 We now move on to the description of the controllers of an excitation system. So, far we have been discussing the issues related to the modeling of the power apparatus that is the exciter. But an exciter in some sense is incomplete without this kind of discussion of the excitation system is incomplete without the discussion of the controllers associated with the excitation system. In the previous class we did the physical models of an excitation system we did not actually derive in detail the model, but listed down the things which ought to be modeled. In today's class in today's lecture we discuss one of the major control systems associated with the excitation system that is the automatic voltage regulator. So, in the previous class we studied two kinds of excitation systems the static excitation system whose basic model is a static one with limits which are determined by the generator terminal voltage. The control signal here is effectively obtained from the excitation system controllers. The brushless excitation system is a bit more complicated in the modeling aspects. The main issue here of course is that the exciter alternator has to be modeled by a dynamic model not a static model. And the diode rectifier which is an uncontrolled rectifier has got a a limit that is its voltage cannot go negative the output voltage to the field cannot go negative. So, there is a limit associated with that the control rectifier associated with the brushless excitation system is modeled in a similar way to a static excitation system that is it is a static model with limits that is given on the left hand side. Of course, IEEE has come up with standard excitation models. In fact, the IEEE type AC1A model which describes the model of a brushless excitation system is given here. The functional block diagram of an excitation system including not only the power apparatus, but also the regulators, limiters and stabilizers which effectively form the control system. Remember a control system is designed by us it is more of a mathematical logic which is implemented using some physical elements. So, you can have analog or digital implementations of the mathematical logic which really determines the behavior of the controller. Now, the mathematical logic in fact is a mixture of you know differential or dynamical equations which have to be effectively numerically or implemented or implemented using analog blocks. Now, the most important or the core function of an excitation system controller is regulation of the generator terminal voltage. So, what it does is essentially keeps the terminal voltage of the synchronous generator the main synchronous generator almost constant. Now, so this is shown in a functional block diagram here you can pay attention to that there is a regulator. Regulator is something which tries to bring in this case the voltage near or equal to a set point value it is called the reference value, but regulation is not the only function which is actually done by any excitation system controller. You also find that there are additional blocks corresponding to what is known as a power system stabilizer and limiters and protective circuits. So, you can have the overall functional block diagram contains something more than a regulator. Now, we will try to understand each of these blocks one by one. First of all is the regulation block. The regulation block essentially takes a reference or set point voltage. You compare it with the measured voltage, measured terminal voltage of the synchronous generator and use this error signal to drive a controller which in fact gives the control system which is used by the control rectifiers of the excitation power apparatus. So, this is a controller. So, this is a controller or regulator you know. Now, this controller of course is a mathematical function. It is some mathematical function like some things we will describe in a few moments from now. The voltage regulator effectively compares the set point which is given by us and the actual terminal voltage. Now, there is one point about the set point itself the voltage reference of a synchronous generator terminals is typically going to be one per unit. We are going to really try to keep the generator terminal near about its rated value, but it is not sacrosanct that V ref should be equal to 1. You must have done in a course on any course on synchronous machines or electrical machines that by changing the excitation system voltage. Remember the change in a voltage reference here finally affects the control signal and eventually affects the voltage which is going to apply to the field of the main synchronous generator. So, by changing this reference voltage say by increasing it you are going to change effectively the field voltage and therefore, the reactive power output of the generator. So, if you recall a synchronous generator in its most simplest steady state model can be represented in this fashion. This is the synchronous machine terminals if the machine does not have saliency you can represented by a synchronous reactance which is nothing but equal to x d or x q because they are equal. So, this is a kind this is a steady state model of a synchronous machine. So, by changing E f d if I change the field voltage I effectively can change the reactive power output of the generator. Remember the real power output of a generator can be changed only by changing the mechanical power or the mechanical torque of the machine. So, that