNR Q divided by, should I do SNR P divided by SNR Q or which is a better, I think SNR P is better, so we will do that SNR P divided by SNR Q will work out to roughly 3 M squared by 2 pi squared, so it is an approximate calculation, so you have to take M to be a power of 2, so first thing to take is M equals 2, if you take M equals 2, you know 4 Pim and 4 Psk have to be exactly identical, there is no difference between the two, do you agree or not, 4 Pim and 4 Psk are exactly the same, there is no difference, but since we did a lot of approximation, here you would not get the exact equal, but it is almost the same, if you take pi to be 3, maybe it is roughly almost the same if you put M equals 2, so you have some problems there, but it is okay, it is just an approximation all over the place, but as M becomes larger, our approximation is going to be more and more true, for M equals 2 pi by 4 is definitely sine pi by 4 is probably far away from pi by 4, reasonably far away, so it is not a good approximation, so if you go to M equals 4 for instance, then this becomes a very good approximation, M equals 4, M equals 6, 8, M equals 16 and on there it becomes better and better, so if you plot this versus M, M 2, 4, 8 I have only 3 points, for 2 it works out to be 0 dB, if you do the accurate plot, it will work out to be 0 dB, for 4 it works out to be roughly around some 4.2 dB, for 8 it is around 10 dB, what do these numbers mean? So if I compare 16 QAM with 16 Psk, I get the same probability of error for 16 QAM at 4.2 dB better than 16 Psk, what happens in 64 QAM and 64 Psk? It is 10 dB better, the same probability of error is roughly obtained, but 10 dB better point, so well I have done a dB conversion as well, I mean if you do plug it in, you will get some number you have to convert to dB, so maybe this is not a good expression to use, you can even directly use the sign expression if you want, it is not a big deal. So these are the things that you can do when you, so you see this is one part of digital communication if you will, I mean you draw a constellation, compute the decision regions, compute the probability of error, then make some comparisons at that level and decide which is a good constellation. So here we are roughly deciding that QAM is better, but what advantage does Psk have over QAM? The envelope is very constant, so constant envelope is always better from so many points of view and design, so in practice maybe that is better, but today, I mean analog digital design has evolved to a stage where pretty much QAM is okay, it is not a big deal, you do not have to worry so much about QAM and QAM is almost a de facto standard, except in very very low power devices that you want to use, QAM is pretty much a de facto standard today. And you see there is an advantage over Psk in terms of probability of error. So this kind of a calculation is interesting and you should know how to do it for any two arbitrary constellations. Suppose I give you two different constellations and then I ask you to find out which is better and what is the actual gain at a fixed probability of error, you should be capable of doing all these approximations and coming up to that, okay? So maybe I would not expect you to know the 2D Gaussian integration and all that, that approximation is slightly more difficult, but to the extent possible, you should be able to approximate it and come up with some number like 1 dB, 2 dB, you should not give a complicated integration expression and say go evaluate it, it does not make much sense. So at some level it should be good. Alright, so that's about Psk and QAM. Okay, so a couple of loose ends which we haven't really looked at closely before winding up this part and moving on to strictly band limited type situations and I want to use maximum bandwidth. I'll go to that soon enough, but before that I want to just finish up a few things. One thing I didn't do is how do you come up with what is called MQAM when M is a power of 2, okay? So we did not do this yet, right? What did we do? We did only M squared QAM where M is a power of 2. So those kinds of constructions are called rectangular, square QAM, okay? Maybe rectangular QAM. So M squared QAM is a very square case, right? So I think it's square QAM, okay? So when you have other powers, odd powers of 2, okay, 8, 32, 128 and so on, you have the rectangular QAM case, okay? So you'll see it's not a big deal. You just try to pick points, enough points around to get what you want, but even this is becoming less fashionable these days. If you look at the latest even wireless standards, people will do 4, 16, 64, 256. No, they just tend to avoid the middle points because it just complicates matters, okay? Right? So there's lots of advantage in doing M squared QAM, the X and Y axis become independent. You can do so many things in a much more free