 We now continue in this lecture with the analysis of the short circuited generator. In the previous class we just started upon what happens when you have the synchronous machine running at a constant speed under open circuit conditions and we apply a step voltage at the field winding. The voltage which we applied the field winding is such that under open circuited conditions we will get one per unit line to line RMS voltage at the terminals of the stator assuming of course, that the stator is connected as a star is connected in star. Now we did not actually complete that part of the exercise I had written a small psi lap program to see the transient behavior of an open circuited generator. So, we continue with our discussion of an open circuited generator by the end of this lecture we would have come to the short circuited generator and the transient response of a synchronous machine which is short circuited. So, actually what we will be doing is starting two transients we will assume the machine is rotating at the synchronous speed then we give a step change in the field voltage. So, that the voltage builds up in a generator the generator gets excited and then we short circuit the generator and see the transients which arise due to that. Now in this particular study we will of course, not be considering the interaction with the mechanical system we will assume that the machine is running at a constant speed we will assume it is running at the rated or the base speed. Of course, as we shall see later under short circuit conditions the non zero torque is created whenever we apply a short circuit. So, in fact, the speed will be affected in a certain way, but for the time being we do not consider that interaction with the mechanical system we will of course, do that later in this course. In fact, a large part of our course will be dedicated to studying electromechanical oscillation which we shall see later. If you look at the synchronous machine equations in fact, just the flux equations since we are not considering the mechanical equation or the torque equations. The flux equations are in fact, linear and these transients in fact, can be studied by a simple linear system analysis that is using Eigen values Eigen vectors and complete response can be characterized and we can plot the response. We do not have to do a numerical integration because this is a linear system and the response of the system comes out to be in a nice closed form. So, that is what we discussed in the previous class we just revise what we did then we have got the equations of the flux of the machine under constant speed it turns out that a 1 a 2 b 2 will be constant matrices. E f d is a step of 1 per unit recall that E f d is proportional to the field voltage in fact, we apply the field voltage. So, that we will have E f d is 1 per unit or in other words in steady state we will have the line to line voltage of a star connected synchronous generator to be 1 per unit. So, that is the voltage we applied the field recall that we have formulated a model in such a way that we do not really specify the field voltage, but we directly, but we say we give the field voltage in terms of what effect it has on the open circuit voltage in steady state. So, again let me repeat that we apply a field voltage such that E f d is 1 or the open circuited line to line RMS voltage of the synchronous machine is star connected is 1 per unit. So, we assume of course, the speed is constant and we order the states in this fashion. I d and I q in fact, you may wonder what is where is V d and V q here V d and V q are the voltages at the terminal of the generator. What we assume of course, is that the terminal voltage of the generator is written in terms of the current I d and I q and since we are considering the open circuited case we will assume that R L is a very large we will take R L is very large. So, open circuited generator is represented as star connected load with the resistance values R L as very large. Now, so in some sense the effect of V d and this V d and V q is got subsumed into A 2 since we are expressed it in terms of I d and I q. A 1 looks like this we did it in the previous class omega is the speed of the machine. We will assume that in the study in the study this speed is constant and equal to omega base or the rated value of the synchronous generator. I d and I q in fact, are related to the fluxes as I had shown sometimes back the fluxes by a matrix of this kind remember that this is this model is a per unit model. We are talking in terms of x d dash x q dash, but remember that earlier we were talking in terms of L d dash and L d double dash in per unit L d dash L d double dash is equivalent to x d dash and x d double dash and so on. So, just remember that this is a per unit model B 2 of course, is this remember we are using model 2 with the assumption that T d c double dash is equal to T d double dash. The data for this generator I will just show it to you on a slide. So, the data for this is given in this slide we will be using model 2 of course, and T d 0 dash will of course, take as 5 seconds not 8 seconds as given before. Now, what we will do is of course, kind of write a program to plot the values of the transient the transients seen in the voltages currents and so on for this particular transient. So, the transient we are considering is a step change in the field voltage we will of course, do the short circuit subsequently. So, right now we will of course, do the step change in field voltage. Recall that once we have got the equations of the machine in this form which is a purely state space form psi dot is equal to a into psi plus B 2 into E f t. Remember that i is nothing but a 3 into psi. So, finally we get this psi dot is equal to a psi plus B 2. This is a typical input output rather state space form of the machine equations. And we saw in the previous