 the last lecture we had looked at the small signal modeling for the DC machines. We derived the small signal models by assuming that there are disturbances to the inputs that are applied to the DC machine namely the voltage applied to the armature, the voltage applied to the field and to the load torque that may be there and in response to that the system responds with small disturbances in the flow of armature current, field current and in the speed and by assuming that the disturbances are small we derived equations that describe relationships between these disturbances and those are then the small signal equations of the DC machine. And we said that these small signal equations can then be used to derive relationships between a response considered as an output and a input that you may be applying for example the speed as the response and armature voltage as the input and we had said that we can look at those equations in the next lecture. So here it is we can see here the small signal equations of the DC machine you have d by dt of ?if that is the field change in the flow of field current d by dt of ?ia that is the change in the armature current and d by dt of ?r that is the change in the rotor speed. This is then related to the change in the field current armature current and the rotor speed and the input disturbances that may be there. So one can see that the change in the field current that flows is purely dependent upon the field excitation itself. So that is d ?if by dt is nothing but –rf by lf into ?if plus ?vf by lf. So that means it is not influenced by what is happening in the armature or on the mechanical system field is so to say independent. Whereas if you look at the change for that is d equation for d by dt of ?ia you see that it depends on the change in the field excitation due to this term –msr ?r by la multiplied by ?if. So the rate of change of armature flow of current depends upon what is happening in the field and of course –r a by la into ?ia which must definitely be there and then also it depends upon the change in rotor speed due to this term –msr if by la into ?r and the change in armature voltage itself ?va by la and you see here another aspect that the way in which d by dt of ia depends upon the change in field flow of current it is dependent on the speed at which the machine is operating. Therefore you have this ?r that appears here and similarly the manner in which d by dt of ?ia depends upon the change in speed is decided by the value of if at which the machine is operating. So these are therefore dependent upon at what speed you are looking at for the small signal model. Similarly d by dt of ?a if we look at then that is also dependent upon the change in field that is happening and it is dependent upon the armature flow of current that was there at steady state prior to the disturbances that are there and then here it also depends upon armature current change and that is dependent on how much field current was flowing prior to the disturbance and then –b by j into ?r itself and of course the input load torque disturbance that is there. So this is then the small signal equation for the separately excited DC machine and this equation has been written in the standard form where you have x dot is equal to ax plus bu. So this matrix then forms the a matrix this vector is the state vector and this matrix is the input matrix or b matrix and this is the input vector u and this is then x dot which is dx by dt in order to solve this and obtain the responses as dependent on the inputs you then recast this equation in the Laplace domain. So x dot then becomes s of x as we mentioned sometime back the initial conditions for the disturbances are 0 and therefore you do not have sx – x0. So the Laplace transform of x dot is s x and then you have ax this x is really a function of s and then bu is a function of s here x is a function of s that is the Laplace domain and then if you bring this term to the left hand side you can write this expression as s times u where u is the identity matrix u is the identity matrix. So s times u – a x is bu and therefore x itself can be derived as su – a inverse. So you take this left hand side term to the right multiplied by the inverse so su – a inverse b x u. So by looking at this expression one can then derive appropriate Laplace transform expressions for various elements of this vector x when subjected to the disturbances that are there in this vector u. So let us look at one specific example so if you look at that expression this is your vector x ?if as we have said before is the change in field current this is the change in armature current and this is the change in speed. So these Laplace transform expressions can then be written as 1 over d times this description that is provided here multiplied by the vector of input u and this expression that shown here is nothing but s times u – a inverse that is what we had seen here su – a inverse so when expanded that is what it looks like this that is what it looks like and d itself is shown here s plus rf by lf x s plus ra by la x s plus b by j plus msr2 x if2 by jla so you have a fairly big looking expression here so from this one can then extract whichever transfer function that we want transfer functions are basically ratio of a specific output variable Laplace transform of a certain output variable for example one can call it as ?if of s by ?vf of s. So if that is the one that we are looking at ?if of s is related to ?vf of s as is here by the first term that occurs in this su – a inverse and therefore ?if of s is nothing but the first term here divided by d x ?vf by lf and one can see that this term divided by d is nothing but the term here is the same as the one that is there in the square bracket and therefore ?if by vf would be nothing but 1 over lf x 1 by s plus rf by lf it is a simple expression even though this looks intimidating when you simplify it because this term what is here is identical to the term inside the square bracket here and therefore that is removed what you are left with is s plus rf by lf. So in this manner one can derive any transfer function that we want for example if you want to look at the transfer function between if and armature voltage as the input you see that the term which relates the 2 is 0 that means if you change armature voltage field current is not going to be affected. Similarly if you look at the relationship between ?if and load torque that is