is one thing you should keep in mind. So, if you look at the steady state model of a non salient synchronous machine by changing E f d you can change the reactive power output. So, of course, we do need to maintain the terminal voltage near about 1 per unit, but in case you change if you change try to keep this voltage slightly greater than this. Remember that this although I have represented the rest of the system as a voltage source it actually is something like this. So, the correct way of representing the rest of the system could be this is at an angle delta the rest of the network. Now, the rest of the network again consists of transmission lines etcetera simplified model of that could be inductances etcetera. So, if I change the magnitude of the voltage the terminal voltage effectively the reactive power output of the machine will change. So, q of the machine can be changed by changing the reference value of the terminal voltage of a synchronous machine. So, of course, we will not want v ref to be very much different from 1 per unit after all the voltage in a synchronous machine has to be maintained at roughly what its rated value is if you have a voltage is much larger than that you will have problems of insulation and more importantly you may have over fluxing in the machine. So, one of the important issues here is what determines v ref well one of the thing is how much reactive power has the generator to be to supply. Of course, by reducing v ref one can make the generator absorb reactive power. So, v ref is determined by how much the reactive power output of the generator should be. So, a plant operator a power plant operator could tweak around this v ref near about the rated value of the synchronous machine. So, you of course, we can expect v ref to be say 1.02 or 1.03, but please do not expect it to be 1.5 or so that would really be much larger than the rated value and that would tax the insulation of the synchronous machine windings and moreover the flux in the machine the fluxes in the machine would become greater the flux densities in the machine would exceed the saturation value in this case. So, remember the flux densities in the core are dependent going to be dependent on the voltage of the machine divided by the frequency of the machine. So, in steady state b will be proportional to v by f. So, one of the reasons why you cannot exceed make v ref too large is that you will go into saturation if you go into saturation the machine becomes non-linear that is one of the it will increase if the flux level increase it will increase the heating in the machine as well. So, that is one of the major points which you should remember when setting v ref of course, v ref can be less you can have v ref as 0.98 per unit that is also feasible like you know you may under low load conditions want the generator to absorb reactive power in that case you would reduce v ref of the machine. So, this is basically what goes into deciding what v ref of the machine is. Now, once you have got v ref you have decided what v ref should be the power plant operator has some freedom in deciding that he compares it is compared with the terminal voltage of the machine. So, now the question which you would you know which needs to be asked is this controller whichever is there is it going to drive this error to 0 that is will it make v ref exactly equal to be the answer is well it depends on the mathematical function used in this controller. Now, another interesting variation which you see other of interesting variation which you see is instead of maintaining v is equal to v ref in some machines or in some situations v ref is you know you try to maintain v ref equal to v in most machines. So, this error is driven either to 0 or near 0 that is what your controller should do it depends on the mathematical function you are using, but the aim of course, would be to reduce the error almost to 0 that is v ref approximately equal to v. So, that is why it is called a regulator a voltage regulator, but in some situations you would not necessarily try to keep v ref v almost equal to v ref, but you can have v ref almost equal to v minus k into the magnitude of current going out of the machine. So, you can have a situation like this is well. So, you can instead of having a control system like what I have shown you before that is just having v what you can have is this is known as a load compensator. So, the thing is that of course, I should this usually is positive the point is that instead of keeping v equal to v ref for almost equal to v ref we will keep v plus k i is equal to v ref. So, in case i is large and depending on this value of k if it is non zero you will find that v ref is not maintained at v is not maintained at v ref, but slightly lower than v ref. Now, the question is why do this at this additional complication here the thing is that if you have got 2 machines which are connected to the same bus. So, this is 2 synchronous machines connected to the same bus and both these machines have an excitation system. So, both these machines have an excitation system both are trying to maintain the voltage of this bus a constant at v ref. Now, what will happen is that in case both these machines are trying to maintain the voltage here at v ref the question arises is how much which each individual machine contribute to keeping this voltage at v ref. Now, normally what we do is we make this change to the automatic voltage