 So, welcome everyone to this next lecture. Today we will spend some time on finishing something small about describing function, namely about sufficient conditions for existence of periodic orbits and sufficient conditions for non-existence of periodic orbits. So, after this part we are going to do some problems on circle criteria, popo criteria, the so called small gain theorem. These are problems we will do immediately after we finish this for existence for non-existence. One thing one should notice in non-linear dynamical systems, in non-linear control, it is very hard to have necessary and sufficient conditions. Even the Lipschitz condition that we saw for existence of solutions to differential equation, there is only a sufficient condition for existence. It is not necessary, there can exist solutions to differential equations even though the Lipschitz condition is not satisfied. In that sense the Lipschitz condition is only a sufficient condition there. So, here also indeed existence of periodic orbits and non-existence of periodic orbits seem like opposite, but we have sufficient conditions for existence and sufficient condition for non-existence. So, in that sense, we have only seen one, we will have seen only one part of it. Why? Because necessary and sufficient conditions for existence of periodic orbits is in need of very difficult thing to have. One requires to do to the best of my understanding, still research is required to be done to find necessary and sufficient conditions and often these are so complicated that at some point it becomes too artificial, a little too artificial to pursue and have as a result. So, recall that we had said that when you have G as a transfer function, a linear system, linear time invariant system and a non-linearity phi and you consider the describing function of this non-linearity phi, a function of A and omega, where A and omega are the amplitude and frequency of this sinusoidal input at the entrant, at the entry of the non-linearity. So, we have already assumed, we have already noted that the higher harmonics of phi, when if phi is time invariant non-linearity, then what comes and if phi is bounded input bounded output stable, then what comes out here is also a periodic signal in the same period, but it need not be a pure sinusoid, it might have higher harmonics also, it might have cosinusoid also, but higher harmonics is being ignored when we define the describing function. Why? Because the describing function was defined using only the first harmonic of the Fourier series, the first Fourier series coefficient and this ignoring of higher harmonics was is justified if G is a low pass filter. So, suppose the Nyquist plot of G looks like this and suppose the describing function happened to be such that this is the plot of eta of A, for convenience we will assume that this is a function of only A, otherwise we do not have this convenience of plotting them separately and just looking at the point of intersection. If eta depended on A and omega, then this point of intersection had to be correspond to the same frequency for the Nyquist plot and also for the describing function eta A, omega. So, now notice that the higher harmonics that are being ignored, that might if they were not ignored, then the intersection perhaps is not here, but inside this ball, inside the circle. This is some circle of radius r, where low pass means r is small, but one can quantify exactly how small r is and one has to look at such a ball. Now what is the sufficient condition for non-existence? A sufficient condition for non-existence would be that this describing function, this is eta of A, this happened to be far such that such a ball, even if you take such a circle, even if you take such a ball and still there is no intersection with the Nyquist plot of G. For example, here you see that this ball is just touching. So, how much big radius ball we should draw along the Nyquist, along the describing function curve, that depends on how much low pass G is. Why? Because the describing function indicates only the first harmonic and higher harmonics have to be ignored or not, that depends on the extent to which G is low pass. So, let me just recapitulate this again. Suppose this is the Nyquist plot of G, the exact transfer function is not being discussed here. Suppose the describing function was like this, that is one eta phi 1 A, another one was like this N phi 2 A. Here this Nyquist, this describing function curve has a function of A and the Nyquist plot do not intersect, while in this case it intersects. So, just because it intersects, perhaps we feel that maybe there exists a periodic solution. I said that this was only approximation method, but does there exist one close by, at least yes or no question we want to have for sure. The value we might have some inaccuracy that is acceptable. Here on the other hand we do not have intersection, but notice that you may not have intersection even in this case, this is another let us say eta phi 3 A, but then here the ignoring of higher harmonics perhaps was playing a role and if you had not ignored, maybe it did intersect. So, it is reasonable to ask if a ball around this intersects