Hello everyone, I am new to sliding mode control and interested in it very much. Today I come here with several questions, I hope you can give me a hand to my study, I am helpless and a little bit upset cos I couldn't slove those problems myself.Thank you very much!!!
As we know, equivalent control is derived from the equation of dot(s)=0.Then we can solve the function of Ueq. But, in real sliding mode, dot(s) would not equal to zero cos nonlinear switching item works on it. I design the control law this way:U=Ueq0+Usw(Usw=K*Sgn(s)),and make conparison with U=K*Sgn(s). Through Matlab simulation, I find that using of the equivalent control method to design control law, the oscillation of dot(s) is smaller than using the control law of U=K*Sgn(s). The equivalent control method I mention here is add an item with K*Sgn(s),so you can see it combines with two parts while another is just a single switching item.
(1) Can we say that the method of equivalent control could ensure dot(s) more close to zero but not equal to zero ? I think in 1-sliding mode control, dot(s) could not be equal to zero, or , people don't have to put forward 2-sliding mode to avoid chattering. But we do get Ueq by letting dot(s)=0.Can anybody explain it to me？
(2)In 1-sliding mode control, we use sdot(s)＜0 to ensure sliding mode existance.However, it seems we can not use this way in 2-sliding mode control. So people have to develop new approaches to give the condition of sliding mode existance. In HOSM, I read two approaches, one is extend former system into a new system with auxiliary equations(Prof.Levant), the other is Prof.Slotine’s approach. I think it plays the role of making sure high sliding mode existance, just like that of sdot(s)＜0 in 1-sliding mode control. But I am not sure if I am right.
(3)For high order sliding mode control,from my point of view, the wisdom is to increase the constrictions, then have the control U worked on the high order time-derivative of s, under the control U, ensure s and its time-derivative of s equal to zero(r-1), and prove it mathematically. s,dot(s),.....dot(s)(r-1)=0 could have more smooth performance. Do I understand it correctly？
Hi Dawn,
You mentioned about two types of sliding mode controllers: (1) u = ueq + usw, and (2) u = usw, where usw = − K sgn(s). The former is a model-based sliding-mode controller because the knowledge of the system model is required to construct the equivalent control, ueq. Mathematically, the latter is regarded as a model-free sliding-mode controller because it is based solely on the manipulation of the switching gain K to satisfy the reachability condition, s*s' < 0, thus, a system model is not required.
Since the type-2 SMC is lacking ueq, naturally, the magnitude of the switching gain K is much higher than the K value in type-1 SMC in order to compensate for the control action of ueq. Since the chattering effect is proportional to the magnitude of K, it also explains why a higher oscillation is observed in type-2 SMC when compared to the type-1 SMC.
From the practical engineering standpoints, the chattering effect is usually undesired. Hence, type-2 SMC is not a feasible solution in most engineering problems and type-1 SMC is preferred. For type-1 SMC, the model uncertainty and model inaccuracy in ueq can be compensated by a proper selection of the switching gain K in usw. We will discuss further about your questions if you understand up to this point.
Hello Dawn,
For the sake of clarity consider system
\dot s = u + f
where $s$ is the sliding variable, $u$ the control input, and $f$ an unknown but bounded disturbance.
Firstly, we state that $s$ evolves in a compact set, id est, a bounded and closed interval of time. Then, the control objective is to drive $s$ to zero as fast as possible, this is, in finite time, even in case of bounded additive disturbances $f$ by means a bounded control $u$.
The solution is achieved by $u = -k sign(s)$ with $k > f(t)$ for any moment. And the equivalent control is $ueq = -f(t)$. However, in this case $f(t)$ is unknown and in addition $\dot s = 0$ is not exact just equivalent. Thus, we have $\dot s$ switching around $\dot s=0$.
If you require $\dot s = 0$ we need both $u$ and $f$ being continuous. A special case, when $f$ has at least a bounded weak derivative, 2-sliding mode are a perfect solution. If disturbance is even smoother, then control $u$ could be also smoother and higher-order derivatives of $s$ would be well-behaved.
