 We continue our discussion on the DQO transformation on the flux equations of a synchronous machine. Recall that the DQO transformation is a time variant transformation and the basic reason why we actually go for DQO transformation is to get a set of time invariant equations as far as the flux relationships are concerned. In the previous lecture we began transforming even the differential equation that is the Faraday's laws apply to the flux of the machine. And we notice that they were whenever we apply a time variant transformation in the flux equations in the new variables we do get what is known as a speed EMF term. We will just recap what we have done first before we proceed further in this lecture. So, today's lecture is we continue to see the implications of the DQO transformation on the machine equations. Now recall that the DQO transformation essentially transforms any A, B, C variables the phase variables A, B, C using a transformation matrix C p into F D, F q and F 0 which are known as the DQO variables. C p is a function of theta theta is of course, the electrical angle which you have discussed before similarly of course, F D, F q and F 0 are equal to C p inverse of F A, F B, F c. So, we assume of course, C p inverse does exist in fact we can verify that it does exist. Remember that by doing a transformation we are just we are going to reformulate the equations in the new variables. So, we are not writing any fundamentally new physical equations, but we are just reformulating the existing we will be reformulating the existing flux relationships as well as the flux differential equations in the new variables. So, once we can do the analysis in the DQ 0 variables which is presumably going to be simpler. Remember we do the analysis we can transform back to the A, B, C reference frame using this transformation. So, that is the basic idea of the transformation. Now, C p of course, is this this K D, K q and K 0 are non-zero constants 3 non-zero constants which are used in the transformation. Now, the reason why I have not put any numerical value on to this K D and K q and K 0 is that it is not essential to the values of K D, K q and K 0 can be arbitrary, but non-zero if there is zero of course, we will not get C p inverse, C p inverse will not exist. So, K D, K q, K 0 are in fact non-zero constants and whatever be their value as long they are not zero, they satisfy our purpose of trying to convert the time variant flux relationships into time invariant flux relationships in the DQ 0 variables. C p inverse in fact, looks like C p transpose. In fact, if you look at the structure you will see that you have got cos theta, cos theta minus 2 by 3 and cos theta plus 2 pi by 3 these constants K 1, K 2 and K 3 are related to K D by this relationship K 1 is equal to 2 by 3 times K D, K 2 is equal to 2 by 3 times K q and K 0 is equal to 1 upon 3 times K 0. So, this is basically the inverse you can just verify this of course, at leisure by multiplying C p with C p inverse and just verifying that it does turn out to be an identity matrix which will of course, verify that C p inverse is indeed the inverse as I have written it down here. Now, the basic idea here is to convert the flux equations. So, what we will do is try to convert the a, b, c the psi s remember is nothing but psi a, psi b and psi c and psi r of course, is psi f psi h psi g and psi k psi d q. So, what we do is try to change the variables a, b, c to d q 0 of course, psi r will keep unchanged. So, the transformation if you look at it is C p this is sub matrix C p the sub matrix of this larger matrix has got this element as C p this is a 3 by 4 null matrix this 4 by into 3 null matrix and this is an identity matrix which is 4 into 4. So, what I will do is I will apply the transformation to these variables remember that the flux current relationship is psi s is equal to L s s of theta L s r of theta L s s of theta into i s into L s r of theta into i r and similarly psi r is expressed in this fashion. Now, if I call this the L matrix I will call this L it is it is easy to see that psi if you try to reformulate this particular equation we are just reformulating it in the new variables. So, psi d q 0 psi r will be this into L L is nothing but this matrix remember of course, these are sub matrices. So, this into L into this transformation into i t this transformation into d q 0 in i r. Now, the good thing which we saw in the last lecture was when we work out this we did work out one particular term in this L matrix, but I encourage you to actually try out to evaluate the complete L matrix the complete L matrix if one evaluates one will find that the of course, this you get this new relationship where this dash I have called this L s s dash L s r dash L s s L s r s dash in L r r L r r force remains unchanged. So, L s s dash. So, this should be L s s dash