 Let us continue with the derivation that we have been doing. We started out with the three phase machine, three phase induction machine where the rotor is in ABC reference frame and the stator is also in the ABC reference frame that was the natural reference frame description of the induction machine. From this we have gone to the two phase induction machine where the rotor is in a beta reference frame and the stator is also in a beta reference frame. We will mark this, this is the rotor and this is the stator, a beta reference frame is the reference frame attached to the rotor two axis and therefore it is rotating a beta of the stator is fixed to the stator and therefore it is stationary and to do this we have used the power invariant transformation which means that if the number of turns here was N ABC per phase and then this then goes to N a beta where N a beta is root three times N ABC by root two. So you have number of turns here that are different from the number of turns in the three phase machine while converting it to the two phase equivalent using this approach and then from this we have gone on to the stator reference frame all referred to the stator that means the rotor is now referred to pseudo stationary coils in the a beta of the stator and the stator remains as it is in the a beta reference frame. In doing this again we have used the condition of power invariancy but in this case the number of turns still remains the same as what was here therefore the number of turns in the windings now is different from that of N ABC we have gone from N ABC to N a beta given by this relation and between these two the turns number of turns remains the same and therefore what we have the resultant system is you have the beta s axis and then you have the a s axis and you have the coil on the a s coil this is the stator coil and then you have the stator coil on the beta s axis and then you have the rotor coil which is now referred to the stator and here again you have the rotor coil referred to the stator. So this is the rotor referred to stator this is also rotor referred to stator whereas this is stator and this is also stator coil note that we have in these between these two coils the dot points are here and the dot points are here which means the current is flowing into the dot point here current is flowing into the dot point here similarly you have current that are flowing into the dot point. So this is now the description of the machine all coils are stationary with respect to each other and for this we have derived the electrical system equations so let us put that down you have v a v beta of the stator and then vr a and vr beta for the rotor so this is the left hand side vector this is then equal to yesterday we wrote down the operational impedance form in the last lecture so let us put that down you have RS plus LS bar p and then 0 here you have MSR bar p and then 0 again 0 RS plus LS bar p 0 MSR bar p and then here you have MSR bar p and MSR bar into omega R R R plus LR bar p and then LR bar r minus MSR bar p and this is for omega R and MSR bar p minus LR bar r multiplied by the vector of currents so I a I beta I R a and then I R beta so this was the description that we had derived in the last lecture now let us look at this description a little more closely we know that LS bar is nothing but the leakage inductance plus 3 by 2 times LMS as we come down from the three phase description to this and MSR bar is nothing but 3 by 2 times MSR similarly you have LR bar is nothing but LLR plus 3 by 2 times LMR so these are the equivalent if you come down from the three phase this is what we get so let us look at this expression a little more closely if we look at the first equation that for V a we are basically writing down the voltage equation for the excitation given here this is applied a voltage V a here and we find that V a is given by RS into I a plus LS p I a this is nothing but the resistance drop plus the self inductance drop and then you do not have any term here which means that there is no drop caused here due to flow of current in beta s coil which is to be expected because this coil is there at an axis which is 90 degrees away from this therefore the mutual inductance term would be 0 mutual inductance is 0 and that is indeed what we see here here you have MSR bar into diR a by dt which means that there is a mutual inductance between these two and that is again to be expected because these two coils are on the same axis and therefore any flow of current here any rate of change of current here would induce an emf here defined by the mutual inductance so that term is what we see here and this term is 0 this refers to the flow of current in the iR beta and this says that there is no effect of current that is flowing here on this coil which again intuitively one may say this coil is at an axis 90 degrees similar to that of the beta s coil and therefore there is no induced emf here and therefore this equation can be written down directly by observation looking at the fact that you have these two coils situated on one axis these two on the other axis similarly when we now move down to the second equation the second equation the second equation also we see is similarly arranged to the first you have RS plus Lsp for the self term flow of current i beta and you have a mutual inductance term