 Welcome to this next lecture about disturbing functions. We are going to continue disturbing function and in particular we will use disturbing functions for finding amplitude and periodic and frequency of periodic orbits. Of course, we should again note that this is only an approximation method. We will also see some conditions under which the disturbing function provides us a guarantee of existence of periodic orbits and some other conditions which is a guarantee for non-existence, a necessary condition for existence and a sufficient condition for non-existence. Sorry, I might have gone wrong in this. We will see a sufficient condition for existence of periodic orbits and a sufficient condition for non-existence of periodic orbits. The last thing we noted was that if we have a feedback loop like this, we had reviewed a disturbing function. We also derived the disturbing function for the jump hysteresis. That was an example of a non-linearity with memory. The memory aspect about it also made brought in an imaginary part of the disturbing function because the other non-linearity is that we had considered namely the dead zone, the signum non-linearity and the saturation non-linearity where all memory less, they were odd. That is why they happened to be real. They had only a real part but then the jump hysteresis also depends on the rate of change, the sign of the rate of change of the input signal, input sinusoid. That is what brought in the imaginary part also. So, today we are going to see how to find the amplitude and frequency of periodic orbits. What we will not see in this course is stability of the limit cycle. Of course, this periodic orbit is already a limit cycle, stability of limit cycles. What this means is, of course, we are seeing some conditions for existence of this periodic orbit whether closely for initial conditions that are perturbed from the periodic orbit. If it is an initial condition close to this orbit but not on the periodic orbit, do those trajectories also converge to this periodic orbit? In that case, we will call this periodic orbit, we will call this limit cycle a stable limit cycle or do they go away. So, let us see what happens in the case of linear systems. There, we had already noted very early that we have a continuum of periodic orbits and that is the situation that we will just verify again using the disturbing function method also. So, please come back to this figure. So, we started with some point here. It is important to note that the amplitude and frequency of the periodic orbit that we obtain, the frequency of course is the same throughout anywhere on this orbit but the amplitude that we obtain by calculations is amplitude of the sinusoid at the input to the non-linearity. It is not the amplitude at this point, it is the amplitude of the sinusoid at this point but the frequency of course remains the same everywhere inside this loop. So, let us take an example. So, consider again g of s with g equal to 1 over s plus 1, s plus 2, s plus 3 connected with constant k in the feedback. Now, this is like solving for 1 plus k g of s r of t equal to 0. Notice a close relation with Nyquist plot. We start from a point here. We call this r of t. It goes through gain k, then minus k after going through this point and then minus gk. That acts on r is same as r. That equation gives us this for non-trivial periodic solution. If this should be a non-trivial periodic solution, let us assume that this r of t is equal to a sin omega t that gives us 1 plus k g of j omega equal to 0. This is like asking for what value of k, the k has to be real. We allow amplification only by real numbers positive or negative. That is a minor issue. Is there some value of k such that this becomes equal to 0? The fact that k has to be real means that that the Nyquist plot of g has to intersect the real axis. If k has to be positive, then it also means that g of j omega real part of g of j omega has to be negative. If k has to be positive for this equation to be satisfied for any omega, if this should be satisfied for some omega, then these conditions have to already be true. So, Nyquist plot of this transfer function, we already noted was equal to this. This is a point minus 1. This is 1 by 6 and this already noted is minus 1 by 6 t and that happens to be at omega is equal to square root of 11 radians per second. Of course, plus minus both, plus square root of 11 and minus square root of 11 both. That is indeed the Nyquist plot intersects at two points here. Now, for k not equal to 60, 1 plus k g of j omega no solutions. For k equal to 60, 1 plus k g of j omega equal to 0, indeed equal to 0 at omega is equal to square root of 11. Let us take only positive frequency. Of course, minus square root of 11 also but so what? It just means that the periodic solution, we have two solutions so that the signal eventually is real. To get a real signal, we need two independent complex sinusoids which when we add by a linear combination, we get a real signal. That is the reason that we have minus square root of 11 also. That is besides the point when dealing with real signals. So, now at k equal to 60 alone, this equation has a solution for some omega. When k is not equal to 60, this particular thing does not have a solution for any omega. So, now let us see at k equal to 60. At k is equal to 60, we get at k equal to 60 and omega equal to square root of 11, we obtain 1 plus 60 times minus 1 by 60 is acting on a sin square root of 11 t equal to 0. This is a differential equation. If there were dependence on omega here, one should interpret it as differentiating this. What we are writing here is a differential equation when applied to this particular sinusoid. When applied to sinusoid is when the frequency very