 We were looking at small signal model for the induction machine in the last lecture. So what we had was the expression for the machine equation was V equals RI plus LPI plus G1Omega s into I plus G2 into omega R into I. This equation being written for variables in the synchronous reference frame and we have dropped the superscript and the subscript here in order to simplify the notations. We understand that it is in the synchronous reference frame. So when you disturb the equations we are let us review what we have done towards the end of the last lecture. So if you disturb this you have V0 at which the system was operating and then you apply a disturbance which is ? V that is then equal to this R into I plus ? I plus L I0 plus ? I L into P of I0 plus ? I plus G1 into ? s0 plus ? omega s into I0 plus ? I plus then G2 into ? R0 plus ? omega R into I0 plus ? I. So here the inputs so to say for the disturbances are ? V and ? omega s because the frequency of the AC supply is also in our hands if you are looking at an induction machine controlled by an inverter. So this equation consists of both steady state terms and the disturbances the steady state terms must by themselves be equal on the left hand and the right hand side and therefore one can remove the effect of the steady state terms this R I0 is it can be removed V0 is then equal to R I0 plus L P I0 plus G1 ? s0 into I0 plus G2 into ? R0 into I0. So these terms on the left and the right hand side they are equal and hence can be removed from the equation which then gives us an expression for ? V ? V is then R into ? I that is from this term plus L into P ? I that comes from here from this term you would have G1 ? ? s into I0 G1 ? ? s into I0 and then you will have G1 ? s0 into ? I into ? I and then you would have another term G1 ? ? s into ? I which we neglect in comparison with these two terms because this is a product of two small disturbances which we consider it to be negligible and then you have G2 into ? R0 ? I plus G2 into ? R into I0 similarly we neglect the term G2 into ? R into ? I so this is an expression that then relates the small signal inputs given as disturbances to the response which is basically ? I and ? ? R apart from this you have torque is equal to I transpose into G2 into I and therefore in order to get the disturbed values you could say T plus ? T equals I plus I0 T0 ? ? T is I0 ? I transpose into G2 into I0 ? I and one can therefore write ? T as being let us say G3 into ? I so this expression would be dependent on ? I term so we write it as some matrix G3 into ? I so this is one part of it and then if you look at the mechanical equation you have JD ? by DT equals this develop torque minus the load torque minus let us say some fluid friction that is viscous friction into ? and therefore this can be written as JD by DT or let us write it in the other notation J into P of ? ? R note that this is ? R that is the rotor speed and this is also the rotor speed therefore this is equal to ? T – ? T L – B into ? ? R so these are your equations relating disturbances you have one equation here and this is the mechanical equation and then you have the electrical equation this ? T is nothing but this expression which goes there and therefore one can now combine all these three expressions and write it in the so called state variable notation which is useful for further simplifications and that can then be written as P vector ? I and ? ? note that this vector ? I itself is not a single element it is it has four currents four into one vector if you neglect zero sequence otherwise it would be a six into one vector so this is then equal to L inverse multiplied by – R – G1 ? S0 – G2 ? S0 sorry G2 ? R0 and then in the second column you have – L inverse G2 into I0 and here you have G3 by J and then here you have – B by J this matrix multiplied by ? I and ? R plus L inverse and then – L inverse V1 I0 0 0 0 1 by J multiplied by ? V ? ? S and ? T so this would then be the state variable representation of the small signal model so here you have the state vector which can be written as P x if you describe this as the state vector x that is then equal to A x where this matrix is A and that is your state vector x plus B x the input vector U this is the input I mean this is B matrix and then here you have the input vector so these are your inputs that are there that is the change in the supply voltage change in the supply frequency and change in the load torque once you have it in this form this is nothing but a series of differential equations first order differential equations and as we have seen in the case of DC machines one can if you want to solve this transform this into the Laplace domain so you have S times the vector x of S equals A times x of S plus B times the input vector U of S and therefore you can solve for x of S as S into identity matrix please note the change in notation here it is usual to write in this expression S times I whereas here we have used I for the vector of flow of amperes so you have S I minus A inverse into X of S plus S I minus A inverse into B times U of S that will then give you the state vector x you can then take the inverse Laplace transform and for any given input x input U you can find out what x is that is what we have