Conclusion. If you have some knowledge of the system, use such a knowledge to avoid the use of a big sliding gain controller, avoiding the "harmful chattering" (not dynamic chattering due to parasitic dynamics) effects. Even more, if you know the worst regularity of disturbances, use the best/smooth possible control to achieve a required performance in those conditions.
Cheers!
Hi Dawn,
You mentioned about two types of sliding mode controllers: (1) u = ueq + usw, and (2) u = usw, where usw = − K sgn(s). The former is a model-based sliding-mode controller because the knowledge of the system model is required to construct the equivalent control, ueq. Mathematically, the latter is regarded as a model-free sliding-mode controller because it is based solely on the manipulation of the switching gain K to satisfy the reachability condition, s*s' < 0, thus, a system model is not required.
Since the type-2 SMC is lacking ueq, naturally, the magnitude of the switching gain K is much higher than the K value in type-1 SMC in order to compensate for the control action of ueq. Since the chattering effect is proportional to the magnitude of K, it also explains why a higher oscillation is observed in type-2 SMC when compared to the type-1 SMC.
From the practical engineering standpoints, the chattering effect is usually undesired. Hence, type-2 SMC is not a feasible solution in most engineering problems and type-1 SMC is preferred. For type-1 SMC, the model uncertainty and model inaccuracy in ueq can be compensated by a proper selection of the switching gain K in usw. We will discuss further about your questions if you understand up to this point.
Hi Aldo Jonathan Muñoz-Vázquez,
Thank you veru much for ur explaination in detail. I know it better than before. In traditional sliding mode, the main task are two jobs, first is to design a sliding surface then design a control law that satisfy s*dot(s)＜0 to ensure sliding mode existance.
I make some conclusion this way, by comparison with 1-sliding mode, high order sliding mode, the sliing surface is not s=0 anymore, it is a function of s and its derivatives, for example, σ=s+dot(s), while in 1-slidind mode is just
s=a1*X1+a2*X2......an*Xn=0.So here people extend it sliding surface, s and its derivatives are used ! and hope that s=dot(s)....dot(s)(r-1)=0. extend the sliding surface to a higher dimension, the result is we could get high order accuracy.
Do you agree with what I said above?
If( you agree with me ==true)
{
After putting forward the concept of high order sliding mode, The following task is to find how to ensure high order sliding mode exist. Then, It comes two main theories. First is Levant's and the other is Slotine's. Finally, under the law of high order sliding mode existance, people could design control law U to achieve 2-sliding mode. Just like we design control law under s*dot(s)＜0 in 1-sliding mode control. But in high order SMC, it would be more complicated.
}
else
{
Where is my misunderstanding?
}
Thank you for helping me!
Best wishes
Hi Yew-Chung Chak,
Thank you for ur analysis. I do really agree with what you said above. The approach of U=Ueq0+Usw is better than U=K*Sgn(s), and I proved it with a Matlab simulation. U=Ueq0+Usw, in this control law, Ueq0 could be taken as the item of "chattering reducing compensation".
Actually, We get Ueq by letting dot(s)=0, and it is, somehow, a state feedback item as a part of the control law U. Let's say, In this way, dot(s) would have a more strong ability to tend to zero than that of U=K*sgn(s). Or rather, we can see that under the law of U=Ueq0+Usw, the oscillation is weaker than U=K*Sgn(s). Do I draw this conclusion correctly? ^_^
If you think what I said above could prove I understand it, I want to further discuss it up to the second order and high order sliding mode.I want to share my uderstanding while askin questions to both 1-sliding mode and high sliding mode, And help those people that have similar problem like me.
best wishes!.
Hi Dawn Ind,
I would suggest the following:
1. If you are looking to study the conventional sliding mode control (first order), then a very good option is to consider the books of Prof. Vadim. I. Utkin: Sliding Modes in Control and Optimization and Sliding Mode Control for Electromechanical Systems
2. As the pioneer of higher order sliding mode is Prof. Levant, so you can seek his publications for this purpose
Look up the pioneering works of Korovin and Utkin on sliding mode control if you want to know the origins of the subject. If you have good modeling knowledge of the system and you have actuators that give you precision control, and sensors with good linearity, sliding mode methods do not give you the best performance, for they were constructed for robustness and not for optimality.
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