 Okay, so this is lecture 15, so the last thing we saw in the previous lecture was what? What is the last thing we saw? Nyquist criteria, right? So the Nyquist criteria is what we saw, okay? So what does it say? If you want to put it very briefly, it says if you want the, if you want a band limited signal g of t and all its shifts by capital T to be orthonormal, right, that's the conditions, right? g of t is what? Band limited to, what did I take, did I take w by 2, w by 2, okay, so minus w by 2 to w by 2, so suppose you have g of f being band limited like this, right, g of t is band limited, okay, so it's, and you want what? You want g of t minus kt for k not equal to 0, okay, k and integer which is non-zero to be, well, I don't have to say anything about k, so let's simply say k and integer should be orthonormal, okay, that orthonormality is not a big deal, orthogonality is major thing, right, the shifts have to be orthogonal with respect to each other, once they are orthogonal, if you normalize one, everything gets normalized, so it's not a big deal, there's no problem with that, they have to be orthonormal and Nyquist criteria gives you a condition for this, what does that condition work out to, okay, so you think of the c of f which is mod g of f squared, then what should the c of f satisfy, okay, summation 1 by t, summation m equals minus infinity to infinity, c of f minus m by t should be 1, okay, so that was the, that's the Nyquist criteria, brief and the proof is quite simple and elementary, there's nothing very big there, but there's some subtle points, the way we convert from discrete time to continuous time and how we use the relationship between the two Fourier transforms, okay, the discrete time Fourier transform and the continuous time Fourier transform for the continuous time signal, okay, the relationship is very crucial there, okay, so these are the various things that play, w comes into the picture through the bandwidth and t enters the picture through the shifts, okay, so from here we can quickly draw several useful conclusions, first conclusion you can draw is, so usually you want to think of w as fixed, okay, w is some bandwidth that is fixed, how do you fix w, sorry, one of the constraints, no, from in a practical situation from the channel, how do you fix w, yeah, so you find the range of bandwidth for which h is, your impulse response h of f, well channel response h of f is flat, right, you wanted an ideal channel response, as in magnitude response should be flat and phase should be linear, you wanted both of that for your channel response, so if you're given a certain bandwidth to work with, you sound your channel, figure out what the channel's frequency response is and then identify that region where it's going to be flat, from there you fix your w, okay, so in practice in the real system when you're designing things, that's how you fix w, so w is fixed and then the next choice is in your signaling scheme, you remember the transmit scheme that I had, right, I convert bits into symbols and then I think of my symbols as an impulse strain separated by the symbol time t, okay, the t is the next choice, so this Nyquist criteria is going to tell you how to choose t and how to choose my transmit filter response g of t, okay, so that's the picture I have in mind, so this picture is clear to you, right, so how that works, okay, the transmitter, I think it must be in my previous, last week's lecture, I don't want to go back to it, so the transmitter has that setup, so that's what we're trying to find, so once you fix w, how can you choose capital T and how can you choose g of t, sorry, yeah, so that's the first condition that I probably wrote down last class, the first condition that you can quickly see is since g of t is, okay, so let me ask once again, let me see if this is clear with people, so why did I want g of t to be band limited between minus w by 2 and w, okay, so from a more model point of view, so what did it mean, once g of t was band limited, what could I do, exactly, so the correlation and all that comes through, so once it's band limited, after it goes through h of t what happens, you get the same thing, okay, so which means on the receiver side, I can, I know what my orthonormal basis is and I can correlate very easily with the same thing, okay, so that's the reason why I want g of t to be band limited within this, because it's very important to understand that, okay, within a flat range of h of f, okay, all right, so the first conclusion you can draw is this 1 by t, okay, so what is this Nyquist criteria telling you, so you have w by 2 and suppose I pick my, so