regulator. So, that if a machine contributes that if a machine contributes more current you know or more load more current in that case this gets modified this additional term essentially reduces v makes v ref slightly less than v rather makes v slightly less than v ref. The reason it is done is that you can then impose a sharing between these 2 generators if one generator supplies more current automatically what is done is that this becomes larger if this becomes larger v ref will not be maintained v will not be maintained at v ref but slightly lower value. So, what will eventually happen is that this synchronous generator will start if this i increases effectively what will happen it will try to reduce this is the exciter power apparatus this is the controller if this one of the generators takes on more load it tends to decrease it will tend to decrease the final exciter voltage. And therefore, that what happens is that it prevents one generator from taking all the load. In fact, by choosing a case appropriately for both machines for both machines if I choose my k appropriately I can actually impose a proper sharing agreement between these 2 generators. So, no one machine pushes in more current than what is necessary. So, what happens is that you make the excitation system or the exciter field voltage a function of the load current. So, if you have got 2 alternators or 3 alternators which are bused together they are connected to the same bus and your automatic voltage regulator is trying to maintain the voltage of that bus at v ref. So, each generator has got an independent voltage regulator which is trying to maintain one common bus voltage at v ref then to impose the sharing of reactive power one may have to put this additional component in the control system which makes the excitation field voltage dependent on i. So, if one generator starts pushing more current it automatically self corrects. So, that is the basic idea. So, the sharing of the generators can be imposed by choosing appropriate values of k for each generator is it ok. But of course, if the generators are not bused together in fact typically what happens is each generator is connected to its unit transformer it is called the generator transformer which steps up the voltage and the generators are bused together at the high voltage side. So, this is the low voltage say 15 k v 15 k v and this is 220 k v and or 440 or 400 kilo volts say. So, if you have got a situation like this each generator maintains its own terminal voltage constant it is not maintaining a common bus voltage at v ref. So, in this case it is not necessary to have a load compensator for sharing of because you are not maintaining a common bus voltage at v ref. So, but in case two generators are bused at a common point directly the terminals of the generator are connected to a common bus and then you step it up say or use it for other purposes. In that case you need to have a load compensator to ensure that while trying to meet a common objective the sharing of the load is done as per our design this k value. So, I leave it as an exercise for you to show that indeed if I choose k 1 and k 2 appropriately for say two identical generators I can or two different generators for that matter by choosing the k value appropriately I can impose a sharing of reactive power between the two generators. So, this is something I request you to try to prove. So, the steps are just to make things simpler assume two identical generators assume that they are supplying the same amount of real power they have got the same mechanical power input say that is the values are equal. Let us assume that the v ref for both the automatic voltage regulators independently I mean the two independent automatic voltage regulators is the same and these two machines are bused together at a common point then by choosing k 1 and k 2 appropriately the reactive power sharing can be imposed as desired by us. So, you can choose k 1 k 2 as per your reactive power sharing requirement in this situation. So, this particular function of v ref is being modified this the summing point being modified by applying an additional correction signal here you can say which is dependent on the loading of the machine alters the characteristic of the machine alters the voltage you are not no longer maintaining v is equal to v ref, but maintaining v plus k i approximately equal to v ref. So, this is called a load compensator it is usually not used in situations where a common objective is not being met by two excitation systems for example, in this case when the machines are bused together at the high voltage side and your voltage regulator is trying to maintain these voltages constant it is not necessary to apply this load compensator and in such situations it may be absent. So, please chew upon this point now if you look at the other functions which are shown in this block diagram there are limiters and a stabilizers, but we will come to that a bit later in fact I have already given you an hint where this limiter may come into into play. For example, if I have given v ref 1.05 and you also see that the frequency of rotation of the machine has come down it is my less than the normal say 50 hertz the frequency has come down in that case the flux level in the core may be higher than what is sustainable by the core in the sense this core will start heating up. So, and will also get saturated. So, in such a situation you may