 been studying the synchronization transient of a synchronous machine connected to an infinite bus. What we saw last time was that you can have a near bumpless synchronization of a synchronous machine to an infinite bus which is essentially a fixed voltage source. The machine locks on to the frequency of the synchronize of the infinite bus in case it is synchronized well. Now, the thing we noticed last time was that if the speed of the machine initially is almost the same as that of the infinite bus and at the instant of you know the interconnection the voltage open circuit voltage of the generator the voltage phasor you can say is practically the same as the infinite bus voltage phasor then you get a kind of a bumpless transfer. Of course, you cannot get in perfect transfer without transients it is practically not possible. There will be some transient because you cannot do this match which I mentioned sometime back exactly. So, what you really see is that when you connect a synchronous machine to an infinite bus you do get small synchronization transient. Most notably you notice a kind of a what is known as a swing mode or a low frequency oscillation which usually damps down and that really after that the synchronous machine is in synchronism with the voltage source. Thereafter if you increase the mechanical torque to the synchronous machine you find that the machine transfers power from itself to the voltage source or the infinite bus. So, what you notice there of course is that as you go on increasing the mechanical power you will find that the angle delta increases and in fact there comes a point after which if you try to increase the mechanical power the synchronous machine in fact loses synchronism there is a it kind of fails to reach a steady state in case you go on increasing the mechanical power beyond the point. Of course, the important thing is that we had in fact seen precisely such a transient in one of our lecture. In fact, the first lecture I had shown you a small demonstration clip it will be good to revisit that clip you can go back to the first lecture and see that clip that is precisely what we have tried to simulate in the 21st lecture that is a synchronous machine connected to a infinite bus and there in thereafter we go on increasing the mechanical power to a point at which it loses synchronism. One of the things which we did not do when we increase the mechanical power was or the mechanical torque to the synchronous machine was that we did not increase the field voltage simultaneously. In fact, a synchronous machine tends to lose synchronism very easily in case you try to load it without a concurrent increase in the field voltage. So, what we will do this in this lecture is we will first of all revisit this earlier transient in which we first give a small step we first synchronize the machine then give a small step in the mechanical power of 0.25 per unit and then we try to increase the mechanical power right up to its rated value of 1 per unit and there we do not increase the field voltage we will see that it loses synchronism thereafter we will redo this simulation with by increasing the field voltage concurrently that is at the same time as we rather at this we increase the field voltage simultaneously perhaps simultaneously is a better word we increase the field voltage simultaneously along with the mechanical torque. You will find that under such circumstances the generator is able to supply the rated power to the infinite bus now one small clarification of course, I hope I did you did not miss it last time was that when we are simulating the equations of a synchronous machine we are neglecting the stator transients or the stator flux transients that is we have replaced d psi d by d t and d psi q by d t by 0. The reason is of course, that the transients associated with these states psi d and psi q are very fast. So, what we have done essentially is replace the differential equations which relate the psi d and psi q fluxes by algebraic equations. One thing you should note here is that an implicit thing is that when I am making this assumption I am really interested in the slower electromechanical transients that is why I can do without the psi d and psi q differential equations replace them in fact by the algebraic equations and still get a reasonably correct result. Now you may ask that well why do I do it anyway you can just as well keep the differential equations corresponding to psi d and psi q the reason why of course, I have done that is if I retain the differential equations in psi d and psi q you get what is known as a stiff system that is a mixture of fast and slow transients are existing in that system and as a result of which simple numerical integration schemes are likely to misbehave. So, what I have really done is I have made this approximation of neglecting the fast transients removed in some sense the stiffness of the system and then used Euler method which is the simplest method to simulate the system. So, that is why I have done this if of course, I had used I had retained the stiffness in the system that is I had retained