 we had commenced the simulation of a single machine connected to a voltage source through a reactor. We had in fact, just started that the main point to be noted in that was the behavior of the AVR. Now, AVR of course, is the automatic voltage regulator. The simulation studies of interest are the step response of the AVR to a step change in the reference voltage of the AVR. Also, we need to see in the simulation how with changing load conditions whether the AVR is able to regulate the voltage at the terminal of a synchronous machine connected to a voltage source or an infinite bus. Of course, the important difference between the simulation we are carrying out now and what we carried out a few lectures back was that the synchronous machine is not directly connected to a stiff voltage source, but it is connected via a very a kind of a toy model of a transmission line. And the transmission line in fact, is a model just like a reactor. Now, if you look at what we are trying to do the synchronous machine was connected via a transmission line to another stiff voltage source is the infinite bus. This is the transmission line, this is the synchronous generator, this is the terminal voltage of this machine was being regulated. So, we took the feedback of the voltage at this point and fed it to a voltage regulator of this kind, simple proportional kind of voltage regulator and the excitation system was assumed to be a static one. So, the model of it with normalized quantities is simply a plant with a gain of 1 and the limits of the static excitation system of course, were taken this is fed to the field winding and of course, the feedback here is the magnitude of the voltage which is root of V d square plus V q square. We assume there is no measurement delay here, we could if we wished a model this delay by a first order transfer function, but we will not do that right now. The simulation was is being carried out with the following steps, first the synchronous machine is synchronized to this voltage source. It is in fact, it is a kind of a bumpless synchronization we have talked about synchronization of a synchronous machine before. So, I will not talk of it again right now, initially the voltage other the power of course, initially is 0 at the time of synchronization and the open circuit voltage is 1 per unit and the machine is running at practically the rated speed. Now in order to create 1 per unit at the terminals of the machine, you need to have E f d is equal to 1. So, the output of the excitation system should be should give a field voltage which results in the terminal voltage being at its rated value at rated speed. So, E f d is 1 now to create this E f d of course, E f d is within the limits because the terminal voltage being 1 itself. So, 7.0 into 1 and minus 7.0 it is within the limits. So, it is not getting clip and to create this E f d you have got an automatic voltage regulator V ref and this is V which is nothing but root of V d square plus V q square. Now, since this is equal to 1 this is 1 the gain is k a. So, it follows that the value of this error here is 1.0 divided by k a. So, V ref in fact is nothing but V plus E f d by k a. So, one thing you should notice this is a proportional controller there is some steady small steady state error. In fact, the steady state error is less if k a is large. So, V ref and V are not exactly equal if you got a proportional controller you would need to have a small steady state error in order to get a non zero E f d. So, this is something which you should keep in mind. So, V ref can be back calculated. So, what is the V ref which results in V being equal to 1.0 under open circuit conditions well it is equal to this. So, V ref is slightly greater than what is required that is what V be required. So, if this is 1 this will be E f d which is non 1 by say if k a is 200 which is a typical gain per unit by per unit then your V ref is equal to 1 plus 1 upon 200. So, that is equal to 1.005. So, you see that V ref and we are almost the same provided k a is large which is indeed the case k is normally quite large. So, in our studies right now we will assume k a is 200 P a the time constant here is we will take it as 0.02 it is a small value and once we synchronize the machine there is no power flow. So, actually nothing changes because you have just had a bumpless synchronization the conditions on the machine do not change. However, if I increase the mechanical power out input for example, initially of course, before synchronization the machine is simply rolling. So, if I increase the mechanical power after synchronization say I give a step change it is not possible to give a step change to the mechanical power, but let us just take it as a mathematical idealization. If I give a step change in the torque electrical torque you will find that the power flow through this increases and presumably it will remain in synchronism. So, the speed of this and this will be the same these two machines this machine and this voltage source is going to be the same that is in spite of the change in power the synchronous machine tends to remain in synchronism with this bus, that is a important change as soon as some current starts flowing that is some power starts flowing the voltage here tends to drop. So, if you did not have a automatic voltage regulator what would happen is this voltage