 In looking at the equations for the induction machine we saw the model of the induction machine in the natural reference frame and then we have derived an approach to transform the natural reference frame model into the aß reference frame which we said is essentially a two phase description. So what we have done is gone from three phase, two phase description. We have seen the form of equations in the three phase case and we have seen the form of the equations in the two phase needless to say the two phase description also includes the zero sequence component including zero sequence which as we said for the most part is not there in a three phase machine if the system is operating in the normal way. So let us look at we have seen how the equations look like let us look at the input variables themselves in the three phase case you have three inputs which is va let us say that va is vm cos ?t then vb would be vm cos of ?t-2 ? by 3 and vc would be vm cos ?t-4 ? by 3. And while going from three phase, two phase what we are doing is transforming all the variables from the three phase to the two phase reference frame and therefore what we have is v a vb and v0 this is equal to square root of 2 by 3 is there and then 1-1.5 and then-1.5, 0, 3 by 2 and-3 by 2, 1 over 2, 1 over 2 and 1 over 2. So if you want to get v a vb and v0 you need to substitute the expressions for va, vb and vc in this vector. So that is what you have here va, vb, vc note that we are substituting the actual values as a function of the variable t and therefore what we are substituting here is not the rms value or a phasor in this place it is the actual function of t and therefore what we end up with is v a is root of 2 by 3 multiplied by vm cos ?t-1.5 of vm cos ?t-2 ? by 3-1.5 of vm cos ?t-4 ? by 3 and therefore this is nothing but or we can say va-vb by 2-vc by 2 which is root of 2 by 3 into va-vb-vc by 2 and vb-vc is nothing but-va because va-vb-vc is equal to 0 obviously it is a balanced excitation that we are giving. So in that case then what you have is root of 2 by 3 multiplied by 3 by 2 into va so this is root of 3 by 2 into vm cos ?t. So v a turns out to be square root of 3 by 2 into vm cos ?t what happens to vb? So vb is nothing but root of 2 by 3 multiplied by vm cos ?t-2 ? by 3-vm cos ?t-4 ? by 3 multiplied by root of root 3 by 2 so that is the expression so that is nothing but root of 2 by 3 into root of 3 by 2 multiplied by vm multiplied by cos of ?t-2 ? by 3-cos ?t-4 ? by 3 and so this expression can be written as cos ?t cos 2 ? by 3. Plus sin ?t sin 2 ? by 3 and then you have-cos ?t cos 4 ? by 3-sin ?t sin 4 ? by 3 that is what is there enclosed in the square bracket cos 2 ? by 3 is cos of 120 degrees that is-1 half cos 120 is cos 90 plus 30 that is-sin 30 that is-1 half and then sin 2 ? by 3 is root of 3 by 2 that is-1 20 is sin 90 plus 30 that is cos 30 that is root 3 by 2 and then cos 4 ? by 3 cos of 4 ? by 3 is cos of 240 degrees that is cos of 180 plus 60 degrees that is-cos 60 degrees so that is plus half and then this is again so plus root 3 by 2 and therefore what you can see is this term and this term cancels so what you are left with is root 3 sin ?t by Vm into root 3 by 2 into root of 2 by 3 that is nothing but root 3 cancels and then you have root 2 so you have root of 3 by 2 into Vm sin ? so this is nothing but V ? and your V a was root of 3 by 2 into Vm cos of 90 so you see that the variables that are exciting the induction machine in the 2 phase are applied voltages that are phase shifted by 90 degrees whereas in the 3 phase machine the applied voltages are phase shifted by 120 degrees. So a 2 phase induction machine whose applied excitations are phase shifted by 90 degrees with respect to time will behave the same way as a case of a 3 phase induction machine whose excitations are phase shifted by 120 degrees with respect to time. The equations of course have appropriately been modified as we saw in the last class and therefore since these 2 machines are equivalent the machine description should also be obtainable by looking at the 2 phase at the beginning itself instead of going from 3 phase to 2 phase we could start writing the expression directly from the 2 phase system so how can you do that and we should also look at another aspect before we go to that V0 is nothing but root 2 by 3 into 1 over root 2 times Va plus 1 over root 2 times Vb plus 1 over root 2 times Vc and therefore we already know that this equation holds Va plus Vb plus Vc is 0 so it is root 2 by 3 into 1 over root 2 into Va plus Vb plus Vc and this is already 0 so when you excite the 3 phase machine with a balanced excitation the 0 sequence component goes to 0 you have only the a term and V a term and Vb term that is why we again call it as 2 phase equation. So how do we get the machine equation directly from the 2 phase case now let us say that we have a 2 phase machine and the 2 phase machine what we have is the a axis and the