 Okay, so this is lecture 6 and like I said we're going to jump into digital communications from this lecture onwards and you might want to turn one of these fans off the other one. I actually had that turned off but if you can hear me it's okay but you might want to turn that fan off. So the general title for the next few lectures can be called modulation and I'll say optimum receiver principles. So as you can imagine this is the very basics and you should make sure you spend a lot of time with this and let it sink in and you can understand what's going on. So we'll first begin by looking at modulation closely and try to understand it and describe it in a compact succinct form and see what it tells about receiver principles and how you go about designing a receiver extra. So the main tool in understanding modulation, what's modulation now? It's what the transmitter is doing, taking a sequence of bits and converting it into a signal X of t that's modulation, that's what I define as modulation. To understand that an important tool is to describe those signals as vectors. So you move from this continuous time X of t type description to a vector description. This is not sampling, it's different from sampling. So don't think it's the same as sampling, not by sampling which is used for signal processing. We don't care about sampling. I don't even know what X of t is, I want to be able to design X of t. So I have to come up with a very smart description and for that we use some basic linear algebra and vector spaces and to describe the whole thing as vectors. Once you describe it as a vector you'll see you get a much better handle on how to go about designing these signals, what it actually means and how the receiver will react to the signal in this vector description. So that's the main overview for the first few minutes of this lecture at least. How do you go from this waveform channel? So right now we have a waveform channel which is Y equals X of t plus n of t. So you see I've slowly moved to a capital notation. Why did I go to the capital letter notation? Because each of these things are random processes now. Initially I've been talking of it as Y of t is X of t plus n of t. In our notation each of these things are random processes. So to be consistent I'm going to say capital letters. So each of these guys is a random process now. So that's my notation. So this is what I would call a waveform channel. So while it's a good powerful channel model it's not clear how the receiver should react to a waveform. So on the other hand vectors you will see are much easier. So the first few minutes of this lecture at least maybe a predominant part of this lecture we will spend in converting this waveform channel with no loss. So remember no loss is very important with no loss of information. X of t is carrying some information about the bits B. Without losing any of that information about B can we convert this waveform channel into a vector channel which I would write as Y equals X plus n. So this is a vector channel. So this vector channel is mathematically much more easier to deal with and even in practice it's much more easier to deal with. So you're much more comfortable saving a vector in a computer as opposed to a waveform. So if you want to process a vector it's much easier. So in design it's much easier. In thinking about the signal it's much easier. In implementation also it's much easier. So this is a very very powerful simplification which is once again at the heart of digital communication. So you want no loss in the process. So no loss in the sense no loss of information about the bits that are being carried in X of t. So this is what we'll do most of this class. So let me go back and describe this modulation once again. So what did we have as modulation? So this is the box we had for the transmitter. So the transmitter is going to take it's just one big box. It takes a sequence of bits. So to be specific I'll say it takes an n bits. So if you have n bits for now we'll say all of them are data. So you have n data bits. So if you have n data bits how many different possibilities are there? 2 power n possibilities and we'll also assume that each of these possibilities are equally likely. So later on we'll see maybe there's need to change those assumptions. But for now we'll say these n bits are equally likely. So what would be the probability distribution on the b? If you think of this b as a random vector probability that b equals a given n bit vector equals 1 by 2 power m. So that's the probability description for b. So the transmitter is going to put out X of t. Now I have to describe what this X of t is. So I'll once again do what I will call a sample function type description. I'll say X of t equals, so I should be, there's no bracket here. Sorry about that. X of t equals X i of t. So I'll use i here. I could as well use b but I'll say i and i ranges from 1 to n with probability 1 by 2 power n. So what am I doing here? I'm saying for each bit sequence that is coming into my transmitter, my transmitter puts out a signal and I'm saying that signal is going to be different for each of those bit sequences. If you want some hope of being able to recover that bit sequence it better be different. And