This is the sandbox, so I'm assuming I can edit this as I please.
Test0
( − 3 d B p o i n t ) = 1 2 π ( R 1 ∗ R 2 ∗ C 1 ∗ C 2 ) 1 2 {\displaystyle (-3dB point) = \frac{1}{2\pi}(R1*R2*C1*C2)^\frac{1}{2} }
x [ n + 1 ] = x [ n ] + γ ( u [ n ] − x [ n ] ) − K P ∑ j = 1 N ( 1 N x j [ n ] ) + K I ∑ j = 1 N ( 1 N w j [ n ] ) {\displaystyle x[n+1] = x[n] + \gamma(u[n]-x[n]) - K_P\sum_{j=1}^N(\frac{1}{N}x_j[n]) + K_I\sum_{j=1}^N(\frac{1}{N}w_j[n]) }
w [ n + 1 ] = w [ n ] − K I ∑ j = 1 N ( 1 N x j [ n ] ) {\displaystyle w[n+1] = w[n] - K_I\sum_{j=1}^N(\frac{1}{N}x_j[n]) }
[ u ˙ x u ˙ y ] = [ 1 0 2 ( R x − x M x ) ( R y − x M y ) 0 0 1 0 ( R x − x M x ) 2 ( R y − x M y ) ] [ k M x 0 0 0 0 0 k M y 0 0 0 0 0 k M x x 0 0 0 0 0 k M y y 0 0 0 0 0 k M x y ] [ g M x − x M x g M y − x M y g M x x − x M x x g M y y − x M y y g M x y − x M x y ] {\displaystyle \begin{bmatrix} \dot u_x \\ \dot u_y \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 2(R_x-x_{Mx}) & (R_y-x_{My}) & 0 \\ 0 & 1 & 0 & (R_x-x_{Mx}) & 2(R_y-x_{My})\\ \end{bmatrix} \begin{bmatrix} k_{Mx} & 0 & 0 & 0 & 0 \\ 0 & k_{My} & 0 & 0 & 0 \\ 0 & 0 & k_{Mxx} & 0 & 0 \\ 0 & 0 & 0 & k_{Myy} & 0 \\ 0 & 0 & 0 & 0 & k_{Mxy} \\ \end{bmatrix} \begin{bmatrix} g_{Mx}-x_{Mx} \\ g_{My}-x_{My}\\ g_{Mxx}-x_{Mxx} \\ g_{Myy}-x_{Myy} \\ g_{Mxy}-x_{Mxy} \\ \end{bmatrix} }