 Welcome to this next lecture on describing functions. So, now we are going to use the jump hysteresis. In particular, we are going to study the jump hysteresis and find out the amplitude and frequency for oscillations when that is connected to a first order system. It is a first order system that we will connect it to and find out the amplitude and frequency. So, while doing this, we should note that the transfer function to which we connect should have some low pass characteristics. This we described only briefly in our previous lecture. So, note that this describing function method is a approximation method. This is a jump hysteresis. What was the jump hysteresis? Let us recall this. Let us call this R, R of t which is equal to a sin omega t. This value is minus b. This value is plus b. This is the output. Suppose this is y of t. This is y of t. What are the two graphs? When R dot is positive, that I mean traverses this curve, then it jumps down. When the rate of change of R changes its sign, this jumps down here and it traverses this path when R dot is negative and when R dot again changes sign and becomes positive, it jumps up. In that sense, this is a hysteresis. This jump amount is 2b. Why? Because there is bias up by amount b when R dot is positive. There is a bias downwards by amount b when the rate of change of the input is negative. Now, we should note that if we want to equate only the first harmonic, after all note that the describing function which is a function in general of both amplitude and omega is only first harmonic. Based on the first harmonic, we decided to call this as a gain. Of course, there is a more systematic method where we can use a higher harmonics also and use that to define the describing function. That is indeed a more advanced topic and that is explained in detail in Hassan Khalil's book on non-linear systems but that is outside the scope of this course. Nor do we teach that. We do not teach that either in the regular IT Bombay non-linear dynamical systems course. So, coming back to the fact that the describing function was only the first harmonic, we will like to have an argument, some justification why the higher harmonics do not play much role. The ignoring aspect of the higher harmonics is what makes the describing function only an approximation method for finding the frequency and amplitude of the periodic orbits. But suppose G was a low pass filter, suppose G was low pass, then we can indeed say that the higher harmonics are going to be amplified less by G and to what extent it is low pass decides the accuracy of the calculation that we that we accuracy of the solution that we get by using this calculation procedure. This describing function that we get by ignoring the second harmonic and onwards has some information which we have ignored that information ignoring is justified if G is a low pass filter. So, please note whenever I forgot a minus sign here, please note whenever we use the describing function method to calculate the frequency and amplitude of the periodic orbit, one should check that G is a low pass filter. If G is a high pass filter, then the calculations are expected to be very wrong. The sufficient conditions for existence and the sufficient conditions for non-existence are very likely to fail. So, suppose G was low pass was a low pass filter, then ignoring higher harmonics justified if G was a low pass filter. Now, we will take an example. Let G of S be equal to minus of S plus 1 over 3 S plus 1. So, this is the example that we will take. So, is this a low pass filter that we can check? So, for S equal to 0, the DC gain is 1, 0 dB as incidental. Then at the first we encounter is a pole, 1 by 3 radians per second, because it is a pole the magnitude starts decreasing at the rate of 20 decibels per second. This is an asymptotic plot until we encounter a 0, which is 1 radians per second and after that it flattens. So, this is indeed some low pass filter. Of course, the duration of decrease is not much, but then if it had been low pass, if the decrease were for a longer duration for a longer band of frequency, then the accuracy of the resulting amplitude and frequency would have been more. It would have been more accurate. So, we have done the needful of checking that G is a low pass filter. So, now consider G of S connected with the jump stresses. So, now one should check that 1 plus eta of A, the jump stresses we saw was again independent of omega. So, check for intersection of minus 1 by eta of A and G of j omega on the complex plane, the Nyquist plot of G and the curve minus 1 over eta of A, these two will plot and we check for the intersection. We can do this separately, notice that only because this is a function of only A and this is a function of only omega. If eta were a function of omega also, then their intersection is not enough. You see, because they also have to intersect for the same value of omega here and here because of this particular convenient form in which one of the parameters A affects only one function, eta of A, the other parameter omega affects only other function G because of this particular separation in which they affect the two functions, eta and G. One is able to plot them separately and look for the intersection. Plotting G in particular is easy because there is nothing but the Nyquist plot. But then we will now ask the question for eta of A, for this jump stresses, we already know how it looks. But instead of plotting minus 