the d psi d by d t and d psi q by d t terms in the equations in such a case you would find that you would need to use a method like trapezoidal rule or backward Euler method. Now, the problem in doing that is once you discretize the differential equations with trapezoidal rule or backward Euler method remember these are non-linear differential equations then what you really get is are non-linear algebraic equations once you discretize the system by a by these numerical methods you get non-linear algebraic equations and for every time step I would need to solve numerically solve for every time step out numerically you have to solve the algebraic equations. On the other hand Euler method is simple and the per time step computation is very very straight forward. So, that is what we have done so far today I will try to simultaneously increase the field voltage along with the torque increase and we can see that in such a case you can in fact run the synchronous machine at rated power. So, let us just redo the transient we did last time and go ahead and simulate the transient with field voltage increase simultaneously. So, let us go on to that so today's lecture of course, we will be revisiting the transient associated with the synchronization of a synchronous machine. Now, what we will do is of course, like in the previous lectures we will do a simulation of this system. So, let me just show you the simulation file here this is gen underscore sink is the file I will be doing a 25 second simulation I will be synchronizing the machine at 0.2 seconds this of course, the you see the data of the machine this something we have done quite a few times before in this in these lectures. So, I just quickly skim through this as the mechanical power is changed to 1 per unit we also double the field voltage. So, simultaneously we double the field voltage and we will redo this transient redo this simulation here and I plot now the speed and you notice that in this particular case the machine regains rather is still in synchronism. So, by increasing the field voltage in some sense we have ensured that the machine is able to deliver the rated power and remain in synchronism and operates stably of course, these there is a electromechanical oscillation or a swing which is seen in the speed whenever we give any such disturbance. Now, we can of course, have a look at how delta looks as well. So, for the first disturbance of course, delta goes from 0 to 0.4 roughly this is radians and the second transient also it is stable there is a swing and there is a gradual rise as well rather a kind of a oscillate decaying oscillation which we see here and also decaying exponent I mean there is also this kind of rise because of which it settles finally near about 1 radian. So, this is basically how the system behaves in case we increase the mechanical power along with a change in the field voltage with if we do not change the field voltage along with the mechanical power the system can lose synchronism. So, that is one important thing you should remember in the next few lectures we will go on to talking about excitation systems in the sense that we will try to understand how the field voltage is changed by a control system or a control system and an excitation system. Field voltage in a synchronous machine changes along with the loading in fact, we engineer it in such a way that whenever the loading of a synchronous machine changes the field voltage is also changed it is in fact, very necessary to do so. So, this is something you should keep in mind now what we will do next is look at another form of analysis what we will do is do a kind of small signal analysis of a synchronous machine. So, far what we have been doing here is in fact, looking at the numerical simulation of the system the reason why we do a numerical simulation is that the system is non-linear. Now, we can in fact, do a linearized analysis around an operating point using Eigen analysis the Eigen analysis tool as we have done before the only difference here is that since this is non-linear we have to first convert the system into a linearized system around an operating point. So, what we will be really doing is since this is a non-linear system we cannot infer the non-linear behavior by Eigen analysis that is not possible, but what we can do is if you are at an operating point that is at an equilibrium point if you give small disturbances you can rewrite the equations in a linear form which is valid only for small disturbances and infer the behavior from the Eigen values of the resulting linear differential equations. So, this is what we will do next at this point we should look at where are actually the non-linearities in our equations let us have a look. Now, if you look at the equations of a synchronous machine we had we have been using these compact form of equations in our analysis that is first of all we have written the flux equations d by d t of the flux equations is equal to a 1 into the flux. So, this is the state space form of the flux equations remember that a 1 is a function of speed. So, actually although this is written