 This is lecture 8, okay, and we saw a whole bunch of very important things in the last lecture. I want to go through roughly the crux of what we saw in the last lecture was how to go from a waveform type channel which is y of t equals x of t plus n of t, okay I want to remind you once again that since the bits are unknown and random, x of t itself becomes a random process, n of t anyway is the noise process about which you can't claim to know and then y of t in effect becomes a random process, so you can't expect any deterministic type reduction. So if at all you reduce it, you will get a random vector type of solution. So we saw how to reduce this. What is the main, what is the most important tool from a purely mathematical point of view, what did we use? Yeah, just simple orthogonalization based linear algebra, right? So we saw if you have a finite number of, you can think of these x of t as vectors, okay, once you fix a basis and how do you find the basis from all possible signals that are going out, you do orthogonalization, find a basis. Once you find a basis, everything becomes very easy, everything becomes a vector, okay. So you went from here using I'll say linear algebra to a large extent, okay, that too very simple linear algebra and actually if you do it rigorously, there's lots more areas involved, okay. So you have to define what happens to random processes when they go through correlators, all these things carefully, you can do all that and you can meticulously make your way towards a vector channel, okay. So the vector channel then becomes y is x plus n, okay. So what's the dimension of this vector space in which this vector channel gets defined, okay. Where does the dimension come from? Number of basis vectors that you got from the orthonormalization, okay. So if you have a whole bunch of x of t's, say you're sending n bits, then you'll have 2 power n different x of t's, okay. So you start from there and reduce it to an orthogonal basis, you might get a certain number of basis vectors, okay. So that is the dimension here, okay. So each of these guys now is a random vector, okay. So from random process, well continuous time random process, we've gone to random vectors, okay. And once you have random vectors, it's also possible to quickly figure out what the joint PDF and joint PMFS, okay. One of the typical assumptions you know, you don't have to make this assumption from in the derivation we saw that n would be distributed as what, assuming this is a white Gaussian process, okay. And you take a PSD to be n naught by 2, okay. So PSD n naught by 2 in all frequencies, you take that, then n becomes what? It becomes IID, normal, zero mean variance, n naught by 2, okay. So the PDF for n is very clear and also clear that x and n will be independent. How do you find the PDF for x? Joint PDF for x. First of all, in most cases, what will be the type of distribution for x? Will it be continuous? Will it be, for instance, n is continuous, right? So will x be continuous or will it be discrete? Yeah, in most cases it will be discrete, right? So you have only one of 2 power n possibilities, okay. And typically you don't think of the 2 power n as a very large number. So usually when you do modulation, you do modulation over very small number of bits together. It's also possible to do it for a large number of bits. Maybe we'll see such a system later on. But for now you can imagine just a small number of bits at a time, okay. So x would be discrete, okay. And it's not, you should know how to find it, okay. So it's usually assumed to be IID over the number of discrete signal vectors that you got, okay. So that's the distribution. But this will be typically discrete, okay. So this x is typically denoted on a signal space, okay. So that's called a constellation, okay. So that's pretty important. What's the constellation? Okay, I'm doing it in 2D here. So in general, it will be m dimensional signal space, okay, where this m came from the reduction that you did, or the normalization that you did. And each possible x, you put a kind of a star here. Each of these guys are signal points now. They can be converted into a signal if you want, okay, using the orthonormal basis. I also made a remark later saying typically we can also do the reverse, okay, to design signals themselves. You can say I start with an orthonormal basis first and define points on that orthonormal signal space, on that signal space, using my orthonormal basis vectors to go back to the signals. That's another thing we did, okay. So in this conversion, there is one thing that is preserved, okay. One aspect of that signal that's very important to you from a constrained point of view, one thing is preserved. What is that? What do you preserve? So from this vector itself, without knowing the basis, you can compute something about the signal. What can you compute? The energy, right? The norm of the vector is going to equal the energy of the actual signal, okay. So the power or energy in the signal is faithfully represented in the vector as well. What is not easily represented without knowing the basis? The bandwidth, right? You know the bandwidth, you should