 In the previous class, we derived voltage equations of a synchronous machine. In fact, we applied Faraday's law and we got the relationship between voltage and fluxes and of course, the relationship between fluxes and the currents. Also, we try to get a relationship the in fact that relationship is the inductance matrix, which it turns out is position dependent that is theta dependent. Remember an important thing is that theta is continuously changing in a synchronous machine. So, that inductance matrix in fact, is time dependent it is dependent on the theta position. Now, in today's lecture what we will do? This is the lecture is continuation of the modeling of a synchronous machine. We will introduce you to a transformation of variables called the T Q transformation or the Parkes transformation. In the previous lecture we kind of concluded at the point where we derived the equation of torque in terms of the current. We will just redo that again at the beginning of the lecture just to refresh your memory and then we move on and I will introduce this transformation, which I just mentioned. Now, if you recall what we did last time for the mechanical equations of a synchronous machine. So, if you have got a synchronous machine moving in this direction with the mechanical speed of omega m and it has got electromagnetic torque which of course, opposes the motion and the prime over torque T m. So, if you look at the equations of this system you have J is in fact, the moment of inertia in kg meter square. So, this is the well known equation which describes the rotation and we the electrical torque by basic electromagnetic energy conversion basic electromagnetic energy conversion formula tells you the electromagnetic torque is given by partial derivative of the co-energy with respect to theta. So, the torque in this direction and remember theta is measured in this direction in the anticlockwise direction. So, in that case you have this the co-energy of course, is a function of the currents. So, when I take the partial derivative I should keep the currents constant. The co-energy is defined as half of I s transpose I r transpose I s and I r remember are the L inductance the inductance matrix of the rotor of the system. We already derived what this L matrix is in fact, some three of its sub matrices are in fact, functions of theta. So, L is in fact, made out of sub matrices L s s L s r L r s and L r r out of which L r r is not theta dependent, but all the rest are theta dependent. So, this is what we have done before I s remember is denoting the stator currents I r is the rotor currents. Now, T e which is nothing, but minus T omega dash by d theta in fact, for a this is in fact, only for a two pole machine remember that the correct formula in general is T e is the partial derivative negative of the partial derivative of d omega dash d w dash is the co-energy with respect to the mechanical angle. So, I this is the please remember that this is the mechanical angle of course, when we derived L the it is a function of theta whereas, this is theta m just remember that. So, what we have is theta m which is nothing, but now if we define T e dash to be equal to minus of this we can show of course, that your T e dash will be equal to minus of half we will get this as minus of half I s transpose. And as I mentioned last time you would not find L r r coming in these equations because, the partial derivative of L r r with respect to theta is equal to 0. And we have already made an observation that L r s transpose is equal to L s r. So, you just get one of them you know this simplifies to this. So, this is what we did last time we got a tor expression for T e dash. And once you of course, do that you can write your equations as j into d omega m by d t just again one small point which I wish to emphasize this is the mechanical speed, mechanical speed angular speed. So, you can write this as 2 times p times the electrical speed rate of change of electrical speed is equal to T m minus T e which is nothing, but. So, this is basically what we have and remember T e dash is this. So, if we know. So, we have in fact got a relationship between the currents and the torques. So, in fact just to correlate something we have done sometime back let me just multiply both sides by half omega m square. So, I will do half into omega m the base value of that speed or the rated value of the speed into j into 2 by p d omega by d t. So, I will just I have just multiplied it there is equal to you multiplied here also on this side also. So, you will get half of omega m b square into T m minus p by 2 T e dash. Now, what I will do next is divide both sides by the volt ampere base of the machine. Remember omega m base is the mechanical base frequency angular frequency. So, actually it is usually the base frequency is the rated frequency of the machine rated the mechanical angular frequency of the machine. So, that is one thing which you should remember. So, actually if you divide by volt ampere base will half j omega m b square divided by volt ampere base into 2 by p d omega by d t is equal to half omega m b square by volt ampere base into T m minus p by 2 T e dash. Now, a few things this is what is commonly known as H or the inertia constant of the machine and its units are joules per volt ampere or mega joules per mega volt ampere. So, this is called H of the machine. So, you have basically if