 Okay, so this is lecture 9 am I right, 9 and the last thing we saw in last class was M PAM right, what is M PAM? So when you do M PAM signaling, so M is typically a power of 2, that is accepted. What is your signaling constellation, your basis is chosen to be? In this case when I have a lot of bandwidth compared to my signaling bandwidth, I have a lot of bandwidth to use, I can take my basis comfortably to be root 1 by T between 0 and T. We will see later on a case where the bandwidth is not so free, so it is a little bit tight, you will have to change your basis, but it is still work okay, so that is the first thing. And then once you have this basis, I can simply do a constellation description, I can, I have only a one dimensional signal space and that one dimensional signal space, I will simply pick out the points which are going to be the scaling factors for my basis okay, so those are what? For the M PAM it is going to be, so if you have 0 here, I am going to pick plus 1, plus 3, so on till what? Plus M minus 1 okay, you can check that you will have M by 2 points in this way okay, on the negative side it is going to be minus 1, minus 3, so on till minus M minus 1, once again M by 2 points on this side, overall remember 0 is not a point okay, so maybe I should just do, just to show that that is not a point, it is just an axis okay, so the other constellation points only, plus 1 through plus M minus 1 in steps of 2 and minus 1 through minus M minus 1 in steps of 2 okay, so all of them you can say occur with equal likelihood okay, I should also provide the information of how bits are being mapped into the constellation points right, so how many bits would you map here? Your N would be log M base 2 okay and your bit vector will be B1, B2 to BN okay, so in this case it is just log M base 2 and you have 2 power N possibilities which is equal to M and you have M points here, so right now we will not worry about what is the best way of mapping the bits to the symbols or some ways of doing it, but let us just say you pick a mapping, any mapping is okay as far as I am concerned okay, so you do that, you get a transmitting scheme okay, so how do you build a circuit which does your transmission? You first read your bits, depending on your bits, you pick one of your say scaling factors okay, so that multiplies your phi 1 of t okay, so you might think of all kinds of resistors and capacitors to build it, so in practice today most of these systems are built with in fact D2A converters okay, so what are you putting out ultimately at the end of the day? Putting out a waveform, so imagine sampling it with a huge sampling frequency okay, so then run a D2A and all you have to do is do a sample and hold on your D2A and pick different samples according to your modulation, so it can be easily programmed into a processor with digital output and converted that into a using a good D2A converter, so these modulators are done very differently okay, you do not have to worry about building complex analog circuits, of course if you build those analog circuits what is the advantage? You will be consuming very little power and all these things in your transmitter but people do not usually care, the flexibility in your transmitter is what is more important today, so you just put a processor, program your processor to put out a waveform like this okay, it is not too difficult to do that okay, so it can be done, so that is how usually you do it, so I can today think happily about my transmitter completely in terms of signal space and points, I do not have to worry so much about going into an actual implementation and all that okay, all that I will imagine is being done in a routine fashion okay, so that is the description, so one thing we know is accurately represented in the constellation is the average energy okay, so you can compute average energy assuming once again that all these bits are equally likely okay independent IID type bits, this will work out to if you do the computation it involves some careful summation right, it is not so trivial if you have to do a careful summation it will work out to M squared minus 1 by 3 okay, so it will work out like a square of an AP okay, so of an arithmetic progression right, so when you do that it is you have to use some tricks to make sure you know what the closed form expression is but it can be done okay, it will come out to M squared minus 1 by 3 okay, so that is the power, so bandwidth like I said I have been saying it is I am assuming it is enough, I am using enough bandwidth, I am using a lot of bandwidth maybe it is a lot of waste, we will not worry about that for now, what about the other things, the other thing I am worried about is bit rate right, so in every T seconds how many bits am I sending log M base 2 okay, so that is a lot of bits okay, so how do you increase that number, you increase M okay, so but you can also imagine for the same power, same energy when you increase M what happens right, energy is increasing right, so if you want to keep energy the same you cannot afford to increase M also, so there is a payoff to be played here okay, so noise I did not really talk about, noise would be again it would have just one dimension, so y is going to be equal to simply x plus n, n would be simply normal with 