I'm calling them X 1 of t, X 2 of t, so on till X 2 power n of t each signal corresponds to a specific n bit sequence. So instead of i I could as well use b which is the vector itself but just for simplification I'll say i. Simple seems to be seems to be okay. So this is the most general description possible. It's not clear how you go from here to a vector type description. I'll talk about it later but before that let's fix something about data rate. So there's some information missing about data rate here. So for that I'll have to say support of this X i of t, what's support? The values of t for which X i of t is not 0. I'll say support is contained in 0 t. So what am I saying now? So you take a total time t seconds to transmit these n bits. So what will be your bit rate? n by t. So n by t bits per second. So this is kind of what I mean by that. So if you take all the supports together you should be totally including 0 t. So bit rate becomes n by t. So it's just I've not done anything here. There's no real analysis yet. I've just given names to the various quantities involved and made things concrete. So we'll have to start doing analysis only now but for now at least this much is clear and it's not too bad. So typically you'll see as you go along into your future careers you'll see notation is half the battle in most cases. So you have to come up with notation, describe everything in notation. After that several times the analysis will be very obvious. So that's just half the battle. So it looks like we have one half the battle. We have to do the remaining half now. So for that we'll use very simple tools from linear algebra. So I have 2 power n signals. Now I have 2 power n signals x i of t. So I'll try to write these signals as linear combinations of a certain basis set of signals. So I'll choose a basis set of signals. I don't know how many elements the basis will have. I don't know it. Maybe it will have a large number of elements but I don't care. Once I choose a basis I can write my signals as a linear combination of those basis elements and the scale as being complex numbers. Once I do that every signal is represented by the vector of complex numbers I used in the linear combination with the basis. Seems very trivial and basic. There's nothing more to it. So that's what we'll do now. So how do I go from a set of signals x i of t to a basis? I've given you x i of t, x 1 of t to x 2 power n of t. How do you go to a basis from here? There is a standard procedure that people use, the Gram-Schmidt orthonormalization process. If you're given a set of vectors which you don't know if they're orthonormal or not, if they don't know they're orthogonal or not, how do you find a set of basis vectors that span them? Whose span will include these vectors? You do Gram-Schmidt orthonormalization process. So that's what you do. So that's the first step we'll do. So we'll do Gram-Schmidt on, Gram-Schmidt is 2 m so 1 m. I think I'm going to use 1 m but maybe that's 2 m here. You do Gram-Schmidt on orthonormalization on these x i of t to get, you don't know how many basis vectors you'll get. So I'll say I get m basis vectors to get m orthonormal basis vectors but I know they'll be orthonormal. What will be my inner product? I've already defined an inner product for functions. That's that l2 inner product. So I'll say all of my x i of t are in l2. Like I said, I didn't say that very explicitly but x i of t will be in l2. They're all finite energy signals. I'm not going to assume I have infinite energy. Since they are all in l2, I'll use the l2 inner product in my Gram-Schmidt. So to get m orthonormal basis, I'll call those basis vectors phi 1 of t, so on till phi m of t. Remember, my inner product I'm using is what? That integral from minus infinity to infinity x i of t, x j of t dt. That's my inner product. It's the same thing we'll hold here. Since it's orthonormal, what do I know about the phi i's? Inner product of phi i of t and phi j of t will be what? The 0 if i not equal to 0. A compact notation for this is this chronicle delta, delta i j. Delta i j is a function which is 0 if i is not equal to j and it's 1 if i equals j. And then what else do I know? What about the norm of each phi i? It's 1. That also I know by just by design. I know I can do it whenever I have an inner product and a set of basis. So one more thing I can say about the phi i of t is their support. Their support, it's enough if it's contained in 0 t. Usually you'll never get anything outside of 0 t if you do Gram-Schmidt. You start with support fully contained in 0 t, you do Gram-Schmidt because Gram-Schmidt is still a linear combination. Each of these phi i's ultimately will be linear combinations of the x i's. So you'll never get anything outside of 0 t. So still the support is going to be 0 t. So this is a very simple process by which I've gone from a given set of 2 power n's signals to m basis vectors. So now what do I have? I have a very simple way of vectorizing my signals. Instead of trying to define my x of t in each and every time instance t, I'll assume that this basis is known to both the transmitter and the receiver. So you convey the basis using some other means ahead of time. Maybe you write a letter and post it. So