 So, in the last lecture what we saw was that if you took a linear plant which is positive real and the definition of positive real has these ambiguities, but what we effectively would mean is that the Nyquist plot lies in the right half plane and the system the given system is linear. Now, if you take such a plant and then such a plant is of course, passive. Now, if you now take nonlinearity such that the characteristic of the nonlinearity lies in the 0 to infinity sector, then one can think of that nonlinearity also as being a passive system. As a result what happens is when you interconnect the linear system given by a transfer function which is positive real with a nonlinearity which is in the 0 infinity sector, then you end up with a resulting system which when you do not give any input is asymptotically I mean of course, the origin is an equilibrium point of this system without inputs and this system is asymptotically stable. In fact, when the transfer function is taken to be strictly positive real, then you can say more. In fact, you could say that the resulting system is exponentially stable. So, let me just reiterate what I have just said in terms of some diagrams so that it would be clear to you what I am trying to say. So, what we are doing is the following. So, you have a linear plant Gs and you have this feedback connection and you have this nonlinearity so this negative feedback. Now, this linear plant that you have, we are saying that this linear plant is positive real. Of course, in the last class I had given various interpretations of what we mean by positive real but what I would mean by positive real here is that the Nyquist plot of this plant is in the right half plane and in addition Gs is stable. Of course, one could also use the definition of positive real to say that Gs plus G transpose S star is greater than equal to 0 for all real for all S whose real part is positive. That is another equivalent definition. Now, by making this assumption, let me write down Gs is positive real and stable. Now, if I call the input of this Gs as U and the output of Gs is Y, then what it means when you say that Gs is positive real and stable is that U transpose Y is greater than equal to V dot where V is the storage function of Gs. We have already talked about what the storage function could be and in order to find out what the storage function could be, we invoke the positive real lemma or the Kalman-Yakubovic lemma from where we get this matrix P and if you recall when you have Gs to be positive real and stable, then that is equivalent to this set of equations. That means if you take the minimal representation for G of S to be Ax plus Bu Y equal to Cx plus Du, then using these matrices you can write down that there exists a P which is positive definite and two other matrices L and W such that first of all A transpose P plus Pa is equal to minus L transpose L and then you have Pb is equal to C transpose minus L transpose W and lastly you have W transpose W is equal to D plus D transpose and so this P that you get which is positive definite, you take the storage function V to be X transpose Px where this P is this P that you get from the positive real lemma. Now what about the nonlinearity? Now the nonlinearity that one picks is of the following kind. So the nonlinearity, so let me draw that diagram once more. So here you have the nonlinearity, we are calling this output of the linear thing U sorry the output Y and the input U. Let me call the input of the nonlinearity psi and let me call the output of nonlinearity phi. Now the nonlinearity is such that if you draw the characteristics of the nonlinearity, so you have psi on this axis and you have phi on this axis then the nonlinearity is something that lies in the first and the third quadrant. Now if it lies in the first and the third quadrant then it is very clear that if you multiply the input of the nonlinearity which is psi to the output of the nonlinearity which is phi then psi or psi transpose phi if one is considering the vectorial, the vector situation that means multi-input, multi-output situation this is clearly greater than equal to 0. Now since this is greater than equal to 0, one can say that this nonlinearity has a storage function V1 which is the 0 storage function. Now earlier in the last slide I had said that you have U transpose Y is greater than equal to V dot where this V was given as X transpose PX with this P coming from the positive real lemma. Now if you look here there are these interconnection equations which is Y is equal to psi and phi is equal to minus phi is equal to minus U. Therefore this psi transpose phi in fact psi transpose phi is really equal to minus U transpose Y and this is greater than equal to 0. So if you add this equation and this equation you end up with 0 is greater than equal to V dot. Now what this means is that if you use this V that is X transpose PX as a Lyapunov function for this closed loop system assuming that this input the external input is 0. So if you have the external input to be 0 this is like a