 In the previous two lectures, we have had a brief overview of power systems and stability problems associated with it. What I really told you in those lectures was a description of the phenomena, the basic phenomena. We now go on to the nitty gritties of trying to analyze this phenomena, trying to understand how they occur and look at it in a more mathematical fashion. Now, mathematicians will scoff at us and probably be horrified by the kind of approximations we will be doing in order to get a basic understanding of the phenomena. You will understand of course, that the nature of the models which we will use will be determined by which phenomena we want to understand. So, today's lecture we will learn about some basic concepts of dynamical systems. We will also understand a few examples pertaining to a power system, but the main issue which we will try to tackle in this particular lecture is a general understanding, to get a general understanding of dynamical systems and how we can attack the problem scientifically. In the past two lectures, I have not really given any mathematical or rigorous explanation for the kind of phenomena we have seen that is voltage instability or loss of synchronism. I just gave you a kind of a hint that the reason why the systems behave the way they do, due to the physical laws which govern their motion. So, today what we will do is try to understand some kind of a general ideas, understand general ideas of dynamical systems. The first thing about dynamical systems is what is an equilibrium? Of course, we would like for example, a power system or any other system to be at an acceptable equilibrium at all times. This of course, will not be true because a power system is always subjected to some disturbance or the other. It could be a minor load change, for example, you switch off a light in your house, it is a load change or it there could be really large faults involving tripping of components due to you know some short circuit and so on, those are large disturbances. So, what we need to do is understand what do you mean by equilibrium? Now, if you look at it using a very simple example. So, what we will do is just look at a very simple example of a ball on a hill. So, you can just concentrate on what I am drawing here. Suppose you have got a hill of this kind, you got a valley here and a ball is somewhere on this hill. So, you got a ball on this hill. The ball will be at equilibrium if it is x coordinate is here. So, this is what is known as an equilibrium. So, I will call this x 1. Actually, what I have told you is not strictly speaking true, what I should tell you is that if this ball is at this point and its speed is equal to 0, then it is at equilibrium here. You can always have a ball which is rolling down and at this point it has got some non-zero speed. In that case, it will continue moving. So, when we say something is at equilibrium, we should really specify all the states corresponding to that equilibrium. So, what I should say is that if x is equal to x 1 and x dot or which is nothing but a notational simplification of dx by dt, the rate of change of x is equal to 0. If these two things are satisfied, you are at an equilibrium. The ball will be at an equilibrium. Is there another equilibrium? There is one more. For example, if I manage to place the ball here, I will call this x 2. x is equal to x 2 and x dot is nothing but dx by dt is equal to 0. This also defines an equilibrium. So, if I place a ball right there at the peak of this hill very carefully so that its velocity is 0 and I am just exactly here, then the ball will stay there. So, that is also in equilibrium. There are two equilibria here in this diagram. As far as I have shown you here, there are only two equilibria here and here and there are two states of the system. What do you mean by state? We will not go into any rigorous definition, but in this context you see that the values of the states are the minimum information I need to give you to know whether you are at equilibrium or not. So, you should you need to specify two things, the value of x and the value of x dot. So, this particular system has two states. Now, you will notice that there is a qualitative difference between the equilibrium here and here. In fact, it would have come to your mind right away. This equilibrium 1 is different from this equilibrium 2. How is this equilibrium different? Well, in case I place this ball at this equilibrium and I give it a small disturbance, what would happen? It would kind of suppose I gave it a kick while it was here. Suppose, somebody came and gave it a kick. So, it would go up, then come down, it would go up again and then come down. There would be forces which you try to pull this thing back to equilibrium, but of course, if there is no damping it will set off an oscillatory motion, but the fact remains that there are forces which are pulling it back to this equilibrium. So, we this particular ball would eventually stabilize at this equilibrium. The point is of course, there should be some friction here, so that it comes back to this equilibrium. Otherwise, it will just go on continuing to oscillate. Contrast this with a situation where the ball is