예제 제어가능 표준형

제어가능 표준형

• $G(s) = \dfrac{5s+4}{s^3 + 2s^2 + 3s +1}$ 제어가능 표준형으로 바꾸면?

해]

$$G(s)= \dfrac{Y(s)}{U(s)}= \dfrac{5s+4}{s^3 + 2s^2 + 3s +1} \dfrac{z(s)}{z(s)}$$

$$\begin{aligned} &Y(s) = (5s+4)Z(s) \\ &U(s) = (s^3 + 2s^2 + 3s +1)Z(s) \\ &u(t) = z^{(3)}(t) + 2z^{(2)}(t) + 3z^{(1)}(t)s +z(t) \\ &x_1(t) = z(t) \\ &x_2(t) = \dot{x_1(t)} = z^{(1)}(t)\\ &x_3(t) = \dot{x_{2}(t)} = z^{(2)}(t)\\ &\dot{x_{3}(t)} = u(t)-(2z^{(2)}(t) + 3z^{(1)}(t)s +z(t))\\ \end{aligned}$$

$$\begin{aligned}\dot{x_1(t)} = 0x_1(t)+ x_2(t)+0x_3(t) \\ \dot{x_{2}(t)} = 0x_1(t)+ 0x_2(t)+x_3(t) \\ \dot{x_{3}(t)} = -x_1(t) -3 x_2(t)-2x_3(t) + u(t)\end{aligned}$$

$$\begin{bmatrix} \dot{x_1(t)} \\ \dot{x_{2}(t)}\\ \dot{x_{3}(t)} \end{bmatrix} = \begin{bmatrix} 0 & 1&0\\0&0&1\\ -1&-3&-2 \end{bmatrix} \begin{bmatrix} {x_1(t)} \\ {x_{2}(t)}\\ {x_{3}(t)} \end{bmatrix} + \begin{bmatrix} 0 \\ 0\\ 1 \end{bmatrix}u(t)$$

$$y(t) = 5 \dot{z(t)} + 4z(t)$$

$$\begin{bmatrix} y(t) \end{bmatrix} = \begin{bmatrix} 4& 5&0 \end{bmatrix} \begin{bmatrix} {x_1(t)} \\ {x_{2}(t)}\\ {x_{3}(t)} \end{bmatrix} + \begin{bmatrix} 0 \end{bmatrix} u(t)$$