Example 관측가능 표준형

관측가능형

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

해]

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

• 양변을 크로스로 곱해준다

$$(s^3 + 2s^2 + 3s +1)Y(s) = (5s+4)U(s)$$

$$(s^3 + 2s^2 + 3s)Y(s) + Y(s) = 5sU(s)+4U(s)$$

$$(s^3 + 2s^2 + 3s)Y(s) - 5sU(s) = -Y(s) +4U(s) = sX_1(s)$$

$$(s^2 + 2s + 3)Y(s) - 5U(s) = X_1(s)$$

$$(s^2 + 2s )Y(s) = X_1(s)- 3Y(s) + 5U(s) = sX_2(s)$$

$$(s + 2)Y(s) = X_2(s)$$

$$sY(s) = X_2(s) - 2Y(s) = sX_3(s)$$

$$Y(s) = X_3(s)$$

• 마지막 식에서 얻은 결과 $X_3(s) = Y(s)$를 위 식들에 대입한다.

$$\begin{aligned} sX_1(s) &= -Y(s) +4U(s) \\ sX_2(s) &= X_1(s)- 3Y(s) + 5U(s) \\ sX_3(s) &= X_2(s) - 2Y(s) \\ \\ Y(s) &= X_3(s)\end{aligned}$$

$$\begin{bmatrix} \dot{x_1(t)} \\ \dot{x_{2}(t)}\\ \dot{x_{3}(t)} \end{bmatrix} = \begin{bmatrix} 0 & 0&-1\\1&0&-3\\ 0&1&-2 \end{bmatrix} \begin{bmatrix} {x_1(t)} \\ {x_{2}(t)}\\ {x_{3}(t)} \end{bmatrix} + \begin{bmatrix} 4 \\ 5\\ 0\end{bmatrix}u(t)$$

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