 a transmission line is a power system component which is a distributed power parameter component and spans a you know the component essentially is you know spread over a large geographical distance and as a result of which the first treatment of a transmission line typically involves the formulation of equation in terms of partial differential equations. So, we did this last time a transmission line is effectively modeled by these equations they are also called a telegrapher equations and these are a partial set of partial differential equations. Now, if resistance and the conductance are neglected I mean they are assumed to be 0 in such a case they are normally quite small in that case the response of the transmission line is given by the very well known travelling wave equation. This is of course, with the assumption that r and g are equal to 0 and small c here is the velocity of light and remember the l and c here are the inductance and capacitance per unit length capital C that is the upper case c and upper case l are in fact, the inductance and capacitance per unit length. So, this is one of the important equations travelling wave equations. Now, in this particular lecture we will continue with our discussion of transmission lines we were kind of poised at a very interesting discussion in the previous lecture we continue with the discussion and hopefully by the end of this lecture we shall come to a fairly usable model of a transmission line not only for high frequency studies, but also for lower frequency study of lower frequency phenomena. Thereafter we will go on to the study of prime movers. So, in this lecture we will begin with. So, we will do the study of we will continue our discussion of transmission lines and sometime at the end of the course of this lecture we will also try to I will try to introduce you to prime mover systems. Now, one of the interesting discussion which we kind of left half way in the previous lecture was the model of a transmission line the dynamical model of a transmission line. Of course, I have already given you a dynamical model, but it is in terms of partial differential equations and the solution surprisingly for a lossless case is very neat the travelling wave equation looks very neat. Now, if you want to actually you know understand how a transmission line behaves. So, you will have to use a partial differential equation solution. Now, we have in our under graduate years come across this model of a transmission line this model is a pi equivalent of a transmission line. This is a pi equivalent of a transmission line, but it is very important to remember that this equivalent using in fact lumped blocks of impedance and susceptance or sorry I should say impedance and admittance are actually valid only for sinusoidal steady state conditions. So, in fact if you look at what z bar and y bar mean in these equations they are in fact impedance and admittance and what you have is gamma here gamma bar in fact is determined by this. So, remember that frequency comes in this and the distance also comes in this, but this is essentially a sinusoidal steady state model as I mentioned in the previous class we may be tempted to you know consider this model as you know correct for dynamical for the study of dynamical phenomena as well. The answer is strictly speaking no remember that this is a lumped parameter pi equivalent which tells you the correct terminal relationships in case of sinusoidal steady state, but the correct solution for a lossless line in fact is given by these equations and in of course, in the case where r and g are not equal to 0 you would have to find ways of solving the partial differential equations you will not get a neat solution like this in case r and g are not equal to 0. So, this is something which has to be made clear interestingly the partial differential equation the lossless case the solution with the lossless case in fact tells us something about the transmission line which is very neat and nice. In fact, a transmission line using these this solution it can be shown that if v k and v m are the instantaneous voltages at both ends of the line and I m and I k are in fact, the currents at both the ends of the line then you can represent I m as I m is dependent on v m by this equation where z c is in fact, the characteristic impedance we have discussed what characteristic impedance is z c is root of l by c where l and c are in fact, inductance and capacitance per unit length small c is the velocity of electromagnetic propagation which for air is practically equal to the velocity of light. So, that is 1 upon square root of l c now, what I was getting at is that the current I m can be written down as a function of the current I m at a time period time instant t can be written down as a component which is dependent on the voltage at that end of the line and a current source I m which is in fact, dependent on the currents and voltages which exist at the other end of the line t by d by c seconds ago t minus d by c actually tells you that I m is equal to I k sometime before what I k was sometime before and similarly, it is also dependent on what v k was sometime before. So, it depends on what was existing at the other end of the line sometime back similarly, I k can be represented in this way where I k of t is equal to dependent on v k of t sorry v k of t and the current at the other end of the line this I capital I k is in fact, a current source which contains what I have known as the history terms. So, if you look at in fact, the first equation