would it would drop and E f d would remain the same, but with the voltage regulator as soon as V drops V ref is of course, the same right now V ref drops E f d is increased. So, this voltage here is regulated and E f d is not a constant. So, an automatic voltage regulator saves us the trouble of trying to you know manually trying to change E f d every time we change the mechanical power it does it automatically by satisfying this regulation function. So, that is the first test signal we will give to this machine the second test signal we will give is give a step change in V ref keeping torque constant. So, what we will do is increase this reference voltage of this A V R that is I am going to increase this voltage. Now, you know that if I increase this voltage will the mechanical power or the electrical power be affected no what instead will get affected is since this voltage is increased the reactive power output of the machine will change. So, this Q here will change. So, this is one important thing which we will see now in our simulation. So, today's lecture we continue the simulation of automatic voltage regulator in fact, it is a system which contains an automatic voltage regulator. So, what we will do is first synchronize the machine it is a bumpless absolutely smooth transfer smooth interconnection there after that is of course, done at time t is equal to 0 and there after we we increase the mechanical torque of the machine we increase the mechanical torque of the machine and there after we increase the reference voltage of the A V R to 1.05. Now, since this is a absolutely smooth synchronization we have assumed that the rated speed of the machine the machine is running at exactly the speed which is equal to the infinite bus and the line to line RMS open circuited voltage is equal to the voltage which appears right at the point where the circuit breaker has to be closed. So, you can see that there is going to be a bumpless transfer of the magnitude of voltages the phase angle also is 0 at the time of synchronization. So, what we will have is a bumpless synchronization first at time t is equal to 0 we will not see any transient because it is absolutely a smooth synchronization we will do the simulation right away. So, I have already written the program. So, this time I will not show you the program it is simulating the system using Euler method Euler method can be used only if I remove the stiffness out of the system remove the stiffness in the system. So, I have neglected d side the stator transients at d psi d by d t d i d by d t and so on. So, this is something we have explained before in an earlier class. So, what I will do is concentrate more on the results this time. So, let us just see what happens to the speed. So, I will plot yeah. So, what you see here is almost a bumpless transfer here synchronization here at this point I give a step change in the mechanical power. If you increase the mechanical power in this system the infinite the machine does not lose synchronism. In fact, the step change in torque is 0.5 per unit from 0 to 0.5 per unit the machine remains in synchronize in the sense that it the speed of the synchronous machine all the way deviates in transients. There is a this swing which you see a low frequency swing it tends to come back to synchronism the speed tends to come back to synchronism at 10 seconds I have given again another step in the voltage reference of the synchronous machine. So, you see that in both cases the machine tends to remain in synchronism that is the mechanical the speed of the machine equals to the frequency of the infinite bus. Now, you can also see various other parameters for example, delta if you look at the delta of the machine this is of course, in radian the machine is synchronized when delta is equal to 0 and speed is equal to 0. So, you have got this bumpless synchronization right in the beginning and as soon as you give a step change in torque the phase angle of the synchronous machine the delta of the synchronous machine rather it changes. You see this swing which is also observable in speed it is also observable here. So, you see the swing this is one particular mode which is associated with the electromechanical variables delta and omega especially also when you change V ref you find that delta actually slightly decreases. Now, why is that so you have not changed the mechanical torque at this point you have changed delta to 1.05 from 1.005. So, if you make the step change in V ref delta reduces because E F D has changed once you change V ref you are effectively changing E F D also. So, that is why delta changes you again see a swing and with settles down to a steady state value. So, the delta settles down to a steady state value. So, if you actually plot the mechanical power rather the electrical power output of the generator which is V D I D plus V Q I Q you will find that we will do it. So, what you see here is of course, the mechanical power changes to 0.5 the mechanical sorry the electrical power follows the mechanical power you should remember you have got an infinite bus the synchronous machine is connected to an infinite bus. So, if the speed remains finally, the machine remains in synchronism and mechanical power becomes equal to the electrical power. So, you see that the mechanical electrical power output P is equal to the mechanical