 we were looking at transformation to the stationary reference frame, we started off with the machine description in the three phase and then from that we went on to the aß0 reference frame both aß0 for the stator and aß0 for the rotor and we said that the 0 sequence is most likely not there so we can neglect the 0 sequence component. And this aß0 reference frame then has two axis one is the aß0 axis that of the stator and then you have the aß0 axis that of the stator. The aß0 axis has an MMF oriented along its direction due to an aß0 coil and then similarly the aß0 axis has another MMF oriented along that and we also then have an aß0 axis and then aß0 axis which is at 90 degrees to that so you have a aßr coil and an aßr coil. So this is a representation of your two phase machine neglecting 0 sequence component and for this we can write the expression in terms of voltage equals r into i plus p of inductance matrix into the i vector v is in the aß0 reference frame for the stator and aß0 for the rotor i is also aß0 here and aß0 there. The inductance matrix we have seen can be written by observation and the inductance matrix can then be written as L let us call it aß0 so that there is no confusion. The inductance matrix can then be written as Ls of the stator to distinguish this from the self inductance in the three phase case let us call this as Ls bar and then there is a question of mutual inductance between the aß and bß axis and we observed yesterday that that is 0 so the mutual inductance is 0 let us give some more space here so mutual inductance is 0 and then you have aßr into cos of ?r representing the mutual inductance between the aß and the aßr axis and then you have –msr sin ?r represents mutual inductance between the coil on the aß axis and the coil on the aßr axis and the second row would then represent the flux linkages or the inductance of the aß axis and since the two coils are identical on the aß and bß axis the self inductance term is the same and then you have between aß and the aßr axis so that is sin of ?r into msr bar and then you have msr bar cos ?r and then here you have msr bar cos ?r and then this is –msr sin ?r and then here you have msr sin ?r cos ?r here you have the rotor inductance there is no mutual inductance between the aßr and bß axis because again they are at 90 degrees and therefore there is 0 here and then lr here so this is the inductance matrix neglecting 0 sequence components if 0 sequence component was there you want to include that then you would have one column here extra one column here extra one row here extra and one row here extra only the term coming here will have an entry and that corresponds to the leakage inductance of the stator and similarly here there would be another entry which corresponds to leakage inductance of the rotor we already know that from the earlier derivations that we have done going from three phase to two phase now we can see that this inductance matrix can be conveniently divided into four sub matrices so this matrix we can call as l11 this sub matrix l22 this is then l12 and l21 obviously we can see that l21 is equal to l12 transpose. So this expression v aß equals ri plus p x li describes the electrical system of this two phase machine and we see that the inductance matrix depends on the rotor angle which is what we observed in the last lecture and in order to get rid of that we said that one can transform the rotor to a stationary axis having observed that the rotor angle arises or occurs in this description because the rotor is moving with respect to stator and looking at the mmf produced by the rotor a r axis and the ßr axis at a given angle ? yesterday we found that the same mmf could have been developed by two coils that are placed here one on the a s axis along the same axis a s another along the ß s axis. So if we consider two fictitious coils one along a s axis and another along the ß s axis both representing together the mmf that would have been produced by the a r and ßr excitations then these two mmf being equal one can define a relationship between the flow of currents here we will call this current as i this is the flow of current even though this coil is along the a s axis this flow of current is determined by both a r and ßr and therefore we will call this as current flowing along the a axis due to the rotor currents both the rotor currents so we call that as i r with the subscript a similarly here the current that is flowing is i r with a subscript ß so we define the relationship yesterday between the two currents i r a and then i r ß this relationship we found that it was cos ßr – sin ßr sin ßr and cos ßr multiplying i a and i ß these are the currents that are flowing