what it tells is the equivalent at one end of the line. So, if you look at I m at one end of the line you can see that this equation effectively tells you I m can be obtained by a circuit of this kind. So, all I have done here of course, is represented this equation I just represented this equation I just represented this equation by this circuit. So, if you manage to solve this circuit you know of course, you need to you know define what else is connected to the system, but the point is that if I look at the equations they are effectively you know representative of this circuit it is a resistive circuit this is a history term. Similarly, you can obtain I k from this circuit. So, whatever if you solve this circuit you will get I k this I m and I k remember capital I m and capital I k are in fact, history terms which depend on the currents at the other end of the line. So, it is a very important thing to note this that these are in fact, history terms which tell you about the current in the other end of the line and sometime ago that sometime ago is of course, d by c d is the length of the line divided by c seconds before. So, in fact it is an interesting point here that the solution of a lossless line in fact comes out to be simply algebraically related to the currents at the other end, but of course, with a time delay it is a pure transport delay. So, in fact this in fact is useful if you want to for example, simulate a transmission line. So, if I tell you you know the conditions which exist at the boundary of a transmission line the boundaries of a transmission line you should be able to tell how the system behaves for other instance of time. So, for example, if you look at this in fact equation if you call d by c h. So, if I call d by c h where h is you know I can evaluate the values of i m t and i k t at discrete instance of time by using these algebraic equations quite easily. For example, I could try to understand the behavior of a transmission line for example, which is connected to a voltage source. So, this is our transmission line I can use this suppose I want to understand how the voltages and currents behave in case I switch on a voltage source at t is equal to 0 and I want to know how this voltage at this end varies and how this voltage here varies under open circuit conditions that is the current at this end is 0 and the voltage here is defined at t is equal to 0 onwards. With this information I could evaluate i m and i k at discrete instance of time in fact, those discrete instance of time are the time required for the wave to travel from one end to the other. So, this is a very interesting kind of equation that we get which is of course, not true in case r and g are non-zero. So, let me just again repeat what I what I said in case I present to you a circuit which needs to be simulated and transmission line behavior which needs to be simulated the dynamic behavior. Then I can use these equations these equations are in fact, algebraic equations with a history term. The history term in fact, involves voltages at the other end of the line voltages and currents at the other end of the line. So, if I tell you the boundary conditions that is the voltage or current conditions at one end of the line and the other end of the line you should be for all other instance of time be able to tell how the behavior is. In fact, it changes at every h interval of time it changes after h a period of h where h is d by c the amount of time it takes for the information to travel from one end to the other. So, this is one interesting outcome of the traveling wave equation that you can actually simulate a lossless transmission line quite easily I mean it looks very complicated, but if it is lossless you can simulate it quite easily. Now, one thing I should mention here is that I am evaluating the currents and voltages at both ends of the line at discrete in I can easily obtain it at discrete instance of time because of the nature of the algebraic equations which I am getting. So, this is somewhat different from the discretization the you know kind of thing we were doing for other circuits. For example, if you have got a lumped inductance if I discretize it using trapezoid rule the if I discretize the equation l d i by d t is equal to the voltage across a lumped inductor then I will get I at k plus 1 into h this is the current at the discrete time instant. So, if I have discretized using trapezoid rule this is k plus 1 of course should be inside a back bracket into h divided by h is equal to v k plus 1 into h plus. So, this is of course into l plus v of k into h into half. So, in fact the discretized equations of for a lumped inductor using trapezoid rule is in fact this out here also you will notice that the current at the k plus 1 instant is dependent on the current at an earlier instant. So, this is an interesting thing. So, if you discretize a continuous time lumped parameter equation again you get basically the answer you get a kind of when you discretize it you get essentially an algebraic equation which relates new value of in this case the current the new value of the current in terms of the history the current and voltage history of this circuit. But remember that this you know dependence on a history term is local in the sense the current at the k plus 1 instant depends on the local current and voltage at the k th instant. But in a case of a transmission line remember that our current you know at one end of the line at the k th instant depends on the current and voltages at the other end of