EP1.23. Relación entre la función de transferencia y la ecuación de estado

Dadas las siguientes ecuaciones de estado, (i) obtener analíticamente la función de transferencia o matriz de funciones de transferencia (si no se especifica la salida, suponer que es todo el estado) y verificar los resultados con las funciones ss y tf de MATLAB; (ii) obtener de nuevo la ecuación de estado a partir de la función de transferencia con los comandos anteriores (aunque las ecuaciones de estado tienen diferente forma, son equivalentes); (iii) simular con MATLAB los tres modelos anteriores y mostrar que dan el mismo resultado.

A. Caso de tiempo continuo

1.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x},\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 1\\\end{array} \right]  $
2.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u,u=1,\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 1\\\end{array} \right]  $
3.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u,u=1,\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ -1\\\end{array} \right]  $
4.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u,u=\delta (t-2),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ -1\\\end{array} \right]  $
5.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 0&  1\\ -1&  0\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 0\\ 1\\\end{array} \right] u,u=1,\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right]  $
6.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 0&  1\\ -1&  0\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 0\\ 1\\\end{array} \right] u,u=\delta (t),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right]  $
7.  $\dot{\mathbf{x}}=\left[ \begin{matrix} -0.0197&  0\\ 0.0178&  -0.0129\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 0.0263\\ 0\\\end{array} \right] u,u=1,\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ -1\\\end{array} \right]  $
8.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 0&  1\\ 0&  0\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 0\\ 1\\\end{array} \right] u,u=1,\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 1\\\end{array} \right]  $
9.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u(t-2),u=u_s(t),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right]  $
10.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u(t-0.6),u=u_s(t),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right]  $
11.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 2&  3\\ 2&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ -1\\\end{array} \right] u(t-1.5),u=u_s(t),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right]  $
12.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 1&  0\\ 1&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] u(t-4),u=u_s(t),\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 1\\\end{array} \right]  $
13.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 1&  0\\ 1&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] u(t-0.8),u=u_s(t),\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 1\\\end{array} \right]  $
14.  $\dot{\mathbf{x}}=\left[ \begin{matrix} 1&  0\\ 1&  1\\\end{matrix} \right] \mathbf{x}+\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] u(t-0.8),u=\sin t,\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 1\\\end{array} \right]  $

B. Caso de tiempo discreto

1. $\mathbf{x}(k+1)=\left[ \begin{matrix} 1&  0.5\\ 0&  1\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 0.125\\ 0.5\\\end{array} \right] u(k),\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 0\\\end{array} \right] $
2. $\mathbf{x}(k+1)=\left[ \begin{matrix} 1&  0.086\\ -0.172&  0.733\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 0.00453\\ 0.0861\\\end{array} \right] u(k),\mathbf{x}(0)=\left[ \begin{array}{c} 0\\ 0\\\end{array} \right] $
3. $\mathbf{x}(k+1)=\left[ \begin{matrix} 0.819&  0\\ 0.234&  0.741\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 0.181\\ 0.025\\\end{array} \right] u(k),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
4. $\mathbf{x}(k+1)=\left[ \begin{matrix} e^{-T}&  0\\ 1-e^{-T}&  1\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 1-e^{-T}\\ T-1+e^{-T}\\\end{array} \right] u(k),T=\{0.1,1\},\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
5. $\mathbf{x}(k+1)=\left[ \begin{matrix} 0.5&  1\\ 0&  -0.2\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] u(k-1),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
6. $\mathbf{x}(k+1)=\left[ \begin{matrix} 0.5&  1\\ 0&  -0.2\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] u(k-2),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
7. $\mathbf{x}(k+1)=\left[ \begin{matrix} -0.5&  0\\ 0&  0.3\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{matrix} 1&  0\\ 0&  1\\\end{matrix} \right] \mathbf{u}(k),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
8. $\mathbf{x}(k+1)=\left[ \begin{matrix} -0.5&  0\\ 0&  0.3\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{matrix} 1&  0\\ 0&  1\\\end{matrix} \right] \mathbf{u}(k-3),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\\end{array} \right] $
9. $\mathbf{x}(k+1)=\left[ \begin{matrix} -0.2&  1&  0\\ 0&  -0.2&  0\\ 0&  0&  0.5\\\end{matrix} \right] \mathbf{x}(k)+\left[ \begin{array}{c} 0\\ 1\\ 1\\\end{array} \right] u(k-1),\mathbf{x}(0)=\left[ \begin{array}{c} 1\\ 0\\ 1\\\end{array} \right] $