beta axis for the stator so it means that you have a winding whose MMF axis is along the a axis the symbolism to denote a winding whose MMF is along a particular axis is in this way so this is the a s and then you have the beta winding for the stator whose which is producing MMF along the beta axis and similarly the a, b, c of the rotor have also been transformed into the a, beta reference frame but however the a, beta of the rotor are rotating along with the rotor and therefore you have an a axis of the rotor and a beta axis for the rotor so to distinguish them we will call it as a s, a, r, beta s and then beta r and you have a coil here in the sense you have a coil on the rotor whose which is producing an MMF along the a axis and then you have a coil on the beta r axis as well. And representing this way we are not making any reference to the 3 phase machine we are starting directly from a 2 phase machine and this angle is the rotor angle ?r and therefore that angle is also ?r now when we write the equations we need to know how the flux lines are going to be oriented and therefore we consider dot points here and we consider that flow of current is into the dot point here that is your i a for the stator and this is the rotor variable i a voltage applied here is v a voltage applied here is v a similarly for the beta winding this is i a and i a is here v a and v a so this is a general representation of a 2 phase machine how does one write equations for this if you want to write the expression for v a that is applied voltage on the stator a phase you would write r s resistance of the stator times i a plus p times ? a and ? a is the flux linkage of the coil on the a axis of the stator and that is nothing but the self inductance of the stator coil multiplied by i a that is the flux which is caused by its own flow of current then of course there will be a mutual linkage between if you want to consider between beta s and the a is there is no flux linkage between the two because beta s is at 90 degrees with respect to a and therefore flux linkage is 0 maybe we can write it as 0 x i a and then there would be a flux linkage because these two axis a axis and the beta axis are rotating there will be a mutual flux linkage between this coil and this coil and if you call that as m msr times cos ? r multiplied by i a that is the current that is flowing here in order to distinguish this msr from whatever we had written earlier as msr in the 3 phase machine case we will simply call this as msr bar and then you have a flux linkage due to this coil here and that is msr again into cos of 90 plus ? r multiplied by i a these two values msr for the a axis and the beta axis are the same because we assume that this winding and this winding are identical in all aspects except that they are displaced with respect to space and therefore if an i beta flows here and if this coil were really oriented along this axis the self inductance would have been the same therefore msr is really the same and therefore this can be written as ls into i a plus 0 times i beta plus msr cos ? r into i a and then cos of 90 plus ? r is minus msr bar sin ? r into i beta and therefore this is what you are going to substitute here now similarly one can write expressions for v ? as well I will leave it to you as an exercise to write down the expression for v ? now let us write the expression for v a r that is nothing but in our notation v a v a is rr okay rr into i a plus p times ? a and where the expression for ? a will then be lr into i a that is self inductance multiplied by the flow of current in itself plus there will be no mutual flux linkage between the coil a r and beta r because they are stationary with respect to each other and separated by 90 degrees and therefore there will be no mutual linkage due to that so you have 0 into i beta there will be flux linkage between these two and therefore that is nothing but the same msr multiplied by cos ? r multiplied by i a and then you have mutual between these two terms that is nothing but msr multiplied by sin ? r multiplied by i beta now in order to formulate the equation you need to substitute the ? a here and if one does that you get the expression v a equals rr s into i a plus ls p into i a plus the differentiation of the last two terms involving msr will give rise to two terms each because the term the derivative of msr cos ? r i a plus or minus msr sin ? r i beta in this expression both i a and the rotor angle are functions of p and therefore you need to differentiate both of them rs i a plus ls p i a plus you have msr cos ? r p of i a you are differentiating this and then you have plus msr into minus sin ? r d ? r by dt into i a so this term is expanded at these two because ? r is a function of time and i a is a function of time and then coming to the second term you have minus msr sin ? r p i beta and then minus msr cos ? r d ? r by dt into i beta and if you look at this expression you will find that this is the same as what this has the same form as what we had derived in the earlier lecture. If you remember we had v a ? 