 Hello everyone, welcome to the final lecture in this NPTEL course on non-linear control. So we had, we were looking at sliding mode control and we sort of looked at last time a way for chattering attenuation, right? And after that we also, we started to discuss the notion of equivalent control, right? What is equivalent control? It is basically the control that you obtain from the fact that your sliding variable goes to 0 in spite of the disturbance beyond a certain time, right? So this is a very important and interesting idea because this equivalent control is obviously not an implementable control and this is the actual control. But the fact is the actual control in a time average sense must go to the equivalent control, otherwise there is no way that the system would remain on the sliding surface, right? So therefore an estimate for the equivalent control is simply a filtered, a low pass filtering of the signum function here and that is what is implemented here. And essentially now what we can say is that from here you can see that if you simply match this guy with this guy, yeah? If you say, as you can imagine this has to match, otherwise there is no way, you understand there is no way that sliding, ideal sliding happens, right? And therefore once you match these two you understand that this quantity and this quantity have to be the same, right? So therefore what we are essentially claiming is that a very good estimate for f is rho low pass filter of signum of sigma, right, for t greater than equal to t r, right? And you understand this becomes a very nice way of estimating not just disturbance but any derivative for that matter, right? Because where was the disturbance? The disturbance was actually in x2 dot, right? So even if there was no notion of control, yeah? Even if there was no notion of control you could in fact compute the derivative using this idea. And because of lack of time we are not going to get into that but this is very closely connected to the notion of a sliding mode differentiator, yeah? It is a very nice way of doing a, obtaining a differential or in fact obtaining any disturbance signal, yeah? So what have we done? We have taken simply the signum of the, this sliding mode variable and we are scaling it by this rho, you know this rho which is obviously greater than l and all of that, right? And so we are simply scaling by rho this low pass filter version of signum of sigma and we are saying that this is a great estimate for any disturbance, right? And that is great, yeah? Because that is what we will sort of use in trying to look at the notion of a super twisting control, yeah? Because until now you have seen that the chattering is not still completely gone, okay? So what is this super twisting control, right? So you see that even though the chattering was gone in the first level, in the second level that is even though the chattering was less because we use an integral control for example in this chattering attenuation kind of idea, still there is going to be chattering, right? At the next level. So it is not like you are going to completely get rid of this but super twisting controls are a sort of quite an amazing way of getting rather smooth controls, right? Are a very good way of getting rather smooth controls, okay? So let us sort of look at what is this super twisting control? So what is the idea here? I am going to go back to the original treatment here. If you see what did we do? We had this half sigma squared as our V, right? And then we constructed our control as minus rho signum sigma, yeah? I mean the V is already cancelled the CX2 and all that so we are not worried about that anymore. So here we have constructed the V as minus rho signum sigma, right? So now we do something slightly different, right? And that is what is the title here, it is a super twisting control. And this is due to some really, really interesting results from Uthkin, etc. So what is the super twisting control? You remember that we had sigma as X2 plus CX1 and we had V as 1 half sigma squared, right? And of course you had, you know what was X1 and X2. So if you look at V dot, we had taken it, it is sigma sigma dot which is X2 X2 dot which is U plus F plus CX2. I am deliberately not writing the arguments of F just to save some space, right? And what did we do? We said that we will prescribe U as minus CX2 plus a new control V and so this becomes sigma times V plus F and this is the point when we had prescribed V as minus rho signum of sigma, right? Now this is where we prescribe a different control, right? We say that our V is not going to be minus rho signum sigma but it is going to be minus rho sigma to the power half sigma sigma, okay, all right? Minus rho sigma to the power half sigma sigma, right? Now let us consider for a moment and say that is no different, no disturbance at all. Say F is in fact 0. Then what happens is that V dot becomes minus rho sigma to the power half times sigma signum sigma which is basically, this is just absolute value of sigma so this is just minus rho sigma to the power 3 by 2, right? This is what you have. This is minus rho sigma to the power 3 by 2, okay? Now so let us see, so right, right, right? So I am just trying to see how we can do this so I want to write this in terms of the V itself, right? And so that is my primary question, right? So this is actually going to be if, yeah, so this implies sigma is somehow equal to square root of twice V is what I get from here and I have substituted here, right, the absolute value of sigma is square root of 2 V. If I substitute it here, I will get minus rho 2 to the power half, so 2 to the power 3 fourth times V to the power 3 fourth, right, okay, great? So it does not matter what I choose my rho as, it is evident that this is less than 1 so implies have finite time convergence, okay, have finite time convergence. So it is not like we have done any achieving new in terms of the convergence, the convergence is still finite time convergence for the variable sigma, the sliding variable sigma, however something really cool has happened, right? My control has changed, yeah, let us not worry about you, let us just look at V because U is just a smooth term along with V. Now earlier the control was minus rho sigma of sigma. So what