fashion, okay? So I'll do this just for completeness. You'll see how it works, okay? So if M is 4, we know what it is, okay? It is a case we considered already. M is 2 I'm not doing, okay? So it's not really QAM anymore. So it's just PAM. So M is 4. You have a simple constellation. What do you do for M equals 8? What do you think you can do? Yeah, you can do anything you want, pretty much nobody's going to stop you. So you can choose. So one thing that's commonly done is either you put two points here or put two points on the Y axis. But you might say I don't want to do it. I want to pick one point here, one point here, one point on top, one point below. You may guess you can do it, okay? So it's okay. If you find that that is easy for you, you can do that also, okay? But this is a very standard 8 QAM constellation. So if you see people talking about 8 QAM, they're usually talking about this constellation, okay? So 16 is once again easy, okay? I'll do 32 also on top of 16. So you'll see I'll do 16 first. And then I show you how to do 32. It's quite easy. Just pick points around the 16 QAM constellation till you get 32, okay? This is 16, okay? So I'll draw a dotted line here to show that this is 16, okay? So this is 16. And for 32, I'll do this. This is again a standard thing to do to get 32. I'll pick four points here. Then I'll pick four points here. What am I leaving out? I'm leaving out that corner point, okay? So then I'll pick four points here. Once again, I'll pick four points here, okay? And then one can join all these guys with the dotted line. Together, now I get 32 QAM. What was, is it, does it make any logical sense to you dropping out the corner points? Why would I drop out the corner points? Sorry? Yeah, though, those are the maximum energy points, right? Let's see why the corner points have maximum energy is the distance is maximum from the origin. So they have a lot of energy. So if at all you want to drop something, you'll be better off dropping maximum energy points, okay? So this is a standard trick you can use. In fact, you can automate this process, okay? So you keep picking one, three, five, seven, so on on the x-axis, same thing on the y-axis and look at the square grid. And then drop maximum energy points till you get to the number of points you want, okay? So that can be like a general algorithm for coming up with MQAM, even if M is not a power of 2. So you might say, why do you want to pick M? Not a power of 2. Maybe, maybe for exam questions, it's a good thing to pick, okay? So maybe for those kind of things, one can do those things, okay? So 64 would be square once again, right? So 8 on each x and y, 128 would be non-square again. So you just keep picking more points and fill out the square grid and then drop the maximum energy points till you get 128. So you would get a nice 128 there, okay? So that's how you proceed for these things, all right? So this is one part which I didn't clearly talk, but you can do that. So what will be one of the complications in, say, 32 QAM and all that, okay? Decision regions will be a little bit more involved, but one can do it. It's not a big problem. You can do it. But there's also approximations to the probability of error for the general MQAM, okay? I'm not talking much about it. Okay, so other things which I've not talked at all about is the other signaling schemes which were popular and I think which are popular among particularly the low power communication devices. So you might say, what is this low power communication device? So usually when you think of sensors and all that, so what sensors would you imagine are sensing some physical quantity, like temperature, whatever, pressure and all that, and you want to put it over a large geographical area and leave it unattended kind of, you know? I mean then maybe you fly an aircraft over it and try to read information from the sensors, okay? So you can't maintain the sensors. So you can't change the battery every so often. So you want to use very little power from the sensors. So for those kind of things, you want to put a very simple communication system out there. For those kind of things, maybe something else makes sense, okay? So the problem with QAM is one of the significant problems in detecting QAM is it needs to be coherent for getting good performance and all that. There are other schemes which give you a good non-coherent detector. So FSK for instance is a very popular scheme, which you can think of in this picture also, but FSK is a popular scheme which allows a non-coherent detector, okay? I haven't talked much about it, but maybe later on if we have time, we'll do it, okay? So there are two types here, one that I left out, one I'll call, one is typically called continuous phase modulation, okay? It's also called CPM, okay? So if you don't like the likes of Prakash Karat and other