How can i linearize the following system ?
State equations:
$ x'_1 = (x_1-2)x_2 - 2x_2 $
$ x'_2 =-(x_1-2)^2 + x_2 + u -1 $
Output equation:
$ y = x_1 $
assume the eq. point : $ \tilde x=[2 ,0]^T$ and $\tilde u=1$
I know the teory, but pratically, how can i find the linearized equation ?
\begin{align*} A &= \frac{\partial f}{\partial x} \Big\vert_{x=x_e,u=0} = \renewcommand\arraystretch{1.5} \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix} x_2 & x_1-4 \\ -2(x_1-2) & 1 \end{bmatrix}_{x=(2,0)} = \begin{bmatrix} 0 & -2 \\ 0 & 1 \end{bmatrix} \\ B &= \frac{\partial f}{\partial u} \Big\vert_{x=x_e} = \renewcommand\arraystretch{1.5} \begin{bmatrix} \frac{\partial f_1}{\partial u} \\ \frac{\partial f_2}{\partial u} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \end{align*} The linearized system is then \begin{align} \dot{x}&=Ax+Bu \\ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} &= \begin{bmatrix} 0 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} u \end{align}