 So, what we saw in the last lecture is the Propov criterion and let me sort of restate what the Propov criterion is, so again we are looking at a situation where you have a nonlinearity and a linear plant and they are interconnected in the following way and we want to talk about the asymptotic stability of this system and there are these conditions. First of all the nonlinearity is locally Lipschitz memory less, so locally Lipschitz is something to do with the continuity of this function and memory less we have already seen that it does not depend upon the on the past, so memory less, so the nonlinearity is all that and the nonlinearity of course also lies in the 0k sector. Then in order to talk about the asymptotic stability of all this what we did was we constructed an equivalent feedback structure where what we did was to the nonlinearity we put 1 plus AS in series and we fed back 1 by k and we had the plant GS, so in series to that we had 1 plus AS and this was fed back negative feedback and here again the same gain 1 by k is put forward here and fed forward positive sign came through. So we did this modification for the nonlinearity in the following way and so we do the modification for the linear part in that following way and then from that we have we showed that this new nonlinearity is still going to be passive that means in terms of this input here I call it sigma 1, so this whole thing is the new nonlinearity and this psi is the output of the nonlinearity sigma 1 is the input and this nonlinearity is passive with respect to the input sigma 1 and the output psi and then that linear plant if we can show that that is also passive then the resulting system is passive and therefore this asymptotically stable and therefore that is asymptotically stable. So the resulting linear plant that you have here is 1 plus AS times GS plus 1 by k and this must be positive real if the interconnection is going to be passive if that is going to be passive then this must be positive real and therefore this must be positive real and this is the Popov criterion. Of course this being positive real is like saying the real part of the trans function 1 plus AS times GS plus 1 by k must be greater than equal to 0 that is the Popov criterion. Now how does one check the Popov criterion? Well one can check the Popov criterion so of course some other things need to be talked about you see if this 1 plus AS times GS this guy is plus 1 by k this is going to be positive real then 1 plus AS times GS this transfer function okay so let us just look at the transfer function GS and make some comments about the transfer function GS. You see given a transfer function PS by QS if you say that this transfer function is positive real then what we are saying is that the Nyquist plot of this transfer function is going to P of j omega by Q of j omega so the Nyquist plot is going to lie in the first and the second quadrant sorry yeah the first and the fourth quadrant okay that means it's real part is going to be positive but if it had to be positive then we can say something about the degree of P and Q we can say that the degree of P P of s minus the degree of Q of s that means the difference in degree this the modulus value can at most be 1 so relative degree can be at most 1. Now when you are looking at transfer functions typically you are looking at proper transfer functions that means the denominator polynomial has a degree which is greater than equal to that of the numerator polynomial so the two situations that you could have then for positive real functions is the degree of PS is equal to degree of Q s or degree of PS plus 1 is equal to degree of Q s so in addition if you put the properness condition okay. Now the Pope of criterion says that given a transfer function G of s 1 plus a s times G of s plus 1 by K is positive real now if this whole transfer function had to be positive real that means the degree of the numerator divided by the degree of the denominator must be having having a difference they must have equal degree or the numerator must be 1 degree smaller than the denominator. Now this 1 by K portion is something that we can forget because this being positive 1 by K being positive you can take it off and so you can claim that 1 plus a s times G s should be positive real or shifted positive real that kind of thing and therefore we can conclude that G of s for this to be positive real this whole thing to G of s must be strictly proper. So G of s must be strictly proper because you are now multiplying the numerator by one more degree in s and the resulting thing continues to be proper therefore G of s must be strictly proper. So going back to the Pope of criterion you always have this G of s interconnected to the nonlinearity the nonlinearity is in the 0 K x sector and this G of s has to be strictly proper. Now how does one check whether this condition is satisfied well the best way to check whether that condition is satisfied is by evaluation. So let us assume that you have this transfer function G of s which is let us say P s by Q s we need not bother about that. We want to check that 1 plus a s times G s plus 1 by K the real part of this whole thing is greater than equal to 0. So the real part when you substitute s equal to j omega so that is like saying 1 plus j a omega times the real part of G j omega plus j times the imaginary part of G j omega plus 1 by K must be greater than equal to 0. It is the real part of this thing. So one could evaluate the real part of this. So the real part of this is going to be the real part of G j omega. So the real part of G j omega then this product is going to be imaginary then you have this product here that gives you minus a omega times the imaginary part of G j omega plus 1 by K is greater than equal to 0. Now how does one determine whether for a given transfer function G this condition is satisfied. So we do the following we do the following. So we make