is nothing but it turns out is a diagonal matrix that is an interesting thing it is a diagonal matrix where L d L q and L 0 are not functions of theta. So, that is the basic beauty of this transformation is that the relationships in the d q 0 variables are in fact, the flux current relationship in the d q 0 variables are in fact, independent of theta not only that as you see L d for example, if I if you just expand this you will find that psi d is equal to L d into i d plus 0 into i q plus 0 into i 0. So, there seems to be a decoupling in the flux relationships that is also an interesting thing which happens because of the fact that this is diagonal. In fact, L s r dash is this m a m a f by k d m a h by k d 0 0 0 0 m a f by k d 0 m a g by k q this is wrongly written. So, I will write this a k q and m a h by k q and the last column of course, is the last row is 0. Now, L s r dash is equal to L r s dash only if this is true. So, it is not true in general see remember L s r was equal to L r s transpose, but this is true only if we choose k d and k q such that they satisfy this relationship k d square is 2 by 3 and k q square is equal to also 2 by 3. Now, what you see here is of course, there is a complete decoupling between what are known as the axis coils what are the d axis coils f h and the stator coil which is transformed into d q frame. So, if you look at this will be f h and d on this and g k and q here on the q axis. So, the basic idea here is that basic thing which comes out is that the fluxes in the coils on the d axis including this fictitious coil we will call it the d axis coil is not dependent on the q axis currents. Now, that is a very interesting thing in fact it is tempting to think that this transformation what it essentially has done is represented the three stationary stator coils in the a b c frame of reference. It has kind of converted this three stationary a b c coils into two rotating d q coils which are also rotating at this angular frequency omega. This is tempting, but as we shall see soon when we write down Faraday's equation we will see that the E M F induced in this will be dependent on the fluxes caused in this axis. So, we should remember that although this gives a nice picture of the decoupling of the fluxes and the currents in the d and q axis the E M F equations of course, have a coupling. So, what do I mean by that we did this in the last class we have seen that d psi by d t psi of course, is nothing but psi s and psi r is equal to minus of r i minus of v this is what basically our flux equations were which we did in you know about two classes back r of course, is a diagonal matrix which has sub matrix r s and r r r s and r r r of course, other it is a diagonal matrix containing the resistances of the f g h and k coils i of course, is i s and i r v is nothing but v s and v r. So, what you have if you just take out the psi s equations you will have d psi s by d t is equal to minus r s i s minus v s of course, psi s you know the subscript s actually denotes. So, psi s is nothing but actually psi a psi b and psi c and similarly i s and v s r i a i b i c and v a v b b c. Now, if you look at this equation we can substitute psi a psi b psi c as nothing but c p into psi d q 0. So, our equations in fact become this. So, I have just rewritten the equations these equations. These ones I have rewritten it like this and the rewritten equation I have substituted for i s which is nothing but i a i d i c I substituted it by d q 0 variables. So, v s is nothing but c p into v d q 0. So, if you recall our discussion in the previous class I had asked what is c p psi d q 0. In fact it is minus of c p is minus of this is nothing but minus of c p this is applying the chain rule plus this extra term. Remember that c p unlike some of the transformations we did in the first few lectures of this course c p is in fact a function of theta. Theta in a synchronous machine is a function of time it cannot be considered even under steady state conditions you will find that theta is in fact varying continuously. So, d c p by d t in fact has to be evaluated you have to apply chain rule. So, what we will get is this is in fact having two terms. So, your flux relationship so I will just this has been rewritten again this particular term d c p by d t can be written as d c p by d theta into d theta by d t. So, this extra term as we discussed in the class previous class comes out because of being mathematically consistent while applying the transformation of variables. This extra term in fact denotes see what we you know this you know rate of change of flux of course is common with the a b c equations, but we have got this extra what is known as p d m f term. So, if you look at this d c p by d t if we actually evaluated in fact. So, we carry on forth from last time d c p by d t d theta is equal to it is easy to