power it eventually becomes 0.5. If I change the V ref of the machine if I change the V ref of the machine the reference voltage of the AVR of the machine you find that the output electrical power P does not change this is not surprising after all by changing the reference voltage of the machine you are not changing the mechanical power. The mechanical power becomes equal to the electrical power in steady state if the mechanical power does not change the electrical power does not do. So, this is an interesting point here what if I plot Q Q is the reactive power output of the machine. If you look at the reactive power output of the machine if I if of course, the machine is initially synchronized there is no real or reactive power output at the point of synchronization. But, if I change the mechanical power the electrical power also changes F D also changes and you will find that the generator in fact, supplies a bit of reactive power there is some reactive power output of the generator. If I change V ref on the other hand if I change V ref what you are doing is not changing much the real power is not changed, but the reactive power changes quite substantially. So, by changing the reference voltage of the AVR we are in effect in effect changing the reactive power output of the generator and of course, the terminal voltage also will change under these circumstances. So, let us just look at the terminal voltage of the machine. Remember that the first disturbance or the first step a change is that of the synchronous mechanical torque at 10 seconds we are given a step change in the reference voltage of the AVR from 1.005 to 1.05. So, let us just plot the terminal voltage I think it is called V G. So, if you look at the terminal voltage well I have plotted something wrong I will just just a moment we will just look at the variable which is corresponding to it just slipped off my mind it is V GEN. So, we will just see what V GEN is this is the terminal voltage of the synchronous machine. You see that the terminal voltage of the synchronous machine initially is 1 when I increased if I increased the real power output of the machine by changing the mechanical power at 0.5 seconds you notice that the voltage slightly drops. Now why does the voltage drop that is the important point which you should ponder upon we will try to address it here. What happens actually is that when you have got a synchronous machine under open circuit conditions there is a certain field voltage which will produce 1 per unit at the terminals. Once you load the machine on the other hand that is you increase the mechanical power and therefore, the electrical power what happens is that the terminal voltage tends to drop now because the terminal voltage tends to drop E F D is increased. So, if E F D is increased V ref is the same. So, let us just do this you know kind of a cause effect kind of analysis mechanical power increases electrical power also tends to increase because of that if electrical power increases there will be from current 0 there is some current. So, current increases because current increases you will find that the terminal voltage tends to drop the terminal voltage tending to drop will force this automatic voltage regulator to increase E F D. So, if E F D is increased it also means the error has changed V ref we have not touched right now we are just giving a step change in mechanical power. So, what happens is effectively if the error in steady state if E F D required is more we will have to drop more this is because we are using a proportional controller and steady state error is non zero. So, to get more E F D you will have the steady state error will have to be more. So, V slightly decreases. So, once you load the machine since you have got a proportional type automatic voltage regulator you find that the voltage slightly decreases of course, if I give a step change in the V ref itself. So, if you look at what I am done here now now E F D is a certain value now V ref is increased if V ref is increased of course, you will find that the error here will have increased if error increase E F D will increase as a result of that V will increase, but remember that V ref and V are not going to be exactly equal that is why even though if I made V ref is equal to point for 1.05 at 10 seconds you find that actually the voltage settles down to around 1.0425. So, this is a kind of interesting look at the simulation it should correlate well with our steady state analysis. If you look at of course, this is something we have discussed even before you have got you should keep an eye on this swing. Now, if you look at this swing which is there which is seen in almost all the quantities it is a very prominent mode you will find that it is an oscillation of roughly 2 hertz slightly less than 2 hertz. Now, we will also look at E F D we have not looked at E F D. So, let us see how E F D behaves. So, look at E F D E F D once you give a step change in V ref sorry mechanical power E F D has increased from 1 which is the value required to get 1 per unit at the terminals of the generator under open circuit conditions or no load conditions. If I increase the mechanical power if I want to get 1 per unit at the terminals I will have to increase E F D and that is what exactly the voltage regulator has done it has increased V F D from around 1 to