conveniently comes as evaluated at j omega. So, this we get is equal to 0 for all A for any amplitude. So, if you ensure that k is equal to 60 and the frequency omega is equal to square root of 11, then the periodic orbit inside this particular figure. So, indeed A sin square root of 11 t is a periodic solution here when k is equal to 60 and this transfer function, this system. For any A, yes. The answer to this is yes, which means continuum of periodic orbits. It is not isolated. It is not for just a very specific value of A that you have a periodic orbit, you have a continuum. In a way, this is not good. What is bad about this? If you are building an oscillator in the laboratory, after sometime you want that it stabilizes to a fixed period, we do not want that any period is possible. Of course, another important, another worse situation is for arbitrarily small perturbations, we might have arbitrarily small perturbations in the plant, in the system information. It might turn out that we have instability. We do not have a sustained oscillation, but we have instability or these sustained oscillations just die down to 0. Depending on whether that point, whether this gain 60 turns out to stabilize or causes instability. Depending on that, these oscillations might either die down to 0 or they might blow up. This is inevitable because this is a linear time invariant system. What about for k larger than 60? We know that again we cannot have periodic solutions. In fact, for any initial condition, we have instability. For any non-zero initial condition, we have instability of the closed loop. So, the next thing to ask is we are going to replace this block by the saturation non-linearity. So, we have the standard saturation non-linearity here. We have a block k. Notice that as soon as we have this, we will in fact be able to set what amplitude we want. So, this is eta. This is the non-linearity which we will try to plot. We already saw last time that this is how it looks. So, now we want to evaluate 1 plus k eta a g of j omega equal to 0. Find k, a and omega such that for the time being, why do not we just take k equal to 1? Let us take k equal to 1. This is our complex plane. Nyquist plot of g, we have already plotted like this. What is conventionally done is that plot g of j omega and minus 1 over eta a on complex plane. Luckily, these both are kind of independent. What is independent about this? This depends only on omega and this depends only on a because this non-linearity is memoryless. It does not depend on frequency explicitly. Hence, this is some locus of points for different values of a and it is some curve in the complex plane that is parameterized only by the parameter a, the amplitude. So, we can plot these both independently and check where they intersect. Look for intersection points. To say that k is equal to 1 in this equation, to say that 1 plus eta a times g of j omega equal to 0 means g of j omega equal to minus 1 by eta a, eta of a. This means that these two curves intersect at that particular complex number. When they intersect, you look at at what amplitude this curve has reached that point, at what frequency omega that curve has reached that point and that gives you the a and omega of the periodic orbit. Only that we should note that the amplitude is amplitude of the sinusoid at the entry to the non-linearity. Please note that this is again an approximate analysis. Why? Because a describing function has taken only the first order harmonic. One might say that if g has low pass characteristics, if the higher harmonics all have very small amplification, then that results in pretty good accuracy of this periodic orbit. If g has a good roll-off, if the difference in the degrees of numerator, denominator is 2 or more, then this indeed is a valid approximation, but it should have low pass characteristics for all these arguments to work and also the frequency omega at which it intersects should be well within the cut-off. These are only just some thumb rules, but then Vidya Sagar's book makes this more precise, rigorous. So, just this argument we will see in little more detail and then we will return to this saturation non-linearity example. So, consider 1 plus eta a g of j omega most convenient. What is the convenient aspect about this? Eta depends only on a and not omega. Plot g of j omega and minus 1 by eta of a simultaneously in the complex plane for different omega and a. So, in the example that we have been seeing, this is g of j omega for different values of omega. What about minus 1 by eta of a? Notice that for our example, we had this as eta of a. So, minus 1 of eta of a. So, this is the point minus 1 for these values equal to 1 until amplitude equal to 1 because we are dealing with the standard saturation non-linearity. The saturation non-linearity that causes amplification 1 exactly 1 for all amplitudes up to a equal to 1 for all a sin omega t inputs up to a equal to 1 beyond that it just saturates to equal saturates to 1. So, for all these points minus 1 over 1 turns out to be this point and for a larger than that eta of a is decreasing. So, minus 1 over eta of a this is some curve like this. So, this for a less than or equal to 1 and then this is for a increasing a greater than 1 increasing. What have we plotted? This is the curve of eta of a it is a curve of not eta of a, but minus 1 over eta of a starting from a equal to 0 up to a tending to infinity. This is the locus of points where the curve is stuck to this point until a is equal to 1 for a larger than 1 it starts moving in this direction. How do we conclude this? We have plotted this eta of a versus a. One can also plot minus 1 of eta of a versus a that we do on this part here. This is how minus 1 over eta of a looks like. So, it