do something like that. You convey the basis by some other means and then what is the only thing I have to specify to specify my x of t, x i of t, the coefficients of the basis. So that's a very simple trick you use to convert a continuous time signal which is possibly infinite dimensional into a simple vector, a finite dimensional vector. So that's what we do here. So once you have that, so I'm not going to write down Gram-Schmidt. So maybe I should write down Gram-Schmidt. I'll write the first few steps because a typical quiz question is to give you a set of signals and ask you to do Gram-Schmidt and come up with the orthogonal basis. So you should know how to do it. It's very simple. So I'm not going to do it formally or properly. I'll just quickly write down the first few steps and I'll say and so on. I'm assuming you've seen Gram-Schmidt before. It's a very simple procedure. So you should know what this is. So this is Gram-Schmidt. Phi 1 of t, I will set to be x 1 of t divided by norm x 1 of t. So you basically normalize x 1 of t. So phi 1 of t is okay. So what will I do for phi 2 of t? I'll do it in two steps. I'll first define a phi 2 tilde of t which finds that part of x of t which is orthogonal to phi 1 of t. How do you find that? You take x 2 of t and then you subtract the part of phi 1 of x 2 of t which lies along phi 1 of t. How do you do that? You take an inner product with phi 1 of t and multiply with phi 1 of t. So you know phi 2 of t, phi 2 tilde of t now is going to be orthogonal to phi 1 of t. The only thing you don't know is whether it is unit norm or not. So then you go ahead and normalize it. So you define phi 2 of t as what? Phi 2 tilde divided by norm of phi 2 tilde of t. So remember this inner product is the integral from minus infinity to infinity x 2 of t phi 1 of t dt. So that's what it is. But what will it be? The inner product will actually be a complex number. It won't be anything else. Nothing more to be worried about. So then what will I do for phi 3 tilde? I have to find that part of x 3 of t which is orthogonal to the entire linear space spanned by phi 1 of t and phi 2 of t. So I do the same thing. So I will take x 3 of t minus what? x 3 of t dot product with phi 1 of t, phi 1 of t. So I find that part of x 3 of t which is contained in the linear space spanned by phi 1 of t and phi 2 of t then subtracted from phi x 3 of t. And then what will I do? I will normalize phi 3 tilde to get phi 3 of t. So that's what you do. So now I am going to say and so on. So you keep proceeding like this to get your entire basis. How can sometimes what will happen is x 3 of t will lie in the linear space spanned by phi 1 of t and phi 2 of t. In which case what will happen? Phi 3 tilde will become 0. So then you don't get anything new. You have to jump to phi 4 of t. So in general m can be less than or equal to 2 power m. So that's what you will get finally. So this is Gram-Smith. It's very simple. So like I said a tutorial problem is typically to give you a set of signals then ask you to do Gram-Smith to come up with a vector representation. So you do it and then represent each signal as a vector with the correct number of dimensions as you get from Gram-Smith. So that's the procedure. But in practice this is usually done in reverse. So you never really design signals first. Sometimes maybe you do but most cases you never design XI first and then do Gram-Smith to find the vector. What will you do otherwise? You start with the basis and then pick suitable vectors to go to any XI you want. So that seems like a much better way of going about doing things and starting with the signal. But there might be some cases where you have you do not have real control over the XI of t or maybe it's there are additional constraints that are placed on it. In that case you might have to pick XI of t first and then think of it as a vector. But typically in practice mostly you pick the basis first and then pick your signals as linear combinations of the basis. So that's the first bit. So then what do we do? I will go ahead and write down how we vectorize each of these signals. Now once you have the basis what can you do? You can write each XI of t as I will write it in a long form but this succinct short form also possible. XI of t phi 1 of t plus XI of t phi 2 of t phi 2 of t plus so on till XI of t phi m of t phi m of t. So to simplify our notation a little bit further by XI j I will denote XI of t dot product with phi j of t for each i and j. So this is short term notation for these dot products. So I will do that once I do that what will happen to XI of t will simply be summation from j equals 1 to m XI j phi j t. So this is true for i between 1 and 2 power n. So remember XI j is a complex number. So it's the result of a dot product so it's a complex number. So already I am thinking of complex valued processes for instance. So there are two real processes here which I think I order them and think of that as a complex process. So this phi j of t assuming the basis is available to both the transmitter and the receiver the only information is carried by the coefficients. So I will say my signal itself XI XI of t my signal XI of t is well represented by the vector XI which is XI 1 XI 2 so on till