 For this DC motor example, we are simply verifying that we have the conditions that are needed. First is that you have g add fg and add f square g to be linearly independent which turned out to be a little bit complicated. I will really ask you to try to verify this numerically. And then the second condition that we needed to verify was that the distribution formed out of g and add fg is in fact, involutive. And that is pretty straight forward turned out to be 0 0 0. So, yeah you have involutivity. Now, we have the, we have by Frobenius theorem that this is in fact, completely integrable right. So, therefore, we have some output beta for which you have d beta multiplied by this terms of the distribution to be 0 ok. This is supposed to give you a partial differential equation. So, I am going to expand this. What is g? It is I will use some 1 over ls and what is f add fg is this guy minus k l r r k l r x 3 k j x 2. So, this is 0 ok. So, how do you find beta? Just write out the partial differential equations ok. Let us see what am I going to get? The first one is going to give me 1 over ls del beta del x 1 is 0 correct right. This means what? What does this mean when you are trying to solve? Absolutely, implies beta equal to in a very very bad notation ok. Yeah, you should not write this like I mean in a paper and all in exam paper it is fine not in an article yeah. But anyway I am just saying beta is independent of x 1 ok. What do I get from the second and third now? So, you see that del b del x 1 is already 0 right. So, when I expand this one I do not have to worry about this term right. So, I am going to simply write what minus k l r x 3 del beta del x 2 plus k j x 2 del beta del x 3 equal to 0 ok. So, this is whatever the k goes away right. So, I have some condition right. How do I go forward now? You can guess yeah. What do you think? What do you think I should have beta I will make this bigger if you do not see it well enough? What do you think you can get beta from here? See there is it is very symmetric right. There is x 3 partial with respect to x 2, x 2 partial with respect to x 3 ok. So, to me it seems like there should be you know x 2 square plus x 3 square something like x 2 square plus x 3 square yeah. Because if I have x 2 square it becomes twice x 2 this gives me x 2 x 3. If I have x 3 square I get twice x 3 I get x 2 x 3 again. Now, this will cancel out modulo some constants right yeah make sense yeah that is how I am guessing yeah instead of trick that is all I am not doing anything magical here yeah. But you can solve anyway you want I guess yeah I do not know again what is an obvious way to solve otherwise yeah typically how you do solve PDEs is you assume to be you know some that the it is the two variables appear independently and things like there are some methods of solving PDEs ok some only in some cases you can analytically solve them yeah. But here I will guess so what is this by the way what is the scaling on x 2 square and x 3 square you think can somebody tell me L r x 2 square plus j x 3 square perfect this will work why because it will give me you can put a half if you want but nobody gets this will give me the first term will give me L r x 2 twice L r x 2 so L r L r goes away. So, this becomes x 2 x 3 this will give me j x 3 2 j x 3 so j j goes away so x 2 x 3 again so that becomes 0 that is a fair choice ok. So, again what did we end up doing what is this beta what was this beta what is all this work for output this is the output with respect to which the system is feedback linearizable. Now go back to what we had done we had taken the system fine the friction was missing no problem I mean there is this friction term that is missing from here that I think has 0 impact on anything yeah it is this friction term is missing in the third dynamics not here no problem. But the output was pre specified to some x 3 alright now what we are saying is that we are not going to pre specify the output we are going to try we are not given an output we are going to try to find the best output under which the I get complete feedback linearizability ok and that is this guy L r x 2 square plus j x 3 square ok and now that we have you know this output why do we try ok try I am going to say h x is L r x 2 square plus j x 3 square ok what about h dot let me put a half if you do not mind yeah I will put a half just what about h dot this is L r x 2 x 2 dot plus j x 3 x 3 dot. Notice that x 2 dot and x 3 dot do not contain the control ok right. So, this is going to give me some big mess I am quite sure yeah I am not going to write it sorry I am not going to write it yeah but I am going to compute h double dot yeah this is L r x 2 dot squared plus L r x 2 x 2 double dot plus j x 3 dot squared plus j x 3 x 3 double dot yeah yeah just product rule used yeah I do not want to write this mess. So, I am avoiding writing it yeah now what do I know I know that x 2 dot square and x 3 dot square are not bringing the control right because x 2 dot is this x 3 dot is this does not have the control on it. But when I take the double derivative of x 2 and in fact the double derivative of x 3 I have to take derivatives of these guys right and here I have x 1 dot appearing here I will have x 1 dot appearing ok and that will give me the control ok. So, what did I end up finding that control appears ok. So, what is it relative degree is what where I do this correct by the way should the control be appearing here on the third derivative no this is fine sorry I got all my coordinates yeah I got all my coordinates right should I have wait wait wait I should be very careful did I get all my coordinates what would be my coordinates in this case they will be h h dot and h double no control should not appear here either no no no no no no no no no no no no no no will the control appear here this computation is so painful that I am not very keen on doing it what is the control term can anybody compute it here and here can somebody tell me what is the control term forget every other term what will be this term what will the control term coming from this L r x 2 and in x 2 double dot I have this guy right I will have so I am going to write it as minus k L s L r x 1 dot x 3 yeah that is the term that is