1 over eta of A, why do not we plot the Nyquist plot of minus 1 over G? So, this which we have written here can also be seen as, alternatively, eta of A and minus 1 over G of G omega. These two intersecting also equally helpful, equally good. It is exactly the same, there is no difference between the two procedures. For the jump stresses, eta of A was equal to m plus and some constants. Let me just look this up. So, I just look this up. So, the constants are 4 here and pi here. The second part, the imaginary part is what we obtained from noting that the jump is nothing but in phase with cos omega t, which is the derivative of sin omega t and that is why we just wrote this. This is nothing but m plus 4Bj over pi A. We are used to writing the constants before the variables and hence this. So, how does the plot of this look? Let us say m is equal to 2, B is equal to 1, amplitude we are not sure. Amplitude is a variable and we have to adjust the amplitude according to the intersection. So, this is m equal to 2, the real part is constant. So, for A equal to 0, it is way up and this is where it reaches for A equal to infinity and this is 2, this is a 0.2. What about minus 1 over G of S for our example is equal to 3S plus 1 over S plus 1. That happens to be, first order transfer functions are necklace plots are nothing but a circle. In this case, these are the two points. For omega equal to 0, for S equal to 0, we see that it starts at 1 for omega equal to 0 and this is where it is for omega equal to infinity. The necklace plot is complete only after you draw the orientations. So, for that purpose we note that this one is we first have a 0 and then a pole. So, in that sense this minus 1 over G is a high pass filter. So, because the phase starts increasing from omega equal to 0, we decide that the arrow should have been like this. So, this is where it indeed intersects. So, it is intersecting at the center of this diameter 1, 3 is 2. That is precisely where m equal to 2 is. One can check that this point where it intersects on the topmost point above the center of the circle has to be the location of the pole. I will just quickly write this in another result. So, this particular point we know what value it is. It happens to be nothing but 2 plus j. The radius of this circle is 1. So, the imaginary part is equal to 1. So, we should put that this is equal to m is equal to 2 m plus 4 b over pi a j which already gives us that m is equal to 2. This of course, we assume that is why we do this line vertically starting with the real part equal to 2. And what about, so we get the, and b was equal to 1, notice that the b was equal to 1. So, pi a equal to 4 b which gives a is equal to 4 over pi. So, for amplitude a equal to 4 by pi, we have an intersection of eta a curve and minus 1 over g of g omega curve. For omega equal, what is the value of omega at which we have this intersection? Eta of a does not depend on omega. But for what value of omega is g of j omega at this point? That is not too hard. We will just quickly write this. So, let me just tell you at this place that omega should be equal to 1 radians per second. The exact reason for this is what we will see in the following slide. So, we have obtained the coordinates of this point because it happens to be on the Nyquist plot of minus 1 over gs and eta of a. We know that there is a value of amplitude a for which it intersects. There is also a value of omega for which it intersects. Those happen to be omega equal to 1 radians per second and a is equal to 4 over pi. Why a is equal to 4 over pi? Of course, we have derived this just now. Why omega is 1 radians per second is what we will see very quickly. So, one can check a very simple property. Suppose, you have s plus a over s plus b, c, c s plus a over s plus b and no pole 0 cancellation. Then the Nyquist plot of this happens to be a circle in which the radius, the center is something. These values are nothing but the values. One of them is for omega tending to infinity. One of them is omega equal to 0. Which one is which? Of course, depends on the relative values of c, a and b. For omega equal to 0, it is a by b. For omega is equal to infinity, it is c. So, depending on between them which is higher, which is lower, it will be one of these. That will also decide the orientation. But this point which is right above the center, one can check that that is one of this is omega is equal to b radians per second. The other one will then be omega is equal to minus b radians per second. So, one of these points happens to be for plus b radians per second, then the other one will be minus b radians per second. So, why that is the point right above this is not too hard to prove. One can easily prove this. If you already know this fact, one can substitute this and see. I also happened to encounter in the case of constructing problems for describing functions. So, can verify that imaginary part of g of j omega at omega equal to b plus minus b is equal to radius. Radius of this circle and the real part of g of j omega is nothing but center, center coordinate. This will prove that it is right above the center or right below and both plus and minus b and where b is the pole. If the pole is at minus b or plus b, that pole is actually on the real axis. But the corresponding frequency, the corresponding value on the imaginary axis is what makes it exactly above or exactly below the center. Of course, this whole likewise plot could have been