Modeling magnetic levitation system with current as the input signal
Well, if your system is described by
$$ m\frac{d^2H}{dt^2} = mg-K\frac{I^2}{H^2}.$$
With the state
$$ x := \begin{bmatrix}H \\ \dot{H}\end{bmatrix}$$
we have that
$$ \dot{x} = \begin{bmatrix}\dot{H} \\ g-K\frac{I^2}{mH^2}\end{bmatrix}.$$
For any \$h&gt;0\$ the point \$x_{eq} = \begin{bmatrix} h \\ 0\end{bmatrix}\$ will be an equillibrium, since \$\forall h&gt;0 \$ we can have \$ I_{eq} := \sqrt{\frac{mgh^2}{K}}\$ to make \$\dot{x}=0\$. So we have the following description for small perturbations \$z\$ around that equilibrium, and small inputs \$v\$ that are added to \$I\$ as
$$ I := \sqrt{\frac{m(gh^2+v)}{K}}, $$
then we will have
$$ \dot{z} = \begin{bmatrix} z_2 \\ g-K \frac{m(gh^2+v)}{K} \frac{1}{m(h+z_1)^2}\end{bmatrix},$$
$$ \dot{z} = \begin{bmatrix} z_2 \\ g- \frac{gh^2+v}{(h+z_1)^2}\end{bmatrix}.$$
Now, linearize around that:
$$ \begin{align} A &amp;:=  \frac{\partial \dot{z}}{\partial z} \Big\vert_{z_1=0,\, v = 0 } = \begin{bmatrix} 0 &amp; 1 \\ \frac{2 \left(g h^{2} + v\right)}{\left(h + z_{1}\right)^{3}} &amp; 0\end{bmatrix}\Big\vert_{z_1=0,\, v = 0 } = \begin{bmatrix} 0 &amp; 1 \\ \frac{2 g h^{2}}{h^{3}} &amp; 0  \end{bmatrix}= \begin{bmatrix} 0 &amp; 1 \\ \frac{2 g}{h} &amp; 0  \end{bmatrix}.  \\ B&amp;:= \frac{\partial \dot{z}}{\partial v}\Big\vert_{z_1=0,\, v = 0 }  = \begin{bmatrix} 0 \\ -\frac{1}{\left(h + z_{1}\right)^{2}} \end{bmatrix}\Big\vert_{z_1=0,\, v = 0 } = \begin{bmatrix} 0 \\ -\frac{1}{h^2} \end{bmatrix}. \end{align}$$
The output matrix \$C\$ will depend on your sensors, but I will assume that you are measuring the height. And changes to the input current will not affect instantaneously the output, so
$$ \begin{align} C&amp;:= \begin{bmatrix}  1 &amp; 0 \end{bmatrix},\\ D&amp;:= 0. \end{align}$$
And that way you get the linearized system
$$ \begin{align} \dot{z} &amp;= Az +Bv, \\ Y &amp;= Cz. \end{align} $$