class that you could get the analytical expression for the response in this fashion and that really boils down to evaluating this expression here. So, we do not do a numerical integration because it is not necessary we have got a closed form expression. We will plug in the values of time in order to obtain the final response. In this particular expression p is in fact, the Eigen vector right Eigen vector matrix which we have talked in the first you know the first 10 lectures of this course we have discussed analysis method for dynamical system where I define the right Eigen vector. And this matrix here is a diagonal matrix. In fact, there are in a's of 6 by 6 matrix. So, actually in this case you will have 6 Eigen values not 3 as shown here. And this is defined in this fashion. So, this is where we were last time. What we will do now is of course, I will show you the program corresponding to this particular. Transient. So, I have programmed this of course, using psi lab and we had seen the program in the previous class. We will just have a look at it fast again and then run it. So, let us go to that program. So, we are at the psi lab workspace at the present time. I will show you the program it is called step dot s c i. We had seen it last time we will just run through it again quickly. So, I have entered the data it is quite intuitive and easy to follow I will just rerun it again read through it. Of course, we have given the standard parameters here. We have been given the some the reactances as well as the time constants. We have been given what are known as the open circuit time constants why these are called open circuit time constants will become apparent soon. Of course, the way we have formulated our equations we formulated them in terms of the time constant t d dash t d double dash etcetera. But remember you can get t d dash and t d double dash using these two equations. In fact, you get a quadratic which we have to solve in order to get t d dash and t d double dash and similarly, t q dash and t q double dash. So, that is what is shown on the screen at present from the parameters. We are in fact, obtaining t d dash and t d double dash similarly, t q dash and t q double dash. We have of course, taken the states in this fashion and we have to we take R L to be a very large value in order to mimic open circuit conditions. And this is the A matrix, this is the A 2 matrix, this is the B 2 matrix, this is the A 3 matrix, this is the final state space matrix is A. The command v d is equal to spec of A is in fact, will obtain the Eigen values and Eigen vectors of A. The final steady state value of the states can be obtained by setting psi dot is equal to 0 in which case of course, the steady state value of the fluxes are given by A inverse B 2, which is implemented in this program here. I will just see if I can move this slightly. So, now we evaluate the time response of the states. So, actually it is a direct evaluation, it is not a numerical integration because it is not necessary to numerically integrate to obtain the answers. We will assume the initial conditions before we apply the step R 0 and we actually evaluate the expression which I had just given some time back. Finally of course, I can get once I get the fluxes which are x denoted by x here, I can obtain the currents and the voltages and the torques and so on. So in fact, we will plot the values of torque and the voltages etcetera shortly. So, I will minimize this and now I will actually run the program. So, what I will do is run the program from the psi lab workspace. So, in fact, we are simulating it for more than 30 around 30 seconds. We shall see why we need to simulate this for 30 seconds. So, it is taking a bit of time because I have evaluating it at a relatively short time interval. I mean the if you look at the time I am evaluating this for, I will just go down scroll down the program and I am evaluating it every 5 milliseconds for 30 seconds. So, because of that the program did take some, it did take a while for it to simulate it. In fact, not simulate it, evaluate it at various time instance. Now, the thing is before I go on and show you how the time response looks like, what we first will see are the Eigen values of the matrix A. In fact, if you will try to evaluate the Eigen values, we are getting two large numbers. These large numbers are indicative of something happening very fast. In fact, these arise because we have chosen R L to be a very large number. So, in fact, R L being a very large number results in a very large negative Eigen value. So, this transient in fact, dies down very soon. It is like having a large an R L circuit with a large R L. So, that is why you get these two large Eigen values. The rest of the Eigen values on the other hand are not very large. They in fact, quite small and one interesting thing is that if you look at 1 upon T D 0 dash, it comes out to be 0.2. So, in fact, one of the Eigen values is 0.2. So, this similarly 1 by T D 0 double dash is 33.33 and you find that Eigen value also is minus 33.33. Similarly, T Q dash and T Q 0 double dash. So, T Q 0 dash and T Q 0 double dash. So, if you look at the Eigen values, in fact, they are very close or practically equal to the open circuit, the reciprocal of the open circuit time constant. So, let me put this in perspective. When you have a synchronous machine which is under open circuit conditions, the Eigen values are in fact, the reciprocals of the open circuit time constant. In fact, that is why they are called the open circuit time constant. In fact, I waited for a very long time to actually explain this particular point about why they are called open circuit time constant. I hope it is clear now. In fact, the Eigen values which you get, you can even prove this analytically. We have done it numerically, but the Eigen values are in fact, the reciprocals of the open