stays at minus 1 until here for values lower for values larger values of a goes on decreasing like this. It is the reciprocal of this multiplied by minus 1. So, that you see it is all real important point to notice it is real and always negative and the magnitude is going on increasing and distance from the horizontal axis is just going on decreasing distance from the horizontal axis is nothing but distance from the origin here that is why we have plotted the curve like this. So, what is it that we have concluded from here that this Nyquist plot for different values of omega and the curve minus 1 over eta of a for different values of a positive do not intersect at all. These two curves this particular equation this equation for the saturation non-linearity with k equal to 1 of course has no intersection then no periodic orbits. This is not too surprising we will see this expected using Popo criteria and the circle criteria. G of s saturation standard why do not we call this standard because we assume so if we have this then no periodic orbits why because 1 plus eta of a G of j omega equal to 0 has no solution a, omega here one solution is one pair a, omega. So, there is no pair a, omega for which this equation is satisfied. So, that is the reason that we had a k there come back to this we have eta of a. So, why do not we pick? Suppose we want a equal to 10 as the amplitude it turns out that if you have these one equation and two variables then there might be a degree of freedom but the degree of freedom is in fact it does not have any solution. But even if there exists a solution the degree of freedom still may not be there there might be a unique point that might also happen because that is in this case happening because this intersects the real axis at only one point and this is always real. So, you can have intersection only on the real axis and that may not be a continuum in that sense there may not be a degree of freedom even though we have one equation and two unknowns a and omega on the complex plane they are supposed to intersect and this eta of a is not arbitrary complex number it is only a real number g of j omega intersects the real axis only at omega equal to 0 omega is equal to plus minus square root of 11. These situations has first of all caused that there is no solution here but for larger for different if we have parameter k which is also again forced to be real perhaps we can get whatever amplitude we want that is the part that we are going to see next. So, we will look at this standard saturation nonlinearity. Suppose we want a equal to 10 and suppose for a equal to 10 this value turns out to be 0.08 this value here suppose what is supposed about it ideally we are supposed to use that particular closed form expression that we saw in the previous lecture which we if time permits we will perform on psi lab and indeed show that suppose for a equal to 10 the value here is 0.08 notice that this curve has this nice property that for any value larger than 1 for every amplitude there is a unique gain that that has that intersects here. So, in other words this map eta of a is 1 to 1 for all values of a larger than 1 it is a 1 to 1 map. So, suppose 0.08 is the value here now we can plug in this so 1 plus 0.08 times k times g of j omega equal to 0 we also know that k is real all this quantity is real for this to have a solution g of j omega has to also be real we know that this can happen only at omega equal to 0 or at omega equal to square root of 11 omega equal to 0 is not interesting because it is a periodic orbit with frequency equal to 0 it is not really a periodic orbit it is DC gain. So, we are going to put omega is equal to square root of 11 by which we are going to now solve this equation for k 1 plus 0.08 times k times minus 1 by 60 why because g of s evaluated at s is equal to j square root of 11 is nothing but our real access negative real access intersection which we saw was equal to minus 1 by 60 this is nothing but this point is equal to minus 1 over 60 if this should be equal to 0 this gives us an equation in k. So, k is nothing but equal to 60 divided by 0.08 let us just calculate this calculation this turns out to be 750 it is little more than 10 times of this because if this was 0.1 this should have been 600 but it is less than 0.1 and has this little more than 600 so it turns out to be 750 of course noting that this is approximate so so we expect that for gain as large as 750 is when we get amplitude equal to 10 for g of s with saturation nonlinearity standard and with k equal to 750 there is a minus sign here then we have 10 sin square root of 11 t as approximate periodic solution in this way we can set the frequency we can set the amplitude frequency we are not able to set because g of j omega intersects the negative real access it intersects the real access only at omega equal to 0 and omega is equal to plus minus square root of 11 there is no choice of frequency for the standard saturation nonlinearity but the amplitude can be set it is it may not be exactly 10 because the deserving function method is only an approximation procedure but there is an isolated amplitude only only for a very particular value of a will this have a periodic orbit that is what makes this method to be useful as for building oscillator of a fixed amplitude unlike linear systems where we end up having a continuum of periodic orbits one can also check that this is pretty robust to perturbations inside the system parameters also also for nearby initial conditions it comes back to this that of course we argued using that for oscillation smaller than this it ends up blowing up because this range 750 happens to be making it unstable but it is chopping it to value 1 that is why it indeed can't become