or not, here such a ball, here such a band. So, instead of a ball, we can move this ball along this particular curve and see if that band intersects. If the band itself does not intersect this, then one might be safe, one might safely say that periodic solutions will not exist. One can go ahead and claim that periodic solutions do not exist, if not just the, not just non-intersection of this disturbing function curve, but also non-intersection of this band in which the thickness of the band depends on how much low pass the transfer function g is. The next harmonic, how much the gain has decreased, how much the gain of g, how much in the body plot the magnitude curve has decreased compared to the frequency corresponding to the first harmonic. So, this gives us sufficient condition for non-existence. Now, we will say that what will happen if the band intersects, even if the curves intersect, it is still possible that when you include that band, that time you have non-intersection, the actual curve, the actual curve might be some curve inside this band. That is why we have to check that the entire band is not intersecting and hence no curve inside this will also not intersect the Nyquist plot. So, now the next question is we are going to ask for sufficient condition for existence. Is the opposite of this a sufficient condition for existence? What is the question? We have obtained a sufficient condition for non-existence, namely the band around the disturbing function curve as a function of A. Look at the band and the band also does not intersect the Nyquist plot that non-intersection of the band has been proved in Vidya Sagar's book as a sufficient condition for non-existence of periodic orbits. What is the opposite of the statement? That the band intersects. If the band intersects for example in this case, it is not necessary that the curve intersects. If the band intersects, it might still happen that the actual curve also does not intersect just like this curve that we have shown here. So, actually there are no periodic orbits. So, intersection of the band with the Nyquist plot is clearly not a sufficient condition for existence of periodic orbits. The opposite of this condition that the band intersects with the Nyquist plot that band intersection with the Nyquist plot is clearly not a sufficient condition for existence of periodic orbits but it turns out that it will be a necessary condition for existence of periodic orbits. But it is still not sufficient clearly because look at this example. Here the actual curve does not intersect but the band intersects. The band is like this and just because the band intersects does not mean that the first harmonic curve intersects, it does not mean that the actual curve intersects either. So, look at this example where the actual curve intersects. This also is not a sufficient condition for existence of periodic orbits because we are not sure that this eta phi 1a is the actual curve. This is just the first harmonic. The actual curve might be something else. We need some condition that the actual curve intersects the Nyquist plot. The intersection of the actual curve which is all the harmonics considered together, loosely speaking, that is the curve that we want to guarantee intersection. Then we will have a sufficient condition for existence of periodic orbits. So, I will construct an example where the first harmonic intersects at two points but perhaps the actual curve is like this. The actual curve does not intersect. And the actual curve is of course inside this band. So, inside this band if it happens that the first harmonic curve intersects the Nyquist plot but the actual one does not. That can happen if this band is sort of just touching or touching and intersecting at two points. If it is touching, then clearly anything inside will not intersect at all. But if it is intersecting at two points, that also is a situation where the actual curve perhaps may not intersect at all. The first harmonic curve might intersect at two points, two points very close to each other because of which the actual curve actually can escape, can escape. It need not. This is only a sufficient condition that we are aiming for. So, let us get a feeling for what is the sufficient condition for existence of a periodic orbit. So, here is an example. This is eta phi 1 A. This is the actual curve that escapes eta phi 1 actual A. That is because there is, it is of course it has to be inside this band. It is inside this band, this band around, around this. This curve should not be so far from this. So, actual curve has escaped without intersecting the Nyquist plot. But we will like to say some, there is some property of the first harmonic that is allowing this to happen. Namely, inside that band, inside any small neighborhood, there are two intersections. But suppose, this having function was such that inside that band, so this band is actually both in the omega. So, uncertainty in both A and omega. So, you look at this point of intersection of the disturbing function. You look at some band around the amplitude parameter, some uncertainty along the omega parameter. This, the Nyquist plot is where omega varies along the Nyquist