0 a ? 0 equals r times i a ? 0 a ? 0 these are vectors plus we had l a ? 0 a ? 0 p of i a ? 0 i a ? 0 then we had a matrix which we called as g multiplied by d ? r by dt g is again in the a ? 0 reference frame multiplied by this vector i a ? 0 a ? 0 elements of this g matrix would then come from these terms these entries would go into the g matrix you have d ? r by dt and the vector i a ? 0 will compose of these things. The inductance matrix on the other hand would comprise of terms like this derivative of p i would come here and the resistance matrix is of course going to remain like this. In this manner then one can write the expression for v a also stator for the expression for the rotor and you would get a similar looking expression which again falls into this format and therefore this equation in the a ? 0 reference frame can be written directly by observation the only thing is that we do not know how this term msr bar is going to be related to what was msr derived from the 3 phase k and that we have seen now that msr bar is nothing but 3 by 2 times what we had written as msr in the 3 phase machine. Similarly this inductance ls what we find is that ls is nothing but we have written the self inductance here which would be a leakage inductance along with some along with the magnetizing inductance and we found that ls is nothing but the leakage inductance plus 3 by 2 times the magnetizing inductance. So if we remember these things then the expression for the 2 phase machine can be written directly from by looking at the 2 phase axis we of course will not have an expression for v0 if you remember v0 for the stator was nothing but rs plus lls p multiplied by i0 i a and i ? do not figure in this expression. Similarly v0 for the rotor was rr plus llr p into i0 for the rotor the a axis of the rotor and ? axis of the rotor i a and i ? here do not figure in this expression as well. So if one remembers this also one can complete the a ? ? 0 description of the machine from the elementary idea. So we have now converted this 3 phase machine to a 2 phase machine and the 2 phase machine will work in the same way as the 3 phase machine provided you give excitations that are 90 degrees in phase to the a ? and ? axis but have we simplified anything. Now in the expression for the 3 phase machine we found that we had an impedance description the operation if you put this expression as v equals some z into i this z was a 6 by 6 matrix and this matrix was a function of rotor angle. What we have done is we have reduced this to another description this v equals ri plus ll into p i plus g ? r d ? r by dt this equation can be compressed and written again in this form which is nothing but r plus l into p plus g into ? multiplied by i. So which can be written as some z into i this could be in the a ? ? 0 reference frame again a ? ? 0 everywhere this description if we exclude the 0 sequence part this impedance matrix is only 4 by 4 so excluding 0 sequence but even though we have sort of apparently reduce the size here the matrix is still dependent on rotor angle. If you want to solve remember our goal is to get hold of a model for a machine using which we will be able to determine how the machine is going to behave with respect to time for any given inputs that are applied as va, vb, vc or v a, vb and so on that means we have we need to be able to solve the set of equations in order to arrive at the dynamics of the machine. So how does one solve these equations these equations are not simple these equations are not linear they are non-linear equations because of this term ? into i and therefore it is quite difficult to solve by hand you would need to take recourse to some digital computer algorithms in order to solve these set of equations. However if you are going to take recourse to a digital computer and this is the equation that you have let us say what will you do you can recast this equation as lvi equals v-r into i-g into ? into i. So having written it in this form this looks like a first order differential equation set where you can write p i as l inverse multiplied by v-r i-g ? i this is an equation for the electrical subsystem part of it and then you have the mechanical system which would say j d ? by dt is the torque developed- whatever load torques are there on the system this torque developed will then be a function of i we have already seen i into gi is the generated torque and therefore one can substitute that expression here. So these two sets of equations or this entire set of equations is what has to be solved in a digital computer in order to obtain the dynamics for a given input vector v and load torque here. But the process is not again very easy because in the digital computer one would solve it by algorithms known as numerical integration algorithm. And these numerical