 So, in the last lecture, what we had started doing was what are called loop transformations. So, I had started initially with a nonlinearity in the k1 infinity sector and I showed how we can convert that nonlinearity or we can think of that nonlinearity as a nonlinearity in the 0 infinity sector. So, in some way the k1 infinity sector you expand it out so that it becomes the 0 infinity sector. So, today I will continue with that, but probably I will also revisit the k1 infinity and sort of try and wrap it around nicely. So, maybe what we will do is we will continue with what we were doing in the last lecture that is after we had finished the k1 infinity sector we had started looking at the 0 k2 sector. So, let us think of a nonlinearity in the 0 k2 sector. So, we are thinking of a nonlinearity which belongs to the 0 k2 sector. So, what I mean by that is so here xi is the input to the nonlinearity, phi is the output to the nonlinearity and we have this line which has slope equal to k2 and what we are saying is that we have a nonlinearity which lies in the 0 k2 sector. So, it lies something like that, that is a nonlinearity. Now, that nonlinearity of course one could think of that nonlinearity as phi by psi that means at any point you take the phi and divide by psi you get this particular sort of triangle and of course the slope, the tangent of the angle that it subtends is smaller than the angle by the slope k2. So, this is less than equal to k2 but it is greater than equal to 0 and then we could rewrite any nonlinearity in the sector by means of a quadratic form. So, quadratic form that we could rewrite this as is the following. So, you take k2 xi minus phi. So, when you do k2 xi minus phi. So, if I am thinking about this particular psi here k2 psi is here and phi is here. So, this quantity here is positive. If I on the other hand take a psi which is negative k2 psi is here phi is here and so this quantity here is negative. So, if I multiply k2 psi minus phi to psi I would get something positive but that is not the quadratic form that I am going to take. What the quadratic form I am going to take is you see phi here corresponding to this negative quantity the corresponding phi is also negative and here the phi is positive. So, what I am going to do is I am going to take the quadratic form given by k2 psi minus phi transpose phi and this is going to be greater than equal to 0 for any nonlinearity that lies in this particular sector 0 k2. Now, so suppose now you have this original nonlinearity and you have the input psi and the output phi. We want to convert it into a nonlinearity which has as its output it has exactly the same output as the original nonlinearity but its input is modified and this new input psi tilde transpose phi greater than equal to 0 this psi tilde is given by k2 psi minus phi. So, how to modify this so that the psi tilde becomes k2 psi minus phi. So, the way we are going to do it is the following. So, let me put a gain here which is k2 inverse. So, if I put a gain here k2 inverse then what should have been here is k2 times psi. So, k2 psi multiplying k2 inverse will give me psi. Now, I have to get this k2 psi and what I do is I take this phi and I feed it back here with a positive sign. So, I am adding phi to something so that I get k2 psi and so the something to which I have to add phi to get k2 psi is k2 psi minus phi. And now if I think of this nonlinearity in the box which is this nonlinearity with this gain and this feedback then this nonlinearity has as its input k2 psi minus phi and it has as its output phi. And therefore this nonlinearity by this equation or this equation is therefore a passive nonlinearity. So, the transformation that I do for something in the 0 k2 sector is in this way and when I do the transformation in this way I get this new nonlinearity which has the original nonlinearity as psi as input and phi as the output. The new nonlinearity continues to have this phi as the output but it has k2 psi minus phi as the input. Now, if we are now looking at feedback structure where you have a plant GS and you have this nonlinearity and this nonlinearity is in the 0 k2 sector and we are looking at this feedback connection between the linear plant and this nonlinearity. We modify this nonlinearity to that new thing. So, the new nonlinearity is obtained in this way. You have a gain here k2 inverse and whatever is the output you feedback a unity feedback with a positive sign. So, that what you have here is really k2 psi minus phi and the output remains the same phi. So, here you have psi phi and this is u the input of the linear plant and the output y. So, let me call this new plant that I have here which would be a modified version of this plant. Let me call that G2S. So, the G2S has as its input the same as the input of GS but it has as its output this k2 psi minus phi. So, let me instead of calling this u let me just call this minus phi because this u is equal to minus phi and here y is equal to psi and here what I should have is k2 psi minus phi. Therefore, the new plant that you get there G2 is output by input. So, it is k2 psi minus phi divided by minus phi. From here you know that G is psi divided by minus phi. So, if you now evaluate this, this turns out to be k2 times G this transfer function plus 1. So, now if you had a linear