plot. Along the disturbing function curve is where the amplitude varies. And inside this, if you consider just these two, there is only one point of intersection. In other words, there is a certain function which is 1 to 1. That 1 to 1 is being noticed by the fact that inside this band of frequencies and inside this band of amplitude, there is a unique intersection. That is precisely what is not happening here. Here along the omega, along the amplitude band, there were two intersections and this is the frequency band. Along the frequency band, also there are these two intersections. Because of two intersections happening both within the same amplitude band and within the same omega band, this situation that the actual curve escapes without intersection is possible. So, if in a band about A naught and about omega naught, unique intersection. A naught and omega naught of course are the point of intersection of the disturbing function as a function of A and the Nyquist plot as a function of omega. But then there is an uncertainty. There is a band around A naught that we should be considering because of ignoring the higher harmonics. That same reason also causes us to look for a band about omega naught. And inside this band also if this is a unique point of intersection, then we will say that this is a sufficient condition, sufficient condition for existence, for existence of what? Of periodic orbits. So, this has been proved in Vidya Sagar's book on non-linear systems that it has been of course proved far more rigorously and the purpose of discussing this was just to give a feeling of what type of analysis arguments are used in proving such statements. So, this is also again only a sufficient condition just because this to say that this is not necessary means I would like to point out that even if there is a situation where it is not a unique intersection, still this periodic orbits could exist. In that sense this condition is only a sufficient condition. It is not a necessary condition for existence of periodic orbits. So, this has been described in detail in Vidya Sagar's book and we will not pursue this discussion further in this course. So, this completes the topic of disturbing function. We have also seen some problems where disturbing function predicts the existence of certain periodic orbits, how to find the amplitude and the frequency. We also related that with how Popo criterion circle criteria also give a very similar band for non-existence of periodic orbits. Why? Because they give a band for absolute stability and absolute stability automatically rules out periodic orbits, it rules out instability also. So, we are going to now spend some time to solve some problems involving Popo criteria, circle criteria and also important is called small gain theorem which we only barely touched in our previous lectures. Problem solving for circle criteria, Popo criteria and the small gain theorem. So, while solving these problems, we will also very quickly recapitulate what these theorems are, what these very celebrated results are. So, these development of theory has happened over the last 50 to 60 years and it is fundamental in the theory of non-linear dynamical systems. In fact, the listeners are strongly recommended to have a look at the selected best 25 papers that appeared in the last century on non-linear dynamical systems. This is in a book edited by Agrawal in which there are many papers that speak about these results and also further development after this. The Kalman Yakubovic Popo lima circle criteria, Popo criteria, the passivity theorem are undoubtedly the strongest best results that has happened in control theory in non-linear dynamical systems. So, let us take an example of a transfer function S plus 1 over S minus 2. So, this is Nyquist plot is a circle. As I said for first order systems, the Nyquist plot is always a circle. It is the orientation that we have to spend more effort. Why? Because the real axis intersections are also very easy. One is for S to infinity and the other intersection is for S equal to 0. So, minus 1 by 2. So, as I said one needs to have a good understanding of linear systems before one comes to non-linear systems. So, one should be familiar with Bode plot, Nyquist plot, root locus for us to quickly benefit by using circle criteria. This is where the pole is and this is where the 0 is. So, we know that for large gain the system closed loop will become stable. Why? Because this is how the root locus is. After some value it will become stable and hence when we blow this Nyquist plot large enough then this particular, right now we have one unstable pole. So, P is equal to 1 and it is not encircling the Nyquist point minus 1. So, Z will also be equal to 1. So, indeed there is one unstable pole for the closed loop. For the closed loop, suppose this is G equal to this. So, there is one unstable pole in a closed loop also for unity gain. But when we take a gain that is larger than 1, that is nothing but making this circle larger and larger and by multiplying this by 2, we know that this point come here. So, we know that for gain larger than 2, this will encircle the point minus 1 and then there is no other change