here. If the ball is placed here at a 0 velocity, in case I give a small disturbance, it will just roll off. So, this particular equilibrium is such that if the ball is at it, a small push can really take you out of the equilibrium. So, in fact, a kind of intuitive definition of stability can be got right here is that if you have got an equilibrium and if I give a small push from that equilibrium, if the ball tends to come back to the equilibrium, eventually of course, it should settle back at the equilibrium, then that equilibrium point is stable. To put it more precisely, it is stable for small disturbances. So, that is one important thing which you should keep in mind is that there could be many equilibria, but if that states are subjected to a small disturbance like the ball being given a small push, are we going to come back to the equilibrium? So, that is the basic concept of stability. In the example which I have shown you, this is an unstable equilibrium and this is a stable equilibrium. One of the ways we can really mathematically define an equilibrium is if we take the states for in this case x which is the position and x dot which is the velocity. If the rate of change of position and the rate of change of velocity, which is nothing but I will call velocity as d v in that case we will have. So, equilibrium is the point at which the rate of change of the states are all equal to 0s. Now, as we saw in the previous slide, in this particular example there were two equilibria. Now, this is the general definition of equilibrium and states. How do you really analyze the small disturbance behavior of the states? So, that is something which we will do slowly and understand it using a power system example, a simplified power system example in this particular lecture. But before we move on to that, let us just look at some more interesting things about stability. You take this example again. These are the two equilibria. This is unstable. This is stable for small disturbances and this is unstable for small disturbances. However small the disturbance may be. The point is if a ball is here and I give not a small disturbance, but a big disturbance. Somebody really kicks it hard. Then you may find that this ball if somebody kicks it for example, this ball will tend to move upward. It is constantly being decelerated because of the force of gravity. But that force is not adequate to stop the ball's motion before it reaches this point. So, what will happen is that the ball rolls over this hill and starts rolling down here. And if that happens, this hill just goes on like this or it goes on like this for example, you will find that the ball will never come back to this particular equilibrium. So, although this equilibrium is stable for small disturbances, it is not so for large disturbances. A big enough kick to this ball can make it roll over this hill and never come back to this equilibrium. So, this is a typical situation where a system is small disturbance stable at this equilibrium, but if a large enough disturbance is there, it will not return to its equilibrium. So, this is what is known as large disturbance stability. So, I hope you got a kind of a feel for these two concepts. Let us do one thing. Let us take a simple power system example and go ahead and try to analyze it. A very similar example, we will have two states. Consider a single machine, a synchronous machine connected say via a transformer to a transmission line and a very large grid. The grid is very large and let us say it is practically a voltage source of constant magnitude E, constant frequency so that its phase angle never changes. So, it is a sinusoid, a three phase sinusoidal voltage source whose voltage magnitude, frequency and phase angle simply do not change whatever you do. So, it is a very large system. This is a transmission line. Now, when we are trying to analyze this system for small and large disturbances, what do we need to do? We need to really form a model. Now, the whole idea of this course in the later lectures is to get good models of all these elements like a transmission line, a transformer, synchronous machine. One important point which you should notice that we will never be getting exact models. We will never be using exact models in the sense that every model will involve some approximation or the other. For example, look at this example. The synchronous machine let me model as a voltage source behind what is known as a transient reactance. How do we come to this model? We will do that in the course as the course moves on. Let me just tell you that this is not a very respectable model of a synchronous machine. I am just using it to highlight a certain phenomena. So, what I will do is a synchronous machine is model as a voltage source E s with a constant magnitude behind a transient reactance. So, although this E s is constant does not mean that the terminal voltage is a constant. E s is a constant. Delta is the rotor position. So, what we have is if the rotor moves it directly reflects in the phase angle of the voltage which appears. So, this is of course, not very difficult at least intuitively to understand that this could be a kind of a model of a synchronous