it looks almost as if it is linear a 1 is actually a function of the state of the states speed is a state of the system remember that this is written in a kind of a composite form i d i q are in fact related to all the fluxes by another set of equations the in fact the inductance effectively the inductance matrices or the reactance matrices in per unit. These equations are of course in per unit form a 1 looks like this it is in fact linear non-linear because a 1 itself is a function of the states. So, a 1 into psi involves effectively product terms of speed and flux. So, that is why it is basically a set of non-linear equations. So, there is a non-linearity here in fact these are of course the definitions of a 1 the a 1 b 1 b 2 a 3. So, one of the things you will notice here is of course I think I have effectively written the torque equation if you look at it in per unit this also involves speed product terms. So, if you look at the basic compact equations of a synchronous machine this is a product kind of term a 1 contains is a function of the states and also the torque equation in per unit is a non-linear equation. So, when we are trying to use Eigen analysis tools remember that the mechanical and electrical equations are in fact coupled and you see the flux and current equations in the mechanical equations and in the flux equations there is a omega dependence. So, because of this non-linearity we cannot directly apply linear analysis we will have to do a linearization around in equilibrium point. So, the basic idea is suppose you are operating at a particular equilibrium point. So, for example, if you look at the mechanical equations 2 h pi omega b d omega by d t is equal to t m minus psi d i q minus psi q i d. Now, if you look at this this is a non-linear equation. Now, if you are operating at a certain equilibrium point and the equilibrium values of psi d is psi d e. Similarly, the equilibrium value of psi q psi q e and of course, i q and i d are dependent on the equilibrium values of psi f e in addition to the psi f e psi g e psi k e and psi h e. So, these are the equilibrium points. So, this e here is denoting equilibrium. So, this is the equilibrium value of these states. So, if you look at you rewrite these equations again if you are spreading the speed equilibrium speed of a synchronous machine connected to an infinite bus is well if you look at d delta by d t it is nothing but omega minus omega 0 this is the frequency of the infinite bus. So, if you look at what is the equilibrium speed of the machine the equilibrium speed of the machine is omega e and by definition of equilibria the value of the states at which all the d by d t is become equal to 0. So, if you look at omega e it should be equal to omega naught. So, the value of omega at equilibrium is omega e it is equal to omega naught. So, these are essentially the equilibrium values of the states. So, we are actually obtained by setting d delta by d t d omega by d t and all the d psi by d t is equal to 0. Now, what are the equilibrium values of what are the actual values of psi d psi q e psi f e and so on. What is the equilibrium of value of delta e for example what is the equilibrium value of delta. So, these are all the states. So, one of the equilibrium value we have of course, got then there are six flux states for which we have to get the equilibrium values and delta equilibrium value has to be obtained. Now, remember that we had obtained the expression for the electrical torque if you recall electrical torque expression. The electrical torque expression was in steady state of a synchronous machine was electrical torque under equilibrium conditions is of course, it is psi d e i q e minus psi q e i d e. But, we can write it down as we have derived this in some of our previous lectures sin delta e upon x d plus 1 upon x q minus 1 upon x d into v line to line r m s square pi 2 sin 2 delta e. So, the equilibrium value of the torque is nothing but e f d this is depending on the field voltage v line to line r m s is the voltage line to line r m s voltage of the voltage source to which the synchronous machine is connected to delta e of course, is the equilibrium angle. Now, the important thing is that this is this if you look at psi d e psi d e i q minus psi q i d this is psi d e minus psi essentially an equilibrium or rather the transient this expression for torque is valid also in transient conditions. But, this remember this expression here which I have written down is only valid during steady state. So, please do not use it for transient conditions. So, under equilibrium conditions this is true. So, if I set d delta by d t is equal to 0 we get the equilibrium value of omega should be equal to omega naught also by setting d omega by d t equal to 0 we get t m should be equal to the equilibrium value of this. Now, if we know what e f d is and what v line to line r m s is from this we can easily back calculate what the value of delta e is. So, I know what t e e is t m. So, if I given you what the mechanical torque is and what the e f d is in such a case you should be able to compute delta e. So, first