 So, you would have already seen how to handle non-linear systems which do not have an input. Now, a very clever thing which was done was that when you try to analyze these non-linear systems which do not have an input, you can actually decompose that into two separate things, one a linear system and the other you pack all the non-linearities together. And then, when you want to talk about the stability of the equilibrium point of this non-linear, the original non-linear system that is equivalent to talking about the stability of this interconnection between the linear system and that non-linearity. So, let me try and explain what I mean by this. So, suppose you were to look at an equation like let us say the second derivative of y plus let us say the first derivative of y squared plus let us say 2 times y equal to 0. Now, one wants to analyze this system. This system is a system without any inputs and there are well known methods of trying to analyze the system. But what I was just talking about was this clever way of thinking about this system as an interconnection between a linear system and a non-linearity. So, now let me try and give you how that is done. So, think of this equation d 2 y dt squared plus 2 y equal to u. This now is a linear system. Now, this is exactly like this equation except that you could think of this d y dt squared as minus u. So, let me put that as the next equation which is u equal to minus d y dt squared and this is a non-linearity. Now, the interconnection of these two systems then would give us exactly the same as this original non-linear system. So, let us look at what this interconnection would look like. So, if you look at that system d 2 y dt squared plus 2 y equal to u, this is a linear system which has transfer function given by ys by us. This is the transfer function gs which is equal to 1 upon s squared plus 2. So, now, let us think of this gs in this way and here is the input that you give which is u and the output that you get for this which is y. You take this y and put it to a non-linearity and this non-linearity has the characteristics that if the input to the non-linearity is psi, the output is psi squared. Now, in this case if we now feedback this thing with a negative feedback, then if this e which is the input of this feedback loop, if this e is set to 0, then what we would be looking at is exactly the original equation that we had which is d 2 y dt squared plus dy dt, the whole thing squared plus 2 y equal to 0. So, talking about this closed loop system being asymptotically stable for example, is exactly the same as saying that the original non-linear system is asymptotically stable. Now, of course, in this system we can clearly make out that the system would have the, I mean, if you draw a phase plane portrait of the system, then the origin is an equilibrium point. Now, the origin is an equilibrium point if one wants to find out whether this origin is, let us say, asymptotically stable that is the same as in this feedback system whether that given system, this feedback system is asymptotically stable. So, if you set up this feedback system and set y with some value and then you let this system evolve and if this system evolves finally, settling down to the value y equal to 0, u equal to 0, then what that would mean is this system is asymptotically stable and what that effectively means is that the original non-linear system is effectively stable. Now, this clever trick is, I mean, once this clever trick was discovered, this was used again and again. So, the non-linear systems were analyzed by looking at splitting it up into a linear system and a non-linear part and looking at the feedback connection of this non-linear system and this non-linear part. Now, once that was done, then people came up with some sort of a conjecture. Now, let me give you what this conjecture was all about. So, suppose you had a system like, let us say, similar to the earlier one, but d 2 y dt squared plus let us say, 3 times dy dt plus let us say, 2 times y plus y cubed equal to 0. So, this now can be split up into a linear system. So, the linear part would be given by a transfer function which is 1 upon s squared plus 3s plus 2 and the non-linearity that you would interconnect with this linear system is given by something where if psi is the input, the output is psi cubed. Now, if one looks at this non-linearity carefully, this non-linearity, so suppose I think of psi which is the input to the non-linearity, then the output to the non-linearity is here and this looks like something like that. So, that is psi cubed. Now, the conjecture which was made was made by a person called Isaman and Isaman made the following conjecture that suppose you have a non-linearity and let us assume that this non-linearity is memoryless. So, what do I mean by memoryless? What I mean by memoryless is that this non-linearity, the output of the non-linearity only depends upon the instantaneous input. Now, the earlier non-linearity that we looked at where the input was psi, the output was psi cubed and this output was independent of what the past values of inputs were. So, it is a sort of the output only depended upon the instantaneous input and so one could think of this non-linearity as a memoryless non-linearity. So, typically if you look at a non-linearity which is memoryless, it might be something like that. So, this here is the input to the non-linearity and this is the output to the non-linearity. So, when you have this as the input then the output is this value here and since it just depends on the, so this curve of course is a slightly twisted curve. So, it looks like with this input there might be multiple values, but let us assume that the curve was straight enough so that such a situation never arose. So, the output purely depends upon the instantaneous value of input. Now, the conjecture was made by this person Isaman and his conjecture was the following. If you look at this non-linearity at each point there is some tangent to the curve. Now, one way that you could talk