0 mean and variance and not by 2 okay, so these are the various parameters, you can do the conditional distributions for y given x very easily, each conditional distribution will be a Gaussian centered at that point with mean as the point that was transmitted and variance being n naught by 2 okay, so you can also do the joint distribution, the PDF for y itself which would be a mixture Gaussian, so all these things are descriptions for M pair okay, so hopefully this is clear okay, this is the rudimentary thing, so if at all you go into, you are given some bandwidth and power and if you have to construct a communication system, the first thing to try would be something like an M PAM okay, assuming you have enough bandwidth or assuming your T is large enough so that you are not taking too much bandwidth, that is the first short implementation without worrying too much about anything else, we do not know how to do the receiver right, we do not know how to do the complete receiver right, what is the first step of the receiver? The correlator right or integrate and dump basically, you integrate for a period of T, in this case the correlator becomes a simple integrate and dump, do you see that? It is just a constant, basis is a constant, so you will be just integrating your received signal from 0 to T and taking that value okay, so you know that step, from y how to go to x also is very clear, intuitively maybe it is very clear to you, just take the nearest point it will be okay, we have not seen that rigorously enough, we will see it so okay, so even at this point like at the ninth lecture, you know how to build a rudimentary communication system right, can you or can you not? At least theoretically you know how to do it, it is not very difficult and all these things like I said transmitters and receivers can be done with processors and completely digital, so you should be able to do it, you should be confident of being able to take a board which has a processor and a D2A converter, being able to do a PAM transmitter, it is nothing in it okay, as long as a transmitter is fast enough you can easily do it okay, same thing you should be able to do at the receiver, all you need is an integrator and dumping which can be done very easily in software also right, so it is just adding or how hard is it, you can do it, so this should be confident of being able to do okay, even in reality alright, so this is a very standard communication system, so M PAM is a very very commonly used system, so you can imagine 2 PAM is the simplest case, PPSK it is commonly used 4 PAM, 8 PAM and all that okay, as you go to larger and larger M, you are sending more bits at the same time but you are also using more energy okay, so if you can do that then it is fine alright, so that is the M PAM, the next thing I am going to talk about is another type of transmitter which is actually 2 dimensional, which is I think the 5th case we are looking at, it will go back once again to phase shift keying okay, we saw binary phase shift keying matched with 2 PAM but you can also expand it in different way to get what is called quadrature phase shift keying or QPSK okay, you can also call it 4 PSK if you want, that is another name okay, phase shift keying okay, so I think the way I have written it down I am going to do it slightly more painfully okay, so I am going to go back once again to the very basics and give you the signals first and then we will work out the basis elements, do the Gram-Smith once again just for completeness okay, N equals 2, so you are mapping 2 bits at a time okay, so you have 2 bits B1, B2 okay, 4 possibilities occurring with equal probability, 1 4th each okay, that is my assumption on this okay, so my X 00 T which is the signal corresponding to the bits 00 is going to be root 2 ES by T okay, I will go back to ES, cosine 2 pi F0 T between once again between 0 and T okay, so once I say this naturally, so the bandwidth assumption should come in naturally into your head, so this is going to occupy a lot of bandwidth okay, so I have to have all that bandwidth to be able to receive this X 00 of T at the receiver without any problems in the channel, so I am assuming that is happening, I have a lot of bandwidth and I am using a lot of bandwidth in this okay, so X 01 of T okay, so this F0, what is this F0, this F0 will take to be some M0 by T okay, some not necessary, some large frequency okay, so we will think of this as the center frequency okay, so for sending other bit combinations we will shift the phase of this sinusoid in the interval 0 to T okay, so we will be shifting the phase, but I will write it slightly differently, it is all the same, I can shift cos, I can get sin right, so I will write it in that form, but just to make the grams with more interesting, but you can also think of it as simply phase shifting with the cosine, there is no problem there, root 2 ES by T sin 2 pi F0 T between 0 and T okay, so you see it is clearly a pi by 2 phase shifted version of the previous one okay, so the X 11 of T is once again going to be minus root 2 ES by T cosine 2 pi F0 T, which is now what, a pi phase shifted version of the first one okay, and then you have X 10 T which is minus root 2 ES by T sin 2 pi F0 T which is a 3 pi by 2 phase shifted version of the so that