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.  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.  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.


There are other other algorithms for public key cryptography. We will in this course look at RSA we'll look at one other a little bit later next year and I'm going to skip over a few slides here about the general requirements for public key cryptography We'll use RSA as an example then come back to them at a later date and and return to those concepts developed by reverse Shamir and Adelman in 1978 and they set up or there's a company now RSA security quite large in terms of cryptography and They sell products eventually using RSA and other algorithms. So it's the most widely used public key algorithm It's a block cipher. It operates on a blocks on a block of data Where the plain text must be represented as an integer So if you have to have a JPEG that you want to encrypt with RSA Then you take a block of bits from that image okay, you take some bits and That's an integer those say you take a thousand bits and you get a one thousand bit integer number and that's used in the RSA algorithm that we just saw plain text raised to the power of E mod n and you get your ciphertext and that's the block of ciphertext a bit more a few more details about the algorithm So we encrypt in blocks Same in a any block cipher if we have a large piece of plain text Then we divide into blocks and encrypt one block at a time and there are ways to combine those blocks together similar to the modes of operation for the symmetric ciphers The block Must be the value must be less than n Okay, n is our modulus and will be chosen. So the block must be less. So in practice what you do is let's say n is a 1,024 bit number Okay, if n is a 1,024 bit number then the block Is a size of i bits where i Is chosen such that it to to the power of i is less than n a simpler example if If n is 100 Not 100 bits, but the decimal value of 100 How big is the block? How big should the block be if if n is the decimal value of 100? Then we'd set the block to be how many bits i bits how many? 6 2 to the power of 6 is 64 no 2 to the power of 7 is 128 That is if n is 100 in decimal Then we'd take a normally six bits of plain text as a block Because with six bits we'll always get a number less than 100 In fact 60 60 less than 64 if we had seven bits we may get some numbers greater than 100 and our plain text must be less than n so that's if we have n is And a 1,024 bit number then we can choose the block size and to encrypt we've seen m to the power of e mod n decrypt same algorithm in fact just Change the values of the input c to the power of d mod n. So the same steps It's just here. We use a different value Then for the exponent same n will come to this part in a moment and each user has a public and private key and We normally denote the public key as the two values e and n the pair of values and the private key the two values d and n but note n is Not private Okay, because it because it's the same n if I choose my values of e What did I choose e equal to 4? d equal to 2 n equals to 20 if I choose those values that bad values, but if I did Then I would tell you my value of e equals 4. I Would tell you my value of n equals 20 because I a public so I can tell you my public key I would not tell you my value of d. I must keep that private So even though we write n as part of the private key n is known to other users because it's in the public here as well So the secret value is d and we must keep d secret. Otherwise It no longer is secure So we need to look at why does it what under what conditions does RSA in work? We saw a case where it didn't work where we Decrypted and we got the wrong plaintext. Of course, we cannot allow that so under what conditions What values of e n and d doesn't work and then that will determine how to choose e d and n? note if We substitute this equation for c into the decryption equation So we have two equations here c equals m to the e mod n and M equals c to the d mod n encrypt decrypt Now replace this c with the first equation and That's what we get on this second line here. We get and We'll write it on the board If we take the decryption equation and replace c with m to the power of e mod n That was the value of c all raised to the power of d That's a d mod n That's this equation or here and And the properties of mod is that we can take it out of the brackets here and they it becomes m to the power of e to the power all to the power of d Mod n mod n when we do multiple mods by the same modulus. It's just the same as doing one mod so simply mod n and again a simple property of our Exponential that equals m to the power of e multiply by d mod n so what that says is that If I encrypt using this equation and then decrypt using m equals c to the power of d mod n Then it means that m equals the same value m to the power of ed mod n Under what conditions is this true? That's what we want to find out. We saw a case where it's not true Because what we did is that we had our original m Let's say m1 here We encrypted using our equation and then we decrypted and We got this other m2 5 we get a different m and That was unsuccessful of course That's because the values of ed and n were inappropriate so We want to find values of ed and n such that when you take a number and You raise it to e multiplied by d and mod by n you get that same number back