about this non-linearity is you could limit the non-linearity. So, this particular non-linearity for example, one could say that 0 is less than equal to f of xi by xi which is less than equal to infinity. So, this is a non-linearity whose slope, instantaneous slopes lie between 0 and infinity. So, of course, I am sorry, but initial curve probably had slopes which were greater than infinity, but this one is some trajectory or this is a non-linearity which has its slopes lying between 0 and infinity. Now, if one looks at such non-linearity, so let me give some more examples of similar non-linearity. So, you could have a non-linearity like that and then clearly all the slopes of this non-linearity lie between this line. So, if I call the input xi, so this line is k times xi or k1 times xi and there is another slope here. So, this is k2 times xi and so this particular non-linearity one could say satisfies the characteristics that f xi by xi is less than is greater than this particular slope k1 and is less than this other slope k2. Now, suppose we interconnect this non-linearity, so this non-linearity is interconnected with a linear plant Gs in this feedback structure and then suppose this linear plant was such that if instead of the non-linearity, you used a linear gain k and this k was between k1 and k2 and the resulting linear system was stable, then Isoman's conjecture was that if you put the non-linearity instead of that linear gain, then the resulting system would also be stable, would be asymptotically stable. So, the idea being that if you are at this point and this was the input, then there is some slope here. Now, this slope when you put that linearity, the resulting system is stable or asymptotically stable. Now, if the resulting system is stable, then the guess was that for the non-linearity because locally it is like this linearity here, therefore it will behave nicely. So, now as a result of course the system evolves and you go to some other xi and at that xi there is some other slope. Now, this slope when you put in as the linear thing, then again you get a asymptotically stable system, asymptotically stable linear system. Now, working in this way, the guess was that if you had a non-linearity which lay in some sector, so this particular non-linearity f of xi is set to lie in the sector, so it is a non-linearity which lies in the sector k1 k2. So, if you have a non-linearity which lies in the sector k1 k2, then if the linear plant is such that when you put feedback gain, any gain between k1 and k2 and the resulting system is asymptotically stable, then if instead of that linear feedback you put the non-linearity in there, then the resulting system is going to be asymptotically stable. So, this was Isaman's conjecture. So, perhaps I should just formally write it down. So, suppose that the linear feedback system is asymptotically stable for values of k in the range k1 to k2, then a non-linearity that belongs to this sector which is equivalent to saying that the non-linearity f xi by xi, this is less than, I mean k1 is less than equal to this and which is less than equal to k2, then a non-linearity of this kind in feedback with the linear system is asymptotically stable. Now, this is Isaman's conjecture and this was given roughly in 1949. Now, initially it was not clear whether this way of looking at non-linearities as approximation of linear systems is going to work. And in fact, in the literature for sometime this used to be called the method of linearization because the idea was that you have this non-linearity and at various instance for given input there is an output and there is also the slope of the output according to the characteristics of the non-linearity. And then Isaman's conjecture essentially said that if for all those slopes that you get in the non-linear function, I mean if for all of those, if the non-linearity is in that particular sector and for those gains it turns out that the given closed loop system is stable, then the closed loop system is stable if you put in the non-linearity. It turns out that Isaman's conjecture is false and this was proved by several people over the years, several people proved this thing. So, some of the more famous names include Hahn who did this in 1963, then Willems. So, Willems showed a counter example that this does not hold. Willems did this in 1966, then there was Fitz who did this also in 1966. So, they constructed all sorts of examples which showed that this conjecture of Isaman is not correct. So, let me now give an example of a system where this actually doesn't hold. So, for this system, I consider the linear plant Gs to be s plus 1 upon s squared. So, if you take this linear plant in a feedback structure, then it should be clear that for all for all feedback k in 0 to infinity, this given feedback system is stable. That means if you take Gs and you give a feedback which is a linear gain k, then the resulting system is stable. We can see this in several ways. I mean, we have already looked at this Nyquist criterion. So, perhaps we can draw the Nyquist plot of this Gs. So, if this is the complex plane and we want to draw the Nyquist plot here, then as you start moving up, let us say from a small value epsilon, you start moving up. For the small value epsilon, you will get something here and you will have a curve like that. So, this goes right up to the top and then you have this infinitely large circle and corresponding to that, you will get something here like this and then you come up the negative slope and when you come up the negative slope, you essentially have the reflection of the original one and of course, this particular thing has double pole at the origin and has a 0 at minus 1. So, because of that to avoid this double pole, we can draw a small circle here of radius epsilon and when you draw the small circle, when you look at its image here, you will end up getting something like this. So, now this contour C, we traversed in this way, that means in the clockwise direction and as a result, the contour that we