circuit time constants, which also means that the kind of responses we are going to get are going to be of the kind T upon, I will just write this again in case it was not very clear. You can expect in the response to have of course, provided that these Eigen values or modes are observable in the output, you are likely to have responses, especially in the voltage to contain this and this. In fact, you can also expect, but actually these are not observable in the voltage, these two modes, these two are. In fact, so that is what we can expect. This is something you can actually prove, I will not prove it here, but you can prove that these two time constant will be visible in the voltage response. So, if I actually plot time versus v d, so I will just interpret this for you, in case it is not very clearly visible on your screen. This is 6 e raise to minus 4, so the 6 into 10 raise to minus 4, so actually it is 0. In fact, v d is if you have under in fact, it is showing it to be not exactly 0, because R L we have taken to be a large number, which is not infinity. So, it is not a perfectly open circuited condition, but this is a v d is extremely 0, extremely small I am sorry and it is practically 0. So, v d is under open circuited conditions is practically 0, even under transient situations. In fact, we have proved this about two or three lectures back, where we had considered the open circuited steady state behavior of the machine. Steady state behavior of the machine is in fact, something we did some time back and we did prove that v d is equal to 0 and v q in fact, will be equal to e f d, which is the open circuit line to line RMS voltage in steady state. So, the open circuited, so since we have in fact, set e f d or we have given the field voltage, so that we will get e f d is equal to 1, we should expect that v q will settle down to 1. So, let us just see that yes it does. So, it settles down to the value 1, starts from 0 settles down to the value 1 and if you look at this time scale it is 30 seconds. So, in 30 seconds it takes about 30 seconds to settle down. In fact, if you see the settling time it is approximately 30 seconds. The time constant recall of the open circuit time constant of a machine can be very large. In fact, here I have taken it 5 seconds. In fact, you can have open circuit time constants of the order of 10 seconds as well, in which case the settling time would be approximately 50 or 60 seconds. So, I have taken a time constant of 5. So, 5 5's are 25. So, approximately 5 times time constant is the settling time. So, this e raise to minus t by t d 0 dash is very clearly visible as a slowly increasing you know exponentially rising response in v q. Of course, from v d and v q assuming that theta is equal to omega t omega 0 t, we can obtain v a as well. So, if I plot v a, this is how it looks. Of course, it is a sinusoid, it reaches sinusoidal steady state. So, it also has the envelope rises in the same way as v q, but this is sinusoidal. In fact, if I try to zoom this, let us see if I can, this is in fact the 50 hertz sinusoid, since speed is constant. So, this is a sinusoidal value, I will just assume this. Of course, you notice that this is tending towards point approximately 0.8 and not 1. That is not surprising, because we have we assume that this is a star connected winding and the line to line voltage RMS voltage is 1 per unit. In fact, we have chosen our V F to be such a value, so that the line to line RMS voltage is 1 per unit. So, one can expect that the phase to the voltage across each winding, which is also the phase to neutral voltage will be square root of 2 by 3 into 1 into 1, which is 0.81. That is why this you see the peak value of this is going towards 0.8. So, you see this going towards 0.8. So, this is basically the open circuit response of a generator. Of course, the point is that somebody may ask well I am not seeing the other time constant e raise to minus t by t d 0 double dash in this. Actually t d 0 double dash t 0 dash is 5 and t d 0 double dash is in fact 0.03. So, in fact the reciprocal of 0.03, which is also an Eigen value as mentioned sometime is much much much faster than 1 upon 5. So, that basically the other mode so to speak is not visible very clearly though it does exist. The reason of course, being that it dies down much faster than the transient associated with t d 0 dash, which is the open circuit time constant. Now, t q 0 dash and t q 0 double dash related modes are not observable in the response. So, this is something I request you to prove that for this transient in which I take 0 initial conditions remember I am taking 0 initial conditions. If you take 0 initial conditions you will not see these two terms you know t this related to t q 0 dash and t q 0 double dash appearing in the response. So, you do not see this for step response you do not observe this for the step response, but you do see these two this is not clearly visible because it is very fast it dies down very fast I can call this the dominant mode. So, we move on now from this point to the short circuit analysis of a machine. So, what we need to do here now is consider that the machine is rotating at a constant speed as we have been doing so far. Then what we do is give a step change in V f so that you know build up voltage. Once you build up the voltage and reaches steady state now give it give a short circuit to the generator. So, what we will do is set R L equal to 0. So, once the machine reaches its steady state we will assume that R L sudden we will take R L equal to 0 suddenly. Changing R L of course, changes the value of the state matrix A that is because A 2 changes if A 2 changes then A also will change. So, you will find that the state matrix changes and the Eigen values also change. So, the modes of the system the nature of the modes of the system are different under short circuit conditions as