 Great, even if you do not completely follow the model and the logic behind the model and so on, we do not have to worry about it. Modeling is not part of this course. You are given a rotational dynamics of a rigid body. This works for any rigid body system, quadrotors, UAVs, satellites, even underwater rigid bodies. Yeah, does not matter. It has to be a rigid body. That is it. If you are flexible, then this inertia may not be constant and there is more complications there. Flexibilities give rise to bigger challenges. This is not a good model for flexible systems. But if you have any rigid body system, this is a good model. Notice because it is a satellite system, so gravity etc. is missing here. So, if you are working with again something that is on earth or underwater or whatever, then you have to have gravity. If you have viscous damping and things like that, those also have to be added in the dynamics equation. But kinematics is exactly the same. And the dynamics is actually evolving on a linear space. No problem, omega is on a R3. So, nice, good. What I have mentioned now is that I am going to take the output as omega. So, because I am looking at a cascade connection sort of a thing. Now, if you notice, omega is an input to this system. So, let us see. We are going back to this cascade of a passive system with some nonlinear stable system. So, omega is an input to this system. If omega is 0, then the system is passive and linear. Sorry, system is stable because if omega is 0, there is nothing on the right hand side. Rho dot equal to 0 is a stable system. Nothing to do. It is a stable system. So, I am already done with my first assumption that if this in output is not there, then my system is stable. So, good. So, ticked one box. The next question is this system passive. We have to check whether the system is passive. So, that is what I am saying in the beginning. With omega equal to 0, Rho dot equal to 0 is stable. So, I can actually choose any function w. With any function w, this is stable. Later on, of course, I will specify what function w we should choose. But because I am left with Rho dot equal to 0, this is stable with any function w. I do not have to worry about what function w to choose. I can choose any positive definite radial and bounded w and I am good. You can have Rho square, Rho 4, anything, anything. Excellent. So, that is good. For passivity of this system, I am going to choose this. This should again remind you of what you did with the robotic system. You had the first term as half q dot as mq q dot, half q dot transpose mq q dot, which was the term corresponding to the inertia, kinetic energy. This is exactly the rotational kinetic energy. This is the expression for the rotational kinetic energy. And this is, of course, radially unbounded and all that. So, what, of course, I have given you this is an exercise. You have to show that v dot is actually less than equal to u transpose y. It is too easy. It is like you have to write one line here. It is just one sentence. You just have to remember that there is this q symmetric, there is a cross product happening here. You have to just remember there is a cross product happening here. If you take a v dot, you are going to get omega transpose j omega dot and you just plug in j omega dot from here. And that is it. So, it is very, very easy to conclude that v dot omega is actually less than equal to u transpose y with y being omega itself. So, that you have to verify. So, obviously, what we have shown is that we have a driver system which is passive in this u y combination. And we of course, have this rho system which is being driven and that is stable in the absence of the output y. So, we are done. These are the only two assumptions that we require for the previous result to hold. So, what do we know? We know that the entire system is passive with this new output v and y where the u that is the earlier output u is just v minus this guy. Where am I getting this expression from? This is just this expression. I have just populated this. The original input is just the new input and this lgv type term. That is what this is. This is in fact, partial of w with respect to rho and then this guy. This is precisely what is multiplying the omega and that is all. And what do we know? We know that the new system is now passive with this input output combination. And then what? Then life is too easy. I have actually asked you to do this as an exercise. I have now given you a choice of w. I am not asking you to choose any arbitrary w. This is simply so that the structure of the feedback looks nice. That is all. If you choose this w which is k log natural 1 plus rho transpose rho, a weird looking choice. But this choice of w actually gives you a nice feedback and also gives you this nice zero state. Zero state observability is anyway there. We do not have to worry about it. But anyway, you have to verify that also. That you have in fact, zero state observability, not necessarily detectability. We in fact, have zero state observability. So, all you have to do is pop. You just have to put this w in this expression here. That is it. Because that is what is going to give you your control. And once you have that, zero state observability anyway, you have to verify with this for this particular output omega. So, what does that mean? It means that if omega is 0, then you want to show that rho and omega are 0. So, omega was anyway 0. If omega is 0, omega is already 0, you only have to show that rho is also 0. So, you want to sort of look at that. Because that will come from your feedback expression. Remember, zero state observability has to be verified in the absence of the input. So, this expression. So, notice that in order to check zero state observability, you typically will say that if omega is 0, we also know that omega dot has to remain 0. This is how we go about checking. Zero state observability is exactly like the Lassalle argument. So, if you want to omega to stay at 0, you need omega dot to be also 0. Which means that the entire right-hand side has to be 0. Control was already has to be 0 to check zero state observability. This term is already 0. So, all you will have is that this term should also be 0. And this term you will notice will bring in the rho. And you will be able to claim that this equal to 0 is the same as rho equal to 0. And you will be able to claim zero state observability like you need. You just have to carefully take the partial and so on. And finally, once you have that the entire system is passive with this input-output combination, you know that I can choose my V as a function of the output. That is minus phi omega. In fact, I can essentially, I am just giving an example. I could for example, choose V as minus k tan hyperbolic omega. And this is enough to give me global asymptotic stability. So, the complete feedback is not just this. Remember, it is this along with this. There are two terms. Because this I already prescribed, the real input that goes to the thruster of the actuator would be U, not just V. So, V is only one piece. So, the V piece is this. And then there is something more which comes from here. This is the actual control law or the actual control that command, control command that you send to the actuators. But you know that with this combination you will have global asymptotic stability. So, this is again something rather nice and powerful. So, you can see that even for this within half a page almost, I can come up with a feedback law, stabilizing feedback for a rather non-linear system like the rigid body attitude dynamics. And this also a saturated feedback by the way. This is bounded feedback. In fact, you will notice that there is some nice properties of the first part also. This part also has some nice properties. You will get some nice saturated feedback. So, again rather powerful result. Within half a page, if you can actually solve this problem, usually when I do it without knowledge of passivity, you will need a little bit more work. It is not this straightforward. But here in this case, because you are actually employing this idea of passivity interconnected with a cascade connection with a stable system, you have this nice simple construction of feedback. And these two exercises that I have given you are actually rather short. Anyway, even for this case, you will have numerics to do. So, I will give you inertia values and initial conditions. Actually, that is why I need to give you some inertia values and some initial conditions for omega and rho. And you will actually be implementing this controller and seeing how it performs. Make sense? Alright? Okay.