people, you might want to avoid this also, okay? The other is FSK for instance, I mean, I think I talked about FSK, but I didn't do it in great detail, okay? So there's another version of it called CPFSK, okay? It's continuous phase FSK. So this continuous phase is considered one more thing which is useful in electronics from an analog electronics point of view, but today once again, it's not a big deal, okay? Even if you don't have continuous phase in your signal, people can deal with it, so it's gone out of fashion, okay? So the CPM has several, I almost said factions to it, it's not factions, several different types. One is called MSK, it's called Minimum Shift King, okay? So another one is Offset QPSK, okay? So all these things are different schemes which I have not talked at all about, but if you pick any standard textbook on digital communication, all these things will be elaborated, okay? So you might want to read through and understand what this CPM is and FSK and CPFSK. All right. So the last piece of missing information is labeling points with bits, okay? That is something I did not talk about, okay? So that's one thing and one more thing I did not talk about is bit error rate, okay? So far we've been only dealing with symbol error rates, right? So what is the probability that my received symbol is not equal to my transmitted symbol, okay? So from symbols, I have to go back to bits and that might cause some differences as far as error rates are concerned. So those two things I'll quickly talk about, okay? So one very popular way of labeling points with bits is what's called gray coding, okay? So the logic in gray coding is adjacent points must differ only in one bit, differ in one bit, okay? So it's possible to do this for almost all of our signaling schemes, okay? So adjacent points should differ only in one bit. So you start with a very simple four PAM, okay? So you have four points, okay? How do you label so that adjacent points differ in one bit? So you can't do sequential. That's the only thing you can't do. You can't do 001, 1011. So because 01 and 10 are different, two places. So all you have to do is 00, 01, 1110. It's a small change to it, okay? So you can also do 8 BAM, okay? So I'll just, there are, in fact, there are several ways of doing gray coding. This is not one way of doing unique gray coding, right? You, for instance, if you flip this or you rotate it, anything you do maintain all of these things, okay? So those, the several ways of doing it, this is just one way. And most, in most cases, you can just by trial and error figure out what the gray coding is. So you don't have to have any gray theoretical algorithm in your mind to do this. So for instance, 8 BAM, okay? So there are four points on the side, four points on the side, okay? So what shall we begin with? We'll begin with 00, what can I do next? 001, what can I do next? 011, okay? What can I do next? 010, okay? What can I do next? 110, okay? 111, okay? 101, 10, okay? So this is, so like I said, I mean, there's no real confusion. There are enough bits, enough length 3 bits, so you can just keep playing around with it and eventually you'll get to the gray code. It's not a big deal, okay? So 16 QAM maybe is a little bit more interesting, okay? So we'll do that next. I'll come to that. So right now I'm just using it as one way of coding. You might choose any other logic if you want, but I'll come to it soon, okay? So you have 16 points, okay? So how do you think you should do it? This time you look at it, it might be scary, but just start somewhere and keep going, okay? So for 2D things, it's good to split it into 2 bits first and then do the 2 bits and then do the remaining 2 bits, okay? So it always works out that way. So maybe I can try, I can try what? I'll try, I'll try the last 2 bits here, 00, 01, 11, 10, okay? And then the first 2 bits I'll keep the same for all 4, okay? And then what can I do? I can do a gray code for the first 2 bits column once, okay? So I'll do 01, 11, 10, okay? So and then retain the same 0000, okay? Okay, 01, 01, 01, 11, 01, 11, 11, 11, 10, then I do 10, 01, 10, 11, 10, okay? This is just once again, one way of doing gray coding in 16 km, you might want to do it any other way, okay? So what's the logic once again? One of the perceived advantages of gray coding is, okay? One of the easily intuitive understandings is, suppose you transmit one symbol, okay? What is your most likely erroneous symbol going to be? The next one, right? So that's one symbol error anyway, you've made an error, but when you make that one symbol error, how many bitters would you make? Only one if you did gray coding, okay? But maybe it doesn't matter in many cases, okay? But in some cases, it might matter for you to minimize the bit error rate given that you made a symbol error, okay? So if you're looking at such kind of things, if