Anyone know when that is true? Look at this equation. Does it look familiar to you? One of the theorems. Okay, if you go back to the theorems We had Fermat's theorem and Euler's theorem and in fact, they're similar Here something is equivalent to the same number raised to the power of e multiplied by d in mod n This was from earlier today We say when we mod n then A is equivalent to age raised to the power of the totion of n plus 1 That was the Euler's theorem Similar structure replace a with m and let's see what Requirements are for that to work. I'm going to go through an example. I'll do it on the board I think you have it printed hopefully at the end of your This topic and your handouts. Let's check So at the end of this topic you have some notes on number theory just to remind you and then a handout on public key cryptography and I'm going to go through that on the board. So you don't have to copy at all You can make notes as you go and we'll turn this off so the the question is when does RSA work and Given the encryption and decryption equations it should work if this is True if m equals m to the power of ed mod n Well under what conditions is that true? That's what we want to find out and we note that There's some similarity between our equation m equals m to the power of ed mod n and Euler's equation here a Equals a or is equivalent to a to the power of the totion of n plus 1 in mod n So let's try and take advantage of that and write our equation in the similar form that is When does this Equation or statement match this one under what conditions? I'll write them both Or we've got the first one This is Euler's equation. I'll just replace a with m We know this is true Okay so Under what conditions is this equation true? when does m equal m to the power of ed Compare the two equations Euler's equation and our one of the RSA Well, maybe easier. What's different between? this and this ed this this exponent here and The totion of n plus 1 okay for them to be the same So for this one to have the form of Euler's equation We would need what we need ed To be equal to the totion of n plus 1 Okay, if it was that would be the same Because if ed equals the totion of n plus 1 then this becomes m equals m to the totion of n plus 1 mod n And we know that's true So this will be true if Ed equals the totion of n plus 1 is that a step clear We want to derive and find out on what values of ed and n to choose So we'll go through these steps to do so and again, this is in your handouts written up a bit more detailed at the end of the the slides there's a handout on Public key cryptography some examples. Okay, so you have it there So we're at the state that we need this to be true when you take What's 8 plus 1 mod 8? what is 9 plus 1 mod 8 a mod 9 Okay, what is x plus 1 mod x one What is the? totion of n plus 1 mod the totion of m one Okay, now let's use that and Here That's mod both sides by the totion of n we can do that we have this Statement here apply the same operation on both sides. So if this is true, that's mod both sides by e times d mod the totion of n So if e times d equals the totion of n plus 1 Then it's also true that e times d mod the totion of n And put it in brackets equals the totion of n Plus one Mod the totion of n. So all I did took this statement and took the mod of both sides What is the right-hand side? From here, it's one the totion of n plus one mod the totion of n is one so now we have the case that our Our original equation will be true if e times d Mod the totion of n Equals one what what does this resemble e times d? Mod the totion of n equals one the multiplicative inverse e times d equals one in the mod of the totion of n Let's make sure everyone can see this So our conditions are now so Again repeat We have this equation. We need this to be true for RSA to work under what conditions conditions is this true well Let's equate it to Euler's theorem and We know Euler's theorem is true So long as the exponent here is the totion of n plus one Here we have e times d So if we set e times d to be the totion of n plus one This will be true So let's do that e times d equals the totion of n plus one in that case RSA will work All I've done now is taken the mod in totion of n of both sides and So ed mod the totion of n and the right-hand side mod the totion of n And we know that the right-hand side can be simplified to just one Totion of n plus one all mod totion of n is just one so now our condition is and we can write this differently e times d is Equivalent to one when we use mod the totion of n That's just a different way that we write the mod we can write it in two ways. Don't write this one down We can write a mod B equals C or A is equivalent to C in mod B Just different different ways to write a modulus here. We say A mod B equals C or we can say a is equivalent to C when we mod both by B That's all I've done here e e times d mod the totion of n equals one or e times d is equivalent to one When we mod both sides by the totion of n What is this and you said it before That's our definition of I don't know Multiplicative inverse if D is the multiplicative inverse of e Then this is true Multiply two numbers and get one as the answer Then they are multiplicative inverses of each other I would have find a good plan what's When does e have a multiplicative inverse when is this true the multiplicative inverse exists when e and the totion of n are relatively prime Okay, because