 In the previous class, we discussed the analysis of a simplified model of a power system. In fact, the model we chose was very simplified one of a single machine connected to constant voltage source and our focus was on the analysis of that system. We continue that today. So, today's lecture is titled analysis of dynamical systems. We continue with our analysis. In the previous class, if you recall, we discussed the concept of equilibrium or equilibria of dynamical systems states small and large disturbance stability and we began on this example which I just talked about the single machine connected to a voltage source. In today's class, we will try to focus on large disturbance behavior of this particular system and then we will turn our attention to the general analysis of linear system which is a certain class of systems. We considered a toy model last time of a power system. So, let us just recap quickly on what we did last time. So, we had a single machine connected to a voltage source. This was a stiff voltage source or we call it an infinite bus. The synchronous machine was modeled as a voltage source behind a transient reactance and the phase angle of the voltage source was in fact delta which is related to the position of the rotating machine. So, the simplified model we did not derive this model. I just gave it to you. This is in fact a very simplified model as I mentioned in the previous lecture. It gives the motion of the rotor angle and the rotor speed deviations from omega naught which is the frequency of this infinite bus. This particular simplified model what we did was first of all we try to study the small disturbance stability of this particular system. So, the first step which we took was finding out the equilibria of this system. In fact, an equilibrium is defined at the as the point or the values of the states for which the derivatives of the states become 0 and that yielded us two equilibrium points. Both have omega e equal to omega. This comes out of this particular equation and delta e was sign inverse this quantity. This is this value of delta e results in this derivative becoming equal to 0. Now, between 0 and 180 degrees there could be two equilibria. So, if this is for example, P m mechanical power which you assume to be a constant then there are two equilibria delta 1 and delta 2. One is less than 90 degrees and the other one is greater than 90 degrees. Of course, you could have this continuing and you could actually have many more equilibria, but you would notice that they are spaced 200 that is 360 degrees away from these. So, actually they are indicative of the same position of the rotor. So, we will not talk of equilibria beyond 180 degrees. We set upon ourselves to find out the stability of this system and we to some extent did an analysis for small disturbances around these two equilibrium. We need to really understand the large disturbance behavior. In fact, the large disturbance behavior really is something which is important in the sense that in the first lecture I told you of the phenomena of loss of synchronism. It is in fact, a manifestation of the non-linear behavior of the system. So, small disturbance stability around equilibria is actually easier to analyze and easier to you know describe. So, that is what we did in the last class. So, just again looking at what we did. We assumed that we had small deviations around the equilibrium. So, we assume that delta omega and delta delta are very small. P m is a constant. Of course, I mentioned it sometime back. Then we got this particular model where k was E s E cos delta E by x where delta E was an equilibrium value. So, if you are studying deviations around an equilibrium delta E omega E then k would be this. So, after that we obtain the response for small disturbance. In fact, whether a system is going to be stable or not after you give it a disturbance that is whether it is going to come back to the equilibrium is determined by the response of the system. So, response of this particular system was written down. In fact, it was guessed from by analogy with the spring mass system. We did not actually derive it. We will do that of course, for general linear systems in the following lectures. So, the response for small disturbances is given by this. You notice it is an oscillatory response where omega n is dependent on this value of k and h. It is the natural frequency of the oscillation in delta and omega. So, for small disturbances around the equilibrium, we have got an oscillatory response. Just one small point this assumes that k is greater than 0. So, this is true actually this response is true provided k is greater than 0. There is a oscillatory response if k is greater than 0 that is delta E less than 90 degrees. The values of a and phi incidentally are obtained from the initial values of delta delta and delta omega. Note that whenever we are talking of a stability of a system, we are really discussing the situation when you are away from the equilibrium. You are not at the equilibrium. If you are at the equilibrium, of course, you will simply stay there. The point is that you are giving push from the equilibrium due to some disturbance. So, this disturbances could be things like step change in the voltage of the infinite bus or something small. So, we will not discuss what the disturbance is actually, but we will assume that you have been given a disturbance around an equilibrium and of course, once you get disturbed, you will oscillate around the equilibrium provided your original equilibrium was less than 90 degrees that is delta E was less than 90 degrees. This is true for the first equilibrium