that occurs due to this so a self term due to this and mutual due to this there is no mutual term between these coils because they are at an angle of 90 degrees. Let us move down to the third equation in this equation we are writing down an expression for the voltage in the coil here the rotor coil referred to stator as seen from the stator and here we find that you have MSRP into I a which we can expect because these two coils are on the same axis and therefore if there is a current flowing here it would have an induced emf here defined by the mutual inductance and therefore this term is alright MSR into P times I a that is d by dt of I a but now you find that there is another term which is an induced emf but occurring due to speed this term is MSR ?r multiplied by i beta this term occurs due to flow of current in the coil which is at 90 degrees to it and therefore it is multiplied by I beta and this emf term depends upon speed and therefore this kind of term is given a specific name there it is called as a speed emf term. So speed emf term is so called because it appears to arise due to speed for if the speed were equal to 0 then this term would vanish and this term does not depend upon di by dt it is simply MSR ?r multiplied by i beta whereas these terms MSRP similarly here and here they arise not due to speed there is no speed term in any of these things they arise due to di by dt if di by dt were 0 then such of those terms would be 0 and therefore those terms are called as transformer emf term so these are multiplied by ?r or some speed these are multiplied by di by dt and then you have the self term which is the resistance drop plus the self inductance drop and then again you have a speed emf term and we notice that this speed emf term arises due to ir beta and therefore you see that this speed emf term as well as this speed emf term arises due to coils that are in beta axis occurring in a coil on the a axis. So the voltage equation of a pseudo stationary coil on the a axis now has speed emf terms due to flow of i in an axis 90 degrees away therefore this has speed emf terms due to this current and this current when we come down to the next equation that is vr beta you see that there is a speed emf term here this again is due to a current flowing in the a axis so if you now look at the equation for this coil this has speed emf terms due to flow of current here and it should also have a speed emf term therefore due to flow of current here which indeed is true there is a speed emf term here and there is a speed emf term here on the other hand due to a another situated on the same axis you have a mutual inductance term which is MSR x p and finally you have the self term which is the resistance drop plus the self inductance di by dt. So what we find from this is that transformation to the stationary reference frame brings about new terms in the impedance description which arise due to speed that is there so you now have speed emf terms and transformer emf term we find further that the speed emf terms arise from quadrature excitation or quadrature currents or quadrature coil and we also note another thing speed emf terms arise from these and are present in pseudo stationary coil for example speed emf terms do not occur here in these two expressions which are not for pseudo stationary coil V a and V a are really stationary because they pertain to the stator coil there are no speed emf term whereas in these two coils that are there are speed emf terms and these coils are pseudo stationary coils having come from the rotor. So you find speed emf terms in the stator stationary coils and those speed emf terms arise due to coils on the other axis a speed emf term arises due to coils on the beta axis beta speed emf term arises due to coils on the alpha axis and we also find that the speed emf term is negative for all terms on this axis not negative for terms on this axis so these are some of the observations that you make this is how this appears to have been formed and having understood this one can extend this for more situation suppose you now have a hypothetical machine let us say it has one more coil here and one more coil here as well and let us say that these are again pseudo stationary coils this coil and this coil. So if we have to consider a machine like this then may be what we could call this as we could call this by let us say V r1 alpha is the voltage that is applied here and let us call this as V r1 beta for the voltage that is applied here so that means you would now have two more equations here V r1 alpha and V r1 beta you will have two more flow of current I r1 alpha and I r1 beta so how would you then augment this operational impedance description to include these two also that is fairly simple having understood the way the operational impedance description looks like one can now augment this if you are to write an expression for V r1 alpha this is a pseudo stationary entity and therefore you will have with respect to I alpha term you will have a mutual inductance term let us call that as MS r1 it may not be the same mutual inductance between these two so MS r1 P I alpha and then you need to have a term with I beta this is pseudo stationary and you have beta on the other axis and therefore there will be a mutual inductance term and a speed emf arising due to that omega r into I