the line d by c seconds before. So, this is something which is fixed by the nature that is the length of the circuit as well as the physical parameters that see which is the velocity of light for an overhead line. Whereas, when you discretize a lumped parameter a lumped parameter device like an inductor lumped inductor you again get an algebraic relationship. But remember that the dependence on the history term or rather the history term is still local in case of a transmission line the current at one end of the line is dependent on the current at the other end of the line the at a previous time instant and the time instant is really dependent on the distance. So, whereas, here it depends on the time step you have chosen for discretizing this continuous time differential equation. So, there is slight difference between what you are getting there and what you are getting here, but it turns out that since this kind of algebraic relationship is obtained for a transmission line directly you can interface the transmission line equations and the equations obtained by discretization of a continuous time lumped parameters element like inductance and do a simulation. For example, it is it is not very difficult to do a simulation now because of these facts of a system like this you got a transmission line and at the other end you have got a lumped parameter inductor. So, when you discretize this you will get basically algebraic equations with history terms and of course, the discretization interval is h out here you again get voltage and current at both ends dependent on one another, but of course, there is a delay element they depend on the history terms with you know the history being relevant to what was the situation d by c seconds before. So, if actually if I choose h to be a multiple of d by c or h equal to d by c it should be easily one should be easily able to interface the discrete time equations which you get here with the algebraic equations which are given by these which are given here. So, this is an interesting thing in fact, this is the way you know transmission lines are represented in programs called electromagnetic transient programs. So, this is how things are done you you have got a transmission line you assume a suppose you assume it is lossless then the the currents and voltages at discrete instance of time can be obtained by these algebraic relationships and continuous time systems lumped parameter systems are connected to the transmission lines. In fact, can be brought to a similar form because of discretization by some numerical integration method and in fact, if the h is either a multiple of d by c or equal to d by c this this becomes even more easier to interface all these equations. So, just chew upon this and you can actually do a simulation of a transmission line connected to a lumped capacitor or lumped inductor and so on it is it is not very difficult to do that. So, only thing is of course, that there is a important assumption here that the line is lossless. So, can you think of some means of you know kind of taking into consideration you know the losses in the line that is a non zero r or g the answer is well the equation which we get the travelling wave equation or this algebraic relationship which we got is true only for r is equal to g is equal to 0 we just cannot use this and if we cannot use this we cannot use the simple algebraic relationship between the currents and voltages at either end of the line which we have just discussed sometime back. So, if r and g are to be brought into the picture the best idea would be to consider a lossless line as consisting of a lossy line as a consisting of a lossless line plus the effect of the resistance is considered as a separate series resistance which is connected separately outside. So, this is a lumped resistance plus a lossless line. So, this is what seems to be the reasonable thing to do I mean of course, the validity of this approximation needs to be checked out I mean this is of course, something which would occur to as a nice trick to you know use these algebraic relationships even when the system is a lossy system. So, you consider a lossy transmission line as a lossless transmission and plus the effect of the resistance is considered separately as a lumped element. So, this is how you would try to simulate or understand the effect dynamical behavior of a transmission line. There is one more small issue which you need to tackle this is something I will not spend a lot of time, but you can just think of in case I want to do a simulation of a transmission line like is shown here and the discretization of the lumped element like an inductor which is connected say at one end of the transmission line is done at a time step of h and d by c that is the length of the line divided by the speed or the velocity of propagation is not an integral multiple of h. In that case it turns out to be somewhat difficult to interface the algebraic relationship which you get of the transmission line with the algebraic relationship which you get by discretizing the lumped parameter continuous time differential equation with a discretization time interval of h. So, what one would normally do under that such a circumstances try to get h to be rather d by c to be an integral multiple of h as close to it as possible I mean it may not be exactly equal to d by c may not be exactly equal to k by h, but you can make it approximately. So, alternatively you would need in case you do not you cannot really achieve this what can you do. So, this is