that is delta E is delta 1 and omega E is omega 0. So, this is the first equilibrium. When I say first equilibrium, I am talking of this particular equilibrium point delta 1 which is less than 90 degrees. Of course, a point which may occur to you would you call this oscillatory behavior is stable or unstable. The thing is that if you look at the response here, it just continues the system just continues to oscillate around the equilibrium. So, would you call it a stable or an unstable response? Actually, we would like the system to come back to the equilibrium. So, if you just oscillates around the equilibrium, you would call it a marginally stable system. It is not really coming back to the equilibrium nor is it growing away from the equilibrium with time. So, of course, if there is some damping in the system, the oscillations will die down. I am not proving this here, but you can well imagine it that if there is some kind of viscous damping or something in the system, it will kind of oscillations will die down with time. In fact, in a real system in a real power system, wherein where you model everything including the damper windings, the field winding, there is usually some damping in the system. In fact, if you have got control systems associated with the control of field voltage, you can even make damping negative that is the oscillations grow with time. So, this is an issue which we will handle later on. Right now, we have not considered any damping in our model. It is a very simple model. We are not going to get any damping. It is just going to be an oscillation around the equilibrium. If you are here the equilibrium delta 1, delta is equal to delta 1, then that is what you will expect. Later on, we will come across more detailed models of synchronous machines in which damping is present. Some kind of damping will be there. The other equilibrium delta 2 which is greater than 90 degrees which is at unstable point. K is greater less than 0. Why is it unstable? That is because the response turns out to be this. And one component where this omega is real, one of the components here is going to be greater than 0. That is this or this can be greater than 0. So, what will happen is that you will have a combination of growing and decaying terms in general. Of course, there is an issue that will R k 1 and k 2 non-zero. The answer is of course, k 1 and k 2 are dependent on the initial value of deviations. And for most situations, for most initial conditions, k 1 and k 2 will be non-zero. So, what you will have is you will have a combination of growing and decaying terms. The growing terms effectively mean that if I give a disturbance from the equilibrium, you are going to move away from the equilibrium. And this is what happens at this particular equilibrium point. So, this is what we discussed in the previous lecture. One important point which you should note when we are talking of operating at unstable. You cannot actually operate at an unstable equilibrium point. If you give a push, you are going to go away from the equilibrium point. So, any small push. So, it is a small disturbance unstable system. And if you have got a small disturbance unstable system, you cannot operate at all. So, you have to be, you really cannot operate at this equilibrium delta 2 omega e is equal to omega naught. So, this is an important point which you should keep in mind. If for example, the earlier equilibrium were to be unstable, that also it would not be feasible to operate there either. So, this is an important thing. A system should be small disturbance stable if you want to operate it at that point. So, just to summarize, if delta e is delta 1 omega is omega naught, you get oscillatory response for a small disturbance. So, this is very important. The other equilibrium could have growing terms. So, it is an unstable system. At the other equilibrium, delta is equal to delta 2. For large disturbances, we cannot make any approximation based on omega or delta delta being small. So, we cannot assume that the disturbance is small. If the disturbance is not or rather if the deviation, initial deviation is small, if the deviation from the equilibrium is not small, you cannot assume that sign of the deviation is equal to the deviation. That is what basically we did in when we got the linear approximation. This particular approximation which we got was based on the fact that sign delta delta was approximately equal to delta delta. This was a key approximation and cos delta delta was approximately equal to 1. This is true only for small disturbances. For large disturbances, we cannot use this model. So, we have to use the full blown model in which this sign delta is retained. This if you look at this particular term here, you can look at it as some kind of a accelerating torque or accelerating power. We are talking in terms of per unit and we will assume speed deviations from the equilibrium are not too large. So, we can per unit torque and per unit power are in fact almost equivalent. So, we will just take this approximation. So, if you take this accelerating power here, this is accelerating power or approximately the accelerating torque. You will notice that as compared to this, in this particular system, this is a linearized system, we are assuming that the acceleration accelerating torque is proportional to the angular deviation. Here, if you do not assume small disturbances, this expression does not tell you that. It tells you it is it is depend on the sign of the angle. So, if you look at how this particular accelerating power or accelerating torque looks like, if you look at the accelerating torque or accelerating