상태변수 방정식

Problem 예제.

Multiple Choice • Views 499 • Comments 0 • Last Updated at 1 month ago  
  • 시스템 응답
  • 상태변수방정식

다음 계통의 상태 천이 행렬 ϕ(t)를 구하면?

[X1˙X2˙]=[01−2−3][X1X2]\begin{bmatrix} \dot{X_1}\\ \dot{X_2} \end{bmatrix} = \begin{bmatrix} 0&1\\-2 & -3\end{bmatrix}\begin{bmatrix} X_1\\ X_2 \end{bmatrix}

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1

[2e−t−e2te−t−e2t−2e−t+2e2t−et+2e2t]\begin{bmatrix} 2e^{-t}-e^{2t} & e^{-t} - e^{2t} \\ -2e^{-t}+2e^{2t} & -e^{t} + 2e^{2t} \end{bmatrix}

2

[2e−t−e−2te−t−e−2t−2e−t+2e−2t−e−t+2e−2t]\begin{bmatrix} 2e^{-t}-e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t}+2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix}

3

[−2e−t+e−2t−e−t−e2t−2e−t−2e−2t−e−t−2e−2t]\begin{bmatrix} -2e^{-t}+e^{-2t} & -e^{-t} - e^{2t} \\ -2e^{-t}-2e^{-2t} & -e^{-t} - 2e^{-2t} \end{bmatrix}

4

[2et+e2t−et+e−2t2et−2e2tet−2e−2t]\begin{bmatrix} 2e^{t}+e^{2t} & -e^{t} + e^{-2t} \\ 2e^{t}-2e^{2t} & e^{t} - 2e^{-2t} \end{bmatrix}

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