system with this nonlinearity in the 0 k2 sector and you look at this closed loop system, this is exactly the same as looking at this modified nonlinearity along with this plant G2S but this G2S is really k2 times G plus 1. Of course, how do we realize k2 times G plus 1 if you had the original G, you can put a gain of k2 in series with it and this plus 1 you can get by having a unity feed forward. So, G2S is really this net transfer function. So, if you see just like in the last case you have some sort of a symmetry because in the nonlinearity you are putting this k2 inverse and then you are having this feedback which is positive feedback and therefore Gs will also get modified but this time the gain that you have in series with Gs is going to be positive k2. I mean it is going to be k2 whereas with the nonlinearity you had k2 inverse and here in this loop in this portion in the nonlinear portion this was a feedback. So, out here it is a feed forward. If you recall in the k1 infinity sector we had used a feed forward in the nonlinearity as a result of which you had a feedback in the linear plant. Here in the nonlinearity using a feedback therefore in the linear part you will have a feed forward. Now this kind of transformations go under the name of loop transformations. Now by doing this loop transformation you have got a new nonlinearity here and this nonlinearity is passive. This whole thing that I am marking this whole nonlinearity is passive. Therefore this G2s this modified linear plant if this modified linear plant is strictly positive real and it is stable then the original plant along with this nonlinearity of course if this is strictly positive real and stable then this resulting feedback system is going to be asymptotically stable and that is the same as saying that this particular system is asymptotically stable. Therefore from this what we can conclude is if you have a nonlinearity and you have a linear plant Gs and you are looking at this feedback structure and if the nonlinearity lies in the 0 k2 sector then this resulting system is asymptotically stable if k2 times G plus 1 this transfer function which is a modified transfer function that Gs becomes because you have modified nonlinearity these two are equivalent and the modified G is k2g plus 1 this is positive real stable. So what we are really doing is when we are looking at nonlinearities which are in sectors which are really in some sense subsets of the earlier situation means initially the passive theorem passive lemma and so on were proved for nonlinearities in the 0 infinity sector. Now we are looking at something in the 0 k2 sector where k2 is strictly smaller than infinity then one would expect more transfer functions to be interconnected to this nonlinearity resulting in something which is asymptotically stable. Of course if you have a nonlinearity in the 0 k2 sector you see this is true that all the nonlinearity is in the 0 k2 sector this is a subset of the nonlinearity is in the 0 infinity sector. So of course if you take a plant here which is positive real and stable then the resulting system is anyway going to be asymptotically stable. But what this result tells you is that you need not necessarily take Gs which is strictly positive real and stable but you could take a Gs such that k2 times Gs plus 1 this resulting transfer function is positive real and stable. So we could have a G which is not positive real or stable and you could have k2 G plus 1 resulting in something which is positive real and stable and if that is true then that G along with the original nonlinearity that will again result in a system which is asymptotically stable. Now I had used some sort of quadratic forms. So let me just revisit these quadratic forms and there are I mean depending on the tastes of people there are new additional definitions given to many of these systems. So when we consider the nonlinearity in the k1 infinity sector when we had looked at a nonlinearity in the k1 infinity sector what this means is the output of the nonlinearity divided by the input lies between infinity and k1 and then this I could rewrite as phi minus k1 psi transpose psi being greater than equal to 0. So this inequality that I have here of a nonlinearity in this class I could write it in this way and this then I could rewrite in this particular way and then when I rewrite it in this way I could think of so the original nonlinearity was there with input psi and output phi. So I retain the input as it is but the output I modify to phi tilde and so I look at it this way where phi tilde is given as the original phi minus k1 psi and then this new modified nonlinearity is passive. Similarly when you take a nonlinearity in the 0 k2 sector then the inequalities similar to this the inequalities that you would get is the output by the input is less than k2 and it is greater than 0 and in this particular case what I did was I kept the output the same so and I wrote this inequality in quadratic form and so the quadratic form that I wrote was phi and multiplying phi multiplied k2 psi minus phi and this is greater than equal to 0 and this k2 psi minus phi I define that as the new input so I kept the output the same but I modified the input and so psi transpose phi greater than equal to 0 where psi tilde was given by k2 psi minus phi. So what