see this you take the derivative of the transformation. So, we will get minus k d sin theta k q cos theta and of course, k 0 when you take the derivative you will get 0. So, I will write 0 here and you will have minus k d sin theta minus 2 pi by 3 difficult to fit it in a paper, but we will just try to do that k q into cos theta minus 2 pi by 3 and this will be sorry this is minus k d sin theta plus 2 pi by 3 and this term here is k q cos theta plus 2 pi by 3. So, this is basically what we have I will just read out this term it may not be very clear here it is k q cos theta plus 2 pi by 3 and this is of course, a distinct term here. Now, it is very easy to see I mean it is not too difficult you know how what c p looks like c p looks like this sorry this is c p inverse I am sorry c p looks like this. So, it is very by inspection you can really make out that this is nothing but c p into a matrix p 1 where p 1 is nothing but 0 k q by k d 0 minus k d by k q 0 0 and this is 0. So, what I have. So, let me just repeat d c p by d t d theta is equal to c p times p 1 where p 1 is this. So, our final equations are here. So, we have this flux equations come out to be minus of c p d psi d q 0 by d t minus d theta by d t. So, I will just call this theta dot c p p 1 psi d q 0 minus r s into c p i d q 0 is nothing but c p into v d q 0. So, your flux equation becomes in fact I will just multiply pre multiply both both sides by c p inverse you will get minus d psi d q 0 by d t minus d psi d q 0 minus theta dot into p 1 into psi d q 0 minus r s of course, remember is a diagonal matrix containing r a r p and r c of course, if all the coils are identical I can just say r a into i d q 0 is equal to v d q 0 which leads us to if I really write this down separately you know because it sometimes is not very evident what we are you know getting unless you write down all the equation separately. So, what I will do is I will write down the d q 0 equation separately. So, the d q 0 equation separately turn out to be minus d psi d by d t minus d theta by d t in fact or theta dot is nothing but it is nothing but omega angular frequency and this is k q by k d psi q minus r a i d is equal to v d minus d psi d q by d t plus omega k d by k q psi d minus r a i q is equal to v d psi d by d psi d by d psi d by d psi d q and minus d psi 0 by d t minus r a i 0 is equal to v 0 where omega is equal to d theta by d t or d theta dot. So, if it is not clear I am just panning this a bit. Now, one interesting thing which you should see here is that there is what I mentioned sometime back this p d m f terms. In fact, although we saw that there is complete d coupling between the d and q axis. So, recall we had a complete d coupling between the d and q axis if you recall this equation or the flux equations. So, in fact we had mentioned sometime back that psi d is dependent on i d i f and i h, but it is not dependent on i q i 0 or i g and i k it is not dependent on the q axis. So, the flux current relationships there is a complete d coupling between the d and the q axis coils. So, I call this the d axis coil. So, the fluxes in the d axis coil are not dependent on the q axis currents none of the currents, but importantly when you look at the flux equation there are p d m m f terms. So, thus the flux Faraday's law when you apply to psi d psi d you have to put this extra term which comes because of applying the correct mathematics to the transformed equations. So, although sometimes it is tempting to start from a kind of physical model of rotating windings when we come to obtaining the flux equations in the d coil or the flux equations in the q coil. Remember that there are p d m f's in the d coil due to the flux in the q axis due to the flux in the q axis. This cannot be explained by just starting off from this model. So, one of the things which you should keep in mind is that it is a good idea as I mentioned in the previous class to first work out the mathematics. The correct mathematics give you the give you gives you these equations where there are extra p d m f terms this p d m f terms come because we apply the derivative to the time varying transformation as well. These are the correct equations. So, please remember that there is coupling coming between the d i and q axis coils because of these extras what I call as p d m f terms. Now, these are as far as the stator equations are concerned the rotor equations of course, we know and d psi g by d t is equal to sorry plus r g i g is equal to 0. So, just remember these are the remaining equations. In fact, if you know the flux if you know the flux and current relationships and these equations relating to the e m f's the rotor flux equations in the d axis and in the q axis we in fact, have got the complete flux description. So, if I know of course, I should the relationship psi d q 0 psi r I know this