More Facts about IO Linearization

We are now aware of how to perform input-output linearization. To summarize:

1. For an output with relative degree $r$, we are able to construct a feedback linearization mapping such that the input-output linearized system is of order $r$.
2. The remaining state will construct a “zero plane” $Z$ where the zero dynamics on the plane will determine the stability of the overall system.

Now, we can draw an obvious conclusion if the zero dynamic is indeed stable:

Theorem: If $z = 0$ is locally exponentially stable for the zero dynamics, $\dot{z} = q(0, z)$, then $u_{IO}, v$ locally exponentially stabilizes $x = 0$.

The proof is as follows:

Proof: The closed loop system is given by
$$
\begin{align}
\dot{\xi} &= A_{CL} \xi, A_{CL} = A - BK \\
\dot{z} &= q(\xi, z)
\end{align}
$$
where
$$
A_{CL} = \begin{pmatrix}
0 & 1 & 0 &\ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-k_1 & -k_2 & -k_3 & \ldots & -k_r
\end{pmatrix}
$$
where $\Re{\lambda_i} < 0$ for all $i = 1, \ldots, r$.
If we linearize the system at $\xi = z = 0$, we get the following:

$$
\frac{d}{dt} \begin{pmatrix} \delta \xi \\ \delta z \end{pmatrix} = \begin{pmatrix}
A_{CL} & 0 \\
\frac{\partial q}{\partial \xi}(0, 0) & \frac{\partial q}{\partial z}(0, 0)
\end{pmatrix} \begin{pmatrix} \delta \xi \\ \delta z \end{pmatrix}
$$
The matrix is Hurwitz, thus the end of the proof.

We notice that the relationship between the new states $\xi$ and original states $x$ is such that
$$ \xi = T(x)$$
To verify if $T$ is a diffeomorphism, we introduce the following theorem:

Inverse Function Theorem: A function $T: \mathbb{R}^n \to \mathbb{R}^n$, $T \in C^1$ satisfies
$$ \frac{\partial T}{\partial x}(x_0) \neq 0 $$
is full rank, then $T^{-1}$ exists, and is continuous and differentiable.

Now let’s connect IO linearization back to feedback linearization. It’s not hard to see that if we can guarantee the zero dynamic to be stable, then the original system will be stable; the best way to make sure the zero dynamic is always stable is that the zero plane shrinks to just a point, and this is done if $r = n$. Therefore we have the following theorem:

Isidori, chapter 4: If a nonlinear system $\Sigma$ has a relative degree $r$ at $x_0$, then on the neighborhood of $x_0$, the functions
$$ \{ h(x), L_fh(x), \ldots, L_f^{r-1}h(x) \} $$
are independent.
Then, we can conclude that, $\Sigma$ is feedback linearizible, if and only if $\exists y = h(x)$ such that the output has relative degree $r = n$.

In short, if the output of the system satisfies that the relative degree equals to the system degree, then the system is always linearizible. However, if the output has a relative degree smaller than the system degree, it’s possible that we didn’t pick a good output – how do we know if a system can in fact be feedback linearizable? We’ll have to introduce some more new concepts to answer the question.

Some Concepts in Differential Geometry

Our audience may find themselves familiar with these concepts, if they have taken classes in general relativity.

Manifold

Let $M$ be a non-empty set of $\mathbb{R}^n$ and let $1 \le m < n$, then $M$ is a n-dimentional smooth manifold of $\mathbb{R}^n$ if, $\forall p \in M$, $\exists r > 0$, $F:B_r(p) \to \mathbb{R}^{n-m}$ such that:

1. $M \cap B_r(p) = \{ x \in \mathbb{R}^n | F(x) = 0 \}$
2. $ F \in C^0 $
3. $ \forall \bar{x} \in M \cap B_r(p)$, $\text{rank} \frac{\partial F}{\partial x}(\bar{x}) = n - m $