Feedback linearization

Remarks

Example: the pendulum $x_1' = x_2$, $x_2' = -\sin x_1 + u$ has $f(x) = (x_2, -\sin x_1)$ and $g(x) = (0, 1)$; with $\Phi = \id$, $\alpha(x) = \sin x_1$ and $\beta = 1$ one gets $z_1' = z_2$, $z_2' = v$, which is controllable. Non-example: with $g \equiv 0$ the input has no effect, so clause 2 fails for every choice of $\Phi$, $\alpha$ and $\beta$. Write $\mathrm{ad}_f g = [f, g]$ for the Lie bracket. The system is locally feedback linearizable at $x_0$ if and only if two conditions hold. The vectors $g, \mathrm{ad}_f g, \ldots, \mathrm{ad}_f^{n-1} g$ are linearly independent at $x_0$. The distribution spanned by $g, \mathrm{ad}_f g, \ldots, \mathrm{ad}_f^{n-2} g$ is involutive near $x_0$. The related input–output linearization only makes the map from $v$ to a chosen output linear, and may leave internal dynamics.