 In this next lecture on modeling and analysis of machines, let us look at how the transformation to this arbitrary reference frame is being done. The arbitrary reference frame as we saw in the last lecture is a reference frame that is rotating. The speed at which it is going to rotate can be considered to be anything. You can fix the speed to be 0 in which case it would then be the stationary reference frame. You can fix the speed to be anything else that is useful. Normally the useful speeds are the synchronous speed in which case it is called as the synchronous reference frame or in some cases in the case of induction machine for example where the rotor rotates at a speed different than the synchronous speed. You could have a reference frame rotating with the speed of the rotor in which case you would say that the reference frame is a rotor reference frame. So this induction machine description that we have seen and we have evolved until now consists of a description in the so called stationary reference frame where you have the a axis and the beta axis. The a and beta axis are both fixed to the stator and therefore they are stationary and in this reference frame you have a stator excitation source which you would then call as the a s coil and then along the beta coil you have another excitation that is the beta s coil and then you have the rotor having been transformed to the stator side. It is a pseudo stationary reference frame for the rotor. So you have an a r coil and then you have a beta r coil. So what you would do is you would then apply a voltage v a to the stator and a flow of current i a is there. Remember we are using the notation that if you have a voltage the later here refers to the stator and the later on top refers to the rotor and therefore we have v a as far as the rotor is concerned both the a beta variables which were in the rotating reference frame of the rotor have an influence on what is v a here. Therefore we denoted this by the symbol v r but then on the a axis of the stator and similarly and you have a flow of current here that is i r a and similarly you have a voltage which is v beta and then r and a current i r beta flowing and then you have a voltage that is v beta and then a current i beta that is flowing and we said that dot points are here so that currents that are flowing in will cause additive magnetic field in the machine. So for this kind of a arrangement we had evolved starting from a three phase induction machine we had transformed the three phase to two phase both the stator and rotor and then the two phase of the rotor was transformed to the stationary reference frame and we arrived at a machine description which then looks like this. So you had v a and then v beta v 0 and then v r a v r beta and then v 0 of r as well. So what we can do is maybe let me put it as a vector v a and then v beta v 0 v r a v r beta and v r 0. So this is the vector of inputs that are applied 0 sequence included and then we had this description as r s being the stator resistance plus p times Ls where Ls is the self inductance and we had an expression for the self inductance Ls is L leakage plus 3 by 2 times Msr where Msr is the mutual inductance between the stator and the I am sorry in this case it is just Ms mutual inductance between the phases. And then you have 0 here because this is a coil that is really stationary it is not a pseudo stationary coil and then you have between the 0 sequence components so that is also 0 and then between the stator and the rotor so you have 3 by 2 times Msr p and then 0 for v r beta and then the last term is again 0. In the second axis we had 0 here and then rs plus Ls into p 0 and then 0 here this is 3 by 2 times Msr into p and then 0. So for the v 0 term of the stator you have rs plus L leakage inductance p and then 0 everywhere else for the v a r term this is a pseudo stationary coil that has been placed the rotor has been transformed to the stator and therefore you will have pdmf terms the first term then relates to the coil both on the a axis and therefore this is a transformer emf term so you have 3 by 2 times Msr into p whereas the second term relates to for this coil what happens due to flow of current in the beta axis and therefore that is a speed emf term so you have 3 by 2 times Msr into ?r and then you have the 0 sequence term is 0 and then you have rr plus Lr into p where Lr is then L leakage inductance of the rotor that being the stator and then 3 by 2 times Msr and then the term here this will be due to the beta axis so you have Lr into ?r and this term is 0 and then v ? axis so this is minus 3 by 2 times Msr into ?r and then this is 3 by 2 times Msr into p and this is 0 and then minus Lr into ?r rr plus Lr into p and 0 and then the last row which corresponds to the 0 sequence of the rotor winding so that would be 0 here 0 here 0 here 0 here 0 here and rr plus Lr into p so this is the so called operational impedance matrix and this multiplies the vector of currents i a i ? i 0 and then i r a i r ? and i r 0 so this was the description at which point we stopped and then tried to see how this description can be used in order to describe DC machines so now we come back to AC machines. What we want to do now if we look at this excitation v a and the flow of current i a they are basically excitations and the response at the frequency of the supply so v a would then have an expression like some vm into cos of ?t and then v ? would then be vm into sin of ?t all the variables here are now going to be at the frequency of the supply the entire machine behavior is viewed from the stator terminals if you remember back on the induction machine equivalent circuit that you would have done in your first machine course you would have drawn the machine equivalent circuit referred to the stator in which all the excitations the responses are all at supply frequency and you have a similar kind of situation here. What we want to see is how this would change if you go to a rotating frame that is now going to move so let us say that you have a rotating frame a frame that is that has a an axis here which we will call as the d axis and an axis 90 