those things make sense to you, then you might want to do it. But actually it turns out today people do something called coded modulation. Suppose you do coding and modulation together, something like gray coding has very deep benefits. There are some benefits because of gray coding fundamentally, okay? So those things, one might say it's still a research type topic, okay? So people are still working on it. There are a lot of advantages here. Maybe if we have time towards the end, we'll see some of these things, okay? So but gray coding is something good. You might also do other things. It's not a big deal, okay? Ultimately, we'll see symbol error and bit error are very closely, they have to be very close. They can't be too far off, okay? After all, you're taking just log m base two bits and mapping it to a symbol, okay? It can't be two way off. We'll do that also later, okay? So but I want to point out that there are other labeling schemes out there and other, in fact, other ways of doing, multiple ways of doing gray coding itself, okay? So all these things have their own advantages. There is one thing called set partitioning, which is also considered a very good way of doing labeling, okay? So which is again something that one can look at, okay? So this said it just for completeness, so that you have a way of doing this. Okay, so the last, well the last and final thing I'm going to do is look at symbol errors versus bit errors, okay? So it's very simple. I'll do it for MQAM. You'll see it's a relatively simple idea. I'll just quickly go through it without any problem. So here's an intuitive way of thinking about it without going into the rigorous probabilities and all that, okay? If you make one symbol error in MQAM, what can happen in terms of bit errors, okay? There are several possibilities, right? If I tell you that the symbol I received is not the symbol I transmitted, right? Each symbol corresponds to log M base 2 bits. So next question to ask is how many bit errors did I make? What are the possible answers? Okay, zero is not a possibility because I already told you I made an error, okay? So it can go from one to log M base 2. So this can result in either one bit error and all the way to log M base 2 bit errors, okay? So this is the intuitive basis for what we'll be doing next, okay? Suppose I say now I transmit n symbols, okay? And the result of the transmission and detection, I made n sub s symbol errors. Suppose this happened, okay? So now if I look at the bits picture, when I transmitted n symbols, how many bits did I transmit? n times log M base 2 bits were transmitted and how many bit errors did I make? I cannot be sure but roughly it will be between, well it will definitely be between ns bit errors and ns times log M base 2 bit errors, okay? So this is the possibility. So now in this situation my symbol error rate worked out to what? Assuming all these numbers are really, really large, I am doing it for a long time, my symbol error rate will roughly be ns by n, right? What will be my bit error rate? Well I cannot be sure but I can bound it. I know at best at worst it will be ns by n and at best it will be ns by n log M base 2, okay? Right? So this is the, this is, this is very true. One can also show it very rigorously if you want. Probability of bit error is bounded between probability of symbol error and probability of symbol error divided by number of bits per symbol, okay? So it is a very simple intuitive thing to remember. So if I look at a case where everything is going to 0, if I force everything to 0 then it does not matter. And symbol error rate goes to 0, bit error rate will also go to 0, okay? Because it is sandwiched in between, okay? So that is the, that is the logic, all right? So we do not have to worry too much about the distinction between bit errors and symbol errors. You can look one on any one thing and build a, build a certain result and that will also be valid typically for bit error rates, okay? So that is, I think that pretty much ties up our loose ends. I want to briefly summarize where we are and then move to the next logical step, okay? So here is the summary, okay? Remember, so far we have been looking at an ideal AWGN channel but also very large bandwidth, okay? So I have been trying to emphasize this very large bandwidth. How did I, where did I make that assumption of very large bandwidth? Okay? Yeah, so I was saying my signals are going to have started 0 and ended T, okay? So once I say that, my bandwidth is going to be very, very large and I love to receive that signal. So I am assuming my bandwidth is very large there, okay? So because of that assumption, everything worked out properly for me, okay? So if I did not make that assumption, so many things would not have worked out. So because of that, I could pick my bases in a very simple way, okay? I can