XI m which belongs to what the m dimensional okay I put a cm there okay so it's not complex. So for some reason you are not able to see it so maybe I will write it down below it belongs to the m dimensional complex vector space okay. So maybe I should put a transpose here just to be consistent with so I think of all vectors as column vectors I will put a transpose there just for consistency okay. So that's it we are done pretty much okay. So as far as the modulation is concerned we have vectorized it okay. So we have said my modulation is completely described in a vector form as long as I can do I can assume the basis is available and for any signals it's possible so it's not a big deal there's no problem there okay. So I want to reiterate once again so we have gone from a random process X of t a continuous time random process X of t to what a random vector X okay. So this is a random continuous time random process this is a random vector okay what is the pdf for this random vector how many values does this random vector take takes 2 power n values each with equal probability what are the 2 power n vectors that it can take XI okay so that's how I'll define my random vector X okay. So my random vector X is defined so that probability of X equals XI equals 1 by 2 power n for I between 1 and 2 power okay. So that's my very similar to the way I define my random process okay so exact same thing that happens yes typically yes how do we decide yeah I'll talk about it as we go on we'll do a whole bunch of examples so you see so typically you start with the smallest possible dimension right so one seems like a good dimension to pick so a lot of a lot of cases people choose one two okay the very few cases where it goes larger two is the most common dimension so so anything else I should add yeah so another thing I should do is to now say right so this random vector is composed of how many components m components right each of which is a each of this is what a random variable okay so how will you find the PDF of each of these random variables how will you find the PDF of each of these random variables sorry from the join PDF right so you know the PDF for the entire random vector okay so from there you find you marginalize and find your PDF for the random variable okay so the way I wrote it down will this XI be a discrete random variable or a continuous random variable discrete right it'll take a discrete set of values at most it can be 2 power n and you'll see typically in the way we choose it it'll take very few values each of these things will take very few values okay together the random vector the random vector itself is discrete it only takes 2 power n values okay so the random variable will also take 2 power n values but it's tough to write down the PDF right so I can maybe give some notation and write it down but from the random vectors it's very easy to figure out how many of them can take okay right so that's that's the modulation okay any questions on how I went about doing it okay so I can write in fact to go back from X to the random process I can write X of t as what summation j equals 1 to m X j phi j of t okay so that's that's the way I define my random process in terms of my discrete random variables X j okay so I've gone back and forth from both okay this is a sample function definition very classic definition all right any questions things that are concerning you things that are not clear okay so this is this is doable for the general case okay so so at this point I could do a whole bunch of examples but I'm going to postpone the example until later till we have completely vectorized the whole thing okay so we only vectorized X of t right we have Y of t equals X of t plus n of t okay you have to tackle n of t next after that you have to tackle Y of t okay so when you tackle Y of t you have to be really really careful because Y of t is what's being used by the receiver the no loss will really come into the picture when you tackle Y of t okay so what kind of information you keep and all that is very important so I'm going to do it a little bit and postpone the example until later okay so I think this is simple enough you don't have to see an example okay so that's my assumption so we'll see examples we'll see tons of examples together so when you see everything together it'll make more sense as opposed to seeing one example after that okay so we're going to now move towards the noise signal okay so what is my model for the noise signal okay n of t is a white Gaussian process random process and I've been assuming what does it mean I'll assume my spectral density is okay not by two right that's what my assumption for the noise is okay so there are quite a few n's here the small n capital n n of t okay hopefully it's it's it's clear okay so I could choose other notation but I think n of t for noise makes a lot of sense if we use something else it's only confusing okay all right so so so so so so so what I'll do next is a little bit of a hand waving argument to quickly get to the answer okay so so I can it's possible to do these things rigorously by defining expansions of random processes using basis and all that but I'll simply motivate it by sample function arguments and say how you can do it