degrees to that which we will call as the q axis this angle is of course 90 degrees and this reference frame that is rotating at this particular instant that we have drawn this figure this reference frame makes an angle of ? with respect to the a axis that is fixed on the stator and let us further say that we now want to find out what will be the mmf generated by let us say the stator excitations forget about the rotor for the time being let us look at the a applied to the stator v ? applied to the stator there is a current flowing i ? and i a as a result of that there is going to be mmf produced around the machine air gap and we want to know what will be the mmf that is produced due to this at some arbitrary angle let us say that this angle is ? so we want to know what is the mmf that is produced at this angle ? so in order to do that what we can do is to find out the mmf distribution due to this excitation in a and if you remember we had looked at the mmf generated by a distributed stator winding it was trapezoidal in nature and then we split it in the fundamental in Fourier series we neglected the other terms in the Fourier series and we took only the fundamental so we are looking at the fundamental component alone and that will then be the number of turns that are available here if you call this as number of turns of the stator or let us call it as it is not really number of turns of the stator let me call it as n a ? ? for the stator and we can call the rotor turns as n a ? that is to mean in the a ? reference frame what are the number of turns the stator turns and the rotor turns may be different so if you now have n a ? as the number of turns in the stator we are considering only the stator for the time being then the mmf generated by this i a that is flowing in so many turns at this point 5 could be written as n a ? multiplied by i a into some term which is a fixed number since it is going to be a fixed number some scale factor will be involved looking at the amplitude of the mmf however this is enough for our equations now this multiplied by cos ? is going to be the mmf due to this current at this angle ? and then you have the mmf due to the flow of current in the other axis which one can write as n a ? again because the number of turns are the same in both axis this multiplied by i ? multiplied by sin ? so this is then the mmf f at a particular angle 5 please note again that this i a is the instantaneous value of whatever current is flowing there so if we have drawn this diagram at a given instant of time t then this current i a is that flow of current at that particular instant it is not the rms value or the peak value so to say but now from the figure we can write 5 as this is the angle that we want to say we can write that as ? plus this small difference and therefore I can write this as ? plus 5 – ? so ? is nothing but ? plus 5 – ? and we substitute this expression in the expression for mmf so what we have is f equals let me take this n a ? out and then you have i a cos of ? plus 5 – ? plus i ? into ? plus 5 – ? so that is what you have and these expressions can be simplified again you have 5 – ? as one group and then ? in the other expression so you have n a ? multiplied by i a cos ? plus cos ? cos 5 – ? minus i a sin ? sin 5 – ? and then plus i a cos ? cos 5 – ? this is a this is i ? sin of this so this is i ? sin ? cos 5 – ? sin a cos b and then plus cos a that is the expression you get upon expanding this mmf term and what we can do in this is group the terms we will group all terms that are having this in one hand therefore cos 5 – ? is here cos 5 – ? is here so we can write this further as n a ? multiplied by i a cos ? plus i a sin ? this whole thing multiplied by cos of ? – ? and then you have – i a sin ? plus i ? cos ? into sin of ? – ? that is what you get so we started out with the mmf that is produced by the coil stator a coil and stator ? coil and that mmf equation is the same as what we have derived here however if you look at this equation this equation says that you can write this as n a ? multiplied by let me call this term as id and let me call this term as iq so I have id cos 5 – ? plus iq sin 5 – ? now let us look at the first expression the first expression says that the mmf is produced mmf at this angle ? is given by some maximum value of mmf around the circumference multiplied by cos 5 it is multiplied by cos 5 because this mmf source which is the flow of current in this coil along the a axis is separated by an angle 5 from the point at which we want to determine the mmf similarly this coil is separated by an angle of 90 – 5 from the point where we want to determine the mmf now that is equal to this and if you look at this expression it says that you can consider the mmf at this point 5 at this angle 5 which is generated as being generated by a flow of current along an axis which is separated by ? – ? from the point where we are determining the mmf and this is 90 – 5 – ? which means that this is equivalent to saying that you have a coil along the d axis and you have a coil along the q axis and let us assume that there is a current here flowing which is id and there is a current here that is flowing which is iq and this axis is separated from the displaced from the place where we want to determine the mmf by an angle which is ? – ? that is this angle and therefore instead of considering the mmf here being generated by this coil and this coil due to a current i a and ib I can get the same value by considering a coil which is now placed along the d axis and q axis which has a current id and iq flowing in it id being defined by this expression i a cos ? plus ib sin ? and iq being defined by this expression – i a sin ? plus ib ib x cos ?. So effectively we are now looking at the system being viewed from this moving set of axis we are no longer looking at this but we are looking at the same mmf being generated from this set of axis which may be moving at this particular instant it is at an angle ? and therefore now we can define a relationship between id and iq as being equal to cos ? sin ? – sin ? and cos ? times i a and ib we are doing this only to the a