that can happen for larger gain. So, for gain larger than 2, we know that this will have the closed loop pole that have come into the left half complex plane. So, we know that Z will be equal to 0 for gain larger than 2. So, n has to be 1 and has to be plus 1 for gain larger than 2. So, if it has to be plus 1, then this orientation has to already be anticlockwise. That is the conclusion that we drew by using the Nyquist criteria where this is the number of anticlockwise encirclements, P s number of unstable open loop poles. Open loop because we have nothing else in the feedback. In the loop, we have only transfer function G and hence it is okay to call this open loop poles, while Z is the number of closed loop poles, number of unstable closed loop poles. The number of unstable closed loop poles Z is what we want to find out by using the Nyquist criteria for stability. So, now we know that these all points are these all points here are the ones that are encircled the correct number of points and so are these points also. This point for example corresponding to plus 1 by 2 that corresponds to a gain of K equal to minus 2. So, what the circle criteria says? From the Nyquist criteria we know that these all points in the interior, the diameter are the ones that are encircled P number of times. If you want Z to be equal to 0, then you should look for those points on the real axis which are encircled exactly P number of times. If P is equal to 1 like in our example, then look for all those points on the real axis which correspond to which happen to get encircled P number of times anticlockwise and that gives us points and from those points we can translate to gains. So, K is equal to minus 2 will correspond to an input output map u, y where this is the K, the input is u, output is y. So, K equal to minus 2 is a line with negative slope like this. So, now suppose somebody says what about this? Suppose this happens to be slope 3, slope equal to 3, then that 3 corresponds to a point minus 1 by 3. Is the point minus 1 by 3 inside the circle? Yes. Is it encircled P number of times anticlockwise? What is P? P is equal to 1. So, it is encircled 1 number of times anticlockwise. So, 3 is also line that stabilizes. If you put K equal to 3, that will also result in close-loop stability. So, the Nyquist criteria very easily gives us a sector corresponding to lines that all cause close-loop stability. Now, this is what we get by the Nyquist criteria already. The circle criteria says that you take corresponding to that same diameter, you construct a circle. If you have to know, construct a circle and that circle also should get encircled P number of times, that circle might, that diameter might have to be smaller. This is easily seen for various other examples. In this case, you make the circle as large as possible and it will indeed come out to be the same circle. So, for first order systems, it will turn out that the sector of linearity is the same as sector of nonlinearities. This is an important conclusion worth noting down. Check that sector of linearities. So, the sector of linearities will turn out to be an open set. When all the slope values that cause close-loop stability, those values, the slopes will be the sector of linearities that will be an open set. The boundaries will not get encircled because the boundaries will cause imaginary access poles of the closed loop. It is equal to sector of, and if we are using the circle criteria, then this is possibly time-waving. When will, for what kind of systems are these going to be equal? For first order. Why would this happen? Because for first order systems, the Nyquist plot itself is a circle and because it is a circle, the circle criteria, the largest circle will also happen to be the same circle. The largest circle will be either the largest sector, the circle corresponding to the largest sector will either be the largest circle inside this or it will be the smallest circle outside this. When will it be outside this? Because that it can also happen that the open loop is already stable. We will see another one such example. Take s plus 1 over s plus 2. Here we have 1 by 2, 1, a circle like this. For s equal to 0, we have that it starts here and then this one is high pass filter. First we encounter a 0 and then a pole. So, the phase would be increasing and hence this is the orientation. So, here when we have n is equal to p minus z, there are some points which correspond to negative gain that cause instability to the system and those points are indeed getting encircled clockwise. Why is it expected that it will get encircled clockwise? If for some point z should be equal to plus 1, we already have this minus sign here and p is equal to 0. So, n would have to be minus 1, n would have to be negative. That is why these points would have to get encircled clockwise to begin with. Why is p 0? Because the open loop is already stable. So, now these are all points for which are encircled p number of times. What is p? p is 0. So, these all the points on the real axis which are outside the circle correspond to those which are encircled the correct number of times. So, circle criteria is, what is the circle criteria? Look