 This is lecture what 12, okay so the last class we saw some basically I think I was spending quite a bit of time with MPAM to establish the decision regions and come up with some expression for the probability of error, okay so the MPAM that we were looking at I want to do it a little bit I think towards the end I was rushing a little bit so I want to do it more carefully this time, okay so MPAM, okay so you have 0 and the way I took it it was d by 2, right 3d by 2 so on till m minus 1d by 2, okay so minus d by 2 minus 3d by 2 so on till minus m minus 1d by 2 that's my MPAM constellation, okay and the decision regions are fairly simple, okay for the outer points you have two different at certain type of a decision region and okay and for the inner points you have a certain finite decision region, okay so that's clear enough so what I'm going to write down so I did some calculations to show how probability of error given a certain transmission point is simply an integral over the decision region with probability of correct decision is an integral over the decision region probability of error will be 1 minus integral over that decision region, okay so I'm going to do that now so suppose you say x is say m minus 1d by 2, okay so you have to take an integral over the decision region for the conditional pdf, right so that's what I want to find out, okay I want to find probability of error conditional on x equals m minus 1 into d by 2, okay so for that what will the conditional pdf look like, okay so this is the conditional pdf f y given x equals m minus 1d by 2 y, how will the conditional pdf look like, it'll be a normal centered at m minus 1d by 2 and variance will be n naught by 2, okay so that's how the so roughly maybe it looks like this, okay the p equal occur here and what is my decision region, m minus 2d by 2 onwards to the right, okay so that's my decision region, okay so this point would be, well that's not a constellation point but that is my so what is this distance d by 2, okay so that's what's important all this m minus 1 m minus 2 and all that is just convenient notation for denoting the point what's crucial is this distance, okay so the reason is I want this integral which is the probability of correct decision but then what will be the probability of error, the other integral, okay so the other integral is much easier to write using q functions once you know the distance from the mean, okay so you see easily probability of error given given x equals m minus 1d by 2 equals what integral from minus infinity to m minus 2d by 2 the normal pdf with mean m minus 1d by 2 and variance n naught by 2, okay so you can either use one of your standard formulas or can do a change of variables convert it into a suitable unit Gaussian and then find the actual distance there, you can do that work and show this is equal to q d by 2 root n naught by 2, okay so root n naught by 2 is the standard deviation and d by 2 is the distance from the mean silver coat as q d by 2 divided by root n naught by 2, okay so okay so if you want if you want a useful trick to remember this the way to remember this is following for instance the general formula for q gives you something for the unit normal unit variance 0 mean unit normal pdf, okay so maybe you don't have that maybe in a general situation you have say mean mu naught and then variance say sigma squared, okay so I don't know why I'm saying mu naught I'll just say mu, okay mean mu and variance sigma squared, okay so q will give you suppose you have this to be a general I don't know some alpha, okay, right the integral from alpha to infinity of this pdf will work out to what? q alpha minus mu divided by sigma, okay so this is the formula, okay, right so if I pick some beta here and ask you for this integral what will it be? Yeah, okay is that clear? q of beta minus mu is 1 minus I'm sorry what will it work out to? Yeah, okay so that integral should also work out to something very similar I don't know why I'm struggling a little bit here what will this work out to? Minus infinity to beta normal mu sigma squared, okay so you can write this as 1 minus integral from yeah beta to infinity normal mu sigma squared, okay this is working out okay am I writing it fine, okay so finally I think the correct answer will have to work out to q of mu minus beta by sigma, okay, right is this fine, okay so there'll be a half term missing so you should use q of 0 is half and all that, okay so it involves a little bit more of work at all it'll be a very simple derivation but anyway you see this will have to work out to q of mu minus beta by sigma is that clear? Okay if you want you can show this you can write this as q of beta minus mu but then it'll be negative q of minus x is half plus q of plus x and all that, okay so you do that you'll get this answer, okay so play around with this so these are the two expressions to remember, okay so these two are the expressions to remember so what matters here you see notice alpha minus mu is what? What is this alpha minus mu? It's this distance right deviation from the mean likewise what is beta minus mu, mu minus meter is deviation from the mean, okay whenever you're calculating what's called the tail probability what matters you can write this as q of deviation from mean divided by standard deviation, okay that's for the tail probability, okay if you're calculating the other probability it'll be one minus the tail probability so this is useful to