was the earlier control going to look at? Look like if I try to plot it, let us see, we take a different color, if, right? The earlier control was depending on the sign of sigma, right? It was going to jump between the, yeah, the earlier control could be something like this, right? Depending on the sign of sigma, it would look something like this, right? So this is the old control, yeah? But the super twisting control is rather nice. It is not just signum of sigma, it is signum of sigma multiplied by the square root of sigma and this makes it rather nice and continuous, yeah? So this is rather cool, yeah? Why the control is not going to be these jumps anymore at all? Because even if sigma becomes, here what was happening was even if sigma became slightly positive, you had this guy. If it became slightly negative, you had this guy. Just moving from this to this, this to this, this to this, it was like a bang bang. Here the movement is much more smooth. If it is slightly positive, then you only have a slightly up curve, slightly negative, slightly up and so on and so forth, right? But this you will have as the super twisting control. Now the only problem is when there is actually the disturbance, right? It is not really cancelling the disturbance term, if you notice. The whole purpose of rho sigma was it was dominating the disturbance term. This guy is not doing that. And so what do we propose? We propose that, so if you look at what happens in the presence of disturbance is that the sigma dot dynamics starts to look like minus rho sigma to the power half sigma because the Cx2 is already cancelled, the sigma dot will just be this plus the disturbance, yeah? So you see the disturbance is not really compensated, okay? Not really compensated. So in the presence of disturbance, you will not get convergence to the sliding surface, right? So what do we do? We actually use the idea of this estimator, right? What is it? Suppose we have this assumption that f dot is also less than equal to some L bar, right? And then we construct Rv as minus rho sigma sigma plus a variable w where the w is now an estimator, right? Where the w is now an estimator of f, right, okay? So w is somehow f hat. Why does this work? It is exactly from the notion of the equivalent control, exactly coming from the notion of the equivalent control, right? If you look at this, what is the equivalent control? We say that the estimate is simply a low pass filter, right? And what is the low pass filter? It's something like this, right? So here, the only thing we've done is we've not put in a low pass filter type term. We've simply kept it at this B signum sigma, right? And this also works, right? So this is from the notion of equivalent control, yeah? You can actually estimate, right? You can actually estimate the quantity f, okay? You can actually estimate the quantity f and that's the whole sort of really, really cool idea, okay? That's the whole interesting, interesting idea. That, of course, you can always have a, you know, there's no problem if you have a minus kw as well, no problem, right? But this works, right? This is becoming, this will be an estimator of f and that is essentially what you apply on top of your super twisting control and this will also compensate for the f itself, okay? This will also compensate for the f itself. Of course, we are not specifying what B is, so we'll just say B large enough, yeah? Because if you remember here also, this rho was large enough to compensate for f, f dot and so on and so forth, okay? So again, we have shown these without proof. So obviously, you cannot say that this is a complete rigorous treatment, but the idea is the control obtained from super twisting is going to be much nicer, much cleaner, right? Than anything that you will get from the typical classical sliding mode control, okay? The final sort of notion that I want to introduce is the notion of second order sliding mode, right? Or higher order sliding mode. You can always have higher and higher order sliding mode, but I'm simply going to introduce, motivate this using the notion of a second order sliding mode, all right? So what is the notion? Earlier we constructed sigma as x2 plus some cx1, okay? And we said this is first order, yeah? Because it, essentially it is, what does it do? It, sigma goes to 0 finite time and slides on sigma equal to 0 for infinite time, right? Until both states actually go to 0, right? You remember the picture? The picture was drawn probably in the first page itself here, first and second, right? You actually reach the sliding surface and then you keep sliding for infinite time until you reach here, right? So you are moving in a one dimensional surface for infinite time, right? And this is why this is the first order sliding mode, right? And this, what we say is when a behavior like this happens for a dynamical system that it starts to move or restrict itself to a one dimensional surface, then we say that there is partial dynamical collapse, right? Which means what, that you started off with a second order system, two state variables, but you collapsed and started moving on a dimension one curve, right? That's a partial dynamic collapse. When you do second order sliding mode, you actually do complete dynamical collapse. So the target is complete dynamical collapse, which means what? Implies x1 and x2 both go to 0 in finite time, okay? And how does one do that? One does that by choose non-linear sliding surface, right? What is this non-linear sliding surface? It's very straightforward, right? I choose my sigma exactly motivated by the super twisting sort of an expression. Instead of x2 plus cx1, I take a c absolute x1 to the power half, sigma x1, right? The purpose of having this super twisting type expression, right? I am going to use this, call this a super twisting type expression because this is exactly what we used in super twisting control, right? The purpose of having this is to make sure that this is a nice continuous function, right? If I just use c signums x1, then it's not continuous, right? And that's, you cannot even categorize that as a surface, right? So as I said, purpose is to ensure continuous sigma, right? Because if it is not continuous, then sigma equal to 0 is not a surface, right? So therefore, the