beta MS r1 and then you are looking at the flow of current here that is the third term due to this so you have the third term arising it is again between this coil and this coil therefore if we call that as MS r2 then MS r2 into P this is a DI by DT term because you are referring to an IR alpha which is on the same axis and then you will have an MS r2 omega r into IR beta IR beta is a current flowing on the other axis and this being a pseudo stationary coil there would be a speed emf due to this and then you have its own so R let me call it rr1 plus lr1 into P times IR1 this is the self term and then it would also have a speed emf term due to the flow of current here and that is lr1 multiplied by omega r into IR1 beta so having understood how this operational impedance matrix looks like one can therefore write down the expression for any number of coils on the alpha axis and the beta axis it allows us to write this down by observation. Now this description of an electrical machine which we have drawn here there is one coil here another coil here may be more coils similarly more coils this is a description this is the name given to this kind of a description is that of a electrical you know why it is now given a name like that we have derived this machine description starting from an induction machine we started from a three phase induction machine and then a two phase and then we referred everything to state but as we go along we will see that this machine description what we have derived can be used to describe any electrical machine that falls in the group of machines that we are looking at for example this machine can also be used to describe a DC machine it can also be used to describe a synchronous machine it can also be used to describe an induction machine and we know that if you look at a DC machine DC machines are again of various varieties you have separately excited machines you have shunt excited machines you have series excited machines and in this case again you have long shunt short shunt and then you have compound machines in this case you have cumulatively compounded you have differentially compounded machine the DC machines itself are of various varieties one can go on to see we will look at that in a little while that all these machine descriptions can be arrived at from a machine description like this where there are some n coils on the a axis and n coils on the beta axis the DC machine can be seen that few of the coils here could be removed few coils here could be removed and you get a particular kind of DC machine depending on the way you interconnect them you get other kinds of DC machine and that is why this is given this name because this can also describe a synchronous machine or an induction machine one can then say that this forms a generalized machine description it is also a generalized machine description from which one can by specific specific constraints imposed on the variables one can arrive at various forms of electrical machine. Now we have been all along looking at the electrical equation let us also see what happens to the generated R now this description the generalized machine description we have written it in the form V equals some Z x I but this can also be expanded so we can write this in the form Vr aß aß equals R x IR aß aß aß aß aß x PIR aß aß aß aß GIR x ΩR so we had indeed written the equation we the derivation that we made in the last lecture brought about an equation in this form and then we compressed it into the operational impedance matrix it helps us again it helps to expand this in a similar form here so this is the matrix that we have now in order to derive an expression for the generated electromagnetic torque let us look at the three phase machine we started out with or rather the machine here the two phase machine we started out with an expression which looks like this V aß aß equals R x I aß aß p of L aß aß aß this was the machine description with which we started out and starting from our basic definitions the way we derived the electromagnetic generated torque you would remember that the way we started this out was in order to derive this expression you then consider the input power input power is given by I aß I transpose into V I transpose V is nothing but I transpose R I plus I transpose p of L this approach is what we discussed right at the beginning the first several lectures in the first several lectures an approach to derive forces approach to derive the electromagnetic generated torque we discussed and that is what we are now trying to recollect so input power is like this and we know that in the machine input power must be consumed in primarily three different ways one is the resistive losses another is the mechanical output and then the third is the rate of change of field energy I am just putting this down so that one can recollect what we had heard several lectures earlier and from this in order to get the mechanical output mechanical power output in order to get this thing we started out with an expression for the magnetic stored energy I think we called it as WF that is given by half of I transpose L into I in whichever reference frame we are considering it and then the rate of change of field energy is nothing but the derivative of this to DWF by dt is equal to d by dt of half I transpose L into I now we can simplify