you look at this particular equation I will rewrite it. So, you can actually write it as you will get 2 times H you get this 2 onto this side 2 times H I got this 2 onto this side. I will take one of these omega m b is onto this side into into p 2 d omega by d t is equal to omega m base divided by volt ampere base into T m minus p by 2 T e dash. So, actually if you look at how these omega m mechanical base into p divided by 2 would actually give you the electrical radians per second base value. So, you will have 2 H by omega base. So, omega base is the electrical this is the radian frequency base value of the machine. If you look at this this is in fact the mechanical torque base. So, you have got T m is the mechanical torque divided by the torque base of the machine minus what you have is I will just rewrite it here. So, it will become very easy to understand what I am doing this into p by 2 into T e dash just check if it is volt ampere base into p by 2 in this. So, finally, what we get is the torque equation the machine is 2 H by omega b into d omega by d t is equal to T m in per unit minus T e dash divided by volt ampere base divided by omega base the electrical base. So, this is the final equation of the torque of the machine. In fact, you will note that T e dash will be a function of theta m theta you know the inductance matrices will be actually functions of theta whatever we have derived earlier functions of theta. So, in fact if you look at interesting thing about this is that the number of poles does not appear in this equation. So, when you write these things in per unit you will not find the number of poles. In fact, we will express T e in terms of the fluxes and the currents and the inductance which is a function of theta volt ampere base and omega b, but you will not get p explicitly. So, actually when you turn when you go to the per unit system normally you know you can actually be blind to the number of poles if you are working in the per unit system. Of course, if you want to actually get the mechanical speed if you know the electrical speed then you do need require the number of poles. So, that is of course, something which you should remember. In fact, you can write this in fact like this 2 h into d omega by omega b. So, this is in fact that is why I called this a per unit equation everything is in per unit rather we have not really x this T e is not in per unit, but it divided by volt ampere base by omega base. So, eventually of course, after some time I will show you how this everything in this also can be expressed in per unit. So, this is the final equation in fact this is the swing equation we used sometime back in our earlier lectures that is this is the origin of this particular equation. Now, moving on a bit before I of course, move on let me tell you that the typical values of h you know the typical values of h for a thermal unit can be between 2 to 6 2.5 to 6 this is of course, from for a 2 pole machine these values I have taken from Kundur's book for a pole machine. And for hydro units it is usually 2 to 10 and of course, I am not written the units they are mega joules per m v or joule per volt ampere. So, the unit of h is mega joule per m v now one of the things you will if you recall the equations of the synchronous machine you will get the rate of change of flux is you know dependent on v. In fact, if you recall the formula which I had written last time the general form is d psi by d t is equal to minus r into i minus v and if you recall psi was nothing but l into i the psi is of course, made out of psi s and psi r and i is also made out of i s and i r and v is nothing but v a v b v c and minus v f is made out of the rotor and stator voltages 0 0 0. So, this is something we did in the previous class in fact, you can replace instead of writing it as this way you can even write it you can replace this by minus r l inverse psi minus v. So, actually this is so the only complication when we are trying to solve this equation this looks actually like a linear equation even if you assume that theta is a constant theta is rather rate of change of theta is a constant the speed of a synchronous machine is constant remember that this l will be time varying. So, if in fact, if you just look at these flux equations that time varying equations the linear time invariant time variant equations. So, if I want to get the solution of this particular equation well I cannot use directly the Eigen analysis techniques to solve this equation again I repeat this inductance matrix is time varying it is not a linear time invariant system it is a function of theta even if you assume that theta rather the speed of the machine is constant still you will get a time invariant set of equations. So, that is why it will be good to see how you can you know explore ways to try to analyze this. Now, one of the most powerful ideas which you have in fact, you used before is to use the idea of a transformation or transformation is like we have used transformations for example, when we are analyzing the system x dot is equal to A x we tried to find out the transformation which would diagonalize the matrix the idea was of course, a diagonal matrix implies there is no coupling between the states. So, if I got x dot is equal to a diagonal matrix into x there is no coupling between the individual states and you can easily get the solution in terms of the exponential