know what the basis is, what the actual signal is, okay. So other than that, the energy itself is nicely represented in this vector itself, okay. So since norm of vector x equals energy of x of t, so I should put small here, doing x of t, okay. So typically the assumption is by the time you come to the constellation, you've already figured your bandwidth out, okay. You've figured out your bandwidth and you've chosen the basis suitably so that it fits nicely into your bandwidth. So you don't have to worry too much about it. So if you don't do that, no point in coming to the constellation, okay. So the constellation doesn't make much sense if you don't do that, right. Because your x of t is what? It's the signal that is received at the receiver. It's already gone through the channel and all that. You've arranged for all those things the way we did that, okay. If you're not suddenly saying bandwidth is not enough, then you have to go back and fix that first, okay. So once you do that, energy is pretty much the only thing you need, right. That's very good. And it's also nice to note that in the noise process, the noise random vector that we got, the PSD of the noise process is nicely represented, right. The variance of your noise is N0 by 2 which is the height of your power spectral density for your noise, okay. Like I said, in practical systems, it's possible to measure this power of the hub, I mean, height of the power spectral density of noise, okay. So this model nicely captures many of the important complications that will arise in the real communication system, okay. So that's what's so nice about this model, okay. There was also another part of the noise which we rejected. The reason why we rejected is, okay. Given this vector y, that part of the noise is irrelevant to x, okay. So it's independent of x. So you don't have to look at it at all. It won't affect any of your detection process, okay. So how did our receiver look now finally? Okay, so you have a bit B. You're doing transmitters doing a modulation to get some vector, some signal x of t, okay. Noise is getting added to it, right. You get r of t. What are you going to do with the receiver? What's the first step? You're going to correlate with the basis that you got, okay. So you do correlate with phi i of t, i going from 1 to m, okay. So you do correlation to get a vector y, and that's enough. What do you do next? You run it through what I called a detector, okay. So at this point, what you have is a detection problem. We'll precisely define it later on. But at this point, once you get your y, you have a detection problem in the sense that y has a certain joint distribution with, if you correlate the x of t, what do you get? Correlate with phi i. What will you get here? You'll get a certain vector x, right. So this x and y have a joint distribution, okay. And the joint distribution also involves n, which is the correlated version of n of t, right. So given a joint distribution, you're allowed to observe only one part of it. You're not allowed to observe x, right, at the receiver. You can observe only y, which has a certain joint distribution with x. Given that, you have to find the best guess of x, okay. So that is the detector. That is a detection problem. We'll see later on how it's properly framed. And it's very easy to do it with the vector formulation. At least the formulation is easy. Solution might be difficult in certain cases, but formulation is typically easy, okay. So this is the picture. So once you have this picture in mind, you can nicely move to a vector picture, okay, where you say, from a bit b, I go to what? Okay, I do a certain modulation. I go to what? The vector x, okay, to which what gets added is what? A vector n, okay, and you get a vector y, which directly goes into my detector. To produce, well, you would like b hat, but you might actually produce even x hat. It doesn't matter, right. So b hat or x hat, whatever you produce. Okay, because b to x is a, hopefully a one-to-one more transformation. You don't want to do anything other than one-to-one, okay. So that's what you're doing, okay. So this is the picture which we will use pretty much, but you should know that where it comes from is this correlation picture, okay. If you don't do these correlators, this picture makes no sense, okay. And auto normalization plays a very important role in the way we got to the vector of presentation, okay. Any questions? People have time to think about this. Any comments, questions about what's going on? Sir, in the sending, what is actually being sent? Well, it's an interesting question. What do you think the answer is? Vector x, in reality, in the real system, you know. I think you should think of it as x of t. Okay, in the real system, x of t is the actual signal that is being sent, okay. So in reality, signals will be continuous time, right. You can't think of real, I mean, see, you're probably used to thinking of signals as vectors. It's okay. It's not a big, it's not a bad approximation given that you have sampling and all that. But in reality here, x of t has to be sent because we have a certain bandwidth limitation, power limitation