something you can think over this left you to think over what would be a reasonable or you know a satisfying way let me say of handling a situation where d by c is not an integral multiple of h and as a result of which it becomes difficult to interface the algebraic relationships obtained from the travelling wave equation of a transmission line with the discrete time equations arising due to the differential use of some numerical integration technique for a lumped element. So, this is essentially how you would you know this is of course, a particular issue which you may need to tackle, but there have been reasonable ways to really solve this problem and I leave this to you to think about. The next issue is something which I actually left you last in the last lecture was you have got this lumped pi equivalent of a transmission line from sinusoidal steady state analysis. The question is would that lumped parameter you know let me call it a model of a transmission line a lumped parameter model of a transmission line obtained from sinusoidal steady state analysis suffice to really simulate or mimic the behavior of a real transmission line that is the question which you would like to answer next. So, as I you know discuss in the previous class I mean I left you with a problem if you recall what we did last time the problem was this is an example from the last lecture from the book by Sauer and Pi where you have got a transmission line which is 100 miles it has got these parameters L is 1.5 milli Henry per mile C is 0.02 microfarad per mile and you switch on a voltage source which has a source resistance of 10 ohm on to this transmission line which is open circuited at the receiving end. Now, the question which I post to you was that if I try to simulate or understand the behavior of this transmission line by using the algebraic equations given here the algebraic equations given here and see how this system behaves in that case how would the response differ in case I took instead a lumped parameter model of a transmission line where the induct this inductance I assume to be L into H remember L is inductance per unit length L into D sorry and this is C into D by 2 and C into D by 2 C is of course, the capacitance per unit length. So, the question is that if I use this lumped parameter model of a transmission line and try to find out how this system behaves in fact you can analytically get the response or even numerically integrate and obtain the response. How does this system behave as compared to a proper simulation of a lossless line using the travelling wave equation. So, you use the algebraic equations with history terms which are actually obtained from the travelling wave solution of a transmission line and see how both of them compare. So, if you do that you get a somewhat surprising result. So, what is that result. So, if you look at this is actually the response of the transmission line the bold what you see as the bold line here which looks a bit like a square wave initially is actually the response obtained from the travelling wave model that is using the travelling wave kind of response of a transmission line that is using the model the algebraic model of a transmission line with history terms as I mentioned sometime back. Remember that for this particular circuit the receiving end reacts or you will see something happening only after the wave reaches the other end. So, this is what you will get using the you know the kind of algebraic equations with history terms which are valid for a lossless transmission line this is also called Bergeron method. Instead if you use the pi model using lumped parameter that is lumped inductances and capacitances you get what is seen here as the dashed line. One of the important differences between the travelling wave response or the detail response of a using the travelling wave model of a transmission line and the lumped parameter response is that the lumped parameter response starts immediately after the disturbance is initiated at the sending end. So, you see the effects right away whereas there is a clear time delay when you consider the travelling wave model the travelling wave model of course, is more accurate. So, what you really although we are talking of a comparison remember that the travelling wave model is actually the more accurate one. An interesting thing of course, which you see here is that although there is of course, a difference between the travelling wave model and the lumped parameter simulation of the system there is some similarity to in fact, if you look at the response it seems to be some kind of a filtered response you know the lumped parameter response is a kind of a more smoother and you can call it a kind of a low frequency part of the response. So, if you look at it is similar it is not the same of course, but it is similar and in steady state if you of course, if you allow this particular system to settle down remember what is the system we have considered please have a look at it again if you look at this again it is a sinusoidal source switched on to an open circuited line. So, if you wait for a long enough time one would expect that the system would reach a sinusoidal steady state. The interesting thing is that if one looks at you know the receiving end voltages after the system is allowed to settle down you will find that in fact, both the lumped parameter model as well as the travelling wave model also called the Bergeron method settle down to the same value. Now why is this so in fact, this is not surprising at all remember