system with no inputs and this system with no inputs will have these conditions satisfied and if you have these conditions satisfied then you have 0 is greater than equal to V dot and if you take V to be X transpose PX and that is then that can act like a Lyapunov function of for this closed loop system and then what we have here is that its derivative is less than equal to 0. So in fact from that you can conclude that this resulting system is stable. Now if you remember we had also talked about the Kalman-Yakobovich Pope of Lemma what we had said was that if you take this thing to be strictly positive real okay. So maybe I will mark that with plaque then the change in the equations are that instead of this you will get minus epsilon P also into this equation into this first equation all the other equations remain as they are. But what this means is when you substitute in here instead of V dot being less than equal to 0 you will get 0 is strictly greater than V dot and as a result because of that epsilon P which is there using the Kalman-Yakobovich Lemma if this Gs was taken to be strictly positive real then we can conclude exponential stability okay. So what we are going to do in this class is we will now explore what new conclusions we can draw using this rather powerful result okay. Maybe before we do that I would like to use this V as which we have got from this positive definite matrix P as a Lyapunov function and show that this resulting system is asymptotically stable. Well in order to show that I essentially only have to show that U transpose Y is greater than equal to V dot and then the rest of it we have already seen. That means if you take U transpose Y greater than equal to V dot and you already know when you take a nonlinearity like this that psi transpose phi is greater than equal to 0 and when you sum both of this you get 0 is greater than equal to V dot. So all you have to show is that U transpose Y is greater than equal to V dot okay. One small adjustment of a constant I have to do I should not be taking V as X transpose Px but I should be taking this as X transpose Px by 2 a half. So now if we take that then so V we are taking to be a half X transpose Px. So then in that case V dot is half X dot transpose Px plus half X transpose Px dot but we know that X dot is AX plus BU from the equation of the system of the linear system. So substituting that in here we will get a half of X transpose A transpose P plus from here similarly so PAX plus I would get half U transpose B transpose PX plus half X transpose PBU. Now this from the Kalman-Yakubovic lemma or rather the positive real lemma this is the same as minus L transpose L. So I can put that in there and if I look at these two terms Pb we know that one of the equations that we have from the positive real lemma is Pb is equal to C transpose minus L transpose W. So for Pb I can substitute C transpose minus L transpose W and similarly for this so I will just expand this out and show what that comes to. So a half X transpose PbU becomes a half X transpose C transpose U plus or rather minus a half X transpose L transpose W U. Now if you look at the output equation of the linear plant you have Y is equal to Cx plus du and so the Cx I can substitute is Y minus du and once I do that I get a half Y transpose U minus a half U transpose D transpose U. So this is from this X transpose PbU from this thing you would get exactly the transposes of this. So when you put them all together then you end up with B dot is equal to minus a half X transpose L transpose Lx that accounts for this and then from both of this you will end up with U transpose Y because this half and then another half coming from there and then you have this term and there will be a similar term coming from the other portion and so putting both of them together you will have U transpose D transpose U and you will have U transpose du and D plus D transpose one of those equations that you had was D plus D transpose is equal to W transpose W the positive real lemma gave us this equation. So using that you will get U transpose W transpose W U a half of that and then these last two terms which so there will be this term and the transpose of this both with an M minus sign and so if I put all of them together then what I end up with is V dot is equal to first of all U transpose by which we had here and then these two terms and the other two terms this and its transpose putting them all together you will get minus half Lx plus WU the whole thing transpose Lx plus WU and now you see this quantity here is positive quantity and therefore you can conclude V dot is less than equal to U transpose Y and this is essentially what we wanted in order to conclude from this sheet that U transpose Y is greater than equal to V dot 0 is greater than and you already have this above the non-linearity and when you put both of them together you get 0 is greater than equal to V dot and therefore you can show stability. Now if one assumes that G of S is strictly positive real then in that case in this equation here A transpose