power, if you are at the equilibrium that is delta delta is equal to 0. At the equilibrium delta 1, the accelerating torque is 0, but it becomes negative that is P m minus E s e sin delta by x becomes negative if delta delta goes away from the equilibrium and delta delta is positive. If it is negative, this is positive. So, it goes negative. So, in some way if you if delta delta is greater than 0, you will find that there is some tendency to get you back to the equilibrium because the torque is restorative in nature. That is a very important point like a spring mass system. You take a spring mass system and you give it a push from the equilibrium. If it is a positive push, the spring gets stretched and is a tendency to pull it back. So, nature of the forces are restorative. Conversely, if you push the spring, the spring gets compressed and as a tendency to push you back to the equilibrium. So, we can say that the torques or the forces are restorative in nature in this particular case, but look at this carefully. It is negative, but after a point it becomes positive. So, if you look at this particular system, it becomes positive after some time. For this spring mass system, if you assume that the spring has got you know is ideal, then you can stretch it and the force is always proportional to the stretch. This is not true here. You will find that the accelerating torque power becomes negative. It is restorative till this point and thereafter it becomes positive. So, for small disturbances, we can take it as restorative, but for large disturbance it is not correct to assume it is restorative. So, in fact, when we did the linearized analysis, what we effectively took was the slope of this particular curve at this point. So, the slope was the k which we talked about sometime back. So, for small disturbances, you can use this particular straight line with the slope of minus k, but for large disturbance it is not correct to assume that the torque or the accelerating torque may not be restoring after sometime. Now, the fact that the torque deviates from this linear or this line has got important consequences. In fact, it goes it kind of comes down after some it is the maximum and comes down after sometime becomes negative. You will find that if the disturbance is large enough, it is a very interesting thing. You have got for small deviations, the force is almost proportional to the stretch the restorative force, but if you give a large enough disturbance it will start of move it will move and then the restorative force suddenly starts coming down and becomes negative. Now, if the restoring force becomes negative, the mass will never come back to the equilibrium. So, what we should you know if if the spring stretches up to a point if it stretches up to a point at which this restoring forces become equal to 0, then you may not come back to the this old equilibrium again. So, that is the important thing. So, if there is a large disturbance, you may on may not come back to the equilibrium, you may actually just go away I mean you may just become unstable. So, that is an important point interesting point is that the restoring torques in this particular system become negative when the angular deviation reaches angular deviation becomes delta 2 minus delta 1 that is delta becomes delta 2. So, this particular point. So, we should ensure that you can kind of formulate a kind of a rule. So, if I give a disturbance because of which the angular deviation is greater than 0 and they are not small and if the speed deviation from omega naught does not become 0 before delta 1 becomes equal to sorry delta becomes equal to delta 2 or delta delta equals delta 2 minus delta 1, then you have reached a place where torques are no longer restorative and you may become unstable. So, that is the basic basic reason why for large disturbances a synchronous machine connected to a voltage source or later on we will see connected to other synchronous machines could go unstable. Now, the whole catch is after a disturbance are you going to be stable or not is something which depends on the response of the system because the rule says that is the speed going to be equal to 0 before the torques become non restorative. So, the thing is now you have to actually see what is the response of the speed when it will become 0 and so on. So, that is something which is complicated. So, let us just take a example of a large disturbance you have got a synchronous machine connected to another voltage source is connected via two lines on one of the lines there is a three phase fault. So, you can have a three phase fault on a transmission line if there is a three phase fault there are fault currents. In fact, the voltage here if it is a three phase bolted fault then you will find that the voltage here dips down to 0 there will be large fault currents and typically there will be relays in the system and circuit breakers. So, relays will detect that there is a fault in the system in this part of the transmission system and trip this these circuit breakers. So, this line gets isolated. So, post fault you just have this particular line remaining in the system this line has gone off. So, this is a typical stability question which you may come across for studying large disturbances is that if there is a fault like a large disturbance like a fault and because of that fault there is a significant deviation in this angles and the speeds from the equilibrium are we going to come back to an acceptable equilibrium after the fault is clear. So, this is the post fault system. So, are we going to come to the post fault equilibrium that is the question which