that the pi equivalent is valid it is valid in fact, for the sinusoidal steady state. So, the travelling wave model is essentially settling down to the same steady state. So, the as far as the sinusoidal steady state is considered the pi model of a transmission line is in fact, correct in fact, there is nothing wrong with it. Only thing is that the initial part of this transient which we saw an expanded portion in the previous slide there is some difference there is some difference in the initial part of the response using a more detailed partial differential equation model as compared to the lumped parameter model. So, this is what essentially you will be losing out in case you know you choose a lumped parameter kind of model. Now it is obvious that if you are not interested in the high frequency transients which are seen right at the beginning of this plot then it appears that you can just as well use the lumped parameter model. So, if you are not interested in high frequency transients in fact, it is a good enough approximation to even to dynamical analysis with the lumped parameter model. But remember the origin of the lumped parameter model original was actually from the two port equivalent under sinusoidal steady state conditions, but you can actually use it if you want to obtain the low frequency behavior of a transmission line. So, it is it is to use a pi equivalent model. So, this is what we get from this particular study what you cannot of course, get from this model is the lumped parameter model is the behavior just after the disturbance then there is a substantial difference. So, this is something which we saw sometime back that right at the beginning there is a difference between the responses although the responses are similar they are still different. Now going on further in fact, our discussion so far has been restricted to a kind of a single phase line a distributed parameter model of a single phase of line and we actually got travelling equation wave equation for it. The question which we can ask ourselves is if you have got more than one line for example, suppose you have got a bipolar line that is the electromagnetic environment consists not only of two lines, but something like this two transmission lines and the ground. So, the ground also could be a part of the electromagnetic environment they could be currents for example, in the ground in such a case our equations the equations of a transmission line are in fact, suppose the current to this is i 1 and this is i 2 then you can show that the at any just like in the previous case you can write down the equations like these are partial differential equations. Remember that what we are doing different from the previous cases that we have introduced another conducting kind of system into the analysis this is a ground. So, the electromagnetic environment is slightly different from the previous case where you just had two lines in this universe and nothing else. Now we have got v 1 say at a distance x as before this is v 1 this is the local the voltage of this point with respect to ground at this point. So, one should be very clear about what we mean by this similarly, the voltage of the other wire at any point on the other wire with respect to ground locally is v 2. So, you have got v 1 and v 2 of course I have shown this slightly tilted. So, you can look at it this way this is ground. So, this is v 2 this is v 1 and this is i 1 and this is i 2. And like before you have got the sending end and the receiving end and so on. So, the equations for v 1 v 2 i 1 i 2 v 1 and v 2 are the voltages with respect to the local ground here are in fact given by I am sorry this should be. So, we will have a matrix here is equal to just a moment. So, what is different in this case as compared to the next remember that v 1 is the voltage of the first at a point on the first wire with respect to the local ground v 2 is the voltage with respect to ground the local ground for the second wire i 1 is the current in the first wire i 2 is the current in the second wire. And you could have in fact current in the ground in fact the ground current in case i 1 if i 1 is not equal to minus i 2 in such a case you can have currents to the ground. So, in some sense what I am doing here is including the effect of the ground or the electromagnetic environment around these two wires as well. So, this is basically the equation which you get and a similar equation exists for the current as well. So, you got L s L m these are inductance you have got an inductance matrix is equal to this for a lossless case of course. Now, the question is we had considered the equations of a transmission line earlier this is of course with losses, but there we had just the voltage across the line and a current there is only the current whatever current flowed here was equal to the current which. So, this whatever current suppose it was 1 ampere flowing this direction 1 ampere would be flowing in this direction. So, this was the situation before now you have got the ground with possible current flow through the ground and you have voltages with respect to the ground v 1 and v 2. So, this is what is different. So, we have got more variables and you also have got a bit of coupling these variables are coupled to each other. So, how does one solve in such a case these are actually a simple case of how you can use mathematical tools to solve this kind of situation. Now, in fact with r and g equal to 0 we know the solution of this equation is actually a travelling wave solution. Can we