idea of sliding more, sliding surface, sliding manifold is not possible, right? If this is not continuous, so it has to at least be continuous and that is what this sort of a term does, right? Now what do we know about this thing, right? What do we know about this thing, right? If sigma goes to 0 in finite time, say, then what do I have? I have x1 dot equals minus c absolute x1 to the power half sigma x1, right? All I have done is substitute for x2 as x1 dot, right? Now if I take v as half x1 squared, I know exactly that v dot is going to be minus, because we did this analysis right here, right? We did this analysis exactly for the similar case here, right? Instead of sigma, there will be just x1, right? So I am going to get v dot as minus c x1 absolute value to the power 3 by 2 and that is just, again, just pulling it out from here, that is just going to be this guy, right? Minus 2 to the power 3 fourth c v to the power 3 fourth, right? And therefore, what do you have? Implies x1 goes to 0 in finite time, right? Because obviously this is less than 1, right? So that is exactly the condition, right? That v dot has to be some negative constant multiplied by v to the power alpha, where alpha is less than 1 and that is exactly the finite time convergence condition. Therefore, x1 goes to 0 in finite time and x2 is simply x1 dot and therefore, x2 also goes to 0 in finite time, right? And so we have what we will have what we want, right? This is what does it mean? We will have finite time collapse, right? We will have complete dynamical collapse because in implies have complete dynamical collapse, right? Why? Because we ensure that both states go to 0 in finite time, yeah? And stay there, right? So it is not sliding on a one-dimensional surface, it goes to a 0-dimensional surface which is origin, right? Goes to origin in finite time and we are done, right? And we are done, right? So that is the very interesting notion. Now what is it that we need to remember? We need to of course find out what you to use for sigma going to 0 in finite time, right? That is an important question, what will be the control, right? Now if you see, if I try to use v equal to sigma square to arrive at a control, I am going to get pretty complicated things, right? Because I have to take derivative of this guy and all that. I have to take derivative of this guy and all that, right? Because if I, so sigma dot complicated, not that it is not possible, but yeah, it is fairly complicated. The, it is fairly complicated, right? Because once I not to do sigma sigma dot, I will have to take derivative of this guy, right? So however, it is well understood that u equal to again minus rho sigma works for large sigma, okay? But again with a lot of chattering, right? This control also will have a lot of chattering, okay? So notice that this looks similar to before, but actually it is different because sigma itself is different, right? Okay? But again this control like I said before also has a decent bit of chattering, right? Even though you have sliding, you have a non-linear sliding surface, but the point is only the sliding surface is non-linear. The fact that if you go across it this way or that way, across even a non-linear sliding surface, your control is still jumping, right? Then the other option is of course, you can or use super twisting, yeah? So whatever be the order of the sliding mode or the slide, you can always switch to a super twisting control, right? Because then you get rid of the chattering issue in a very, very nice and efficient way, right? Because this is definitely going to jump across the sliding surface, right? Even if the sliding surface is linear, non-linear, it doesn't matter. It's still going to jump. It is still going to move along around it and if it moves across it, it's the control going to jump and therefore there's going to be high frequency chatter. On the other hand, the super twisting control because it's not going to be rho sigma, it is going to be something like minus rho sigma to the power half sigma sigma, yeah? We are going to have a much more smoother performance, okay? And that's sort of what you're looking for, all right? So that's sort of all we wanted to discuss in sliding mode control. We did first order sliding mode. We did second order sliding mode. We sort of understood how using the notion of equivalent control, you can actually obtain the value of the disturbance itself. We understood the limitations in the sense that there is a lot of boundedness, uniform boundedness assumptions. We understood that there is chattering issues that we need to resolve. Super twisting control is one of the really cool developments in sliding mode control area, which actually gets rid of chattering very efficiently, right? And this can be combined with the function, the disturbance determination scheme to actually get to the sliding surface, yeah? So of course, we did most of this without proof and with a very nice double integrator example. Things are significantly more complicated if you do more elaborate things for more general systems. And of course, there is enough literature and books and texts in this area. So I have particularly been referring to the book by Yuri Stessel and others, yeah? Sliding mode control, yeah? There is quite a few authors. All the most of the pretty heavy lifters in the area are sort of co-authors in this book. And so it's one of the better written books. Most of the material that we've covered is from the first chapter, yeah? And so as you can imagine, this is what you've covered is literally just the first chapter introduction, so sliding mode. And there is so much more than this in sliding mode control, yeah? That we have not actually covered here. So that brings us to the end of this NPTEL course. I really, really hope that you enjoy learning through this course. And I also really hope that some of what you learned ends up being useful for you in the real world. And I would be very happy to hear here and on YouTube, you know, please tag me, please tag the course on your social network pages. And I mean, so that more and more people can actually take the course, learn from the course. And there is more interest in the topic in general, yeah? So that's all from me. Thank you so much again.