 Last lecture we had been looking at transformation to the arbitrary reference frame, we derived the equations for the transformation and that can then be applied to the equations of the induction machine. I had suggested that you do it as an exercise, so before we go ahead let us start once again, I mean let us take a stand back and look at what really we have done before we proceed further. We started out with the induction machine equations in the natural reference frame, you had Va, Vb, Vc of the stator, Va, Vb, Vc of the rotor remember that in our notation, Ab and C mentioned as the superscripts refer to the rotor, Ab and C mentioned in the subscript region refer to the stator, so all electrical variables voltage V and I will have subscripts if we are referring to stator variables, we will have superscripts if we are referring to rotor variables. On the other hand as far as resistances and inductances are concerned which one they belong to whether they belong to the stator or to the rotor will be referred to by the subscript alone, the subscript of S in the machine data resistance and inductances, the subscript of S refers to the stator whereas the subscript of R refers to the rotor that is the notation that we are following. So here we have the induction machine equations in the natural reference frame, so you have three quantities on the stator, three quantities on the rotor, applied voltages and then you have Ia, Ib, Ic of the stator and those of the rotor, so this is the operational impedance description which arises from V equals Ri plus d by dt of ? which is the flux linkage which can then be written as L x I, so this expression arises from that, so you have three stator equations, three rotor equations and you can see that the mutual terms this area and this area refer to the mutual terms between the stator and the rotor and they obviously depend on the rotor angle, these angles are again defined here, ? R is the rotor angle, ?1 and ?2 are defined in terms of ?R and this enters the picture because the rotor is now rotating with respect to the stator and we said that this is rather involve description because as the rotor rotates the angles are going to change at every instant the angles are going to be different and this description has to be evaluated at every rotor angle which means it is going to make the numerical evaluation quite difficult. And therefore we look for schemes of simplification, of course the schemes of simplification have other uses as well as we have been saying all along but at the outset that appears to be that you want to simplify the description and let us in order to do that in order to simplify the description you move on to a three phase to two phase conversion and how to do that we looked at MMF invariancy, so let us to recollect what is it that we have done let us consider we had three axis on the stator you have the stator A axis and then the stator B axis and then the stator C axis remember that these are axis of the windings that are there on the stator, so the three phase windings axis are separated by 120 degrees and that is what we have drawn here. Now if you energize this three phase system with a three phase voltage or V or I we have we can look at I for some flow of current is there in the three phases now these three together will then generate a net MMF and let us say that we draw the net MMF generated by the three phases as an arrow and we said that this net MMF which can then be defined by the length of the arrow and the angle that it makes with the stator A phase axis which we take as a reference in order to describe this net MMF it may be sufficient to take one component along the horizontal axis and another component along the vertical axis. So if we describe these two it is sufficient to describe the net MMF sufficient to give all details about the net MMF. So based on this we have in instead of describing the MMF as an FA and FB and FC we then choose to describe this as an F a and Fb two components of the MMF and then because of requirements of inversion and seeing what additional information was missed out we then added the F0 component as well which is a representative of the zero sequence component of the excitation or of the MMF. So we have in this manner succeeded in splitting or in representing the three phase MMF in terms of two phases. In other words what we are saying is that instead of three excitation terms which are caused by an excitation winding lying along the A axis another winding lying along the B axis another winding lying along the C axis instead of these three what we have done is we have switched over to two sources of MMF alone one MMF source along the F a axis along the a axis and one along the B axis and another for to cater to unbalance systems of excitation another along the zero axis the zero axis is deemed to be perpendicular to both a and B axis. So it would be an axis which is mutually perpendicular in that manner. So instead of these three A, B and C you now have these two and perhaps the zero sequence. Now though this has been described with respect to the stator you could also do a similar transformation with respect to the rotor as far as the stator is concerned the three A, B, C are stationary in space and therefore a and B axis pertaining to the stator are also stationary in space as far as the rotor is concerned the A, B, C axis of the rotor are rotating in space and therefore the a and B axis are also rotating in space. So you now have instead of the three A, B, C phases you have a stator a axis a stator beta axis and then you have a rotor a axis and then a rotor beta axis call it beta R and that makes some angle at a given instant which is equal to the rotor angle that is the ?R that you are talking about. One thing we must understand very well at this stage that let us say you are looking at the stator a and beta axis so you have the stator a let us call it