unstable that is what makes this limit cycle as a stable limit cycle before we see another example in which the frequency also can be set we will proceed a little further with this example and we will relate how for k less than 60 we already know that there can't exist periodic orbits using the circle criteria using the circle or the popo criteria both these criterias are relevant in this particular case so coming back to this particular figure g of s this the saturation nonlinearity is pretty easy to visualize but we have defined a word called standard that is actually not so standard across the literature that is why I will just define this again input u output y in which up to 1 between minus 1 and 1 it has amplification by exactly 1 this is a line with slope 1 beyond that it has slope 0 in other words it has saturated this is what we refer to as a standard saturation nonlinearity now we can ask the question this minus sign is also important that is that we have 1 plus k eta a times g of s g of j omega equal to 0 one might ask for k less than 60 for k less than 60 this describing function you see one can directly multiply the k with the describing function also we already have a here eta of a we can multiply this by let us say 50 times eta of a so that it is a disturbing function of this nonlinearity as a whole these these all these two blocks together so this times starts from 50 and comes down so that so that minus 1 by this so Nyquist plot is like this g of j omega what is this point this point here is minus 1 by 60 on the other hand it is here it starts and this is minus 1 over 50 times eta a the plot of this whole block is this plot of 50 or times eta a has been plotted here so minus 1 over that starts from a little closer point minus 1 is here it has come to minus 1 by 50 it has come much closer than what it was but the point for intersection is minus 1 by 60 so we can conclude that for k less than 60 no intersection no intersection what is the meaning that there is no intersection no solution for no solution to the equation 1 plus k eta a g of j omega by solution we mean here a omega k triple and there is no k a and omega value that satisfies this for k less than 60 of course for a real greater than 0 for omega real we are also working under this constraint what is the meaning that omega is real and positive it means that it is a sinusoid plus or minus omega is less important but omega is real is required for it to be pure sinusoid a is positive that is because it is an amplitude and it is real number and k less than 60 is what this condition says from the Nyquist plot intersecting at minus 1 by 60 and eta of a by itself starts at equal to plus 1 and only decreases after that this equation has a solution only for k larger than 60 k greater than 60 has unique solution what about for k equal to 60 has unique solution k a and omega of course omega is always fixed equal to square root of 11 radians per second but what is the amplitude that you can get that amplitude depends on the value of k there is a 1 2 1 corresponds between the amplitude and k for each amplitude that you want you have a value of k if you want amplitude less than 1 there is a non uniqueness there is for amplitudes larger than 1 that you have some uniqueness coming from the game k why because for k equal to 60 continuum continuum of periodic orbits and all these periodic orbits have a less than or equal to 1 so this is not hard to check I think that this is a very good exercise to pursue the how for k equal to 60 it is indeed expected that you can have many periodic orbits continuum of periodic orbits all of them having amplitude at most 1 what is continuum about it a equal to 0.65 works as an amplitude a is equal to 0.65001 also works as an amplitude in other words is a continuum why because for k equal to 60 there is a range for which the saturation non-linearity in the linear range itself already has a continuum the amplitude can't be larger than 1 because that is when the non-linearity starts having its effect and then it will clip so this is something that one can check as an exercise now we are going to use the popo criteria to say that for k less than 60 of course no solution is expected yeah all these periodic orbits can exist only for k larger than 60 so for that purpose we are going to say for k less than 60 yeah saturation non-linearity with saturation non-linearity not the standard one now we are going to say the standard one together with this block this whole non-linearity we can think of yeah as non-linearity in the sector 0 to k yeah this is some non-linearity inside this sector 0 to k k is of course positive we expect k is positive in this arguments input u output y here is our sector this is a line with slope k and of course the line with slope 0 is this horizontal axis so this is our sector 0 to k does the saturation non-linearity line inside the sector very much for each k it is equal to this so the saturation non-linearity indeed lies inside this yeah in fact it lies inside this because for some range it is equal to this line for some range of the input for input in the range plus minus 1 so now we want to ask for up to what value of k will you have stability for all non-linearity inside this range inside the sector yeah for what value of k will you have stability stability absolute stability means that there for every initial condition the trajectories go to 0 and what is absolute about it for all non-linearity inside the sector yeah all non-linearity is one might further distinguish between time invariant or time varying which makes a distinction between the circle criteria and the popo criteria so popo criteria applies to only sectors of the type 0 to k so this is how the Nyquist plot is the circle