we need to ask. So, just putting this let us try to formulate this as a kind of a problem at time before t is equal to 0 and t 1 the system say is in the pre fault condition. So, this is in the pre fault condition in the pre fault condition you are operating at delta 1 with a superscript p. So, this is the equilibrium point when you have a fault let us say this fault occurs at t 1 the electrical power becomes 0 why does it become 0 because the terminal voltage has dropped down to 0 we will assume it is a three phase fault. So, the voltage drops down to 0. So, electrical power this is in fact electrical power p e it becomes 0. So, there is suddenly mechanical power has become much greater than the electrical power and you will find that the machines accelerate both delta and omega will suddenly start increasing. At time t is equal to 0 the fault is cleared by the fact that relays detect the fault and trip the faulted component that is the transmission line at t is equal to 0. So, at time t is equal to 0 the electrical power will become this this is the electrical power p e the equilibrium now are delta 1 and delta 2. Remember that p e the electrical power after the fault is different from the pre fault electrical power because you have tripped one line. So, now your electrical power initially was E s e by x pre fault sin delta and now it is E s e x sin delta in the pre fault system x p superscript p is given by x 1 x dash plus x e by 2 after the fault is cleared it is simply x dash plus x e. Now of course, we have these equations these equations define the movement of the rotor angle and we have also seen that due to the fault your angles and speeds have become greater than 0 they have deviated from the equilibrium. So, there is a there are substantial deviation there is substantial deviation from the equilibrium. Now how do you know whether the system is going to be stable or not well you could numerically integrate the system. So, if you numerically integrate the system you can actually find out. So, numerically integration means this particular integration is performed by discretizing this set of equations. How to do this is something we will discuss later in the course, but soon enough if you do a numerical integration of the system I will just show you a you know numerical integration if the clearing time of the fault that is the fault duration that is t 1 to t 0 that is the fault duration is small the behavior mimics that of the small disturbance behavior it is oscillatory. If the fault duration is increased you will find that the oscillation magnitude increases and it still looks like a sine wave I mean that is what the small disturbance behavior is like, but if now the fault clearing time and therefore, the deviation from the equilibrium increases beyond the point you will find that the nature of the curve is like this it is no longer looking like a sine wave this is in fact a numerically integrated solution of the system. So, I actually use a numerically integration technique to find out this how to do it let me assure you will learn its sometime soon in this course. One important point is if the clearing time becomes greater than a certain value you become you lose synchronism you become unstable you are no longer going to come back this angle just grows with time. So, this is a typical situation in which because of a large disturbance you are not going to reach equilibrium if the fault duration is large then this is what will happen remember the equilibrium is stable you know small disturbance stable, but rather we can say that it involves oscillatory behavior around the equilibrium, but for large disturbances it never reaches the equilibrium again. So, the question is of course, this is why do we need to do non-linear you know why do we need to do numerical integration in order to find out this particular response the answer is this whenever you have a non-linear system unfortunately the response cannot be written down in terms of simple functions which you know. So, but this is not true for linear system. So, if you take a linear system you can write it in terms of it turns out that the response is in fact in all linear systems you will find that a response contains terms like this e raise to lambda t e raise to minus lambda t t e raise to lambda t sin. So, typically linear systems are a superposition of terms of this kind you will find that this is how most linear system. So, if you take out the any linear system response you find out this, but for non-linear systems we cannot write down the response in terms of simple well known functions and as a result of which you may have to in other than in very special cases you may actually have to you can find out the answer only if you numerically integrate the system. So, the response in general for a non-linear system for example, the one we are talking of for large disturbance you have to consider the non-linear system will have to be obtained from doing a numerical. So, we have to rely on a computer typically if the system is very large. So, that is one of the problems which you will face in this particular course that even if you get the model to get to infer the responses is not always very easy. In fact, when you talk of a system which has got hundreds of generators and tens of thousands of lines and buses and it is a non-linear system to infer the behavior of the system becomes very tough because the response is a complicated function and it can only be obtained by doing a computer study of the system. And this is a bit worrisome actually philosophically worrisome also it basically says that unless you work out or really use very