a S then and ?S for the stator we had looked at the net MMF generated by the three phases of the stator winding and whether you consider the MMF source now to be along the a and beta axis or a, b and c axis the net MMF is always the same we are talking about representing this vector or a phasor either in terms of a, b components or in terms of a, b, c components only the set of axis that are different but the MMF phase phasor is the same. So having defined this for a, b, c to a, beta for the stator one can define similarly for the rotor and therefore as far as all the six phase currents are concerned this is how the relationship would be this matrix M enters the description here and here so I a, ?0 for the stator is M times I a, b, c I a, ?0 of the rotor is M times I a, b, c of the rotor note that you have superscripts here and you have subscripts here. So this defines the three phase to two phase transformation and similarly one can define a transformation for voltages as well and then when these two are applied to the induction machine equations what we get is this you have V a, ?0 of the stator and that of the rotor transpose that means this vector is given by this operational impedance description multiplied by the vector of I a, b, etc. Now again in this case because the rotor has been split into two phase components that are still rotating with respect to the stator the mutual terms M1 and M2 which arise here are still dependent on the rotor angle we have not yet got rid of the rotor angle dependency. So we still have to simplify it further so what do we do we then transform to the stationary frame in this only the rotor variables are transformed the stator is already stationary there is nothing further that needs to be done the rotor on the other hand is rotating and if you make that stationary then all MMF sources are stationary with respect to each other and therefore the impedance description is not likely to have terms that are varying with respect to the rotor angle. So how do we do that you have F a, ? and 0 that is now transformed into these terms these are the rotor variables actually as mentioned by the superscript R but then this first term is on the a axis of the stator the second term is on the ? axis of the stator and the third term lies along the 0 axis. So in order to apply this for all the variables in the machine you then arrive at a matrix description like this where I a, ? 0 of the stator remains the same U is an identity matrix so I a, ? 0 is identity matrix into I a, ? 0 that means no transformation for the stator whereas for the rotor I a, ? 0 of the rotor is multiplied by C1 that is the matrix that you see here and it results in the rotor variables being referred to the stator a, ? axis the superscript R denotes that it is the rotor. So if you use this description apply this to the machine equations remember again what we are trying to do you have the a, ? 0 axis of the stator this is a, s you have a, s and then you have the axis for the rotor let me draw the rotor axis here this is a, r and you have a ? r here let me rectify that line ? r and the rotor generates an MMF let us say that the rotor generates an MMF the space phasor of MMF generated by the rotor is here now earlier this MMF was described by two components one along the a, r axis another along the ? r axis now instead what we are doing is we are now converting it to MMF components that lie on the stator so we are referring to this component and this component so you have an f, r, a here and then you have f, r, ? here again one needs to see or understand from the figure that the MMF space phasor is really the same earlier we were viewing that from the a, r, ? r axis we were viewing that from the a, r and ? r axis and therefore you had an a, r part and a ? r of this f component now on the other hand you choose to view it from stationary reference frame that is choose to view it from this axis so what we are really referring looking at is obtaining this component and this component here these two still refer to the same rotor MMF space phasor the stator MMF space phasor and hence the a, v remain unchanged so if we apply this to the induction machine equations what we land up with is an equation set that looks like this you have v a, ? 0 of the stator v a, ? 0 of the rotor now in the stator reference frame is equal to this impedance description multiplied by the vector of I and here you see that the angle dependency in these mutual terms is now no longer there this is because now what we have is you have the a and ? axis a, s and the ? s axis and you had the stator coils here already and now in addition you have the rotor coils as well the rotor coils are pseudo stationary coils that means actual coils are rotating but the same MMF would be produced by imagining these two coils to be stationary and a suitable current therefore goes through that so with this then you have the impedance description as it appears here and the expression for the generated electromagnetic TE is given by 3 by 2 times MSR into I, ? of the stator into I, ? of the rotor – I, ? of the stator into I, ? of the rotor so this is the description that you land up with. In this description then the induction machine behavior is seen through completely stationary coils on the ? axis and a and if a stationary coil has to produce a rotating MMF then these coils must be excited with voltages that are at supply frequency so we have also seen how these relationships affect the V and I terms we find that when you look at stator referred stationary frame excitations we found that the variables are all at supply frequency and therefore your I, ?, I, ? all these vectors are supply frequency AC terms. In many occasions we said then that it would be useful if these variables are DC or therefore it is better to look at this from a rotating reference frame and therefore you transform this further which was what we had