P plus PA there will be one additional term here and this additional term if I call it epsilon P this additional term this additional term will finally end up in the last equation and here you would have minus epsilon x transpose Px and this will now let us prove that the resulting system is in fact asymptotically stable. Now what I am going to do is I am going to make use of this rather powerful theorem and I am going to look at all kinds of non-linearities and derive new results about which non-linearities with when put in a feedback connection with certain kind of linear plant would result in an asymptotically stable system. So let me conclude about this first. So what I am saying now is suppose you take a non-linearity let me call it f that belongs to the 0 infinity sector. Now when I say that a non-linearity belongs to 0 infinity sector what I mean by that is that if psi is the input of the non-linearity and this is f of psi the output of the non-linearity then this non-linearity lies in the 0 to infinity sector that means it lies either in the first quadrant or in the third quadrant. If you have any such non-linearity any non-linearity which belongs to this class and you take a G of s which is stable and strictly positive real then this feedback connection of the strictly positive real stable plant with the non-linearity results in something which is asymptotically stable. But suppose now the non-linearity that we consider is this particular non-linearity. Now if you consider this particular non-linearity then you can see that of course it is true that this non-linearity lies in the 0 infinity sector but in fact you can say more things about this non-linearity because suppose I draw slope like this call it k1 and suppose I draw another slope let us say like that call it k2. Then in fact this non-linearity f lies in the k1 k2 sector. So what I am trying to say is that the non-linearity is such that k1 is less than equal to f xi by xi which is less than equal to k2. So in some sense by declaring f to be in this sector from 0 to infinity we are doing overkill because we can get a much tighter bound for the non-linearity and in fact we can say that the non-linearity lies in the sector k1 k2. In fact out here as you can see so here the way I have drawn it of course the k2 there is some portion here of the characteristic which has the slope k2 but instead of writing this sometimes people would write something like this k1 k2 an open interval k1 k2. Now when you mean an open interval k1 k2 or a closed interval k1 k2 the difference between them is essentially to do with these inequalities whether they are strict inequalities or just you know non-strict inequalities. So when you have non-strict inequalities then you would put the closed bracket when you have strict inequality you would put the open bracket yeah and of course there would be semi open semi open semi closed kind of intervals also for the non-linearity. Now fine because this non-linearity lies in the 0 infinity sector therefore we know that if you take any stable strictly positive strictly positive real transfer function and connect it in this feedback loop with this particular non-linearity you will get an asymptotically stable system. But since we can put a tighter bound on this non-linearity that means this non-linearity actually lies in the k1 k2 sector therefore one would expect that apart from these plants there are other plants also which you could connect with this non-linearity and these plants may not be belonging to this class that means they may not be strictly positive real or stable but despite that the resulting system is asymptotically stable and the reason why one would believe that is because it is true that this particular non-linearity is in the 0 infinity sector but in fact we can say that it is in the k1 k2 sector and so if it is in the k1 k2 sector it would be very surprising if there are no extra plants that one can connect to the non-linearity resulting in the system being stable. I mean one would naturally expect that there would be more plants that you can connect to the non-linearity and the resulting system is asymptotically stable. So now what we are going to do is we are going to explore this situation where you have the non-linearity given in a certain sector and one wants to characterize all the transfer functions which you can connect in this feedback loop with those non-linearity such that the resulting system is asymptotically stable. So now in order to do this what we would do is we will do some characterization of the various kinds of non-linearities. So let me begin by looking at certain classes of these non-linearities and I will show that these non-linearities can be transformed into a non-linearity in the 0 infinity sector. Let me make this clear by some examples. So suppose you have a non-linearity belonging to the k1 infinity sector. What do we mean by saying that a non-linearity belongs to the k1 infinity sector? What we mean is here is this line k1. Let me use a new sheet rather than okay. So we want to