high level of complication of computation you cannot actually infer how the system behaves and that is although you know the model. So, it is very interesting because you know the physics of the system the physics is very well known, but still you do not know how the system is going to behave unless you actually do a study which probably requires a lot of computing resources. This is of course, not a system like this a single machine infinite bus can be you know you can even do a numerical integration without a lot of computer resources, but a realistic system requires a lot of computation. So, that is something which is philosophically interesting about the behavior of system. So, even if you know the physics of the problem you know the you know the equation is described each particular component once you integrate all of them thousands of such components how they will behave requires a lot of computation. Interestingly enough this particular system this you know this particular system I mean only this system. In fact, only this system or systems which have similar equations the even though computation of response requires numerical integration techniques stability evaluation although it is not easy for this particular problem it can be obtained. For example, look at this particular problem of a ball in a valley we have talked about this problem in the previous class. If I know because of certain disturbance this particular ball has deviated from the equilibrium it has got displaced by an equilibrium by a certain amount and it has also acquired some velocity because of the disturbance. Then the question is is this ball going to roll back or no. So, one way of finding out you can make a rule that if this ball speed becomes 0 before it reaches this point then it will roll back. So, at the limiting condition the ball will just go right up to the ends just reach here it will just become it will reach speed equal to 0 just at this point is the kind of limiting condition. Intuitively in this particular case if I tell you energy is conserved there is another way of finding out whether this is going to happen or not. So, I am not going to evaluate the response, but still from a certain criterion I can tell you whether the system is going to be stable or not. If I compute the kinetic and energy and potential energy of this ball at this point at this initial condition if it is greater than just the potential energy at this point because you know we assume kinetic energy is 0 at this point and it will be just be stable that is the speed becomes 0 just at this point it will be stable. So, the criterion is that the kinetic and potential energy here is equal to the potential energy at this point I will call it delta 2 then if it is greater than it is unstable, but if it is equal to you will find a ball just goes and sits here, but this is of course true only if energy is conserved. So, I am talking of a very limited kind of situation a very you know specific situation. In fact, in our undergraduate courses we learn about a criteria called equal area criterion. In fact, it is based on this particular feature. So, the only difference is that we will have to define what is conserved actually you can actually show that w which is equal to this particular quantity half of 2 h by omega b omega square minus p m into delta plus this into cos delta is the quantity which is conserved. How can you find it out just try to evaluate d w by d t. So, shall we do that. So, this is what it is. So, d w by d t is equal to 2 h by omega b into omega minus omega naught into d omega minus omega naught by d t minus p m into d delta by d t plus well minus of e s e sin delta by x d delta by d t. And d delta by d t of course, is equal to omega minus omega naught from our equations. So, what we have is d omega by d t d w by d t is nothing but 2 h by omega b d omega minus omega naught by d t minus p m minus let us call this and this there is some mistake in the sign. So, obviously I have written something wrong here this should be minus and this should be plus. So, you look at this. So, if I define my w to be this, this is equal to 0. Why is it equal to 0? Because of the second differential equation this we have we know that this is equal to this. So, what we have is in this particular system the single machine connected to an infinite bus this is conserved. So, that is it is a constant. So, if at time t is equal to 0 your deviations are delta and delta delta and delta omega that is the values of delta at 0 and omega 0 are these. Then the condition for stability is that if the energy evaluated at the fault clearing time is greater than the energy evaluated at delta 2 omega 0 because this is omega 0 the first term will become 0. So, basically we are using the same criteria as we are using here. So, in fact this is the basis for equal area criterion which we have learnt in our undergraduate years. Now, we will have a quick recap the aim of this particular these couple of lectures which we are having is actually talk about general systems. So, I have taken an example of a single machine infinite bus system this is because we will get a kind of a bird's eye view of power system dynamics. We do not get lost in the detail of general dynamical systems, but of course, a study of that is interesting and important. So, just quick recap a single machine connected to an infinite bus the two equilibria one of the equilibria if you give a small disturbance you get oscillatory behavior the other one is unstable for small disturbance you cannot operate there at all. For the first equilibria for large disturbances you can become unstable, but large