this expression and apply it to this derivation to this equation in the a beta reference frame and if you remember the expression that we landed with was the generated electromagnetic torque is half of I transpose dL by d theta multiplied by I this was the expression that you would get if you proceed all that we had to do is differentiate this that gives the rate of change of field energy the expression for input power is there we know that this term is the resistive loss component and then after having differentiated this if you subtract that from this expression the remaining term would then basically give you the mechanical output that you have but our goal now is not to derive the expression in the a beta reference frame we want to derive the expression for the generated torque in the stator reference frame that is pertaining to this equation. So in order to get that we now need to basically replace this terms I with the flow of currents in the stator reference frame and for that these currents are now I a beta a beta for this we use the relationship I R a beta a beta is equal to the transformation matrix we called it C I think so a beta a beta this was the relationship that we wrote and therefore we have I a beta a beta equals C inverse I R a beta in both both are now in the stator so we replace these expression these vectors I with this expression let us see what we get so you have DWF by DT equals D by DT of half of C inverse I a beta a beta C inverse is what we have to put and we know that C inverse is nothing but C transpose so you have C transpose I R a beta a beta and maybe I will put a flower bracket here since there are now more brackets so transpose of this matrix multiplied by L a beta a beta by C transpose I a beta a beta R so this is your expression for DWF by DT in the stator reference frame so let us expand this this is nothing but D by DT of half of I R a beta transpose multiplied by C L a beta a beta C transpose I R a beta now in this expression we have seen in the last lecture that this term C L C transpose is the same as L a beta a beta and therefore we can write this as D by DT instead of D by DT I will simply say P half I R a beta a beta transpose L a beta a beta multiplied by I R a beta a beta and now we need to differentiate this and since there is one term second term and a third term we have to differentiate them one by one so this differentiation gives us half of P I R a beta a beta this is one term multiplied by L a beta a beta into I R a beta this is the first term and then you have the second term which is half of I R a beta a beta multiplied by P of L a beta multiplied by I R a beta and then the third term gives us half of I R a beta L multiplied by P times I R a beta these are the three terms that come about on differentiation in these three terms if you look at the first term and the last term this matrix L a beta a beta is a symmetric matrix and therefore if you are going to multiply this matrix with P I R a beta a beta here and P I R a beta later these two being the same this term and this term therefore would be identical and that therefore gives us I R I have missed a transpose here so there is a transpose here and the transpose here so this is I R transpose a beta a beta into L a beta a beta into P times I a beta this is the sum of the first term and the last term you look at the second term this is the derivative of L a beta a beta and from our derivation yesterday and indeed what you see today this L a beta a beta does not have any term that depends on T this is a all terms are basically constant and therefore derivative of this matrix is 0 and therefore this term goes to 0 therefore DWF by DT is only this part so if you now apply the same logic whatever we have done here that is apply obtain an expression for the input electrical that is given electrical input power then you have to multiply this by I transpose you have I transpose here I transpose here and I transpose here and then you know that the resistive loss component is due to this and the remaining term must now be the mechanical output plus the rate of change of energy and this rate of change of field energy is what we have derived now and that is I transpose L into P I and that indeed would be your second term if you multiply by I transpose and therefore I transpose G into I into omega R gives you the mechanical out and therefore one can say that the torque that is generated in this expression is nothing but I transpose one can call it I R transpose I into G multiplied by I so this is the expression for the generated electromagnetic torque. Now one can expand this and try to simplify this expression in terms of the actual flow of I alpha I beta and so on so this expression for torque can be written as follows I alpha beta transpose is nothing but I alpha I beta I R alpha I R beta the matrix G is nothing but 0 0 0 and 0 MSR bar 0 LR bar – MSR bar 0 – LR bar and 0 this is your G matrix and then you have again this vector I alpha I beta I R alpha I R beta. So upon simplifying this what you have is I alpha I beta I R alpha I R beta multiplied by the first row multiplied is obviously going to give you 0 second row would obviously give again 0 then you have MSR bar multiplied by I beta plus LR bar multiplied by I R beta and then the last row gives us – MSR bar multiplied by I alpha – LR bar