and all that. All that is only satisfied by x of t. If you say x vector, what about bandwidth for it? I mean, what do you think of that? Okay, so all those things are not true. So what's actually being sent physically through the pair of wires is x of t, okay. So this is a model which nicely fits that reality, okay. Everything about that reality is nicely captured in this model, okay. So it's called the ideal AWGN model, okay. So that's an important question because, so the reason why that question is important is, so see, as you keep learning digital communications a lot, you'll pretty much only do vectors, okay. So in fact, it's also become to a stage where all your signal processing algorithms sit on processors and computers on these boards that you have. So I don't know if you're familiar with building of systems and all that. Today people make boards, okay. What are those boards? Who do they make them? What's that called? It's a nice, it's called PCBs, right. So you always do PCBs, okay. Printed circuit boards on which all your circuits are now mostly chips, right. So there's no resistors and capacitors of course there, but most of the important things that are done inside chips, okay. So it's all fabricated chips. And it's just a question of when you build a system going there and getting the right processor, getting the right A to D converter, D to I converter, getting this, putting them together on a PCB, making the layout, maybe so many layers, you send it out, PCB comes back, you populate the board and you test it out, right. That's how you make systems today, okay. So what is the communication engineer see mostly? You write pretty much C programs, right. Which get compiled into some processors, assembly code and it's running there, right. So for you, you can even think that the only thing you're worried about is this vector X. Whose problem is it to convert that vector X into an actual signal that's going out? Somebody else's problem. Maybe somebody who's doing analog or RF or BLSA or something. It's not your problem. If you can make some system work with this vector model, it will convert faithfully into a continuous time model, okay. I can say that up to certain approximations, right. So obviously some problems there, okay. There's so many other issues, but those are not very important for communication engineers, okay. So it's enough if you work with this model. So you can even think of it as what's going out is X, okay. And say, once it leaves my processor, I don't care. Whatever happened to it can happen to it. I don't care, okay. The reason is even the energy of the signal is accurately converted is converted in this X, okay. So it's okay. It's not too bad to think about it. But if you want to be a complete systems kind of guy, you want to know everything that's happening in your board, you better know how the whole process is and how it's happening, okay. But yeah, I mean, in reality today, most communication engineers works with this model. Nobody really goes into the actual conversion, okay. So there are some real problems in that conversion. For instance, before you put it out on the pair of wires, these signals will have to be driven into the wires, right. So you have to drive up something, okay. Some part of the signal, like the current or voltage, you have to drive up, okay. So when you drive it up, you have to use some power amplifiers and all that. And this power amplifiers are not linear, okay. So what's the problem? Suddenly you have a non-linear component somewhere. Yeah, your whole assumptions will go for a toss, okay. So all your basis function you chose carefully for an X of t will not be basis functions anymore, will not be orthogonal anymore, okay. So there are several ways of dealing with it, okay. You'll see when you learn, people try to minimize peak power as opposed to average power. See, the X, vector X will represent only average power, right. So you might want to reduce peak power. At that time, you'll have to worry about the X of t, or figure out how to change X, the vector so that the X of t can change. So all those things are real problems when you convert them into practice. So I would say, don't ignore the X of t aspect, okay. So keep that also in mind, even though mostly what you deal with is the vector. Okay, any other question, comments? So if you're not thinking about this seriously, it's a good thing to think about, okay. So it's the foundation, right. Once you accept this, everything else will seem like tutorial problem, okay. But this is the foundation. So you should think more about these things. Okay, all right, so let's move on now. From here then, we saw a few examples of specific modulation schemes. And I can now describe everything in signal space. I think I first give you an example of how it happened from X of t, how you went back. So maybe we'll do that a little bit more, but I'll describe the first thing we saw, which is BPSK in signal space. So it's very easy in signal space, right. So what do you do in signal space? Simply say it's a one-dimensional signaling scheme with my bit zero being plus one and my bit one being minus one, okay. So that picture