relationship as well d q 0 and i r and this is nothing but l s s dash l s r dash l r s dash n l r r. So, we have in fact, got a complete picture of the system. In fact, you can substitute for i g i k i f i h as well as i d q 0 by psi d q 0 and psi r using this relationship and you will get finally, things in the state space form just do it on a separate sheet. So, you will get you can of course, write it as psi dot is equal to some a matrix into psi plus b into v. So, we can get it in this form where psi is nothing but psi d q 0 and psi r. This a is in fact, does not have time coming in explicitly you do not have time or theta coming in explicitly. Of course, a does contain omega remember that because of the fact that the equations of flux have this p d m f term a is a function of omega. So, but if omega is a constant then this a becomes linear this particular equation become linear time invariant. So, that is the beauty of applying the d q 0 transform if your constant speed of course, this becomes a linear time variant equation. There is no explicit dependence on theta which itself changes with time. So, this is one of the main you know important things which really comes out of of applying the d q transformation. Now, in general of course, whenever you have a linear time variant system it is not obvious that by transforming it in a certain way you will get a linear time invariant system. It turns out that the machine equations have special a special structure which permit the use of this transformation C p which make the system time invariant. Now, one of the things we have not discussed now we will shortly again discuss it. This feature of making that you know equations of flux equations of a machine time invariant is not dependent on the specific values of k d k q and a k 0 it could be absolutely arbitrary choice of course, k d k q k 0 should not be 0 none of them should be 0. So, there is some flexibility in the choice of k d k q k 0. Now, before we go on to discussing this particular point again let us just look at what happens to the torque equations. Now, if you look at the torque equation just recall what was the equation for torque. If you recall what we did sometime back I will just show you the equations first. So, you can recall what we have done this was the torque equation in the previous lecture where T e is nothing but the minus of the partial derivative of co energy with respect to the mechanical position torque T e of course, is in this direction theta is measured in this theta and theta m are both measured in this direction the anticlockwise direction here the co energy is expressed as a function of currents. So, you have got this particular formula and of course, eventually we did get this. So, T e which is the electrical torque is nothing but minus of p by 2 into p is the number of poles into d by derivative partial derivative of co energy with respect to theta which is the electrical angle. So, we call of course, T e dash as this. So, T e dash turns out to be this. So, this is what our equations are I basically took out the derivative partial derivative of co energy with respect to this. So, now if I want to get this expression for torque T e dash remember T e the actual torque is actually minus p by 2 times this. So, we will have T e is equal to T e dash the actual torque remember is p by 2 minus p by 2 p by 2 times T e dash. So, T e dash is nothing but 2 by p times T e half. So, I substitute here. So, I have got I s transpose here. So, I will have to write I s remember is I a I d I c. So, I can write this as I d q 0 transpose into C p transpose into the partial derivative of L s s with respect to theta L s s remember is the sub matrix of the original L matrix into there is not much space here. So, I will just we can write it here I guess I d q 0 I s here is C p into I d q 0 inadvertently we have just written I d q 0. Remember I d q 0 is a row or rather a column I d I q and I 0. So, this is a neat way of writing it plus in fact, this is 2 times I d q 0 transpose C p transpose d L s r by d theta into I r closing on the curly bracket. So, what we get is this you can just work out we have already about 2 lectures back we took out what L s s is actually. So, I will just write down what d L s s by d theta is nothing but minus of 2 L a a 2 this is a time varying part or the theta varying part into sin of 2 theta sin of 2 theta minus 2 pi by 3 sin of 2 theta plus 2 pi by 3 sin of 2 theta minus 2 pi by 3 sin of 2 theta plus 2 pi by 3 we have sin 2 theta and of course, this will be I will just since it may not be very clear I will just read it out sin of 2 theta plus 2 pi by 3 this is not very clear here. So, maybe we will just write it down again I will just write sin of 2 theta sin of 2 theta sin of 2 theta sin of 2 theta sin of 2 theta. This is this L s s by L theta d L s r by d t we will