I am really doing is I take a nonlinearity in the sector there are these inequalities which are satisfied but that is equivalent to saying that it is this particular quadratic form that is satisfied and if this quadratic form is satisfied and I think of a new nonlinearity which has phi minus k1 psi as the as the new output I keep the input the same I change the output then this new nonlinearity will actually be this new nonlinearity would actually be in the 0 infinity sector. Similarly if I take something in the 0 k2 sector this is one way to define it but I am redefining this in form in the quadratic form and if I redefine this in the quadratic form I think of this here as the new input so I keep the output the same I change the input into this new input and then the resulting system is again in the 0 infinity sector. Now if you look at this case nonlinearity in this case then what we are doing is we have kept the input the same but we have modified the output. On the other hand if you look at a nonlinearity in this sector we have kept the output the same and we modified the input. Now the way we modify the output in this particular case is we give a feed forward the original plant we give a feed forward. So such systems are also sometimes called input feed forward passive and similarly these systems what we did was we kept the output the same but we gave a feedback from the output and therefore these are called output feedback passive. Now these are all definitions but the important thing to realize is that whenever you have some nonlinearity in some sector like k1 infinity you can convert it using this quadratic form into something which is passive and similarly if you have some nonlinearity in the 0 k2 sector then by modifying the input I mean this is the quadratic form that it satisfies and so if you modify the input then the new nonlinearity that you create is in the 0 infinity sector. And once things are in the 0 infinity sector then you know that if you put a positive real stable plant in feedback loop with this such a nonlinearity you get asymptotic stability. So whatever was the linear plant that you connected that will undergo a transformation to be a new linear plant which you associate with this particular nonlinearity and that new linear plant should now be positive real and stable and similarly in this case. Now of course we could also look at a nonlinearity which is in a bound k1 k2 sector. So one could look at m some nonlinearity in the k1 k2 sector. So what do we mean by that? This is the input, this is the output let us assume this has slope k1 and you have another line which has slope k2 and what we are saying is that we have a nonlinearity which lies in the k1 k2 sector. Now for a nonlinearity that lies in the k1 k2 sector we can again write some sort of a quadratic form but and then use that quadratic form to convert this k1 k2 sector into the 0 infinity sector that means passive. Now one way to go about doing this is what one could do is we could convert this k1 k2 into a 0 k sector and how does one convert this into a 0 k sector? Well you have this nonlinearity with input psi and output phi and now if you think of input as this psi the output is this particular phi. What one does is we modify the input to the new input psi tilde which is given by no sorry what we do is for the same psi you have a new output phi tilde which is given by the old phi minus k1 psi. How to do it here? Well what we are doing is the new so this is the phi tilde. Now what can we say about this new nonlinearity that means this nonlinearity which is there in the box? Well what we can say is that this new nonlinearity whose output is phi tilde and the input is psi this new nonlinearity belongs to the 0 k sector where k is k2 minus k1. Now it should be clear why this is k2 minus k1 because you see we are talking about nonlinearity in the sector. So one of the worst cases would be I mean we could really think of the nonlinearity as a linearity which is output is k2 times the input. Now when we put this transformation then the output becomes k2 times psi minus k1 times psi which is k times psi and so what we have effectively done is we have rotated this round and so the new nonlinearity that you get is psi is the input phi tilde is the output and you will have slope k which is k2 minus k1 and the nonlinearity now will lie like that. Now once you have something in the 0 k sector we can use this particular thing to convert it now into something in the 0 infinity sector. So what do we do? Well to the nonlinearity you put k inverse here and then whatever is the output you feed it back and you feed it back with the positive sign I believe be sure here. And now the phi tilde but now we have modified the input also and it has become psi tilde. Now this new nonlinearity with psi tilde and phi tilde is in fact in the 0 infinity sector. So here are some transformations that are going around. You have a Gs and you have this nonlinearity and this nonlinearity is in the k1 k2 sector. So you do the first round of transformations and so you get this new nonlinearity. So the new nonlinearity you get is the old nonlinearity with a gain of k1 put here. So this is the new nonlinearity instead of this we have put this additional thing and as a result this Gs gets modified and how does Gs get