the following plot on the x axis you plot the real part of G j omega and on the y axis you plot omega times the imaginary part of G j omega. So at any omega you evaluate the real part of G j omega and it has some value let us say something and omega times the imaginary part of G j omega it has some value. So these two will give you some point there. Now for each omega you evaluate these points and you plot this thing. Such a plot is called the Popov plot. Now what we need to check on the Popov plot is that the real part of G j omega minus a omega times the imaginary part of G j omega plus 1 by K is greater than equal to 0. But this now with the axis this showing the real part of G j omega and this axis showing the imaginary part of G j omega this is essentially an equation of I mean if you if I think of this axis as y and this axis as x then what we are saying is x minus a times y plus 1 by K is greater than equal to 0. So of course if you look at the equality you get this equation which is x minus ay plus 1 by K equal to 0. So this equation of what line? This line is going to pass through the point minus 1 by K and it is going to have a positive slope and the slope of so you have a line like that and the slope of this line is 1 by A. Now x minus ay plus 1 by K greater than equal to 0 is essentially this portion of the plot. So in order to check whether a given transfer function is going to satisfy the Popov criterion what you have to really draw is the Popov plot where on the x axis you have the real part of G j omega on the y axis you have the omega times imaginary part of G j omega for example the Nyquist plot you would be plotting real part of G j omega and imaginary part of without the omega. So you get the Nyquist plot but this is not the Nyquist plot but the Popov plot. So you on the x you have real part of G j omega on the y you have omega times the imaginary part of G j omega you get the Popov plot and because you wanted to check 1 plus As times G s plus 1 by K is positive real this is what you wanted to check. So this 1 by K so minus 1 by K you take as an intercept of a line with slope 1 by A and then the Popov plot that you have plotted must be such that it should lie to the right of it that means if it lies to the right of it of course it satisfies this particular inequality and therefore this transfer function is going to be positive real. So the way you check whether something is positive real or not is by drawing the Popov plot and you have the line and then you check whether it is positive I mean it is lying on the appropriate half space. Okay so we have now seen how one uses this Popov criterion through the Popov plot you can get the idea whether the resulting system is actually asymptotically stable or not. Now initially when a lot of these results were arrived at in those days itself for the proofs linear matrix inequalities had made an appearance but at that point in time maybe because the computational power was not that high these linear matrix inequalities stayed restricted to the theory in the sense that they used these linear matrix inequalities in the proofs of theorems and so on but was not practically used because it was considered not very efficient. But nowadays of course there is a lot more computing power and linear matrix inequalities have come in a big way so now people do use the linear matrix inequality so what I will do right now is I would show how a lot of these results like the circle criterion and Popov criterion and so on do come up in the context of linear matrix inequalities and what is the relationship between them. Of course this whole thing centers around a very central theorem which was proved by Yakubovic sometime in the early 60s so let me now motivate how this comes about. So what we will do is we will assume that we given the state space equations of a linear system so you have x dot equal to A x plus B U let us say and y equal to C x plus D U. All right now this is the linear part so you have the linear part so the linear part and you have the input you have the output. Now you are going to connect a nonlinearity to this thing yeah how we connected is right now not very important but we are going to connect a nonlinearity to this linear part. But ultimately what one wants to do is find a storage function for the net system so let us make the assumption that the storage function or the Lyapunov function is dependent upon the states of the linear part and let us say it is something like x transpose Px of course if this had to be a Lyapunov function this P had to be positive definite but so the question is whether one can find a positive definite P such that x transpose Px this is positive and if we think of this as V of x and you evaluate V dot of x then this V dot of x must be negative yeah and if that is true and this is true that means P is a positive definite matrix and its derivative is negative over the trajectories then what you really have is an asymptotic I mean you can conclude asymptotic stability but if you evaluate this V dot x I mean in this particular case one could write this V dot x as something like 2 x transpose P times Ax plus Bu yeah I mean so of course there is something here something there but I have just put them all together so this is the expression that you get for V dot and it is in terms of the states of the linear system and the input u okay and so what you want to really do or what you want to really find is in this x u space so the space of states and inputs you want this expression to be negative for all values of x and u this expression must be negative and for all values of x and the P matrix that you have must be positive definite so this this expression must be negative and P must be positive definite if you can do that then you can say that the that the net system is asymptotically stable okay of course when I write such an expression down I have written this purely in terms of the linear