is why it is called phase shift keying okay, so each bit combination tells you how much you have to phase shift by okay, so once again the bandwidth is large, assume my bits are uniform, so if you do Gram-Schmidt what will you get, yeah two basis functions, one will be the cosine, other will be the sin normalized right, so everything else is a multiple of that okay, but it is two dimensional, it is not one dimension okay, so that is the final result that you get okay, so if you do Gram-Schmidt the basis are okay phi 1 T which would be root 2 by T cos 2 pi F0 T okay, of course once again between 0 and T and phi 2 is root 2 by T sin 2 pi F0 T between 0 and T okay, so you get two basis and in terms of the basis every signal now becomes a two dimensional point right, so the signal space is two dimension, so if I were to write down the constellation you would have four points okay, so this is root ES minus root ES minus root ES and root ES again and the way I wrote it down if this represents phi 1, this represents phi 2 okay, so you are going to have this being 0 0, this being 0 1, this being 1 1 and this being 1 0 okay, so that is it, this is your constellation, this is QPSK, so I can write down the vector version of each of my signal points okay, so x 0 0 vector is what root ES 0, of course there is a transpose okay x 0 1 is 0 root ES okay x 1 1 is minus root ES 0 x 1 0 is 0 minus root ES okay, so there is nothing wholly about the way I have chosen my assignment 0 0 0 1 when you can choose any way you want, I just picked it like this, you can pick it in any other way, the constellation does not change, many of the properties do not change, maybe some minor properties change, it is not very relevant to us okay, so it is fine, we will keep it anyway we want okay, there is also another version of QPSK called offset QPSK, in that case the constellation is rotated by pi by 4 okay, so if you rotate this constellation by pi by 4 what will you get, these points will be on the four corners of a square okay, so that is offset QPSK, so if you go back using the same basis you will see your signals will all be shifted by pi by 4 phase, that is okay, that is not a big deal and that is supposed to have some advantages in some practical implementations in terms of signals that is supposed to be better, but phase is just relative, so what matters really is the difference between two phases, absolute phase itself is really no significance but the offset QPSK is supposed to have some advantages okay, so you can see why it is called phase shift king and all these things and you can also see it is different from four PAM right, it is very different from four PAM okay, it is first of all two dimensional things change a lot okay, so it is very different from four PAM hopefully you see that okay, one more thing you see is each signal has the same energy okay right, all the four signals you are transmitting, all of them have the same energy ES and the average energy is also ES, but in four PAM it did not work out like that, some signals have higher energy, some signals have lower energy and average works out to some number okay, so that is another point which is different about phase shift keying, that is one more reason why phase shift keying sometimes preferred okay, you have all the signals being over four PAM for instance okay, you might have a reason why you might want to send QPSK better than four PAM because all your signals will have the same energy, you do not have to worry about suddenly some signal having a larger energy okay, you can plan for the for every time properly okay, so that is PSK, so if you do for instance the model completely you will get the vector y equals the vector x plus n where x is uniform here, the reason why I am saying uniform is I am assuming my bits are uniform okay, so I am saying uniform on that, what is n? It is IID normal two dimensional which means zero and variance n naught by two, so once you know this you can do conditional PDFs for y1, y2 okay, so what will it be? It will be a circularly symmetric Gaussian centered on the transmitted point okay, so all that can be done, it is quite standard as well alright, so there is also a generalization for QPSK okay, so there is no reason why you have to divide the unit circle into just four different, pick only four points in the unit circle, you can pick any number of points, you can pick eight points, you can pick sixteen points without any problem, you can keep on increasing the number of points you pick and you will get more and more signal points in the constellation, you can increase m in that way okay, so that is the next step which is called MPSK okay, here m will take to be a power of two, there is no reason why it should not be a power of two okay, once again the basis vectors are the same as 4PSK okay, you have just cosine and sine okay, is that a question floating around, happy okay alright, so you have cosine and sine and I will just give you the constellation okay, so the same basis vectors phi 1, phi 2 as we had before and the constellation it is good to draw a circle so that you can pick out the points there okay, so I will take so I do not know if I have actually taken a proper, so I do not know I think okay, so I will just take four okay, so we will keep picking