modified? Well the Gs gets modified by feedback. So now this is equivalent to this where you have modified the nonlinearity and you have modified the plant. This nonlinearity which is there in this dotted box that I am talking about if I call that nonlinearity NL1 this NL1 belongs to the 0 k sector where k is equal to k2 minus k1. Now this is further modified now in the following way. So you have nonlinearity, you have this k1. So this is nonlinearity and now you put in k inverse here and you take positive feedback and this new nonlinearity that one is having let me call this new nonlinearity NL2. This belongs to the 0 infinity sector and of course when you do this you have to do the corresponding change in the transfer function and the change would be so you have the linear plant, you had the feedback and then you do exactly this but here you multiply by k and then you feed forward, unity feed forward and this is the resulting linear plant that you have. So this is equivalent to this is equivalent to this. So if Gs with this nonlinearity in the k1, k2 sector is to be stable then this linear plant with this nonlinearity this must be asymptotically stable. But what is this? This one could get from this by the transformations that it has to go through. So if I call this linear plant G1 then I know G1 is equal to G upon 1 plus k1 G and then this G1 gets converted to this one. So if I call this linear plant G2, so G2 I know is k times G1 plus 1. So substituting G1, this is the same as k times G upon 1 plus k1 G plus 1. This is the same as 1 plus k2 G upon 1 plus k1 G. So if one is given nonlinearity in the k1, k2 sector and we are asked to find out all the linear plants which when interconnected with this nonlinearity gives rise to a system which is asymptotically stable then we could do a loop transformation on this k1, k2 and convert it into a nonlinearity in the 0 k sector and that is this thing. But what that would mean is this G will get modified to this G1 which is G upon 1 plus k1 G and then something in the 0 k sector you can convert it into something in the 0 infinity sector that means you can make this nonlinearity convert this nonlinearity into a passive nonlinearity by this additional thing that you do here, some output feedback. Now once you do this, the linear plant will also have to be modified accordingly and when the linear plant is modified accordingly then this linear plant is modified to this linear plant but that is saying that this linear plant G2 is k times G1 plus 1 but then G1 itself was dependent on G so when you put all of them together you get 1 plus k2 G upon 1 plus k1 G. Now so theoretically if you given this nonlinearity in k1, k2 sector for any given plant G you could calculate 1 plus k2 G upon 1 plus k1 G and check whether this plant is positive real and if it is positive real then the original system is asymptotically stable. So this transformation that we have, this goes under, I mean this is a theorem on its own and it goes under the name of the circle criterion. So let us find out what this circle criterion is all about. So what we had said that if you have a nonlinearity in the k1, k2 sector and you have a linear plant connected to the nonlinearity, so linear plant G linear connected in feedback with nonlinearity nL is asymptotically stable if, so this is what we had shown the last time 1 plus k2 G upon 1 plus k1 G is positive real and stable. So given a G what one could do is one could calculate this 1 plus k2 G upon 1 plus k1 G and then check whether this transfer function is positive real. But checking for positive realness, one way to check for positive realness is by using the Nyquist criterion. Now is there a way to check whether 1 plus k2 G upon 1 plus k1 G is positive real but we still want to use the Nyquist plot of the original G. Now it turns out that this is possible and this way of predicting whether 1 plus k2 G upon 1 plus k1 G is positive real and stable by using the Nyquist plot of G, this is what circle criterion is. So we will look at this transfer function and we will use the Nyquist plot of G and try to say whether this transfer function is positive real and stable. So this is what we do. So for that let us look at the complex plane. So we are given the Nyquist plot of G of S. So the Nyquist plot of G of S essentially means that so maybe we have something like that. So this is G j omega. So at various omegas we have evaluated G S and we plotted that that is the Nyquist plot. So let us take some particular point here. So this is let us say G at j omega 0. Now we are interested in finding something out about this transfer function which is 1 upon k2 G S upon 1 plus k1 G S. So let us just look at the denominator. So this denominator I could pull k1 out and this is the same as 1 upon k1 plus G S. Now if I am going to evaluate this at omega 0, well this vector here is G j omega 0 and if this point here is minus 1 by k1 then this vector is minus 1 by k1 and therefore this vector here is this vector here is G j omega 0 minus minus 1 upon k1 which means this is really G of j omega 0 plus 1 by k1. So whatever is in the denominator is obtained by looking at this particular vector. Now in the same way one could also look at the numerator and for the numerator if one pulls out k2 you have 1 by k2 plus G S and so G j omega 0 is