on the left side depending on the relative values of a, b and c. If they are all negative, then it could be here or it could also be encircling the origin that depends on the signs of a, b and c. I am just telling you this property that the point right above the center will always be at the frequency corresponding to the pole even though the pole is on the real axis. So, this is the property that we used in our previous example for this jump hysteresis. Our minus 1 over g of s happened to be 3s plus 1 over s plus 1. So, we said at 1 radians per second is where it should have been. Whether it is plus 1 or minus 1, we decided that because for omega equal to 0, it increases like this and for omega equal to infinity. So, this point is for minus 1 radians per second. Omega equal to minus infinity decreases, it increases from minus infinity radians per second, minus 1 radians per second and finally at omega equal to 0 radians per second. So, this point here in this case is minus 1 radians per second and this is 1 radians per second. So, that is the property of first order transfer function that we used. So, one can ask, can we have different omega values? So, we can set the different omega value by just changing the value of m. This is the whole family of, once you have plotted minus 1 over g of s, Nyquist plot, all you have to do and notice that omega equal to 0 to omega equal to infinity, this is how it is. You can take different different values of m and check that it will intersect this Nyquist plot at different points. Of course, calculating the precise value will be a little more effort, but if somebody comes to us and tells us that this is the frequency of oscillation I need, the frequency is what they can specify, the amplitude can also be specified in fact. Why? Because if the amplitude is specified, one can design B accordingly. Is this a good problem to work on? So, here is an exercise problem. Consider g of s equal to minus of s plus 1 over 3 s plus 1, use jump hysteresis in feedback, negative feedback. Design the nonlinearity, design the nonlinearity means to design m and B. These are the two parameters that play a role. Inside the nonlinearity itself, the parameter a is not a property of the nonlinearity because a is the amplitude of the incoming signal. While b is a parameter, by how much it will jump is a property of the nonlinearity and so is the slope. To have frequency 4 radians per second and amplitude 8, amplitude 8 to have periodic of the periodic orbits. So, one can ask inside the close loop at which point is it 8? So, 8 sin 4t at input to nonlinearity, this is understood. If you want the amplitude to be 8 at some other point, one would have to redesign a by the gain of g, the gain of g corresponds to 4 radians per second. The magnitude of the transfer function g of j omega when evaluated at 4j, the magnitude, please note it is not this real part or imaginary part, but the magnitude because amplitude gets magnified by the magnitude of the transfer function. So, this 8 when somebody specifies, if you want to use this 8 for a, the parameter a, then this is at the input to the nonlinearity. Our a sin omega t, r of t, the reference signal r of t equal to a sin omega t is precisely contributing the parameter a into the deserving function provided that is at the input to the nonlinearity. So, one this is a problem one can pursue and a very good exercise problem that we typically have in our exams here. So, we now come to another problem that we pursued only partially in our previous lecture. This is about how the saturation nonlinearity happens to be periodic orbits for k greater than 60. So, consider again the transfer function g of s over s plus 1 times s plus 2 times s plus 3 and we had saturation nonlinearity. We had saturation nonlinearity, but more generally it is a sector nonlinearity. So, we asked the question, we had only plotted the Nyquist plot of this, at that point we had only estimated the rightmost vertical line such that the Nyquist plot is still further to the right of this, but rightmost such vertical line. While doing that, we had used some ad hoc estimates, the point minus 1 is here, this intersection point of course we said is minus 1 by 60, this is 1 by 6 for this particular example. Now, let us see a systematic method to calculate the real axis intersection of this vertical line where this vertical line is satisfying the property that it is tangential to this curve and the Nyquist plot is to the right of this. This property is satisfied for this vertical line also, but we want the rightmost such line. So, we are going to calculate this particular vertical line. So, for that purpose we will find out at which point the real part reaches its extrema. So, this vertical line satisfies the property that because it is vertical, the real part of real part of G of j omega is leased at omega naught. Which omega naught? Omega naught is a frequency correspond to this point when the vertical axis which is rightmost is tangential to this Nyquist plot. And why is it leased? Because real part of G j omega is negative already, it does not become further negative. This is the leased value of real part of G j omega with sign. So, why do not we just try to differentiate this with respect to omega and put that equal to 0, that would give us this point and this point. And maybe actually it will also give us this, but had omega been finite at this point. So, we will do this to find the value omega