directly apply this to this situation the answer is no we cannot directly applied because now you have got a coupled set of you have got two sets of equations and each of them is in some sense a vector equation. So, how do you solve this problem in such a case if you have you define the difference voltages v 1 minus v 2 and the common voltage as v 1 plus v 2 say divided by 2. Similarly, i difference and v common is equal to v i 1 plus i 2 divided by 2. In fact, the common current is in fact proportional to the ground current. So, if I transform v 1 and v 2 to v def v com i def i com I can reformulate these equations this is something I leave for you to do you can reformulate the equations in i def i com v def and v com. And one very interesting thing occurs is this something you can it is very very easy to prove. So, you have got for example, d i def by dot t d i com by is equal to you can reformulate the equations in terms of these new variables and the surprising thing is that this is a diagonal matrix in case you use this transformation of variables. In fact, the terms are in fact, L s minus L m and here you get L s plus L m 0 0. Similarly, you will have C s minus C m 0 0 C s minus plus C m dou v def. So, what you have essentially is this kind of decoupling takes place between the i def and i com variables and as a result of which you can take the pair i def and v def and v com and i com and essentially since v def the equations between i def and v def are completely decoupled from the equations of v com and i com and the equations in fact, are very similar to these ones of course, with r and g equal to 0. You can actually get a travelling wave solution for v def i def v com and i com separately because there is complete decoupling between these and get the solution for this system as well. So, it is a very interesting thing which I have tried to tell you is that you can use transformations this is a basically a linear transformation of variables in order to get decoupling. Once you get a decoupling you can get the travelling wave solution for the diff variables and the com variables separately and if you want to get v 1 and v 2 eventually in the end you just super impose the solution you just add up the solutions you know by using the reverse transformation I should say add up with the reverse transformation to get the original variables. So, it turns out that even in cases where you have got a system you know you can imagine that if you have got a three phase system. So, if you have got a three phase system and with ground as well this is a typical electromagnetic environment which you will see in fact you can have a client in close proximity to each other etcetera. It turns out that although there are coupled partial differential equations in the original variables like v a g that is the voltage of a phase with respect to the local ground. If you formulate a equations in that way using the phase variables with respect to ground the voltages of the phase variables with respect to ground or the individual phase currents will turn out that in fact you will get some kind of coupling. But, if it is a symmetrical system as in that you know the case which I showed you sometime you had a symmetry C s C m C m C s the matrix which related all the original variables was symmetric. If the system is symmetric it turns out that some of the transformations we have studied before can be applied to the three phase case and we could actually do a kind of we can get a neat model in terms of the new variables. For example, you may ask can I apply the d q transformation which I have used for studying a synchronous machine to a three phase transmission line. The answer is yes you can provided it is a symmetric transmission line. So, if you have got for example, a transmission line which is which has got parameters like this it could be a lumped parameter representation. In which case of course, a transmission line becomes a for example, if you are studying a transmission line using the lumped parameter approximation we have already discussed this in the beginning of rather sometime ago. So, you will have d I a by d t for a three phase system is equal to V a 1 minus V a 2 V b 1 minus V b 2 and V c 1 minus V c 2 and of course, so this is 1 and 2. So, actually V a 1 means the voltage a with respect to the local ground at this end of the transmission line V a 2 is the voltage with respect to the local ground at the other end of the transmission line. So, see this is a lumped parameter model of a transmission and which you have suppose I wish to use. Now, the question which is the limited question we are asking here is not about lumped parameters or distributed parameter lines is whether, but can we get a transformation to make this a decoupled set of equations or rather I can ask you a very more specific question if I transform I a I b I c to I d I q and I 0 and V a V b V c also to I you know the d q variables what will happen of this you know equation. So, question is the differential equation in fact, will look like this eventually where this is nothing but what obtained by applying the transformation of variables. So, what I have done is applied the time variant transformation C p do you recall that the time invariant time variant transformation C p which is a function of theta theta is the angular position of a machine of course, now the question arises is which machine or there is no machine here, but suppose I take any machine and use its theta and transform this these set of equations in that case you will