Normal form of a non linear system

Question

Let the system: $\begin{cases} \dot{x_1}=a\sin{x_2}\\ \dot{x_2}=-x_1^2+u\\ y=x_1 \end{cases}$
that is:
$\begin{cases} \dot{x}= \begin{bmatrix} a\sin{x_2}\\ -x_1^2 \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}u=f(x)+g(x)u\\ y=x_1 \end{cases}$
Since the relative degree is equal to $2=n$, I think, looking at the system, that since the input appears only in $\dot{x_2}$, the system is in normal form. But to be sure I have tried to write how it should be the system in normal form, so: $\begin{cases} \dot{\xi_1}=\xi_2\\ \dot{\xi_2}=\mathcal{L}_f^2h(x)+\mathcal{L}_g\mathcal{L}_fh(x)u\\ \end{cases}$
(where I have used the lie derivative for the notation and $y=\xi$)
The $\mathcal{L}_fh(x)$=$a\sin{x_2}$ and so $\mathcal{L}_f^2h(x)$=$a\cos{x_2}(-x_1^2)$.
Then $\mathcal{L}_g\mathcal{L}_fh(x)u$=$a\cos{x_2}u$, so $\dot{\xi_2}=a\cos{x_2}(-x_1^2+u)$.
So my question is: why there is the factor $a\cos{x_2}$ in $\dot{\xi_2}$ that there is no in $\dot{x_2}$, even if the original system seems to be normal? The system written in $\xi$ should be exactly equal, right?

Answer 1

They are equal. What you basically do is a nonlinear state (and input) transformation to obtain a linear system basically. So you define a new input $v$ as $$v = L^2_fh(x)+L_gL_fh(x)u$$ such that $$\begin{align} \dot\xi_1 & = \xi_2; \qquad (1a) \\ \dot\xi_2 & = v \qquad (1b) \end{align} $$ So if you draw this in a block diagram (I'll do it in text for now), you obtain:

--> [ NL transformation ] --> [original system] --> [NL transformation]$^{-1}$ -->

which is equal to

--> (1) -->

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edit: What you do to obtain a normal form is you first differentiate the output, such that you can obtain something like $y^{(n)}(t)+\cdots+\dot y(t) + y(t) = v(t)$. So: \begin{align} y &= x_1 = \xi_1 \quad \text{(first 'normal' state)} \qquad (2a)\\ \dot y & = \dot x_1 = \xi_2 \quad \text{(second 'normal' state)}\quad(2b) \end{align} We only have two state, so two 'normal' state should suffice. Furthermore, the jacobian of $[y, \dot{y}]^\top$ w.r.t. the state is full rank around the origin, so this is fine.

Next you calculate your dynamics of the normal system, which requires to determine the time-derivative of (2). This implies taking the derivative w.r.t. $x_2$ of $a\sin x_2$ (which is obviously $a \cos x_2$)

I hope this explains your question