naught and then we will evaluate G of j omega, real part of G j omega at this omega naught. That will be a procedure for finding the rightmost vertical line such that it is tangential to this Nyquist plot. G of j omega equal to 1 over which gives us minus omega square plus 3 j omega plus 2 which becomes equal to minus j omega cube. Omega square comes from this term and this term. So, minus 6 omega square j omega comes from this and plus 9 plus 2. So, plus 11 j omega and the extremmost term come from just one, extremmost degrees come from just one. Just like this came from only the product of this and this. Like that constant term will come from this product of these both. So, plus 6. So, now we are going to find out the real part. So, real part of G of j omega is equal to 6 minus 6 omega square 11 omega minus omega cube plus sign was then the denominator. So, now we have minus sign in the numerator because we are going to write its complex conjugate. 6 minus 6 omega square square plus 11 omega minus omega cube square. These are the real parts and imaginary parts. So, we have multiplied by the complex conjugate so that the denominator becomes real and the numerator becomes equal to this. So, now we are ready to split the real part and imaginary part. So, the real part of G of j omega becomes equal to 6 minus 6 omega square divided by plus 11 omega minus omega cube whole square. So, this real part we want to know for what value of omega this one reaches its least value. We expect that this becomes negative and the least value it attains is the value itself is of relevance and the value of omega at which this least value is attained will give us the point at which it is tangential to that particular curve, vertical line. So, because this has a very high degree in the denominator what is easier is we will differentiate this with respect to omega. This gives us this is like differentiating numerator by denominator. So, this is nothing but denominator whole square. The denominator itself omega minus omega cube whole square times derivative of the numerator which is nothing but minus 12 omega plus minus 6 minus 6 omega square times derivative of the denominator again divided by the denominator square. Which denominator is this? Is this the denominator of the transfer function G of s? But it is not the denominator of the transfer function, but it is this particular denominator that we have written here. One can of course use some simplification and differentiate the inverse of this rather than this itself, but then we decided to just do the routine procedure. So, we have to differentiate the denominator here which is 2 times 6 minus 6 omega square times the derivative of minus 6 omega square which is minus 12 omega plus this seems to be a very lengthy calculation. So, at the end we have d by d omega real part of G of j omega equal to 0. We want to extract out only the numerator you see. So, the numerator we will get by picking out terms, this entire part we will pick. There is still something continuation, which we will write directly here minus 12 omega times 6 minus 6 omega square plus 11 omega minus omega cube square minus 6 minus 6 omega square times minus 24 omega times 6 minus 6 omega square plus derivative of this term that is what remain 2 times 11 omega minus omega cube times the derivative of this term itself which is 11 minus 3 omega square. This is multiplied to this. So, this equal to 0 is what we want to solve. So, let us try to get rid of various factors. There are just too many things to cancel. I think it is easier that we differentiate the inverse of this. One can try to pursue this and check that we get the same answer as this other thing that we will pursue. What is this other thing that we will pursue? We have that we want to find out when this real part reaches its extrema. Instead of checking when the real part of this reaches its extrema, we will differentiate the inverse of this. Inverse of the real part of G of j omega is equal to 6 minus 6 omega square plus 11 omega minus omega cube square divided by 6 minus 6 omega square. This is the inverse. How did I get this? I just took the inverse of this expression. Take the inverse of this means we cancel of one factor here and we have this square divided by 6 minus 6 omega square. Now, we are going to differentiate d by d omega of this, 1 over real part of G of j omega and we get this equal to minus 12 omega plus 6 minus 6 omega square whole square times denominator times derivative of the numerator. 6 minus 6 omega square, the derivative of this is 11 omega minus omega cube times 11 minus 3 omega square minus this factor times the derivative of this. Minus 11 omega minus omega cube whole square divided by 6 minus 6 omega square whole square times the derivative of the denominator which is equal to minus 12 omega. It appears to be only slightly simpler. First thing we will notice, we can extract out a factor omega and cancel off from everything. So, indeed one of the extrema is at omega equal to 0, that we knew. After canceling off a factor omega, let us simplify this. So, this comes to minus 12 times 6 minus 6 omega square whole square. We have plus 6 minus 6 omega square times 11 minus omega square. We have cancelled a factor omega here. This term comes as it is. Then here we have cancelled off this omega here plus 12 times 11 minus omega square whole square times omega square. This is a pretty big long calculation that we have got which we will now