remember back to our original definition Some number has a multiplicative inverse if that number is relatively prime with the modulus So because not all numbers have a multiplicative inverse So now we say for rsa to work our condition is that e is relatively prime With a totion of n that is the the greatest common divisor of e and the totion of n equals one So that's one of our conditions for rsa to work choose a value of n and E such that e is relatively prime with a totion of n No, this is not n. This is the totion of n because if this is true Then we know that we can find some multiplicative inverse. We can find D If it's not true, then we may not be able to find D Okay, so we need e to be relatively prime with the totion of n if that's true There is a multiplicative inverse there is a D that exists and we can find D there are algorithms to find D and We haven't covered it, but Euclid's algorithm Can be used to find D? That is if you know the totion of n and If you choose some e Which is relatively prime with it? There's algorithms to reasonably quickly find the multiplicative inverse. Okay D is The inverse of e That is multiply them together you get one So we've reached one of our conditions for how rsa will work For this to be true Choose e relatively prime with a totion of n and then once you've got that e calculate D Such that e times D was one and then this equation is Equivalent or is in the form of Euler's equation and it will always be true So what we're going to arrive at is the steps that the user must use to choose e, d and n Okay, if you remember back to the example I had before I had an e of 4 a d of 2 and an n of 20. I Chose them. I effectively randomly they didn't work So in fact we need an algorithm for choosing e, d and n Well, what are the conditions of that algorithm? Here's one of them so that when you generate your keys you use this condition Let's stop that part there. There's some conditions Come back to the totion of n is How do we calculate the totion of n? The totion of some number Go the end of our number theory slides. We mentioned today Calculating the totion of a large number is Hard if n is a very large number Then finding its totion is practically impossible Unless we have some Specific value of n. Okay, let's say we choose a large n a random number and it's a thousand bits long Then there are no known ways to calculate the totion of it in a reason reasonable amount of time But we need the totion of n Because we need to find e which is relatively prime with it So now the question is for RSA we need to choose a value of n and We need to choose a value such that it will be easy to calculate the totion of n So what should we choose? How should we choose n? Normally calculating the totion of n is hard except in some cases when is it easy? I didn't write it up there But we saw it recently when is it easy to calculate the totion of n When it's the multiplication of two prime numbers if n equals p times q Where p and q are prime numbers Then we know that the totion of n equals p minus 1 Times q minus 1 so if I choose two prime numbers p and q Multiply them together I get n Then I can quickly calculate the totion of n. It's a simple calculation p minus 1 minus multiplied by q minus 1 so What we need for RSA there are other ways to choose n We do this or we set this requirement so that it's practically easy to generate the keys The user that creates their keys must choose values of n e and d We have a requirement on e We set a requirement on how they should choose n So that it's easy to calculate the totion event of n because we need that here So we require the user to choose two prime numbers p and q multiply them together and Then you get n you can easily determine the totion of n and And then from that you can just choose a value e which is relatively prime with the totion of n and Then there's algorithms to calculate d the multiplicative inverse of e and we know That there will be an inverse of e because it is relatively prime with the totion of n This condition so that's how you choose e d and n to work in RSA You choose two prime numbers Calculate n calculate the totion of n Choose e which is relatively prime with that and then calculate d and you have e e d and n and Because we've followed through these steps our encryption and decryption will always work because we're relying on this form of Euler's equation that is these are equivalent so let's Let's try it with another example and see how to generate those keys So this is really the key generation steps How does a user create their public and private key? We have an example on the handout Okay, we have an example that I'll go through again on the board. It's in the handout So you have it in front of you So instead of choosing any value for e d and n we need to generate values So what we do is we choose p and q Which it must be prime numbers Okay, so in this example, I'm going to choose p equal to 17 and q equal to 11 We'll we'll return later to the security of RSA and they should be large prime numbers But of course in this example, I'm using small ones so we can calculate on the board, but choose two prime numbers Calculate n What's n? Well 17 times 11 187 So now we have our value for n Here n equals p times q next calculate the totion of n the totion of 187 if I give you that in the exam If you try and manually calculate you'll take forever not forever, but you'll take a long time But if