okay so theoretically I can't think of n of t as one function right it's a random process okay so there's no there's no meaning in taking inner product of n of t with phi one of t and phi two of t and all that but I have to define carefully what that means okay so but even if I think of n of t as a bunch of sample functions maybe this process is not so clear but but it can be made clear okay so it can be made proper and rigorous and I don't want to spend too much time doing that okay so welcome to read the books that I suggested to you Proacus has a very nice section on how to make that formal it's called a certain expansion for the random process okay so it's very rigorous even from a random process point of view okay but I'm going to kind of hand wave loosely and use sample functions and random variables random processes interchangeably and come up with the answer okay so that's what I'm going to do now okay so what I'll do is I'll imagine a particular sample function of n of t okay and try to expand it in the basis that I got for my modulation functions okay so suppose you take a sample function for n of t okay so you look at a sample function okay n of t okay I'll try to expand it now using the basis that I had before but what does it mean do we know do I know that n of t will lie in that space I don't know okay so there might be something extra okay so I'll have to account for that also okay so eventually you'll see that something extra doesn't really matter okay so eventually we'll show that but right now when I write it I will have to include that extra term as well okay so I'm going to write n terms m terms sorry which are n of t phi j of t okay phi j of t okay so this is the part of the sample function of the noise process which lies in the linear space spanned by the basis that I got from a Gram-Schmidt run on the modulated signals okay then on top of this I will have an extra n2 of t okay which is orthogonal to that entire space okay so this is the proper way of doing it in general there can be something like that n2 of t is no guarantee that a Gaussian noise process or any noise process should fall within the linear space spanned by all modulated signals there's no reason that should happen okay so this guy which does fall this part which does fall in the space I'll call it n1 of t okay right okay so this is what I've done all right so now if I imagine doing this for each and every sample function right I'll finally end up with a situation where I can do this for the entire random process itself imagine doing this for the entire random process itself but there's some rigor there that one needs to carefully prove can be done you can see the books it's doable okay so if I do that if I do this for every sample function you can see this guy will turn out to be a random variable right right now for a particular sample function it is one complex number if you change the sample function what will happen I'll get another random number so I'll ultimately get a random variable I'll call that as nj okay that's my random variable that I get by running sample processes to the through this correlation okay it's a correlation right this inner product is a correlation with phi j of t okay so if I run my random process through a correlator I get a I get a sample function I get a random variable okay so how do I get this nj I take my n of t then run it through correlation with what phi j of t okay I will get a random variable okay one needs to carefully study this there are so since phi j of t is unit norm you can show very easily that this nj will in fact be a proper random variable in fact it'll be a normal random variable okay it'll be a Gaussian random variable right you can show further that it'll be it'll have zero mean and variance n naught by 2 okay so we'll do this I mean it's not too difficult I'll I'll do once again a simple little proof of something like this I'm sorry is there a question okay so so what am I saying here so once again the noise process can be split into two parts one part which lies in the linear space span by the phi j of t another part which is orthogonal to it when you do that you have to do a correlation with the basis vectors so the noise process correlated with with the basis function will give you a random variable nj okay so that's my nj okay so once I do that I get this nice picture where this n1 of t will become a random process which I will call okay so the set of all n1 of t defines a random process which I will call n1 capital n1 of t which is summation from 1 to m nj phi j of t okay so once again these things require careful proof but I'm going to say it's possible okay and then argue that these things are possible okay so likewise this collection of n2 of t is another random process which I will call capital n2 of t okay so in effect what's possible always is the following okay so you can take the random process n of t and decompose it into a sum of two random processes n1 of t and n2 of t okay where n1 of t is given by this simple summation where each nj is a normal random variable and in fact more is known I'm going to write down that soon enough okay so we'll see that okay so finally what we have is n of t can be split into two forms n1 of t plus n2 of