functions. Now, here our aim here we can use again the idea of a transformation, but right now we will limit ourselves to trying to make these time invariant set of equations or other time variant set of equations into time invariant set of equations. So, what we will do is try to get a transformation of variable. So, that we will get this our objective. Now, of course, how do you get the transformation? Well, what I will do is introduce the transformation directly and see what its consequences are. So, I am not going to derive the transformation for you, but so just let me introduce it to you directly without any further delay. This transformation is called Parkes transformation. So, it is defined as so instead of looking at the variables or looking at the differential equations in the variables f I a, I b, I c or V a, V b, V c or psi s, I b, psi c this f could denote any three phase variable. I look at the variables f d, f q and f 0 using a transformation C p this is a matrix which is dependent on theta. So, that is an important point you should keep in mind. Now, what is this transformation? Of course, the whole key is what is this transformation? It turns out that this is really a very useful transformation. So, C p is equal to I will just define this transformation. It is a 3 by 3 matrix as you may have guessed something you may have guessed that I am applying the transformation only to the a, b, c variables and not to the field you know I f, g, h and k variables. So, only the stator quantities in fact, the a, b, c quantities we are applying this particular transformation. Now, in fact, just recall I mean it is just a interesting correlation which you should recall that the currents on the rotor become constants in steady state. Whereas, you know that currents in the stator or the voltage in the stator are in fact, three phase sinusoids. In fact, for balance conditions you will get three phase balance sinusoids. So, if I am going to try to make what you call linear time invariant set of equations one can kind of guess very intuitively that you have to apply the transformation only to the a, b, c variables. So, indeed that is the case. So, this particular transformation which I am going to tell you talk about is actually applied only to the stator variables a, b, c variables. So, an interesting thing about this transformation even without going any further is that its inverse c p inverse always exists unless of course, you choose k d, k q, k 0 or you know any of them 0. So, if k d, k q and k 0 are non-zero then you can actually get a transformation c p inverse which is defined like this. It has this form k 1 cos theta plus 2 pi by 3. I will just show it this side slightly. This is k 2 sin theta minus 2 pi by 3 k 3 and k 1 sin theta plus 2 pi by 3. This is k 3 where k 1 is equal to 2 and 3 k d. This is something you can please work out. I am not working it out for you, but you can just in fact, one simple thing you can do is just multiply c p and c p inverse and just verify that it is actually giving an identity matrix. Note that there is a minor error here. The 2, 3 component of c p inverse should be k 2 and not k 1. So, c p inverse comes out to be this. Why do you need c p inverse? See, the point is that whenever you use any transformation, any kind of transformation analysis, remember that if I define a transformation of variables, do my analysis in the variables. If I want to know what the old variables were eventually, I would need the inverse transformation. So, in fact f d, f q and f 0 is equal to c p inverse theta. So, actually a transformation makes sense only if you can go back and forth between the old and the new variables. So, it is important that c p should be invertible, but it turns out that c p is actually invertible unless of course, you choose k d, k q and k 0 is equal to 0. So, c p inverse turns out is invertible. Another interesting thing you will notice, I mean before we go ahead is that the structure of c p and c p inverse look similar. I mean if you look at the structure of c p and c p inverse. So, look at c p. There is constants along this column. You have got your sin theta's along this cos theta's here. Of course, the constants are k d and k q and k 0. Whereas, if you look at this, the cosine functions are here. The sin functions are here. This is the constant. So, you can guess that for some you know c p inverse looks like the transpose. At least in structure, it looks like a transpose of c p inverse. In fact, for special values of k d, k q and k 0, this is indeed true. So, you will actually for example, let me just let you out on this. This k d if k d and k q if the root 2 by 3 and k 0 is 1. So, if k d is equal to you know root 2 by 3 and k 0 it turns out is equal to 1 by root 3. In that case, c p will be equal to c p transpose. So, that is one important result which you should keep in mind. So, some special values of k d and k q c p transpose and c p inverse are equal. Now, the main utility of this transformation can be seen if you look at the transformation of flux linkages. So, if for example, if you have got you know we have done this before psi s and psi r is equal to these are functions of theta. Now, if I use a transformation, remember I only have to transform the stator, stator fluxes and the stator currents. So, suppose I define as before, this is a 3 by rather 3 by 4 matrix full of 0s and this is an