compared to the open circuit conditions. As you may guess well there is nothing to guess really once you have a short circuit there will be a current and there will be the voltage of course, will be 0 at the terminals. So, that is what you will expect to happen. Now what we will do is of course, we will not simulate the complete open circuit conditions. In fact, we will not reevaluate the system under open circuit conditions. What we will assume is that we have given this step and the system is now in a steady state corresponding to open circuit conditions. Then after about 5 seconds we will give the short circuit. So, we will not build up the voltage again we will assume that it is already built up and the system is in steady state under open circuit conditions running at a constant speed and having E F D is 1 per unit that is the line to line RMS voltage is 1 per unit. So, what we need to do is the system we assume just a moment we will just scroll back. This is another program this is not the same program as I showed you sometime back the initial part of the program is the same. We assume that under open circuit conditions RL is a large value. What we do initially is obtain the steady state value of the states for E F D is equal to 1. So, what we will do is assume that the machine is up in steady state. Then evaluate the response assuming that the initial conditions are equal to the steady state values. If we do that of course, if you are in steady state remember that the rates of change of the rates of change of the states are 0. So, we if you are in if we plug in the initial conditions corresponding to the steady state into our expression for the time response we are just going to remain steady. So, after sometime of course, this is for around 5 seconds we remain steady at the open circuit conditions. After that we apply a short circuit remember when we apply a short circuit RL becomes equal to 0. So, that is what given here wonder if it is seen there RL is equal to 0 and then we of course, re evaluate A 2 and the state matrix. We re evaluate the state matrix A and of course, from 5 seconds onwards we use the new A matrix. So, E raise to what we do is really E raise to the A nu from t is equal to 5 seconds onwards. So, if you look at the analytical expression suppose you have got the initial condition rather if you want to express x in x dot is equal to A x plus B u I will just write this again slightly in a large and you have got the value of x at 5 seconds is given. In that case x of t after 5 seconds is actually can be written as E raise to A into t minus 5 into x of 5 plus I have to t E raise to A t minus tau B into u into d tau u of tau into d tau. So, this is basically if you know the initial conditions at 5 seconds and you want to evaluate what happens after at time is equal to 5 seconds onwards. So, what we will do now is actually evaluate the short circuit the short circuit conditions. So, in fact, if you take out the Eigen values of A under short circuited conditions the Eigen values turn out to be like this. In fact, if you look at the Eigen values there is a term here which is minus 3 plus or minus 3 1 3.95 this is of course, a reminiscent of omega in fact, in radians per second. So, if you look at the imaginary part well it is reminiscent of that we will discuss this a bit this issue in fact, this point a bit further later. The other time the other Eigen values are minus 2 minus 34.1 minus 43.1 and minus 1.2. In fact, if you look at the time constants in fact, 1 by T d dash 1.2085 in fact, the reciprocal of the T d dash is in fact, one of the Eigen values of this system 1 by T d double dash is also an Eigen value. So, the 4 in fact, the last 4 Eigen values are the reciprocals of the time constants T d dash, T d double dash, T q dash and T q double dash. In fact, you will expect in the response to see terms like e raise to minus T d dash and e raise to minus T d double dash and in fact, that is why these terms are called a short circuit time constants of the system. So, that is the reason why the time constants are called a short circuit time constants. So, if I actually plot V a or so if I plot V a initially of course, the system is under steady state and under open circuit conditions at time t is equal to 5 you find that this becomes 0. So, you find that at time t is equal to 5 seconds the voltage V d sorry V a for phase a becomes 0. In fact, it becomes 0 for phase b and c also which I am not showing right now. If you look at V q, V q also becomes I am sorry we will have to close this figure first and I will plot it again. So, actually it is not seen very clearly it goes from 1 to 0 at 5 seconds. So, since there is a short circuit this happens V q in fact. So, if you look at V q it becomes 0 from 1 at 5 seconds. So, in fact, if you plot I a on the other hand for example, first what we will do is plot the current I d if you look at plot of current I d this is how I d looks like. So, it is of course, 0 in initially because you are under open circuit conditions and then there is a big jump it in fact, the current becomes a very large negative value and then slowly it settles down to a value here which is approximately steady state value is in fact, approximately 0.55. Whereas, so this is the response under short circuit conditions. So, what you notice of course, is the response consists of several modes. In fact, some the modes of course, are not clearly seen all of the modes are not very clearly seen, but one of the dramatic things you see is that there is an oscillation this is not surprising because one of our Eigen values is a complex Eigen value with a imaginary part equal to omega in fact 3 1 3 you recall that Eigen value. So, this is what is it is manifest here you also see that there seems to be some kind of a exponential decay. So, you know you find an oscillation which decay. So, if you just look