actually in the rotor of course in the two phase axis of the rotor not the three phase machine and these can then be equivalently thought of as being some currents the same mmf being produced due to these two currents which are flowing now these currents are flowing in the pseudo stationary fictitious s winding and the ß s winding so having defined this we can now observe that if you know these two currents you can get these two currents and similarly if you know these two currents it is feasible to get those two currents how can you do that you need to invert this and bring it over here and you see that the determinant of this determinant of this is simply cos ßr – sin ßr which is 1 and therefore the inverse of this is simply the transpose of this and therefore if you call this by the matrix C22 then we can say that i a ß is equal to C22 transpose ßr a ß so this enables you to go from a ß to fictitious coil or fictitious coil to a ß and as we observed yesterday this does not have zero sequence if we have zero sequence that does not go through this equation zero sequence is retained as it is so you need to add one row of zeroes and one two zeroes here with the diagonal entry being one and the same zero sequence therefore comes here. Now this defines a relationship between i variable here and i variable here we would like in this case in order to transform the variables from the rotating reference frame to the stationary reference frame then we would like to have power invariancy which then means that v a ßi a ß transpose x v a ß you would like it to be i r a ß transpose x vr a ß this is the power in fictitious coils this is the power in a ß coils in the rotor and therefore if we have to get this we can now use this relationship and then say that replace this expression that is C22 transpose x i r a ß this whole thing transpose x v a ß equals i r transpose a ß into vr a ß which then means that i r transpose a ß into C22 transpose transpose a C22 into v a ß equals i r transpose a ß into vr a ß so looking at this expression we then conclude that vr a ß equals C22 into v a ß therefore the transformation for voltages is the same as the transformation for this i a ß here and i r a ß here are related by this C22 similarly here also it is related and therefore we have defined that relationship between this voltage and fictitious coil voltage. Now having defined all this we now need to see how we can apply this to this expression for the two phase machine. Now we want to transform only the rotor to fictitious coils here the stator is already stationary and therefore if you transform the rotor to stationary here then all these four are stationary with respect to each other we hope therefore that the rotor angle will not appear in the equations that is what we need to verify now and therefore we will define an augmented matrix C which is 1 0 or let me put it in a short form this is the identity matrix U which is 2 by 2 and then 0 matrix and C22. So having defined this then we can now define these two that is i r a ß equals a ß of this at the rotor and a ß of the stator this is therefore equal to C ß i a ß and a ß similarly you would have the relationship v r a ß of the stator a ß of the rotor is equal to C ß v a ß and a ß or if you want to put it in an expanded form what this matrix simply means is that v a v ß v r a v r ß is 1 0 0 0 0 1 0 0 0 0 cos ß ñ time ß r 0 ß ß ß ß so that is what this relationship means and therefore we see that we are going to retain the stator voltages as it is there is no change in the stator variables the rotor variables however undergo a change from the a ß axis of the rotor to a ß fixed on the stator. So these two variables are pseudo stationary variables in the stator reference frame these two variables are in the rotor reference frame similarly one can define this these two equations are similar now using this we would like to see what would happen to the machine description if we use these two equation and substitute it back in our machine description that we wrote. So our machine description was v a ß a ß equal r i a ß a ß plus derivative of the flux linkages now substitute these equations V aß is nothing but C inverse x Vr aß aß and therefore we can substitute it there this Vr aß aß is this matrix where aß is stator variable and the rotor is also referred to aß so we have C inverse Vr aß aß equals r into C inverse Ir aß aß plus derivative of L aß aß multiplied by C inverse Ir aß aß so having substituted this expression in the original machine equation voltage equation this is what you get and in order to get Vr aß aß you need to multiply by C matrix throughout and therefore you can multiply this by C and multiply this by C