simply window the bases and pick it, okay? So the first thing we spent some time understanding is the equivalence between real base band and one of the very crucial things and complex pass band, okay? So this is a very, very, okay, sorry, real pass band and complex base band, okay? This equivalence is very, very crucial and how did we go back and forth? By up conversion and down conversion by multiplying suitably, okay? You have to take a real part root 2 times and all these things, okay? So that was one of the most important concepts that we introduced in this context, okay? So real base pass band and complex base band and that's very useful, okay? And so what else did we do? Okay? And then we assumed that the bandwidth I have is much, much larger than 1 by T, okay? So what is T now? Okay? So once I assumed this, I was able to pick a very simple signaling scheme, okay? So signaling became very, very simple because of this. In what sense, okay? I can now pick just one basis function, phi of T, okay? Which is what? It's simply constant between 0 and T, say 1 by root T, okay? And then what can I let every other signal to be? If I want to send anything a simple multiplication of this, okay? The reason why I was able to do this is because since my bandwidth is much, much larger than 1 by T, I know that this signal will come through unaffected, no problem. It won't exactly come through as it is, okay? And if I now send another symbol after that, what will happen? That will also come through exactly, right? So I can even string symbol after symbol in this way and get them faithfully at the other end because I have enough bandwidth, okay? I assumed a huge bandwidth was there. So even if I send symbol after symbol, none of them will cause any problem. So I can repeat this for every 0 to T and I will get it, okay? So in practice, if your bandwidth is not very large, what would you do? Yeah, just increase the T, keep on sending it for a long time till you are sure that it will definitely come through, okay? So don't change it too fast, okay? So this is definitely an implementable scheme for any system except that you won't be using the bandwidth optimally, okay? So we did that, signaling was simple and I want to write down how that signaling worked out, okay? So this is how this complex baseband QAM worked out for us, okay? So we started with B which was 2 times log m base 2 bits, okay? This is going to give me what? What is the first step? The way we picture it? Or do you take the bits and then pick a point from my constellation, okay? So it gives me this a plus jb which is the point in the constellation. Then what do you do? You multiply this a plus jb by this basis to get my complex baseband signal, okay? This is my complex baseband signal. Then what do I do? I upconvert, okay? So once I do upconversion, I get my real passband signal xabt, okay? So this is going to go through some additive noise and they receive something and then what do I do? I do downconversion to get something and then I do a correlation receiver. In this case, the correlation simply becomes an integrate and dump, you have to integrate over a long time to make sure you are using the bandwidth and all that, but this becomes an integrate and dump, there's no problem, okay? So it's actually, it comes from, we know that this is the optimal thing to do from the correlation receiver ideas, okay? So we know the optimum receiver has to do the correlation and which worked out to this. So once you do integrate and dump, what do you get? You in fact get a plus jb plus noise, okay? Two-dimensional noise and what is the pdf of this noise? iid normal means 0, variance n0 by 2, what is this n0 by 2? PSD of the n of t, okay? So in our model, what do we do? We jump directly from here to here for probability of error calculation and all that, but you can choose to jump from anywhere to anywhere, right? Go in this scheme anywhere and see the correspondence, but for various calculations, some things are easy. So here then you do a detection, okay? And your detection is, okay? The various types possible, MAP, ML, but the simplest thing is what? Minimum distance, okay? And whether it is optimal or not depends on several conditions, but that might be okay, okay? And you get your b hat, okay? So this is the picture that we have been looking at so far, okay? And we did several calculations with it, tried to convince ourselves that it works and it is good enough and people who are doing 471, this might be a first-cut model to implement and see whether it works or not, okay? So that is a very simple nice model, all right? So now we have to start using the bandwidth better, okay? So you might say, why do I want to use the bandwidth better? For one, you might want a faster rate of communication, okay? Given a certain bandwidth, you might want, for instance, the maximum