at me here you will find that once there is a step there is a step there is a oscillatory value which decays, but also there is a kind of a there is another exponential you know decay here. So, there is a decay of this kind and there is a decay of an oscillation as well. So, actually it is not very surprising because you are really seeing several modes it is a response is a superposition of several modes and the key modes of course, in this case are 1 by T D double dash and 1 by T D dash and of course, the oscillatory Eigen value which we have seen sometime back. So, there is a oscillatory mode we will just you know get the Eigen values again. So, you can expect to see these things I am sorry. So, you are getting going to get a complex pair of modes of course, the whole system is stable because it has got Eigen values with negative real parts. So, in fact, if you plot I a which is the phase current phase current you will find that we will just redraw this we will redraw this I a yeah. So, what you see in I a is in fact is the phase current remember in the phase current you see that although in steady state if you look at things in steady state it is absolutely symmetric the wave form is symmetric it is a symmetric sinusoid as one would expect in transient during transients you see a DC offset as well as the envelope of this I a is large. So, what you have as the response of I a if you look at me you will find that the envelope of I a decays with time the envelope of I a decays with time, but also the envelope itself has got a DC offset. So, you see a DC offset like this and you see an envelope which is also dying with time. So, you have a huge envelope in the beginning that envelope itself decays and also the envelope is offset and it is coming down like that. So, that is the typical response of the phase currents in case you suddenly short circuit generator. So, this is what is a typical response of a short circuited synchronous machine. Now, one of the things which we need to which we can of course, see here is that we have done this before the steady state value of I d is nothing, but minus of E f d in steady state divided by x d you know this is what one can easily prove for a short circuited generator in steady state this I leave this as an exercise and I q in steady state is going to be equal to 0 this is something which you can prove very easily. So, let us just verify that this is true. So, we will just check out that I d if look at the value of I d although there is an initial we will have to we will just close this and redo it. So, the steady state value of I d is approximately minus 5 5 and if you actually look at E f d is in fact 1. So, 1 divided by x d is also 0.5 5. So, what we are seeing is that this of course, the transient finally, becomes the steady state and in steady state x I d is minus E f d by x d dash x d I am sorry it should be x d. If you look at I q the steady state value is 0 does not mean of course, that the transients values are 0 and you see this here you see that there is some transient and this becomes finally, 0 in steady state. Now, we move on to trying to understand some other aspects of a short circuited generator. In fact, if you look at the initial value just after the transient has begun that is you know just at the time of short circuit the currents are very large. If you look at I d or I q is of course, 0, but I d in fact is quite large you look at this the steady state value is only 0.55, but if you look at the initial transient you see the peak value goes right up to minus 8. So, initially the current is extremely large. So, this is in fact called a sub transient period this initial few cycles of the fault just after the short circuit has occurred is called a sub transient period of the machine. In fact, if you look at you see that the peak value becomes as low as 0.8 and the mean value is approximately if you look at this figure this the mean value you can say since this is a oscillation or talking terms of a mean value the mean value is approximately minus 3. In fact, it is approximately like this. So, if you take out the mean value something like this. So, if you look at if you look at the curve of I d you will find that it is initially 0 and then it becomes and then there is an oscillation the envelope of the oscillation dies down there is a transient and then it settles down to the steady state value which is E f d x d. In fact, the mean value of this initial transient let us call it this is roughly E f d by x d double dash. So, let us just verify that what we are seeing from this figure here is approximately 3.5 minus 3.5 this is approximately minus 3.5. So, actually let us see what 1 divided by x d double dash is. In fact, this minus it is 4.3 this is of course, because I cannot make out the mean value very easily from this figure, but what you notice is that in the initial period the current has got an oscillatory component and the mean value is approximately 3 to 4 minus 3 to 4 I d value is minus 3 to 4 this is because of this in fact is the initial behavior just after a short circuit. In fact, just after a short circuit one can say that the machine currents are determined roughly by the sub transient reactance of the machine. In fact, x d double dash is called the sub transient reactance of the machine. So, the sub transient period the machine experiences a much larger current compared to what it sees in steady state. In steady state in fact, the current is 0.55 it is even lower than 1 per unit the rated current of the machine, but under sub transient conditions it can be quite high. Now, one of the things you will notice in the Eigen values corresponding to this we will just see the Eigen values again. You find that there is a clear separation of Eigen values. If you look at the Eigen values the first two have a very large magnitude. In fact, the large they have a