been seeing in the last few lectures couple of lectures that you now transform these variables a, ?, 0 now these are stator variables where this you now have the stator axis a and ? and then further you define another set of axis which is really rotating with respect to time let me use this color so you have a d axis and then a q axis this makes an angle ? at the given instant and d, ? by dt which is the speed at which the new set of axis is going to rotate we simply call it by ?x if that is the case then the same MMF produced by the same MMF produced by excitations on the a and ? will now be produced by excitations along d and q if those two have to produce the same MMF then the equations are relating a, ?, MMF sources and d, q, MMF sources is given by this expression so when you now both stator and rotor excitation sources are along the a, ? axis so you have to transform both of them to rotating axis and if you need to transform both the relationship between the a, ? axis terms and the d, q, 0 axis terms now looks like this where d, 1 is the matrix that we see here again I would ask you to remember that what is really happening is that we have a net MMF that has been produced which is result of excitations along the a and b axis say for example stator if you take the stator then this is the net stator MMF that will have a b axis component and that will have an a axis component again what we are doing is the MMF remains has to remain the same we now want to locate a d axis component and a q axis component that is all that we are trying to do we are looking at the same MMF from a different set of axis so we are now looking at this d axis obtaining the d axis component and q axis component which is what this vector is the MMF is still the same the space phasor of MMF in reality remains the same we are choosing to look at it from a new set of axis so if you now apply this to the induction machine equations this is where we stopped in the last lecture which I had suggested that you do it as an exercise so if you apply this to the induction machine equations then the equation set that you land up with is like this you have vdx vqx and v0x these are the voltages along coils that are now imaginary coils rotating along with the dq axis lying along the d axis q axis and the 0 axis similarly rotor coils d axis q axis and the 0 axis these are the voltages that are applied those are related to the flow of idx iqx i0x etc by this description note that now you have a term here this is an equation for vdx which is rs plus lssp which is the self that is the resistance drop plus the self inductance di by dt but now the q axis also influences the voltage along the d axis even though they are at 90 degrees with respect to each other now if you remember and look remember back on the stationary reference frame equations when we derived them it we talked about pdmf term which arise in pseudo stationary term that is the a beta coils of the rotor when they are referred to the stator they become pseudo stationary and in those equations you had pdmf term appearing now this dx coil is also a pseudo stationary coil and therefore it has a pdmf term arising from all coils on quadrature axis which means the q axis now it has an omega x into lss and similarly here you have another pdmf term which arises due to rotor q axis term and this is a di by dt term that is arising due to the same coil on the same axis coil that of the rotor but so you have di by dt terms for the self flow of current and then a mutual current but flowing along the same axis this is multiplied by idx this is multiplied by iqx this is also qx term similarly if you look at the q axis you have a pdmf term due to the d axis current a self di by dt term and a resistance term of course due to its own current and then another pdmf term due to the d axis current d axis coil of the rotor and q axis coil of the rotor contributes a di by dt term. So the rules look the same in order to formulate this impedance description one can do it by observation also if one remembers that in pseudo stationary equations for pseudo stationary voltages you have pdmf term arising due to flow of current in the other axis here on the other hand if you look at the pdmf term they arise due to the difference of speed between the axis and the actual coil speed omega x now is the speed at which the axis rotates omega r is the speed at which the rotor is actually rotating and so the pdmf term now depends upon the difference between omega x and omega r similarly here here and here all the pdmf term depend on the difference between omega x and omega r these are simply di by dt ms which are self and those along the same axis the expression for the generated electromagnetic torque is given by this iq of the stator multiplied by id of the rotor minus id of the stator multiplied by iq of the rotor. So here this x or rather omega x can be chosen as per our requirement we have chosen d a d delta by dt that is the speed of the axis that could be anything you could even fix it at 0 in which case it becomes a stationary frame and therefore one can arrive at various simplifications of that equation if you want reference frame rotating at synchronous speed you would then put omega x equal to omega s which is the synchronous speed if you want a frame rotating at 0 speed that stationary frame you would put omega x equal to 0 in some cases it may be good to do the whole analysis in terms of reference frame fix to the rotor in which case you would put omega x equal to omega r if you look back at the earlier equations if you put omega x equal to 0 you see that this term goes away and this term goes away omega x equal to 0 would mean stationary reference frame that means the stator coils are really stationary now they are no longer pseudo stationary and therefore there is no speed induced emf which is what we had earlier also if you put omega x equal to omega s that is