write everything in terms of q, okay that's one expression, okay you can also use this to write integral from beta to alpha of the normal PDF what will this be, okay so anyway this is an assignment for you, okay try this it's not too difficult write it in terms of the q function, okay so it's okay to write, alright, okay so that's the that's the probability let me go back once again to that page and remind you what we got this is conditional on transmitting one of the right most extreme points probability of error given x equals m minus 1 d by 2 this has to be equal to q of d by 2 divided by root n naught by 2, okay, alright so you see once again the formula at work, okay so what's the distance from I mean I'm interested in this tail probability, right and what is the distance from the deviation from the mean it's d by 2 and divide by 0, okay so use that it's easy to work this out, okay, alright so the next thing is to look at probability of error conditional on the other extreme point is very easy what should it be should also be the exact same thing q of d by 2 divided by root n naught by 2, okay so then I'm going to argue now I'll do for one interior point and everything else will be the same, okay so I will do probability of error conditional on x equals d by 2, okay so this is what I'm going to do let's draw the conditional PDF this is f y given x I think I've been using all kinds of different notation, okay hopefully it's clear to you, right it's going to be mean d by 2 and variance root n naught by 2 and what's the probability that I'm interested in I'm actually interested in two probabilities, two different areas one is this area, okay and what's the other area after d, right so I'm interested in this area, okay so you can visualize this from various points of view from the constellation if you see if you transmit d by 2 if your y is less than 0 then you're closer to minus d by 2 if your y is greater than d then you're closer to 3 times d by 2 in both cases you'll make an error, okay so I'm interested in the integral over that area what is that integral going to be there are two tail probabilities, right so it's going to be q of but you notice the distance is the same, okay the deviation from the mean is same for both, right d by 2 so it's going to be 2 times q d by 2 by root n naught by 2 is that okay, right simple enough formula, okay so now that we've found the conditional PDFs we can find the probability of error itself, okay so how many extreme points do I have I have two of them okay so it's then and they occur with probability 2 by m 2 by m q d by 2 by root n naught by 2 plus what m minus 2 by m q of okay there's a 2 I'm sorry 2 q of d by 2 divided by root n naught by 2, okay it's a simple simplification from this point to show this will be the same as 2 times 1 minus m q d by 2 divided by root n naught by 2 is that okay is that right is that okay all right so this argument of q is quite important to me okay so for m when m is large or even if for any m what's outside the q is not going to matter that much because it's a constant it doesn't change if I change my power at the transmitter it doesn't change if I change my bandwidth it doesn't change if I change any of my other parameters right so remember why did I calculate probability of error I want to understand all the tradeoffs very carefully what's the tradeoff between power of the transmitter bandwidth of the transmitter and all that okay so bandwidth turns out we can't really understand okay so I was trying to write down expected value of n square of t but it turns out since we assumed a huge bandwidth that doesn't make much sense okay so we simply define something based on n naught and we'll proceed for for now later on I'll fix the bandwidth thing so for now we'll we'll ignore the bandwidth part of it I want to understand how the power of the transmitter and power of the noise okay power of the noise is represented simply by n naught the level of the noise all these things tradeoff and give you an expression for probability of error that's what I want to understand right so I should clearly look at this expression very closely and figure out what is important and what's not important okay outside on the left hand side you have coefficients which don't change when any of those things change when will it change when m changes right when m changes it will change okay why is m crucial for me what does m determine number of symbols or the rate okay the bit rate is going to be determined by m okay so that's a that's an expression which you might want to write down so so maybe I'm sorry let me just write it down here I don't want to lose focus of this on this expression it's quite important okay so this log base to m is the bits per symbol okay so m is crucial but you see I don't have to worry too much about it why is it that I don't have to worry too much about it yeah so as one one reason is that when m is large you can definitely ignore it it's no problem but also look at the other term multiplying it it's q okay q is going to decrease exponentially in its argument okay and for any reasonable value for its argument it's going to be far far smaller than this expression and you can conveniently not care too much about it up to some order you can find the expression just with q okay so q is very important okay what whatever shows up inside q as an argument of q is extremely crucial for me and whatever design I do I do in my transmitter and