and b terms we are not doing anything with respect to the zero sequence term the zero sequence term as we saw sometime back is deemed to be along an axis that is 90 degrees to both a and b that means out of the plane and that is not really going to affect the mmf here because this angle is equal to 90 degrees that is on a different plane and therefore we transform only the a to this dq this is for the stator we could have done the same thing for the rotor as well because the rotor also has one coil now those pseudo stationary one coil on the a axis and one coil on the a axis we if we want to determine at this angle ? what is the mmf generated due to this current in the a coil and this current in the a coil you would proceed in the same way and you would get a similar expression for the equivalent flow of currents along the d and q axis. So you would have put d as a superscript q as a superscript and then this would be ir a and ir ? so both the stator and the rotor have now been referred to this so called arbitrary reference frame now again why is it arbitrary because the speed of rotation that means d ? by dt ? is the angle made by this d axis with respect to the a axis and if this is going to be rotating ? is changing with time and d ? by dt we will simply call it as ?x we do not know what that speed is the speed in general the useful values of speed are it could be ?s which is the synchronous speed it could be ?r which is the rotor speed it could be ?0 which means 0 speed these are three useful values that are usually then considered in the literature let us look at what is the implication of this kind of referring. So in this reference frame we know the expressions for i a and i ? we know that i a has this form some amplitude multiplied by cos ?t where ? is then the synchronous speed because the frequency of stator excitation is going to determine the speed at which the mmf revolves around the air gap which is nothing but the synchronous speed so that is ?s and similarly you have i ? equals im sin ?s in t. So if you substitute these two into this expression what do we get you get id equals im cos ?st cos ? plus im sin ?st into sin ? which is nothing but im into cos of ?st – ? similarly you have iq which is im – im sin ? cos ?st plus im cos ? sin ?st this is then equal to im into sin of ?st – ? now let us first look at the case where d ? by dt is equal to ?s that means the reference frame is rotating at synchronous speed so if it is rotating at synchronous speed d ? by dt equal to ?s means that ?s or ? equals ? integral of this so that is nothing but ?s into t but in order to satisfy the integration we need to add an arbitrary term ? which will have to be evaluated based on some initial conditions of the system that means at t equal to 0 ? will be equal to ? so ? then represents the angle which the d axis of the reference frame makes with the a axis of the stator at t equal to 0 ? may be 0 or ? may be something else. So if we substitute this expression here what you get is id equals im into cos of so when you put this here ?st – ?st – ? so you get im of – cos of ? which is im cos ? and iq is im sin of ?st – ?st – ? which is im sin of – ? that is – im sin ? that is what you have what is interesting here is that this expression id and iq are not dependent on the variable t that means these are basically constants they are not functions of time being that that also means now therefore id and iq are dc quantities which means now that the excitation provided to this d axis and q axis is no longer ac but it is dc that is very interesting and has very useful implications when you are trying to design a control system I think sometime back I mentioned that it would be good to have dc variables in the loop in order to have your PI controllers they work well with dc and you can be assured of 0 error if it is designed appropriately. How much is the dc value however depends on the angle which the d axis makes at t equal to 0 one can choose that angle appropriately you can choose it based on any condition that you want and if you choose ? equal to 0 then id becomes equal to im and then iq is 0 so that is again useful to say that if I orient my frame d and q axis the reference frame if it is oriented at some particular angle then you can see that only d axis component exist and there is no q axis component. So this is one interesting aspect of the development on the other hand suppose I choose d? by dt that is the speed of the reference frame as being equal to ?r then what happens then id becomes equal to im x cos of ?st-?r t-? which is nothing but im cos of ?s-?r x t-? and therefore this now ?s-?r x t is the frequency of flow of excitation flow in the rotor. So this is nothing but the slip frequency one can alternatively write this as s x ?s x t and therefore that is nothing but the slip frequency that is there. Similarly iq is im x sin of ?s-?r x t-? and this is therefore in the slip frequency that means if you transform yourself to a frame that is going to be rotating at rotor speed that means with respect to the rotor the reference frame is stationary because both are rotating at now at the same speed and if they are rotating at the same speed you are then observing all the dynamics of the machine from a frame that is rotating along with the rotor or so to say sitting on the rotor then what you see is the excitations that are applied to the d and q axis are of slip frequency in nature that means anything that you see from the rotor all the MMFs that are there in the air gap they all appear to be rotating at slip frequency. The actual MMF is rotating at synchronous speed but then sitting on the rotor you are also rotating at rotor speed and therefore with respect to you the MMFs are rotating at synchronous speed how about d ? by dt equal to 0 that means if the reference frame is stationary then d ? by dt equal to 0 and therefore ? is a fixed number and therefore the expressions will simply be id equals im cos ?t-? where ? is a simple number not a function of time and therefore id will be at the frequency of the supply iq is at the frequency of the supply it simply amounts to rotating this a and ? axis by some angle and this is also stationary. So this is just if the reference frame is not going to rotate and the speed is 0 then it amounts to simply shifting the axis by some angle and therefore since it is stationary the excitations are all at supply frequency. So with this reference frame then one can choose whatever you want as I said most for most analysis the synchronous speed reference frame is the one that is of use and that is therefore called as synchronous reference frame this as I mentioned before it is called rotor reference frame and this is then the stator reference frame. So these are the cases now we have dealt with i a and i ? being transformed to id and iq how about the supply voltages. So in order to get that what we say is let us assume that power is invariant which means that the net power that is given in the a b reference frame can be written as i transpose a b reference frame i transpose v is the input that is given in the a b reference frame and what we have here is this idq is equal to let us call this as the C matrix or let us call it more as let us not call it as C we have been using all along let me call it as D matrix let me call it as D11 multiplied by i a beta. So if I now want to find out i a beta having known id and iq I need to pre-multiply this by D11 inverse and what would be the inverse of this the inverse of this is nothing but the transpose of this because the determinant is 1 cos2 delta plus sin2 delta is 1 and therefore we can say i a beta equals D11 transpose into idq. So let us substitute this expression in our expression for the total input electrical power and therefore this becomes D11 transpose into idq and transpose of that multiplied by v a beta this is nothing but idq transpose multiplied by D11 into v a beta which we can now designate as idq transpose into vdq where vdq is then this expression and one can see that vdq is similar to the definition for v for a beta to dq and therefore we write vdq equals D11 into v a beta so the manner in which the i a beta gets transformed to idq and v a beta gets transformed to vdq they are identical. So having defined this these equations we now need to see how that impacts our machine description. So the machine description we have already written down in the stationary reference frame what we have basically this equation can be written as you would have seen now many times can be written as v equals r into i plus l into vi plus g into i into ?r that is our expression. So all these are in the a beta frame lr and then g so on. Now what we want to do is to use those equations remember these vectors have both stator terms and rotor terms you have the stator terms here first three of them are stator and the next three of them are the rotor terms and what we want to do is define a relationship between we want to transform this to the dq reference frame. So we want to define a relationship from the stator to rotor you have vd vq so we have a relationship which we want to define from the a beta to the dq reference frame. So let us say vd vq v0 these are the stator terms and then you have vd vq and v0 which are the rotor terms so that is what we have and we need to define them in terms of v a vbeta v0 for the stator and the rotor now obviously we have seen that the stator equations depend only on stator variables stator equations do not depend on the rotor variables and therefore what we can say is here you have cos ? and then sin ? – sin ? cos ? 0 sequence as I mentioned is unchanged we do not touch the 0 sequence it is not necessary to touch that so that remains the same and this does not depend on the rotor terms and therefore these three are 0 when you come to the rotor you have it does not depend on the stator 0 sequences or the a beta so that is not there similarly here you have cos ? sin ? 0 – sin ? cos ? 0 0 1 so this is the transformation that is then used from v a vbeta v0 v a but r vbeta but the rotor and v0 rotor so that is the expression and that is why we called this matrix as d11 so that is the same as d22 also these are 0 matrices so what we can say is vdq let us now have this notation dq for the stator and dq for the rotor as well this is equal to d times v a beta on the stator and the rotor is also referred to a beta of the stator so this vector we let us call it as a beta and then r similarly then we can define idq dq is d times ir a beta using this we need to do we need to look at what happens to the equations and if we want v a beta then we have to pre-multiply this by d inverse and because this is a matrix that has entries only in the block diagonal and this term inverse of this itself is equal to the d11 with a superscript so what we can say is d transpose into vdq dq is equal to v a beta r and similarly idq dq and d transpose is equal to i a beta r so it is these two equations that we need to substitute in this in order to find out our new machine description which we shall proceed to do so now so what we have is we substitute so dd transpose pdq dq equals r d transpose i dq dq plus l times p d transpose i dq dq plus g times d transpose i dq dq omega r and therefore if I want to find out vdq dq you have to multiply by the inverse of this matrix throughout and again inverse is nothing but the transpose and therefore this is d times here d times here and d times here and therefore this simplifies as dr d transpose i dq dq plus now in this term you have p and then d transpose i dq dq this d is also going to be dependent on t with respect to time that is and therefore one can write this as dl d transpose p i dq dq plus dl pd transpose i dq plus dg d transpose i dq into omega r now in this term pd transpose you are taking the derivative of d and d depends on delta therefore you will get a d delta by dt term which is then your term which depends upon the speed of the reference ring. So this is the expression that you get and one has to substitute the terms and simplify it to see what is the effect it is going to have on the equations of the machine itself which I will leave it to you as an exercise to do you will look at the final equations in the next lecture you will stop here for now.