integration algorithms essentially take the access of with respect to time let us say you want to compute the solution may be some current i with respect to time what they do is the numerical integration algorithms consider discrete instance of time along the axis and try to estimate the solution at each instant of time. So maybe at this instant of time solutions here value of i is here at this instant it is somewhere else this instant it is somewhere else here it is somewhere else so we could say that the current order flows like that. So you discretize the axis and then try to compute the solution at each instant of t that you are willing to consider which would mean that this inductance matrix has to be evaluated all these have to be evaluated at every instant of time r matrix is a fixed matrix you can evaluated once and store it g matrix is not a fixed matrix because it is going to depend on angles inductance matrix is again not going to be a fixed matrix because it depends on the rotor angle and the inductance matrix you have to compute either the inverse or do something in order to get the solution and therefore evaluating the solution is going to be a difficult job because these g and l are going to change at every instant they are going to change because the rotor angle as the rotor is going to rotate the rotor angle obviously is going to change and therefore l matrix and g matrix are changing with respect to time this makes the solution a little more involved it is not as easy as if g and l are not going to change the time had l been a fixed matrix g another fixed matrix solution would have been easier to do. So this is one difficulty with respect to solving these equations so how does one eliminate this difficulty one has to see why this difficulty arises why is it that l matrix and g matrix are going to change with respect to time and if you look at the origin of the angle expression that appears in this one can see that it arises because of the situation here you have the a s axis which is fixed in space because this pertains to the stator and the stator does not rotate whereas a r axis is moving in space it is moving in space because this pertains to the rotor and the rotor is rotating and therefore whenever you have an excitation on the a r axis and something on the a s axis the angle between the two is going to continuously change with respect to time as the rotor rotates and therefore the mutual inductance between these two is bound to change one cannot avoid that at all. And therefore if you want to have a description that is independent of the rotor angle that is feasible only in a system where all these are fixed with respect to each other suppose you had a hypothetical case let us say you are not having this a r axis you do not have the b r axis let us say you had a coil another coil here and another coil here this coil does not move that coil does not move these two also do not move if there is not anything that is going to move obviously all the inductances are fixed there is no variation in the inductance and therefore the flux linkage expression would not involve the rotor angle if the flux linkage expression does not involve the rotor angle the derivatives and so on all of them do not involve the rotor angle and therefore this L and G etc they are likely to be fixed they would not vary with respect to angle and with respect to time. And therefore we would like to see if we can again do some kind of algebra transform the set of equations that we have got in order to reflect a system like this where these coils are fixed with respect to each other. So how does one do that again one has to start with the mmf being equivalent because any machine that we are any route that we are going to take must establish or must result in the machine behavior being unchanged. So if this if we have the a beta axis here you have a coil here and you have a coil here any mmf that is produced by a r should be produced by any other excitation system that we consider only then the two are equivalent. So let us look at that situation and see what happens by doing so the a s coil is already fixed in space so we need not do anything to that it is only this coil and this coil that are moving. So in order to see how we can proceed let us consider the mmf produced by a r and beta r. So let us consider some axis which is located at an angle ? with respect to the a s axis we want to find out the mmf along an axis along an angle ? at an angle ? due to the a r coil and beta r coil there is some current that is already flowing here there is some current that is already flowing here let us find the expression for mmf. So if we say that the mmf at this angle ? if we call that as okay f ? f at the angle ? this can be written as the number of turns of this coil n a we will put n a here multiplied by i a x cos of remember this angle is the rotor angle and if you want to find the mmf component at this angle