t where both these guys are random processes and this n1 of t is given by a simple expansion of this form okay all right so the only difficulty is some technical language of random processes and sample functions so in a product I can strictly do only for random sample functions but anyway we know correlation is like filtering so you can see why all those things should work out finally okay so you can define in a product like things for random process as well all right so so so so so so just make sure I'm doing this somewhere okay doing this somewhere good so so let's look at these nj's more closely okay so we'll try to figure out so what I'm going to try next is to figure out the joint distribution of the nj's okay so the reason why I'm interested in that is if once I do that then I can write n of t I can represent n of t as what a random vector n which is n1 n2 to nm and n2 of t which is another random process which is in some way orthogonal to n1 of t okay so this is the this is the best I can do with the noise process okay I know the noise process does not lie completely in the linear space spanned by the modulated signals so but but what can I do I can break up the n of t into a vector which completely describes that part of the random process which will lie in the linear space and then some continuous time random process which is orthogonal to it okay so eventually what we'll see is when we look at the receiver we'll see n2 of t contains no information about the bits okay that seems to be clear right at least intuitively it seems to be clear all the information about the bits well you should be very careful here n2 of t does not affect any of the information in the bits okay so you might as well throw it out okay so we'll see that okay so it's possible to nicely throw it out okay and the n part which lies in the linear space is the only thing which will affect your bits and it's enough if you decode against that if you receive against that so n2 of t you can reject in your receiver okay so we'll see that so which makes my entire random process n of t at least the relevant part of it a vector okay so you don't have to worry about the continuous time so we'll see that eventually for now we'll just keep both of these things around and figure out how to do it okay so what is once again what's ni okay so before that let me write down the statement the claim here is this n the vector n is iid normal with zero mean and variance n0 by 2 okay that's the claim it's identically distributed and independent each of these things are normal with mean zero and variance n0 by 2 okay so that's the claim so once I show that you see the noise process decomposes itself to a into a very very simple probability distribution okay so not not discrete definitely it's continuous but still it's an independent Gaussian which is as simple as it gets and you deal with noise okay so let's do a spoof of this okay so what is my ni to remind you once again it's integral from minus infinity to infinity n of t phi i of t dt okay so this form might look strange to you but this is actually a random variable like I argued okay when this is a random process what you get out is a random variable okay so you can define this integration with the random process inside okay like you do for filtering of random process okay it's possible to do this okay so so so so where am I okay so so the next argument for is for why it should be Gaussian okay so once again I'll simply appeal appeal to your filtering notions to say that this will be Gaussian okay so when you filter a Gaussian process you get a Gaussian process at the output so when you do this correlation also this should be Gaussian okay so we won't rigorously prove that we'll just assume by an extension of the linear combinations filtering argument that this will also be a Gaussian okay so only thing we'll show is that the mean is 0 and the covariance or the cross correlation is goes to 0 and then the variance is n naught by 2 those are the only things we'll show okay so it's also easy to see it's not too difficult once you buy the remaining things okay what's expected value of n i okay in the integral when you take the expected value inside okay so once again that's a little bit of less rigorous thing but we'll take it inside once you take it inside phi phi i of t is deterministic it'll come out so only thing you take expectation where for us n i of t which n of t which is 0 okay so it's a very simple proof you can show very easily that this will work out to 0 okay so this will work out to 0 because this guy is 0 okay so for the covariance a little bit more work is required n i n j is in this case it simply becomes expected value of n i n j right the mean is 0 right so it's simply expected value of n i n j which is nothing but expected value of you have to write two integrals and multiply them so one thing to be careful about whenever you write two integrals is what and multiply them what should you do you should use different dummy variables okay otherwise you'll end up doing all kinds of strange stuff it's not correct okay i'll use tau here okay so now i'll write this as a double integral okay i'll i'll drop the limits it's all minus infinity to infinity and you see the expected