you recognize like you know, there wasn't a question in the exam this semester yet if you recognize that 187 is in fact p times q two prime numbers then we know immediately that the totion of 187 is 16 times 10 17 minus 1 times 11 minus 1 160 So this is the steps that you take when you create your own keys Choose the two primes calculate n calculate the totion of n What do you do next? We have n we have the totion of n now choose an e which is relatively prime in this There are multiple values that so they're a multiple ease that you can choose here in this case I'm going to choose seven Small value we can calculate the answer Is it relatively prime with 160? It's a prime number. So the factors of seven or one and seven So is 160 divisible by seven No Therefore they are relatively prime Because remember relatively prime the greatest common divisor is one the divisors of e are one and seven There are many divisors of 160 Any of them seven? No, therefore the greatest common divisor between these two is one they're relatively prime Now you calculate d and running out of space Seven times something equals one in mod 160 We can get rid of this and All right, you can trial try different numbers. So seven times Something equals 100 equals one seven times some number mod 160 Equals one. That's what you're trying to solve find that number or the other words seven times some number either equals 161 is there such a number That is 161 divided by seven. Do we get an integer or you use your calculator? No, I don't think so Actually, maybe you do No, you do 23 seven times 23 140 161 easy one in this case Seven times 23 at a hundred and sixty one hundred and sixty one mod 160 is one Yeah, that's why I had a hundred and sixty one here because when we mod by 160 we want the answer to be one easy D was 23 and Now we have our keys E is seven n is one eight seven That's the public key e and n private key D is 23 n is one eight seven so the generation of the keys and No, the keys in fact used in the example on the in the lecture notes and You can check you can check when you choose any value of m. That's less than n You encrypt and decrypt you always get the plaintext back any questions on how to generate keys so we've gone through two things why or a derivation of the rules for generating the keys and Then we gave an example of how to apply those rules to generate keys You should understand both I think The derivation and the the example that the applying the steps and you'll see past exam questions or even quiz questions Generate the keys for RSA, but for small values like this. This is not so hard With larger values, of course, it's much harder. We need computer support So in RSA Think every user generates their key pair so what each of you does is Follows these steps generates your key of public private key and then you can advertise your public key to everyone else You keep your value of D private So you can tell everyone PU is seven one eight seven But of course you don't tell anyone that D is 23 and Then you encrypt or if someone wants to send to you they encrypt with your public key and you decrypt with your private key Any questions before we look at the harder part? You would follow those steps. So the steps for key generation Are these? Very large one you'd have some software that would do it for you Okay, but exactly those steps for a very for real keys. We need to use larger values We'll see why later, but that is you choose two large prime numbers P and Q So your software chooses two large prime numbers multiplies them together gets n Calculates the totion of n which is very easy. It's just multiply two numbers and then there are algorithms for Finding a number relatively prime so testing if a number is relatively prime with the totion of n So choose an e and in fact in practice You can use the same e as other people So one common value of e is three Okay, even if you have very large values of the prime numbers and n You can choose e would have it e to be three and Someone else can also have the same value of e so long as the values of n are different It's still reasonably secure secure on except in very special cases So you can choose your own value of e but there are also some recommended values because they're better for Implementation and there are algorithms for given e find e Okay, and they're reasonably fast You can write software quite easily to do this. It's just that as the numbers get bigger and bigger It gets slower and slower There are some difficulties in there Choose two large prime numbers is the problem Choosing prime numbers is not easy large prime numbers. Okay Tell me a prime number which is 500 bits in length Then it's not easy to determine Whether a number is prime or not But it turns out that in most cases there are ways to test with reasonable confidence that a number is prime So you randomly choose a number test if it's prime If it and it gives you enough confidence that it's prime then use that and then just choose another one modify them together Any other questions before we? remove some of this okay, the question or the comment is if We calculated D In generating my key and it's in particular my secret value D. I calculated it from e Okay, in this case. I knew e was seven and I calculated D was 23 So if I can do that, then why can't a malicious user and attacker also do that? Because remember a malicious user knows my public values. They know e and n and They