identity matrix I this is a 4 by 4 identity matrix psi I will call this d q and 0 and psi r. So, what I am doing is I am not transforming psi r this just say psi r is equal to psi r. So, I am only transforming psi a psi b psi c into psi d q 0. So, that f variable is I mean I just applied whatever transformation to flux psi a psi b psi c. So, if I define a transformation of this kind, how do in fact a similar transformation can be applied to i s and i r. So, with the same matrix. So, in that case it is easy to see that the final equations relating psi s psi d q 0 and i d q 0 and psi r and i r r psi d q 0 psi r is equal to L s s L s r L r r into into c p this is the matrix of 0 this is 0 and this is identity matrix of psi 4 into 4 into i d q 0 and i. So, what I have actually done is I have replaced i s and i r by this i d q and i r. So, i s and i r is equal to c p into i d q 0 i r is equal to i r. So, I just replace this by this then you have got this L r r then you have got this L matrix. So, I just repeat what I said I have replaced i s and i r by this then of course, L s s L s r also come here I mean I will just show this again. So, this is what we get and of course, we will get c p inverse here. So, actually if you get this on to this side you will get psi s and i r. So, basically what I have done is I have taken this particular equation here you have got a b c a b c this has been replaced by d q 0 here and d q 0 here. So, what you need to do is of course, I will rewrite this we will get psi d q 0 psi r is equal to finally, c p inverse L s s c p c p inverse L s r L r s into c p L r r into this is multiplied by i d q 0 and i r. So, this is what we get finally. So, you are now the whole equation whole let me just give a sneak preview to what we are going to get what we intend we intend to do is try to see if this actually is not going to be a function of theta. So, what we are trying to do is see the relationship between i d q 0 and i d psi d q 0 i d q 0 and i r and psi r and i r. Now, a nice thing it would be nice if all these terms came out to be independent of theta. So, in fact, they do in fact, let me just try to show it at least for one term for example, if I want to do I will just show it for one term I request you to just go and do it for each and every term. So, what it turns out is that for example, c p inverse L s r just let me try to compute this now what is L s r. So, do you recall what L s r is L s r is actually c p inverse please look into what we did sometime last in the last class L s r is equal to L r s transpose which is nothing but L r s partitioned this L s r into two parts L r s d and L r s q L r s d q is this as this form and L r s d has this form. So, for example, if I use c p L s r. So, what do we have is L s r d is C and L s r q. So, if I actually apply this transformation what is L s r d I will just write it down again m a f cos theta m a f cos theta minus 2 pi by 3 m a f cos theta minus 2 pi by 3 m a f cos theta plus 2 pi by 3 then you have got another column I will not write down that column this is nothing but L s r d. So, in fact there is another column I am not written it down. So, you have got terms here here and here also. So, if I do c p inverse of L s r d you know what c p inverse is this is what c p inverse looks like. So, I have a good look at this because we need to fit all our manipulations on a small sheet of paper. So, if you have got c p inverse into L s r. So, c p inverse into L s r d I am not showing the derivation for all terms I will just do it for one term. So, if you do c p inverse into L s r d the first component of c p inverse L s r d the first component of L s r d the first component that is the 1 1 component you can easily you know find out that it will be basically the first row of c p that is k 1 cos theta k 1 cos theta minus 2 pi by 3 and k 1 cos theta plus 2 pi by 3. So, this is in fact. So, the first term 1 1 term of this matrix I am just talking of the 1 term of this complete matrix the first term of this complete matrix is nothing but the first row of c p inverse into the first column of L s r d. So, the first column of M s r d is M a f cos theta M a f cos theta minus 2 pi by 3 and M a f cos theta plus 2 pi by 3. So, if you multiply these 2 you will get the first term of this. So, I am just doing it for 1 term I hope you will be able to follow up on the other terms. So, what you will get if you do that will be so c p inverse L s r d the first term is nothing but I will just work it out. So, you will get k 1 which is going to be common M a f which is going to be common and you will get cos square theta plus cos square theta plus cos square theta minus 2 pi by 3 plus cos square theta plus 2 pi by 3. So, what you have got is this is the first term of c p L s r d. So, remember just before you lose track let me just recall what we are doing this is the final relationship between the d q fluxes and the d q o currents and of course, the rotor currents. This has to be evaluated remember that c p is a function of theta L s r L r s and L s s r functions of theta. What I am just trying to show you is the derivation of 1 term in fact, the 1 the first term or rather the 1 comma 1 term of this matrix and I leave of course, the computation of all other terms to you as an exercise. So, I will just do it for 1 term. So, that term comes out to be k 1 M a f in 2 yeah what should it be. Now, this you must have