nicely represents everything that my modulator is doing, okay. It might actually be some complicated PCB with several layers, all that is relevant to me. What do I care about? What it's actually doing in terms of the detection problem. Okay, so ultimately the detection problem is what's important to you when you try to decode, right. So what it's doing is only this, okay. I have bit zero and bit one. I take bit zero and my vector or in this case scalar corresponding to that is plus one. And if I'd have bit one, scalar corresponding to that is minus one. I might have used root ES or something and it's okay. Plus one and minus one is general enough, okay. You'll see the reason is as long as you keep N naught by two as general, you can take this to be the same, okay. It's not a big deal. All right, so from this, the important step that you have to do in the model before we go to the detection problem is what? This is X, right. You have to go from X to Y and then figure out what the PDF or the joint PDF for X and Y is, right. In this case, since X is discrete, Y is continuous. Typically the conditional PDF will nicely capture the joint PDF, okay. So what are you only about the conditional PDF? What will be the conditional PDF? Y given X is plus one will be normal mean plus one variance and not by two, okay. So those are the things you worry about, okay. So conditional PDF, okay. I'm not writing everything here, okay. So I think I used a different notation, right. So I did this, I'll stick to that. This becomes what? One by root two pi sigma, what's sigma? Y minus one square by two sigma square. What's sigma now? Yeah, root N naught by two, okay. You can write a similar expression for Y given minus one. So now your detection problem is all set. You can say it's a question of just doing detection. Okay, so we'll do it. We'll see how to do that. That's also very easy. I mean, even the answer there, we could guess, right, very easily. How will you detect an arbitrary Y? If I say Y is 0.5, what will you say X was? Plus one, right. So it's just based on some very simple intuitive notions, symmetry of the noise, et cetera. You can say it has to be the closest point, okay. So we'll justify that later on, rigorously. Okay, so that's BPSK. The next example we saw where we kind of, I mean, we were halfway through that example. So I'll pick it up from the beginning. What's this, FSK, okay. Frequency shift k-ing, okay. So I'll do it once again like I did before. I said in this case, we choose N equals one, okay. So we do one bit at a time. So my bit B becomes zero or one, okay. And corresponding to bit zero, I had this signal to send out, okay. So root two ES by T cos two pi M0 by T, okay. This is between zero and T, okay. I said F0, I will let as M0 by two, M0 will be a positive integer, okay. And then I said X1 of T is going to be root two ES by T cosine two pi M1 T by T. Again, the same time interval. Never let F1 be M1 by T, okay. So I asked you to figure out something. If you want to even plot this, KM0 and M1 will be integers because of that. In fact, these two signals themselves will be orthogonal, okay. So if I integrate from zero to two T and multiply, you'll get zero, okay. So if you do basis vectors, you'll get two different basis, okay. So you can think of it as cos two pi M0 T and cos two pi M1 T by T, okay. So you'll get two different, so it's a two dimensional signaling scheme. What is the thing that I asked you to figure out? Yeah, so the complex envelope for X0 of T and X1 of T, okay. So what's the answer? What's the complex envelope of X0 of T? Give me a description if you don't have an exact answer. It's okay, I mean, if at all you've tried it, you can give the answer. We'll see if it works out or not. What's the first choice you have to make when you figure out something called a complex envelope? What is involved when you go from a pass-band real signal to a complex envelope? There's one quantity which you have to fix. The center frequency, right? So what's a good choice for center frequency for X0 of T? F0 itself, right? So you maybe pick center frequency as F0, okay. So what does this fix? When I say center frequency is fixed, what does it mean? Okay, my up conversion and down conversion circuits are fixed now. So what is the whole point of doing complex envelope? I want to process my signals in baseband, right? So if I want to process it in baseband, then I have to down convert with this F0. If I down convert with some other F, then I'll get something else, okay. Then all my processing might have to be different, okay. So I'm fixing F0 as center frequency for X0 of T. So it has a real meaning, okay. So even though I call it the complex envelope, it has a real meaning, okay. So once you down convert, you'll get two signals, the real part of X0 tilde of T and the imaginary part of X0 tilde of T, which you can process in baseband in your receiver, okay. So what's, suppose I choose it as F0, what is my X0 tilde? I'm sorry, it's not of T, you're giving me the formula, okay. No, I mean, this may not be that difficult to figure out, right? So in this case, it's really simple. Go back and think about what X of