want to find D So let's look at how or what the malicious user needs to do to break this By break it. I mean for example find D Was that should be secret or later? We'll see to Find the complaint text given ciphertext Remove this Before we go through that. Let's make a another point. What should be secret D? Of course should be secret When you generate keys you also choose prime numbers p and q They should be secret as well. That's it an important point P and q the values you chose you should not tell anyone and you can but p and q you cannot Now the question is If the attacker knows Let's say the attacker What information do they know? Well What's public they know e equals seven? They know n equals 187 And in the else they know at the start They know the algorithm so they know The steps that were applied to generate the key they don't know D If they find D, then this is insecure. So how do they find D? What's some steps? What I get why find p and q? So we can find the totion five n. Okay, so let's go backwards in our steps. What do we do? To find D to generate D We had e and we had the totion of n and Given those we said that if we have e and the totion of n, then it's relatively easy to find D Okay So if the attacker they know e To find D then they must find the totion of n Because if they can find the totion of n Then give an e they can find D Okay, how do they find the totion of n? Find p and q okay or Two ways So we know n equals 187 The question is now. What is the totion of 187? from the attackers perspective and Two ways to find that what we say the manual approach Let's look at all the numbers from one up to 186 and Compare if they're relatively prime with 187 because by definition we just count those numbers and that gives us the totion of n. So we can manually Let's say we manually calculate With 187 it's easy in fact. It wouldn't take long for a piece of software or even you could do it in an exam just list the numbers from one up to 186 and Determine which ones are relatively prime with 187, but if this was a large number Then and we said at the end of the number theory slides today that Determining the totion manually of a large number is practically impossible So if n is very large 187 is not large, but any of say 1000 bits long Then manually finding the totion of n is considered impossible So if we cannot do that what else can the attacker do I think you said it before Well instead of manually determining the totion of n Find p and q Okay, because once we know p and q it's easy to calculate so To find d there were two approaches one manually determine the totion of n to find p and q How do we find p and q? We've got n. How do we find p and q? What's the the process called and if I don't have the size? Integer factorization as the name Let's show the last slide of our number theory said that there are certain problems in number theory which are Practically impossible to solve if the numbers are large enough and we cannot see it. Yeah First one integer factorization we said that if We have two unknown primes p and q which are multiplied together then given n It's practically impossible to find p and q Which is the challenge for the attacker here? They know and they need to find p and q When n is 187 it's easy Okay, you can try different numbers, and I think some of you did that in the mid-term exam But when n is a 1,000 bit number It would take too long to find p and q. Okay, there are no way No known ways to do that in reasonable time So integer integer factorization Factoring a large number into its primes is considered impossible if the number is large enough If we could do that then we could easily break RSA Because we'd find p and q we'd find the totion of n then we'd find d. Okay. It's no longer secure But we cannot do that if it's large enough similar Manually determining the totion of n calculating Euler's totion With large enough n Manually calculating it is considered impossible Okay, there are no known algorithms that would do it in reasonable time so the attackers task of Solving these two problems becomes very hard if we increase the size of those numbers in This simple example you could solve it But with large numbers you would not be able to and that's where the strength of RSA comes in That we can generate the keys but without knowledge of pq the two primes it's Practically impossible to find out what the secret value d is so the user knows d and The attacker cannot find d can the attacker do anything else What else can they try? So here we tried to find d Two ways we we see if the numbers are large enough they won't work Anything else? Show what we've done Remember how we generated the keys okay p times q totient Relatively prime well That didn't work for the attacker because we couldn't find the totion of n And if we can't find the totion of n Then we can't find d Okay, so that approach doesn't work if our numbers are large enough Yep Will there always be a p and q you choose p and q so yes your first step Here key generation your first step when you generate your own keys choose two to large prime numbers So yes, that's what you start with Then you calculate n So you know p and q as the attacker doesn't know it anything else that we can do log n What about from the encryption equations? again Let's write down our our known Algorithms for encryption and decryption and see if the attacker can take advantage of that c equals m To the e mod n and