disturbance behavior requires you to consider the non-linear behavior of the system that is the restoring torques are not linear functions of the angular deviation. So, what you have is basically a large disturbance behavior is much more interesting in the sense that it shows that if you have got a large enough disturbance you no longer get an oscillator response, but something like an unstable response and actually this unstable response in this particular situation is in fact the loss of synchronism phenomena which you have learnt before. So, if you recall in the first lecture I had said that machines which are connected to each other synchronous machine connected to another synchronous machine or a synchronous machine connected to a good voltage source it tends to remain in synchronism with that with the rest of the system. So, if you give a small pusher disturbance it tends to oscillate and come back to an equilibrium come back to the equilibrium, but if you give a large disturbance it may lose synchronism. So, I will just one important point is the inherent characteristic of the system is like this the physics of the system is such that it behaves in such a way. So, you give a push for small disturbance it will oscillate and come back to the equilibrium you have a large push it loses synchronism. So, this is as much true for this toy example as it it is for a large system. Of course, before we get a loss into a lot of mathematical detail involving you know variables like x and y let us just look at you know what happens in a large system if you have got a very large system you got many machines connected to each other. So, what happens for large disturbances in a large system the thing is that you have this loss of synchronism phenomena, but as I mentioned in the first lecture you have got groups of machines their angles they move together their angles move or increase relative to the angles of another group of machines or many other they can be three groups of machines also. So, you know this particular phenomena occurs in realistic large power systems and you know it is not easy to analyze it is to analyze this small system by equal area criterion, but it turns out it is a big challenge to analyze a large non-linear system. So, coming back again linear systems we have not actually defined linear systems in a very rigorous fashion I in this course in fact a lot of things are imbibed and not defined. So, in for example, linear systems are the other equations of this kind wherever constant coefficient the rate of change of a state is equal to a constant coefficient into a state. So, of course, this is a single order system is a first order system you can have a higher order system as well basically of two sets of states non-linear systems this function can be a bit more complicated it is not simply a constant coefficient into the state it is it can be a complicated function like sin x as we saw in a single machine infinite bus system. So, for example, you can take a higher order system as well for example, this is a higher order non-linear system rate of change of x 1 this is a coupled system rate of change of x 1 is minus x 1 into x 2 plus x 1 square x 2 itself the rate of change is again dependent on x 1. So, this is a example of a non-linear system you can have similar coupled equations of the order 3, 4, 5, 1000, 1 million and so on. A linear system a coupled linear system can be of this form in fact when you are talking of a linear system this is how it is in fact in these two systems I have not given an input you can have of course another input quantity plus u something into u and this also could be a function of u as well which is an input like P m is an input in our single machine infinite system infinite bus system. So, you could have inputs as well, but if you do not have an input it is called an autonomous system. So, these two are autonomous systems without any input u in general linear systems can also you know if you are having a higher order linear system a good way of writing this is a into x. So, a is a matrix x is a vector x is this d x by d t is also a vector of the individual derivatives of the states. So, this is a general way of writing a higher order linear system a higher order non-linear system would be d x by d t or we will call it x dot because x dot is easier saves less saves a bit of space a non-linear system could be f x u where x is a vector f is also vector. So, you have got in fact like x 1 dot is equal to f 1 x 1 x 2 u 1 u 2 and x 2 this is a possible system f 2 x 1 x 2 u 1 u 2. So, this is a coupled non-linear system of two states contain having two states. Now linear systems when we are talking of may be you know you may come across a physical system which is inherently linear that can happens most physical systems incidentally turn out to have some non-linearity or the other, but we can actually have some systems which are inherently linear for example, I can design a system which is for all practical purposes linear. So, some systems are inherently linear other linear systems are obtained see when do you really come across a linear system they may as I said it may be inherently linear or it may arise due to an approximation. For example, original non-linear system when we consider small deviations from equilibria of non-linear systems we end up with a linear system. So, this is what in fact we did for our single machine infinite bus system we started off with a non-linear system this is a non-linear system and because you are analyzing small disturbances we could get a small disturbance model around an equilibrium point. So, this is a linear system