state second state is computed this is the equilibrium values of omega e and delta e. There after once you have you know the equilibrium value of omega is omega e also you have got the equations of the differential equations corresponding to the fluxes. So, if you look at them here I have written them down in compact form here. So, you set this equal to 0 a 1 is a function of the time constants as well as omega, but you know that the equilibrium value of omega is omega e. So, you get an equation. So, you can evaluate what a 1 is this is set to 0 a 2 into i d i q i d i q itself is a function of psi d and psi q it is a linear function. So, a 2 into a 3 into psi basically because i d i q itself is related by this expression. There after if you look at b 1. So, b 1 is also a constant matrix v d and v q. In fact, if you look at what I am writing here v d we have seen before by the definition of the voltage source we have we wrote down the time of the voltage of the infinite bus in terms of v a v b v c as per that definition v d will come out to be minus v line to line r m s into sin delta e and v q comes out to be v line to line r m s into cos delta e. So, this is of course, obtained we have obtained this before in our course. Remember that the rotor position is nothing but omega naught t plus delta. So, this is the definition of the rotor position omega naught is the speed of the infinite bus and because of that and the definition of v a n v b n and v c n the phase to neutral voltages of the infinite bus we obtain v d and v q in this form. So, what I wish to say what I wish to say here is that first of all we have got the equilibrium value of the speed then we have got the equilibrium value of delta e from equilibrium value of speed and delta e. We can from the flux equations which are shown here by setting the right hand side sorry the left hand side here equal to 0 that is put setting d by d t equal to 0 obtain the equilibrium values of psi d psi q psi f psi h psi g and psi k. So, this is basically what is involved in getting first the equilibrium values. Once you have got the equilibrium values in order to do a small signal or a small disturbance analysis around an equilibrium point what we really need to do is reformulate the equations or linearize the equations which is suitable for such small disturbance analysis. So, for example, we have got d by d t of the flux is equal to a 1 into the flux and so on. Now, what we do is we assume that is the fluxes we are considering only small disturbances from the equilibrium. So, we will assume that psi d e plus delta psi. So, this is the equilibrium value of the flux and this is the deviation from it. So, we can reformulate the differential equations in terms of only the deviations around the equilibrium. Now, if these deviations are small one can do a procedure called linearization. So, I will just show you the linearization procedure for a simple case 2 h we just look at this equation you can actually linearize every equation equal to t m minus psi d i q minus psi q i d this is of course, a non-linear term. Now, we assume omega is equal to omega equilibrium plus the deviation from the equilibrium this is a known quantity. So, we can reformulate these equations you can write this as 2 h by omega b d of omega e plus delta omega omega e is a constant it is the equilibrium value of the speed. So, what we will do is is equal to t m we will assume that t m is a constant minus psi d i q minus psi q i d now psi d i q minus psi q i d in fact, can be written down as. So, it is psi e psi d e plus delta psi d e plus delta into i q e plus delta i q minus psi q e plus delta psi q into i d e plus delta i d. So, this becomes if you neglect well first let us write it down. So, what you can write it down is psi d e i q e minus psi q e i d e plus psi d e delta i q plus psi q e delta i d sorry this should be a minus plus delta psi d e sorry I am sorry delta psi d e into i q e minus delta psi q into i d e plus what are second order terms that is product of two delta kind of terms. Note that psi d e into i q e minus psi q into i d e plus what are second order terms that is product of two delta kind of terms. Note that psi d e into i q e minus psi q e into i d e is in fact, equal to the equilibrium value of the torque electrical torque which is of course, going to be equal to the mechanical torque T m. So, psi d e into i q e minus psi q e into i d e is in fact, T m. So, what we have essentially is the linearized equation 2 h by omega b into delta if you assume that T m is a constant in fact, you can write this as psi d e delta i d minus psi q e delta i q plus delta psi d i d e minus delta psi q i q e. So, this is just an example of how you can do linearization. So, you can in fact, linearize the d psi d by d t equation and d psi q by d t equation in a similar fashion. So, what you eventually get is you know you have to couple of course, the mechanical and electrical equations the flux equations are coupled. So, finally, what you will get is going to be delta x in the linearized states assuming of course, delta T m is equal to 0 and delta e f d is