suppose this is w, okay and then if I pick my 1 by t here, okay, I have to shift g of f to 1 by t and then it will die down, when will it die down, it will die down before the w by 2, okay, so which means you will definitely have a part where it is 0, where it cannot be 1, okay, so 1 by t cannot be to the right of w, okay, so the best you can do is if 1 by t equals w and in general you get a simple bound which says 1 by t less than or equal to w for Nyquist criteria to be satisfied, okay, so it's also called this one, so this 1 by t equals w is called the Nyquist rate, Nyquist symbol rate, okay, so if you're band with this, so base band band width is w by 2 implies, w is the Nyquist symbol rate, if you go about this Nyquist symbol rate, what can you not guarantee, okay, so finally what does it mean in terms of the receiver, it will have to deal with, there was a term that I used, there was an acronym, okay, so it's a very popular acronym with the defense forces, so you should also know it, okay, so inter symbol interference can be completely avoided if you, it can be completely avoided only if you choose to signal at less than or equal to the Nyquist rate, okay, so this is the Nyquist rate and that's the main deal, okay, so you can get rid of inter symbol interference, your S of k, right, will be received as S of k plus noise, S of k minus 1 or S of k plus 1 will have no part to play when you receive S of k, okay, so that's the main thing, okay, so ISA can be eliminated in this case, all right, so the next thing we're going to see, okay, any questions on this, anybody go back and think about this and have some questions on the derivation, how it worked out, okay, so here are the three cases that we'll consider, okay, so all of them are basically inspired by the Nyquist criteria, if 1 by t is greater than w, you can quickly conclude ISI free communication is not possible, okay, so that's fine, so what if 1 by t equals w or you decide to signal at the maximum possible rate that is permitted, okay, so you only have one choice for C of f, okay, do you see that, because look at what happens, 1 by t equals w, my C of f, if I try to draw C of f now, okay, it's going to be strictly band limited between minus w by 2 and w by 2 and then my 1 by t equals w, okay, so the only way I can make C of f after shifting constant is to have C of f equals correct, what will be the level of C of f, what will be the height, okay, I said it's not very relevant but to be very rigorous, yeah, it'll be t, okay, just write t, okay, because I'm doing a 1 by t, you want it to be equal to 1, so just to be very perfect, okay, so that's the only choice, okay, so what can I choose G of f to be, if I choose this is C f, what is the possible choice for G f, remember C of f is mod square, right, so the magnitude at least you can definitely pick, the magnitude will be, well if you want to be very rigorous about it, it'll be root t, okay, okay, but the phase is ambiguous, so you can for instance choose zero phase and be happy about it, that's one thing you can do, all right, so what's the, so what will be G of t or C of t for that matter, okay, so my C of f in the case when 1 by t equals w becomes t times rect minus w by 2 w by 2 f, right, so if you do a Fourier transform here, C of t will become what, yeah, what's in, okay, so you can show it'll become simply sinc t by t in my notation, so what does sinc t by t in my notation, sin pi t by t by pi t by t, okay, so G of f is only slightly different, instead of t you had root t, so because of that there'll be a small modification, you'll get G of t working out to be 1 by root t sinc t by t, okay, so the extra factor comes about because of, so what's t, t is 1 by w, okay, so I don't have it here, remember t is 1 by 2, signaling at the, you're saying there is some other G of f, I'm sorry, I have not talked about any problem for 1 by t equals ws, I'm sorry, 1 by t, I'm sorry, no I didn't understand the question, please repeat the question, okay, so what is, so okay what am I trying to do here, let me, let me answer, let's say something and then we'll see if it answers your question I'm trying to find a C of f or a G of t which will satisfy the Nyquist criteria when 1 by t equals w, so what does it mean once my G of t satisfies the Nyquist criteria, there is no is, what's the doubt about it, right, for 1 by t equals w I'm trying to find a G of t which satisfies the Nyquist criteria, which means as long as my channel is ideal and flat there is no is, what was your question, is it the question or you're saying there is some, there might be some problem with picking this as a sync function