applied there again it is 0 indicating that they are completely uncoupled whereas if you want to look at the term ?ia a change in ?ia will occur due to a change in ?vf because this first term is not 0 it can also occur due to a change in ?va because the second term is not 0 it can also occur due to a change in load torque because the third term is not 0 similarly one can also look at ?r moving on further if we look at a specific case which is what we wanted to look at in the last lecture. So this gives us the relationship between a change in the rotor speed and change to change in the voltage applied to the armature and that is described like this. Now you see that here this is dependent only on if that means the field current that was flowing prior to the disturbances if you say that the field current is kept fixed that it is not going to then this is independent of where it operates and therefore if you apply a change in armature voltage irrespective of the operating speed of the machine the manner in which the speed is going to respond to that change in armature voltage will always be same independent of at what speed the machine is operating you can see that this is independent of speed it is also independent of armature current so which therefore is a reflection of what is the load the machine is operating at at the time when you give the disturbance. So whether you are talking about a low load situation or a high load situation if there is a change in the armature voltage that is applied the machine will always behave in the same way. Now looking at this expression we know from the understanding of Laplace transforms that we have the manner in which the response is going to happen there are two aspects to it one is the variation with respect to time that means whether the response is going to be oscillatory that means if you change armature voltage a little whether the speed will oscillate and settle down to a new value or whether it will simply rise like that and settle down to a new value all that is the manner in which response is going to change with respect to time and that we know from our understanding of Laplace transforms that this is dependent on the poles of this transfer function poles of transfer function. So one looks at the roots of this equation and then dependent on the root depending on the roots root locations one can conclude how this system is going to behave and that root as we see is dependent on what field current machine is operating in whereas the other aspect is the amplitude of variation and that has is not only dependent on the roots it depends on the numerator as well in this case there are no zeros the numerator that is going to be there and here there are no zeros if there are zeros then it would depend on the location of the zeros also. So one can get a considerable amount of information by looking at this so let us look at another example from the same system so here we are looking at the relationship between the speed and the field voltage that is applied as input. So if we go back one can see that the relationship between the speed that is here and the field voltage that is applied would then be given by this term the term in the third row first column of Su-A inverse and that is exactly what we have taken out and written here in a simplified form so that is what it is and one can see that this expression depends upon the steady state armature flow of current that was there prior to the disturbance it depends on the field current it depends on the speed at which the machine is operating and therefore it is really dependent on what is the system circumstances at the time of applying this disturbance. So I have an example here so for example let us take this DC machine in order to understand what the small signal disturbance means and how the response is really related to the large signal equations of the machine. So here we have a DC machine it is a 5 HP DC machine the armature voltage the rated voltage is 240 volts the field voltage rated is 300 volts the rated speed of the machine is 1750 rpm that means under rated voltage and field conditions if you apply rated load that is the speed at which the machine is expected to operate that has an armature voltage armature resistance of 2.581 ohms armature inductance of 28 millihenry and a field resistance and a field inductance LAF represents the mutual inductance between the field and the armature that is the same as the term MSR that we have been using J refers to the moment of inertia of the load as well as the machine both referred to the shaft of the machine B is the viscous friction term and TF is the friction term and to this a load torque has also been applied and here we look at the large signal model response. So if you remember the large signal equations of the DC machine we are looking like this VF and VA are the inputs that are applied to the machine this was equal to RF plus P times LF 0 and then here you had MSR into omega R and then you have RA plus P times LA this multiplies IF and IA these were the equations for the applied voltage as far as the generated electromagnetic TE that is the torque is concerned we know that that is MSR into IF into IA. So this is the expression for the generated electromagnetic torque and the mechanical equation would then be written as J times d omega by dt is equal to TE minus the load torque these are if you remember the large signal equations of the DC machine these large signal equations of the DC machine are valid under all conditions where the assumptions having derived these machines are met which is by and large under all situations. So one can study what is happening to the DC machine with these equations itself but these are non-linear in general if you are considering both variation of field current and variation of speed and the small signal approach what we have used helps us to derive linearized models and then relationships between certain disturbance inputs and disturbance out. So let us look at how these two are to be used so this is the large signal response itself in this what we have done is applied a large step voltage to the armature that means armature voltage goes from 0 to 240 volts in one single step and there is also a load on the machine right from time t equal to 0 and with that what we are looking at is the response of speed in radian per second with respect to time. So we have drawn a graph of how the speed is going to change with