T is in terms of the complex envelope, okay. Right, what is it? It's XI of T times cos 2 pi F0 T minus XQ of T times sin 2 pi F0 T. But that cos 2 pi F0 T and sin 2 pi F0 T are for all time, okay. So here, something else is slightly different. So how will you translate this? What's the only thing you have to do? What do you have to do? Okay, take some time, right. Put pen on paper, figure it out. Tell me what XI of T is. It's not really that difficult. Any answers, any ideas? If somebody has any ideas, it's also okay. Yeah, so what is it? You're right. I mean, so what's, yeah, you're right. I mean, X0 tilde of T can be real, okay. It's not something to be so often. Delta of T, there's no delta of T. What will it have to be? Route 2 is S by T between zero and T, do you agree? Does it seem fair? Okay, so can I say this is just simply 2S by T between and say maybe a suitably shifted rect, right? So I'll have to say T minus T by 2. Is that fine? That's not fine. It's not? Yeah, why? Oh, where is the band limitation coming from? I don't care about, you see, first of all, X0 of T. I made a lot of remarks about band width of X0 of T. What did I say band width of X0 of T is going to be? Theoretically, it's going to be infinite, okay? But in practice, you can approximate and say if I allow about F0 and what F0, if I allow, say 10 times 1 by T on, say 5 by 5, 5 by T or 6 by T or 7 by T, so enough lobes of the sink. If I allow, then I will get this signal almost, okay? So that's the assumption. So my bandwidth is supposed to be very large. So if we take such a large signal and do baseband equivalent, we'll once again not get a band limited thing, right? Okay, so it'll be once again for a large band, but assuming you have a large enough baseband processing bandwidth, so it's okay. So it'll work out, okay? Are you able to make sense of this in the frequency domain? Are you happy with this, the Hilbert transform type picture? Is it working out? Okay, right? Yes or no? No? I don't know. Think about it, okay? So try to make sense of it in the frequency domain and see if it makes sense, okay? So I think this is enough, like he points out. Yeah, yeah. So you'll have to think of some approximations and all that. So this is a good thing, a good way of doing it. But what about X1 of T? How will you deal with X1 of T? Okay, one way of doing it is, say I'll change my center frequency now. I'll say I'll change my center frequency to F1, but is that a fair thing to do? Do you think that's, at least from practical point of view, is that reasonable? What will it mean if you do think of X1 of T now with a different central frequency? First of all, in practice, then you'll have to have another down converter or up converter with a different frequency. Maybe that's not too bad. You'll say that's just a cost I'll add. I have enough money. What else is a problem? Yeah, that's a problem, right? Signal is random. You don't know whether it's being F0 or F1, okay? So you won't know what to do with your down converter. Okay, you can't keep doing both and trying to figure out what it actually was later, okay? It'll be disaster. You might as well do just one, okay? So the whole philosophy of the thing will change once you realize the signal is random. So you don't know what you have to fix some frequency. So a good choice is to fix F0 for both, okay? You fix F0 as the center frequency for both. If you do that, what will happen for X1 tilde? Okay, same center frequency. What will happen to X1 tilde? Okay, anyway, I'll put a question mark next to it since enough people have questions. Think about it and you guys have to come back with me and tell me if you're happy with this answer, okay? So make sure the mathematics works out. Read up something about the complex envelope. See if it works out like you've seen it before. So what will you do for X1 tilde of T? So I think the zero to T is what's confusing people. So if it was just delta for all time, then yes, answers will come out, okay? So that's a good way to start. I'm assuming T is very large, okay? So you have the whole cosine. Then what would you get if you take cos 2 pi f1 T and downconvert it to the center frequency of zero? What will you get? You'll get E part J, 2 pi f1 minus f0 T, okay? That's what you would get. So here, what can you expect change to be? It's going to be defined for, multiplied by the rect, okay? So pretty much that's what will happen, okay? But once again, go back and make sure all the mathematics of the definition of upconversion and downconversion are working out properly and everything that you can do, you can do that, okay? There'll of course be a problem with this band limitation. So you approximate the band limitation to, say you band limited to some large enough thing and make sure it works out, okay? So it has to work out to something like this, E part J, 2 pi f1 minus f0 T between T and T. Okay, so you see the complex envelope works out to different types of numbers, all right? So you can also pick other center frequencies and make both of them complex if you want, okay? So you can make, pick something like f0 plus f1 by 2, okay? Maybe then both of them become