that was a non-linear system. So, linear systems are obtained as approximation that is small deviations from the equilibria of non-linear systems of course, there is a actually a formal procedure to get a linear system a linearized system of a originally non-linear system. So, if you have got x dot is equal to this is a non-linear system you assumed as small deviations from the equilibrium you substitute this and this into this get a Taylor series approximation this particular term is equal to 0 because at equilibrium the definition of equilibrium x dot should be equal to 0. So, at equilibria x dot should be equal to 0 therefore, f of x c u e should be equal to 0. So, what you get eventually is delta x is a partial derivative of f evaluated at the equilibrium point it is very important you are evaluating this at the equilibrium point plus d f by d u partial derivative of f by sorry dou f by dou u evaluated at the equilibrium. So, of course, you know this is an approximation because in neglecting higher order terms in the Taylor series expansion. So, there are higher order terms like delta x square delta u square etcetera that kind of those kind of terms we are actually neglecting. So, what we end up is a linearized system with the understanding that it will probably give you good results or give you correct results near about the equilibrium point. So, if your disturbances are small this should give you a good idea actually this should be delta x dot this should give you a good idea of how the system is going to behave. In fact, there are you know kind of exceptions to this particular rule for example, if you try to apply this kind of procedure to evaluate the small disturbance stability of the system x dot is equal to minus x square around the equilibrium x is equal to 0. See actually the this as an equilibrium x e is equal to 0 that is obtained simply by putting the derivative here equal to 0. So, how do we get the equilibrium set this equal to 0 x e becomes equal to 0. So, the equilibrium of this is this. So, if you linearize this system you will have x e plus delta x dot which is nothing but delta x dot because x e is a value. So, derivative of it does not make any sense in fact it is 0 this is equal to minus of x e plus delta x square. So, that becomes equal to this is equal to x e minus of x e square plus 2 x e delta x plus delta x square. Now, we neglect this term I told you we neglect higher order terms of small small disturbances when we evaluate small disturbance stability x e because actually it should be like this x e is equal to 0. So, the small disturbance model of this system is by the by the procedure I have told you is in fact since x e is equal to 0 it is 0. So, it says that if you have if you have got a small deviation from the equilibrium there is no movement this is clearly not true. So, this is an exception to what I said the formal procedure which I told you before. So, there are certain systems in which you cannot apply this formal procedure in order to tell how the behavior for small disturbances is going to be for small disturbances. In fact, this particular system if x is is negative then you will get this x dot is positive. So, if you have got this is 0 the equilibrium if you are here the system tends to move to this equilibrium. So, in fact today if you start from here you may actually come back to this equilibrium if you are here on the other hand if x is positive x square is positive minus x square is positive is negative and you will come back to this I am sorry I made a mistake here if x is negative you tend to move away from the equilibrium. So, if x is negative since minus x square is negative you move further away from the equilibrium. So, actually the non-linear behavior or even the small disturbance behavior of this system does not seem to be obtained from the simplified system which we obtained by using the procedure which I described in the previous slide. So, obviously there are some exceptions to trying to understand the small disturbance behavior by neglecting higher order terms, but these exceptions are very very rare and for all practical purposes for all most of the systems we are going to encounter. In fact, this simplification will tell you everything about the small disturbance behavior of the system. So, please do not get ruffle too much by the example which I gave you that is a pathological example for most systems by this formal procedure you can get the small disturbance behavior of the system. One important point to get the equilibrium we need to solve this particular equation if this is a non-linear equation in that case you will have to use a numerical technique like Newton-Raphson or Gauss-Seidel method to obtain x c and u v. So, even this is an interesting step when you are going to linearize a system in order to obtain the small disturbance model. So, just to recap a linear system may be obtained from the a non-linear system and that linear system is obtained by neglecting the higher order terms in most situations the system so obtained will give you the correct small disturbance behavior for most systems it is going to be that way. Now, coming to a very very important topic now the analysis of general linear systems. So, we are going to talk about general linear systems now if you look at this particular example x dot is equal to a x the solution of it is very simple you can verify that x of t is equal to e raise to a t x 0 you take the derivative of this it turns out to be a into x. So, obviously this is the solution of this this system now another consistency check we should see is that at x is t is equal to 0 is x is equal to