require to have some overriding functions which prevents the flux in the machine from becoming too large. So, what you have to do is if effectively under such situation if you see that the flux limit is being exceeded what you would do is override what you want to have rather override the regulator in the sense that do not aim to achieve v ref is equal to say 1.05, but reduce this v ref to a point at which the flux is not exceeding its rated value. So, this kind of predictive and limiting functions can be there basically what they do is override v ref or at least change v ref. So, what you have usually is the stabilizing and limiting functions. So, what you have normally is v ref and in case the generators are not bused together right at the terminals this is the structure of the summing point of the voltage regulator, but v ref itself may be overridden in the sense that some additional controls if you for example, if I put some additional input here say 0.05 per unit I add at this point. So, what will happen is that if I if the controller here the controller here drives this to 0 what you will have is v ref is equal to v minus 0.05 or in other words v ref minus v plus 0.05 is equal to 0. So, what in effect by doing this I have done is increased v ref. So, if you have any additional signals adding on to v ref the in effect compromise on this voltage regulation function or we will no longer be equal to v ref. So, for example, if there is an overfluxing situation you may adjust this becomes minus you may adjust this v ref by an additional signal which adds on to this summing point. So, I may add on minus 0.05. So, as to so I augment this summer what comes out of this summer. So, finally, v ref v will not be equal to v ref, but slightly less than v ref. So, this is what effectively a limiting functions does the other function which we have seen here is. So, actually I will read all this. So, you have got v ref this is the structure of a voltage regulator summing point without load compensation. So, we just have v here this is what is normally required you may have overriding additions or subtractions from limiters and you may even seek to modulate you may wish to modulate you may wish to modulate this voltage reference by some external stabilizers Now, we shall discuss these stabilizers later in the course, but remember you can modulate these things. Now, one of the important things about limiters and stabilizers is which you should be very clear about is that limiters normally gives 0 output if the machine has not hit any limit. Similarly, a stabilizer gives 0 output when the machine is in steady state. So, the overall regulation function of this voltage regulator eventually boils down to this and this the summing point in this regulation function boils down to taking the difference between v ref and v in steady state if no limit is violated. So, the voltage regulator in some sense gets in is completely enabled when no limit violation is taking place. However, if any limit violation is taking place some input will manifest at this point due to the limiting function and we will no longer be equal to v ref it will get overridden in some sense by this additional superimposed signal. During transients the stabilizing function augments this v ref say you can be during transients you could be sending in a modulation signal how to derive this modulation signal is something we will discuss later. We may modulate this point why we need to do it well by using this modulation function we expect that we will try to improve the dynamic performance of the synchronous machine. So, this is what essentially is the things you will see at any summing point one of the questions you may ask is well you have got this controller or regulator which takes this error does some mathematical function on it and derives the control signal which goes to the power apparatus the excitation power apparatus. The question is can we not shift these limiting functions to this point. So, can we not instead of changing the summing point here can we have these overriding functions acting here the answer is yes you can. So, in some implementations of regulators this limiting the output of limiting functions and stabilizing functions are in fact added at the control signal level the control signal which is sent to the excitation power apparatus that is to the control rectifier of the power apparatus excitation power apparatus. So, you can have this kind of alternative structure where you are having the summing junction here for the limiting and the stabilizing function. Now, before now the main important point here is now what is the structure or mathematical function which are typically used in the controller block which is shown here and also what are the mathematical functions which are used to implement these limiters and stabilizers. Now, to understand this we will actually have to understand what are known as transfer function blocks. Well you do not have to understand this in terms of what are known as transfer function blocks, but most of the excitation system control system representations which are available in the literature come in the form of block diagrams. So, you should be able to write down the dynamical equations or the differential equations or the mathematical functions which really correspond to these block diagrams. So, mainly there are what are known as transfer function block diagram. In fact, I have already shown you