approximately 1.3 or 1.2. Now, if I give another step if I give the step change in V ref you find that E F D again increases of course, that is true because if I want to get a higher voltage I will have to increase my E F D. Now, remember that once say the moment I give a step you see that the change in E F D is quite large why is that. So, that is because this is extremely high gain system remember I told you in order to get a very good response of the voltage regulator we do use a very high gain in our automatic voltage regulator this is because in order to offset the relatively slow response of the field winding what we do is we give a big push initially to the field voltage. So, that to get the field winding moving in the sense that you change the field current. So, that is why you provide for a very large when you are designing an excitation system in AVR you provide for large gains and a fairly large range in which E F D can vary normally E F D need not vary more than under steady state conditions more than 3 per unit 1 to 3 per unit is all what is required. But normally the ceiling is kept at more than 7 or 7 or around 7 times the terminal voltage. So, the ceiling voltage is around 7 per unit plus or minus 7 per unit. So, of course, if I give a sudden change like a step change in V ref the field gets forced, but of course, it cannot exceed 7. So, it kind of gets clipped at near about 7. So, what you see here is in this simulation is that the E F D is try to increase to a very high value in order to get a very good response from the sluggish field winding. But it is clipped at 7. So, you see this almost flat top here you see this and of course, once the field winding gets going there is no need to have such a large E F D and E F D finally, settles down to a more comfortable value which is roughly around 1.7. So, this is an interesting point. So, this is the showing the effect of the limiter the clipper which is there. You can also have a situation if I give a step change in the reverse direction of course, E F D will decrease. So, I will not do that simulation, but you can well imagine what will happen and in fact, you can try out this doing this particular simulation is very it is an interesting simulation. Now, let me tell you something which is very very interesting and important. Suppose, I if you recall I have written down this program and disturbances which I have given are at t greater than 5 seconds I give a step change in mechanical power at t greater than 10 seconds V ref is given a step. So, what I will do is I will not give a step change in V ref I will keep it as it is, but at 10 seconds I will give an additional step to t m. So, I will put mechanical power is 1 per unit which is at rated speed equal to the rated power output of the machine. So, what I am done is I am trying to load the machine completely. Just one small point which I need to reiterate here is that normally one cannot give such step changes in mechanical torque this is just an idealization it is just a kind of a toy simulation in that sense. In later classes we shall see about the modeling of the prime over systems themselves and you will note that you cannot give such step changes normally. So, let us just try to redo this simulation in the meantime I will just close this graphics graphical window which is of the previous simulation. So, if you look at the simulation now if you look at for example, just look at speed there is an interesting problem which seems to have arisen now is that the speed does not seem to settle down. In fact, the machine to for all practical purposes seems to have lost synchronism you see that the speed is increasing with time and if you plot for example, delta also we will of course, close this and redo this. So, you see delta it does not seem to be reaching steady state it seems to be growing with time. So, these oscillations are growing with time. So, why is that happening I mean that is an interesting point which you should ponder on I mean it is going on increasing with time. In fact, if I simulate for a longer time this is a simulation for 20 seconds if I make it a 25 second simulation let us see what happens. Simulate for a slightly longer time and redo this close this. The delta seems to have kind of is not settling down it in fact, if you look at what is happening is that it is increasing. So, this looks as if this particular operating point or equilibrium point which it should settle down to does not seem to be stable it seems to be small signal unstable. So, that is an interesting thing. So, the feedback system which we have got seems to seems to be suggesting seems to be creating a situation for certain operating points which leads to a oscillation which does not seem to be dying down. So, if you in fact, maybe if I give a make V ref also increase at the same time let us see what happens let us do another simulation. So, what we saw in the previous simulation was that there is a oscillation in delta, but it is it is kind of increasing with time it is not you know usually we expected the delta decrease and you reach steady state. So, by giving step changes etcetera we were expecting that things would reach steady state, but it appears that there are situations in which steady state is not reached. In fact, this seems to be even worse your angle seems to have increased even more. So, or