I want to find the transfer function for a system described by this non-linear state-space model:
\begin{equation}
X(t) = \begin{bmatrix}
y(t)\\
v(t)\\
\varphi (t)\\
\omega (t)\\
T_d(t)
\end{bmatrix} \Rightarrow \dot X(t) =
\begin{bmatrix}
\dot y\\
\dot v\\
\dot \varphi\\
\dot \omega\\
\dot T_d
\end{bmatrix} =
\begin{bmatrix}
v\\
y\omega ^2 – g\\
\omega\\
\frac{T_d}{my^2} – \frac{g}{y} – \frac{2\dot y\omega}{y}\\
\frac{1}{\tau}(k_m u(t) – T_d(t))
\end{bmatrix}
\end{equation}
\begin{equation}
y(t) = \begin{bmatrix}
1 & 0 & 0 & 0 & 0
\end{bmatrix} x(t)
\end{equation}
The operating point for equilibtium state is given by:
$(\varphi _0 = 0)$.
\begin{equation}
T_{d}0=mgy_0
\end{equation}
This is how I obtain the transfer function:
\begin{equation}
s\Delta X_{1}(s) = \Delta X_{2}(s)
\end{equation}
\begin{equation}
s\Delta X_{2}(s) = \Delta X_{1}(s)\Delta X_{4}(s) – g)
\end{equation}
\begin{equation}
s\Delta X_{3}(s) = \Delta X_{4}
\end{equation}
\begin{equation}
s\Delta X_{4}(s) = \frac{\Delta X_{5}(s)}{m\Delta X_{1}^2} – g\frac{1}{\Delta X_{1}(s)} – \frac{2\Delta X_{1}(s)\Delta X_{4}(s)}{\Delta X_{1}(s)}
\end{equation}
\begin{equation}
s\Delta X_{5}=\frac{1}{\tau}(kmu(t) – \Delta X_{5}(s))
\end{equation}
\begin{equation}
\Delta Y(s) = \Delta X_{1}(s)
\end{equation}
\begin{equation}
\Delta X_{5}(s) = \frac{kmU(s)}{s\tau + 1}
\end{equation}
\begin{equation}
\frac{K_{m}}{T_{d}(1 + \tau s)}
\end{equation}
Best Answer
Your system has no equilibrium point, so a transfer function from a linearization might not represent the actual system behavior even when the system is arbitrarily close to your point of linearization.
Note that you divide by $y$, so we need $y\neq 0$. However, this means that $\dot{v}=0$ only if $y=g/\omega^2$. Since you divide by $\omega^2$, it follows that $\omega\neq 0$ is necessary. However, $\dot{\varphi}=0$ if and only if $\omega=0$, a contradiction. So there is no operating point that satisfies the requirement of an equilibrium, namely that $\dot{x}=0$.
Note that this argument assumes $g\neq 0$. I guess $g$ stands for gravity of earth, so this assumption should be justified. If this is not true and in fact $g=0$ then you have an equilbrium at $v=0,\omega=0,T_d=0,u=0,\varphi\in R$ and $y\in R\setminus\{0\}$.
However, the transfer function in this case is $G(s)=0$ (independently of the choice of $\varphi$ and $y$), so not very useful either.