respect to time. So you see that the speed there is a slight dip in speed it goes negative for a small interval before which after which it becomes greater than 0 this dip is because of the fact that load torque is already applied on the machine attempting to rotate the shaft in the other direction. So before the machine can generate sufficient amount of generated torque by itself the load would have rotated the machine in the other direction and that is why the speed has reversed and then when the machine itself generates enough of torque the machine accelerates and you see that there is a considerable overshoot and then the machine settles down to a steady operating speed there are no oscillations here and what is done is we are applying a small change in the field voltage a field voltage change of 1 volt is applied that means the field voltage rated value was 300 volts as you can see here the 300 is made 301. So that is a really small disturbance and that is applied at t equal to 6 seconds that means at t equal to 6 seconds the machine has already settled down into a steady state there is a particular value of rotor speed and because load torque has already been applied there is a certain armature flow of current that is there certain field current is already flowing one can find out what they are by looking at the steady state equations also. But what is done here is a small change in field voltage from this graph one cannot see any effect of that small change in field voltage but let us zoom into this area of the speed graph and what we see here is how the speed is going to change with respect to the small voltage that is applied to the field one can see that the speed was operating at near about 209.22 radian or so at this point instant there is a slight change increase in the field voltage that is been applied and one can see that as a result the speed drops continuously and finally settles down at a speed near about 208.62 or so 208.6 radian per second there are no oscillations here and the speed response appears to be typical of an over damped system the fall in speed is rather slow it is not really happening fast and there are no overshoots and oscillations and so on. Now this is the response that has been predicted by the large signal model that means how we do it is you solve the set of equations until t equal to 6 seconds you solve it with a field voltage equal to 300 and after t becomes greater than 6 you make the 300 as 301 and continue solving so that is what you get from this on the other hand the nature of this response you see that initially the speed was operating here it was at this level and finally it has settled down at this level so this is then the change in speed the Laplace transform expressions that we derived they were expressions for change in speed for change in field voltage that is the expression that we have derived a few minutes ago. So if we use those expressions what we expect to see is this shape we do not know where it was operating initially from these expressions right this expression only gives you the manner in which speed is changing with respect to the change in the field voltage. So if you now use those Laplace transform expressions and substitute if required the value of speed at which the machine is operating value of armature current at which the machine is operating field current certainly those are required because the transfer function as you see from here depends upon the speed of operation field current of operation and armature current of operation so substitute all that and apply this 1 volt change in the field voltage that is nothing but the unit step response for this system and what you see is this. So here this graph shows you a comparison of the response that has been produced due to the Laplace transform description and that which has been obtained from the large signal model we have done since we are only looking at ?o what has been done in this plot for the large signal response is the actual value of ? – ? at steady state so having subtracted ? at steady state the response will therefore start from 0 change at 0. So this is the time that then is equivalent of t equal to 6 seconds and as far as the small signal plot is concerned we are applying a unit step input at t equal to 0 and the response is shown by this graph here it is actually both the responses are shown here since they are overlapping exactly it is difficult to make out that there are two graphs here there is one graph which is green in colour and one that is blue in colour the blue one is the result of the simulation made using large signal model so this is the large signal model and the green one has been generated using the small signal model. So one can see from this graph that the response predicted by the small signal model is exactly the same as the one that you get using the large signal model there is no difference at all but one must note at the same time we are applying only a 0.3% disturbance on the field the field was earlier 300 volt and you are applying 1 volt disturbance that means 1 upon 300 one third 0.3% is the disturbance that you are applying for which it predicts exactly what the large signal model predicts and therefore one can therefore go ahead and use the small signal model itself in order to predict the response. But if you now increase the disturbance let us say you are doing a 5 volt step in the field voltage here again you see the difference between the two responses the blue again represents the response due to the large signal model the green represents the response from the small signal model and you see that while initially there appears to be no difference as time goes by the difference between them slowly increases and it settles down to an error of 1.2% between the two and therefore as the disturbance magnitude increases the response estimated from the small signal model is likely to deviate from the response that would actually be there in the machine. The large signal model response is the actual response whereas the small signal model has been obtained by making certain approximations we have neglected certain terms saying that they will be small but if they are not small this is the kind of error that one is going to land up with. But it is also important to note that initially the responses are almost identical and if one does not if one says that 1.2% error is acceptable then even a 5 volt step is a small signal disturbance