Example 제어가능 표준형

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  • 제어가능 표준형
  • 상태변수방정식

제어가능 표준형

  • G(s)=5s+4s3+2s2+3s+1 G(s) = \dfrac{5s+4}{s^3 + 2s^2 + 3s +1} 제어가능 표준형으로 바꾸면?

해]

G(s)=Y(s)U(s)=5s+4s3+2s2+3s+1z(s)z(s) G(s)= \dfrac{Y(s)}{U(s)}= \dfrac{5s+4}{s^3 + 2s^2 + 3s +1} \dfrac{z(s)}{z(s)}

Y(s)=(5s+4)Z(s)U(s)=(s3+2s2+3s+1)Z(s)u(t)=z(3)(t)+2z(2)(t)+3z(1)(t)s+z(t)x1(t)=z(t)x2(t)=x1(t)˙=z(1)(t)x3(t)=x2(t)˙=z(2)(t)x3(t)˙=u(t)(2z(2)(t)+3z(1)(t)s+z(t)) \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}

x1(t)˙=0x1(t)+x2(t)+0x3(t)x2(t)˙=0x1(t)+0x2(t)+x3(t)x3(t)˙=x1(t)3x2(t)2x3(t)+u(t) \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}

[x1(t)˙x2(t)˙x3(t)˙]=[010001132][x1(t)x2(t)x3(t)]+[001]u(t) \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)=5z(t)˙+4z(t) y(t) = 5 \dot{z(t)} + 4z(t)

[y(t)]=[450][x1(t)x2(t)x3(t)]+[0]u(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)

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