similar m equals c to the d mod n What can we do? Let's say we have In the past we've obtained a cypher text value and The corresponding plain text we don't know the key But we've been lucky enough that we've intercepted some cypher text and we've somehow found out the the plain text Okay, it's an old message. It doesn't give us any value But what if we know m and c as the attacker What if they know m so See and of course they still know e and n they want to find d Well look at this equation here's an equation with four variables m equals c to the d mod n three of them are known One is unknown easy What is it d Equals the discrete log in base c of of m mod n Here we have an X Exponentials c to the power of d mod n equals m the inverse operation that is find the exponent Find d given m c and n Which is a discrete logarithm? The discrete log base c mod n of m equals d So you just need to solve the discrete logarithm How do they know m? Well, maybe they've determined a they found a plain text From an old message. It's no longer relevant, but they found the plain text ciphertext pair So that's possible in some cases so if they find m c and n solve a discrete logarithm and We said that's impossible another challenge in number theory Is I'll show you that on the board in a moment a third challenge in number theory is to solving discrete logarithms It's practically impossible to factor a number into its primes It's practically impossible to manually determine Euler's totion and it's practically impossible to solve discrete logarithms Practically impossible when the numbers are large enough, okay? so the challenge Solve a discrete logarithm But we understand that if the numbers are large enough we cannot solve that okay, so we see that these these difficult problems in number theory are what make RSA strong in theory There's a solution in practice. It takes too long to find that solution anyone else want to break RSA Any other approaches Brute force what is brute force involved with RSA? How would you brute force? different approaches All right one approach. We're trying to find D Or we want want to decrypt some information for example, so we have What do we have we have this equation we we know a ciphertext we know and The N If we don't know the plaintext a brute force would be try all possible values of D Okay Because and the same with a brute force in any cipher if we just try the different values Eventually, we'll get a value of the plaintext that makes sense So we take our ciphertext Raise it to the power of D Mod by N Where we just choose a random value of D We get some plaintext M If it makes sense Success if it doesn't make sense try a different value of D and keep trying So there's one brute force attack try different values of D How do we prevent that make sure D is large enough? Okay Make sure D is large enough such that we must make many attempts and it takes too long to try them all to find The plaintext so the size of D is important in this case Anything else that we can try similar here We've got the inverse operation We know C We know E we know N find M, which is a discrete logarithm except it's slightly different We're trying to find the base when we know the index same difficulty as it is as the finding D So this equation we can write We have a challenge find M given the ciphertext E and N and again It comes down to solving a discrete logarithm, which is too hard if the numbers are large enough There may be a few others that we've we haven't covered the idea is to look at what the attacker knows They know the algorithm They know the public information and then what steps do they have to take to try and find the secret information the plaintext or the key or the private key in particular D and It turns out in all cases So long as we have numbers large enough There are no known techniques which which will do it in reasonable time and RSA today is considered secure with large enough numbers any questions on Those things we've gone through for RSA today What you should take from this today. Okay understand the basic equation Understand how to generate keys the steps The lecture notes describe those steps in more detail. Okay, you have them in the slides Be able to generate the keys. I'll give you some quiz examples of generating keys simple ones understand Although it's removed how Why we need those steps to generate keys Why is it that we need to Choose e relatively prime with the totion of N So there's steps that we derive there And then eventually we will return next In two yeah next week What can the attacker do to try and break it So what Opportunities does the attacker have to try and break RSA and eventually in all cases you lead to you'll find problems Which are too hard so long as the numbers are large enough and that is enough for today on RSA Let's stop there. I Hope what you see so the last thing I hope what you see is the the the beauty in RSA in that Here's the encryption algorithm The most simplest equation you can see in any cipher almost simpler than Caesar cipher in writing it down and Yet secure Let's stop. Well, I'll give you some examples next week, but How long will it take to take if the numbers are large enough it will take you forever? Hundreds of years thousands of years, okay? On the simpler method you'll do that in an exam. Okay on the simple numbers Yes Yes, yes, sim simple simple algorithm, but it's as secure as the other algorithms. Okay, correct