machine, but there is absolutely no rigor. We are just I am just putting it forth to you. The transformer is modeled just by its leakage and the transmission line and this voltage source. The transmission line is just one lumped inductor x. This is x transformer and I will call this x line. So, this is the electrical circuit model. We have really made a lot of approximations to bring a transmission line model from something very complicated. In fact, transmission line is modeled by Maxwell equations. This is described Maxwell equations. So, it is a big long story of how we come to a lumped reactance equivalent of a transmission line. I hope I will get some time to tell you how it comes about, but right now let us just take this model of a transmission line as simple lumped reactance x. So, I will just simplify this. I will just make it a bit. So, this circuit as far as this circuit is concerned is how it looks. So, x c is nothing but x dash plus x t plus x l. Now, what are the equations which describe this system? Now, those who are purists would say well this transmission line has to be modeled by a partial differential equation. This transform model probably again one has to use Maxwell equations. This synchronous machine well we have just given you a model and not told you how you know it really comes about. It involves of course, Maxwell equations and Newton equations which describe the motion of the rotor. So, just let me put forward a particular model. We will be doing the origin of this model later on in the course, but right now you just take it from me. I will just explain each element. This is omega b which is the base radian frequency that is 2 pi into the frequency base. H is nothing but the inertia constant of the machine which is actually equal to half into j into the mechanical speed square divided by the volt ampere base. So, its units are actually mega joules per MVA. So, these are the units. In the expression for H, we should have omega m b square that is the mechanical speed base square. Out here I have written it as omega m square, but what really needs to be what is correct is in fact, omega m base square. So, please note that minor error here. Omega is the electrical frequency, electrical radian frequency of the generator, the electrical radian speed I should say. So, it is 2 if there are 4 poles of course, the mechanical motion into p by 2 is equal to the electrical speed. So, this is a mechanical speed is the electrical speed. Omega naught is 2 pi into f 0 where f 0 is the frequency of the infinite bus that is the voltage source whose frequency and voltage do not change at all. So, this is f 0. Now, this is basically P m is the mechanical power input. So, this is the mechanical power input. This is the electrical power output of the machine. Now, one important point is remember that P m and P e are in per unit. So, that is a very important thing you should note this P m and P e are in per unit. Otherwise, you will not find this dimensionally. So, P m and P e are in per unit. In fact, the mechanical and electrical powers this is in fact, an is an approximate equation what should have been correctly written is T m minus T e in per unit. So, in fact, this is equal to T m minus T e in per unit or both in per unit. But, I have written it as mechanical power minus electrical power. The approximation here is that mechanical power in per unit and mechanical torque in per unit are practically the same. So, this is one approximation which we have made right away. So, let me just rewrite these equations again 2 h by omega b d omega minus omega naught is approximately equal to the mechanical power minus the electrical power in per unit. So, that is one important thing you should remember. In fact, the rotor position the rate of change of the rotor position is given by omega minus omega naught. Now, the rate of change of the rotor position as seen by a frame which is rotating at omega naught. So, if the machine is rotating exactly at 50 hertz and your omega naught also corresponds to 50 hertz, then this angle delta will be a constant. However, if the frequency of this even transiently changes the position of say this mark will appear to move. So, let me just put it you know in a kind of physical way. If I am rotating at 50 hertz and I am looking at this mark which is also rotating at 50 hertz this mark will appear stationary. However, if there is a transient change in the speed I will see this mark moving. So, that movement of this mark is captured by this equation. Remember the position of the rotor of the rotor of a synchronous machine also determines the phase angle of the voltage which appears at the terminals. That is why what you see here is this delta also has a bearing on the phase angle of the voltage of this voltage source in this model of the synchronous machine. This again not difficult to understand intuitively because the position of the field which is on the rotor will really determine the kind of the phase angle of the voltage which appear in the stator winding say the a phase stator winding. So, you know the point here I wish to make here is that it is not difficult to see intuitively. We will do this in a bit more detail later that