talk about this f which is in the k1 infinity sector. So what we mean by that is so this is the line with slope k1 and if the non-linearity is in the k1 infinity sector that means it lies here and here. So it could be something like that. In other words the non-linearity is such that f psi by psi, psi is the input of the non-linearity and f psi is the output. This is less than infinity but greater than equal to k1. Okay. Now if you have a non-linearity like this, we can convert this into a non-linearity in the 0 infinity sector. Okay. Now how does one convert this into a non-linearity in the 0 infinity sector? Well, if you have a non-linearity in the 0 infinity sector, then the following is true. Psi times h of psi is greater than equal to 0. Now for a non-linearity which lies in this sector, this is true. But this is true, is the same as saying f of psi minus k1 of psi. Okay. So for example, if I look at this particular psi, so this is f of psi and so f psi minus k1 psi is this portion here which is positive. On the other hand if I take a psi which is negative then f of psi is here and this is k1 psi and therefore this quantity here is in fact negative. This quantity here is negative. So if now this thing let me put a transpose is multiplied to psi, then here you see you get a positive quantity multiplying positive. So the resulting thing is positive. In the left half you have a negative thing multiplying negative. So the resulting thing is positive. So for any non-linearity which lies in the k1 infinity sector, we can say that this condition is always satisfied. Now if this condition is satisfied, then this is what we do. So this is slope k1 and you have a non-linearity like that and what we had just written was, so the input of the non-linearity psi and f of psi is the output of the non-linearity. So if you think of this non-linearity like this and let me call the input psi and the output is f of psi, then one way we could characterize such a non-linearity is by this quadratic form which is f psi minus k1 psi transpose psi is greater than equal to 0. Therefore now if you think of, so this is the output phi, if you think of a non-linearity, some new non-linearity which has the same psi as the input but it has phi tilde as the output where this phi tilde is really f of psi minus k1 psi, then this new non-linearity will have the property that phi tilde transpose psi is greater than equal to 0 and therefore this new non-linearity, let me call it NL1 is passive. Now how does one obtain NL1 from NL? Well one way to obtain NL1 from NL is the following. So you have the original non-linearity and it has the input psi and if it has the input psi it will give the output which let me now call it phi but you want a new non-linearity which doesn't give phi as the output but gives phi tilde as the output. So how to get phi tilde as the output? You want to use the same psi as the input. Well if I take a k1 here, so k1 times psi is what I will get and I use a feed forward then what I will have here is phi tilde which is phi minus k1 psi and so now if I think of what is in this dotted box as my new non-linearity it has as its input psi and it has as its output phi tilde and this phi tilde transpose psi is greater than equal to 0. So this resulting non-linearity is in fact passive. Now therefore this resulting non-linearity if it is now connected in a feedback loop with a plant which is strictly positive real then the resulting system is asymptotically stable. So let me draw that. So you have this non-linearity. So you have psi and you have k1 of psi fed forward. So it is still psi here. So the original non-linearity had this output phi but this new non-linearity has the output phi tilde and now if this phi tilde is connected to some gs where this gs is strictly positive real and stable then the resulting system is of course I mean using our earlier results whatever we had concluded earlier using the positive real lemma and so on we can conclude that this resulting closed loop system is asymptotically stable. But now we are interested in knowing not what can be connected to this new non-linearity but what can just be connected to this non-linearity what plants can be connected to this non-linearity and the resulting system would be asymptotically stable. So what I am now going to do is I am going to do a set of transformations and these transformations go under the name of loop transformations and by loop transformations we can actually conclude or we can find a much larger class of plants which you can connect to the non-linearity and the resulting system would be asymptotically stable. So this is what we do. So what we are going to do is so here is a linear plant and here is a non-linearity and we are assuming that this non-linearity is in the k1 infinity sector and we would like to know that when you do this feedback connection what are the linear plants? So the question we would ask is the following how to characterize all linear plants which when connected to non-linearity in the k1 infinity sector results in asymptotic