 in the last lecture, we were looking at rewriting the equations of the induction machine in a different frame of reference. The original induction machine equations we had written in the so called natural reference frame where the stator variables being 3 voltages Va, Vb and Vc and the stator variables Ia, Ib and Ic, these are all written in the so called ABC reference frame that means we look at the actual voltage applied across the stator a phase, the voltage applied to the stator b phase, voltage applied to the stator c phase, the flow of current in the stator a phase, b phase, c phase and so on. And similarly we have rotor variables which are again the rotor a phase, a phase voltage, b phase voltage, c phase voltage and the flow of current in the 3 phases of the rotor. So with this we had a fairly large expression describing the voltages, the relationship between the voltages 3 here and 3 here and the flow of current 3 here and 3 here, a fairly large description we were thinking of how one can try to simplify these expressions and for that purpose we then try to look at the MMF created by or generated by 3 phase winding, the net MMF generated by the 3 phase winding and we found that the MMF created by the 3 phase winding could be approximated or could be equivalently thought to be generated by a 2 phase winding itself. So the 3 phase winding MMFs are generated by windings that are situated actually separated by 120 degrees whereas the equivalent 2 phase MMF is generated by windings that are situated such that their axis are separated by 90 degrees in space. So we call these as the a axis and the b axis whereas the original axis of the machine is then a axis, b axis and c axis. So we describe or we derive the way of transforming variables from the a, b, c or the natural reference frame axis to the a, b axis there are only 2 axis here. And in that we found that if MMF has to be invariant that is the MMF generated by windings that are there on the a, b, c axis is the same thing that is thought to have been generated by windings on the a and b axis so it should generate the same MMF. And if that is going to be the case we found that the ratio of turns between the 3 windings on the a, b, c axis and the 2 windings on the a, b axis that ratio of number of turns was something that could be decided based on some requirement of the other. We also found that if you want to go from a, b, c to a, b and then also go back having found out variables in a, b frame of reference if you now want to find out a, b, c you need to do a matrix inversion and that was not feasible unless you introduce a 0 sequence component as well. So we introduced a 0 sequence component to complete the description thereby from the a, b, c 0 reference frame so to say reference frame meaning it is just some place from which you observe instead of observing the mechanics of the machine electro dynamics nature of the machine from a, b, c frame of reference that is where you describe these variables you might as well describe it from the a, b and then now added 0 frame of reference. In order to describe the ratio of number of turns we found that there are 2 ways that are generally used one is then known as the power invariancy approach power invariant transformation and another is then the non power invariant transformation and we had seen how these 2 work and what happens to the ratio of number of turns if we use either one transformation or the other and we saw that if you are using the non p i method then we have the vector i a, b, 0 being described by 2 by 3 times the matrix m into i a, b, c this is a vector containing where i a, b, 0 is a vector of i a, i, b and i 0 and i a, b, c is a vector i a, i b, c and then if you want to find out the inverse that means having found or known i a, b, 0 in some way if you want to get what would be the equivalent 3 phase current we have to do the inversion and then you have 3 by 2 times m inverse i a, b, 0 should be equal to i a, b, but then we also know from the description in the earlier class that m inverse is equal to 2 by 3 times m transpose and therefore i a, b, c is 3 by 2 into 2 by 3 into m transpose into i a, b, 0 which means this is simply equal to m transpose into i a, b, 0. So one can transform a, b, c to a, b, 0 or a, b, 0 to a, b, c and we also have used the requirement or we have used an idea that in order to derive the number of turns that the way in which the vector i is getting transformed should be the same in the way in which the vector v is getting transformed and therefore we had defined the relationship v a, b, 0 is 2 by 3 times m into v a, b, c which therefore would mean that v a, b, c is m transpose into v a, b, c. So this we had seen in the last lecture. We see that the transformation from a, b, c to a, b, 0 is not really the same as transformation from a, b, 0 to a, b, c not really the same in the sense here there is a factor 2 by 3 coming and here there is no factor 2 by 3 coming. Other than that you have m and m transpose we know that m transpose and m inverse are related in some manner. On the other hand if you use the other approach