criteria I will draw a more correct figure where it is tangent more properly consider g of s equal to 1 over s plus 1 s plus 2 s plus 3 what is the problem that we are trying to solve now for this example non-linearity specifically what non-linearity are you talking about sector non-linearity in 0 to k this fixed find k such that absolute stability for all for for sector non-linearity in 0 to k first time varying time invariant for here we will apply the circle criteria for here we will apply the popo criteria yeah these are the two things that we will do now so for this we will plot the Nyquist plot in little more detail this is where 1 by 6 is this is Nyquist plot this particular axis is the imaginary axis to apply the circle criteria for the sector 0 to k means to of course this requires that the open loop transfer function g is already stable yeah why it requires g is already stable because the non-linearity the particular linearity with slope 0 is also inside the sector and if 0 has to cause stability it means that the open loop also has to already be stable point minus 1 is somewhere there so circle criteria says find right most find right most line yeah right most line such that Nyquist plot is still to the right of this yeah we cannot put we cannot put the right most line here or here the Nyquist plot has to still be to the right of this of this particular line as fine right most line L such that Nyquist plot is to the right to right side is still to the right to right side of L so now this says that this point this real axis suppose this particular point is some point P P we will call this PCC for circle and the circle criteria is the one that tells us that this is the condition yeah suppose PC is this point then this k max that we can get from the circle criteria is nothing but minus 1 over PC the real part PC we are looking at the real axis intersection of this vertical line so we push this line as much to the right until it touches then PC will extract out the real axis intersection it will be a negative point for this example importantly minus 1 by 60 is further here yeah so we will get that by the circle criteria PC suppose it is approximately equal to minus 1 by 50 yeah why do we know that this point minus 1 by 50 it should be some point to the to the left to the left of this point yeah this point is minus 1 by 60 yeah and the Nyquist plot of course this point can't be to the right here it would have intersected already it would be further to the left because this point this curve is intersecting this at an angle here if you draw a vertical one then it will indeed be tangential here but here it is coming and cutting and going in it is not coming in vertically here but going inwards hence it would have come further out and then gone in hence the point that is tangential to this would be tangent at some point here and now it would not have been tangential at the real axis intersection which would have happened here so we know that this point minus this PC point is to the left of minus 1 by 60 yeah that is why I have taken minus 1 by 50 yeah then k max by circle criteria circle criteria equal to 50 then yeah and we know we already knew we already knew no periodic orbits for saturation non-linearity we already knew that there are no periodic orbits for saturation non-linearity for k less than 60 we have got a number which is going to be less than 60 we have got just 50 why is it expected that we will get a lower number the circle criteria and also the popo criteria are are not conservative yeah in some slightly more generalized sense they are in fact necessary and sufficient so this number 50 says because we have to guarantee absolute stability for all sector non-linearities in the range 0 to k max and these non-linearities can be time varying also the circle criteria gives us absolute stability for all non-linearities inside the sector which could be time varying also on the other hand here we have fixed the non-linearity to be a saturation non-linearity yeah it is an extremely conservative very fixed non-linearity and hence we can get a number as large as 60 here on the other hand we are supposed to guarantee absolute stability for all non-linearities inside the sector both time varying and time invariant even time invariant there are many other non-linearities which are time invariant other than the saturation non-linearity for them also the guarantee of absolute stability has to be provided of course time varying non-linearities are another big class of non-linearities again inside that sector hence we are getting a smaller number 50 so up to 50 no periodic orbits we already are able to get from circle criteria up to 60 no periodic orbit for the saturation non-linearity we can next investigate what does the popo criteria tell us yeah what is the number that we get from the popo criteria and before we investigate the popo criteria we can check up to what range of linearities up to what sector largest sector of linearities what is that number it will turn out that that number is also exactly 60 so we are going to next see what is the maximum value of k we get for taking a sector of the type 0 to k yeah this is a 0 this is a line with slope k this is a sector but inside this sector we can include either linearities linearities are nothing but lines inside the sector all passing through 0 of course lines with slope any slope between 0 and k non-linearity is if we say time invariant time invariant non-linearities and time varying linearities of course is a small set why these are lines these are just lines of course they are time invariant also the input output law does not change with time because it is a line time invariant non-linearity at least it may not be a line but at least the dependence