the actual position of this will determine the phase of at a given time the position of this will determine the phase of the voltages which are induced in the stator winding say of the a phase. So, we will do this a bit later we will keep this for later to show this actually is true when we actually derive the synchronous machine model. This is of course, let me re-trade a very very simplistic model of a synchronous machine. This in fact is not even a respectable model but I will use it to highlight a certain phenomena which is actually seen in practice. So, it is only used to show that a certain phenomena actually exist. Now, if you look at the system here what is P e what is P e now P e is equal to E s this is a power transfer between two voltage sources which are quite familiar with. So, this is the you can say the sinusoidal steady state power flow formula for a three phase system a three phase balance system. Now of course, again somebody may raise an objection that well you are studying system dynamics are not you. So, why give a sinusoidal steady state expression for power actually you should actually write the differential equations corresponding to this inductance here is after all this is an inductance x c while. So, this we cannot just treated as a reactance then uses sinusoidal steady state formula and then use it in our differential equations. So, unless I actually make this justification you will probably be unconvinced. So, I agree that I am not given you any such justification. Let me just tell you right now that if the phenomena we are trying to study using this model is much slower than the transients associated or the natural transients associated with this electrical network here. It is just a simple transmission line we can use a kind of a quasi this is a kind of a steady state formula for analyzing this equation. So, what appears actually a transmission line a transformer a synchronous machine it should be made out of many many many differential equations and it in fact, does we have reduced it to just two differential equations and this is you can call it a toy model. It is not a very respectable model it is a toy model which will help us to understand certain phenomena. In fact, it highlights the phenomena I wish to tell you quite well. Now, how do I analyze this system? So, you have got d delta by d t is equal to omega minus omega naught and 2 h by omega b d omega minus omega. When you first look at this equation and I tell you well is the system stable or not stable? Well the correct way to approach this problem is first of all take out what are the equilibrium conditions for this system? The equilibrium condition for this system is remember how do we get the equilibrium conditions by setting the rates of changes of the states which are in this case the speed deviation and the rotor angle delta equal to 0. In the in which case you will get the equilibrium or omega is equal to omega naught and well p m is equal to e s e sin delta by x e. So, at equilibrium I will call this may be omega e omega e and delta e are the equilibrium values of this particular system. Now, so just to look at it graphically this is p e p is nothing but e s e sin delta e by x e. So, this is electrical power versus delta and this is p m say a certain value of p m then the point at which they are equal defines the equilibrium. In fact, from 0 to 180 there are 2 possible equilibria of delta here this is 1 and this is another for this particular value of p m. So, let me just so omega e is equal to omega 0 and delta is equal to sin inverse e s e sin delta sorry rewrite this p m x e. Now, the point the what we are trying to do here is analyze the stability of the system. Now, of course there are 2 equilibria. So, there is one here and there is one other here. So, the first thing which I can try to understand let us not try to understand large disturbance stability right away. Let us talk about the small disturbance stability of the system. How does the system behave in case you give small disturbances if it is initially at this equilibrium. So, the point here is suppose I am at this equilibrium. If I am at this equilibrium I will simply not move the reason is d delta by d t is equal to 0 and d. So, if these 2 things are satisfied I will just stay where I am. So, the only way we can get a transient is to give it a small disturbance if as a result of a disturbance if as a result of a disturbance the system slightly gets deviates from one of the equilibria. So, it is at one of the equilibrium let us say delta 1 itself and it slightly deviates from there. So, let us call this deviation to be small. So, this is a small deviation. So, let us say omega is a small deviation from omega e the equilibrium value and delta is a small deviation from its equilibrium value. So, these are small deviations why are we talking about small deviations. We shall see later that if we talk about small deviations the analysis is tractable. You look at this the set of differential equations which describe the motion of the system. How do you solve this? Unfortunately, it is not easy to solve this you will have to actually kind of integrate these set of equations, but without actually doing