as well and here you have C inverse appearing if you look at the matrix we have here C inverse is going to be C transpose as you can see from this therefore that should be easy to observe otherwise I would leave you to verify that C inverse of this is indeed C transpose and therefore this can be written as Cr C transpose Ir aß aß plus C times P of L aß aß C transpose Ir aß aß now this is what we need to expand and we will try to do that now this matrix Cr C transpose is fairly easy to simplify I will leave that again to you as an exercise it is easy to show that this matrix is nothing but R and therefore this part of the expression is fairly easy we need to focus on this part of the expression therefore that can now be written as I times R aß aß plus this expression can be written as C l aß aß C transpose P of Ir aß aß plus C times P of L C transpose into Ir aß aß so we need to evaluate this matrix C l C transpose and C P l C transpose which we will try to do now let us try to simplify the expression we will consider the term l aß into C transpose so this can be written as l11 l12 l21 l22 l21 is nothing but l12 transpose so we will make use of that here into C transpose is nothing but identity matrix 0 0 C22 transpose so this is l aß aß C transpose so this can be written as l11 l12 P22 transpose l12 transpose and l22 P22 transpose so what we need to evaluate are l12 C22 transpose and l22 C22 transpose so l12 P22 transpose is nothing but you can see that l12 has MSR dash that is common so we would take that out and the remaining terms are cos ? R – sin ? R sin ? R and cos ? R multiplied by C22 transpose and we do not have the C matrix here so let me write the C22 matrix here C22 was cos ? R – sin ? R sin ? R and cos ? R and therefore the C22 matrix is cos ? R sin ? R – sin ? R and cos ? R this whole thing multiplied by MSR bar is this term l12 into C22 transpose so this can this is therefore equal to MSR bar the first term is cos ? ? plus sin ? that is 1 and the next term here is cos ? sin ? – sin ? cos ? so that is 0 and here you have sin ? cos ? – sin ? cos ? is 0 and sin ? plus cos ? is 1 so l12 C22 transpose reduces to a fairly simple matrix that is what you have. The next term that you need to get is l22 C22 transpose l22 is not depending on the rotor angle it is simply LR LR00 and therefore this can be written as LR multiplied by the identity matrix 2 by 2 into C22 transpose which is simply LR – into C22 transpose so it is the same C22 matrix that you have multiplied by LR having defined these two then if you look at the expressions that you have there are two things that we need to do one we have determined l ? ? C transpose and we now need to on the one hand pre-multiply by C on the other hand you need to differentiate this and then pre-multiply by C so let us first pre-multiply by C and then we will do the differentiation so if you now pre-multiply this by C what we are trying to do now is C L ? ? ? C transpose so this is identity matrix 00 C22 multiplied by this term is what we have determined earlier so that is l11 l12 C22 transpose is what you have as MSR bar into identity matrix U and then you have l12 transpose l22 C22 transpose is nothing but LR bar into C22 so that is what you have and if you multiply this what you get is l11 the first term reduces to l11 the second term reduces to MSR multiplied by identity matrix the third term reduces to C22 into l12 transpose and the fourth term here reduces to C22 LR C22 transpose remember LR is a scalar it is not a vector and therefore this reduces to the scalar LR dash into C22 C22 transpose and we know that C22 transpose is nothing but the inverse of C22 and therefore this term reduces to LR into identity matrix here you have C22 into L12 transpose we have already determined L12 into C22 transpose so this term is nothing but the transpose of this and this is already MSR into U and therefore this is nothing but MSR into U so we have determined what is CL ? ? ? ? into C transpose that is nothing but this matrix so we will now keep this aside for the moment and focus on the derivative of LC transpose so we will write this somewhere on the board we will write it here so CL ? ? ? C transpose is nothing but L11 and then MSR bar into identity matrix MSR bar into identity matrix and this matrix is the same as L22 so we will retain it as L22 and we can remove this expression now we need to find out the derivative of LC transpose so let us try to do that the derivative of LC transpose so yes L ? ? ? ? into C transpose is L11 L12 transpose L12 into C22 transpose was this so that is MSR into U and this one is LR bar into C22 transpose now if you take the derivative of this expression each of these terms has to