part and so you might wonder how is the nonlinearity going to come in at all but you see when we write out the equations to for this x and u this x and u are not independent of each other but they are related to each other and how are they related to each other well this u is a function I mean u is really coming from the nonlinearity and this nonlinearity uses the y as the input and so this u is actually dependent on y which in turn of course is dependent on x and so so so so really this x and u are not completely independent but they are dependent and somehow that dependence has to come in so now the suggestion that had come about is that instead of asking for this v dot x which is this expression to be less than 0 Lure suggested that you should okay Lure suggested that instead of okay so showing that v dot is less than 0 is the same as showing minus v dot x is positive definite but instead of showing minus v dot x is positive definite Lure said that minus v dot x positive definite minus some some quadratic expression so let me call it q which depends on x and u so this this quadratic expression it depends on x and u and this quadratic expression is in some sense a relationship given by the nonlinearity and this resulting thing must be positive definite okay so what I am trying to say is that instead of asking the question for all x's and u's you must find a P which is positive definite such that for all x's and u's this expression is negative definite but of course this expression is dependent on x and u so it might not turn out to be negative definite yeah and you might not get a positive definite P but this x and u when you just ask that question you are thinking of x and u as being independent but they are really not independent but are dependent on each other through this nonlinearity and so instead of looking for this expression to be negative definite or the negative of this to be positive definite what one does is one looks at that expression and some more expression and this expression depends on the nonlinearity and this net thing should be greater than 0 and this is in fact what comes out in the in the frequency theorem okay so maybe I would I would state the frequency theorem right away okay so what the frequency theorem states is the following so again consider state space equation A x plus B u let us say and and let us say that we have a nonlinearity and the nonlinearity satisfies a quadratic a quadratic expression like q x u so the nonlinearity satisfies this quadratic expression like q x u is greater than equal to 0 for all x and u that appear on the for the nonlinearity okay and then the necessary and sufficient conditions for existence of such a P that we are looking for for existence of P for which that expression that we were talking about that is the real part of x transpose P A x plus B u plus g times x u being less than equal to 0 for all x and u this for the existence of such a P the necessary and sufficient condition is that sorry I am calling it g I am calling it q so q x u is that q of for x I would use this equation and for x I would write down j omega i minus a inverse B u u this quadratic expression is less than equal to 0 for all u and if this is true okay so if this condition is satisfied so this is this is also called the frequency theorem theorem so what the frequency theorem is saying that the necessary and sufficient condition for existence of positive definite P so P which is positive definite for which the real part of all this is less than equal to 0 for all x and u okay and of course here now we think of this x and u not as time signals but as frequency signals so in the frequency space you could think of these things and then in that case this this thing is satisfied if you just take the quadratic part and for the x u substitute j omega i minus a inverse B u which is what you would get from the linear equation of the linear part and this expression must be less than equal to 0 for all u yeah this is called the frequency theorem and this theorem was in fact proved by Jacobowich okay so I will not give a proof of this theorem but we can straight away see that the necessity will always be there I mean this condition that this thing is less than equal to 0 for all u this is a necessary condition and one way to see that is see if you if you are thinking about the linear part I mean the linear plant then you know you can always write down j omega x equal to A x plus B u of course here let me just put hats because we are talking in terms of frequency in fact even in the even in the previous expression I could put just hats just to differentiate the fact that this u and this x hat we are thinking of is really in the frequency domain okay so we have this okay now if you are going to look at the expression so if you are going to look at the expression the real part of x transpose P and then you have A x plus B u okay plus Q of x u and we want this to be less than equal to 0 but out here because A x hat plus B okay sorry all the hats are there okay but because A x hat plus B u hat is equal to j omega x hat so this is the same as the real part of j omega x hat transpose P x hat plus Q x hat u hat okay but you see this expression here is a purely imaginary part so when you are looking at the real part this doesn't appear so it's just the real part of this and the real part of this and for x hat now using that equation we can substitute and you get j omega minus A inverse B u hat that's equal to x hat just substitute that in here and whatever you have would be Q of j omega minus i inverse minus A inverse B u hat u hat this must be less than equal to 0 for all u hat so you see the necessity of this of the condition that we were talking about in the frequency theorem is clear immediately that means you just assume the linear part and just take take it in the frequency domain and just substitute in here this expression will become a purely imaginary