value is only n of t n of tau okay phi i of t phi j of t tau dt d tau okay so now what is this guy 0 that's the autocorrelation function of the Gaussian white noise process so it will be n naught by 2 delta t minus tau okay so you put in this guy here and argue use your knowledge of delta functions to simplify the first integral you can simplify it as in a very easy form okay so eventually you'll get this to be n naught by 2 the chronicle delta okay so it's a very straightforward thing see remember phi j's are also orthonormal right so eventually you'll get this result very easily okay so the first thing you simplify you'll get n naught by 2 phi i of tau phi j of tau d tau okay dt will go away and that is orthonormal so you get the delta ij okay so it's a very simple proof to show that okay so this shows everything it shows if i is not equal to j the cross correlation the correlation is 0 if i is equal to j correlation is n naught by 2 which is what i wanted to prove okay so my noise vector is iid Gaussian with mean 0 and variance n naught by 2 all right but still I have not gotten rid of the annoying n 2 of the 2 of t part okay so I'm going to make an argument next to get rid of that as far as the receiver is concerned okay so let me just make sure I have everything in place okay yeah so so so so where am I where am I so what about n 2 of t you can show the following for n 2 of t okay so this is the next claim n 2 of t and n i are independent for all t you fix a t for all t okay so you fix one t and look at the random process n 2 of t and then this n i you can show they are independent how you just simply show the correlation is zero okay so you show expected value of okay so here's the proof I won't write it in great detail because it's just a simple repetition of the previous things you show this is zero how will you show it n i is an integral this whole thing is an integral so you write it down you'll get a delta and you'll see because of that orthogonality of v j's it will vanish okay n 2 n 2 is what n of t minus n 1 of t okay so you write it carefully you'll get just busy okay so this is a important result to know about n 2 of t okay so finally we have managed to write my noise process n of t as represented by a vector plus something which is independent for every t okay so that's the important all right so the next thing so so what have we accomplished till now we've finished the transmitted signal which we have successfully vectorized without any problem and then for the noise signal what have we done we've vectorized partly except that there is this independent component sticking around okay which we will get rid of when we look at the received signal okay so let's go to the received signal and make a careful argument to show why this n 2 of t is completely irrelevant okay so remember y of t and n 2 of t are they independent or not no they're not independent right why how do I write y of t y of t equals x of t plus n of t okay which now I know I can write as summation j equals 1 to m x j phi j of t this is the x of t part for n of t I can write j equals 1 to m n j phi j of t and then plus what n 2 of t so clearly y of t and n 2 of t are not independent okay so I can't just happily throw it away in the beginning and say just because it's independent with this it's out okay y of t depends on n 2 of t it affects my received signal so who knows only thing we will argue is n 2 of t does not affect the information part of okay so but let's proceed from here so you see from here I can write join the first two summations and write this as x j plus n j phi j of t plus n 2 of t okay so the crucial argument is these two guys are independent for every t okay I know n j is independent what about x j okay there's no reason why I'm going to assume the noise depends on the data okay so I'm going to assume noise is independent of the data anyway so any so these two will become independent okay so that's the assumption now I'm going to define y j to be equal to x j plus n j okay and then define a vector y to be equal to y 1 y 2 okay so you should all let through why I'm going to do these things okay and then I'm going to call y tilde of t as y tilde of t is summation j equals 1 to m y j phi j of t okay so finally what can I write after all these notation can say y of t is y tilde of t plus n 2 of t and y tilde of t is what x of t plus n 1 of t okay so okay and these two guys are independent right okay all right so this is the this is the picture finally all right so I have a couple of okay so what do I have for the waveform model finally I have x of t coming in okay I have n 1 of t adding first I'll imagine that this is what's happening I'll tell you why this is justified and get y tilde of t and then I'll imagine n 2 of t is being added to get y of t okay so mathematically this is valid right so it's no problem here it's exactly wrote down what my assumption was and did it but I'm going to claim so right now we've been thinking the receiver is here right the receiver has access to y of t I'm going to claim the receiver is actually in fact even here receiver has access to y tilde of t why yeah so you simply do correlation with you take y of t you correlated with all my phj of t I'll get my y tilde of