points like this, maybe you do not get a point and pi by two I do not know, so it just keeps on going like this, you pick at angles two pi by m, so this angle would be two pi by m okay, same angle here also okay, so on okay, so so an arbitrary point okay, so basically you pick m points on the unit circle, the unit circle, my circle got totally messed up okay, what are those m points, maybe root es okay, e power j two pi small m by capital M and m goes from zero to m minus okay, so those are my m points and notice how I wrote it down, I wrote it down as a complex number okay, so this is again very common way of referring to constellation points on in two dimensions right, in two dimensions you can think of a constellation point as being referred to by two coordinates okay x and y okay, so I can always write it as x plus j y which is which I think of as a complex number okay, so this is also very common when people write 2D constellations particularly, always think of your 2D constellation points as a complex number okay, so for the case when m equals 4, you will get back the original case okay, so two pi by 4 is pi by 2, so you will get the four points okay, so how do you think of the four points and four PSK in terms of complex numbers, zero j no, one j minus one minus j, so there are four points plus or minus one plus or minus j okay, so in the MPSK case you have e power j 2 pi m by m okay, so of course the root t s is floating around usually forget the root t s in most cases okay, is that clear, so then you have to map your log m base two bits into each of these constellation points and you can do that in any way you want okay, so this seems to have an advantage over m-pam what's the advantage, when you increase m in m-pam what happened, the energy increased but here the energy doesn't seem to have increased, so why would you ever pick m-pam over MPSK, yeah so okay, so yeah, so eventually you'll pay, you'll see I mean even though you've not paid in terms of e s, you'll pay in terms of some other quantity which is more fundamental, so actually it turns out the e s by n not, that ratio is what's important okay, so even if you don't pay in terms of e s, you'll pay finally in some way or the other okay, so another thing you've done here is you packed the points closer together okay, the distances between points also gotten very very low, so you're paying in terms of that also okay, so we'll later on see how to analyze it at that time, that time maybe it will be clear okay, what I want you to do now is to spend some time and write down the transmitted signal corresponding to the m-th point here in the constellation, what is the actual transmitted signal xm of t corresponding to the m-th point in the signal constellation, is that clear, every point in the signal constellation has to correspond to an actual signal right, what is the actual signal that is transmitted when you're sending the m-th point, that's what I want you to do, you know the basis, you know the factors, so you can multiply and find out what's the answer okay, so there's a possibility to simplify that, you can write it down as one cosine right, can you not, cosine 2 pi f naught t minus 2 pi m by m, do you see that okay, so this of course will have a root 2 e s by t on the outside to make sure things work out okay, so this is the signal corresponding to root e s e power j 2 pi m by m okay right, so this happens for m equals 0 to m minus 1 okay, so this defines everything, is that okay, that fine seems to be some, are you guys happy okay, it will work out like this okay, anything else they wanted to say, you can be smart about it and write it as real part, so that's what I'm going to ask you to do next, which is a little bit more confusing, I want you to write down the, once again I mean this is between 0 and t you know, okay I've been saying bandwidth is large but, assume bandwidth is large but finite, okay don't think of infinite bandwidth very strictly okay, so it's a passband signal, do you agree around f0, so assume around f0 some 10 by t is available for me, 10 by t or 20 by t and that's my bandwidth, I'm restricting it to that but, approximately I'm writing the answer as this okay, so that's what I'm doing, assuming that what is my complex baseband equivalent for this, so the reason why I'm asking you to assume that is last time people said it's not strictly passband, it's not divine, it's assume something like that and then give me an answer, yeah you'll have a minus, minus is quite irrelevant, you can even think of it as plus but anyway it doesn't matter, we'll retain that, so you'll see xm tilde of t will work out to something like this root 2 s by t e power minus j 2 pi m by m, this minus is quite I mean it's okay for m equals minus m 2 pi minus it will work out to the same thing, so minus is not so important but the fact is this complex number representation I had for each of my constellation points actually is what, it corresponds to something in the complex envelope okay, so it has a real signaling meaning also, so that's why in 2D it's good to think of it as complex numbers okay, so when you're processing in baseband, you're processing this complex number, you can think of it okay, so this is my complex envelope okay, of course between 0 and t to make sure