be differentiated derivative is with respect to time and you see that L11 the inductance matrix here that is the first sub matrix here this contains only fixed elements there is not anything that is varying with respect to time and therefore the derivative of L11 is 0 so the derivative of this is equal to 0 in the first place this does not vary with respect to time MSR into U MSR bar into U you can see this matrix there is nothing again varying with respect to time therefore that is also 0 L12 transpose L12 is a matrix that is varying with respect to time because there is a rotor angle and rotor angle is going to change continuously as the rotor rotates this will vary with respect to time so this is P L12 transpose and here you have C22 again has the rotor angle here therefore C22 also varies with respect to time so therefore this is P LR bar into C22 transpose and now let us evaluate those expressions may be we will do that here we do not need these anymore so we will do this here P of L12 transpose is nothing but the derivative of this expression so if you look at this MSR bar is there everywhere that can be taken out so MSR bar and then you have this term derivative of cos ? is sin ?r and then you have derivative of sin ?r that is cos ?r and then you have derivative of – sin ?r that is – cos ?r and then you have derivative of cos ?r which is – sin ?r so you have – sin ?r multiplied by d ?r by d so this is then the first term there and the second term there involves derivation derivative of C22 transpose C22 is already there and we need to look at C22 transpose so P LR bar into C22 transpose is then LR bar multiplied by cos ?r derivative is – sin ?r and then cos ?r here – cos ?r and – sin ?r multiplied by d ?r by d we need C22 transpose so I have taken the transpose cos ?r becomes – sin ?r this sin ?r would go there and the derivative is cos ?r and the other two terms so this is what you have and what we need to do now is to evaluate the last term we need to find out C of P L ? ? ? into C transpose and therefore let us do that C is identity matrix 00 C22 multiplied by the derivative of this as we have already written down the first two terms are 0 and then you have P L12 transpose and then you have P LR bar P22 transpose this is what you have and this term then simplifies to 0 multiplied by 0 is 0 and then 0 multiplied by 0 again is 0 the third term is 0 plus C22 into P L12 transpose and the last term is C22 P LR C22 transpose so we now need to evaluate this term and this term so let us do that we do not need this expression anymore so let us get rid of that so essentially then we need to multiply this matrix and this matrix by C22 that is what this expression say so let us do that C22 is cos ?r – sin ?r sin ?r and cos ?r that is C22 multiplied by P L12 transpose P L12 transpose we already have here and therefore that is MSR bar multiplied by – sin ?r cos ?r – cos ?r – sin ?r multiplied by d ?r by d so that is going to give us MSR bar into the first term is cos ? – sin ? and plus sin ? cos ? therefore this term is 0 the second term gives you cos ? plus sin ? and that is 1 this term gives us – sin ? – cos ? therefore this is – 1 and the last term gives us sin ? cos ? – sin ? cos ? therefore this is 0 so we have found out this term this is nothing but P22 P L12 transpose that we have found out the next term is C22 P Lr C22 transpose we have to make use of this so C22 P Lr bar C22 transpose is nothing but cos ?r – sin ?r and sin ?r and cos ?r multiplied by Lr bar you keep that separately and multiplied by this term which is here so – sin ?r cos ?r – cos ?r – sin ?r multiplied by again d ?r by dt so that gives us Lr bar multiplied by the first term is cos ? – sin ? plus sin ? cos ? therefore this is 0 and then you have cos ? plus sin ? that is 1 – sin ? – cos ? becomes – 1 and then you have sin ? cos ? plus sin ? cos ? – sin ? cos ? therefore this is 0 so we have now evaluated all the sub matrices and we need to put them together to see what it looks like so let us do that assembly here you have C P L ? ? ? into C transpose is what you are going to have and this is if you see the first two entries are 0 matrices so you have 0 here and 0 here the next two entries are available in these two places the first entry is Msr bar I missed this d ?r by dt is Msr bar into a description like this 0 1 – 1 0 so let me write this slightly bigger accommodate that this is Msr bar into 0 1 – 1 0 and then here you have Lr bar again into 0 1 – 1 0 so we have evaluated all the sub matrices that are of interest so let us now put all these equations together and see what it looks like which we can do it again here so what we now have is Vr ? ? of the stator and ? ? in the