receiver I should take care of that argument whatever shows up inside my q is what's going to give me my performance okay so that quantity is extremely crucial okay so hopefully you're convinced of that okay nothing else appears in the probability of error only thing that matters is what's this argument of q okay so we're going to look at that argument of q very very closely and come up with some expressions for it and relate it to the various quantities that we're interested in okay that's what we'll do okay all right so let's let's see what what shows up inside is d but d is a little bit misleading the reason is what we are more interested in is in transmit power and yeah d is also related to the transmit power okay so I I think I wrote down this relation yesterday but I may not remember it very exactly so signal energy e s worked out towards some m squared minus 1 d squared by 12 am I right okay okay so you notice this is this is very interesting so this on the left hand side you have energy of the signal on the right hand side you have two quantities that are of very great importance to you the quantity d which is the minimum distance in your constellation which plays a crucial role inside my argument of q which determines my probability of error and then also m which is related to my rate okay and the next quantity is n naught by 2 I'm going to make a definition and say en is n naught by 2 like I said at this point I can't exactly relate it what is n naught by 2 n naught by 2 worked out as expected value of n squared some n squared what is that n after the correlation okay so it's not clear what it relates to in terms of n of t okay the only thing I can say is it's the level of the power spectral density okay just take it as a level of the power spectral density for now okay later on when we put in the bandwidth assumption very strictly and try to fit everything properly we'll see we can definitely talk about energy in the energy or power of the n of t itself okay so at that time we'll make it more concrete and all that for now let's take it as level of pst okay so let's try to write it in terms of these two quantities which are definitely measurable in my transmitter and receiver so you can put a power meter or something of the signal that you're putting out you can definitely know what es is it's no doubt about that likewise you can also know the level of the pst so these two are realistic quantities that I can relate to in a physical system okay in my model also it's good to relate everything to these guys okay so let's see what happens okay so the probability of error works out as what 2 1 minus 1 by m okay q what okay so let's try to write this the denominator I have root en case no problem the numerator I would have yeah root I think it should be 3 something no 3 es by m squared minus 1 is that okay okay so another way to write it is 1 minus 1 by m q root 3 by m squared minus 1 I will call I will first I'll define this quantity for the first time I'll call it signal to noise ratio okay so that's my snr okay it is signal to noise ratio is to be es by en okay so it's very clear why this quantity signal to noise ratio is so crucial okay because my probability of error is a function of the signal to noise ratio and not es separately and en separately okay so es by en is what really matters to me okay so that's why I've been saying always it doesn't matter you can pick a d arbitrarily okay as long as you allow n0 to be orbit I mean you can fix your d and then let your n0 be arbitrary okay if you do that what can you get you can get all possible snr's if you can get all possible snr's you can get all possible probabilities of error so your tradeoff is not limited by that in your model your tradeoff is not limited by keeping d as a general d okay you can pick d gen d fixed but what should you do you should allow n0 to vary as long as you allow n0 to vary it's okay so this quantity is very very crucial and you have to pay attention to this in all your design yeah yeah so okay so it's right so what he's saying is in reality you might be able to tweak only some things I'm not talking about what I can tweak okay so I'm trying trying to tell you whatever you tweak I can model that just by this okay so it's just in the model in the model it's enough if I fix d and vary n0 if you vary things in some other way you'll give me some snr right ultimately and my probability of errors is a function of only that snr I can achieve that same snr with a different setting here it doesn't matter only the snr matters okay so that's the reason for fixing d arbitrarily as 2 or something okay so that you don't have to worry too much about all these variables floating around but if you want to have a d you can have it it just doesn't make any difference okay it's the same thing all right so snr so but like he points out in reality what will happen in the receiver so I'll tell you how the receiver will work you know you're going to put a power amplifier in the transmitter and you're going to pump it okay so it's going to go out and the receiver usually you put something called a low noise amplifier after you filter out some things you put a low noise amplifier and after that you'll get your signal okay so that signal will have a certain energy that is my es right at the receiver it will have a certain energy that is my es so maybe based on your power amplifier setting at the transmitter based on your low noise amplifier setting