the same and 1 by k2 will be again a point here minus 1 by k2 will be a point here in the negative axis because we are assuming this k1 and k2 are both positive and of course k2 was a larger number than k1 and so 1 by k2 would be a smaller thing and so let us say no sorry k2 is larger so 1 by k2 is going to be smaller. So this is minus 1 by k2 and so you will have a similar vector here okay. So now if we wanted to evaluate this at j omega 0 what we are really evaluating is this vector the magnitude of this vector in the denominator and the magnitude of this other vector 1 upon k2 with plus G j omega 0 in the numerator and of course because I have pulled out this k1 and k2 this will be k2 by k1. So the magnitude of this magnitude of this and this but what we wanted to know was this transfer function is positive real but what would that would mean is that the resulting Nyquist plot should have an angle which lies between plus 90 degrees and minus 90 degrees but this angle of this transfer function is essentially the angle of this which let me call it alpha and the angle of this let me call it beta so this angle is alpha so this is alpha here and this angle is beta and so for this transfer function 1 upon k2 G upon 1 upon k1 G the angle of this particular transfer function is really equal to alpha minus beta. Now if this transfer function were to stand for something which is positive real then this angle alpha minus beta must be less than equal to okay. So the magnitude will give us a magnitude and this angle must lie between plus 90 and minus 90 or plus pi by 2 and minus pi by 2 if you are thinking of this in radiance so this must be less than equal to pi by 2 and greater than equal to minus pi by 2. So now what that means is if you take any point then if that point is to be a point on the Nyquist plot of GS then to know the angle corresponding to that point we draw these lines from minus 1 by k2 and minus 1 by k1 and look at the angles so you have alpha and beta and if the difference between these two angles alpha minus beta lies in this range then when you do the transformation then the resulting point is going to lie in the right half plane. So everything essentially depends on these two points minus 1 by k2 and minus 1 by k1 and so let us now look at how those two points get related so let this point be minus 1 by k2 and let this point be minus 1 by k1. So if we are interested in let us say some point here then what one does is we look at this vector and we look at this vector look at this angle alpha look at this angle beta and we are saying that alpha minus beta should be less than equal to pi by 2 and should be greater than equal to minus pi by 2 and this guarantees if alpha minus beta is in this range then it guarantees that this point z so suppose I call this point z then it guarantees that 1 plus k2 z upon 1 plus k1 z the point to which z will map under this bilinear transformation lies in the right half plane. So all the points z such that alpha minus beta is in this range are permissible points where the Nyquist plot of the original plant G s could exist but now how do we find all those points where alpha minus beta satisfies this inequality. Now if you recall high school geometry then you might remember that if you have a circle this might not really look like a circle but let us assume this is a circle whose diameter is this distance between minus 1 by k1 and minus 1 by k2 and if you take any point on the circle and you look at these two lines then in high school you would have learned that the angle subtended by these two this angle here is pi by 2. Now if this angle is pi by 2 then what can we say about this particular angle alpha minus this particular angle beta well we know beta plus this angle if I call this angle delta we know beta plus delta is equal to pi by 2 but we also know alpha plus delta is equal to pi. So if you subtract the second one from the first one you get alpha minus beta is precisely equal to pi by 2. So this is something that we would have learned in a high school geometry that if you draw this circle then any point on the circle if you subtend it subtends an angle 90 degrees. As a result this quantity alpha minus beta for any point on this circle is going to be precisely pi by 2. So then it turns out that if you take any point outside the circle then the angle will be less than or the modulus of the angle would be less than pi by 2 and if you take any point inside the circle then the angle that is going to get subtended its modulus is going to be greater than pi by 2. So this is really the circle criterion. So what it says is that for so given g if the Nyquist plot of g so given g so suppose you want to find something out about this transfer function 1 plus k2g upon 1 plus k1g given g then from the information about k1 and k2 you can plot these two points minus k 1 by k1 minus 1 by k2 and you can look at the circle and if the Nyquist plot of g does not enter the circle then the Nyquist plot of this transformed transfer function is going to lie completely in the right half plane and that is the circle criterion. Now this sort of throws up a lot of very interesting things which one would like to talk about. So I would talk about what are the various kinds of interpretation that you can get with the circle criterion in my next lecture.