but that we'll come to later, those are not what I'm talking about, I'm just trying to find G of t which will satisfy Nyquist criteria, so for 1 by t equals w this is pretty much unique, okay, so this is a unique answer, you can't get any other answer, right, so I think you started off by saying any other G of f with a discontinuity at w by 2, no, no, no, there's no other G of f, it has to be flat it has to have a certain value, okay, that's the only G of f that will satisfy the Nyquist criteria, okay, well of course phase is ambiguous but phase is not a big deal, any other question, okay, I'm just trying to find, so G of t which will satisfy the Nyquist criteria, basically G of t and G of t minus k t will be what, orthonormal, that will be an orthonormal set, so if you do G of t and G of t minus t, if you do multiply the two of them and integrate from minus infinity to infinity, what will you get, you get 0, do you want something, okay, thought you were going to talk to him, all right, so let's try to, so the couple of things I want to point out, if you try to plot C of t, okay, so why is C of t important first of all, so why am I so bothered about C of t, why can't I just be happy with G of t, okay, so that's the first question you should ask, where is C of t show up in my picture, okay, so maybe I should draw the picture, so once I have this system how will I signal, okay, so I have a sequence of bits b, okay, what is this b0 to bl minus 1, okay, I think this picture for some reason people seem to have forgotten, okay, so I have a first a map according to some alphabet and that size is 2 power n, okay, I get my sk, okay, so k is between 0 and l minus 1 and then what do I do, I filter with G of t to get the transmitted signal, which is what, summation sk G of t minus kt, k equals 0 to l minus 1, then I send it through a channel with response H of t such that what happens, it's flat between minus w by 2 and w by 2, okay and suppose I chose G of t like this and capital T to be 1 by w, okay, so here you can imagine this is one such choice I can make, okay, then I add, then noise gets added to it, it's not that I add noise, noise will get anyway added to it, okay and then what can you do now here, you can correlate with G of t and its shifted versions, okay, so that's one way of doing it or equivalently you can achieve that correlation by filtering with what, G star of minus t which is the matched filter corresponding to G of t and then what should you do, exactly, so when you filter before sampling what will you have here, yeah, so X B of t, yeah G of t convolved with G star of minus t, so those will be C of t sampled, right, so basically you will be sampling C of t here, okay, so one can imagine this being C of t and its delayed versions multiplied by S of k, okay, so you pretty much what you have here will be what, S of k filtered with C of t, okay, do you see that when you correlate you're convolving, right, so it will be C of t and C of t's shifted versions and then you sample, okay, so that's what you get, sample every t, what will you get once you sample every t, okay, since this G of t satisfies the Nyquist criteria, S of k plus n of k, okay and what is this S of k, S of k will be some complex number a k b k and what will this n of k be, n k 1, n k 2, what will this be, i i d normal, all that is true, okay, zero mean variance n naught by, okay, so in spite of the channel being low pass, right, minus w by 2 to w by 2 and in spite of me signaling at 1 by w which is much, much faster than what I had before, what did we have in the previous systems, if my channel was flat between minus w by 2 and w by 2, what will I choose my t to be, much, much smaller than 1 by w, I want to go anywhere near 1 by w, I chose t to be much, much smaller than 1 by w and then what did I do, what did I pick G of t to be, yeah, just constant between 0 and t, okay, that was enough once I, if I signal very, very slowly, now I want to push my signaling rate to as high a rate as possible and through the Nyquist criteria I find that the highest rate that I could ever do avoiding ISI totally is 1 by w but then what, what am I paying with for, what is the change I have to do, I cannot use my simple G of t, right, I have to use the sync as my G of t and then I have to filter with G star of minus t at the receiver, once I do that I still have what, no ISI, simply it's as good as before, right, then if you look from here to here what happens, right, if you forget about what's happening on the right hand side and only view this, this is exactly identical to the previous case that I had just QAM with noise being added, absolutely