the integration say numerically or by some other technique. Can we try to understand the behavior of the system? The answer is yes for small disturbances yes quite easily. So, let us talk about only very small disturbances around the equilibrium. Why are we talking of small disturbances? Because once we have small disturbances we can make certain approximations. So, you take our original equations. So, what I do is d delta by d t is equal to omega minus omega naught which implies d delta e plus delta delta e plus delta delta delta is equal to omega e plus delta omega minus omega naught. So, actually omega e and omega naught are the same delta e is a particular value. So, the derivative of that is equal to 0. So, you have got finally, for small disturbances fine. For small disturbances the other equation this is the first equation this is the second state equation. You have 2 h by omega b into d this is nothing but 2 h by omega b into is equal to p m minus p e. Now, p e is e s e we will assume these to be constant all these things are to be constant sin delta plus delta e plus delta delta. So, that becomes equal to p m minus p e minus equal to p m minus well we can apply the formula for sin delta plus delta delta e plus delta delta. So, we will go to p m minus e s e by x e sin delta e cos delta delta minus sorry plus cos delta e sin delta delta. So, what we have is p m minus e s e by x e sin delta e plus sorry minus e s e by x e cos delta e. So, what we have is p m minus e s e by x e sin delta e plus sorry minus e s e by x e cos delta e. Is this a variable cos delta e and sin delta e no these are cos delta e e is a particular value. So, it is evaluated it is not a variable into delta delta. So, this is using the formula that cos delta delta is approximately 1 if delta delta is very small and sin delta delta approximately delta. So, what we have here is the second differential equation becomes if p m is equal to is a constant then these since at equilibrium e s e sin delta e by x e should be equal to p m these two can get get cancelled. So, what we have here is basically s e by x e cos delta e delta delta. This is a constant value evaluated at the equilibrium point. So, what we have I will call this capital K. So, what we have here is the differential equations for small disturbances around an equilibrium are d delta by d t is equal to delta omega and d 2 h by omega b d delta omega by d t is equal to minus K delta delta. Now, the reason why I got you up to this point is this particular differential equation is actually easy to solve for small disturbances one can get a kind of an exact solution for this differential equation. So, what is the solution of this now how to get the solution of this in this particular case or in a general case is something I will try to tell you later on in this course right now let me just suggest a solution. This looks very much like the equation of a spring mass system m and K where m is 2 h by omega b and K is the spring constant of the spring the differential equations which you get for this are very similar to this differential equations which are obtained. So, let me not derive any solution for this let us assume that the solution for this particular system of equations let us guess it. So, let us say delta delta is equal to a sin omega t I will call this omega n t this is a different omega n from the omegas we have been talking of plus phi. So, what will be delta omega if this is delta will have since this is the derivative phi. So, these are the well we do not know what a we do not even know whether this is the correct solution or not, but if you plug it in it actually satisfies this equation the first one is obviously satisfied if this is true, but you just plug it in this equation. So, what you have is 2 h by omega b you will have omega n square a sin this will be minus is equal to minus K a sin omega n plus phi. So, obviously this is satisfied if omega n square is equal to yeah. So, omega n square is equal to K K by 2 h and this is omega b. So, this is true of course, these equations get satisfied and indeed our guess solution is correct if omega n is taken to be this. So, let us just move on and so have we got the solution yet no we have just said that out of this we have got what omega n should be that is it should be square root of omega b K by 2 h can be positive or the negative square root the things we do not have yet are a n phi we only know that this particular solution this solution is in fact correct. So, how do we get this a n phi well let us say that at time t is equal to 0 we are not at equilibrium we have been slightly displaced from the equilibrium. So, at time t is equal to 0 suppose remember that the whole idea of doing this analysis is to see how we are displaced from the equilibrium if you are not if you are at the equilibrium of course, there are no transients at all because the rates of change of all the states are equal to 0. So, if suppose at t is equal to 0 we have in that case just substituting it in this particular equation delta is equal to a sin phi. So, actually if you look at the equations of this this is delta delta this is delta omega. So, what we have here is time is equal to 0 this is