complex. So that's just a change of basis type thing. So it's not a big deal, okay? I'm sorry? Times? Oh, the root 2 ES by T. You're talking about the root 2 ES by T? Yeah, yeah, that'll be there. That'll be there. I didn't write it. Yeah, yeah, you'll have the root 2 ES by T. Sorry. So I mean, once you think of going back to practice, right? These constants are not too crucial, okay? Ultimately, there'll be some variable resistor or capacitor which you'll tune to get whatever constant you want, okay? So don't worry so much about the constant. Yeah, you have to worry about these things when you do the detection, okay? And the PDF, when I write down the signals, it's quite irrelevant what these constants are, okay? Of course in the exam, it's very relevant. So can't say changing by constant, okay? All right, so convince yourself that this is right or if you have more thoughts or if you think the answer is wrong, something else is right, figure it out and come back and convince me. But I think this is a good enough baseband signal to process. If you don't convert, you'll get something like this with large enough bandwidth, okay? All right, so that's the FSK part of it. And let's go back now to the signal space. How did it look in signal space now? So the basis we picked where? The basis you can pick is root 2 by T cos 2 pi F naught 2 pi F naught T. And again between the rect multiplication is always there and F1T is root 2 by T cos 2 pi F1T, okay? Between T and T, okay? You have to pick F0 and F1 such that, I mean, I'm assuming F0 and F1 have been picked so that these two are orthogonal, okay? Or orthonormal, okay? So you have a integer difference in terms of frequency, so that'll work out, okay? So once you do this, your signals become what? Your vector X0 becomes, right? It becomes a two-dimensional vector. What is a two-dimensional vector? Root ES and? Zero. Okay, and X1 becomes again a two-dimensional vector, zero and root ES, okay? So this is nicely plotted in a 2D signal constellation. Okay, maybe I'll do it right here, okay? So this is the FSK constellation, root ES here, root ES here, say maybe this is zero and this is one, okay? So bit zero goes to root ES and bit one goes to the other root ES, okay? So if I define my random vector X, okay? How will I define it? Equals X0 with probability half and X1 with probability half, I assume my bits are equally likely, right? Zero and one are equally likely with probability half, okay? So I can also think of this X as a random vector with two components, X1 and X2, okay? What's the PMF for X1? Root ES with probability? Half. Half? Zero. And zero with probability? Half. X2 also is identically distributed as X1, okay? But are they independent or X1 and X2 independent? Clearly not independent, okay? So together they take only, X takes only two possible values, everything else is zero. So it's clearly very, very dependent, okay? In fact, totally dependent, okay? But they are identically distributed, okay? So if you wanna compute energy, what will it work out to? Okay, you can compute energy in this form, right? Expected value of mod X squared, okay? This will work out to ES, okay? So that's the motivation for this notation, root ES by T, two ES by T and all that, okay? So usually it's very common to fix ES as one, okay? Like I said, those constants don't matter, but you have to keep N0 as variable to make sure the detection problem is being faithfully captured. All right, so that's FSK. And so once again, so if you go towards the detection problem, so I wanna spend some time on that because this is a strange situation. So what do you do with the receiver once again? Your R of T, okay? Your R of T is gonna go through two correlators, okay? One correlator is with respect to cos two pi F0T, another one with respect to cos two pi F1T, okay? So what will be match filter and all that? So if you want to match filter, so you should wanna implement the correlation as a match filter, you want a filter with response, impulse response cos two pi F0T, can you think of a very simple circuit which will have that impulse response? What kind of circuit will have that? Very simple RLC circuit. So LC, yeah, that's the LC oscillator will have that, okay? So you have an LC and make sure you drain off all the, there's a thing and then turn it on, it'll go to that. So an LC is a nice way of implementing this, okay? So, but you have to be careful with the timing, okay? So if you wanna do it continuously, you have to drain off, shut off and all that. So if you can manage to do that, you can do these correlations with a simple LC. So you'll see some very, very simple systems, these RFID tag communicators, I don't know if you're familiar with RFID tags, they'll use systems like that. So you'll see in those readers and all, people will use simple LC stuff to do reception, it's possible to do such things, okay? So you have, strictly you have correlators with, I don't know what I called it. They call it F0, F1 or F1, F2, F0, F1, okay? F0, T and the other correlator is with F1, T, okay? And this gives you Y1 and Y2, okay? I've been doing F1, F2, I think in the rest, so I'm sorry for the 01 conversion, okay? And