perhaps it is more or less the same, but you see that this is also unstable. So, actually what we are effectively coming to a situation where the equilibrium point under our study appears to be unstable. So, what we should expect is that at this operating point when we do a linearized analysis of the system we should get Eigen values with positive real part. So, that is what we need to correlate in our analysis now. So, let me just again point out what I want to say we are having a situation when we try to change the power from 0.5 to 1 we see that the system is not settling down to the new equilibrium point. Now, what it what this means is that why is it. So, actually why should the behavior depend on the equilibrium point well one of the important points you should note is that this is a non-linear system. So, when we do a linearized analysis we will be linearizing the system around each operating point. So, when we do the linearized analysis around different operating points our Eigen values going to be are going to be different. So, while step changes to 0.5 T m is equal to 0.5 or v ref is equal to 1.05 which we did in the previous simulation were all stable. We see that in trying to come close to T m is equal to 1.0 there seems to be a problem. So, the operating point to which we are trying to get to appears to be not stable not small signal stable itself. So, what we need to do here is try to understand this instability of the swings you see this low frequency swing which we expect to be stable and to die out is growing with time because the feedback feedback control system of the AVR. So, if we kind of try to do a linearized analysis how do we go ahead. Now, linearized analysis is something we have done before I will just quickly retrace some of the steps. So, what we have in our equations of a synchronous machine if you look at the equations of a synchronous machine these are in fact non-linear. So, we have done the linearization of such systems before. So, what we need to do is instead of having a set of system x dot is equal to f of x what we need to do really is do a linearized analysis around an equilibrium point of this system. So, if we have written down our equations in this form we should now see we can do a linearized we have done a simulation of this system this is a full non-linear system. We can try to understand the stability of the equilibrium point by doing a linearized analysis in this form. Remember that when doing the linearized analysis we have to take we have to linearize it around an equilibrium point. So, there is one important thing is first thing is have to linearize. Linearize in the sense use Taylor series expansion of the non-linear functions which appear here that is called linearization and evaluate those you know after truncation of the higher order terms you evaluate this at the equilibrium point. So, second thing is you have to take out the equilibrium points. So, linearized analysis would require first of all you to take out the equilibrium point and the second part is you linearize around an equilibrium point is it. So, this is what you need to do how do you compute the equilibrium point. So, in fact long ago I mean I am sure it is not too long ago you must have done the load flow analysis. So, load flow analysis in fact is a kind of an equilibrium analysis of the system, but of course that is the in some sense the starting point of computing any equilibrium. In our case we have got a very simple system in fact computing the equilibrium points is very very easy. So, let us just go ahead and see how you can do it. Suppose I want to find out this the equilibrium values of the states corresponding to this scenario. You have got this transmission line which has got a plane reactance x this is your infinite bus which we take as 1 angle 0 that is I have already defined what the infinite bus voltages are V a n, V b n and V c n in the previous class. Let us say the terminal voltage of the machine is 1 per unit it is near about 1 per unit and T m is equal to 1. If T m is equal to 1 and this is at rated speed then it also means that electrical power output of the machine we have course neglecting the losses in the resistance of the synchronous generator they are really very small. So, compared to this p. So, we will assume that electrical power output at the terminals of the machine is same as in per unit as T m because the speed is equal to the rated speed. We will assume that the frequency of the infinite bus is also equal to the rated frequency. So, if p is the output of the machine this is 1 per unit what is the phase angle of the terminal voltage of the synchronous machine that can be easily found out you know that for the system with reactance x voltage magnitude 1 on both sides the phase angle of the voltage here theta is such that p e is equal to 1 per unit is nothing but voltage magnitude at both ends divided by x this is of course, a multiplication sign and this is it looks the same. So, we will just make this larger here into sin of theta. So, theta you can calculate theta from this. So, what let me just remind you what we are doing we are computing the equilibrium point corresponding to this operating condition and thereafter doing the linearized analysis. So, theta is sin inverse p e this is at equilibrium divided by 1 into 1 divided by into x. So, this