that is the power invariant approach then what you have is i a, b, 0 as we had derived in the earlier lecture is root of 2 by 3 times m into i a, b, c and therefore this means that i a, b, c is root of 3 by 2 times m inverse into i a, b, 0 and since m inverse is equal to 2 thirds of m transpose we can substitute that expression. So this is root of 2 by 3 into i a, b, c is m inverse of this this is sorry this is root of this expression is wrong this is root of 2 by 3 times I have made some mistake let me check yeah i a, b, 0 is root of 3 by 2 times m inverse into i a, b, c which is root of 3 by 2 m inverse is 2 thirds of m transpose so that is 2 by 3 times m transpose into i a, b, 0 and this in turn gives us this is root 2 by 3 m transpose into i a, b, c. So what we have is if you define that i a, b, 0 is root 2 by 3 times m into i a, b, c the inverse transformation root what you have is i a, b, c is root of 2 by 3 times m transpose into i a, b, 0 similarly therefore you would also get v a, b, c is equal to root of 2 by 3 times m transpose into v a, b, c here what we see is this factor root 2 by 3 occurs both in the transformation from a, b, c to a, b, 0 and also from a, b, 0 to a, b, c and what was m here now becomes m transpose so this transformation is a little easier to have in mind it is easier to remember because root 2 by 3 factor comes always and it is either m or m transpose whereas in this earlier case in one root you have 2 by 3 in the inverse you do not have 2 by 3. So in this manner one can transform variables in the a, b, c reference frame to the a, b, 0 reference frame. Now we have not at this point of time said whether this a, b, c are belonging to the stator or belonging to the rotor they could be anything any 3 phase variable set can be transformed to an equivalent 2 phase so to say 2 phase variable set we normally refer to this as 2 phase variable because in most cases this i, 0 term happens to be 0 because in most cases electrical machines are excited by balanced ac supplies in which case va plus vb plus vc is equal to 0 and if you remember i0 or v0 is equal to whatever scale factor here into 1 by root 2 times va 1 by root 2 times vb plus 1 by root 2 times vc which means va plus vb plus vc divided by root 2 and since this is equal to 0 v0 will be 0 and if v0 is 0 then i0 would also be 0 and therefore this variable is invariably 0 and therefore this is referred to as a 2 phase set of variable so this is normally called as 3 phase to 2 phase and this transformation is also known as clocks transformation so this 3 phase to 2 phase could mean 3 phase of the stator or 3 phase of the rotor it could be any one of those the stator flow of currents generates its own MMF the rotor currents also generate their own MMF and each of those could in turn be thought of as being generated from 2 phases in the case of the stator the since a b and c axis are fixed in space the a and b axis are also fixed in space in the case of the rotor the a b c axis are fixed to the rotor and since the rotor is going to rotate the a b c axis are rotating in space and therefore this a and b axis are rotating along with the a b c axis so the 2 phase system on the stator is fixed to the stator and a stationary whereas the 2 phase system of the rotor is fixed to the rotor and hence it is rotating so both these can then be described by a similar set of equations you can have either stator or the rotor here. So having said all this how do we apply this to the case of induction machine now at this point before we get into the induction machine let us introduce define certain notations now we are going to have stator variables and rotor variables sometimes we may have we may want to write the stator variables in a certain frame of reference rotor variables in certain other frame of reference and therefore the notation can become very involved so we will define certain ways in which we are going to write these variables. Now if we have an input variable or an output let us say v which is the phase voltage that is applied now this letter itself stands for voltage now this may refer to stator or to the rotor and in each case we have an a phase b phase and c phase or a beta 0 and therefore what we will do is we will attach a subscript and a superscript this in turn will refer to we will mean it to refer to stator variables the superscript will refer to rotor so which means that if I write v a since this is coming below and it is stator we will infer that v a refers to the a phase voltage of the stator on the other hand if I write v a it would then refer to a phase of the rotor if I write v a it would refer to the a phase of the stator and v a would refer to the a phase of the rotor similarly we will define variables i a and i a, i a, i a needless to say you could have v b, v beta, v c, v 0 and so on the location of a beta or 0 or a b and c will refer will determine whether we are talking about stator or whether we are talking about the rotor so we will keep this in mind later on we will introduce some more notations when we have occasion to refer to more frames of reference right now we have