for us. Now let us look at the same system that is we are looking at the transfer function between ?r and ?vf so one can see here that the numerator is a first order numerator there is only one s term whereas the denominator is really having order 3 this is a second order and then you have a third order denominator as well. So you have 1 0 and 3 poles so we are here looking at the 0s and the system poles you obviously see that the description depends upon the speed armature current and field current and we are here we have listed how the 0s of the system and how the system denominator roots which are the poles how they are going to vary when speed is going to be changed speed changes we have speed change we have looked at from 100 rpm to 1700 rpm you see that these three which represent the poles of the system they have not changed at all which stands to reason because here the denominator does not involve ?r therefore change in ?r obviously does not impact the roots of the denominator whereas for the numerator you see that as the speed is going to change from one to the other the numerator roots are initially negative in the low speed region and then become greater than 0 that means these two roots are on the left half of the s plane whereas all these roots are on the right half of the s plane. By how will that affect the response you see that ?r by ?vf can be written as some probably gain multiplied by s plus 0 divided by s plus pole 1 into s plus pole 2 into s plus pole 3 there are three poles and 1 0 now as far as the denominator is concerned all the roots are on the left half of the s plane and therefore in the form in which it will occur here p1 will be greater than 0 p2 is greater than 0 and p3 is greater than 0 in this expression the actual pole is negative or maybe if you put it as s minus p1 then the term will become greater but now here looking at this if we want to find out what will be the steady state of this that means having applied a disturbance where the speed settles down at what value the speed will settle down that can be obtained by looking at the relationship here and then trying to let s tend to 0 if you look at s tends to 0 and write down this expression what we see is that the impact of the 0 location being negative or greater than 0 is that for all these 0s which are on the rhp these are all on the right half plane in order to get a 0 on the right half plane the numerator should be of the form say s minus 42.135 for this particular 0 or maybe if you want to look at this particular 0 it should take the form s minus 2022.2 that is what you are going to have and in this if you now let s tend to 0 so what we see is that the numerator now becomes negative whereas the denominator is going to be greater than 0 and therefore ?r by ?vf will then be negative as long as the roots are on the right half of the s plane that is exactly what we have seen here that if you change vf if you change the field voltage by some value the speed drops ?omega is negative is it agreeing with our intuitive understanding if you increase the voltage applied to the field more field current will flow and as a result of higher field current there will be higher magnetic flux generated within the machine and if higher magnetic flux is going to be there the induced emf will increase and if the induced emf increases then for the same applied voltage in order to meet that speed can drop in order to generate similar induced voltages if armature current is not sufficiently disturbed so it appears that this is alright that is ?omega being negative agrees with what we see from the what we see from the 0s of the machine and our simulation output but you get a totally different situation if the 0s are on the left half of the s plane then the numerator will look like if we take the first term for example take the first term it would be s plus 51.541 on the numerator and therefore this is now greater than 0 if you let s tend to 0 denominator if you let s tend to 0 even here it would still be greater than 0 therefore under such circumstances you would find that ?omega r by ?vf as s tends to 0 that means as the speed change approaches steady state is now greater than 0 that means what you are saying is if you apply a small voltage change to the field the speed will actually increase rather than decreasing that is what we understand from the root locations and indeed from simulation you see that speed change is greater than 0 speed is increasing for a unit step change this of course happens at low speeds that is what we see from this near about 100 200 rpm or even lesser this kind of behavior is obtained and that means the machine is going closer and closer to stall conditions so near about stall then this is the behavior that is up to is this explainable well if you look at the DC machine speed torque curves so if you plot the speed on the y axis with respect to the torque on the x axis torque and speed well we should really be plotting it in a slightly different manner so let me extend this line over a long distance so that is what we have and then if you plot the speed torque curve of a separately excited DC machine that is then going to look something like this so that would then be your no load speed and this would be the stall torque that means if you apply that much amount of torque the machine will not rotate at all this no load speed depends upon what is the field that is available because at no load armature current drawn is 0 because the machine is not going to generate any load torque that no load includes no friction also so no load means the machine is really running free under such circumstances the speed which the machine will develop will depend upon the field excitation because the field flux that is there multiplied by the speed at which it is rotating is then responsible for the induced EMF and because no flow of armature current is there the induced EMF must be equal to the armature voltage that is applied so if you increase the field voltage more field current is going to flow and therefore the no load speed of the machine will then come down a little whereas if you look at the stall torque of the machine if you increase the field excitation that is given to the machine more field current flows and for a given armature current the generated torque really increases so the stall torque of the machine would increase if the field excitation is more so with increased slightly increased field