no change from what I had before, except that my signaling rate now has gone up to 1 by w which is the highest possible rate I can have without ISI and everything else remains the same except that your transmit filters become a little bit more complex or you see correlators become a little bit more complex, previously what correlator did I have, I just integrate and dump, it has become a very simple correlator, now I have to do something more fancy but I have the exact same thing, okay, so as far as this is concerned, so what can you do here, just run a detector, okay, for each K you can run it independently, K you don't have to worry about S of K minus 1 when you are detecting S of K, because you know all of those things are independent, you can do minimum distance MAP etc, and then finally produce a B hat, okay, so this is the picture I am talking about, okay, the Nyquist criteria tells you something meaningful in such a system, okay, it tells you if G of t and G of and it shifts by capital T are orthonormal, then if G of t occupies a certain bandwidth and if your channel is flat in that bandwidth, you can achieve ISI free communication with such a setup, okay, you filter the transmit filter G of t and you filter in the receive side with G star of minus t and sample every t, you will pretty much get rid of ISI and you can run your same detector as before and you can signal much faster than what we did, okay, so this is the significant next step that we have done, okay, but this this is, okay, first of all, is there any question something that is disturbing or anything I said which is not quite clear, you are not convinced about, you think it cannot be true, just cannot happily increase your symbol rate without paying for it, this thing is okay, it's fine, okay, so let me ask another question now, what is this N of K, can I define N of K to be a complex Gaussian process, white Gaussian process, right, it is a white Gaussian process, right, what will be the autocorrelation function for the real component of N of K, we N naught by 2 times delta, right, so you see all that, so all those things are correlations when you go back to random process and convince yourself, so this is N of K is a white Gaussian process, okay, so that is very crucial, okay, what else, what other comment can I make, okay, so this is my H of F, okay, so the next thing to, yes, but then you have to do correlation, you cannot do filtering, see when I put a box and write a function of time inside, what I mean is what, saying you're filtering by that, it's an impulse response for the filter, if I put G of T and G of T minus KT, T minus T etc, what am I saying that I should do, should correlate, okay, so that's something else, that's correlation with G of T, G of T minus T, it's this, but these two are the same, okay, but there is a problem with this, what is the problem, what are the various problems that you can think of, if you try to implement such a thing, what is the first problem you can think of, sorry, causality of G of T, yeah, you're disturbed by it, yeah, you have to be, right, G of T is going off on both sides and G of T in fact, decays very, very slowly, how does it decay, 1 by T, okay, in fact, even C of T decays as 1 by T, okay, that's a very, very slow decay, okay, so there are several confusions here with C of T decaying very slowly, first thing is, since it's cost, okay, so how do you, okay, first let's address the causality problem, how will you get over the causality problem, yeah, you just delay the G of T, okay, so instead of G of T, I'm going to put G of T minus T naught, since everything is LTI, everything will get delayed, okay, and somehow I have to estimate that delay or I should know that delay at the receiver and compensate for it, I can do that, okay, so the first problem that people said G of T is non-causal, okay, so this is the first issue, so I say issues, G of T is non-causal, it can be overcome by delay, okay, right, so that's one thing that you can somehow get over without too much of a problem, do you have any other question, yeah, that also you can delay, everything, so you take care of the total delay from G of T to G star of minus T, so you have to delay everything, so even G star of minus T, so you're saying, you're pointing out G star of minus T, well G star of minus T is G of minus T and that's also non-causal, how will you get over it, you once again delay, you know, I mean, so basically the filter that you'll be implementing will be something like this, okay, so this is the filter that you'll be implementing, this is a very implementable thing, definitely in digital, you