expression and so you just left with this expression that's less than equal to 0 yeah of course while stating this I mean V wanted we wanted to find out this V dot is less than 0 and this expression is really a time domain expression and we move from a time domain condition to a frequency domain condition and the frequency domain condition essentially was that this expression that you get so as I told you earlier it was suggested that to try to find this to be strictly less than 0 is not possible because there is a relationship between x and u so you add an extra term which x and u and so you want this whole term to be greater than 0 and this you can just convert this into the frequency domain and the expression that you would end up with is this expression okay and the condition necessary and sufficient condition for this to be true is that you just take the quadratic satisfied by the nonlinearity and for the x hat you get the expression coming from the linear hat and substitute in that and this resulting thing must be less than equal to 0 okay now we can we can see immediately that this particular frequency theorem in some sense gives us all that we saw using the circle criterion and loop transformations and so on okay so here is here is an example of this particular result so suppose we look at the linear plant x dot equal to ax plus bu y equal to cx plus du let us say okay or this is of course the state space the state space equations one could also write the frequency domain equations and you could just say g j omega u g j omega u hat is equal to y hat okay and so think of this as the linear part and you have u hat here you have y hat here okay and now suppose you are going to attach a nonlinearity and something like this but let us not be bothered about whether it is a negative feedback or anything like that if you are going to have an interconnection like this then what you will have here is y hat and what you will have here is u hat if it was a negative feedback of course you would have had minus u hat here okay now this nonlinearity let us assume this nonlinearity is a passive nonlinearity that means it belongs to the zero infinity sector okay what that means is it is either in this quadrant or this quadrant so it is like some something like that and here what you have this is the input output diagram of the nonlinearity so the input of the nonlinearity is really y hat and the output of the nonlinearity is going to be u hat or minus u hat depending upon how you want to see it yeah whether it is a negative feedback or not okay so just to avoid any complications let me just assume that this is in fact minus u hat so that you have a negative feedback and so on okay so if I call that u hat I will have to call this minus u hat so you have minus u hat there okay now a quadratic expression which is satisfied by such a nonlinearity one quadratic expression okay let me call this minus u hat psi then the quadratic expression satisfied by this thing this nonlinearity is input multiplied by output must be greater than 0 because that is precisely what happens in this quadrant and this quadrant okay which means I could write that down as y hat psi it is greater than y hat psi is greater than equal to 0 okay so this y hat psi is really it plays the role of q x psi that we are talking about in the frequency theorem so in the frequency theorem we said there is a nonlinearity satisfies a quadratic expression like that so that quadratic expression is here then further the frequency condition said that the necessary and sufficient condition for existence of this p is that the real part of this for which the real part of this is less than equal to 0 yeah thus simply you just take q that means whatever is the transfer function that comes from the linear part and that must be less than equal to 0 okay so if you if we look at this expression yeah this is a quadratic expression and in this quadratic expression this y hat is something that comes from here so for y hat I could just write down g j omega u hat so what I will have here is g j omega u hat and the psi is really minus u hat so it is u hat squared so g j omega u hat squared and this expression that we are talking about is q this this and this expression is essentially g j omega u hat with a minus sign because the psi is really minus u hat so this is what you get and the frequency theorem says that this must be less than equal to 0 now this being less than equal to 0 well this u hat squared is anyway positive is the same as saying g j omega must be greater than equal to 0 for all omega yeah but of course g j omega of course is a complex expression so what we are really looking for is that the real part of this must be greater than equal to 0 but the real part of this being greater than equal to 0 is in fact our definition for positive real yeah and we do know that if you put a positive real function along with a nonlinearity which has which is in the 0 infinity sector then that that was our basic theorem for that was that was the basic passivity theorem you see yeah so you arrive at the same conditions that means you the linear plant will have to be positive real you arrive at the same condition by just using using this frequency theorem so just to reiterate the frequency theorem it says that if you have this in order to want this have this expression to be less than equal to 0 it is enough to substitute for the x I mean whatever is the quadratic expression satisfied by the nonlinearity for that for the x part you substitute from the linear equation and u hat and the resulting quadratic must be less than equal to 0 yeah actually perhaps you have to say the real part of that because after all this expression would be a complex expression and less than equal to 0 for a complex expression really does not make sense so the real part of it must be less than equal to 0. Now we can we could also look at other cases and you would immediately