상태변수 방정식

Problem 예제.

Multiple Choice • Views 346 • Comments 0 • Last Updated at 1 month ago  
  • 안정도
  • 상태변수방정식

다음 계통의 상태 방정식을 유도하면?

x...+5x¨+10x˙+5x=2u\overset{...} x +5 \ddot x + 10 \dot x + 5 x = 2u

(단, 상태변수를 x1=x,x2=x˙,x3=x¨x_1 = x, x_2 = \dot x, x_3 = \ddot x로 놓았다.)

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1

[x˙1x˙2x˙3]=[−501−1010−500][x1x2x3]+[020]u\begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{bmatrix} = \begin{bmatrix} -5 & 0 & 1 \\ -10 & 1 & 0 \\ -5 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} u

2

[x˙1x˙2x˙3]=[010001−5−10−5][x1x2x3]+[200]u\begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -10 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} u

3

[x˙1x˙2x˙3]=[−500−1010−501][x1x2x3]+[200]u\begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{bmatrix} = \begin{bmatrix} -5 & 0 & 0 \\ -10 & 1 & 0 \\ -5 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} u

4

[x˙1x˙2x˙3]=[010001−5−10−5][x1x2x3]+[002]u\begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -10 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} u

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