everything works out okay, is this clear, of course for this complex, this number also you'll have a huge bandwidth around 0, maybe convince yourself that that's true okay, try to find the, well it's quite simple to see what the Fourier transform of this will be right, it's a constant thing, so you'll get a rect sync multiplied by some phase shift, so it'll be, since the in time domain it's complex and frequency domain you won't have symmetry and all that, it'll be a little bit weird, it'll be a shifted type thing, it won't be a proper sync around 0 okay, it'll be shifted by some arbitrary slight phase but you can work it out okay, it's not a big problem, try to spend some time on this and convince yourself you understand what I mean by all these bandwidth being huge okay, so in future we'll, what we will do is in fact you can see what can happen here, so what we'll do is in future we'll write down the signals pretty much in baseband okay, we don't even have to write down the passband equivalent signal okay, so I'll write down the signal in complex baseband, if I write down in complex baseband what is my basis right, my signal space is the regular complex plane, I don't have to worry about anything else okay, so I don't have to worry about the Gram-Schmidt, I don't have to worry about all this ortho normalization I did okay, so this will be a standard trick that people will use, it's always assumed that your bases are cos 2 pi f0 t and sin 2 pi f0 t for large enough f0 and then you think of your signal space and everything as complex numbers which represent the complex envelope in complex baseband okay, so you can also think of it that way, so in fact this basis will not even enter your picture, you won't even keep track of the actual signal and cos 2 pi f0 t and sin 2 pi f0 t, you don't have to worry about that at all, just worry about baseband okay, complex baseband in which case all you have to worry about is just the complex plane, all your signal constellations are in the complex plane okay, so that's a very common way to think about these things okay, but be very very careful when you think like that okay, every point actually corresponds to a actual signal that you are transmitting and you should know how to go to that okay, so you should know that but since this complex number and the complex envelope are all making a lot of sense, as you get used to these things you will tend to think of your entire signal constellation as just a complex number okay, it's made of complex numbers, but be very careful about the bandwidth that it implies what it means in terms of so many things okay, so be very careful about that but you can think of it as a complex number for ease of use and all that okay, so that's the motivation for this complex envelope thing all right, okay so let's do the final example, this is the final example yeah, final example which is also very very common and popular which is just the seventh one, which is called M squared QAM okay and M is the power of 2 okay, so it's also possible to have M QAM where M is the power of 2 okay, so for so what is the difference between M squared QAM where M is the power of 2 and M QAM where M is the power of 2 yeah only you have even power, so 8 for instance I won't have 8 QAM, 32 QAM, 128 QAM, I'm not allowing when I say M squared QAM, I'm only allowing what 4, 16, 64 and so on okay, it's also possible to have 32 QAM, it's used in several practical systems okay, I'll talk about that later okay, so it's just a minor modification of what is there okay, so we'll talk right now about M squared QAM and the best way to think about it is it's the pass band version of M PAM okay, so think of several analogies here if all these things are not hitting your heads or several analogies to keep in mind okay, as long as you're in base band you're doing base band real signals which means you roughly have only one dimension to work with okay, you don't have this complex number stop, so by going to pass band like in the PSK case, we somehow get two dimensions and we're able to deal with complex numbers in base band okay, right, I kept talking about it when I talked about base band equivalent, so all those things are happening now okay, so you can think of 4 PSK or QPSK as the pass band version of BPSK okay, so likewise here this M squared QAM will be the pass band version or the complex number version of M PAM okay, so the constellation set will look like this, I'll write down the complete one okay, it will be two dimensions again the same phi 1 and phi 2 okay, what is phi 1 and phi 2 cos 2 pi f not t sin 2 pi f not t for some large f not between 0 and t okay, so it works okay, so I'm going to do a few things here, so first I'm going to mark out points on the axis, these are not constellation points, these are just markers to keep me scale okay, so I'll mark out multiples of root ES okay, 1, 3 okay, no let me not do, let me not do root ES, this is too complicated, I'll just do 1, 3 so on till M minus 1 okay, likewise in the negative direction I'll do minus 1, minus 3 so on till minus M minus 1 okay, I'll do the same thing for the y-axis as well okay, 1, 3 so on till M minus 1, minus 1, minus 3 so on till minus M minus 1 okay, and all the points where I have these both these coordinates happening like