learnt of a trigonometric identity cos square theta plus cos square theta minus 2 pi by 3 plus cos square theta plus 2 pi by 3 is a constant. In fact, it is equal to 3 by 2. So, what we find here effectively is that the first term is this in fact, we have already defined what k d k 1 is k 1 is nothing but 2 by 3 into k d k d is of of course, defined in the transformation c p into M a f into 3 by 2. So, what you get is a very convenient kind of number that is equal to M a f divided by k d. So, this is the first term of this matrix. So, let me just show this to you again. So, what we are doing is we have got the first term of this. In fact, it is not a function of theta it is not a function of theta it is a in fact, M a f by k d it is a constant. So, it turns out that all the terms of these this matrix are in fact, not functions of theta. So, that is one very interesting effect of this transformation. So, if I look into the flux and current relationships in these new variables in the d q variables you will find that in fact, they are not functions of theta. So, if you look at this particular matrix here I will just maybe call it some name let us just say its terms are L s s dash this is L s s dash this is L s r dash L r s dash and L r r dash of course, L r r and L r r there is no dash here this simply L r r because L r r does not get transformed at all. Then it turns out that all these all these sub matrices are in fact, not functions of theta. So, an interesting if you really sit and derive it you will get L s s dash is equal to in fact, it turns out to be a diagonal matrix where of course, L d is nothing but, written in terms of the what I can call as the primitive parameters L a a 0 L a b 0 plus 3 by 2 L a a 2 what are these L a a 0 L a b 0 etcetera. Please recall that L s s was made out of you know this L a a 2 L a a 0s and L a b 0s this whole L s s matrix was made like this. So, L s s dash which is the relationship between the transform flux and currents are in fact, related by this and L q is equal to L a a 0 minus L a b 0 minus 3 by 2 L a a 2 and L 0 is nothing but, L a a 0 plus 2 L a b 0 what you notice here of course, if L a a 2 is equal to 0 L d and L q are equal. So, in fact, when does this occur this occurs if there is no saliency. So, in fact, L a a a 2 will become equal to 0 if there is no saliency. So, what you get is L a L d and L q will become equal of course, in general it is not true because this is non 0. So, one of the important things is it is not only time it is not a function of theta L s s is not a function not only not a function of theta, but it is also diagonal. So, there is a that makes it very neat. Similarly, L s r dash is nothing but, M a f by k d M a h by k d 0 and 0 and you will have M a g by k d M a k by k q I am sorry k q and 0 and L r s dash is equal to L s r dash transpose no it is not in general it is not that is an interesting point. So, although L s r L r s and L s r are transpose of each other this is not true in general I should show you that it can be made true if you for a certain choice of k d and k q. So, in fact I am not defined what k d and k q are. So, there is this arbitrary constants. So, actually L r s dash comes or turns out to be 3 by 2 times M a f k d and 3 by 2 times M a h k d and 0 0 0 0 0 3 by 2 times M a g k q and 3 by 2 times M a k k q and this becomes 0 0 0 0. So, if you look at this and this it is not true in general. So, in general this and this will not be true they will not be transpositions of each other, but if I choose if I choose k d k d is equal to k d square is equal to 2 by 3 and k q square is equal to 2 by 3 in that it turns out in that case only it turns out that L s r is equal to L r s transpose. So, this is special values of k d and k q. So, what we have done is in fact of course, one thing you should see right away that L s r dash and L r s dash are not functions of theta that is something which you should immediately see and L r s dash is equal to I am sorry L s r transpose dash only if for a certain values of k d and k q. Now, you should recall that or just remember that the fact that the relationship between psi and I is not dependent on theta psi d q 0 I d q 0 is not dependent on theta is not affected by what value of k d and k q or k 0 you use. So, this value of k d k q and k 0 whatever value you choose except 0 of course, if you choose it as 0 you will not be able to invert c p. So, in that case whatever I am saying is not true, but any non-zero value of k d and k q will lead you to this relationship matrix this is the inductance matrix in the d q 0 variables which is not a function of theta. So, the fact what value of k d k q you choose or k 0 you choose does not alter the fact that by using this transformation you can get flux and current relationship which is not dependent on theta. So, that is one important thing which you should keep in mind I am not actually derive all the terms of this matrix it can be quite tedious, but remember that in case you are I encourage you to sit and derive every term of this matrix. I just showed you the derivation of the first term of L s r dash, but you can actually take out the whole matrix which is in fact a 7 by 7 matrix which is 49 terms from first principles and you actually do each obtain the expressions for each element in that matrix. In fact, what will help you in doing that would be the trigonometric identities cos theta