defined or described two frames of reference one is the so called natural reference frame the natural reference frame is the actual frame of reference when you can measure the variables that you are applying to the machine so the natural reference frame we will always call it as the a, b, c reference frame because generally machines are three phase machines and we will refer to the three phases as a phase, b phase and c phase so the natural reference frame is the three phase a, b, c reference frame. We also have the a, b reference frame or the a, b, 0 reference frame the a, b, 0 reference frame is always a reference frame which we will consider that is fixed to the member which we are looking at if it is a, b, 0 for the stator it is fixed to the stator if it is a, b, 0 of the rotor it is fixed to the rotor and therefore the a, b, 0 of the rotor is going to be rotating a, b, 0 of the stator is going to be stationary that is that arises from the member to which this a, b, 0 is attached so these are two reference frames that we have so far introduced. So now how are we going to apply this to the case of induction machine in the case of induction machine we have this vector v which is the vector of voltages at all phases of the machine both stator and rotor and therefore this vector we will denote as v, a, b, c, a, b, c to denote that this vector contains both stator variables and rotor variables the stator variables are in the a, b, c reference frame the rotor variables are also in the a, b, c reference frame similarly you have the vector i which again is in a, b, c reference frame a, b, c, a, b, c. So the induction machine description that we have is a relationship between v, a, b, c and i, a, b, c which we wrote as this is equal to r into i, a, b, c plus p times psi which is a vector of flux linkages that is again in the a, b, c reference frame which we have expanded it as r into i, a, b, c plus p times l into i, a, b, c if you remember in one of the lectures we had written down elaborate expressions for r and l a fairly big description. Now what we have done is we have tried to transform this a, b, c to a, b, c and we have defined it for a set of three phase variables which could be interpreted as either stator phases or the rotor phases and therefore we can now define the vector v which is a, b, c in the stator a, b, c in the rotor this can be thought of as being derived from the vector v, a, b, c, a, b, c by a transformation matrix c. How do we find out what goes into c that is easy to find out since we know how that has been derived. So let us now define this matrix c we know that the three phase variables of stator voltages for example stator is denoted by the subscript so v, a, v, b and v, c we know that this is equal to root of 2 by 3 times we will use in this course transformation that is invariant with respect to active power so we will use this factor root 2 by 3 times 1-1.5-1.5, 0, root 3 by 2 and – root 3 by 2, 1 by root 2 times I am sorry this is v a, b, c is equal to this matrix times v a. So this is the description that we have this relates the stator a, b, c similarly therefore we can also write v a, v, b, c now we are writing variables in the rotor therefore I am putting this in the superscript this is also equal to root of 2 by 3 times 1 the same –1.5-1.5, 0, root 3 by 2, – root 3 by 2, 1 by root 2, 1 by root 2 times v a, a, b and v. So the transformation for the three phase stator variables and the rotor variables is the same. Now what we have what we need to do in order to get a matrix description c like the one that we have written before that is we want to say a, b, 0, a, b, 0 is equal to c times v a, b, c, v a, b, c then what we can do is this matrix we will define as a matrix c11 and then therefore we can combine these two descriptions into one equation where we write a sub vector v as v a, b, 0 and this sub vector v as v a, b, 0 these are the three vectors consisting of three variables of the stator vector consisting of three variables of the rotor this can as a matrix c11 which is this matrix 0, 0 and c11 again multiplied by v a, b, c, v a, b. So the rotor variables vector that comes here the stator variable vector comes here therefore now you can see that v a, b, 0 is c11 times v a, b, c that is what you have here and v a, b, 0 of the rotor is also equal to c11 times v a, b, c of the rotor that is what you have here. So effectively we have combined these two into one description here similarly then we can write an expression i a, b, 0 and a, b, 0 is equal to c times i a, b, c and a, b, c the same kind of description also apply. So now that we have these two equation which say how to transform a, b, c to a, b, 0 we can substitute this in our expression equations for the induction machine and try to see what happens. So if we do that from this equation we can write v a, b, c a, b, c is equal to c inverse times v a, b, 0, b and c. Now c inverse is nothing but the inverse of this matrix and this matrix is in the block diagonal form and therefore we can write this as c11 inverse c11 inverse and we know what the inverse of c11 is because we have