t so y tilde of t is definitely something that is accessible to the receiver so the receiver is actually in fact here okay so the receiver can potentially take y of t how is that possible now the receiver can take y is this true okay the receiver can take y of t and send it through a bunch of correlators okay so the first correlator is correlation with what y 1 of t okay I'm not writing the d t okay so just assume the d t is there okay the second one is correlation with phi 2 of t so on till the last one which is correlation with phi m of t okay remember the basis is known to the receiver right so you get what here y 1 2 y m which which faithfully represents y tilde okay so y tilde is something which is in the linear space so it's enough if I get y 1 through y m I know my y tilde there's no problem so y tilde is definitely accessible to the receiver okay but once this processing has been done n 2 has been lost okay n 2 is gone okay so n 2 is not there anymore okay so have we lost something is the question okay so it turns out we have not lost anything about x of t okay so we worried only about x of t so what you can show is given y tilde of t x of t and y of t become independent okay so that's what we show technically to claim that y tilde of t is enough for me to get all my information about x of t so that's what we'll show and from this picture it's very clear if you once you're given y tilde of t you don't care what x of t was y of t is simply y tilde of t plus n 2 of t I know everything right so given y tilde of t x of t and y of t are independent okay so once I found my y tilde of t in the receiver x of t and y of t are independent I don't care about anything else that was possibly there in y of t I've gotten everything about x of t in y tilde of t okay so it makes a lot of intuitive sense also but mathematically to rigorously show it you have to show given y tilde of t x of t and y of t are independent okay so in this picture also it seems to be clear enough okay so given the y tilde of t x of t and y of t will be independent okay of course y of t and x of t are not independent okay in general right y of t is x of t plus n of t is very much you're hoping that it's not independent hoping to get information about it but y tilde has all that dependence in it okay once you're given y tilde x of t and y of t become independent okay so if you're worried about only x of t you can throw away everything else that is in y of t and worry only about y tilde okay so so from I'm going to just do proof by picture okay so there's nothing more required if you want you can write it down it's not a big deal it's simple proof by picture it works very clearly all right so finally what do I have finally I have a very very simple vector model without any n twos of t floating around because I know I don't care about it as far as recovering everything is concerned okay so here's my final vector model okay which is very very powerful I'm going to say I have a vector x to which a noise vector n gets added and I get a vector y okay so this is my entire model okay so what happens the transmitter takes in a sequence of bits and puts out a vector x okay so what is x it's an m-dimensional vector and what's the pdf of x right I'm going to say it's equally likely maybe I'll change these definitions later for now it's equally likely to be any one of my original design things what about n n is this once again nm dimensional and it is iid normal with zero mean and variance n0 by 2 this n0 by 2 is the exact same power spectral density for my noise process n of t okay so you might think it's a theoretical quantity but in practice you can measure it at the receiver okay you don't transmit anything you will get some signal right measure its power that will give you my power spectral density for the noise okay so there are machines that will give you there are equipment which will give you that n0 by 2 so it's a very measurable quantity and in your model the same n0 by 2 appears okay and then y is what y1 through ym and how do I define y each yi is xi plus ni okay so this is pretty much a scalar model okay so from from a vector model we've come down to even a scalar model okay if for instance I know the xi's are independent then the scalar model is good enough I don't even care about the vector model okay so I don't know for sure from the vector pdf if the xi's are independent the xi's are independent the scalar model is good enough okay but in general I have the vector model with x1 through xm and y1 through y so given the pdf for x and pdf of n can you find the pdf for y you should be able to find okay so you should say confidently that I know how to find it once you find it your entire problem is solved it becomes a very classic detection problem which we'll talk about as we go along okay so this process is very important go back and look at this lecture once again very carefully and convince yourself we've not made any assumptions or we've not given up anything okay we've kept everything that we possibly need and finally we have a very very simple probabilistic vector model with no random process floating around very simple nice process to model my y of t equals x of t plus n of t which was a waveform model okay so we'll stop