then you go through our detector, produce X hat, okay? So what does the detector need to know once again? The joint PDF of Y and X, okay? So how will you do that? Like I said, since X is discrete, what can be readily described as the conditional PDF, Y given X, okay? So what is F, Y given X, Y given, let's say, okay? So I'll specify B here, okay? B, specifying B is the same as doing, okay? Anyway, I'll do the whole thing, it's not a big deal. I'll say root ES0, okay? What is this distribution? Gaussian, 2D Gaussian with mean? Root ES0 and variance or covariance function, which is, which we'll work out to? Yeah, N0 by 2, identity matrix, okay? So that's how it will work out. So it'll be circularly symmetric about root ES0. So that's how the PDF will look. So if you plot this, if you do a 2D plot of this PDF in MATLAB or something, it'll look circularly symmetric about, it'll die down in all directions about root ES0, so it'll look, okay? What about the PDF for Y itself? Half of this plus half of the other normal PDF, 2D PDF, centered at zero root ES0, okay? So that's how the joint PDF will look, okay? So this is the information that the detector cares about. We don't need anything more for the detector, okay? So you have this, you can do a detection. Okay, what do you think the detector will do? Yeah, so you'll split about Y equals X and if you fall on this side, you say it is that point, if you fall on that side. Since it's all circularly symmetric, it has to work out eventually, okay? So once again, we'll see a rigorous form of this as we go along later, okay? So this is a good reason why I'm repeating myself in these things. I'm going much slower than I usually should maybe, but I think it's good to spend time with these things. These are very basics, very important things. You shouldn't go back and question yourself on these things later on, okay? So make sure you understand, actually the complex envelope picture is also reasonably important, okay? So make sure you understand that stuff very well in your head. But finite bandwidth, yes, it'll be much simpler, but even in this case, you should maybe know what the idea is behind that, okay? Think about that. Okay, so let's go to the next example that we'll see, which I will call four PAM, okay? Four PAM, okay? So here, we'll start doing the, we'll start doing the, what? Just give me one minute. Okay, we'll start doing the reverse. So I'll pick my basis first, and then I'll give you a constellation, okay? That completely gives you everything you want, okay? So just to give you more to it. So I'll say N equals two is what we pick. So my vector B is going to be two bits, B1, B2. So there are four possibilities, all with equal probability, I'm assuming, okay? 0, 0, 0, 1, 1, 1, 1. I'll say my basis is like we did for BPSK before. So BPSK can also be called two PAM, okay? It's also called two PAM. So here, we're doing four PAM. So the basis we pick will be just like the baseband type basis, one by root T between zero and T, okay? So my constellation with respect to this basis will be, I'll have zero here, but zero is not a signal point. I'll have four different points, okay? So I'll drop my root ES stuff from now on, okay? So get used to this root ES being one from now on, okay? So the reason I stuck on to that root ES is most books that you see will always do ES, okay? So I don't want you to get confused by that. But I will drop my root ES, I'll say ES is one from now on, okay? So we'll see later on in the detection problem that that's good enough or maybe in the detection problem we'll do root ES and then show you why ES can be one without any problem, okay? So we'll say this is plus one, this is plus three, this is minus one, this is minus three, okay? Obviously this picture is not to scale, okay? Why is this not to scale? Yeah, right? So plus three is suddenly much compressor. Maybe I should make this to scale, okay? We'll do that, okay? It's not too bad, so it's good to do this to scale. This is not too bad, plus three, it's not too bad, minus three, okay? So basically four equidistant points on the real axis, symmetric about zero because typically you might want your mean to be zero for your random process that you put up, okay? So this constellation and the basis completely defined all my actual signals. So how will my signals look? I'll have four possible signals, one corresponding to each choice of bits, zero, zero, zero, one, one, zero and one, one. I haven't put out what that is. So maybe one way of doing that is to say zero, zero, zero, one, one, one, zero. Maybe this is the way I do it, okay? So what will be the signal corresponding to zero, zero, okay? So if you want to plot, say for instance, X zero, zero of T, what will it be? It will be root one by T between zero and T, okay? What about X zero, one of T? It will be minus root one by T between zero and T. What about X one, one, three times or minus three times, and then X one, zero will be plus three times, okay? So it's amplitude modulation in the sense that you're changing the amplitude of the signal between a certain time interval to figure out, to send different bits. So it's amplitude modulation