of output on input does not change with time this is the output of this non-linearity this is the input the dependence of output on input if it is a line then it is a fixed line there is no option it has to be at the slope only that is a linear system the next option is out dependence output on input does not change with time but it is a non-linearity for example saturation non-linearity the last option is a time varying at some point it could be like this after sometime it can become something like this again inside the sector so when you inside the same sector when you allow linearities you can vouch for a larger value of k we expect that k k k max for linear the max is the maximum sector of the type 0 to k for which you have absolute stability will be greater than or equal to k max for time invariant and that will be larger than k max for time varying why is this true because for a given value of k max the time varying non-linearity set will include time invariant also and that will include linears also after all linearity is also time invariant non-linearity it is also time varying non-linearity in which the time varying aspect is negligible in fact it is equal to 0 yeah so these because of those one set sitting inside another set if you make the set of non-linearity allowed within a sector to be very small then you can vouch for a larger sector yeah that is a meaning of robust way how much how much large type of non-linearity you want to allow within a sector if you want to allow time invariant non-linearity is do you want to allow time varying also depending on that the sector will have to become small so that you have absolute stability for that larger sector so now coming back to this like this plot we will first find k max lin yeah this point is minus 1 by 60 point minus 1 is somewhere here and this is equal to 1 by 6 real part of g of j omega and this is the imaginary part of g of j omega each line inside this sector we are speaking with respect to a very specific type of feedback configuration in which this is where the non-linearity is there is this negative sign because of the negative sign this minus 1 assumes its importance now any point here suppose this corresponds to minus 1 by 2 it means that it is a line corresponds to a line with slope 2 different different lines inside a sector correspond to just points on the negative rear axis because it is a point which is not encircled by the Nyquist plot yeah this is the so g is our famous transfer function it is open loop stable the number of poles in the right half plane is equal to 0 if you want the number of closed loop poles also to be equal to 0 the n had better be equal to 0 so this point minus 1 by 2 instead of instead of checking the number of encirclements of the point minus 1 we have now conveniently started asking number of encirclements of this point minus 1 by 2 because this corresponds to minus 1 by k point k where k is the gain that we introduced to g or to this non-linearity anywhere in the loop it is a magnification by k and k has to be positive for us to be considering the points on the negative real axis so you can take any point to the left of this minus 1 by 60 and it will not be encircled in other words it will be encircled 0 number of times so they will all correspond to linearities which cause stability yeah so any any k less than 60 will not be encircled will imply n equal to 0 what is this n this n is number of encirclements yeah number of encirclements of which point of the point minus 1 by k yeah if correspond to gain k we have minus 1 by k as a point to check instead of minus 1 so this is the number of encirclement and as long as n is equal to 0 p was already equal to 0 that gives us by the necklace criteria z also equal to 0 and hence we have closed loop stability closed loop stability we can use necklace criteria precisely because we are interested in just taking points on the negative real axis points correspond to linear lines and hence linear system and necklace criteria is indeed applicable for linear system now what does the circle criteria say circle criteria said that we will take a yeah so what is this value now k max lin it is nothing but equal to 60 yeah we cannot have k max we cannot achieve for 60 we already have periodic orbits which is instability which is marginal stability for linear systems so we cannot achieve this max we can call it k lin sup k linearities sup why sup instead of max because we cannot achieve k lin sup equal to 60 at 60 we have marginal stability we do not have asymptotic stability in other words we do not have absolute stability what did we get for k time varying which means circle criteria sup again that we got approximately 50 that requires to requires us to take the vertical axis that is tangential to this yeah this tangent point we said is to the left of the point minus 1 by 60 we said it is approximately minus 1 by 50 that corresponds to 50 as a sector now our next we are going to use the popo criteria and we might get some number that is between this might be strictly less than 60 might be strictly more than 50 or it could be equal to one of these that we are not sure at this point for that purpose we will plot the popo plot yeah we will sketch the popo plot for this particular transfer function so recall that the popo criteria involved sketch popo plot which amounts to plotting like the Nyquist plot plot the real part of g of j omega but on the vertical axis plot omega times imaginary part of g of j omega this particular quantity is the even function of omega because the imaginary part itself is an odd function to that we have multiplied by omega so it is enough to plot this for only positive values of omega for this particular transfer function Nyquist plot was