containing the control is going to contain the control. Similarly, if I look at this guy this is J x 3 x 3 double dot will be k L s over J x 1 dot x 2 okay yeah this x 1 dot is going to bring the control I am just writing the terms in the control my feeling is these two terms are cancelling aren't they these two terms cancel out isn't it yes minus k L s x 1 dot x 2 x 3 k L s x 1 dot x 2 x 3 these terms cancel out right okay that is that is the right thing to happen okay. Now not going to write it but h third derivative has control okay okay it look like this second derivative contains the control but those terms will cancel each other out. So control does not appear it can't because I need three coordinates right I started with the three dimensional system. So my coordinates are now these three h, h dot and h double dot yeah unfortunately as you can see very very ugly looking yeah nothing much you can do yeah doesn't look very nice but nothing much you can do if you folks are enter about it we can try the other example also do you want to try the other example or do you want to do it yourself why don't you do it yourself then we will make it an exercise yeah I mean we will try actually we don't know so that was the other system which was what was the other system we were looking at we had this other example right this guy even worse no no no we are not trying this you can see that the G is not even a constant no no maybe something else yeah so I mean see fortunately or unfortunately you can do you know analytically you can work this out for very specific examples only yeah but you can also see that you are you know you are sort of coming up with some rather unusual transformations okay which you would not never have guessed otherwise without looking at this structural details you will never have guessed that this sort of an output would give you complete feedback right we were working we did work with some other you know y equal to x3 and some something arbitrary right yeah you would never have guessed that you would get feedback but yeah with this very very crazy looking coordinates yeah you will get a linear system in the end yeah and this is not like I said this is not linearization or approximation of any kind this is actually the system is linear in these coordinates yeah it is it is almost like I think a lot of people say that if you if you if you squint your eye carefully enough every nonlinear system is linear yeah so if you look at it very carefully every nonlinear system is linear it is like you know that is basically the idea here yeah so a lot of them do have this condition satisfied as you can see it is not easy to verify yeah some of these I would still recommend that you verify numerically yeah but for systems where you can do this analytically you have a pretty powerful result alright a lot of times it is not a scale issue or anything it is not like it is like n becomes large there is a computational issue it is not a computational issue it is just that for example if you have a thousand of the same kind of dynamical systems and you can sort of by some suitable transformation you can make each of these dynamical system appear linear then you have now thousand linear systems that you are working with yeah so it is not going to affect you in terms of scale yeah if you have some you know robot dynamic like a mobile robot dynamics which is which you can feedback linearize yeah then you can say a lot of things right even if you have thousand such mobile robots you can still control them with easier control loss seemingly easier loss yeah of course this nonlinear transformation will sort of play big role but the good thing is you have it is just a transformation now it is just going from one function forward and one function backward yeah everything can be transformed like that so yeah I mean there is a little bit of a pinch of salt associated with it and this is also I believe yeah there is a little bit more on okay I mean there is just a little bit more on zero dynamics and so on if you notice in the last slide of this what I have done is I have looked at the rigid body dynamics yeah and I mean although I have written it as an exercise it is almost like I have done a little bit of the work myself you can or probably the entire bit of work actually at least a little bit of work I have done myself so please take a look at this exercise yeah there this is the rigid body dynamics again yeah in terms of some parameterization you already seen the rigid body dynamics it is part of the current assignment also and this is in some parameterization do not worry about what this row is but what I have asked you to do is to see if it is you know full state feedback linearizable and find the change of coordinates yeah which is this lambda yeah so that is what I have sort of asked you to do and I have actually checked a little bit of the you know involutivity and things like that I don't know if I actually found the coordinates doesn't look like it as of yet so yeah you can probably give that a shot yeah basically the idea is to find an output for which you have feedback full state feedback linearization yeah actually I think I have this this I would encourage you to look at this although it is like a solved exercise yeah but I would encourage you to look at this that basically it is I have not used the integrability condition here because it life becomes very hard with that condition I have simply guessed that y equal to rho if I take my output as rho itself that is a that is going to make give me complete feedback linearization because the second derivative will contain the control yeah so I have sort of guessed at it again instead of actually going by the integrability condition that you have yeah but this also works so a lot of there is a lot of literature in spacecraft community where they use you know this this instead of using the rotation and the angular velocity as variables they take the rotation and its second derivative as sorry as its first derivative as the variables yeah because in those variables you can claim some kind of linear system yeah though this is little bit more complicated than that yeah anyway maybe we will give you some simulations based on this all right