you need to multiplied by cos of ? – ? r that is the component of mmf produced due to the a r that is there and then you need to find out what is the component of mmf produced due to this ? at this point that again is n ? and we know that n a and n ? are the same let us have that in mind then multiplied by i ? multiplied by we can say ? of ? – ? r. So let us expand this expression n a and n ? are the same so let us denote that by simple n multiplied by i a x cos ? cos ? r plus i a sin ? sin ? r plus i ? x sin ? cos ? r – i ? cos ? sin ? so this is the mmf that is generated by a r and ? r at a given angle ? now we can regroup the terms there are 4 terms here so we will rewrite it as n x i a cos ? r – i ? sin ? r multiplied by cos ? plus i a sin ? r plus i ? cos ? r multiplied by sin what we have done is taken the first term and this last term into one set because both multiply cos ? and then taken the second term and the third term as one set because both are multiplied by sin ? but however now this expression looks like some term multiplied by cos ? plus some other term multiplied by sin ? so if I now denote this as some current let me call that as i r a x cos ? plus i r ? multiplied by sin ? this expression means that suppose I had a coil here let me draw that by a different colour suppose I had a coil here through which this current i r a is flowing and I had a coil here through which a current i r ? is flowing then the mmf produced by this coil and this coil at a given angle ? is the same as what was produced by i a flowing here and i ? flowing on the ? axis that is exactly what we have written we started out with mmf produced by mmf at this angle ? produced by the a coil on the rotor and ? coil on the rotor and we find that that mmf is the same as that which will be produced by a current i r a flowing on a coil having the same number of turns but placed along the a s axis and another coil placed along the ? s axis since the a s axis and ? s axis are fixed with respect to space this coil is now fixed with respect to space similarly this coil is also fixed with respect to space therefore these coils this one and this one they are called as pseudo stationary coils why pseudo stationary because the actual mmf is not being produced by these two coils but are being produced by the rotating coil it is the rotor coils that are producing the mmf and we find that we can imagine two more coils on the a and ? axis which are fixed in space but having a fictitious flow of current through them which equivalently produces the same mmf we give a notation as i r a to this why we call this as i r a let us understand that now i r a is i a cos ? r – i ? sin ? r that means this flow of current a fictitious current flowing into this fictitious stationary coil or the pseudo stationary coil is a result of both the currents a i a and i a the flow of current in the rotor is now mapped to a current flowing along the a axis of the stator and another current flowing along the a axis of the stator so we will use this notation to denote that the rotor currents are mapped to the a axis similarly you have the rotor currents mapped to the ? axis as i a sin ? r – i ? cos ? r so the flow of currents in the rotor both a and ? are now mapped to the ? axis of the stator here this is mapped to the a axis of the stator this can be represented also in the following form using the matrix notation as i r a and i r a is equal to cos ? r – sin ? r sin ? r and cos ? r times i a i b so if we do this then we find that these currents are really flowing in stationary coils albeit pseudo stationary coils and these are fictitious current these are flow of currents in the two phase machine which we have arrived at from the three phase machine. So in this manner one can now try to transform the flow of currents in a rotating reference frame that is a ? b reference frame into a pseudo stationary reference frame which is a ? b again but these are fixed to the stator this is only with respect to a and ? how do we deal with the other term which is v0 or i0 we do not do anything to that we keep the i0 term as it is and therefore this would be 001 and 00 here and therefore this equation now defines a relationship between the a b reference frame and the pseudo stationary reference frame so we now have a description as i r a ? 0 equal to another matrix let us call that as d22 multiplied by i a ? 0. Now we would also like to have power in variance here so if we use this idea we can try to get a relation between v a ? 0 r and v a ? 0 I will leave you to look at this and see what could be the equation relating this and this pseudo stationary term we will look at that in the next lecture and then see the implication of going from a rotating reference frame to a pseudo stationary reference frame how does this affect the machine machine equations and whether this will have any implication on the dependency of rotor angle you will see that in the next lecture.