the synchronous speed then you have speed emf terms here this term omega s minus omega r now becomes a slip dependent term omega s minus omega r is nothing but slip times omega s if you put omega x equal to omega r then these two terms which are actually coils of the rotor your reference frame is also a rotor rotating reference frame reference frame rotating with rotor speed and therefore they are now no longer pseudo stationary coils are really moving at omega r and therefore this speed emf term becomes 0 similarly this speed emf term would also become 0 on the other hand these two will have speed emf terms because you are now moving the stator to a reference frame which is rotating at rotor speed so in this manner then one can simplify these equations to look at it from any reference frame that you want but as far as the equations are concerned though the transformations have been done to the same reference frame now we have transformed all the equations to the arbitrary reference frame the rotor turns are still different from those of the stator the stator turns if you assume that it is NABC remember we went from 3 phase to 2 phase at that time NABC would get converted to some other number which may be root 2 by 3 times NABC or some number like that so the stator turns in the dq reference frame are not exactly equal to the stator turns in the 3 phase reference frame they have been modified and so also the rotor turns have also been modified but however the rotor turns are no longer are not really equal to the stator turns one can refer all the terms to the stator all the electrical variables of the rotor to stator turns itself just like the way you do for single phase transformer you refer the secondary side to the input side by multiplying it by the turns ratio in the same way what we can do is the dx rotor coil for example if we take all the variables associated with rotor coil voltages V and I can be transformed to stator turns itself which means again you go by the mmf root so N aß is the number of turns of the rotor coil in the aß reference frame multiplied by idx which is the current flowing through the dx coil dx that is in the arbitrary reference frame the aß reference frame number of turns to the stationary reference frame number of turns to the arbitrary reference frame number of turns there is no change so this still remains the same this must then equal to the stator turns multiplied by whatever is should be the current that is flowing through we know N aß we know idx that is the current that is flowing actually in the dx coil of the rotor we know the stator number of turns and therefore we can find out what idx dash is or we can choose to express idx in terms of idx dash by this equation similarly if we say that input power is invariant we can derive an expression for Vdx as well in this format so this can now be substituted in the machine equation what happens there is that this idx that is there has now been rewritten as N let us look at that equation again Ndx by N aß into idx dash so this is Ndx by N aß into idx dash that is what this would be the qx term would also have Ndx by N qx by N aß into iqx dash Nqx by N aß into iqx dash dx dash should have been on the numerator so let me delete that and rewrite it this would be Ndx idx dash similarly these voltages Vdx would also be equal to N aß by Ndx multiplied by Vdx dash and similarly the qx also from the previous equation we write it. Now you see that this term Ndx by N aß which is multiplying idx is really going to be multiplied by the terms that are along this column so you have 3 by 2 Msr P idx now instead of idx you have Ndx by N aß into idx dash and therefore what we can do is this can now be multiplied by Ndx by N aß and relieve this term from here you can remove Ndx by N aß from here and simply keep this as idx dash that Ndx by N aß is absorbed into the set of entries here similarly this would also have to be multiplied by Nqx by N aß and then that term can also be relieve from here and this would then be made qx dash. Now you have this vector V here and the dx axis and the qx axis these two voltages have been multiplied by N aß by dx and therefore one can take that on to the right hand side therefore multiply now let me remove this so multiply this row by Ndx by N aß multiply this row again by Nqx by N aß which means all the terms here they are going to get multiplied by Ndx by N aß or Nqx by N aß all terms here are again going to get multiplied by Ndx by N aß which is the same as what is happening here but all the terms here are going to get multiplied once because of multiplying by Ndx by N aß for terms along these two columns and again by the same number for terms along these two rows therefore all these terms are going to get multiplied by Ndx by N aß whole square that is what is going to happen if we simplify it to the number of turns as well and therefore we can give these new definitions so let us go to the next slide. Now MSR always gets multiplied by Ndx by N aß which is what we see here Ndx by N aß whereas all the terms here are getting multiplied by Ndx by N aß whole square so it is like referring the secondary side impedances to the input side by multiplying it by the square of the turns ratio so we denote all those terms as Rr dash Lr dash and so on whereas these terms they are 3 by 2 times MSR into Ndx N aß we will call that as Lm so all the terms here now become simply Lm. Now this Lss now it was Lss plus 3 by 2 times magnetizing inductance of the stator phase now that the mutual inductance has been multiplied by the number of turns and is also multiplied by 3 by 2 that is the same number that is going to come here so Lss can then simply be Lss plus Lm whereas the rotor self inductance is then given by this term here you had 3 by 2 MSR multiplied by now that you are referring it to stator number of turns itself this coefficient for the generated electromagnetic torque