partition into d L s r d by d theta and d L s r q by d theta. So, that turns out to be m a f sin theta minus m a h sin theta. So, bit of book keeping is required, but as we shall see soon then finally, what we get as a torque expression after using these formulae is quite neat. So, this is what d if we know d L s r is this. So, we have found out the sub matrices which I mentioned sometime back. Now, you can see that d L s s by d theta into C p is nothing but minus of 3 L a a 2 into C p into p 2 where p 2 is 0 k q by k d 0 k d by k q 0 0 and 0 0 0. So, you can rewrite this. So, if you look at the structure of d L s s by d theta if you look at the structure of it you can write it as in this fashion. Therefore, and further you will find it C p transpose of d L s r by d theta is nothing but in fact, it gets stripped of all the theta is eventually 0 0 3 by 2 k d m a g 3 by 2 k d m a k and this becomes minus 3 by 2 m a f minus 3 by 2 sorry there is a k q here. So, there is a k q k q m a h 0 0 0 0 0 0 and 0. So, bit complicated but this is how it comes out to be you can just verify this. So, eventually we get we will just write down the torque equation T e comes out to be 3 by 2 times k d k q into i q into m a f by k d i f plus m a h by k d i h plus 3 by 2 L a a 2 i d. So, close this bracket then continue minus i d times m a g by k q i g plus m a k by k q i k minus 3 by 2 times L a a 2 i q close this bracket and close this. So, this is the expression for torque, but we know that. So, I will just partition this here we know that psi d is nothing but L d i d this comes out from the flux equation psi d is equal to L d i d plus m a f i f by k d plus m a h by k d i h and of course, psi q is nothing but L q i q plus m a g by k q i g plus m a k by k q i k. So, what do we understand from this if you look at these equations look at this and you look at this. So, one can directly infer from this that T e is nothing but 3 by 2 times k d k q again will put a curly bracket here i q into psi psi d minus L d minus L a a 2 will rewrite this again you can rewrite T e is equal to 3 into k d into k q into a curly bracket here into i q into will put a square bracket here psi d minus L d minus 3 by 2 L a a 2 into i d and I close the square bracket minus i d into psi q minus L q plus 3 by 2 L a a 2 into i q. So, I close this bracket and I close this bracket. So, if you from this it is very easy to see you know I you know you can cancel of this term here you will get really after all the manipulations finally, it yields 3 by 2 times k d k q psi d i q minus psi q i d. In fact, this is a small correction here. So, I should be calling this T e dash this T e dash nothing but T e dash T e dash this. So, what we have here is basically a very neat expression which is again devoid of any of the theta's they all in terms of d and q variables only. So, that is one interesting thing. So, our equations are turning out to be extremely neat now. In fact, if you write down the equations in d q variables you will find that your equations no longer will have theta explicitly appearing in them they are much easier to handle. The flux current relationship is brings out a decoupling between the d and q axis fluxes and currents. The flux differential equations of course, bring out a coupling between the d and q axis fluxes there is a coupling. We now get the torque equation remember what the torque equation was if you recall where we were sometime back this. So, T e dash we have actually found out right now which is which I said was a function of theta. But of course, because we applied the d q transformations it turns out that it is not explicitly coming the theta does not come out explicitly if you express T e dash in terms of the d q variables. So, it is a very very interesting and you know very neat expression we get for torque. Now, one thing about the choice the last topic in this particular lecture what about the choice of k d and k q we saw of course, that if you put k d square is equal to 2 by 3 and k q square is equal to 2 by 3 one interesting thing happens that is L s r dash becomes equal to L r s dash transpose. Also if in fact, we saw sometime back that when we take out C p inverse it is in terms it has got elements k 1 k 2 and k 3 where k 1 is nothing but 2 by 3 times k d and k 2 is 2 by 3 times k q in fact, k q is in the denominator and k d is also in the denominator and k 3 is equal to 1 upon 3 k 0. So, these are the coefficients of the signs and cosines in the C p inverse matrix. If you look at you know these relationships it appears that it is not a bad idea to choose k d equal to k q is equal to root 2 by 3. So, that is an interesting and important point here of course, there is absolute freedom in choice of k d and k q in so far as our main objective of getting a time invariant set of equations. Remember that k d k q do appear in the