intersections of all these points will be my actual constellation points, so what are my constellation points here, I'll pick all these guys, then those guys, these guys entire points okay, that's the first quadrant okay, I'll do the same thing for all four quadrants okay, so you see I get M square points right, I have M points on each axis, M times M would be M square okay, so you do it for all the quadrants okay, so I'm doing it just painfully once to write down the whole thing, those are my M square constellation points all right okay, so this is M square QAM okay, remember what are my phi 1 and phi 2, cos 2 pi f naught t between 0 and t, sine 2 pi f naught t in between 0 and t, and I'm justifying that by saying I have a lot of bandwidth so this basis will nicely fit in, there's no problem okay, in future we'll modify that okay, so we'll change to fit into a much tighter bandwidth, at that time this will make much more sense, but for now we'll simply take it as a large, so don't think of sine and cos as some, so the Fourier transform for the phi 1 will not be a delta okay, so I think roughly layer but actually be a sinc shifted at that post point okay, so it's a large bandwidth that you're occupying okay, okay, so that's the first point, so how do we describe arbitrary points in this constellation okay, so I had an index M for MPSK and I wrote e power j 2 pi M by M, here the best way people do it is a plus jb okay, a and b belong to what, plus or minus 1, plus or minus 3, so until plus or minus M minus 1 okay, so that's an arbitrary point in my M square QAM constellation, once again I'm thinking of a complex number okay, so you can motivate yourself about why this complex number is very meaningful by going to the complex envelope, you'll see the signal corresponding to a plus jb will have a convex envelope, in this case it'll be a minus jb or something, so it's every complex number there is a meaning in that, so that's okay okay, so what else, what else, this is an arbitrary point and what will be, so my x which is the transmitted vector will be actually 2x1 and x2, each xi is what, what's the pdf for each xi, I'll say it's uniform in plus or minus 1, plus or minus 3, so on okay, each xi is actually an M-PAM signal right, you can think of it that way okay, so it's, we have two of them, we'll say they are iid, so the distribution for x is clearly defined, it's 1 by M squared each point taking with equal probability okay, what will be expected value of mod x squared, yeah it'll once again be M squared minus 1 by 3 right, there's just two M-PAMs happening, so each of them will have an average of M squared minus 1 by 3, so it should be also an average, is that clear okay, so it's this analogy between M-PAM and M squared QM is very very useful, so you can in fact think of it as two PAM signals, but they will be sitting in one real passband signal as opposed to a M-PAM signal which usually you think of as a one real baseband signal right, so that's the difference between M squared QM and M-PAM okay, so and you know why this is possible right, why you can have a complex baseband being just one real passband, of course you are actually using both sides of the bandwidth when you go to the complex passband, you know to go to the passband version of a baseband signal okay, so those are all these things are possible, so all these correspondences you should make in your head, so in practice typically if you have a passband bandwidth it always makes sense to do QAM okay, some quadrature modulation, there's no point in doing a real modulation okay, so your complex envelope should actually be complex, so no point in keeping that real okay, you'll be wasting bandwidth if you actually have passband bandwidth okay, so you'll be wasting half the bandwidth if you force your complex envelope to be real okay, so you have to keep it complex so that your passband signal becomes real and still you work it out okay, so that's M squared QAM, any questions anything that's that I'm saying disturbing you for this baseband, passband, if it is then go back and read also, now we can also ask, it's a good time to ask something which is not clear okay, so maybe we'll do that once again, we'll do the passband signal and then the complex envelope for the QAM case assuming a large enough bandwidth around f0 okay, so what's my signal corresponding to a plus jb okay, so maybe I'll write it as xab of t, what is this signal okay, it'll work out to a times root 2 by t cosine 2 pi f0 t plus what, b times root 2 by t sine 2 pi f0 t between t between 0 and t right, this will be the actual signal that I'm transmitting corresponding to a and b chose a plus jb chosen as the constellation point, how will I choose the constellation point, yeah the bits right, the bits will be mapped to constellation points, how many bits can I map log M squared base 2 okay, so that many bits I'll take and map it into a constellation point, I should fix that map ahead of time okay in some way I fix it, fix my constellation point, once my constellation point is chosen corresponding to that I transmit this signal, like I said this signal will occupy a large bandwidth around f0, so I'm assuming it is possible like have a lot of bandwidth around f0, so I have that okay, so I want you to write down the complex