got here using the Nyquist plot was something like this. And then the critical point is the point minus 1 and we find that this Nyquist plot, this contour, the image of the contour f of C, this does not enclose minus 1 and therefore, the resulting system is stable for all gains from 0 to infinity. Now, what I am going to do is I am going to demonstrate a particular nonlinearity which lies in the sector 0 to infinity. Now, what do we mean by saying that a nonlinearity lies in the sector from 0 to infinity. So, earlier I talked about the slope, but it is not really the slope that we are looking at because the slope could very well be negative, but when we are looking at f psi by psi and this is psi and let us say the nonlinearity is like that, so let us say this was f psi, then when we are looking at this f psi by psi, we are really looking at this divided by this. So, when you say f psi by psi lies between 0 and infinity, what you are really saying is that the characteristics of this nonlinearity should lie in the first quadrant, that means when the input is positive, the output is positive or the third quadrant, that means when the input is negative, the output is negative, so the first quadrant or third quadrant. Now, the linear system that we looked at s plus 1 upon s squared, we saw that this particular linear system is stable when you put this in a feedback connection, when you put this in a feedback connection with this k lying between 0 and infinity. Therefore, if now because this is true and if Isomans conjecture were true, if we put in a nonlinearity here, whose characteristics lie in the first quadrant and the third quadrant, then the resulting system should be asymptotically stable. So, now what I am going to do is I am going to demonstrate a particular nonlinearity which lies in the first quadrant and the third quadrant, but the resulting system is not stable. So, here is the nonlinearity, so the nonlinearity is given by the following equation. So, f of u, so if u is input to the nonlinearity, f of u is given to be u upon e plus 1 times e, here e of course is the number e, so e is that 2.7 something something something. So, this is the equation for f of u for u less than equal to 1, but this if you look at this, this is like really linear. So, if you are going to draw the characteristics, you are going to get a line with slope 1 upon e plus 1 by e and this goes right up to 1, so this being the input u input. So, this is the first part and now the second part for u greater than equal to 1, what you have is this is exponential of minus u divided by 1 plus exponential of u. So, here the characteristic that you are going to get is like that. So, now if you look at this nonlinearity, it lies in the third quadrant and the first quadrant. So, this nonlinearity given by this f of u in feedback connection with that plant, so s plus 1 upon s squared and here you have the nonlinearity f, well by Iserman's conjecture because s plus 1 by s squared is stable when you put any feedback k lying in the range from 0 to infinity. Therefore, this linear plant with this nonlinearity defined in this way should result in asymptotic stability. Now, what I am going to show is that this resulting system is not asymptotically stable. I am going to demonstrate a particular solution to this particular system which is actually growing with time and if you have a trajectory which is perpetually growing with time, obviously that system cannot be asymptotically stable. So, here is the particular solution that I am going to give you. You see when you put this s plus 1 upon s squared and you feedback through this nonlinearity, then you know I reverse the earlier trick. This is the same as looking at a system with the equation d2 y dt squared plus f of dy dt plus y equal to 0. So, looking at this system is the same as looking at this system. Now, for this system, I look at a particular solution which is given by, so y of 0 is specified as e minus 1 upon e and dy dt at 0 is specified as 1 by e. So, one wants to solve this differential equation with initial conditions y at 0 is e minus 1 by e and dy dt at 0 is 1 by e. So, I claim that for these initial conditions, the solution for this is given by the following. So, dy dt is equal to exponential of minus y t minus dy dt dt. So, I am claiming that this particular equation gives us a solution to this original system with initial conditions y at 0 being this and dy dt at 0 being this. So, how do we show that this is a solution? Well, let us differentiate it. So, y double dot, that is the second derivative. This is equal to exponential minus y minus y dot and then I have to take the derivatives of these. So, the derivatives of these will give me minus y dot minus y double dot. So, this y double dot I take to the other side and I end up with 1 plus exponential minus y minus y dot. The whole thing multiplying y double dot is equal to exponential minus y minus y dot times, so there is a minus times y dot and so y double dot is going to be I take this below 1 plus exponential minus y minus y dot. But now if you think about this, I sort of this exponential of some negative quantity is 1 upon exponential of the positive quantity. So, if I normalize this, I mean or rather if I simplify this, I will end up with minus y dot upon 1 plus exponential minus y plus y dot. And for y dot, we had said that this is a solution. So, for y dot, we can substitute this and if you now think of y plus y dot as u, then this expression is the same as exponential of minus u upon 1 plus exponential of u. This minus sign is appearing essentially because when you take the f to the other side, that is a negative. So, this particular solution is indeed a solution to this differential equation with initial conditions given by y at 0 is e minus 1 by e and d by dt at 0 is 1 by e. Now, if you look at this solution, then it should be clear that once this and this are the initial conditions, this exponential of some quantity, so d by dt is going to increase. So, when d by dt increases, I mean d by dt is