coming down so this means that we are taking the Nyquist plot and just distorting it along the vertical axis only along the imaginary axis we are doing some distortion along the horizontal axis there is no because we are still plotting the real part so perhaps it looks like this it still goes to 0 in this case because imaginary part of g of j omega goes to 0 pretty fast and multiplying by omega can't change the rate at which it goes to 0 this requires a careful check we can verify this after some time so the what does the popo criteria say take any line with positive slope yeah that this line should have positive slope and this popo plot should be to the right of this line yeah because it has positive slope to its right is an amygdia it is easy the real part of every point on the popo plot should be greater than this should be greater than the this line with positive slope should be above and to the left of the popo plot name in other words the popo plot should be to the right and below of this particular line now we can ask as long as this line has positive slope it it is fine but then the this intersection of the negative real axis gives us the range for which we have stability absolute stability for that sector 0 to k for time invariant non-linearities only yeah suppose this corresponds to point p p this p for point and this subscript p for popo then suppose p p equal to minus 1 by 20 then we have by the popo criteria from 0 to 20 absolute stability absolute stability for this sector of non-linearities but they have to be time invariant yeah now comes can we make the sector larger can we take this line just to have positive slope but further to the right yeah so this point the intersection of the negative real axis does not change because the real axis does not change at all real part of the plot that we are plotting has not changed only the imaginary axis has been imaginary part of j oj omega has got scaled by omega it has become larger for omega larger than 1 it has become smaller for omega less than 1 so how much to the right can this point for this particular example it can come as much right until it intersects this yeah the right most the right most line would be tangential yeah it is allowed to be tangential at this point that is what makes us so k k sup again it is sup because we cannot achieve this by popo which is same as time invariant will be equal to 60 because the real axis intersection of the line that comes as much to the right as possible and still the popo plot is further to the right and the right most we can bring is when it is tangential at this point and that time the slope the precise value of the slope is not important that time the real axis intersection again be equal to minus 1 by 60 so this is what the popo yeah and of course 60 is not achievable because at that point there will be an intersection where we are going to allow intersections of this line with the popo plot so the popo criteria also tells us that for all time invariant non-linearities in the range from 0 to 60 open bracket because 60 is not allowed yeah for all time invariant non-linearities so let us just write this conclusion from popo criteria absolute stability for all for all time invariant non-linearities in sector 0 to 60 yeah from popo criteria we obtain we obtain we obtain deduce deduce that we have absolute stability for all time invariant non-linearities inside the sector yeah is the saturation non-linearity scaled by 60 inside the sector no it is just outside the saturation non-linearity but scaled by 60 saturation non-linearity if it is scaled by 60 then the slope of this is 60 and hence it is not strictly contained inside the sector 0 to 60 yeah it lies on the boundary yeah so if we take square bracket here of course it will be inside the sector but if you take scaled by 59.99 then it will be strictly inside the sector and we had only verified using the disturbing function that there is no intersection and hence we will not have periodic orbits we will have periodic orbits only for k 60 or more okay equal to 60 of course we end up having a continuum but for k strictly larger than 60 the popo criteria already said that perhaps it is not going to work popo criteria was again only a sufficient condition after some after allowing a slightly larger class of non-linearity that becomes necessary and sufficient condition for absolute stability so for time invariant non-linearities in the range 0 to 60 describing function also told us that we do not have periodic orbits popo criteria guarantees absolute stability which also kind of which automatically rules out periodic orbits for any value of k strictly less than 60 yeah this is the conclusion that we got okay and for linearity of course we already know that for k larger than 60 for k equal to 60 we have continuum of periodic orbits because the amplitude will not matter for a linear system for k equal to 60 and for k larger than 60 we have instability for the linear system but for saturation non-linearity we have a unique amplitude that we obtained from the disturbing function and that is precisely when the popo criteria face to guarantee anything in this sense we are able to see that the disturbing function is able to give us for what amplitude and frequency we have intersection of the minus 1 by eta a plot and the g of g omega plot and the popo criteria is not able to guarantee absolute stability for good reason these are all complementing each other in that sense. So, we will continue in the next lecture with another example where we are able to set both the frequency and the amplitude by choosing the non-linearity parameters carefully in particular we will worry we will look into the jump hysteresis non-linearity this ends today's lecture. Thank you.