envelope for this okay, assuming you're taking a large but finite bandwidth around f0, so that this approximately comes through without much help, what is the complex envelope okay, in other words what am I saying you should do, you should come up with some signal here so that xab of t becomes real part of that complex envelope multiplied by e part j 2 pi f0 t okay, so that's your xab tilde, what will that work out to, a minus jb times root 2 by t, so root 2 by t will always come and then you have a minus j, so you can do that okay, so if you want a strict correspondence, if you want a plus jb to correspond to an a plus jb complex envelope, what should you do? Yeah, so you should think of sine as minus, you should pick your basis as minus okay, it doesn't change anything as far as power and bandwidth and nothing changes, everything is the same except mathematically it will be neater, it will be cleaner, so you pick one basis to be cos 2 pi f0 t, pick the other basis to be minus sine 2 pi f0 t okay, so once you do that you will get a very clean nice complex envelope correspondence okay, but it's okay, I mean we can live with this a plus jb a minus jb makes no difference okay, right, so I am going to remind you once again about this i channel q channel business okay, so how would you actually implement this in practice, so you would have in base band two signals right, what would those two signals be, in fact that would be this right xab tilde equals root 2 by t a minus jb for what time? 0 and t okay, so spend some time and find the exact Fourier transform for this okay, I want you to spend some time and do this okay, I think this is a good lesson you might have forgotten, anyway we have a lot of about 10 minutes left and I don't want to do anything else this class, so I want you to spend some time and find the exact Fourier transform for this, yeah it is base band right, it's constant between 0 and t, but it's a complex number between 0 and t, what is the expression? So you are saying this is what xab tilde f is what, root 2 by t a minus jb okay, e power minus j, ft by 2, t sync, ft okay, so are you convinced, is everybody happy with this answer, okay, so this is how it has to look okay, is it symmetric about f equals 0, does it satisfy your real and imaginary symmetry, so is the real part and the imaginary part right, I shouldn't say symmetric about f equals 0, what I want is are the real part and imaginary part right, so if xab tilde of t was real what would you expect, real part will satisfy even symmetry, imaginary part will have odd symmetry, so my question is check that xab tilde of f does not have those properties, clearly it cannot have those properties right, unless you have b equals 0 okay, so can you check that, what will go wrong, check very carefully, yeah the a minus jb will kill you right, so once you have that it will kill you, you will see everything else will be fine, if you didn't have that a minus jb it will be fine okay, once you have the a minus jb you will see the, it will give you a phase shift which will shift the things around and you won't have that real and imaginary part, so that's good enough, so having just, I mean it has to work right, that's the property of Fourier transforms, you must have known this for a long time, so it works that way, so check all those things just to be comfortable, a minus j, there is a pi okay, there's a pi f t, oh I missed a 2 pi okay, I'm sorry, so that will cause no problem, that pi f t will cause no problem, if it is there you will still have a real signal only right, so real signal shifted, it's no problem it's still real, but this a minus jb will kill you, you will get a real and imaginary part being okay, so hopefully you also know how to compute xab f okay, right from this part you can easily do it, from the, it will also occupy a lot of bandwidth around f naught, but it's okay, so this can also be computed, these are important exercises, don't turn in your answer script saying it's infinite bandwidth, so I won't do it okay, so if I ask question you do, in practice something will happen right, when you do this, so is it all interesting questions, people in interviews ask these questions and most of our students give such answers, they say this is a, so for instance I think the question was something like, it was a capacitor being charged, there was no resistor in the picture, the question was what will happen if I close the circuit, something like some such answer, some infinite answer was given okay, something has to happen in practice right, so as engineers you have to figure out what will happen, you have to make some approximation and say something will happen okay, so always here also assume such things are given to you, don't make, don't give me answers in terms of the other thing okay, so the last thing I want to point out with this QAM, you should draw a picture of how the transmitter will look okay, so you have two times log M base two bits coming in right, this is the bits that are coming in okay, what are you doing, you're first doing what's called a mapper okay, so people call it a bit mapper, bit to symbol mapper okay, you're mapping from bit to symbol, so what comes out here, some A plus JB okay right, so this is just one signal at your point, so what people