a transfer function block diagram we call this this IEEE model of a brushless excitation system is in fact a transfer function block diagram you see this S function here. Well this S function is in fact the Laplace variable we have used of course, we have without actually explicitly defining these we have already used the summation blocks this is a product block. So, these are what are known as typical block diagrammatic you know components which are used in trying to model an excitation system of course, this transfer this block diagram here which uses Laplace variable S. In fact, 1 upon S is what is known as is essentially an integrator in integration function. So, this 1 upon S here represents an integration function. Now, this of course, was a model of the power apparatus. So, what we did was roughly model the excitation alternator by a differential equation and then converted it into a block diagrammatic representation of this kind. We now move on to the block diagrammatic representation not of the power apparatus, but of the control system. Now, control system are something which we will be implementing. Now, what should be the structure of any control system? Now, it is usually made up of gain blocks, integration blocks, summation blocks and so on and compactly written down in terms of input output relationships or what are known as transfer function relationships. So, let us just take a minor diversion we need to build up this kind of expertise and try to understand some basic transfer functions which you will see. For example, consider the block 1 upon 1 plus S t this is a transfer function block. In fact, the simplest block 1 can have is of course, a gain block. A gain block is a simply multiplies u by k to get y. A transfer function block on the other hand means something more. In fact, in a transfer function block inherently or you know implicitly there is some numerical or actual implementation integration being done. So, this S in fact, is a Laplace variable 1 over S represents an integration function and S of course, is a derivative function. So, what is 1 upon 1 plus S t derived from? In fact, if you look at the dynamical equations if you look at these dynamical equations I have written them in. In fact, this dynamical represent dynamical model using differential equations is actually how this has to be interpreted. Why is well why does this mean this? It is not very difficult to see. Suppose, I do replace d x by d t here by its appropriate Laplace transform value. What you will get is if I apply Laplace transform on both sides of the equation you will have S into x S is equal to minus 1 upon t x of S plus 1 upon t u of S and y of S is equal to u of S. So, what you will have here is you get this on to this side you will have x of S into 1 plus S t is equal to u of S and as a result of it of these equations you will have y of S is equal to 1 upon 1 plus S t into u of S. So, that is why this differential equation after doing Laplace transform transformation of this differential equation you get effectively this input-output relationship. In fact, input-output relationships have been discussed in the past. In fact, our when we talked about the modeling of a synchronous machine when we actually correlated what we get from measurements the frequency response obtained from measurements we did use a transfer function representation of the synchronous machine. Therein also the basic differential model differential equation model of the synchronous machine was reduced to an input-output relationship in by using this Laplace transform variable S. So, remember that 1 upon 1 plus S t represents this transfer function of course, one should be a bit careful and precise well there is no 1 to 1 mapping between this and this. In the sense that a different model of you can have a different state space or different differential equation model to get this input-output relationship. For example, d x by d t is equal to minus 1 upon t x plus u and then y is equal to 1 over t of u will also give you the same transfer function can you verify that you will get S of x S is equal to minus 1 upon t x of S plus u of S y of S is equal to 1 over t u of S. So, you will have so what you will get eventually is 1 plus S t of x of S is equal to t times u of S y of S is equal to 1 upon t u of S. So, from these equations what we can get effectively is y of S is equal to 1 over 1 plus S t u of S is that y of S is equal to 1 plus S t upon t into u of S. So, we will have equal to I am sorry. So, u of S by t is equal to y of S is equal to 1 of we will just redo this. So, what you will have here is a y of S there was one small error which I did in our previous this should have been x this should have been x. So, please note this error. So, this should be x of S in that case of course, you will get this particular relationship. And just to recall what I was doing just now I was just trying to show that this is an alternative representation of the same transfer function. So, what you will have is it should be x of S it should be x of S and as a result of this we will get. So, what I wish to say here is this is one possible representation of the transfer function, but this is another representation of transfer function. So, there is no unique way of representing transfer functions by state space. So, this is one example of that, but usually in this particular course I will just take this particular representation is very