get basically this matrix will get transform to this matrix it in fact, turns out to be diagonal. How will you get it basically I have replaced I a I b I c by I d I q and I 0. So, what you have to do is suppose I call this the L matrix. So, what you have is L d I this is just written in a compact fashion. So, I am just writing these equations in a some kind of it is just a compact form of a way of writing these equation these are all vectors this is L is the matrix. So, when you transform you will get L this is C p. So, this is what I get is this correct is it fine. So, in fact, if you evaluate C p inverse L C p and your L matrix is symmetric it turns out that this new L that is C p inverse L C p it turns out to be a diagonal matrix. So, it is an interesting and nice thing for nice thing to happen in case we did use a d q transformation using the Parkes transformation of some machine. Now, is this equation correct well the answer is no there is a error which we have actually done. So, this is in fact, not true remember that d of C p into I d q 0 is will have another component d C p by d t into I d q 0 C p is also function of time. So, actually this equation is not correct. So, there is going to be an additional term here which has to come which is in fact this something you can just try out yourself it is going to be. So, it is going to be this into this matrix. So, it is suppose I call this matrix L dash this will become L dash. So, basically what you will have is this equation. So, remember that this is because you have to take the derivative of the time varying transformation whenever you are going to rewrite the equations in terms of the d q variables. So, you are going to get this extra term. So, actually you can in fact write down the equations of a transmission line element. For example, the series reactance of a transmission line in case you are talking in terms of the lumped equivalent you can even write the partial differential equations in terms of the d q variables there is nothing you know special about the lumped equivalent here. So, you can actually get the equations in the d q variables. So, you can actually interface the transmission line equations with eventually a machine equation which are there in the d q variables this can be done. So, this is an important point. So, the final equations which you get for the transmission line are in fact can be written down in the d q variables in this form. In fact, one important point which I must emphasize here is that in case this matrix L is not symmetric that is there is some unbalance or asymmetry in all the phases. In such a case applying the d q transformation may not be of will not be of much use because you will not get this kind of nice time invariant and diagonal form of the matrix L dash. So, in case L is not symmetric it is not it is it really arises from unbalance kind of configuration of the transmission line conductors. In such a case d q transformation will not yield you a useful set of equations, but in case there is symmetry you find that the d q transformation in fact gets you nice decoupled equations in d q and nice diagonal form of the inductance matrix. So, this is an interesting point here about the d q transformation the equation of a transmission line in the d q frame of reference. Now, there are few more things I need to tell you about a transmission line. So, my earlier promise of introducing you to the prime mover systems will have to wait a little bit. So, in the next class we will just continue our discussion of a there is some remnant discussion about transmission line modeling which will continue and then go on to prime mover systems. Incidentally in one of the examples we did to show you the behavior of an AVR the automatic voltage regulator I did model the interconnection of a generator to an infinite bus by a reactor. In fact, if a transmission line is quite short it turns out that you know you can model a transmission line for slow frequency transients by a lumped element. This is what in fact this lecture told you that you could use lumped pi equivalent of a transmission line to even get the dynamical response of the system to a very good approximation if you are not interested in the very fast transients which occur just after the transient is initiated. So, in fact we did we just did show the kind of behavior of the model using a detailed traveling lay wave kind of model as well as a lumped parameter simulation. So, we did actually do that and show you that a lumped parameter model does give you a response which is reasonably for slow transients. So, it gives a reasonable dynamical response. So, you can under certain circumstances depending on what you are really interested in looking at represent a transmission line by a lumped parameter model. So, that is what you can take back from this lecture. In addition the d q model of a transmission line can also be derived and it yields a neat model of a transmission line. In fact, which this would not hold true in case you have got an unbalanced parameter kind of transmission line. The d q model will not be very useful under such circumstances, but of course, a transmission line which is reasonably long is also transposed. So, to again a good approximation the system is balanced and you can apply the d q transformation and get d q model of the transmission line. So, these are the few things which I would like you to take back from this lecture. We will continue a bit about transmission lines in the next lecture and then move on to prime mover systems.