rate at which you can communicate, okay? If such a thing exists, okay? So there are all, so there are some several problems with that. So once you do that, once you say, I am going to also occupy a lot of bandwidth, what will happen? What will happen? One of the crucial things that will happen is, I can't just think of each of these symbols separately, okay? So right now I can run this continuously, why? I can send the first symbol, wait till 0 to t and then send the second symbol and I know I will receive all of them properly and my entire thing will work because my bandwidth is so large compared to this 1 by t, I know everything is going to go through, there is no problem. Now if I start occupying larger and larger bandwidths or decreasing t, okay? So I am increasing my bandwidth, first thing that will happen is, right? What do you think will happen? First thing that will happen is one symbol will spill over into the other symbol, okay? So I will maybe draw another picture and show you how it can happen. So as you guys keep occupying more and more bandwidth, okay? So your time for the signal is becoming shorter and shorter, right? So before the first symbol kind of dies out and it's enough, you'll be starting the next one, okay? So eventually you will have spill over from one symbol to another. So you have to worry about continuous operation, okay? That's the first thing you have to worry about, okay? When you go and want to fill up your bandwidth and occupy a lot of the bandwidth and communicate at a faster rate with limited bandwidth and you do that, you have to worry about continuous operation. You have to look at one symbol to the other, okay? So that will be one significant change that will happen. But beyond that, most of the things will be similar. The same model we'll try to carry over, okay? So we will not drop this correspondence between real pass band and complex base band. It's really useful to us. We want to retain it. We will not suddenly drop the signal constellation. We will still retain it, okay? We will not drop this view of signal space, correlation receiver and detection principle. We will not drop all of that. All of them will be the same. We will try to manage with the same type of setup but change some things crucially here and there and try to make the same thing work. And you'll see it's possible, okay? Definitely for ideal AWGN band limited channels, you can do it with very simple changes. When you don't have an ideal channel, you have to do more tweaks, but it's possible, okay? So that's where we are headed in the next few classes, okay? So the next thing we'll do is we'll look at what are we doing next. Ideal band limited AWGN. Not really large bandwidth. I don't have too many too much bandwidth to burn, okay? So that's the next thing we'll see. Okay? Once again, there are two ways of viewing it. I'm saying band limited. Well any system is really band limited. But what did I do in the previous scenario? I did not use all my band. I used a very small bandwidth, okay? So now there are two ways of viewing it. I'll either say I want to occupy all the bandwidth or signal at a very fast rate. I want those constraints, okay? Those conditions. Those conditions are important, okay? So right now, so you see, what was missing in the previous picture was this trade-off. The bandwidth never really entered the picture because I assumed I'm not going to worry about it. Now we're bringing bandwidth in, okay? So that will give us a nice complete picture of the entire trade-off between bandwidth, power, probability of error, noise power and everything, okay? So this will kind of complete the picture, okay? So this is ideal band limited AWGN. So what is the picture that I'm going to have? Well, so the old picture still remains, okay? So the problem I have is still the same. It's not in any way different. I have n bits, which I want to transmit. So I'm going to use a transmitter and put out some XB of T, which is going to go through a channel, okay? And I'll get a Y of T, which I want the receiver to process and get a B hat, okay? So this is my picture and how am I going to model the channel? Once again, I'm going to model the channel with an impulse response H of T, okay? And that impulse response will correspond to a Fourier transform H of F, okay? So now I might either have a pass band channel or a base band channel, okay? Both possibilities are there. My channel could be either a pass band channel or a base band channel. What do I mean by that? At my disposal, I either have a bandwidth which is in pass band or base band, okay? Usually in wireless and all that, you always have pass band, okay? So but in wire line in some situations, it's possible to imagine having a base band channel, okay? So but I'm going to say my channel right now is ideal. What do I mean by ideal? Within that band, it's