positive. So, d by dt is positive means y increases. So, y increases, d by dt increases and this is exponential. So, it will continue to have d by dt to be positive and therefore, y would continue to increase as time goes on, which essentially means that this system is not asymptotically stable. So, this particular example is an example which tells us that the Isomans conjecture is wrong. So, now at this stage, there was Isomans conjecture which was promising, but it was shown that this Isomans conjecture is false. So, now what can we do about analyzing general systems and what sort of general theory can be obtained, which at least makes use of Isomans conjecture and Isomans idea. And so, now we are going to talk about some things which essentially uses Isomans idea, but gets past the conjecture and gives an actual answer as to which systems when you interconnect, you will end up with something which is asymptotically stable. Now, the inspiration for this thing comes, this particular technique comes from electrical engineering and it comes from electrical circuits which are called passive. So, what do we mean by passive circuits? Now, you must be knowing very well that if you have a circuit made up of passive elements, whatever that passive element is, if you have a circuit made up of passive elements, then the resulting, I mean, then the circuit is passive. This really does not give you an answer. All it says is that I am using the same word again and again. I am saying, if it is made up of passive elements, then the resulting circuit is passive. What exactly is passive? Now, a sort of very layman kind of definition for a passive system is the following. A passive system is something that does not generate energy. So, what it means is, so if you have a passive circuit and you supply energy to the passive circuit, then that energy either gets stored in the circuit or it gets dissipated or lost. But a passive circuit does not generate its own energy. So, maybe we start by looking at some examples to get a generic idea of what passivity is. So, suppose we look at a circuit which consists of just a resistance. So, the voltage that we apply, let me call it V and the current flowing in, let me call it I. Therefore, the power that is fed into the circuit or the supply is V times I. Now, when you supply this, this current I goes through this resistance and it gets dissipated as I squared R, that is heat or something. So, if you look at V dot I in this particular circuit and you look at over a long period of time, that means, say starting from time t equal to 0 to sometime t, then this resulting thing is always going to be greater than 0. Now, one could use this as the definition for passive. So, when we talk about a passive system, then what we are talking about is that the amount of energy supplied to the system is positive. Of course, this definition of passivity may not, I mean this may not always strictly hold the moment we use some element like a capacitance or an inductance. So, let me try and explain what I mean by that. So, suppose now, instead of a resistance, you have a capacitance. And suppose you also have a resistance here and then we apply a voltage V and the current going in is let us say I. Now, let me also additionally assume that this capacitance is charged and so at time t equal to 0, the charge on the capacitance is Vc. And then let me short this. If I short this, then V that means the terminal voltage is 0. Therefore, if you want to talk about supply of energy, well the energy that is supplied in is V.I which is 0. But of course in this circuit, we know something is happening. Of course, even in this particular case, if you of course take the integral from 0 to t of Vi dt, this is 0 which is greater than equal to 0. So, we could still continue to call it passive. But now, suppose we apply a voltage V. So, instead of a short, let us assume that we apply a voltage V1 where V1 is less than Vc. Now, if V1 is less than Vc, then what is going to happen is from the capacitor, the current would flow this way. And so now, if you look at the supply, it is going to be V1 multiplied by I. But I, the actual current which is flowing is in the opposite direction to the convention that we use for I. And so this quantity here is negative. And so, when you take this integral, you are going to end up with something less than 0. But of course, you would already know that we call a capacitance a passive element because the capacitance does not generate energy of its own. But what is happening in this particular situation is that this capacitance at time t equal to 0 was already charged to some voltage Vc. And so the stored energy is being dissipated. So, if we want to also incorporate the stored energy, one way we could do this is instead of taking this integral from 0 to t, we take the integral from minus infinity to plus infinity. And if you take the integral from minus infinity to plus infinity of V dot I dt, then any circuit for which this integral V dot I, this is the power supplied to the circuit from minus infinity to plus infinity, if this is greater than 0, then we can call the circuit passive. So, how do we capture this particular notion of passivity? So, the basic idea is this. So, suppose we start off with a circuit and we assume that the circuit is at rest. What do I mean by the circuit is at rest? So, the circuit could have several energy storing devices like capacitors and inductors. The circuit is at rest by saying that the circuit is at rest, what I mean is that all the capacitors are discharged. And similarly, all the inductors have no current flowing through them. So, there is no stored energy in the circuit, that is what I mean when I say a circuit is at rest. Now, suppose to that circuit, you now supply a certain amount of energy. If you supply a certain amount of energy, what do you think would happen in the circuit? Well, this energy which comes into the circuit, part of it of course, will get dissipated, part of it will get dissipated in the form of heat in the