multiplied by I R alpha. If you multiply this out the first two terms are going to multiply with 0 so there is no problem in the next two terms you see that I R alpha multiplying here LR into I R beta similarly you have here these two terms so this term and this term will contribute to equal and opposite term and they simply nullify each other and therefore the resultant is nothing but MSR bar multiplied by I beta into I R alpha – I alpha into I R beta. So this is then the expression for the generated electromagnetic torque in the machine. So this equation and this equation together now describes the full electrical machine electrical system and the mechanical system one can use this to solve for the dynamics of the machine in whatever manner we want. So this then completes the machine description in the stator attached reference frame we have seen how the machine description can be extended for a machine with more terms in the alpha axis and in the beta axis. We started out with the induction machine so let us now complete one more aspect we have seen how the machine description is going to change when you switch the reference frames let us also see what happens to the supply variables. In the induction machine if we look at the induction machine as the example we started out with the two phase I mean three phase was converted to two phase and therefore let us look at the two phase machine this has two windings on stator and two windings on rotor. Now because it is an induction machine the way you would normally operate an induction machine is to give excitation on the stator and the rotor will be shorted do not give supply to the rotor winding the rotor of an induction machine is normally shorted and the stator is given the electrical supply. So you would give you would apply a source voltage is applied on this it is an AC voltage and we will therefore say that AC voltage of supply frequency supply frequency is let us call it as omega s that is the supply frequency since this is going to be a short there is no applied voltage there. Now in an induction machine you know that the rotor rotates at a speed omega r which is less than the synchronous speed of the rotor and the rotor is therefore going to have currents in the rotor are therefore at the frequency of the slip which is r at slip frequency which is nothing but omega s – omega r. So flow of currents in the rotor is likely to be at a much lower frequency normally the slip is small and therefore this frequency is quite small this is the current that is going to flow in the rotor. In attempting to go to the stator reference frame what we have done is we have transformed these currents that are flowing in the rotor we are trying we try to determine an equivalent current that will flow in the stator that will produce the same MMF and in order to do that what we have done is that use this expression I r a and I r ? is equal to cos ? – sin ? sin ? and cos ? multiplied by your I a and I ?. Now this I a and I ? are the currents that flowing in the rotor and therefore I could simply say that this current is some I m cos ? s – ? r x t plus probably some phase angle as well and therefore this current would be I m sin of ? s – ? r x t plus the same angle ? it would be the current. Therefore if you evaluate the expression for I r a that is then I m multiplied by cos ? cos of ? s – ? r x t plus ? – sin ? sin of ? s – ? r x t plus ? and this can therefore be written as I m into this is of the form cos a cos b – sin a sin b therefore this is nothing but cos of ? s – ? r x t plus ? this angle if you remember is the angle between we had drawn the rotor axis here the rotor coil was actually fixed to the rotor and this angle is the angle made by the rotor and this is what is now going to rotate along with the rotor and therefore this angle is nothing but ? is nothing but ? r x t and therefore if you substitute ? equal to ? r x t this expression reduces to I m cos of ? s x t plus 5. This is interesting because what it says is that slip frequency currents that are flowing in the rotor when you now look at it look at the flow of current in the pseudo stationary stator referred axis that now looks like the supply frequency current right. So this is intuitive to understand because what we are now doing in the induction machine is that you are attempting to look at the mmf that is equivalently generated by currents that are flowing here and you know that the field that is generated by the rotor in the induction machine if the rotor is fixed the field moves at the frequency of the slip but the rotor itself is now rotating at ? r therefore the net speed at which the field is rotating as seen from the stator will be at the frequency of the supply itself will be at the frequency of the stator field and that indeed is the reason why now you have the equivalent flow of current now is having supply frequency. Similarly one can derive an expression for I r ? as well you would find that it is I m sin ? t ? s t plus 5. So essentially now we have derived a representation of the induction of the machine in the stator reference frame that means all the dynamics of the machine are seen from the stator and in the next lectures we will see how this can be adapted in order to form various electrical machines and to describe their behavior as well we will stop with this for today.