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 (-3dBpoint)={\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}}}
![{\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])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e58e1097d8562a5e6a095386e2ada7b0393c040)
![{\displaystyle w[n+1]=w[n]-K_{I}\sum _{j=1}^{N}({\frac {1}{N}}x_{j}[n])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3658661f0dc61ceb76cf33f6fb3506b7f6b8b489)