x. So, if I have told you that at time t is equal to 0 x is equal to x 0 this also should be satisfied as far as the response is concerned. So, you plug in t is equal to 0 you will find that x of t at t is equal to 0 is equal to x of 0. So, this is also consistent. So, this is a solution of this system for this initial condition the interesting thing about the linear system which I have shown you here is that first of all the response is written nicely in terms of a function we well and we understand fairly well that is if a is greater than 0 you will have a growing response a is less than 0 your oscillations die out with time. So, if I start off from a system from an initial condition x of 0 if a is less than 0 you will find it is going down to 0 which incidentally is the equilibrium value equilibrium value of the system is x is equal to 0. So, if I if a is less than 0 this is how the response is going to be if a is greater than 0 this is how the response is going to be just from the value of a you can tell whether the system is stable or not. So, if I start from a initial condition which is not equal to the equilibrium and a is less than 0 your stable the system comes back to the equilibrium if it is not true of course, a is greater than 0 you are going to go away from the equilibrium. So, this is unstable this is stable. Now, so this is what I meant when I said that linear systems are a bit easy to understand the nature of responses are in terms of well known functions. You look at this system now this is a second order system the response is very easy to evaluate because x 1 is dependent on x 1 x 2 is dependent just on x 2. So, the solution of this system is very easy you just have to apply this individually whatever I did in the previous slide to the individual states, but what if I have a system which is coupled this is a second order coupled system. In that case the solution no longer seems easy in the sense that I it does not you know I could not infer it right away this is certainly not a solution of this you can just plug in the value of x 1 and x 2 in this system and you can find out that this is not going to be a solution of this. So, the solution of this requires you to do a bit of bit of algebra and actually by trying to understand this particular solution we introduce ourselves to a very very important tool in engineering which is the idea of a transformation. So, I will give you the kind of a simple system in which a transformation can be used to in order to make it simplified. So, the idea of a transformation is something which I will discuss more in detail in the next class more formally, but just look at this simple system x 1 dot is equal to this and x 2 is this some system I want to analyze. This is a coupled system again the solution is not obvious just by inspection it is not easy to find out I can write this in this fashion as a matrix x 1 dot is equal to x 1 plus 0.5 x 2 x 2 dot is equal to 0.5 x 1 plus x 2. So, this is what I have written in this matrix form. Now, to analyze the system let us just do one thing we do not look at how x 1 and x 2 digitally behave we rather look at how the difference and the sum of these states behave. So, I just subtract these two equations this is a linear system I subtract just simply subtract and this is what I get and when I add them up this is what I get. So, for this particular system I am not saying it looks neat, but this is true only for this particular system I have given the simplified system. So, as to introduce to you the idea of a transformation. So, instead of looking at the variables x 1 and x 2 look at these variables. So, what does it lead you to it leads you to x 1 minus x 2 is 0.5 into this x 1 plus x 2 dot sorry x 1 minus x 2 dot is this x 1 plus x 2 dot is this. So, if I call this as y 1 this variable x 1 minus x 2 as y 1 this is what you get and the solution in y 1 is very very simple right it is got a it is become a decoupled system like the one we have considered some time ago. So, this leads you to a simple solution of course, I have not got the solution in terms still of x 1 and x 2 we will do this formally in the next class. So, this is something we will do it will it will lead us to the general analysis of linear systems. One small point I wish to make here is that if you have got a system x e x dot is equal to a x your solution is x of t is equal to e raise to a t x of 0. If your system is x dot is equal to a x plus b u what is the general solution the answer is that you can write down the solution again very neatly in the sense that you can easily infer the properties of the response, but we will do this in the next class. So, we leave two things for the next class one is a more formal way of understanding the linear transformations to analyze these linear differential equations and also what happens when you have got a you know some input a forcing function like u. So, this these two things we will do it in the next class in the next few classes we will kind of be overwhelmed by a lot of analysis and we will be doing a more of analysis of general systems, but all this these things which you are going to do are going to be useful later or when we apply them to realistic power systems and that is something we let us keep it at the back of our mind. Sometime we will also find out that some of the modeling principles also require you to do a bit of analysis that is why I have decided to tell you a bit about analysis first before going into modeling that is the reason why we have got into the analysis of general systems with that we this is as far as this particular lecture goes we will meet next time. Thank you.