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Given the system: $$\begin{cases}\dot{x_1}= x_2\\\dot{x_2} = -10x_1+1.8{x_1^2}-0.25x_2 +u,\end{cases}$$ where $$u=-1.8{x_1^2}+v,$$ I get the system: $$\begin{cases}\dot{x_1}= x_2,\\\dot{x_2} = -10x_1-0.25x_2 +v,\end{cases}$$ with the matrix $$\mathbf{A}=\begin{pmatrix}0&1 \\ -10&-0.25\end{pmatrix}$$

Is this a good way to feedback linearize a given system? Is there anything I should be careful about?

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    $\begingroup$ It is perfectly fine, although I would ask you to clarify what you mean by "good way" to feedback linearize the system. Is there are more specific query you have that underlies your concern? $\endgroup$ Commented Mar 19, 2023 at 16:59
  • $\begingroup$ With this $u$ the system is very poorly damped, so I would recommend to use $x_2$ for feedback as well to increase the damping. For example $u=-1.8 x_1^2-5 x_2+v$. $\endgroup$
    – SampleTime
    Commented Mar 19, 2023 at 17:23
  • $\begingroup$ @RollenS.D'Souza 'cause i know that i have to calculate the lie derivative, but i haven't is it wrong? $\endgroup$
    – Marià
    Commented Mar 19, 2023 at 17:27

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Your state and feedback transformation is globally well-defined, so there are no issues. In particular, you've used the state-transformation, $$z := \Phi(x) = x,$$ and feedback transformation, $$v := \Psi(x, u) = 1.8\,x_1^2 + u.$$ to render the dynamics of your original dynamical system to a linear controllable system in state $z$, $$\dot{z} = \begin{pmatrix} 0 & 1 \\ -10 & -0.25 \end{pmatrix}\,z + \begin{pmatrix} 0\\ 1\end{pmatrix}\,v.$$ It is globally valid since the state and feedback transformations have a smooth inverse globally.

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  • $\begingroup$ thk so much! so no need to compute the lie derivative ? $\endgroup$
    – Marià
    Commented Mar 19, 2023 at 17:26
  • $\begingroup$ @mdeli, no you do not need to. As long as you found teh state and feedback transformation, you are ok. Lie Derivatives are needed to (1) check existence of such transformation and (2) help you compute the state and feedback transformation if you don't know how to find it. $\endgroup$ Commented Mar 19, 2023 at 17:29

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