can happily sample this and implement this, there's no problem, okay, so that's the delay, okay, so even this can be delay, you can solve this problem, okay, so you simply delayed and then you have to adjust for the delay, where will you adjust, where should you adjust for all these delays at the receiver, which point will have to take care of all these delays, the sampling instant, okay, so sampling instant should be suitably delayed and you should pick the right time to sample so that it corresponds to the undelayed version, okay, the same point, okay, so that might require some careful study, okay, so you should know what the delay is and adjust for it and sample at the suitable time, okay, if you're doing the lab, you'll see this will be one of the first things you'll have to face, okay, in the real signal you have to, if you're doing the advanced communication lab, think DD students are doing it, you'll see this will be one of the first things you'll have to decide when you actually get a received signal, it'll be definitely delayed, okay, then you have to figure out where to sample, okay, so that's the first issue, there's a second issue which I will address a little bit later also more formally, but the thing is if you pick G of t as sink, sink decays, okay, so all these things are t equals 1 by w, G of t is a sink, okay, G of t decays as 1 by t, okay, so if you sample 1 by t at some interval say capital T, capital 2 t so on and then add up all the samples, okay, 1 by capital T but plus 1 by 2 t plus 1 by 3 t, is there a convergence series? No, it doesn't converge, it can be arbitrarily large, okay, it will keep on increasing, okay, so why will that cause any problem in my receiver? Is that something I should be worried about? Okay, so it can get large, so this can get large, so what will happen is suppose my sampling period t is not exact, it's off by say some point 0.1 millisecond or something, okay, then what's happening? All my previous sinks, right, they'll all be having now non-zero components, okay, right, and all of them will add as 1 by t only, so I can get larger numbers, okay, okay, but if we're to decay faster than this then it won't be a problem, okay, so because of this, there seems to be a sensitivity to 2, okay, this is very, so receiver becomes sensitive to, okay, so let me write it that way, receiver becomes very sensitive to errors and t, okay, if you make a small error in t, you can suddenly be off by a lot, okay, so let me draw one more picture to illustrate what is going on here, so you may not appreciate the sink being orthonormal, okay, so let me draw a few pictures then convince you about how this is working out in such a precise fashion and without any, without more, some more pictures, okay, so the first picture I'm going to draw is, draw a sink and a shifted sink, okay, and then ask you to convince yourself that indeed if you multiply the two of them and integrate you'll get 0, okay, so it's not so easy to quickly see from the picture, okay, so what does the sink look, okay, so suppose I take sink, okay, so these are all some illustrations just to expand on this zero ISI which somehow happened by magic, okay, so the first thing is orthonormality for the sink, okay, so let's draw a picture to see how it works out, okay, suppose I draw sink t by t, okay, what is this, sine pi t by t divided by pi t by t, so all of you should know how to plot this, okay, so what's the value at 0, 1, okay, and then it decays and it has zero crossings at t, 2t, 3t, so on, okay, and minus t, minus 2t, minus 3t, so on, am I right, okay, all right, and then it's kind of like a sine wave except that it gets multiplied by 1 by t, so 1 by t goes off like this, then so you should, should also happen here, okay, so kind of works out, okay, so now what happens if I do shift it to the right by capital T, okay, so this whole sink shifts and then I can draw this guy here, okay, so I am drawing it slightly poorly, so hopefully you can visualize how this works out, okay, somebody were to draw these two pictures and say these two signals are orthogonal, it's not, it's not so clear, right, so I don't know, maybe you believe it if I tell you, but it's not so clear, I'm sorry, no actually these two signals are orthogonal, what do I mean by orthogonal, integral sink t by t, sink t minus t by t, right, if you do this what will you get, you'll get 0, so this is true, that's what I mean by orthogonal, not that at each point they have to multiply and give you 0, that won't happen, so this is what I mean by orthogonal, okay, so but you're right about the zero crossings, I'll come to it, it's a little bit