excitation the new stall torque would have moved a little to the right and the new no load speed would have moved little below and therefore the speed torque curve of the machine would then look something like this and then if the machine is actually operating at some low rpm and a very high torque you can see that if you increase the field voltage the speed is then likely to go up and that is exactly what is being shown by these plots now however this kind of operation is invariably not seen anywhere because you rarely approach stall torque of a particular machine the useful range of operation of a DC machine would probably stop at some point here which is about one third of the stall torque would be the rated torque probably in that range right so now all this exercise serves as an example to understand what is the relevance of the small signal model of a machine where it can really be applied how does the response of the small response predicted by the small signal model compare with that of the real large signal model of the machine the large signal model is something that is always applicable under whatever be the magnitude of the disturbances whether they are small whether they are large whatever they are but the small signal model gives us a simpler and linearized system model which can then be used for many other application many other in many other situations where you want to for example design control systems and so on this small signal model is of use. So that is then the utility of the small signal model so we have seen how the DC machine has been handled so let us take a step back and see what all we have done so far. So we started with the induction machine really the three phase induction machine in the natural reference frame of description and this three phase induction machine we then took it through a three phase to two phase transformation and we arrived at the so called two phase induction machine remember that the two phase induction machine really consists of two phases plus zero sequence description and only under situations where the excitation is balanced the two the zero sequence part can be neglected otherwise you will have to consider the zero sequence also. This two phase induction machine also had a rotating rotor and a stationary stator of course and then we went from this to the stationary reference frame in which the rotor was also transformed or the rotor was looked at from the stationary frame and the description here in the stationary reference frame we found that it could be used to describe a variety of machines one of the cases is the induction machine as seen from the stator everything appears to be in the stationary reference frame we saw that the flow in the rotor currents also looks as if it is of supply frequency the stator currents are of course at supply frequency so everything all the input variables excitations look like supply frequency effects and we found that the same description can be used even to model DC machines which is what we have been seeing all along so this description on the one hand can be used to model induction machines we saw that it can be used to model DC machines in its various forms DC machines are separately excited series excited shunt excited and then with an extra field as well all those can be represented and it can also model alternators that is synchronous machines in the case of synchronous machines one can you it would be then transforming the rotor on to the stator because on the stator you really have a field that is rotating the field is excited by DC in the synchronous machine but if you look at the stator and see what is happening to the field you really see a varying field a varying field happens because in this case the rotor is rotating physically and therefore you see a varying field in whichever manner it is what as the stationary observer sees is a field varying with respect to time and such a field can also be produced by excitations from the stator which are varying with respect to time and therefore one can also look at the behavior of the alternator using this description. So in a sense the stationary reference frame description is a generalized description of these three varieties of machine but however this description also has its own limitations because still we find that the excitations that are given for example if you want to describe the induction machine behavior you find that the stator voltage that is applied is of supply frequency stator currents are supply frequency so if one wants to do induction motor control for example you will have to deal with variables that are of supply frequency and that used to be a difficult thing because we are used to dealing with closed loop systems which have this well known PI controllers, PI controllers work the way they do only if the closed loop has DC signals that means under steady state the signals must settle down to some DC value if you have time varying signals under steady state in the PI controller is not expected to give 0 error. So in order to simplify the machine descriptions and indeed this kind of description started with the alternator analysis but has been adapted to the induction machine analysis for ease of control loop formulation in order to facilitate all that the stationary reference frame description is further transformed into another reference frame which in general could then be an arbitrary reference frame the arbitrariness of this arbitrary reference frame refers to the fact that the speed at which this reference frame rotates could be anything for example if you are able to formulate a set of equations where the speed of the reference frame is some let us say omega x then these equations that we have obtained so far which are the stationary frames equations should be able to get them by setting the speed equal to 0. We can then get speed equations in some other reference frame where the speed is equal to may be say synchronous speed or may be equal to the rotor speed or may be equal to some other speed so reference frame itself rotating at some speed which is not 0 has certain advantages with respect to handling the equations and therefore this is what we are going to see in the next few lectures how to transform the machine variables or the system description to the so-called arbitrary reference frame and why is that description useful for us so we shall see that in the next lecture and we shall stop here with this lecture.