start simplifying further. So, first thing to notice is that whatever this whole thing is equal to 0. This is all the numerator. At the denominator we have 6 minus 6 omega square whole square and that is anyway in the denominator. So, first thing to notice is this is an even power of omega. I mean only even powers of omega come into this expression. That is also expected because if some value of omega is a root, minus of that is also a root. So, why do not we first replace omega square by some variable x? That will indeed reduce the degrees and give us more courage to go through this expression evaluate further. So, now we will open these brackets. So, we have 6 x minus 6 whole square times 12 plus 6 minus 6 x. Perhaps we can cancel off 6, this 12, this 6. There are enough constants also that we should be cancelling before we too many large numbers only causes more calculation mistakes. So, there is quite some laborious calculations involved. So, let me just show what I have been doing. So, this is where we last obtained and I said that this is an even function, only even powers of omega are going to appear in the resulting expression. So, why do not we replace omega square with x and then there is some simplification I have been doing and after expanding all the brackets, we eventually get this third order polynomial. So, this polynomial is expected to have 3 roots, at least one of them is real, at least one real root. So, it will require some effort, it will require some solver to find out this real root, whether it is positive or negative is the important thing for us. So, need to check, at least one positive root. Why do we need at least one positive root? At least one real root is guaranteed because this is a cubic polynomial. So, for very large values of x tending to plus infinity and x tending to minus infinity, it is going to change its sign. Therefore, x tending to plus infinity because of this minus sign here it goes to minus infinity and for x tending to minus infinity, it becomes plus infinity. So, somewhere in the middle it will cross for some value of x that is why this whole polynomial has a real root, but that real root is for x. So, for omega square, if omega should have a real root then this polynomial should have at least one positive real root. So, once that is obtained then that we liquid for omega square and we will take square root of that positive root and that will give us the value of omega where it is coming leftmost and we can evaluate the real part of g of j omega at that particular value and that will give us the leftmost point. Leftmost meaning the real part of g of j omega is a least. It is expected to be negative and it will be the least. This can be verified using psi lab for example. So, we will do a very similar calculation for a simpler transfer function also for the same purpose. So, at the end of this calculation one can check that this is less than minus 1 by 60. So, what is it that we had already used in our previous lecture for g of s equal to 1 over s plus 1, s plus 2, s plus 3. We already saw that this point is minus 1 by 60, this point is 1 by 6 and minus 1 is somewhere here. So, we expect that this real-access intersection at when it is tangential happens to be left of minus 1 by 60. For example, minus 1 by 50 of course, exact value needs to be verified by the procedure that I have said. That value we have also seen is related to the circle criteria that is the largest sector for time varying non-linearities, sector bound time varying non-linearities in the sector 0 to k. That k value can be found by this, you find the vertical intersection of the negative real axis and take the inverse of that, that will also be negative, multiply minus sign to that, that will give us this value. We expect that this value will be less than 60 for that same reason, but of course, we expect it to be much more than 1, much more than plus 1. We will take another example for which we will find a similar value and also for that we will use a jump hysteresis and evaluate if there are oscillations, if there are, if there is an oscillation, if there are periodic orbits and what is the frequency and the amplitude. So, take G of s equal to 1 over s square plus s plus 2. So, the angle is product of this. So, this is for omega equal to 0 degrees like this, omega tending to infinity becomes like this. So, notice that the point minus 1 is here, this is equal to 1 by 2. The entire negative real axis is not encircled, which means for the whole sector of linearities in the range 0 to infinity sector of linearities. Each linearity in this range, in this sector means any line of slope, any positive slope, take any line, this is the input, this is the output and take any line with positive slope. Suppose this line of positive slope, slope is equal to let us say 10, then this point corresponds to minus 1 by 10 here. This line of slope 10 plus 10 corresponds to point minus 1 by 10 is minus 1 by 10 is not encircled and this has poles in the left half complex plane. Because it has poles in the left half complex plane, p is equal to 0, n is equal to 0 for the point minus 1 by 10. Hence, the number of closed loop poles is also equal to 0. This will happen for every point on the negative real axis. Every point on the negative real axis just corresponds to a line inside this sector. So, for sector of linearities we conclude that we have closed loop stability. Now, we want to ask the same question for sector of time invariant nonlinearities and for the sector of time varying nonlinearities of the form 0 to k. So, when we ask the question for sector of time varying nonlinearities, that is when we have to apply the circle criteria and again find such a line which is tangential to this curve, tangential to this Nyquist plot. We are going to find real part of G of j omega. We expect the calculation will be simpler because there is only a second order system. 