for segment encircled p number of times anticlockwise. What is p about it? p is the number of unstable open loop poles. So, if you encircle the point any point on the segment p number of times, that will be a linearity that causes closed loop stability. So, that why because it ensures that z is equal to 0. But then of course, a sector of non-linearities will also have linearities. So, that is why this segment had better be encircled. Now, not just this segment, it is not enough that this segment is encircled, but one should also have take a circle with this segment as a diameter. Even the circle has to get encircled if you have to allow not just linearities, but time varying, possibly time varying non-linearities. So, circle with this segment, circle with such a segment as diameter might have to make, might require shortening of segment. So, what about in this case? See, these are the segment is everything outside, except this diameter. So, one can in principle actually take such a circle. So, here if p is equal to 0, then all these points should not get encircled. So, one such circle is this. This segment is not encircled by the Nyquist plot and hence one can take this circle. One can take a larger circle also. One can take this point left and left and the same thing is allowed here also and in fact, one can take the two things together also. Then it will be that this Nyquist plot is encircled inside such a circle. When would this happen if the open loop is already stable and that sector contains a vertical axis also? In such a situation, this would get allowed. So, here is an example where the largest sector, that largest sector corresponds to everything from, everything except this. This is the line with slope minus 1. This is the line with slope minus 2. Everything excluding this, excluding this, please note. So, one can take anything inside this sector and that will cause closed loop stability even if the nonlinearities inside this sector are time varying. So, now we will see how the sector of linearities could be strictly smaller than the sector of nonlinearities. If you allow time varying, for example. So, consider, so I should note that this has to be a second order system. So, we had considered this example in a similar context yesterday. This is 1 by 2. This is a point minus 1. So, is this, so the open loop is already stable? We always have this. So, with phi equal to 1, with phi equal to the unity feedback with the negative sign, is the closed loop stable? Yeah, it is stable because the open loop is already stable and the point minus 1 is n-circle 0 number of times and hence z is also equal to 0. In this equation n is equal to p minus z, p is equal to 0 because the open loop is stable, n is 0 because we are considering unity feedback and hence we are considering n-circle means of the point minus 1. So, it does not n-circle the point minus 1 hence n is 0 and that gives us z also equal to 0. What about some other point? What about this point? It is a minus 1 by 3. Minus 1 by 3 point is also n-circle 0 number of times and hence the closed loop is stable. What about this sector? What about the sector of linearities? Sector of linearities just requires us to check whether this segment is n-circled 0 number of times. Why 0? Because p is equal to 0. Yeah, the entire segment is not encircled. So, n is equal to 0 for that segment also and that gives us that we have closed loop stability for a sector of linearities for any line inside the sector also. So, one can consider making this left and left until it goes in principle to infinity. So, one can take this entire sector 0 to infinity. This is u, this is y. For any line inside this sector, it is some line with positive slope. That slope suppose it is k, then you look for the point minus 1 by k and that happens to be a point on this negative real axis and it is n-circle 0 number of times. So, does that mean that we can take a circle? A circle cannot be taken for arbitrary. Yeah, you see this if this circle comes more and more to the left then at some point intersect this. If you take a k that happens to be very large, let us say k is equal to 1000, that will correspond to the point minus 1 over 1000 which is far inside this k, let us call this a k and then when you construct a circle, it will intersect the n-circle plot. So, the entire circle is not being encircled p number of times. The n-circle plot should not be touching such a circle. So, clearly there is a, there is the largest sector cannot be too large. If you are concerned with 0 to infinity, then it can go only up to a value k max, k max correspond to the circle criteria and it will be open at some point intersect. So, this how to find this k max that is nothing but you can take such a vertical line. If you take such a vertical line, then you can get from 0 up to this particular k max. That k max will be found as you take this point, find its real part and 1 over that will give us k max circle, circle for circle criteria. If one wants to go for higher sector, if one wants to go for larger, then one has to, one cannot start from 0. So, let us draw another figure to explain this. So, if you use the technique that we discussed