how things will look like. So, if I am given if I give you this delta and omega 0 that at time t is equal to 0 can you take figure out what is phi yes you can find out what a is and what is phi. And therefore, we get the complete solution because if I know what a is in that case and if I know what phi is you know what omega n is. So, I have got the complete solution for how the system behaves. So, this is what we get as the final solution of this particular system. Now, what I have shown you is that this particular system is stable well not really stable it is got an oscillatory motion is it really surprising the answer is no suppose I am at this I am at this equilibrium this is P m this is P e this is delta and I am at this equilibrium. If I give a small push if I give a small push to this system that is if this gets slightly away from this equilibrium because of a disturbance and I am suppose at this point let us say delta is slightly away from the equilibrium and say speed deviation is 0. In that case P e becomes greater than P m the machine will decelerate and try to come back to this equilibrium. Similar thing happens here if you are here P e is less than P m. So, the machine will accelerate and come back to this equilibrium. So, however the motion is kind of oscillatory the restoring torques which are there are in fact proportional for the small disturbances they are practically proportional to the angular deviations and therefore you get a kind of spring mass behavior of the system. Of course, what about this equilibrium in this equilibrium it is not oscillatory if I for example my delta is here speed deviation let us say 0, but my delta deviation is there and slightly deviated from the equilibrium. In that case P e is less than P m. So, the machine will accelerate if it accelerates it actually goes away from the equilibrium point. So, one would expect that this particular equilibrium point is not stable it is unstable. So, how come I got this solution it seemed all fine this is my solution where a and phi are obtained from the initial conditions of that is from this you get a and phi. So, everything is fine. So, this seems to say it is oscillatory where omega n is. So, where am I going wrong I mean this particular equilibrium point is unstable remember that this k if you recall our derivation this k is nothing, but e s cos delta equilibrium upon x e this delta e here is dependent on the equilibrium point. In fact, this particular equilibrium point is greater than 90. So, this k is negative if k is negative omega n is a complex number. So, what you have is omega n is some complex number. So, in that case things get a bit complicated because now our solution is delta delta is equal to a sin a complex number omega n I will call it let omega n be. So, at the other equilibrium delta delta is a sin j remember omega n is complex plus phi which is nothing, but a we can expand this nothing, but e. So, this is nothing, but small values sorry it should be j into j omega t here and it should be again j into j omega t. Remember that this sin x is nothing, but equal to e raise to j x minus e raise to minus j x upon 2 j. So, this is what I have used. So, it is nothing, but so now your solution of delta delta if you are at the other equilibrium in which omega n turns out to be complex the solution comes out to be like this where omega of course is a real number, but omega n of course is a complex number. In fact, it is a purely imaginary number not only a complex number, but it is purely imaginary 2. You look at this solution if this is a real number either this or this is going to be positive. So, minus omega will be positive this or this will be positive. Now remember that e raise to 5 t for example, grows with t. So, and of course e raise to minus 5 t decays. So, your solution for delta delta at the other equilibrium for a small disturbance is going to be a super imposition of these two terms. And if the initial conditions are such that k 1 and k 2 are non zero remember k 1 and k 2 are obtained from a and 5. And if this k 1 and k 2 are non zero you will find that there will be one component of this solution which blows with time it increases with time. So, if you are at the other equilibrium point what you have really shown is that you will be unstable you will really grow with time. So, to summarize at least the small signal part remember what we have done really is analyze the small disturbance stability of a single machine connected to an infinite bus using a very simplified model. We have seen that for a particular small disturbance the system not for a particular small disturbance for small disturbances the system gives an oscillatory behavior for the equilibrium which is lower than 90 degrees. And if it is of course the equilibrium point which is greater than 90 degrees which we have shown between 90 degrees and 180 degrees the system is unstable that is if I give a small push the deviation grows with time. And if the deviation grows with time means it will not come back to that particular equilibrium. So, of the two equilibrium which we have got between 0 and 180 for delta for a given mechanical power we see that one of the system is one of the equilibrium