convenient of 1 upon 1 plus S t. In fact, I can split up this if you look at this particular differential equation this is one possible representation of this transfer function. So, if you look at this differential equation I can rewrite it as kind of a integrator this is u and this is y. This is x, x is equal to y, this is x dot if you integrate x dot you get x, x dot is equal to 1 upon t u minus x. So, this is what exactly this says the x dot that is d x by d t is equal to u minus x divided by t. So, this is what effectively is a block diagrammatic expansion you can say of this transfer function. So, if you see this remember this now see this what we have drawn. So, right now is actually quite interesting the transfer function itself that is what I mean 1 upon 1 plus S t is an interesting transfer function if you look at some simple things it does. If you have got u and y here if I give a step change in u what will y be in fact you can prove this by actually solving the state space equation and prove this that if u is a step then y is going to be something like this, this is y and this is u. So, the step change has some interesting features you see that at the instant of the step this transfer function has an output 0 it does not change. Suppose the initial conditions associated with this state x is 0 then if I give a step change in the transfer function we will find that x remains 0 and as a result y also remains 0. So, there is no instantaneous change in y given an a step change in u. So, x y just remains where it is but as time passes you will find that y tends to become equal to u. So, the steady state gain of this transfer function is 1 that is y becomes equal to u. Now, of course the same thing can be got the same you know inference can be obtained directly by setting S is equal to 0 in this transfer function. So, a shortcut method of finding the steady state gain of any transfer function is to put S is equal to 0 in the transfer function. So, if you give a step input the steady state value or the steady state gain for a step input can be obtained by simply putting S is equal to 0. So, you can say that this transient gain or the high frequency gain of this transfer function is actually 0 it does not instantaneously respond but in steady state it takes on the value. Now, what is the nature of this curve? In fact, you will find it y is equal to u 1 minus e raise to t by capital T. So, if this step occurs at time t is equal to 0. So, this is a step change of magnitude 1 occurring at time t is equal to 0 the responses like this the time constant t appears here in the response. So, it is a good idea whenever you come across any transfer function block you should be able to correlate it with its time response under various test inputs. If you look at another interesting and important way of looking at transfer functions or what interpreting how they behave is to look at the transient behavior. So, the transient behavior or the what you call the frequency response we have actually seen the transient behavior for a step change the frequency response of the transfer function can also be obtained that is you vary the frequency and see the gain that is u y upon u in steady state. So, what do I mean by a frequency response of any transfer function? You take this block feed it by inputs of varying frequency in steady state if it is a transfer function normally a transfer function in steady state would at the output if it is actually a stable if it is a stable system that is it is associated with the state space whose Eigen values on the left hand side of the imaginary axis then what you will see is that the output you will get also a sine wave, but it its magnitude would have changed and it would have also got some phase shift it could have got some you know it can have some phase shift. So, you take any transfer function this is how you would measure the steady state frequency response. So, whenever people say a frequency response it is implicit that they are talking of the response under sinusoidal steady state conditions for input sinusoids inputs u which are at different frequency. So, what you do it you do this experiment which I have shown you on the sheet here at a particular frequency record the gain that is how much is the amplitude with respect to the input amplitude. So, that is the gain and also measure the phase shift. So, the phase shift of this there will be some phase shift between this signal and that. So, you measure the phase shift. So, that is what will give you what is known as the frequency response. Remember frequency response implicitly is the steady state frequency response of this system. So, you will get a gain which is a function of frequency that is one important thing. So, the gain is a function of frequency the phase also is a function of frequency it turns out for 1 upon 1 plus s t the steady state gain is 1 steady state frequency what you call the 0 frequency when I say steady state actually it means 0 frequency the 0 frequency gain is 1. Whereas, the high frequency gain is low how do I know this phase shift and the gain well you can do it very simply using a transfer function. If I give you a transfer function if I want to know what is its gain and phase shift all I need to do is put s is equal to g a omega corresponding to that frequency which you are interested in getting the frequency response and you can get the answer. So, this is something of course, you have done