1 over 2 minus omega square plus j omega, real part of this, which is equal to 2 minus omega square minus j omega divided by 2 minus omega square square plus omega square. The real part of this which is equal to, so we get this as a real part. So, we are going to again now conveniently differentiate the inverse of the real part of G of j omega. That will just give us of 2 minus omega square plus omega square over 2 minus omega square. This is what we are going to differentiate with respect to omega and equate that to 0 to get the value of omega. So, what is the derivative? The derivative this is nothing but minus 2 omega plus 2 minus omega square times 2 omega minus omega square times minus 2 omega, whole divided by 2 minus omega square, whole square. This is the derivative of this expression. So, this we will evaluate this minus 2 omega plus we get here 2 omega we can take in common and when we do that we get 2 minus omega square plus omega square. If I have not done any calculation mistakes, 2 omega when we take common what remains is plus omega square and here it is nothing but 2 minus omega square and this is equal to, we can now again take 2 omega common minus 1 plus something over omega square minus 2 whole square because it is being squared we can reverse the sequence here and in above here we have only 2 which is equal to 2 omega over omega square minus 2 whole square. Here we have 2 minus square of this which is nothing but omega to the power 4 plus 4 omega square minus 4. So, this is what we get as the derivative. So, we are going to equate d by d omega of 1 over real part of j of j omega equal to 0 gives us either omega equal to 0 or omega is a root of or omega is root of what omega to the power 4 minus 4 omega square plus 2. It is a root of this polynomial. So, finding the root of this polynomial is what remains is what we will do next. So, let us find the root of x square minus 4x plus 2. So, we want this to have real roots first of all. Secondly, we want at least one root to be positive. Why we want one root to be positive? Because we are going to take square root of that to get the root in omega eventually. So, this indeed satisfies the property it has both real roots because the discriminant is not negative. So, x is equal to 4 plus minus square root of b square minus 4ac that is 16 minus 8 minus 8 divided by 2 that is equal to 4 plus minus square root of 8 which is 2 square root of 2 while 2. This is 2 plus minus square root of 2. So, for two values of x, we are getting this root. So, we need to be able to interpret for each of this. So, notice that we have plotted one over real part of j of j omega that procedure itself might have introduced some more roots. We do not, we only expect that omega equal to 0 of course and at another value of omega which will be negatives of each other. We do not expect so many. So, since we have got so many, what is easiest is to just remove the spurious ones by evaluating the real part of j of j omega at each of these and we expect one of them to be negative. When evaluated at square root of these. So, that would give us correct one. That is what we can do right away. So, we will take omega to be equal to square root of 2 plus square root of 2 and then we will also take omega to be equal to square root of 2 minus square root of 2. Both are positive. You see 2 is more than square root of 2 because we are taking square root of a number larger than 1 of course the square root will be lower than that number. So, both of these will be quantities that are positive that are real. So, we will need to evaluate j of j omega real part of j of j omega at each of these, at both and we expect that the real part will be a minima at one of these. The fact that it is a minima can also be verified by finding the second derivative of real part of j of j omega with respect to omega. When differentiate with respect to omega, if the second derivative is positive then we expect, then we know that this is a local minima. Alternatively, for one over real part of j omega, we can check that it is a maxima at which of these two points. One also needs to see carefully why the extra roots have come. What is the reason that the other roots, how does one explain the extrema at the other values and both omega values. So, let us assume that this is being done. This requires, this is just some routine calculation which I will do after this lecture, but this procedure will give us this point. Once you get this point, we know the largest sector. Suppose this turns out to be equal to minus 1 over 70, 70, 71 approximately. So, we know from 0 to 71, circle criteria gives time varying nonlinear, sector for time varying nonlinearities. So, what about time invariant nonlinearities? That is what we will get from the popo plot. That is what we will sketch now for the same transfer function. So, for g of j omega equal to 1 over 2 minus omega square plus j omega which turns out, which is equal to 2 minus omega square minus j omega divided by, so the popo plot