stability. So now let me call the input to the linear plant u and the output y and let me call the input to the non-linearity psi and the output phi. Then of course by the feedback connection then you know u is equal to minus phi and you also know that the transfer function g is really y by u but that is the same as psi by minus phi. Now so earlier we saw that given this non-linearity with input psi output phi such that this non-linearity belongs to the k1 infinity sector you can get a new non-linearity. So let me call it NL1 such that it has the same input but now the output is phi tilde such that this NL1 is really it belongs to the 0 infinity sector. Now how does one get this NL1 from this NL? We just saw earlier that one way to get this NL1 from this NL is by using this feed forward where you take this k in k1 and feed it there and therefore now what you have will be phi tilde. Now you originally had a g of s here which you had interconnected with this NL. So what I am trying to say is that you had this gs interconnected to the non-linearity NL and we call this u, we call this y, we call this psi, we call this phi. Now one could modify this transfer function g of s into some new transfer function g1 of s such that this g1 of s will take this modified input but give the same output as the original g of s. So what we are trying to say is that we modify this g of s in some way such that it takes this u tilde as the input rather than the earlier u as the output but gives the same output as g of s would give for u. Then we can calculate what this g1 of s would be because this g1 of s, g1 the transfer function g1 of s is given by y divided by u tilde but this is the same as saying psi divided by because y is equal to psi and u tilde is equal to phi tilde minus phi tilde but this minus phi tilde we know is the same as psi, this minus phi tilde is the same as minus phi. So we had taken this phi tilde to be phi minus k1 psi. So I am just substituting for phi tilde so I have minus phi plus k1 psi. Now if you remember in the last slide I had written that the original transfer function g is really psi by minus phi so dividing by minus phi in the numerator and the denominator we get this g1 is really the original g upon 1 plus k1 g. So what we are trying to say is the following. So what we are trying to say is the following. So if you had a nonlinearity psi phi and you had this g of s then this interconnection is exactly the same as the following interconnection where you have this nonlinearity, the same nonlinearity at the before you keep the input to the nonlinearity exactly the same but for the output what you do is you put a k1 here. So you fed forward this k1 psi and so now you have this new input phi tilde. Now this new input phi tilde you are feeding back then the resulting transfer function that you would have here which one called g1 of s as we just derived in the last slide would be g of s upon 1 plus k1 g of s but that is the same as saying you have g of s and then you have 1 plus k1 g of s what that means is you have this feedback structure g of s with fed back with k1 s and this resulting system. So when you put gs with this nonlinearity and this nonlinearity this nonlinearity is in the k1 infinity sector and you have this nonlinearity. Now you can convert this nonlinearity into something in the passive that means you can convert it into something in the 0 infinity sector by doing this feed forward. So you have done a feed forward and what have you done you have forwarded the input to the output by this loop. So along with the nonlinearity you have given a feed forward but when you do this feed forward on this loop on the open loop what you do is you give a feedback with the same k1 on the original plant that you had and then this resulting plant along with this new nonlinearity. So the new nonlinearity is what is inside this dotted box that I am putting along with this new linear plant which is what is there in the dotted box there these two this combination is exactly the same as this combination. And now if this combination is exactly the same now by using this feed forward what we had done is this nonlinearity in the k1 infinity sector you had converted it into a nonlinearity in the 0 infinity sector. Therefore you know that for anything in the 0 infinity sector making use of the positive real lemma and the results about passive systems and so on this resulting system here if this is strictly positive real and stable then that along with this this resulting system is asymptotically stable but since this system is equivalent to this system that is the same as saying that this system is asymptotically stable. Therefore for a nonlinearity in the k1 infinity sector given a gs that gs with that nonlinearity in a closed loop system this would result in asymptotic stability if when you modify g to this thing that means when you modify g to g1 where g1 looks like this gs upon 1 plus k1 gs and if this is strictly positive real and stable then the original system is going to be asymptotically stable. So I hope this is