is actually unstable. So, to summarize a single machine infinite bus behaves like more or less like a spring mass system at the equilibrium point delta 1. Now, large disturbance is something we will learn in the next lecture, but let us do one practical in this particular lecture we have been looking at more of the mathematical treatment of a particular set of differential equations. I said at the beginning that although we are going to use a very toy model of the synchronous machine it is actually going to replicate what we actually see in practice a particular phenomena which we see that is the oscillatory behavior etcetera is actually seen in practice. So, if you recall in my previous lecture I had shown you a small a particular graphic of a disturbance it had occurred in tata power system and you could see that there were oscillations. So, these oscillations actually occur in practice one good check we can do right away you know with this toy model is what is the frequency of oscillations what are the frequency of oscillations which we are likely to face. So, for example, you take this particular system we call this E s angle delta this is E angle 0 this is the infinite bus a constant voltage source. We saw that omega n is equal to root of omega n is equal to root of omega b capital K upon 2 h and this is nothing but omega b this k is nothing but. So, if I take omega b for 50 hertz system the base frequency is nothing but roughly 2 pi into 50 which is nothing but roughly 314 roughly h the typical value of the inertia constant of a machine let us say it is 4 mega joules per m v a let us say E s and E b are 1 cos delta E let us say delta E is 50 degrees. So, cos delta E would be cos 50 is approximately equal to cos 45 is approximately equal to 1 by root 2 x c let us say it is a cumulative impedance of the transient reactance of the generator the transformer as well as the transmission line let us say it is 0.5 per unit. So, in that case what we have is 314 by 2 into 4 1 into 1 into cos 50 by 2 into 1 into cos 50 by 0.5 which is nothing but 304 by 8 this is 1 by root 2 that is 0.7 very roughly this is nothing but 314 into 0.7 divided by 4 half of root of 0.7 which is nothing but roughly very very very roughly. So, this is nothing but let us say 15 divided by 2 around 7 7 0.5 radians per second. So, the frequency of these oscillations are going to be roughly around 1 say 1.2 hertz. So, this is basically the natural frequency of oscillations which you will see if the system is subjected to small disturbances. So, if you are at the stable equilibrium point and if I give a push in that case you will get a oscillation. So, the machines are rotating at 3000 rpm for a 2 pole machine or 1500 rpm for a 4 pole machine, but over and above that you will find that the speed of the machines over and above the 3000 rpm or 1500 rpm depending on the pole number you will find that there is an oscillation taking place of 1 around 1 to 2 hertz. So, actually these oscillations have been observed in practice. So, give a disturbance to a machine it slightly oscillates according to the solution which we have got for this particular system. If the system oscillates it keeps on oscillating because you have got a sin omega t omega n t plus 5. So, it just keeps on oscillating, but actually there are reasons for which because of which usually these oscillations die down usually. So, these oscillations die down how do they die down unfortunately to the equations which I have given you in the toy model do not tell you how they will die down because that particular component of the model has not been included. So, the point which I wanted to say here is we have taken a very simple system and shown that give a small disturbance to it and it oscillates. Now, the real intention of doing this particular example was not so much to trying to tell you about the behavior of the electromechanical behavior of a single machine connected to a voltage source, but the real intention was tell you about how we can actually systematically analyze the system. In fact, we wrote down the differential equation we got the equilibrium linearize the system or what you what we say is we saw how we can analyze the system for small disturbances. We guessed the solution now this is one thing which we will try to do later again where we will not guess the solution, but actually derive it we just guess the solution is a sin omega t n omega n t plus 5, but in the next several lectures we will actually see how we can derive the solution for such a system. Interesting thing which came out of this without having to go into very detailed models we could infer that if such a phenomena does occur for typical values of system parameters you will get oscillations of around 1 hertz and these are actually seen in practice. Now, in the next lecture we will just look at the equations for large disturbances I told you large disturbance behavior can be quite different from the small disturbance behavior and we will try to see whether we can infer certain phenomena for large disturbances. So, to sum up in this particular lecture we have really try to see the concepts of equilibrium and small disturbance stability using an example.