here you can vary the power a little bit that's how it will work okay but typically there will be a target power level after the lna which will be which will have to be met lna is by the way low noise amplifier okay so you have that okay so that's how your signal power will work so you won't know what your d is right it's very difficult to know what your d is you can figure it out by doing some processing at the receiver okay in general if you don't know how far your transmitter and receiver are separated by you don't know all these settings it's not possible to know an exact d and you don't have to know it you only have to know the ratio okay so all these things don't matter too much right but you can figure out the d using some careful single processing at the receiver it's possible okay so that's the those are the things to keep in mind so all these models are nice in your head but in practice they will come up in a very different way okay so you can plot the constellation and all that and figure out how it works okay anyway so this expression is very very crucial when m is large you can make a very convenient approximation what is that you can say probability of error is approximately 2 times q 2 3 s n r divided by m squared minus 1 okay all right so so I want to I want to point out the first of all the simplicity of this expression okay well there is the q function which you can't get a good handle for but again think of it as e power minus x squared okay q of x is of approximately e power minus x so it's an exponential fall and look at the simplicity of the expression it's got probability of error on the left hand side a quantity which you really really care about and inside the q argument you've got the other two quantities you care about okay in fact maybe three quantities power of the signal power of noise and the rate that you're transferring so in just this one simple expression all the tradeoffs are accurately represented okay so so that's the power of modeling things in a very simple way okay if you don't want to do that if you want to take all the complications of your transmitter and receiver and mind hold on to continuous time and do all these things you will never get a nice expression like this okay so you'll never understand how the tradeoffs actually work okay what happens when I double m okay suppose I want to increase my bit rate by one okay right if I want to increase my bit rate by one what should I do I should double m right so what happens when I double m it's a question that can be beautifully answered once you look at this expression okay otherwise if you just you have to do too many experiments to figure out these things things like that are very crucial in practice and let's show you answer okay so there's another is it okay is this expression fine any comments okay all right so there's another related quantity which I'm going to define as normalized SNR okay so that's my definition it's just take it as a definition for now later on maybe we'll look at deeper meanings of this normalized SNR so what's normalized SNR okay I'll denote it SNR norm this is SNR this is for actually m PAM okay SNR divided by m squared minus one okay and you see the motivation is very clear why am I doing it I just don't I want to get rid of m okay I want to normalize by SNR with rate okay so that's my normalization so you see SNR norm becomes viable okay so I'll make more comments I'll show you some plots and then make a lot of arguments about why this is working out why that is working out okay so this is SNR norm and in terms of SNR norm and for large m you see probability of error has this very very simple expression 2 q root 3 SNR norm okay I put equality here so maybe you want to put it wheel that shows approximately that all right okay so let me show you a plot of this hopefully this should work out okay is this picture clear is it okay okay most of you see it so if the guys at the back can't see it I'm not going to make any excuses because there are like tons of seats in the front and you're choosing not to sit in front okay so can you see it this plot is extremely crucial okay whatever else you don't understand you should understand this plot in this course because every digital communication system today one always provides a plot like this okay so what do you have well the title is MPAM so this plot is for MPAM okay on the x-axis I have SNR norm in the db scale okay and on the y-axis I have probability of error which I've called p sub e in what scale logarithmic scale okay so one linear unit is 10 power it's a it's a detection by 10 okay you divide by 10 when you go down one linear okay so it's that's the that's the log of them log of the mic unit and I have two curves there one in red and one in blue okay so the one in red is for m equals 2 okay and the one in blue is for large m okay so you notice it's okay right it's an okay approximation the approximation that we made by ignoring m even for very small m it seems to hold it's roughly the same okay so all these differences it might seem like exactly not the same but in practice when you're dealing with a system these differences don't make much of a difference okay so they don't kill you all that much okay so it's not a big deal so you can you can accept them okay so so this point I don't know how well you can see it maybe if you're close you can see it this this point is 10 power minus 6 okay and that's a crucial point okay the reason