Hello,
I am studying sliding mode control and have found that many references distinguish mismatched uncertainties from matched ones. I know that mismatched uncertainties do not enter the input channel. For example, we consider the system:
dx/dt = A*x + B*(u+z) + p,
where x is the state vector, A and B are known matrices, u is the control input, z is the matched uncertainty, and p is the mismatched uncertainty.
I am wondering why z and p should be handled separately. The above equation can be rewritten as
dx/dt = A*x + B*u + (B*z + p),
and we can handle (B*z + p) as mismatched uncertainties, and there are no matched uncertainties now.
Also, researchers say that mismatched uncertainties are much harder to handle than matched ones. I don't know why. For example, please see the attached file where I derived a simple control law that mitigates the effects of both matched and mismatched uncertainties. Could you explain what is wrong in the derivation? It seems to me that matched and mismatched uncertainties can be handled equally.
Dear David,
Mismatched uncertainties be treated separately from matched uncertainties to reduce the conservatism. In other words, the assimilation of these two quantities weakens the robustness of control generaly and also of Sliding-Mode Control.
Please look at folowing links and attached files in topics.
-Variable-Structure Control of Complex Systems: Analysis and Design
https://books.google.dz/books?isbn=3319489623
-Decentralized Control and Filtering in Interconnected Dynamical Systems
https://books.google.dz/books?isbn=1439838178
-Modern RF and Microwave Measurement Techniques
https://books.google.dz/books?isbn=1107245184
Best regards
Dear Dr. Lafifi,
Thank you very much for your answer.
Then, originally we can handle both matched and mismatched uncertainties equally (as I derived the control law in the attached file), but in order to guarantee higher robustness, we distinguish the two, right?
Also, could you let me know if there is any error in my derivation?
Thank you.
With best regards,
David
Dear David,
I saw your derivation. The problem is that both your uncertainties Z and P are MATCHED ones...
Indeed, you will find in the literature two definitions for "mismatched uncertaintiy". Some authors say that it is the perturbation entering directly in the input channel (I dont like this definition). And other authors say that it is the uncertainty that is not in the same equation that is the control input signal (I prefer this one). If you consider this last definitions both Z and P in your derivation are MATCHED uncertainties. Look this paper below. It explains in detail all this ! And it also shows why SMC is not good to compensate the mismatched case. Regards, Ely Paiva.
https://www.google.com.br/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0ahUKEwjm3ouG6OvWAhUDiZAKHe69CmYQFghKMAQ&url=http%3A%2F%2Fwww.aas.net.cn%2FCN%2Farticle%2FdownloadArticleFile.do%3FattachType%3DPDF%26id%3D19099&usg=AOvVaw05AY7WIzp6mp4VSGslGiQp
Ely Carneiro de Paiva
hello
Why do you say p is matched uncertainty?
matching Condition does not exist for p.
I feel p is mismatched uncertainty.
Where did you get the second definition for mismatched uncertainty?
Thank You
Vector P include both matched and mismatched uncertainties.
and using control input u, you can achieve the condition S = 0. But that does not mean you are handling mismatched uncertainties.
Usually S is designed as a Hurwitz surface and once on the sliding surface S, the system states converges to equilibrium. No such information is present on how you designed your sliding surface S= GX. if Hurwitz still you need to compensate for mismatched uncertainty . The way you have design the sliding surface is very wrong. Try to read my paper how to design sliding surface for mismatched uncertain system .
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