typically think of is from here you have to convert this into XAB tilde of T okay, so this process is typically not given any name okay, so you have to think of it as XAB tilde of T, what is this XAB tilde of T you know, it will actually have two components, the I channel and the Q channel, what will the I channel have, simply A Q channel will have B okay, so this I and Q will now go into what, go into your up conversion circuitry to take it to whatever band of interest that is to you okay, so typically like I said what what goes into this question mark, I gave you a brief some two minutes lecture of how this typically is done, how is it done in today's systems, what would sit in here okay, like I said it's some processor okay, so XAB tilde of T is just a sample and hold type thing for some processor here, so what will this processor do, in this case it's very very easy, nothing to do, simply keep sending A at lots of A's okay in that particular time interval, lot of B's in that particular time okay, so imagine some sample and hold sitting on the other side and it works, in this case it's even ideal right, it's all constant, so it's ideal, so everything will work out okay, in future we'll see, we might have to put some filters here to make things work out properly, because you don't want infinite bandwidth, at that time you'll have to do something okay, so you'll have to do something, at there I'll talk more about it okay, so hopefully all these stages are clear, you have bits coming in, you map them into a complex number typically, which is a symbol okay, and those symbols are mapped into the complex envelope signal itself okay, which typically is done in baseband okay, it is baseband, what's typically it's, it is baseband and like I said usually people use digital processors and sample and hold type digital to analog converters to do that okay, and then you have the up conversion which is truly RF right, it's some oscillator, fair shifter all these things sitting inside, some RF goes on, XAB of T goes on okay, at the receiver what do you do, okay of course some noise is getting added here, I'm not showing it, the receiver what do you do, what do you do, you do down conversion first okay, so what all do you need for down conversion, what should you know for down conversion, the center frequency, the accurate version of the center frequency, the reason I'm saying accurate is, oscillators are notorious for being inaccurate and they also drift with temperature etc etc, so you need a lot of stuff there okay, so in most cases what's done is people derive the center frequency from the signal okay, it's possible to do that, then in that case everything will be accurate, if the signal drifts your center frequency will also drift, all those things are done, so you do that okay, then you get the INQ back okay, which is hopefully an accurate representation of your XAB tilde of T okay, like I said in practice today all this is done digital, so you down convert and you sample at a very high rate, so that you can think of this as a nice digital signal and then what do you do, even in the base, the regular PAM case what should you do here next, what's the operation, you have to do the correlation right, you have to do the correlation with your fee, so in baseband what is my fee, it's just averaging it's a constant right, remember the cost 2 pi of not T, sine 2 pi of not T is gone okay, that's a crucial difference okay, so far I've been talking about the signal in passband and I've been talking about fee 1 and fee 2 as cost 2 pi of not T, the reason is what, what's the advantage in talking about the signal in passband okay, I know the accurate bandwidth and the frequency and everything I know okay, but I can get rid of that and deal with everything in baseband and I'll be fairly accurate, I'll know the power, I'll also know the bandwidth right, I'll know the bandwidth also, the complex baseband bandwidth, it's just shifted when I go to up conversion and down conversion, so everything is captured nicely in baseband itself from XAB tilde of T2, that's the XAB tilde of T at the receiver okay, I don't have to worry about this up conversion and down conversion and I can think of my fee 1 and fee 2 completely in baseband okay, in the QAM case right now, it's just constant 1 by root 1 by T between 0 and T that's my fee 1 okay, so I'm going to just do integrate and dump in this case and I'll get my INQ back okay, so this is A hat, this is B hat, so I'll have to then estimate, then comes my detector right, so I'll have to do detector, I'll get back my actual bits okay, so this is how the whole system looks and what goes here is the correlator okay, I want you to be particularly comfortable with this complex baseband and the shift to the real passband picture and back and forth and why it's all the same okay, so why complex baseband is good enough, you don't have to deal with anything in passband as far as your detector and correlator is concerned, so even these correlations are done in baseband, you can imagine doing it in baseband, there may be some advantage to doing it in passband, I don't know maybe in some cases it's good, but typically you imagine doing it in baseband itself okay, all right, so that's where we'll stop today, you'll have to do your attendance