sensitive to that also, okay, so that's a surprising property it requires some illustration, okay, so that's fine, now what are we doing when we suppose we do BPSK with this as my G of t, okay, suppose in my communication system I'm choosing to do BPSK for the mapper, so either multiply the sink with plus 1 or minus 1, okay, right, is that clear, okay, so notice what's happening, suppose I multiply this sink with plus 1, okay, or say some s of 1, s of 0, okay, and then I multiply this sink with s of 1, and then the next shifted sink if I can draw it with, maybe I'll draw it with a different color, what do I change color, no, it's not color, how should I change color, I don't know, okay, maybe I'll do this medium point, it seems to be red, okay, so if I look at this guy, okay, I'm drawing it very badly, but okay, let's do it, let's do a better picture, okay, so if you do this sink now, okay, so there's a curious phenomena that's happening, what is happening, can someone say, make some observations, okay, suppose I multiply this with s of 2 and so on, and then add up all of them, okay, so what am I doing when I do my modulation, I'm saying s of k g of t minus kt, okay, just for concreteness, assume we're just doing 0 to 2, okay, 0, 1, 2, and bpsk, so that s of k is just plus or minus 1, so you can easily visualize it, what's happening, okay, so I'm going to multiply the first sink I drew with say s of 0, and the second sink with s of 1, third sink with s of 2, and then add up all of them, okay, right, and then what do I do with the receiver, okay, what do I do with the receiver, I'm doing some correlation and all that, but it seems all that complicated, but even at the transmitting end, one nice thing you can observe, okay, what is, so suppose I call this x of t, what is x of 0, s of 0, okay, do you see that, what is x of 1, s of 1, what is x of 2, s of 2, why did that happen, okay, because the 0 crossings, all of them match, okay, all of them match for all of the signals, and only at 0, which is a multiple of t, you have something non-zero, so what multiplies that shifted version alone contributes to x of nt, so you can show very easily x of lt equals s of l, okay, so that's already a nice property that you're seeing, okay, so even without worrying about orthonormality and all that, okay, so now what happens when you do a correlation with the receiver, right, you're doing g star of minus t and sampling at t, which is the same as correlating with g of t or g of t minus t, so on, what happens when you correlate, all the other things in the summation pretty much vanish if your shift was exact, okay, if your shift is not exact, what will happen, it will give you a heavy non-zero contribution because it decays, doesn't decay all that much, it will be still low, but it will be a little bit not all that low, okay, it will be significant, so everything else goes away and you happily get your s of k back, okay, so that's the nice thing about this thing, so it's good to write down those expressions, but it's also important to visualize that these are the actual signals that are going on, x of t looks like this, it's a sum of sinc which is multiplied by plus 1 minus 1 and all that, okay, so if you actually add it up and it will look almost like this, okay, this looks very different from the picture we had before, when in the previous case when our signaling time was really really large, what was the picture, it was more like a rectangle of pulse, this is very much undulating waveform, okay, so this also is another way of visualizing the whole thing and understanding that this is what's happening, okay, so anyway this 1 by t decay is a bit of a problem, you want to get rid of it, you want to reduce that and the only way, one way of nicely reducing that is to signal at a slightly lower symbol rate, you decrease t a little bit, turns out you can afford to increase that rate of decay, if I insist on t equals 1 by w, this is the unique signal, I can't do anything else, so I'll have to reduce my signaling rate t to less than 1 by w, then hope to change my g of t and hope that it will decrease, okay, so that's the next case that we'll consider, okay, so any comments about this, okay, so what should be, I mean there's one question you can ask from, okay, for several reasons you can ask this question, so one of the reasons you might ask this question is I'm adding noise to x of t, right, my channel is bandwidth is within the bandwidth of x of t, x of t is going through, okay, can I have a receiver where I simply sample y of t at multiples of t, okay, I know already that x of l t is only s of l, okay, is that a