equal to 0. So, we will be assuming the inputs are constant will be only giving changes in delta x that is the giving the initial condition or initial disturbances to the states. So, around the equilibrium point the behavior of the equations for small disturbances can be studied using this. So, I have just shown you how you can do a linearization. So, I request you to go ahead go back and just try to do the full system linearization and get it in the form delta x dot is equal to a into delta x. The properties of the transients for small disturbances around the equilibrium can be got just by analyzing the Eigen properties that is the Eigen values and Eigen vectors of the resulting state matrix the A matrix. Of course, A now is dependent on the equilibrium point because it really has terms like psi d e and so on, but is a constant for that equilibrium point it is a constant, but it is a function of the value of the states at the equilibrium point. So, one of the things you should remember is that a non-linear system which is linearized and then we if you do non-linearized rather linearized analysis on it Eigen analysis on it what we will see is that the Eigen values which we get for different equilibrium points are going to be different. So, the transient behavior around different equilibrium points in fact is going to be different it is not going to be the same. Did we actually notice this in our simulations? The answer is yes. So, let us look at the simulations again carefully. So, back to the simulations if you notice even one of the plots which had shown you may be we can have a look at the speed transient. Well, you may say well there is no difference in the kind of behavior around an equilibrium point you have these swings, but one of the simple things you can have actually look at look for rather is whether there is any change in the frequency of the swings. For example, the frequency of the swings we will just do this again one second we will just get this back. If you look at the frequency of swings around this equilibrium point I will just yeah you look at this frequency of the swings around this equilibrium point you will find that it is well almost half a second. So, this is almost a 2 hertz oscillation around the equilibrium point corresponding to I am sorry yeah around the equilibrium point corresponding to T m is equal to 1 and E f d is equal to 2 the frequency of oscillations is in fact frequency of oscillations is in fact around 2 hertz. What about the frequency near about the equilibrium corresponding to T m is equal to 0.25 and E f d is equal to 1. So, let us just have a look at that here. So, if you look at that this is a roughly 6.1 seconds and this is roughly well this is near about slightly more than the frequency the period is more than 0.5 it in fact the frequency seems to be lesser near about T m is equal to 0.25 per unit and if you look at the behavior just after synchronization when mechanical power is in fact 0 the frequency of oscillation is again roughly 0.7. So, with T m is equal to 0 the frequency slightly lesser in fact the period is larger and with larger values of T m with of course, a simultaneous increase of E f d we see that the frequency in fact slightly increases in fact it is higher here compared to here. So, as the operating point changes we get a change in frequency that is not surprising because the Eigen values change because the A matrix whose Eigen values we are computing is a function of the equilibrium values of the states which are obtained after linearization. So, that is basically why this happens in fact one more striking thing which you can observe from this plot in fact all the plots reveal a great deal of information you will notice that the rate of decay of this oscillation is much much slower compared to the decay of oscillation here in no time you will find that this reaches equilibrium whereas, this takes several seconds to reach an equilibrium. So, what one can expect after doing a linear analysis an Eigen value analysis we will do that shortly of course, is that we can infer that whether this frequency is going to change whether the ramping is going to change and so on from that analysis itself. So, let us just verify that by actually writing down an Eigen value analysis program. So, as I mentioned some time back of course, it involves some deal of effort in the sense that you will have to linearize the set of equations form the A matrices or the A matrix of the system which is a function of the equilibrium point. So, this is something you will have to do now one of the points which I need to of course, clarify this point the A matrix which are going to get here would be initially we will include the d psi d by d t and d psi q by d t transient we will not convert the d psi d by d t and d psi q by d t differential equations into algebraic equation we will first consider all of them together. So, what we are really looking at is a model in which you are going to have this set of equations. So, this set of equations is what we are going to consider with the differential equations in psi d and psi q retained