also simply becomes Lm now this is a simpler equation to write because all the terms have been referred to the stator you no longer have to worry about what are the actual number of turns in the rotor all these terms can then be determined by your normal induction motor blocked rotor test and no load test where the equivalent circuit is determined as seen by the stator. So this form is then very useful and in this case we have used ?x equal to ?s that means we are talking about the synchronous reference frame now these equations can be we can use the notation of vectors to simplify writing this equation Vdqdq is then the vector of voltages on the stator and rotor Idqdq is the vector of flow of I along the stator and rotor the notation simplifies I mean notation signifies that the rotor variables are in the dq reference frame stator variables are also in the dq reference frame we will leave it to understand by the equations whether we are talking about the dq reference frame rotating at synchronous speed or rotor speed or whatever arbitrary speed that one may have at which speed it is rotating is known from the entries on the speed matrix that is g1 and g2 in this case because you have ?s you are talking about synchronous reference frame so one can write that equation as V equals r x i plus l x p i that is di by dt plus g1 term multiplied by ?s x i plus another speed matrix multiplied by ?r x i rlg matrices are shown here rlg1 and g2 are shown here and the generated electromagnetic torque is dependent on g2 as I transpose g2 x I now why this notation is useful this can then be used to look at what your small signal modeling can be now we may just like we did in the case of DC machine if you remember we looked at equations in the stationary reference frame and then we said that from these equations one can even derive expressions for DC machine and then we found that those equations are in general non-linear and we try to linearize those equations by applying small disturbances the same thing can be done here as well this is the original equation what we saw on the last slide V equals r i plus l di by dt the g1 ?s x i I have removed the dq dq subscripts for simplicity we understand that we are talking about the synchronous reference frame that is understood one of the advantages of the synchronous reference frame for this particular application to derive the small signal model is that all variables are DC and therefore any just you can now look at applying a step change etc and under steady state because all variables are DC d by dt of the steady state variables would still be 0. So if you take this equation and apply a small perturbation to the input variables that is V and to the load torque which we will see then this becomes V0 this 0 the O in the subscript here this refers to steady state term that means we are looking at the machine operating under steady state and then you apply a certain disturbance to it. So if you apply a certain disturbance delta V in response to that all the response variables that is i ?r and the generated electromagnetic torque which is a function of I again they all undergo disturbances again so you have I0 plus ?i here I0 plus ?i here ?s which is this frequency of the supply that you give could also be change if you are talking about an inverter driven induction machine for example the inverter can feed the induction machine with any frequency supply and therefore you may choose to vary the supply frequency as well that which means that the operating frequency can also undergo a change. So if you then apply disturbances to all the rotor speed undergoes a disturbance this is the same expression rewritten with all the variables as being composed of a steady state part and a disturbance part but however the equations for the steady state itself should be valid and therefore V0 must be equal to ri0 plus lpi0 plus g1 ?s0 into i0 plus g2 ?r0 into i0 and therefore those terms can simply be removed from the expression so V0 cancels with ri0 and then lpi0 g1 ?s0 i0 and then g2 ?r0 i0 and therefore the small signal quantities themselves are related by ?v equals r ?i plus lpi ?i from the third term here you would have g1 ?omega s0 into ?i but this being a product of two small disturbances is neglected so you have only g1 into ?s0 into ?i plus g1 into ?omega s into i0 similarly from these two terms so this is a equation describing the relationships between the disturbances and the response to that disturbances the generated electromagnetic torque is given by i transpose g2i so torque will then undergo a small change which is given by i0 plus ?i transpose g2 into i0 and therefore ?t itself can be written by this expression and you have the mechanical equation the mechanical equation is nothing but j times domega by dt equals the generated electromagnetic torque minus the load torque minus perhaps b into omega and therefore this can be written as j into d by dt of ?0 plus ?omega equals t plus ?t minus tl0 minus ?tl minus bomega 0 minus b ?omega and as usual if we remove the steady state part of it then domega 0 by dt gets cancelled with this t0 tl0 and bomega 0 and what is remaining is d ?omega by dt which is given by this expression ?t minus ?tl minus b ?omega the j is brought to the denominator so this is how it is ?t itself since it is a function of ?i can be written as some g3 into ?i minus this expression. Now this set of expressions which are voltages and due to the speed can then be combined into a state combined and rewritten in the form of the state variable description which can then be used for further determination of transfer functions and responses due to that pole 0 analysis and so on just like in the manner which we did for the DC machine. So I leave that again to you as an exercise to rewrite this in the state variable format and see how the equations look like. We shall stop here for today and continue in the next class.