is in most practical systems it's accepted as a rule of thumb that 10 power minus 6 is a very low bit error rate or bit error rate or error rate in general an error rate of 10 power minus 6 is considered very very good okay so at least symbol level symbol per symbol symbol error rate of 10 power minus 6 definitely a very good point okay so where is that achieved okay roughly about okay 9 dB okay so this point is roughly about 9 dB 9 dB of what normalized SNR okay so you see the advantage in normalized SNR what is the advantage though in terms of normalized SNR why did I have to divide by m squared minus 1 okay now I can talk about this 9 dB independent of m okay and well actual SNR it's dependent on actual SNR but I can talk about this 9 dB as a requirement for 10 power minus 6 symbol error rate independent of m whatever m you have what should you look for your actual SNR normalized by m squared minus 1 should be 9 dB for for achieving 10 power minus 6 probability of symbol error okay so that's a good number to keep in mind so what's dB conversion I'm assuming you've done enough dB conversions in your head so 9 dB is roughly what 10 times right so 10 times 10 dB is 10 times 9 dB is also you can take as 10 times if you want you can reduce by 1 and say 9 times okay so it doesn't matter okay so there's another thing 3 dB is considered 2 times so maybe it is 8 okay so maybe 8 is a good number for you okay so 8 or 10 times so what it says is what this plot tells you finally is finally it tells you everything that you need to do okay right suppose you know what your receiver you have a certain n naught by 2 you have measured your measured your power spectral density as n naught by 2 okay in your transmit power and your LNA at your receiver you have to keep knobbing moving the knobs till you have a signal power which is roughly 10 times if you if you're interested in a 10 power minus 6 symbol error rate suppose you're interested in a 10 power minus 5 symbol error rate what would you do yeah I just go up and then find the related SNR norm remember this has to be SNR norm okay so it has to be normalized with respect to m squared minus 1 okay so once you normalize it it doesn't really matter this plot captures every single m-pam that you can think of okay so that's the nice thing about this plot any questions about this plot and it thinks the thing that is disturbing people everybody understood so this is very crucial if you understand this plot it's okay it seems very simple right so most people on the field most engineers working on the field work with plots like this case this is a language that's very understandable okay and this 9 dB is a very realistic number right it's an actual number so it has a meaning in reality in a physical system it's not just the number that you got from your from your model all right so I'll tell you it's just to just I mean I know this is too early but even at this point you should know something okay so this 9 dB is not an optimal number okay suppose I ask you a question how low can my SNR norm be so that I can still achieve 10 power minus 6 is that a reasonable question to ask yeah you should ask that question as engineers right you want your SNR norm to be as low as possible and still achieve 10 power minus 6 symbol error rate right is that a reasonable question to ask that's a very reasonable question to ask and that's a question you should ask and most systems when you evaluate it you'll have to find its SNR norm and then find out how far it is from this optimal value what do you think that number is the optimal value of SNR norm at which you can achieve 10 power minus 6 error rates just take a guess what do you think it can be since it's 9 dB sorry 3 dB, 4 dB okay so in in 1949 when Shannon published his paper on communication mathematical theory of communication he gave an answer to this question okay an accurate answer and that answer turns out to be 0 dB okay in fact even 10 power minus 6 is not there the proof the way it works you can achieve arbitrarily low probabilities of error okay but it's not at the symbol level well it's also at the symbol level I should say arbitrarily low probabilities of error can be achieved at 0 dB normalized SNR okay it seems reasonable to you okay so somebody gives you a plot like this and says hey I'm at 9 dB I'm doing 10 power minus 6 what should you tell him well you're doing really bad right that's what you should tell him reason is you can achieve that 10 power minus 6 at 0 dB okay and this was known at the very start of the whole communication problem people were trying to solve this people knew that but it was very very tough and today we can get very very close to this number you can get like at 0.51 dB or so you can get 10 power minus 6 very easily okay without too much of a problem okay so in one way the problem of digital communication has been solved you know we're solving we're looking at a problem that's been almost completely solved and still there are some unsolved issues here and there and of course there are other areas related to it which might be useful for you okay so the ideas that are involved in getting from 9 dB to 0 dB are these powerful ideas of error control coding and all those pretty much you have to pick a very powerful error control code and do them so that's where coding fits in but we won't see coding pretty much till maybe till the end of the class maybe a few examples okay so that's the power of this