see that all the results that we had obtained by loop transformation and so on they they just fall out from the frequency theorem so let us let us now of course use the same linear part but the nonlinearity let us assume is in the 0 k sector yeah now if it is in 0 k sector so what that means is here is a line with slope k and so the nonlinearity lies in there okay now if the nonlinearity lies in there then if you think of okay so let me call the input of the nonlinearity sigma and the output let me call it phi okay then clearly this expression this expression k sigma minus phi this expression k sigma minus phi so for this particular sigma what you will have is this length here which is positive okay so if this multiplies phi so phi is positive and k sigma minus phi is positive if sigma is negative if sigma was negative then what you would have is phi would also be negative the output will also be negative and this here k sigma minus phi this quantity here would also be negative so this expression is going to be greater than equal to 0 for all nonlinearity is that lie in the sector and so this will okay so I could rewrite this as sigma minus phi by k times phi and this is going to be the quadratic that that is satisfied by the nonlinearity so so I could say that this is q of x hat u hat this is the quadratic q x hat u hat so now if we are thinking of the linear part and here is the nonlinearity with the input to the nonlinearity being sigma and the output being phi and you connect this up with a negative feedback so you have minus phi here and you have sigma as the output then using the fact that sigma hat is going to be equal to G j omega times phi hat with a negative sign now if this is now substituted in there then we would have effectively done the done the substitution asked by the frequency theorem so if you do that what you end up with is minus G j omega minus 1 by k times phi hat squared this should be less than equal to 0 but this is the same as saying G j omega must be greater than equal to 1 by k and this is in fact the result of course the real part because this so this translates to the real part of this being greater than equal to minus 1 by k and this in fact is what we earlier saw that the so here is minus 1 by k and you have this line and the Nyquist plot of G j omega should lie to the right of it so we get that straight away by using this quadratic and using the frequency theorem yeah so similarly of course when you have you know the circle the circle criterion will also come straight away from the frequency theorem in the circle criterion for example we are looking at nonlinearity which lies in between these two slopes so let us say this is k 1 and this is k 2 and we are interested in the nonlinearity lying between the sectors k 1 and k 2 okay so the one way you can give a quadratic expression for such a nonlinearity is the following so it is k 2 so if the input of the nonlinearity is called sigma and the output is called phi then then we have k 2 sigma minus phi so k 2 sigma minus phi this quantity here when sigma is positive or this quantity here when sigma is negative okay and then this multiplying phi minus k 1 sigma so phi minus k 1 sigma is this quantity here so when sigma is positive both these quantities are positive and so their product is positive and sigma is negative both these quantities are negative and so their product is positive so this is the quadratic that is going to be satisfied by all the nonlinearity and if this is satisfied by all the nonlinearity is then taking the frequency domain I mean after all the interconnection was this is linear part and you have the nonlinear part and this is the input of the nonlinearity this is the output of the nonlinearity and you are connecting this up with negative feedback so that is minus phi hat and sigma hat so just substituting this is minus g j omega times phi hat if we substitute that into this expression what we end up with is k 2 minus k 2 times g j omega minus 1 this and you can pull the phi out and then the other expression that you have is 1 plus k 1 g j omega this whole thing the real part of this must be greater than equal must be this must be sorry less than equal to 0 but then these negative signs you pull out and you will get k 2 g j omega plus 1 times 1 plus k 1 g j omega the real part of this must be greater than equal to 0. But earlier from the circle criterion what we got is 1 plus k 2 g upon 1 plus k 1 g this must be positive real but that of course translates to this. So, as you see this frequency theorem is very powerful theorem and all these results that we talked about using circle criterion, popoff criterion and so on all of them will fall straight into this category by using this frequency theorem and of course the way the way Yakovic arrived at the frequency theorem is because of several developments by other people like Lure who suggested this other modification and what exactly one would seek to be negative definite and so on. Of course, there is a lot of other associated literature which I have not gone into like for example, there is a case where you could be attaching several nonlinearities to a linear plant and it turns out that for each of these nonlinearities you have a quadratic expression of the nonlinearity and finally, when you are going to check for v dot being less than equal to 0 where v is the Lyapunov function kind of thing. So, what you do is each of these quadratic which is satisfied by the various nonlinearity. So, each of those quadratics you add and this net expression you want to be less than equal to 0. Now, this is what is called the S procedure and yeah. So, there is a lot of such associated literature which I am not going into I mean the set of lectures that I was planning to give as far as this course is concerned is sort of over and the rest of the course is handled by Professor Madhu Bailur. So, that is all I have to say. Thank you very much.