defined all the transformations between the a, b, 0 to a, b, c. This is nothing but root 2 by 3 times m and c11 inverse is root 2 by 3 times m transpose and therefore the c11 inverse is nothing but c11 transpose and c11 transpose which is same as c transpose. Therefore we know c11 inverse c or c inverse is easy to write. Similarly we can write i a, b, c, a, b, c equals c inverse times i a, b, 0, i a, b, 0 and our earlier expression says that the induction machine equation v a, b, c a, b, c equals r times i a, b, c plus p times l into i a, b, c. But now we have written i a, b, c in terms of i a, b, c in terms of this so we substitute these expressions here. So what you have is c inverse v a, b, 0 is equal to r times c inverse i a, b, 0 a, b, 0 plus p times l c inverse i a, b, 0 a and you can remove this inverse by multiplying throughout by the matrix c and therefore you have v a, b, 0 a, b, 0 equals p times r c inverse i a, b, 0, 0 plus this is c times p l c inverse. Now this is derivative of a product of these 3 and being vectors or square matrices one cannot, one has to maintain the order in which these occur and therefore this can be expanded further as c r this term remains as it is you can write this as c l c inverse p of i a, b, 0 plus c times p of l c inverse multiplied by i. Now remember that the operator p stands for the derivative operator it is d by dt and both l and I mean c is not dependent on t because c is something that is fixed matrix so that does not depend l alone depends on t because l has the rotor angle inside it and as the rotor is going to rotate the rotor angle will change with respect to time and therefore this l depends on t and if you differentiate therefore since the only term that depends on t in the matrix l is the rotor angle you will then have derivative of the rotor angle which is nothing but speed and therefore this term can be further expanded as expanded so this is c r c inverse i a, b, 0 i a, b, 0 plus c l c inverse derivative of i a, b, 0 plus this term alone can be expanded as p of l into c inverse plus l into p of c inverse and l into p of c inverse is 0 because c inverse does not depend on the rotor angle and therefore this term simply evaluates to that and therefore that is c times p of l into c inverse i a, b, 0 so this is the expression that we land up with. Now one has to simplify these terms in order to find out what this description is going to look like and that simplification is not a simple algebra why it is fairly involved algebra so we will just indicate how to go about doing that simplification so if you take this matrix c r c inverse the matrix r consists of or can be written as r a, b, c, 0, 0 and r a, b, c of the rotor stator off diagonal terms are 0 and r a, b, c itself is the stator resistance r is multiplied by identity matrix u which is identity matrix which is 3 rows and 3 columns 1, 0, 0, 0, 1, 0, 0, 0, 1. Similarly the rotor resistance is then r r multiplied by u which is r r multiplied by the same identity matrix so that is the description you have. So if you now write the term c r c inverse is c 1, 1, 0, c 1, 1 and then you have r a, b, c, 0, 0, r a, b, c, r c inverse and we know that c inverse is nothing but c transpose so c 1, 1 transpose 0, 0, c 1, 1 transpose so this is equal to c 1, 1, 0, 0, c 1, 1 multiplying these 2 so you have r a, b, c, c 1, 1 transpose and then 0 and then you have r a, b, c, c 1, 1 no this term is 0 and here you have r a, b, c, c 1, 1 transpose and this can further be simplified is c 1, 1, r a, b, c, c 1, 1 transpose and this is 0, this is 0, I have c 1, 1, r a, b, c, c 1, 1 transpose and r a, b, c is nothing but a scalar multiplied by identity matrix and therefore this is nothing but r s multiplied by c 1, 1 into identity matrix multiplied by c 1, 1 transpose which is nothing but c 1, 1 into c 1, 1 transpose which is an identity matrix again so you get r s into u, 0, 0 and by similar argument this is r r into u and therefore this matrix is the same as what we started out with. So as far as this term is concerned this matrix is the same as r so there is no difference here. Now how to find out this so let us look at that matrix we will not complete the algebra but we will see how that has to be done. The matrix L can similarly be subdivided into matrix L11, L12, L21 and L22 we have seen what the form of L11, L21 and all that look like and we also notice that L12 is equal to L21 transpose or L21 equal to L12 transpose that also we have seen. So similar to the way we had done CRC inverse this also has to be done so you would then have c 11, 0, 0, c 11 multiplied by L11, L12 we can write this as L12 transpose L22 and then you have c 11 transpose so this becomes your CLC inverse and one can expand this. So this would then be c11, 0, 0, c11 and then L11, c11 transpose and then L12, c11 transpose and L22, c11 transpose and then if we expand this also further you get c11, L11, c11 transpose and then e11, L12, e11 transpose, e11, L12 transpose, c11 transpose so this is the expression but what do each of these terms look like one has to expand this and try to find out what it looks