okay the reason I gave you this information is you should know that what we're studying right now is not the optimal thing it's just one way of doing it so optimal scheme will be a modification of this where you'll introduce something called coding to get very low symbol error rates okay so that's the that's the theory there all right so something else I wanted to do yeah so one some some more computations in terms of actual SNR as opposed to normalized SNR okay so roughly you should know so this once you know these 9 dB numbers you're comfortable with normalized SNR but you should know what it means in terms of actual SNR okay so that that we'll see one rule of thumb which is which is quite useful okay so it's very easy to see okay so I want you to think about this the probability of error expression 2 times q root 3 SNR by m squared minus 1 so roughly for large m you can think of m square minus 1 as m squared itself so the denominator works out works out like that okay so so what rate does this achieve okay so from a useful rate to think of is in terms of bits per symbol okay bits per well yeah bits per symbol I think this is good okay there's also a bits per dimension type quantity I'll introduce that later but anyway bits per symbol is the is a good unit to have okay so suppose I want to increase my rate by 1 bit per symbol what should I do I have to double my m right if I multiply m by 2 I get an increase of 1 in rate if I double my m by 2 to get the same probability of error how much should my SNR increase do you see that do you see how I'm making this argument this is a very crucial argument okay so you see inside q there is an argument and that completely captures all of my probability of error okay when I want to increase my rate my 1 I have to double my m but I want to keep my probability of error is the same okay which means my SNR has to increase four times do you see that it's a very simple argument but still the way it works out is is very nice to see okay so if I want to increase my rate my 1 increase rate by 1 I have to double m okay and to preserve my same probability of error SNR has to increase by how many dB increase by 6 dB okay so the 6 dB number is often quoted as a rule of thumb for doubling your signal constellation okay so increasing your rate by 1 so remember all this is in a very simple m-pam situation the place for m-pam there's no coding or anything it's just simple m-pam it's not optimal or simple simple m-pam you do it okay so it's possible to do that at very very low SNRs without coding okay that's also coding one might say but there's some one can think of it close to modulation but yeah any reasonable rate you have to do code okay so so this is a good good rule of thumb to remember okay so any questions are okay all right okay so so a couple of things to do next is to go to m squared qam okay so you'll see m squared qam and m-pam are very very closely related and one probability of error completely pretty much controls what the other probability of error will be okay so that's that's one thing so let me do that first then I'll do the other thing okay so let's do m squared qam okay so so what's the constellation constellation oh my goodness okay my constellation is actually made of two different m-pam constellations I'm gonna really draw the first quadrant okay so you fill out the other quadrants if you're interested simply draw the first quadrant okay so that's my m squared qam I have say d by 2 3d by 2 m minus 1 d by 2 likewise here d by 2 m minus 1 d by 2 okay so once I assume all my points are independent okay right once I assume all my points are independent my x-axis and y-axis symbols are also chosen independently I can think of an m qam m squared qam symbol right symbol x as 2 p-am symbols this is a p-am symbol m p-am symbol and this is a also an m p-am symbol and once I assume all the m squared points are independent these two will be independent as well so in my detection I can detect my i-channel separately using an m-p-am detector and q-channel separately using my m-p-am detector again okay this is only if I assume they are independent okay if I don't assume they are independent then the whole decision regions will change it's a big headache okay so but it's very common to do this okay so even if you know they are dependent it's just common to finish this off then take care of the dependence in some other way okay so for instance I've been talking about coding I want to briefly point out when you do coding this dependence will be independence will be lost okay but people typically make a sub-optimal decision and take care of the dependence in the decoder for the code okay so that's how it works okay but anyway for now let's assume it's all independent once you assume it's all independent there's no big deal you don't have to build a separate m-squared q-am detector okay you just build two m-p-am detectors one for your i-channel and one for your q-channel then run it okay so your probability of error will be controlled by probability that either x1 is in error or x2 is in error okay and those two are simply probabilities of error for the m-p-am detector okay so this is a very simple way for the independent case if it's not independent what do you have to do find the actual decision regions in two dimension right which is quite complicated okay if you're not independent you have to take care of the probability you can't do ml anymore if you want it to be optimal