pulse because some reason we'll come to it later, okay? All right, so my random variable X that is going out, random vector X is just one variable, right? You have only one basis vector. So it's just one, it takes four different values all with equal probability, minus three, minus one, plus one, and plus three, all these things with probability one by four. Okay, so that's my, the distribution for my input random variable, okay? You have only one basis vector. So at the receiver, R of T should go through only one correlator, okay? So it's just one correlator, you'll get only one Y, okay? Okay, that's there. What about N? You have only one N, okay? So the Y becomes simply X plus N where N is normal with zero mean and variance and not by two, okay? So what would be the conditional PDF of Y given X is plus one, be normal with mean plus one variance and not by two. If X is plus three, it'll be normal with mean plus three and variance and not by two, okay? So it's just a simple reputation of how that happens. If you want to do the joint PDF of Y, how will that look? There'll be a mixture Gaussian, right? Mixture Gaussian, there'll be four Gaussians, one fourth weighted with center at minus three, minus one, plus one, and plus three. Okay, that's how it looks. It's a very simple extension. Based on this also, you can do a detection process. What do you think the detection process will be? You'll have several different thresholds. Closest point is what you'll pick as your perceived point. For instance, if you receive 2.5 plus 2.5, what do you think you'll say? Plus three, minus 2.5, it's gonna be minus 1.5, it'll be something close to one. So it's a very simple detection process that you can do, okay? So even though I receive an R of T, which could be a complicated signal, I'm gonna correlate it with phi one of T, get only one Y, and I can make a decision based on that. So I went over this point before, okay? So if you transmit this X, zero, one of T, you will get what receive vector sum. Squiggles will be there, right, around this, okay? So you don't have to worry about this value of that receive signal at every time. What can you do? You can simply correlate with phi one of T, all that information about X. The bit is there and that after the correlation, okay? So that's good enough, that's the motivation for this. Okay, so that's four PAM. Now I can also do a more general version, okay? So one more thing we have to do, sorry, I forgot that. The power of the signal, we have to compute. What is the expected value of X squared? Five by four, sounds too low, man. Between plus three, minus three. Five. Five, right, five, okay? So it's not really normalized. Maybe you want to normalize it. How will you normalize it? Divide everything by root five, you'll normalize. If you want, you can do that, but it's okay. So yes, we can live with five, okay? Really doesn't matter, it's just a number, okay? All right, so this is some computation you can do, so I'll quickly introduce the most general version of four PAM, what's the most general version of four PAM? You can say M PAM, okay? Where M is a power of two, okay? So four PAM, you can start with two PAM. Two PAM is the simplest case. Then you have four PAM, eight PAM, 16 PAM, 32 PAM, so on, okay? So again, I'll do a very simple definition for my set of transmitted signals, okay? So if M become 64, I'm not going to write down all the 64 transmitted signals. I'll say basis and a constellation, okay? So I'll give a general definition for how that constellation will look. My N, which is the number of bits I'm using will be log M base two, okay? Does that make sense? Should it be log M base two? Yeah, you can map five points, one, two, a 32 PAM constellation, okay? So you have 32 points, you can do that, okay? So my vector B is going to be B1, B2, so on till B log M base two, okay? Right? And my basis is the exact same as before. Just one basis element root one by T between zero and T, okay? And my constellation point is going to be plus one, plus three, plus five, so on till M minus one. On the negative side, minus one, minus three, so on till minus M minus one. That's my constellation. So I've defined my signaling completely. If you go ahead, I'll assume all my bits are with equal probability. Do this computation, this will be useful. You can compute it to be M squared minus one by three. And if you do Y equals X plus N, you will get mixture Gaussian with all of these guys. What's one more thing to figure out? There's one more thing you have to figure out, which is how do the set of bits map to each point, right? I have a set of say, for instance, if I pick M equals 16, I have to map four bits to 16 points, okay? How will I do that mapping? There are several ways of doing it. One way of doing it is to simply start at the left and go from all zeros to all ones. You can do that. There are some people who studied such things and said maybe that's not the best thing to do. There are all kinds of other mappings out there. But we'll ignore that for now. Come back to it later, maybe. Okay, so you can do it in any way you like. All right, so it's top here now. You'll have to do your...