done in the case of DC machine and seen certain examples of what would be the response if you apply certain disturbances now that kind of an analysis can be done for the induction machine also in this reference frame the advantage is that all the variables here that is I and V they are basically DC so as far as the equations are concerned in the way you analyze use those equations are concerned all these variables are DC and therefore it sort of looks similar to a DC machine from that viewpoint. So this is small signal analysis one can then go about deriving the relationships between for example ?r of S how it is going to change with respect to let us say ?vd S that means the stator d axis input if you change that what happens to ?r so this is then described in terms of the Laplace variable S and so you get some numerator by denominator both as a function of S one can then look at the roots of the denominator in order to find out how this is going to change as with respect to time as usual the numerator function also gives you more information about the amplitude of the overshoot etc. So one can do similar analysis as we had seen the DC machine using this small signal representation but this is not all in the case of induction machines in order to do close look control it then becomes necessary to look at how to simplify these machine equations themselves if you see for example in the case of induction machines we have seen here this expression is the differential equation describing the electrical subsystem and you basically have PI here which is what we express the order of the equation itself even if you neglect these terms the order of the equation is you have four variables here you have one variable here so it is a fifth order differential equation and therefore it becomes fairly involved in order to design your close look controllers for this arrangement. So one can then do something to simplify these equations a little more so how does one go about doing that that is what we will see now in the machine equations that we have derived so far what has happened is that you have the stator access this is your a stator access and then you have the ß stator access and we have transformed the equations to the synchronous reference frame which means that you now have a reference frame which is let us say aligned at some angle this is your d synchronous access and then you have the q synchronous access this angle is 90 degrees because all the access are like that way and then this angle is what we have called by the symbol ß and we have said that d ß by dt is the speed at which this reference frame now rotates for the synchronous reference frame d ß by dt is nothing but the synchronous speed. So having said that we have been seeing all along that you may have you initially had coils that were here this was your stator coil and this is the rotor coil on the a access in the stationary frame and then you had the stator coil here and then the rotor coil here what we have now done is transform these things to have a stator coil on this synchronous frame and then a rotor coil on the synchronous frame so you have a stator coil here and then a rotor coil here if you look at the equations themselves let us write the equation for the stator voltage Vds so if you say that this is your stator plus and Vds is applied here and this is Vds that is the rotor coil and then Vqs is applied here and this is Vqs which is the rotor coil remember s stands for synchronous speed and q stands for the access of course as long as we write it below it is a stator as a superscript it is refers to the rotor. So Vds if you take that is a stator d axis that we have seen is RS plus Lssp multiplied by Ids and then you have minus ?s Lss into Iqs and then you have plus Lmp Ids – Lm ?s Iqs this was the equation for the d axis voltage of the stator neglecting 0 sequence components so let us this is neglecting 0 sequence mostly that is true 0 sequence component does not flow in the case of an induction machine because invariably the stator have gets 3 wires applied and 0 sequence components are in phase which means unless the neutral wire is there there is no return path and therefore 0 sequence components cannot flow this is what you therefore have and if we look at this one can see that this term is a di by dt term and this term is a di by dt term it sort of looks like a d by dt of self inductance of the stator multiplied by the direct axis the ds flow of current plus a mutual inductance with the d flow of current in the rotor if you have a system like this you have one coil here and one coil here and one way to write an expression for the flux linkage in this coil let us say ?ds how would you write it the flux linkage in this stator coil arises due to a self flux linkage and a mutual flux linkage there could also be in general flux linkage is from other coils around but in this particular case the other two coils are at an axis which is 90 degrees away and therefore the flux component produced by this coil on this axis would be 0 and therefore if you want to write down the expression for flux linkage in the ds stator coil you would then write it as lss multiplied by ids plus the mutual inductance which is now lm if you look at this expression