yesterday for finding the leftmost point on the n-circle plot, the part that causes real part of j of j omega to reach a minima. If we plot that, then we will get this and suppose this value is equal to minus 1 by 50, then we will get 0 to 50 as a sector of time invariant time varying sector, sector time varying non-linearities. Of course, time varying non-linearities includes time invariant non-linearities also, it includes lines also, but time varying means possibly time varying and possibly non-linear. Suppose you want, does this mean that 50 to 55 sector is not allowed, minus 1 by 55 might be here, such a circle is also clearly n-circle 0 number of times. So, if you do not want to start from 0, what I mean to say is, if you want the sector of non-linearities to be varying only between 50 and 55, then that circle is also n-circle correct number of times. So, the lower value also plays a role in how much high you can go. This 50 cannot be made larger because you are starting from 0, because the non-linearities could be time varying and could be varying from 0 up to 50. But if then if you say that the time varying non-linearities will vary from only let us say 49, then perhaps one can go up to 55 and this depends on the Nyquist plot to what extent high you can go and that depends on from where the non-linearities, from what value, from what sector it starts, from what slope it starts. So, this is an example where the sector of linearities is strictly different, strictly larger than the sector of non-linearities for using the circle criteria and there we saw that the lower value also plays a role as to how high you can make this slope. What about the sector of 0 to infinity we already verified yesterday, sector of time invariant non-linearities. For the same example, what is the sector of non-linearities you can, what is the sector of time invariant non-linearities we can have up to 0, from 0 to infinity. The highest value can go as large as infinity because the popo plot, using the popo plot we verified that a line with positive slope can be taken to intersect very close to 0 at minus 0.0001 also and still we can have that the line is to the left of the popo plot. So, we will now see an example of where the small gain theorem can be used. The small gain theorem for stable gs only, the circle criteria allows for unstable transfer functions also, the sector of non-linearities can be found even when g is unstable because in that case when p is not equal to 0 and p is 1 or 2 or more, that time such a circle has to just get encircled p number of times. This is what the small gain theorem does not allow because the small gain theorem considers sector of the type minus gamma to gamma. It considers such sectors only. In other words, these are sectors which are symmetric about the horizontal axis. This is a line with slope minus gamma, this is a line with slope gamma and any sector within this, any non-linearity possibly time varying inside this only is allowed. So, in particular the line with slope 0 is also allowed, which means that open loop already has to be stable. If you want closed loop stability for this sector, then you are allowing your insisting on closed loop stability for line for the line with 0 slope, which is nothing but open loop. Because of the symmetry of this sector, this plus and minus sign does not play a role as far as the small gain theorem is concerned. So, if you allow a line with slope 0 also inside this sector, whenever a line with slope 0 is allowed inside the sector, that time that is like saying y equal to 0, even when u is non-zero, which means that g has to already be stable. Now, only when you have open loop stability, you can ask for absolute stability of the sector that also has 0. So, that is why small gain theorem is relevant only for stable g because we are considering some such sectors which have 0 also inside it. In fact, here it is symmetric about the line with slope 0. So, what does the small gain theorem say? The small gain theorem says that absolute stability for g between and sector, sector bound time varying non-linearities in the range minus gamma to gamma in this slope, if g, the induced L2 norm is strictly less than 1 by gamma. What does this mean? Of course, this already and g is stable. Writing this L2 here, induced L2 norm already implies that g is stable. There is no pose on the image axis, it has no pose in the right of plane. So, what does this mean in terms of the Nyquist product of g? The Nyquist product of g is at most gamma away. In fact, it is no point on the Nyquist plot is equal to gamma or more away from the origin. So, if this is a circle of radius 1 by gamma, then the Nyquist plot is contained inside this, Nyquist plot not of g. The Nyquist plot of g need not be a circle because g could be of any order. It could pass through the origin, etc. But it should be contained inside a circle of radius 1 by gamma and centered at the origin, centered at the origin. So, a statement that the L2 norm of g, which is nothing but the supremum overall omega in R of g of j omega. This is what we already saw in a previous lecture that was about norms of signals and of operators. There we saw that the