we are not neglecting or replacing d psi d and d psi q by d t equal to 0, but we can always do that in case we do that we will be converting two of these differential equations to algebraic equations allowing us to eliminate two of these variables. So, two algebraic equations when we get we can write psi d and psi q in terms of the other states and get rid of them and reduce the number of differential equations. We in fact did this when we are doing the numerical simulation in order to remove the stiffness when you are doing an Eigen analysis there is no compelling reason to remove the stiffness of the system removing the situation where they are fast and slow transient that is what I mean when I say removing stiffness when you are doing Eigen analysis there is no compelling reason whereas, for simulation if you want to use simple numerical integration methods then you have to remove the stiffness in the equations. So, let us now do an Eigen analysis of the system I have written down a program again in psi lab for doing the Eigen analysis we will just run through it the important differences between a Eigen value analysis program and a simulation program is that of course, we are talking in terms of a system operating initially at an equilibrium. So, for example, if we took a look at the equilibrium T m is equal to 0 and E f d is equal to 1 from this. So, I have saved this from this we could we would need to calculate the equilibrium value of this state corresponding to speed, but we know that omega e is equal to omega naught which is equal to the infinite bus frequency that is of course, the equilibrium value of the speed the equilibrium value of delta has to be obtained by solving this equation omega naught and omega b will of course, assume to be the same the speed of the infinite bus will assume to be equal to the base value of the frequency. So, we can solve for delta once we solve for delta we can also using the algebraic equations using the algebraic equations solve for fluxes and once we solve for fluxes we can get the equilibrium values once we get the equilibrium values we have to form the A matrices from a linearized equations. So, for example, if you look at this equation or let me just in this equation realize that the components of the A matrix eventually what I call is A f is consisting of the equilibrium values delta 0. So, remember that linearized matrices are a function of the states the equilibrium values of the states. So, what we will do is now run this Eigen analysis program I encourage you to try it out yourself you try to create write a program to do the linearized analysis. So, I will just run it for you here remember that we are doing the Eigen analysis of a system around T m is equal to 0 and E f d is equal to 1. So, one thing what we will do is what is the equilibrium value of delta 0 the answer is 0 of course, if mechanical power is 0 T m is equal to 0 you can directly infer from the steady state torque equation that delta 0 is equal to 0. Now, if I find out the Eigen values of the system the Eigen values of a system linearized around the equilibrium point corresponding to T m is equal to 0 what we find is what we find is the Eigen values are as seen here what we see is you have got these two Eigen values which are very large in magnitude they are almost 314 radians per second as discussed in the short circuit analysis of a synchronous machine these are actually corresponding to the d psi d by you know they are very much associated with the psi d and psi q states you will find that these two sort of new Eigen values are coming because of the inclusion of the electromechanical equations the mechanical equations contribute the equations corresponding to omega and delta. So, basically you get these two extra Eigen values compared to the Eigen value analysis of a short circuited synchronous machine these two extra Eigen values in fact, indicate that you should be having a damped oscillation of frequency 9.4 radians per second. So, 9.4 radians per second corresponds to divided by 2 pi corresponds to 1.49 hertz. So, 1 upon 1.49 hertz is in fact 0.67 seconds which is the period of the oscillation corresponding to T m is equal to 0. So, if you look at this again this frequency out here should be around 0.7 that is what Eigen value analysis predict. So, is it actually true look at it again this is around 1.56 and this is roughly 2.4. So, that is around near about 0.7 to 0.8 seconds. So, actually what we are getting through Eigen analysis is in fact, quite accurate also you can see the rate of change of the peaks rather you can see that the peak value of this oscillation keeps decaying with time. I leave this as an exercise to you to verify that the real part of this Eigen value corresponds to the rate of decay which is seen in this figure. We redo the Eigen analysis with E f t is equal to 2 and T m is equal to 1. So, we save this and redo it oops what we need to do it of course. So, if you take out the Eigen values you will find that the Eigen values the most striking thing you will notice is that the frequency of oscillations has increased the