right you have to do m-a-p okay so if you do m-a-p for optimal detection it becomes more complicated all right so anyway so let's write down probability that x bar is in error okay this will be what okay yeah so the way I can write it is one minus one minus probability that x1 is in error squared is that okay right so this is the probability that either x1 or x2 will be in error assuming their iid right iid distributed okay so this is the probability of error for m squared qam this is the probability of error for m-p-am is that okay all right okay so let's let's write down probability of error for m squared qam it's going to be one minus one minus what did we know probability of error for m-p-am is two times q d by two divided by root and not by two okay so this is again rough because I'm not taking care of m if you want to take care of m it'll be two times one minus one by m okay so I'm ignoring that okay so once again ignoring certain terms this will roughly be four times q d by two root and not by two is that okay okay what term did I ignore q squared I ignore because I know it's going to be power minus so if I need reasonable value it's going to go much faster so I can ignore okay all right so now we have to do the conversion back from d2 esn en okay so I'm going to say en again is n not by two there's no problem there for es I have to be slightly more careful what will this work out to two times m squared minus one d squared by 12 so okay okay what will be my rate two times log m base two bits per symbol okay so those are my various quantities of interest okay so now if I go through and do a conversion what does it work out to probability of error for m squared p a q a m okay I'm doing all kinds of I think my bracketing is gone it's got all kinds of errors okay so it doesn't matter let's say all brackets match each other okay this is for m squared q a m looks out to what the four outside is okay you know I mean it doesn't change it doesn't even play a role in my trade-off arguments for s n r and all that okay so inside there'll be a small change what will it be three by two square root of give me don't don't normalize three times three by two s n r by m squared minus one is it okay okay okay so you see this is small small change in the way the q works but it's okay I mean it's pretty much the probability of error for the p a m it's not a big deal so don't have to worry too much about this so even here if you want to say increase your rate by two okay if you want to increase your rate by two so sort of you want to add two bits per symbol then you have to double m and when you want to double m you pay a six db penalty okay so that's okay those are things you can do you can plot this but typically this is not it's not a big deal okay so you can see why m p a m is more fundamental okay so it's much more fundamental than the way this is gonna work okay so so so so so so so the next thing I want to do is do a similar calculation for m psk or m squared psk and compare it with m squared q a m and see if which is see which is better so for instance you were asking a question if I have two different types of constellations right instead of just independent signaling with constellations how do I decide which is better yeah that's an important question to ask right you might might be given several choices and you have to make a choice of constellation if you're given two different constellations how do you choose which one is better so the way to do that is a very similar approach so you go through and find probability of error in terms of these q functions okay and simplify it come to a certain stage where you can write it as just one q function with a certain type of an argument okay do it for both schemes and simply compare the arguments of the q functions and you will know the answer okay right that's the that's the trick to doing it and that's an important thing to know also okay so if you do for say I'll do for m psk okay so if you go ahead and do it what I'm expecting finally is probability of error for m psk it's a little bit more difficult okay so those are eight regions are I mean the decision regions are a little bit different and you can't split it into two mpsk so you can't do that right m psk how does it look right it's going to be on the circle okay all these points okay okay so you can't you can't nicely split it into two different things that are independent so you can't do anything simple that way so it'll be a complicated expression but you can do a series of simplifications and I'm not going to write down the final answer right now but you'll have some constant q square root right SNR times some other constant okay okay this constant will not really matter unless these other things match okay this constant will not really matter it's not a big deal and then by comparing this constant with the constant you had for the other case you can decide which is better okay so it turns out you can make a simple decision right so for instance if m is 2 psk will agree with 2 p a m if m is 4 4 psk will be the same as 4 q a m it has to be the same right 4 psk and 4 q m are the same but when you go to 8 psk and say 16 psk 16 psk and 16 q a m 16 q a m will be better so we'll see that so I'll give you the expression maybe we won't even see the derivation the derivation is a little bit painful and lots of approximations there okay so I'll just give you the expression for the next class and we'll stop here okay