rotor referred to the stator side is nothing but R into I R a ? a ? plus C L a ? C transpose into P I is a matrix which is nothing but L 11 and then Msr bar into U and this is again Msr bar into U and then L 22 P I R a ? a ? plus this matrix which is 0 0 Msr bar into 0 1 – 1 0 Lr bar into similar 0 1 – 1 0 this multiplied by d ? r by dt multiplied by I R a ? a so now you see that this matrix does not have the rotor angle in it this matrix also does not have the rotor angle in it the resistance anyway does not have the rotor angle in it therefore the electrical description of the system now does not have the rotor angle involved unlike this inductance matrix where you had the rotor angle inside so we can now try to compress all these equation and try to write this as the in a form which is similar to what we did earlier the so called operational impedance form operational impedance matrix where we will expand this term V a V ? remember the stator variables have been retained as it is no transformation to the stator variables and then you have V R the rotor variable variables have been transformed on the one hand to the a axis of the stator and on the other hand to the ? axis of the stator we have we are going to write this as V equals some Z multiplied by I so we need to pick terms from this description and then put it here so the first term here will come from the first term of this R matrix first term here and the first term there so the first term here is the resistance of the stator that will come and then you have the self inductance of the stator L11 matrix remember has an LS bar and LS bar so you have plus LS bar into P it is going to multiply the vector I then the second term here will not have any entry the resistance has entries only along the diagonal of diagonal entries are 0 so there is nothing due to the resistance here also there is nothing here the 2 to 1 to element is 0 so the second term would again be 0 and then there is nothing coming from here as well therefore this term is 0 the third term here nothing would come from the resistance matrix whereas here it is MSR bar into identity matrix and therefore this is MSR bar multiplied by P and then the last term is again 0 there so going in this manner if you select the appropriate terms from all the matrices and combine them what you would find is here again it is 0 so this entry is RS plus LS bar into P and then 0 here again and then MSR bar P and here you have MSR bar P and then MSR bar into Omega R this term comes from the matrix here MSR bar d theta R by dt I am writing as Omega R and here it is RR plus LR bar P and here you have LR bar multiplied by Omega R and here you have minus MSR bar into Omega R coming from this term and then you have MSR bar multiplied by P coming from this term and then you have minus LR bar multiplied by Omega R coming from here and then the last term here is RR plus LR bar multiplied by P so this is then the operational impedance form of the electrical equations zero sequence component as we have been mentioning earlier has been left out if the zero sequence component is going to come in that we have left it untransformed and therefore you will have another row here and another row here one column in the middle here one column in the middle there would be only the diagonal entries here and here which will correspond to RS plus the leakage inductance times P and similarly there RR plus the leakage inductance of the rotor times P. Now this matrix is then the operational impedance matrix in the stator reference frame why stator reference frame because the stator coils are already there in the stator of course the rotor coils have now been referred to the stator by having fictitious coils a and b so all the coils are now in the stator and this description will then give you the behavior of the machine as seen from the stator this description is more easier to handle because the impedance is here are not dependent on the rotor angle and therefore if you can write this expression again as the same equation as R into I plus an inductance matrix L alpha beta alpha beta PI plus another matrix which you call as g omega R into I g matrix L matrix and R matrix are not functions of the rotor angle and therefore it is easy to solve them on a digital computer these matrices being fixed. So that is one of the advantages of having the description in this form now it so turns out that by remembering a few rules this matrix can also be written by observation and by understanding those rules this matrix can also be extended to different situations we shall see how to do that in the next lecture and more utilities of the system description of this kind we will stop here for today.