damping has come down this is real part negative real part magnitude has come down. So, what one can get from the simulation essentially is what one should see in the simulation is that the damping is this of course, the plot of delta you see that the damping is much faster near the T m is equal to 0 case as compared to the damping near T m is equal to 1. So, this is the plot of delta remember frequency of oscillations here is higher than the frequency of oscillations here. So, of course, we can also see this in omega the the simulation of omega. So, I will just redo this and plot and plot omega. So, I will just remove this and plot omega. So, the same thing is of course, seen in omega remember the same modes are being seen in different all the all almost all the states. So, what you see in delta the same model characteristics are seen in omega. So, you see that the frequency at different equilibria are different and the damping also is is different I mean this takes longer to damp out compared to the damping here. So, the real part of the Eigen value near T m is equal to 0 can be expected to be larger than the real part negative real part of course, the magnitude is lower in this this is actually true. So, this is basically a good correlation between Eigen analysis and the simulation. So, this is what we see here. In fact, if you look at the value of delta 0 here from the Eigen analysis sorry the computation of I will have to redo the Eigen analysis program. So, the value of delta equilibrium value of delta when T m is equal to 0 and E f d is equal to 1 is 1.09 radians. So, actually we can convert this radian value to a degree value in fact, it is 62 degrees. So, in fact, if you look at the simulation will redo the simulation and plot the value of delta. So, what we will do is this of course, omega will plot the value of delta with T m is equal to. So, if you look at the simulation the simulation also if you look at the value of delta which to which this settles down is also near 1.09. So, what you are really seeing is a good correlation between the numerical simulation as well as the Eigen analysis of a system linearized around an equilibrium point. So, this is an interesting study in which we have actually done the linear analysis also of the synchronization transient. So, I hope you got the hang of what was what you are trying to do we could really it is very interesting you know when you do a simulation you get a certain time response plot, but you feel a great deal of joy when you are able to correlate it with Eigen value analysis which you want many of us feel that is a more analytically and it gives better insight. So, it is important whenever you do a numerical simulation whenever you study a transient it is very important to interpret the results which are coming correctly and you should be able to correlate it with the kind of inferences you get from other tools. So, I hope you got a flavor of this in this lecture in the next lecture we will try to do a simplified I will I will try introduce you to some simplified models of a synchronous machine. In fact, we have used what is probably a full blown model in all our analysis. So, far we will try to do the reverse thing what we will do is try to make more and more simplifications and come down to a bare bound model or rather the classical model with which we had actually done some very simple studies right at the beginning of our course. So, we will really know by looking at what kind of simplifications we have done we can arrive at the toy model which we considered right at the beginning of the course that itself is a nice exercise where we come to know what all we have actually brushed under the carpet in order to get a simple toy model before we move on we will discuss again the general linearization procedure for small signal analysis of a non-linear system around an equilibrium point. So, although I have done the linearization of this system it would be good for you to be familiar and at ease with the general linearization procedure. So, if you have say a non-linear system of this kind x dot x 1 dot is nothing but d x 1 by d t and you have got n variables of this kind then the general non-linear equations are given as shown in the screen on the screen. Now, the equilibrium points for this system are obtained by setting x 1 dot x 2 dot and so on equal to 0 of course, in this non-linear system u denotes the inputs. So, if I have given the inputs that is u 1 0 u m 0 are specified those are the nominal inputs which are given for this equilibrium point then the equilibrium values of the states may be obtained from these algebraic equations. These again are non-linear algebraic equations the equilibrium points are obtained by solving these non-linear algebraic equations. Remember that more than one set of equilibrium states may satisfy the above equation. So, you have may have more than one set. Now, in the general linearization procedure we will try to we will choose the set which is of interest I mean you are really going to do small signal analysis around an equilibrium point. So, you have to decide the