flat. H of F is flat. That's what I mean by ideal. Whatever bandwidth I have within that bandwidth, my channel is flat, okay? So that's my assumption, okay? So if you think of H of F, since it's flat, maybe if you say pass band, okay? It's pass band, then it's going to be some W frequency in pass band. So if you, since I'm going to say H of T is real and all that, it will also have a W on the negative side. So I know if I think of down conversion, I can down convert this pass band real channel response into a complex envelope equivalent. If I do that, what will be my H tilde of F? It's going to be between minus W by 2 and W by 2. And since I'm assuming it's ideal, right? What will be the H tilde of T? That will also be real, okay? So you see the complex baseband picture actually becomes a real baseband picture because of my ideal channel assumption. So I don't have to worry too much about complex in this case, okay? So I might as well take without losing any generality about the channel, a real baseband impulse response, okay? Since my channel is ideal, that is good enough, okay? So that doesn't mean my XB of T will be real, okay? This is only for H of T, okay? Why will my XB of T still have to be complex? Yeah, but I still want to use the entire bandwidth. If it's pass band channel, right? I want to use the entire bandwidth, which will mean I have to make my XB of T still complex so that when it goes to pass band, I will be using the entire bandwidth. So my XB of T can still be complex, but I'm going to say since it's ideal, my H of T will be baseband real, okay? So I can make that assumption because even if it is pass band real, it's going to be baseband real anyway and baseband I'm assuming it's ideal. So it's all ideal. So without loss of generality, we'll take H of T to be baseband real, okay? So when I introduce non-ideal channels, maybe at that point you'll have to assume H of T can be baseband complex, okay? But still baseband is good enough, okay? Because I know I can down convert and view the whole thing in baseband, okay? So that's the first thing we'll do, okay? All right? So now this is simple. It's not too bad. So let me go back and look at requirements on XB of T, okay? So did I have the picture here? Okay. So let me just cut and paste this picture. I don't want to once again draw this. Oh, I'm running out of time. So let me see, okay? All right. So this is my picture now. I'm my H of T. What am I going to assume my H of T to be? I'm going to say H of T is baseband real, okay? So I can do that without any real problems here because of my ideal assumption, okay? So once I do that, so what should my XB of T necessarily be, okay? So the first constraint you need is yeah, it has to be band limited between minus W by 2 and W by 2, okay? Also, I want to be able to repeat this, right? I mean, I want to do it at a reasonable time. So XB of T should also have finite support. It should also be time limited, okay? So this XB of T is seemingly, I mean, the requirement is it has to be both time limited and band limited, okay? Band limited between minus W by 2, W by 2, time limited say maybe 0 to Tn, okay? Some Tn, okay? So picking that we use some time limit, okay? We know definitely that these two are strictly, fully, rigorously not satisfiable. So we'll satisfy them only approximately, okay? So usually the time limit is strictly satisfied because you know in your system you can't keep transmitting one signal forever, okay? So time limit is strictly satisfied which means the band limitation will only be approximately satisfied. So approximation in the sense to an arbitrary accuracy say for instance outside of W by 2, the spectrum will go down to say minus 50 dB, minus 40 dB, minus 30 dB, whatever. Some very low numbers so that you're not interfering with anybody else who wants to use that spectrum, okay? So outside of W by 2 my spectrum will have to drop down to really, really low. That will be my, that will be the way I'll be enforcing this limitation but usually like I said time limitation is, will have to be strictly enforced, okay? So today that's how people do it. Alright, so that's the first point to note but the second point is more critical, okay? There's another requirement on XB of T which comes because of the requirement of continuous operation, okay? So what is the second requirement? This is the, this is, this is because of continuous operation and we'll see it in the next class, okay? The fact that I want to send the next symbol as well. I send the first symbol, first set of n bits then I want to send the next set of n bits as well and if I'm using a lot of bandwidth I'm going to get interference between these two signals and how do I avoid that? Is it possible and all that is what we'll look at next, okay? So because of continuous operation we get that. We'll see that beginning next class.