good receiver to have, sorry, yeah, assume, I've already told you I'm able to synchronize and all that, so what's the problem with that, will you come up with any argument against it or will you say it's perfectly fine, so what if I simply sample y of t at, can I do that, exactly, so noise we have, it's so, noise is going to have huge bandwidth and you're not doing anything to it and you're directly sampling, so maybe I will low pass filter my noise, so say, I'm sorry, well, but you're sampling, so all kinds of crazy things will happen if you don't cut down the bandwidth, suppose I say I cut down my bandwidth to a suitable size and then I sample, can I say this receiver is identical to the previous receiver or do you still see some advantages to the previous receiver, okay, anyway, some food for thought, think about it, okay, if you think you've solved this question come and talk to me, we'll have a discussion about it, okay, so think about if this is a good solution or it's a correlation fundamentally important and you have to do it, okay, of course from a noise point of view it seems obvious that you have to at least low pass filter, so don't directly sample this, so maybe you low pass filter at minus w by 2, w by 2 and then sample, okay, so that the noise also gets filtered and it's reasonable and then you sample, okay, is that a good enough thing, what are the problems, what may or may not be the problems, there's something to think about, all right, so the next thing I want to do before we close for today is to look at the case, third case where I let t to be, where I let 1 by t to be less than w, okay, so this is an interesting case and now you can have several possibilities for, for signals that will satisfy the Nyquist criteria, okay, so if you choose 1 by t to be less than w, okay or if you want you can write t to be greater than 1 by w, okay, then there are many possibilities, many choices for g of t, okay, so one choice which is very popular and used all the time, g of t and I'll say c of t also, right, so one choice which is very very popular is what's called the raised cosine choice, okay, so that's the choice which we will study, okay, so there are several other possibilities, by no means it's unique, if you let 1 by t to be less than w you can have all kinds of choices, but the choice that we'll be studying is raised cosine c of f, okay, so this is the choice we'll study, one of the properties it has is c of t will decay as 1 by t squared now and that's convergent and it falls very fast, okay, so it's very good, okay, so it's also very very popular, okay, so c of f is raised cosine, g of f will become what? square root of raised cosine, so that's the name for g of f corresponding to this, square root raised cosine, okay, so here's the definition, I'll do it with a picture first and then write down the expression because the picture is much much easier, okay, so the choice we make for 1 by t is basically w by 1 plus beta and we let beta be a parameter from 0 to 1, okay, so this is some parameter you have to choose, it's called the roll off parameter, okay, beta equals 0 coincides with the old case that we had, okay and beta equals 1 is when 1 by t is w by 2, half of the Nyquist rate, okay, so that's the range, so we're going to choose my signaling rate to be between w and w by 2 in this fashion, w by 1 plus beta, okay, so that's my choice, okay, or you can view it as t equals what? 1 plus beta by w, okay, so 1 by w plus beta times 1 by w, okay, so and then what do I do? Once I do this, I'm going to draw c r c of f, okay, so I'll call it c r c, okay, so that's the raised cosine f, okay, so 0, I'll draw 1 by t first, okay, so I'll just draw these points so that we have some scaling and idea of what's going on. 1 by t is my shift, okay, with respect to this shift, c of f should add up to a constant, if you alias with respect to this shift, c of f should come up to be a constant, okay, what's happening, okay, little question, okay, so the next point I'll write down is 1 by 2t minus 1 by 2t, okay, and the next two points I'll write down will be 1 plus beta divided by 2t and 1 minus beta divided by, okay, all right, can you see it, I hope you can see it, on two sides of 1 by 2t, I'm marking 1 plus beta by 2t and 1 minus beta by 2t, okay, so like same thing I'll do on this side also, minus 1 minus beta by 2t and then minus 1 plus beta, okay, so this is what I do, okay, between these two guys, I'm going to keep it flat at, okay, and then from here to there, there will be a cosine, so what's cosine, cosine looks like this, no, see it, maybe I'll draw it on the left, okay, so cosine is going