this expression says mutual inductance between these two coils because we have now referred everything to the stator turns as well in addition to referring to a common reference frame this inductance if you remember in the last lecture we had derived this so that is lm into ids similarly one can write expressions for the other flux linkages as well so you have ?qs as lss into iqs plus lm into iqs this is the flux linkage in the q axis coil of the stator is self inductance multiplied by the flow of q axis i and then q current of the rotor multiplied by the mutual inductance. So these are flux linkage of the stator now this flux linkage of the stator you see that there is one part which is along the d axis and this is along the q axis so which means having designated the axis here as d and q the flux linkage of the stator has a certain d component something here and it has a certain q component which is something here and therefore the net flux linkage of the stator can now be drawn as a space phasor in this manner where this length is then your ?ds and this length is your ?qs similarly one can define ?ds which is nothing but the rotor d axis flux linkage that would then be the self inductance of the rotor lrr dash because it is now been referred to the stator turns multiplied by ids dash that is now the current flowing in the d axis coil of the rotor plus lm x ids so this is the mutual inductance and the flow of current in the stator d axis so that is lm x ids together it then contributes the d axis component of the rotor flux linkage we can write ?qs which is the q axis component of the rotor flux linkage as lrr dash multiplied by ids dash plus lm x ids so now one can just like we represented ?ds and ?qs as a space vector this also has a d axis component and a q axis component therefore this can also be represented as a space phasor it could be some other space phasor may be like this so this is then your this length would be ?ds and this length would be ?qs now you see that you have 4 variables here ids iqs of the stator ids iqs of the rotor and you have 4 variables here as well ?ds iqs of the stator and ?ds iqs of the rotor therefore one can instead of describing this equation for vds in terms of the variables i one can choose to describe it as instead of i you can choose to have the independent variable as ? or the dependent variable as ? so either what you could do is use these 4 expressions in order to eliminate all these variables ids iqs for the stator and the rotor or you could choose to retain ids and iqs of the stator but replace these 2 variables in terms of these 2 variables or you could retain these 2 and eliminate the stator id and iq and replace it in terms of stator id and ?q one can therefore have many different ways of writing this expression for example let us say that we choose to eliminate the rotor d axis of current and the rotor q axis current and instead replace this by ?ds and ?qs how does one do that what we will have to do so if we look at that expression what you have is you can write ids as ?ds –lm x ids divided by lrr that similarly you can write iqs as ?qs –lm x iqs divided by lrr that not sure if that is really visible because of the older. So what one has to do is replace the variables for the rotor id and iq by these expressions so that those variables are eliminated so let us look at the first equation again you have vds for the rotor is rs – lssp x ids and then you have – ?s lss x iqs and then you have lmp x ids that now becomes ?ds –lm x ids by lrr that and then you have –?s x lm x iqs so that now becomes ?qs –lm x iqs divided by lrr that so in this way we have now eliminated these two variables ids and iqs so let us simplify this expression so what you have is you get one ids term here there is already an ids here so what you have is rs – lss p ids and then here you get –lm x lrr x ids so it is lss –lm x lrr – p x ids so this term has been taken care of and then you have iqs here similarly you have iqs here as well so you have – ?s x lss and then here so –lm x lrr – x iqs and then you have plus lm by lrr – p ids and then here –?s lm by lrr – x iqs so we had an original equation let us say vds is this we had the equation in terms of id iq of the stator and then id iq of the rotor now we have eliminated id iq of the rotor using the definitions of flux linkages that we had written down and now the expression is id iq of the stator but ?d ?q for the rotor now this can be done for all the equations and when you do that the resulting set of equations look like this what you have is vds vqs vds vqs 0 sequence component has been neglected these are the voltages applied to the stator dq coils these are voltages applied to the rotor dq coils the term s refers to synchronous reference frame this is written as rs – s lss p and then ?s x s lss and this is –rr lm by lrr – 0 and then –?s s lss p 0 –rr lm by lrr – lm by lrr – p –lm by lrr – ?s rr by rr – p and this is ?s ?r and then here you have –lm by lrr – ?s – ?rr – p multiplied by the current vector that is ids iqs ?ds ?q so this now becomes the voltage description of the voltage subsystem in the induction machine there is also the expression for the generated electromagnetic torque which then becomes lm by lrr – multiplied by iqs x ?ds – ids x ? so we have arrived at a slightly modified induction machine equation we have replaced some variables in terms of some other variables one might ask what is the interesting aspect of this well in this particular case if you look at induction machine in general normally the rotor is going to be shorted so vds and vqs for the rotor are invariably 0 unless you are looking at induction machine fed from the rotor as well a doubly fed kind of arrangement otherwise these two are 0 now if you look at these equations one can now try to do something else that is if you look at the figure here we said that there is a stator flux vector and there is a rotor flux vector as well now all these vectors this reference frame is rotating at synchronous speed and then this the rotor flux vector is also rotating at synchronous speed and then the stator flux vector is also rotating at synchronous speed all of them are rotating at synchronous speed and therefore this angle under steady state the angle that the flux vector makes with respect to the d axis or the angle that the stator flux vector makes with respect to the d axis is going to remain fixed it is not going to change with respect to time because all of them rotate at synchronous speed however this angle which the d axis makes with respect to the stator a s axis at t equal to 0 could be different in other words for the same stator flux vector and the rotor flux vector which are rotating at synchronous speed I could have a d axis here or I could choose to have a d axis here or I could choose to have a d axis here all of them rotating at synchronous speed of course one can look at it more from the view point of equation you are saying d ? by dt is ? s which means that ? is equal to ? s x time plus an arbitrary ? 0 how do you fix ? 0 ? 0 is the angle that the d axis is going to make with respect to the a s axis of the stator at t equal to 0 so one can now choose the d axis of the reference frame anywhere you want by choosing the angle ? appropriately or in this case by choosing ? 0 appropriately and since you can now locate this reference frame anywhere you want with the restriction of course that it is going to rotate at synchronous speed because it is a synchronous reference frame. I could choose this d axis of the reference frame along this vector ? of the rotor itself I can choose this itself to me my d s reference frame I can select it anywhere I want here or wherever it is but supposing I choose it here then the q axis would lie at an angle 90 degrees to this which means somewhere here and this phase phasor ? of the rotor would not have any thing along this q axis no part along the q axis it entirely lies only along the d axis so this is choosing a particular orientation for the reference frame right in order to do that I would need to know by some means where this ? of ? vector for the rotor is and having decided having located that vector I now choose a d axis which lies along that vector itself if we do that and we look at this expression for the generated electromagnetic torque e you find that ? d s is nothing but ? of the rotor in the synchronous reference frame ? q s is 0 in other words one could write the phase phasor ? as being composed of ? d s plus j times ? q s and if the reference frame is so aligned to be along the space vector ? itself then ? q s becomes 0 and therefore this ? is just equal to ? d s of course this ? d s in general would be different what we can see is depending on where the d axis is located the magnitude of the d component magnitude of the q component is going to vary which is what happens now if the reference frame is along this itself q component becomes 0 so ? d s is then equal to the magnitude of the ? phasor itself which means that now torque is nothing but lm by lrr dash into ? let me call it ? s into iq s how is this interesting if you look at the case of dc machines let us draw that figure somewhere may be I will draw it here in the case of dc machines what we have is the field generated by the stator poles is there and then you have the armature that is going to be rotating and you place a brush which is in the magnetic neutral axis and you have the armature conductors that are rotating here because the brush is placed in the magnetic neutral axis the mmf produced by the armature the field produced by the armature is oriented along that particular axis and the field produced by the main stator that is the main field is going to be oriented along this axis which means that if you load the armature and armature produces its own field that field is not going to interfere with the main axis field the magnitude of the main axis field under normal circumstances because these two axis are separated by 90 degrees no component of this will be present along this axis which is a great advantage when you want to try and control the speed of the machine or to be more specific in order to control the generated electromagnetic torque this torque is if you remember back the dc machine equations it was a mutual inductance multiplied by the field current multiplied by the armature current that is true because the field current is producing a field in this direction armature current is producing a field in that direction and therefore