large magnitude and this omega the imaginary part is in fact, equal to omega the rated speed of the machine it is in fact, roughly 340. So, that is the radiant frequency corresponding to 50 hertz. Now, it is not surprising in fact, that the magnitude of it is quite large compared to the other Eigen values the magnitude of the first two Eigen values the complex pair. So, in fact can we you know do a fast and slow state decomposition of this particular system it is an interesting possibility right. So, if you look at the state matrix A 1 in fact, you have got omega minus omega and omega here. So, you can guess I mean it is just a guess that the Eigen values corresponding to the first two the first two Eigen values in fact, which have got an imaginary component omega are probably associated very much with the first two variables that is psi d and psi q. In fact, you can formalize this by looking at the participation factors corresponding to the states which we discussed this in the first in around the 8th or 9th lecture of this course. So, you can actually look at the participation factors corresponding to this these Eigen values see the participation of various states, but we can guess we will not get into the complete you know study of the participation factors, but you can guess that psi d and psi q are primarily associated with this Eigen values corresponding to the complex pair of Eigen values the first two ones. And that is because just a bit of guess work if you look at the state matrix it is got in fact, for the first two states this omega minus omega and omega in the off diagonal elements of the first two states. So, that is just a guess, but it is an interesting exercise to actually prove this formally using participation factors and I encourage you to really do it. So, actually an interesting possibility is of course, to do a simplified analysis and assume that these two states you know you set to 0 you know you have got in fact, we will call these the first states because they are associated with we guess that they are associated with the Eigen values of large magnitude and these in fact, are the slower states. In that case we can do a fast and slow decomposition as I had mentioned in the first 10 lectures when we are studying the dynamic analysis of the analysis of dynamical systems. We had discussed this issue of fast and slow states and one of the things I told you that if you have got a system consisting of fast and slow states x fast and x slow then one can actually get the response of the slow states without much error if one assumes that you can set this equal to 0 you set this equal to 0. So, you can in fact, express x f in terms of x s and finally, x s dot will be a s s minus a s f a f f inverse a f s. So, you can in fact, study this system as a lower order system of the slow states in fact, you can we do this in this case yes you can in fact, I will just run through the code in which I show you that this I just programmed it will try to understand it a bit more in detail in the next lecture. I will show you the code for it I have basically I am doing the simulation of a reduced order system in which I have treated psi d and psi q as fast variables and basically converted the differential equation corresponding to these fast equations into algebraic equations and then studied a lower order system. So, I will just look at the code which does this same code what I have done is we will scroll down it is the same code except under short circuited conditions what I do is I form a reduced a matrix by neglecting the fast transients. So, this is something I do so I call this a old as I just store the old a matrix in a old and I form the reduced a matrix this is just a 4 by 4 matrix in the slow variables which are the variables from states 3 to state 6 that is from psi h psi f psi g and psi k and I do the simulation of this system. So, I will just slowly go through this code and I will run it once and we will discuss it again in the next class. So, what I am doing is using a reduced order system to do the simulation I would not call it a simulation the evaluation. So, for example, I can execute this program this is a modified short circuit program with a reduced order state space by eliminating the state equations corresponding to the psi d and psi q. So, wait for a while the Eigen values of the reduced model are these in fact they are not very much different from what we had got earlier from the larger model. So, our fast and slow decomposition and simplification does not cause a very large loss of accuracy as far as the slow states are concerned and of course, if you plot for example, time versus I d what you notice is we will of course, have to do the old figure I will just redo it again we get a response which is sure enough the oscillatory transient which we had seen earlier. So, we actually get the same responses before, but you initially recall that there was an oscillation of frequency omega b now you do not see it. We shall use this model in the next lecture we will just start of the next lecture with this model with this reduced order model we will re look at what results we have got and thereafter we will consider a few more cases of a synchronous machine. So, with this we stop today's lecture we have really this is in fact the first real power system analysis dynamic analysis analysis we have done in this course using the model which we have developed. So, we will of course, continue on this on this theme in the next class with the reduced order model thereafter we will also in this course go and model a few more elements before we can actually talk in terms of doing an analysis of an integrated system and mind you the integrated system though it will be much much more complicated it is amenable to a you know sustained and scientific attack as we have done for the short circuited synchronous generator.