clear so what you are really doing is you are doing some sort of a loop transformation. So what you did is on the nonlinearity you get a feed forward loop and as a result on the linear plant you get a give a feedback loop both these loops have the same gain. So this is a feed forward loop with negative gain here you have feedback loop with a negative feedback. Now as a result this modified linear plant along with this modified nonlinearity they both together now you invoke the positive real lemma or the Kalman Yakovish lemma whichever one is applicable and then you can talk about the stability of that resulting system and this resulting system is exactly equivalent to this system. So therefore you can say something about all the transfer functions which when interconnected to a nonlinearity in this particular sector will give asymptotic stability. So what is the result? The result is that if you take any plant and you do this transformation of the plant this transformation of the plant should result in this new plant G1 of S being strictly positive real and stable and if that is true then the original plant G of S with this nonlinearity in the K1 infinity sector is going to give you stability. Now that was one special case where you took a nonlinearity in the K1 infinity sector. Now one could also think of taking a nonlinearity in the zero K2 sector. So if you take a nonlinearity in the zero K2 sector so what will its characteristics look like? So if I call this psi the input and this the output phi and I draw this line which has slope K2. Since the nonlinearity lies in the zero K2 sector that means nonlinearity is something like this or in other words phi by psi is less than equal to 0 0 is less than equal to phi by psi which is less than equal to K2. Now if you have a nonlinearity of this kind then this particular inequality I could rewrite this in the following way. You see if I take K2 psi minus phi so suppose I take some point psi so this point here is K2 psi and the corresponding phi is this so this quantity here is positive and for this particular case psi was positive. Now similarly if I take psi to be negative I get K2 psi to be here and I get phi to be here so K2 psi minus phi this is negative and psi is negative. But more importantly K2 psi minus phi this quantity here is positive so I could write the following. You see when psi is positive phi the corresponding phi is also positive and here when psi is negative the corresponding phi is also negative. So I could write the following down K2 psi minus phi transpose multiplying phi the output when this is positive this is positive and when this is negative this quantity is negative this quantity is negative so this product is greater than equal to 0. And so what I have really done is I have looked at this nonlinearity and I have looked at certain quadratic inequality which gets satisfied by any nonlinearity which lies in that sector. And now what I am going to do is I am going to use this nonlinearity this particular quadratic relation to modify my original nonlinearity in such a way that now it becomes a nonlinearity in the 0 infinity sector. So how do I do that? I do something very similar to what I did the last time so here is a nonlinearity here let me assume is the input and here is the output phi. Now what I am going to do the last time what we did was we kept the input the same and we modified the output. This time round what I am going to do is I am going to keep the output the same but I am going to modify the input so the new input psi tilde is going to be equal to K2 psi minus phi. How do I do this? Well what I could do is I could take this nonlinearity the output is still going to be the same phi but the input the new input is going to be this psi tilde but let me do the following. Let me put a gain here which is K2 inverse. So now if I give K2 psi as the input here then because of this K2 inverse I will get psi here and then I have the original nonlinearity. So what I am trying to do is I should get psi here such that I have the original nonlinearity. So the original nonlinearity I am multiplying before you reach the nonlinearity I am multiplying the input by K2 inverse. And so now I would use this output and I feed this back. So now if I am to get psi here what I should get here is K2 times psi and so what I should get here should be K2 psi plus phi. So I want K2 psi minus phi here because that is what I want here and so what I will do is I will feed this back use a positive feedback. So if I do K2 psi minus phi here if I have K2 psi minus phi here and I am feeding back this phi this phi will cancel out that phi and I will have K2 psi here and then K2 inverse will give me psi here. So I have the original nonlinearity modified in this particular way should give me a nonlinearity like this and this nonlinearity. So now if I box this off this new nonlinearity has the same output as the original nonlinearity but as the input it has K2 psi minus phi. So it looks like I am out of time so we will continue whatever we are doing in the next lecture.