plus cos theta minus 2 pi by 3 plus cos theta plus 2 pi by 3 is 0. Similarly, sin theta plus sin theta minus 2 pi by 3 plus sin theta plus 2 pi by 3 is equal to 0 and cos square theta plus cos square theta minus 2 pi by 3 plus cos square theta plus 2 pi by 3 is in fact a constant not dependent on theta equal to 3 by 2 the same applies if you replace this cos by sin. So, these are the 2 these are the 4 in fact if you take the sin identities also these 2 cos cosine identities and the corresponding sin identities in fact will help you to get you get all the terms in this matrix and you will see that they are not functions of theta they are functions of k d and k q and for special values of k d and k q you can in fact ensure that the L s r dash and L r s dash are transpose of each other. So, what we really see is we have applied the d q transformation this called in fact the Parkes transformation or the d q transformation matrix and what we have achieved here is make the flux and the current relationships in the d q 0 frame independent of theta. Now, of course, that does not end our work our work is in fact to look at also the differential equations as defined by Faraday's law. So, what was the differential equations of our machine in fact they were d psi by d t is equal to minus r i minus v. So, that was our equation in fact this is in fact composite equation it is I have not I told you that in fact this i is nothing but i s and i r. So, this is i is nothing but i s i r v is nothing but v s and v r and so on. So, if I just write the stator equations they will be d psi s by d t is equal to minus r s i s. In fact, r is a you know a diagonal matrix consisting of r s and r r. So, you will get r s i s. So, I can write the component I mean this is actually I am just writing down the stator equations minus v s. So, if I apply suppose I want to write this of these are of course, derived in the a b c frame of reference. So, these are a b c variables. So, if I want to rewrite these equations they will be I will rewrite this equation first I will write it as. So, minus d psi s by d t minus r s i s is equal to v s. So, I can rewrite this as minus r s i s is equal to minus of d by d t of c p times psi d q 0 minus r s into c p times i d q 0 is equal to c p times v d q 0. So, all the a b c variables are converted to the d q 0 variables is that ok. So, I just read out this this is c p into v d q 0 this is r s into c p into i d q 0 c p is a matrix r s is a matrix. Now, before we you know do further manipulations one interesting point which I need to emphasize here. What is minus d c p psi d q 0 minus d psi d q 0 it is nothing, but minus of c p times d by d t into psi d q 0 right this correct no it is not correct remember that c p is also a function of theta. So, the correct expression would be this plus minus of d by d t of this c p itself. So, that I will rewrite again. So, what you will get is I will just rewrite it minus d c p psi d q 0 just remember that you have to just take into account this is equal to minus c p minus c p into d psi d q 0 by d t minus d c p by d theta and then do d theta by d t and this is multiplied by psi d q 0. So, what we have this last term which I wrote down here you know will be in fact, equal to this this last term of the previous equation will be equal to this. So, what we get is the final relationship which we get is this d theta by d t is in fact, the electrical speed of this machine. So, the key difference between what we have done the transformation which we have used before and now is that it is a time dependent transformation this dependent on theta itself is dependent on time because it is a rotating machine theta is continuously changing it is a function of time. So, what we see is that when you are taking the derivative of c p into psi d q 0 we have to take the derivative of the transformation as well. This was not the case when we did use the linear time invariant systems with constant transformation matrix this is not a constant transformation matrix. So, you get this extra d theta by d t term when we take this derivative this is not very sorry this should be by d t. So, this is in some kind of you know extra term or extra speed dependent term which comes as a result of applying a transformation. In fact, one important point which you should notice that the if you consistently and correctly apply the mathematics you get this term. In fact, it cannot be reasoned out you know why we get a speed dependent term speed dependent term in the differential equations when you do the transformation cannot be reasoned out in any other way than mathematically. So, although one can try to give what is known as the physical interpretation to the d q transformation it is a good idea to first of all work out all the mathematics correctly and then interpret what we are getting. So, let me just again repeat when you apply a time invariant time variant transformation like C p on a differential equation. Remember that you may you will have to take the derivative of the transformation itself and as a result of the derivative of this transformation C p you get a speed dependent term which also called as p d m f. So, we will continue of course, with this derivation we are coming to the close of the basic modeling of a synchronous machine, but still we have a few things to work out. So, we will have we will work out those things in the next class.