if you want to control the generated Te since it is proportional to If and If you can control If alone independently and since If does not influence the main field you can keep this If constant as soon as you change If immediately it will reflect on the generated Te in the case of induction machine normally if you look at the three phase reference frame itself this is not true in the induction machine you have three currents these three currents are responsible for the production of magnetic field inside the machine and also the generated electromagnetic torque Te so if you want to change the generated Te if you change this current it will result in changing the field as well as changing Te and there is a big interaction that going on you will unless both of them settle down you will not be able to control Te really well this approach if you now look at the induction machine in the synchronous reference frame you are able to identify which component of the flow of current in the stator is able to affect the generated Tor and if you choose your reference frame appropriately in that reference frame one can see that the generated Tor is simply a product of the rotor flux vector multiplied by the stator q axis current therefore if you can maintain the rotor flux vector independently stable at a fixed value then as soon as you change q axis i that is iqs it will immediately reflect in a change in the generated electromagnetic torque this is to say that this expression means that we are now by transforming to the synchronous reference frame with the d axis aligned along the rotor flux vector we are looking at an induction machine which behaves essentially like that of a DC machine and therefore in this reference frame the machine should be much easier to do closed loop control one can look at these expressions further and one can see that from these machine equations which we have written from the equations that we have we have here one can look at these two equations and it can be simplified what you have is vqs which is nothing but 0 that is equal to Rr x Lm x Lrr dash x iqs ids and then vds which is again 0 because the rotor is shorted is – Rr Lm Rr dash that is by Lm Lrr dash x ids x Rr x Lr dash x p x psi ds so if you look at this equation one can recast this equation to say that psi ds is nothing but Rr dash Lm by Lr dash x 1 by p plus Rr dash by Lr dash x ids in terms of the Laplace variables one can then call this by the Laplace transform of ids as a function of s this p can be written as s as the Laplace variable now you see that the flux vector psi ds if you oriented along the reference frame psi ds is nothing but the rotor flux vector psi s itself that is just straight away dependent on ids if you control ids you determine what psi s is this is related only as a simple first order equation which is again similar to that of the field in the DC machine. So we can say that ids controls the flux in the machine whereas iqs determines straight away the generated electromagnetic torque ee and one can get from this equation if these two are met then this expression gives an idea of the slip in the machine this says that omega s minus omega r should then be equal to Rr dash Lm by Lr dash x iqs by ids so in turn is nothing but the slip so if you have a machine where you can directly control or where you can directly give as an input ids and iqs these equations say in this particular reference frame when you are viewing the machine it should be feasible to control the induction machine just like you control a DC machine. So incidentally then this forms the basis of what is called as field oriented control of induction machine field oriented control sometimes it is called as vector control of induction machine and here the orientation is done along the rotor field vector and therefore this is called as rotor field orientation one can derive other forms of expression where you can do a stator field orientation or magnetizing flux vector orientation there are so many different ways in which these have been described in the literature. So this is again one of the other useful aspects of looking at models of the induction machine in various reference frame these are all in the synchronous reference frame but one have we have replaced some elements some unknowns or dependent variables by certain other dependent variables you can for example choose to as I said retain ids iqs here but then convert these two to ids and iqs of the stator derive more equations and then try to look at machine description in that I would urge you to look at the literature for other ways of describing this and other ways of field oriented control that this sort of a description makes feasible. So we have looked at induction machines over the last several lectures and then seen how this ideas of machine modeling have made feasible different ways of analyzing the induction machine small signal analysis kind of thing is feasible because of this kind of a representation. There is much more in the literature and I would urge you to look at literature for more elaborate arrangement from the next lecture we will start looking at models for the synchronous machine and see how they can be represented and analyzed. We will stop here with this lecture.