to look like this, right, right, I'm drawing it a little bit poorly, but you see this, this is 0, this is pi by 2, this is pi, right, that's how cosine looks, raised cosine is this shifted up, okay, and I'll only look from 0 to pi, okay, I'll take that part alone and try to fit it into here, raise it and fit it into here, okay, so let me draw that, I'm going to make a bit of a mess, but hopefully we'll get it right, okay, so that's the cosine and this side is simply the shifted version, oops, okay, so it's become totally asymmetric but it should be symmetric, all right, so this part, this part corresponds to 0 to pi of a raised cosine, okay, which is this guy, okay, 0 to pi of the cosine, okay, so I'll have to scale it suitably to fit there, okay, so that's my raised cosine response, okay, so what will happen if I alias, first thing is, first thing is that when alias the flat part will fit in nicely and on the cosine part this thing will come and the two of them will add to give me a flat, okay, so when alias what will happen, something like this will happen and these two cosines will add nicely to give me a, okay, so let me write down precisely what this is, write down an expression for it and the expression is what's usually important, CRC of F equals capital T for 0 mod F between 0 and 1 minus beta by 2T and T by 2 cos squared, okay, so let me write down 1 plus cos and then we'll write cos squared so that it's CC, okay, 1 plus cosine, okay, so the next thing what I'm going to write inside the cosine is just scaling and shifting to squeeze it into that 1 minus beta by 2T and 1 plus beta by 2T, okay, so it'll look a little complicated but it's just to squeeze the cosine into that, okay, so nothing more to it, okay, pi capital T by beta mod F minus 1 minus beta by 2T for 1 minus beta by 2T mod F between 1 minus beta by 2T and 1 plus beta by 2T, okay, and then 0 for mod F greater than 1 plus beta by, okay, so 1 plus beta by 2T equals W by 2, do you agree, that's how I chose my W, right, I chose my 1 by T to be W by 1 plus beta, so that works out to W by 2, so this is W by 2, okay, all right, so this is the raised cosine so you should know the shape, so it's 1 by T, 1 by 2T, T is 1 plus beta by W and then 0 to 1 minus beta by 2T, you choose flat and then you come down in a cosine, okay, so this is not too crucial to remember that the thing that you have to look for is C of T and G of T, okay, so you have to do Fourier transform of this, how many of you think you can do Fourier transform for this and find C of T in 30 seconds, what's the answer, okay, suppose I do a Fourier transform, what's my C of T, how quickly can you find it, okay, so I won't let you do it, I'll give you the answer, it's sinc T by T multiplied by cos beta pi T by capital T divided by 1 minus 2 beta T by capital T squared, okay, so this is the C of T and you see it nicely decays as 1 by T squared, okay, which is what I wanted, okay, so now I have to go to G of F and then go to G of T, okay, so now how do I go to G of F, I have to take square root, for the first part it's just root T, it's okay, for the second part how can I easily take square root, yeah, you write it as 2 cos squared that by 2 and then you take square root, you'll get a cosine, okay, and then from that expression you can do a inverse Fourier transform also very easily, okay, it is just that multiplied by several functions, very easy to do the inverse, well, not very easy, you have to work on it, you can eventually get that, at least conceptually there is nothing more to it, okay, so if you do that the square root raised cosine pulse, okay, cosine pulse, okay, works out to be G of T equals, I'll simply write it down just for telling you what is, a lot of books have this, so it's quite standard, 1 plus alpha pi T by capital T plus T times sine 1 minus alpha pi T by capital T divided by 4 alpha T whole thing divided by 1 minus, okay, so I suddenly shifted to alpha, I don't know why, okay, so all the alphas are betas, okay, sorry for that, 1 minus 4 beta T by capital T square, okay, so that's your square root raised cosine pulse, okay, so the first thing to test is what happens when beta equals 0, okay, so you should convince yourself that beta equals 0, this becomes the sink, okay, it should become the sink, if it doesn't become the sink then something is wrong, okay, so that's not too difficult, one can show it, okay, without too much of a difficulty, we'll get that, okay, so what happened, should divide, okay, so this is what I copied down from a book, I think I might have made a mistake, so anyway, I'll check on this, we'll meet on Monday and we'll discuss.