resistances. But a part of it might get stored either as electrical energy in a capacitance or as magnetic energy in an inductance. So, in very broad terms, what we can say is that the total supply of energy, this would be equal to total energy dissipated plus energy stored. So, now the total energy, the total supply of energy of course, the expression for this would be V dot i. So, V dot i is the power dt if you integrate this from minus infinity to infinity. So, this is the mathematical expression for the total supply of energy. The total energy dissipated, well, the energy dissipated or the power that is dissipated is given by half, is given by i square r. And so, if you integrate this from minus infinity to infinity, that gives you the total, I guess there is no half here, it is just i square r. i square r is the total amount of energy that is dissipated and you put this integral from minus infinity to infinity, this is the mathematical expression for the total energy dissipated. And what about the total energy that is stored? Well, it could be stored in a capacitance and the energy stored in the capacitance will be something like this and there could be energy stored in the inductors, it could be something like that. So, mathematically this is the expression that we will get that the total supply, so your integral is equal to this integral which is the total energy dissipated plus the energy stored which is this. Of course, this is an integral form, so this equation that I have written down can be called dissipation equality. And I can state the same dissipation equality, this is the integral form, I can state it in the differential form which means I just take the derivatives in each of these cases and I would end up with an expression like v dot i is equal to i square r plus dv dt where dv dt is the rate of change of energy stored in the circuit. So, then coming back to an electrical circuit, we call an electrical circuit passive if the supply or power is, now the supply in terms of energy of course is the integrated part, power is the derivative part. So, perhaps let me talk about it as the supply in terms of energy, the energy supplied is greater than the rate of change of stored energy. So, of course, is the rate of change of stored energy, so actually I am talking about the power. So, the supply power is greater than the rate of change of stored energy than such an electrical circuit is called passive. Now, this stored energy is actually quite important, of course electrical circuits which have an input and an output and so when you talk about the stored energy, this stored energy is something that you have in a system with an input and an output and the stored energy plays exactly the same role as a Lyapunov function plays in a system without inputs. So, let me use an electrical circuit example to show that the stored energy plays exactly the same role as a Lyapunov function in a system without inputs. So, let us consider an RLC circuit, so this is R, this is L, this is C and let us assume that there is some stored energy which is half C Vc squared plus half L Il squared. Now, of course, this is a system which has an input and an output. Of course, one could think of the voltage as the input, one could say that voltage is the input and this is the voltage that we are applying and the output is the current. One could also think of the current, I mean one could think of a current source here and the current pushing in, being pushed in and the voltage that is generated as the output. So, it really does not matter whether you think of the voltage as the input or the current as the input, but here let us assume that the voltage is the input and the current is the output. So, this now is a circuit with an input and an output and there is a stored energy which depends on the current I L and the voltage across the capacitor Vc. Now, here the input is V. If you set the input to 0, that means you just shot here, then we have an autonomous system in the sense this is a system without any inputs. Now, when this is a system without any inputs, what is going to happen? Well, what is going to happen is there is going to be oscillations set up in this circuit and during this oscillations, the electrical energy which is stored in the capacitor gets converted into magnetic energy in the inductor through half the cycle, but of course the current, this transfer is taking place through the current because of which there is some energy being dissipated and then the magnetic energy in the inductor is going to return that magnetic energy, convert that magnetic energy back to electrical energy stored in the capacitance and again some more dissipation is going to take place and in this way this keeps oscillating until all the energy which was stored in the capacitor and all the magnetic energy which was stored in the inductor gets dissipated and then you would have no current in the circuit. But now, if you write out these equations, the equations for this particular oscillation that would take place, well, one way to write out is you just think of the current in this circuit and you can write down the write down the n drops in terms of the current. So, you will get the voltage drop here to be L dI dt plus the voltage drop here to be R times I plus the voltage drop here to be 1 by C integral I dt, this is equal to 0. This is a second order differential equation and this second order differential equation, I mean, take derivative once and you will end up with LC, the second derivative of I plus RC times the derivative of I plus I equal to 0. Now, for this particular differential equation, if you use this V as the Lyapunov function, then this Lyapunov function is positive definite and if you take the derivative of this Lyapunov function, you would find that the derivative of this Lyapunov function is negative which essentially proves that this circuit is asymptotically stable when you convert it into a system without inputs. So, let me stop here right now and we will carry on about this in the next lecture.