before, but we are just revising that. So, this is how it is represented. So, this is called a frequency response or rather the transfer function evaluated at this frequency. So, what is the magnitude gain well that can be obtained simply by obtaining the magnitude of this complex number. So, that is what you will give you the gain and the phase shift is again obtained by looking at the argument of this complex number which results. So, what one thing you will notice is that this will always give a negative phase shift this transfer function has a negative phase shift. In fact, at omega is equal to 0 the phase shift is 0 phase shift is 0 degrees at omega tending to a very large value a large value in this context means that omega t is much greater than 1. In such a case you will find that omega t omega t or t omega much greater than 1 you will find that the phase shift is minus 90 degrees of this transfer function. Similarly, at omega is equal to 0 the gain is 1, but the high frequency gain is 0 which correlates well with the step response. Just at the instant at which the step is given you can say that the input signal in some sense is kind of changing very fast or it has got a high rate of change or it has got a very high frequency component. The transfer function does not respond very well or it has a very low gain to high frequencies. We will move on to one more transfer function that is it is a kind of a important transfer function just as the previous one was this is s t upon 1 plus s t this is an interesting transfer function again. You can verify that a steady state space representation of this is y is equal to u minus x. So, this is a change. So, this is a steady state representation of this system or rather state space representation of this system. So, if you look at this can you guess what is going to be the response for a step change in input. Well, you can actually work it out, but it is not difficult to see that first thing is that the high frequency gain of this is going to be 1. If you look at the frequency response with omega large the gain of this is going to be 1. For low frequencies the gain is going to be equal to 0. So, in fact we will not work it out, but a step change for a step change the response of this y is going to be like this. The final value is going to be 0 the initial value is going to be equal to whatever this value. Suppose this is 1 then this is also 1. So, this is known also as a wash out transfer function. It is called a wash out transfer function 3. The characteristic is that it is got a characteristic of a in a sense a high pass filter. It prevents slow changes from going through, but it allows fast changes to go through. So, its steady state gain is 0 its high frequency gain is 1 and this is the state space representation. Now, from state space representation one can easily get a block diagrammatic representation. So, it is quite simple. So, if you look at this it is simply what it was before the only thing is this is an integrator. You can represent a integrator is in Laplace domain is represented by 1 upon s. So, this is u this is y and oh this is not y. y in this case is this is subtracted u minus x this is x, but there is an additional u y is equal to u minus x. So, this is y this is u this is x this is x dot and this is u minus x. So, this is known as a wash out block remember this extra thing here completely changes what you get. In fact, you can easily see that s t upon 1 plus s t is equal to 1 minus 1 upon 1 plus s t. So, it is not surprising you have got a block diagram of this kind in which in fact, this is nothing, but this whole thing is nothing, but 1 upon 1 plus s t. So, you got 1 minus 1 upon 1 plus s t. So, this is as far as one of another of the transfer functions goes. In the next class we will look at a few more transfer functions talk about what are known as what is a p i a proportional integral regulator and a proportional regulator. We look at as I mentioned some time back the lead lag blocks these are also special transfer functions and we will also look at the effect of limiters there are two kinds of limiters. So, far I have implicitly assumed you know what a limiter is, but you can actually have some variations of limiters themselves and they eventually affect how the system response is going to be. Typically a control system will be made out of either plane gain blocks in addition to summers multipliers and so on and transfer functions. When I say transfer function blocks the transfer function blocks themselves represent a linear system, linear state space. Of course, if you put limiters in the transfer function in that case or limiters in transfer function blocks the blocks which I have shown you if I insert a limiter it no longer is a linear system, but if you just remove that little bit complexity transfer functions implicitly are obtained for linear systems or linearized systems non-linear systems are linearized around an equilibrium point. So, using these blocks or these mathematical functions or these you can say mathematical dynamical functions we can actually try to design a control system which meets our objectives. So, I have not I have just discussed some of the control systems which are used of course, rather the transfer functions block diagrams which are used we can use these block diagrams selectively to get the dynamical as well as steady state response that we desire. So, that is what we will do in the next class.