equilibrium point. So, choose the set which is of interest after that you need to take each differential equation each non-linear differential equation and linearize it. So, what this involves I mean what eventually you need to do is obtain the partial derivatives of f 1 with respect to the states x 1 x 2 to x n as well as the inputs u 1 to u m. Now, remember that these partial derivatives are evaluated using the equilibrium values of the states. So, these partial derivatives are essentially constants. Remember of course, that delta x 1 here is a small disturbance from equilibrium. So, delta x 1 is equal to x 1 minus x 1 e or the deviation from the equilibrium. Similarly, delta u is equal to u 1 minus u 1 0 u 1 0 is the nominal input. So, this is what you get as a linearized equation. Now, in my previous development we just for example, linearized psi d i q minus psi q i d you may say that well this looks different from that no actually if you take out the partial derivatives of the of the functions psi d i q minus psi q i d and plug in the equilibrium values you are going to get exactly the same linear model as we got some time back. So, in fact this is a more general and direct you know representation of what we will get if we apply this linearization with delta x and delta u very small. So, this is how you would normally linearized set of non-linear differential equations around an equilibrium point. So, this fact will become clear shortly. Remember of course, that you have to linearize each differential equation right from x 1 to x n and therefore eventually you will get in fact a Jacobian matrix. So, you know you do this for x 1, but you can also do it for the other x 2 x 3 to x n and eventually you will get this delta x is equal to a into delta x plus b into delta u. This a and b of course, are these matrices. Note that these are effectively constant matrices because these partial derivatives are evaluated at the equilibrium point. So, equilibrium point essentially is the denotes the values of the states. So, I have to plug in the values of the states. Now, this will become clear if you consider this example x 1 dot is equal to x 2 and x 2 dot is equal to u minus sin x 1. So, this is a non-linear set of equations. The non-linearity obviously is because of the sin term. So, the first step in doing this is find the equilibrium points. For the nominal input 0.5 the equilibrium point is sin inverse 0.5 or sin inverse half which is 30 degrees and x 2 e is equal to 0. So, x 1 e is in fact 30 degrees. This is of course, the equilibrium points are not unique. For example, x 1 e could also be pi minus that is 180 minus 30 degrees that is also an equilibrium point. So, remember that there is in the non-linear systems there may be more than one equilibrium point. Let us say the equilibrium point corresponding to sin inverse sin inverse 0.5 is equal to 30 degrees and x 2 e is equal to 0 is the equilibrium point of our interest. So, the equilibrium point of our interest is 30 degrees and 0. Let us say suppose this is our equilibrium point of interest. So, you can easily obtain the small signal model. The small signal model is essentially a linear model around this equilibrium point. So, this is eventually the small signal model of the system. One could also in fact try to find out the equilibrium point you know the linear model around another equilibrium point. So, in this case for example, the other equilibrium point is x 1 e is equal to 150 degrees x 2 e is equal to 0 and u naught of course, is the nominal value 0.5 and interestingly the properties of the small signal stability properties of different equilibrium points can be different. So, that is a very interesting and important thing which you should keep in mind. So, this is the general procedure when you want to linearize a non-linear system around an equilibrium point. So, you can directly take partial derivatives and form the Jacobian matrix. Now, I wish to point out again that in the example we have considered and where I have shown you how to linearize a particular differential equation or this you know the torque equation effectively. I have linearized the electrical torque, but it may not be directly evident that you are in fact doing exactly this. In fact, I have in fact taken out the partial derivative of the expression of T e with respect to psi d psi q i d and i q and plugged in the equilibrium values. So, this is, but a more general way of means general way of putting it that is you take out the partial derivatives or evaluate this Jacobian at the equilibrium values of the states. So, eventually what you get is a linear model of a system with constant coefficient matrices and another point to be noted is of course, that if there are multiple equilibrium points you could have different small signal models around different equilibrium points and of course, the properties the stability properties of the A matrix matrix A matrices around different operating points or equilibrium points could be different. So, this is something you should keep in mind. Now, we go back to our example again of a synchronous machine.