상태변수 방정식

Problem 예제.

Multiple Choice • Views 344 • Comments 0 • Last Updated at 1 month ago  
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  • 상태변수방정식

그림과 같은 회로도에서 상태 변수를 각각, x1(t)=ec(t),x2(t)=i1(t),x3(t)=i2(t)x_1(t)=e_c(t), x_2(t)=i_1(t), x_3(t)=i_2(t)로 잡았을 때 벡터 행렬로 나타낸 상태 방정식 X˙=AX+BU \dot{X}=AX+BU에서 A행렬은 무엇인가?

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1

[0−1C−1C−1L1−RL101L200] \begin{bmatrix} 0 & -\dfrac{1}{C} & -\dfrac{1}{C} \\ -\dfrac{1}{L_1} & -\dfrac{R}{L_1} & 0 \\ \dfrac{1}{L_2} & 0 & 0 \end{bmatrix}

2

[01C1C1L1RL101L200] \begin{bmatrix} 0 & \dfrac{1}{C} & \dfrac{1}{C} \\ \dfrac{1}{L_1} & \dfrac{R}{L_1} & 0 \\ \dfrac{1}{L_2} & 0 & 0 \end{bmatrix}

3

[01C−1C−1L1−RL101L200] \begin{bmatrix} 0 & \dfrac{1}{C} & -\dfrac{1}{C} \\ -\dfrac{1}{L_1} & -\dfrac{R}{L_1} & 0 \\ \dfrac{1}{L_2} & 0 & 0 \end{bmatrix}

4

[0−1C−1C−1L1−RL10−1L200] \begin{bmatrix} 0 & -\dfrac{1}{C} & -\dfrac{1}{C} \\ -\dfrac{1}{L_1} & -\dfrac{R}{L_1} & 0 \\ -\dfrac{1}{L_2} & 0 & 0 \end{bmatrix}

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