consists of plotting real part of g of j omega on the horizontal axis just like in the Nyquist plot, but on the imaginary axis we are going to plot omega times the imaginary part of g of j omega. So, we are going to plot, this is going to look very much like the Nyquist plot, only that the vertical axis is going to be shifted, but whether it comes and goes to the origin eventually or some other point on the vertical axis requires a little more careful calculation. So, the imaginary part we can check is nothing but omega over this whole thing and when we multiply that by another omega, we get omega square. So, omega times the imaginary part of g of j omega will be equal to omega square divided by omega square minus 2 plus omega square, square of this plus this. So, as omega tends to infinity, check that this is going to 0. This going to 0 is nothing but to say that this goes and meets the origin. The real part is going to go to 0 because it is a strictly proper transfer function since the Nyquist plot goes to 0 as omega tends to infinity because g is a strictly proper transfer function. If the real part for the Nyquist plot goes to 0, real part for the popo plot will also go to 0 because the popo plot differs from the Nyquist plot only as far as the distortion along the imaginary on the vertical axis is concerned. But the imaginary part need not go to 0 again because we are multiplying by omega that is increasing that is becoming infinity. So, if the imaginary part were going to 0, at least very slowly the product would go to a non-zero constant. So, anyway in this case we have verified that it goes to 0. Now, the next question is to find a line that is of any positive slope such that the popo plot is to the right of this line, to the right and below. So, this is one such line. Of course, another such line is this. Another line is this. But then we do not have to make this vertical anymore like we had to do in the circle criteria. The important thing is that the horizontal axis intersection should be as right as possible. So, notice that because of the property we are able to take this particular line that almost comes and touches this popo plot at the origin and is a line of positive slope, any positive slope. So, asymptotically when we try this, we get this particular point on the horizontal axis intersection to tend to 0. So, let us say minus 0.000, this is also okay for the intersection of the line with positive slope such that the popo plot is to the right of this particular line. So, this gives us that the sector can be of the form for popo criteria and of course for J of as we are speaking for this transfer function can take k equal to k tending to infinity. We cannot take the vertical line itself. So, 0 to infinity is also the largest sector for time invariant non-linearities. So, for time invariant non-linearities also, so we get that the sectors as large as 0 to infinity. We already verified for this example that for sector of linearities also it is the sector 0 to infinity and this also refers to the fact that not just for sector of linearities, in other words for lines of any positive slope, but also for the sector of time invariant non-linearities where we are allowed to use the popo criteria, we obtain that this is the largest sector. So, notice that this is strictly larger than the circle criteria, largest sector. If you say that time varying non-linearities is also going to get allowed, then the sector will be strictly smaller. How small? That depends on that, that depends on this value as I said. So, one thing we can, one other thing we can conclude is with respect to the saturation non-linearity, we will not get periodic orbits for any gain k. Why? Because for any gain k, it will also be inside the sector 0 to infinity of time invariant non-linearities. The saturation non-linearity is time invariant and for how much of a large gain k you multiply, it will still be inside the sector. So, we do not expect periodic orbits for the saturation non-linearities for this particular transfer function. What remains to be checked is for the jump hysteresis, one can calculate whether we will have, this I would leave as an exercise. So, g of s is equal to 1 over s square plus s plus 2. Check if eta of a equal to m plus j 4 b over pi a intersects, which two intersects minus 1 over g of j omega and eta of a or equivalently and minus 1 over eta of a. One can check these two intersect. It is not hard to plot the inverse of this, because inverse of a line, inverse of a line in the complex plane will just become, will just become a semicircle, because this is only a half line. So, for this particular line, it will become, suppose this is m, this will be minus 1 by m. For very far off points, it will tend to the origin. So, this is how the plot of minus 1 over eta of a is. As a, from a is equal to 0 to a equal to infinity, one can check here also that a tend to infinity is going to be here. This point that is farthest from the origin is going to be here, when you take the inverse. So, this is where a equal to infinity is and a equal to 0 is here. So, one can check if this intersects with the Nyquist plot of this and if it intersects at what value of omega and what a value and what amplitude value. That will give us the amplitude of the periodic orbit, if any, at the input to the jump hysteresis non-linearity. Since I would leave as an exercise, this ends this lecture. So, we will see another topic in the next lecture.