like and similarly this term here try to find out what it looks like. I will just show the net result of doing all this algebra I would urge you to verify yourself by performing this equation expanding this form in terms of individual element of L and c. So let us look at how this algebra is going to work out so here we have the expression for the induction machine written in the natural reference frame and as we had seen the matrix L can then be written as L11 instead of L12 I have put it as M12 so M12 and M21 L22. This matrix L11 looks like this you have leakage inductance plus magnetizing inductance this is the mutual inductance of the stator phases and then again leakage and magnetizing it assume to be a three phase balance machine therefore these inductances are the same. We had also seen why MS is equal to minus LMS by 2 we had looked at that and then the matrix M12 is MSR which is the maximum value of mutual inductance multiplied by some function of the rotor angle and then all these are functions of rotor angle so that is M12 M21 is the transpose of M12 matrix and the rotor inductance that is L22 now looks like this very similar to L11 and where MR is again minus LMR by 2 this is the form of the original description of the machine in the natural frame of reference. And now having transformed this to the two phase variables or the aß0 reference frame the matrix L so this is what we have in the aß0 reference frame this can then be written as the same matrix R multiplied by aß0 aß0 so this term is going to multiply the derivative of aß0 therefore this is another inductance matrix let us then call it as L aß0 aß0 this matrix is going to have derivative of the inductance matrix and we said that in the inductance matrix only term which will give a non zero derivative are the terms that involve the angle of the rotor and therefore you would then have a dθ by dt term that comes out of this so you are going to have a dθ by dt multiplied by i aß0 aß0 and this term out of which dθ by dt has been taken out the remaining we will call it as the matrix G and this matrix G is then called as the speed matrix because it is being operated on it is being then multiplied by this dθ by dt term which is nothing but the rotor speed and therefore G is called the speed matrix so what we need we have already seen what is the structure of R we need to find out what is the structure of L what is the structure of G so if you look at the structure of L you see that you have a self inductance term consisting of a leakage inductance and a magnetizing inductance but that magnetizing inductance is not just LMS but 3 by 2 times LMS here again you get a same term and the zero sequence only the leakage inductance enters the picture similarly this part has LLR plus 3 by 2 times LMR similarly here and only the leakage inductance for zero sequence part of the rotor side these two matrices is 3 by 2 times MSR into cosθ and then sinθ that is how this matrix looks and this matrix is then the transpose of this one so this is how the inductance matrix itself looks like and the speed matrix you note that in the speed matrix all these terms are zero the speed matrix is likely to have all the term all these terms zero because the derivative of the inductance matrix in the inductance matrix these terms do not depend on the rotor angle and therefore in the derivative all these become zero similarly here all these terms are also zero note that the zero row has all zero entries both for the stator and as for the rotor these terms are again dependent on the rotor angle so this is how the matrix L and G is going to look like. Now once we know this we can write down the full expression L aß0 G aß0 we can now write this full term now this particular expression though we have derived it from the equation of the three phase induction machine by applying this transformation from a b c to aß0 we will see in the next lecture that this equation could have been written down directly by observation without having going required to go through this elaborate conversion from a b c to aß0 why it can be written down by observation because this expression now refers to a machine which has two windings two windings on stator and two windings on rotor and these two windings are located along axis that are 90 degrees displaced these are again 90 degrees displaced with respect to each other and therefore one can see that in the inductance description L matrix you see that the mutual inductance term here this term would refer to the mutual inductance between the a axis and the a axis that term is 0 why is that 0 because two coils which are located 90 degree with respect to each other flux linkages will not happen by exciting one and saying in the other so mutual flux linkage or the mutual inductance is 0 when the two axis are displaced by 90 degree similarly here also you find that the mutual inductance term is 0 you will see why this term happens to be 0 in the next lecture. So in the next lecture we will start by looking at how this expression can be written down directly from observation so with that we will stop for this lecture.