flux current relationship as well as the flux differential equations as well as in the torque expression here sorry this should be k q, but the fact remains that although k d k q do appear theta does not appear. So, whatever be the value of k d and k q as long as a non zero we do end up with what we have set out to do that is get a neat set of equations which are invariant with respect to theta. So, if we choose k d and k q root 2 by it is a very special choice it will lead us to L s r dash being equal to L r s dash transpose which is convenient it is very convenient also we will find that C p inverse will be equal to C p transpose. So, it is of course, it is easy to remember that could be one advantage of choosing k d and k q in this fashion. So, it is easy to remember that C p inverse is equal to C p transpose, but let me repeat k d and k q could be arbitrary in so far our main objective of getting time invariant equations is concerned. So, let me just summarize, but before I do that it is k d is equal to k q is equal to root 2 by 3 is a kind of a is our it is a special choice and that is the choice which I will really follow in this particular course it is in fact, the same choice is using k r Padhyas book. In fact, if you look at the other literature or literature or many industry papers they follow a convention k d is equal to 1 and k q is equal to minus 1. So, they follow this choice in fact, of course, I did not talk about k 0, k 0 if I want this to be true in fact, I should have told you before you have to choose k 0 is equal to 1 by root 3. So, k 0 also gets defined if I want to have this to be true. So, just remember this again that many industry papers as well as you know in books by Kundur and many other books they follow the convention k d is equal to 1 and k q is equal to minus 1. In this particular course I shall follow Padhyas convention which is or the IEEE convention actually which is k d is equal to k q is equal to root 2 by 3 because it you know gets us some benefits. In fact, you will notice that in this talk expression if I choose k d and k q as root 2 by 3 this you know this whole coefficient gets becomes equal to 1. So, I do not have to remember these coefficients because this if I put k d is equal to root 2 by 3 and k q is equal to root 2 by 3 this coefficient effectively becomes 1. So, there are certain advantages in choosing this, but as I mentioned back the choice is arbitrary it does not hamper or refer to get a time invariant set of equations. Of course, the question then arises if I have if you have got industry papers or papers or books which follow this convention I will call this people who follow this convention as using a transformation C p 1 and v in fact, apply a transformation C p where k d and k q are root 2 by 3. In such a case one can transform the equations which they have used the people who are using these this set of values of k d and k q by again a kind of matrix T k. So, I can actually the variables which they have defined are different from my variables because the coefficients k d k q have been chosen by them to be 1 and minus 1 and we are going to choose root 2 by 3. So, whenever I am going to use their equations and I get answers with their variables I would need to use this transformation matrix T k to transfer from their variables to my variables. So, I leave this as an exercise for you to find out what this T k is it is a very simple constant matrix. So, to summarize this particular lecture what we have done is finally, obtained the equations of the synchronous machine in the d q frame of reference and the equations are neat in the sense that they are not functions of theta. In fact, you can if in fact, the speed of a machine is constant that is d theta by d t is equal to constant the flux equations the flux differential equations are linear time invariant equations of course, if speed is not constant in fact, in general it need not be a constant you have a coupling between the mechanical equations and the flux equations flux differential equations. In fact, the coupling is non-linear remember the torque is psi d i q minus psi q i d. So, there is a product terms there also there is a product when you are talking of the speed m f terms in the flux differential equation there in also there is a product. So, the mechanical equations and the flux equations are in fact, coupled in a non-linear fashion, but the flux equations themselves for a constant speed are in fact, time linear and time invariant that makes at least some of our analysis much much simpler. So, now we move on to you know trying to interpret the equations which have come correlated with parameters obtained by measurement and thereafter the most important thing of course, in this course is to draw inferences from what equations we have got and correlate them to actual power system behavior.