L2 induced norm is nothing but the farthest point of the Nyquist plot from the origin. And to say that this is strictly less than 1 by gamma means that the Nyquist plot is contained inside this. So, the small gain theorem involves if you are asked, find the largest sector using the small gain theorem such that your absolute stability of interconnection of b of g and sector bound nonlinearities of this form that involves finding the smallest circle, smallest circle that contains the Nyquist plot of g and of course, g has to already be stable. So, let us write this as a problem. Consider g of s equal to 1 over s square plus s plus 2. Find largest sector of the form minus gamma to gamma such that absolute stability. What is absolute about it? The stability we are able to conclude, we are asked to conclude stability for not just one nonlinearity inside the sector, but for a whole class of nonlinearities. That is what is absolute about it for not just one nonlinearity, but a whole class of nonlinearities. Absolute stability for all nonlinearities, possibly time varying inside the sector. In sector minus gamma to gamma. So, if this is the question that we have been asked, we notice that the sector that has been specified is of the form minus gamma to gamma. Hence, it is small gain theorem that we will use. The Nyquist plot of this transfer function looks like this. This is 1 by 2. Now, we notice that any point here, for example, on the real axis also corresponds to a point line inside such a sector. So, we are now going to look at the point that is farthest from the origin. So, perhaps the point that is farthest from the origin is here. So, then the smallest circle will be just tangential to this. Let me draw another figure where this is better visible. This is the smallest circle with center at origin and encircling. Of course, if we did not have to circle this, one could make it even smaller and containing, containing Nyquist plot of G. So, where this, of course, this is the G, G is Nyquist plot. So, how will we find how will you find the radius of such a circle? All one has to do is look at the farthest point. In this case, it might be here. Look at the distance of the farthest point from the origin. Do one over that. And that will give us the radius of the smallest circle centered at the origin that contains this. So, clearly this forces, one way to find this is consider Bode plot, Bode magnitude plot of G of S. The peak value, the peak value, peak value, yeah, suppose this corresponds to some minus, minus 2 dB, for example. This peak value is equal to G l2 norm. Of course, when G is stable, this we had seen in detail in our previous lecture on norms of signals and of operators. So, when G is stable, so this, of course, we are not interested in dB. We want the exact value. This peak value is also the farthest point on the Nyquist plot, farthest from the origin. The distance of the farthest point from the origin on the Nyquist plot. So, this will decide. Suppose this is equal to, suppose this is an absolute value, this is equal to 0.9. Then I am taking the values, yeah, I could be wrong here. So, one will say that largest sector minus 1 over 0.9 to 1 over 0.9, yeah, and open brackets. One cannot take equal to 1 over 0.9 because that will cause the circle to touch the Nyquist plot. It will intersect, yeah, that is not allowed. So, that is why we are saying only open bracket. Largest sector using small gain theorem. So, coming back to this example, what would the circle criteria say in this case? Circle criteria is not forced to have the circle, the smallest circle centered at the origin. The circle criteria also allows the smallest circle to contain the Nyquist plot. Why is it allowed to contain? Because G is already stable. Since G is stable, the circle does not have to be encircled, but the circle can in fact encircle the Nyquist plot of G. So, but the smallest circle is not forced to have its center at the origin. Hence, the smallest circle could have this point, this as the diameter, yeah. So, one can think of the smallest circle like this. So, if one removes the constraint that the center of the smallest circle should be at the origin, it could be somewhere to the right of the origin for this example, then the circle can become smaller. So, in such a case, of course, the sector that you get, the largest sector will not be symmetric about the horizontal axis, yeah. So, this is an example where the circle criteria is closely linked to the small gain theorem. The criteria of the largest sector that we get from the small gain theorem, only that the small gain theorem forces that the sector is of the type minus gamma to gamma. In other words, it is symmetric about the horizontal axis while the circle criteria allows from minus gamma min to gamma max. It allows of this type also where it is not symmetric about the horizontal axis, this value and this value need not be equal. So, we will see some more examples if possible before we start our last topic which is namely tangent spaces and manifolds. That is what we will see in the following lecture. This completes several problems on small gain theorem on circle criteria and POPO criteria. Thank you.