is theta we also can compute the current flow through this once you have got theta you can get the current. So, the current phasor can also be obtained. So, once you have got the current phasor. So, current is equal to v angle theta minus e angle 0 e is 1 per unit e is the infinite bus voltage divided by j into x. So, this is effectively the current I once you get the current I we are effectively can get the what the current looks like. So, what we have effectively done is if you actually look at if you look at what we have done in another way what we have done effectively is set the right left hand side d by d t is equal to 0. And therefore, we have got i d i q i d i q. If you look at the equations which we saw on the screen you can have a look at them again 0 minus x x and 0 is equal to minus of v d v q minus e d e q. Now, this is effectively the same as saying i q you can just write this in very compact form i q plus i d is equal to v q plus j v d minus e q plus j e d divided by j of x. You can just expand this and show that this is the same as this. So, what let me tell you what we know that we know what we effectively get is this particular equation. That is why it appears like this. So, the point here is that you have got one angle 0 these are the voltages of the infinite bus e a n e b n and e c n. So, what does it mean that e a n is equal to root 2 by 3 sin omega naught t and e b n and e c n is equal to root 2 by 3 sin omega naught t plus theta minus 2 pi by 3 and e c n is minus 4 pi by 3. So, it means v a n is equal to root 2 by 3 sin omega naught t plus theta and v b n is of course, root 2 by 3 sin omega naught t plus theta minus 2 pi by 3 and e c n is minus 4 pi by 3. So, the point is that if I know theta I can tell you what v a n, v b n, v c n look like as a result of it we can get what v d and v q look like. That is the most important thing which you should get out of this. See if you know e a n I told you that e d is equal to minus e times e is 1 in this particular case sin delta and e q this is plus e cos delta. So, v d v q is therefore, if v is equal to 1.0 angle of theta what it follows is that v d is equal to and v q is equal to 1.0 sin of theta minus delta and 1.0 of cos of theta minus delta. So, we have got v d and v q and e q and e d. How did we get v d and v q? We computed theta, theta is obtained from the initial condition. So, in effect we have obtained what v d and v q should be. If I know v d and v q I can get i q plus j i d or equivalently i d and i q separately from this. So, if I have got i d and i q v d v q and e d e q the next step is to try to back calculate what is the value of delta omega and all the other fluxes. Now delta omega is equal to omega naught in steady state. So, if we are talking of an equilibrium point omega will be equal to omega naught. So, we are taking out an equilibrium we are doing an equilibrium analysis we are back calculating the equilibrium conditions. Under equilibrium conditions psi d is equal to x d into i d plus e f d. How does one get that? Well, what you have to do effectively is set d f dot d psi f by dot by d t is equal to 0 and d psi h by d t is equal to 0. So, if you set this equal to 0 you will get psi f and psi h in terms of psi d. After that replace it in the algebraic equation which relates psi d with i d you will effectively get this particular equation. Similarly, in steady state this is not true in general this is true only in steady state please remember psi q is equal to x q i q similarly you will get psi q. So, under steady state conditions this is what you get. Now, we know that in steady state condition d psi d by d t is equal to 0. So, what we will get is 0 is equal to omega into psi q under equilibrium conditions omega is equal to omega naught this is equal to the base value minus omega b r a i d minus omega b into v d. Similarly, you will have sorry this should be minus omega square plus omega psi d this is got by putting d psi d by d t and d psi q by d t equal to 0 which is true under steady state conditions. This is minus I am sorry this should be minus i d minus omega b v q. So, what happens is if you assume resistance is very small of course, if you have the value there is no harm if you assume that it is very small what we will get is v d is equal to v d is equal to omega b and omega are the same we will assume that the speed of the infinite bus is the same as the base speed in such a case v d is equal to minus psi q and v q is equal to psi d. So, this is one important thing which we get. So, what we have is v d is equal to minus x q i q and what we have here is v q is equal to x d i d plus e f d. So, if we have this particular equation what a simple thing we can do is we have got v q plus j v d is equal to e f d plus yes you can write this is j times x d into x d into v d plus v d plus v d plus v d plus v d plus v d plus v d plus v d plus i q plus j i d minus. So, this should be minus here e f d minus j into x d into i q plus j i d. So, what I have done is just added up these equations and written them in a particular compact fashion. So, what we have here is so what you have is here is j this is plus j this is plus j x d x. So, what you have is v q plus j v d under steady state is equal to e f d minus j x d i q plus j i d minus j x q i q plus j x d i q. An alternative way of writing is this is e f d minus j x q into i q plus j i d and instead of so instead of x d you write it as x q and as a result of which the additional term you will get is. So, you can write it in this fashion also. So, the basic crux of the matter is that if I know v q plus j v d. So, I will just write this down again we will have v q plus j v d is equal to e f d minus j x q plus x d minus x q into i d. So, just so what we get from this is that if I know this and I know this. So, if I know this and I know this I can get this, but remember I do not know delta remember I do not know delta. So, I cannot actually find out what v d and v q is even if I know what theta is. Similarly, I cannot get e d and e q. So, what I do is a very interesting thing e f d I am multiplied by e raise to j delta. So, I will get v q plus j v d into e raise to j delta on both sides I just multiply e raise to j delta I multiplied by e raise to j delta. Now, the question is do I know this the answer is yes, because I know theta. So, just you can just try to prove that if I know v d and v q sorry if I know that v d and v q have this form v q plus j v d will e into e raise to j delta will simply give me what 1.0 and theta is something I know. So, what we have essentially is that I know this I also know this. So, if I know when I know this I mean I know this complete quantity here and I know the complete quantity here as a result of which I know this complete quantity. Now, this is a real number. So, if I take out the magnitude of this something I know and this is something I know I can take its magnitude and angle and equate it to this as a result of which I will be able to get e f d and I will also be able to get what delta is. So, the thing which is important to note is v q plus j v d e raise to j delta is nothing, but 1 angle theta. Similarly, actually if you look at the infinite bus voltage also it is very easy to show that this will be 1 angle 0 as per our initial situation which we have talked of. Similarly, i q plus j i d into e raise to j delta is nothing, but this minus this divided by j x. So, we know this complete term and this complete term we know the value of x q as a result of which we can back camp compute the equilibrium value of delta. So, you can directly get what the equilibrium value of delta is once you got the equilibrium value of delta. The next step is get i q plus j i d remember we know what i q plus j i d into e raise to j delta is what is that equal to 1.0 angle theta minus 1 angle 0 divided by j x. So, if I know delta I can actually find out what i q and i d are once I find i d and i q the next step is from i d you know this value the magnitude of this by equating these 2 compute e f d. So, you can compute e f d so now, you have got e f d initial value delta and once you know e f d and delta. In fact, you can compute all the equilibrium conditions. So, what we have is let me just recapitulate what we have done it may have sounded a bit complicated, but it is not that as complicated as it sounds it will just recapitulate quickly. If e a n is this it means of course, that e d and e q are like this which also means e e q plus j e d e raise to j delta is nothing, but 1.0 plus j 0.0. Similarly, v a n is this so what you get essentially is theta is something you get from the condition that 1 per unit power is flowing through a reactance of x the terminal voltage here is 1 point angle 0 and the terminal voltage here is angle theta. So, this theta is obtained that way so once you have got theta the next step is of course, getting the equilibrium values of delta etcetera remember that once you have got theta you know this you do not know what delta is, but interestingly v q plus j v d into e raise to j delta is nothing, but a function only of theta. So, if you know theta you know v q plus j v d into e raise to j delta. So, this is how you get v q plus j v d into e raise to j delta i q plus j i d into e raise to j delta is nothing, but this. So, we get the values of i d and we can get because of this from this particular equation because we know this and this. And the value of this parameter x q we get this value that is the magnitude and angle can be obtained. So, we know the magnitude of this is equated to the magnitude of the left hand side the right hand side and left hand side magnitude equate. So, we know what e f d plus x d minus x q into i d is there after we also know from the angle of this what delta is once we get this delta we can compute we know what this is i q plus j i d into e raise to delta j delta we know once we know delta we can find out what i q and i d are. So, in that fashion we can get the equilibrium values of all the states corresponding to this situation where the terminal voltage of this is 1 the terminal voltage of the infinite bus is also the voltage of the infinite bus also is 1 p e is 1 also. So, we can back calculate all the values of states. So, that you get this particular situation once you have computed back calculated all the values of the states you can actually do a linearize analysis by plugging in the equilibrium values wherever necessary in the linearized state space equations. Once you do that one can get the Eigen values there is two little time for us to actually compute the Eigen values and I show you everything or the important points corresponding to these Eigen values. So, we will do that in the next class.