상태변수 방정식
Classroom

Problem 예제.

Views 130 • Comments 0 • Last Updated at 1 year ago • Full screen  
  • 시스템 응답
  • 상태변수방정식

A=[01−5−2],B=[01]A= \begin{bmatrix}0& 1\\ -5 &-2 \end{bmatrix}, B = \begin{bmatrix} 0\\ 1 \end{bmatrix}인 상태방정식 dxdt=Ax+Br \dfrac{dx}{dt}=Ax+Br에서 상태 천이 행렬 ϕ(t)는?

Subscribe to see the correct answer.
1

[e−t(cos⁡2t+12sin⁡2t)12e−tsin⁡2t−52e−tsin⁡2te−t(cos⁡2t+12sin⁡2t)]\begin{bmatrix} e^{-t}(\cos 2t + \dfrac{1}{2} \sin 2t ) & \dfrac{1}{2}e^{-t} \sin 2t \\ - \dfrac{5}{2}e^{-t} \sin 2t & e^{-t} ( \cos 2t + \dfrac{1}{2} \sin 2t ) \end{bmatrix}

2

[e−t(cos⁡2t−12sin⁡2t)−52e−tsin⁡2t12e−tsin⁡2te−t(cos⁡2t+12sin⁡2t)]\begin{bmatrix} e^{-t}(\cos 2t - \dfrac{1}{2} \sin 2t ) &- \dfrac{5}{2}e^{-t} \sin 2t \\ \dfrac{1}{2}e^{-t} \sin 2t & e^{-t} ( \cos 2t + \dfrac{1}{2} \sin 2t ) \end{bmatrix}

3

[e−t(cos⁡2t+12sin⁡2t)−52e−tsin⁡2t12e−tsin⁡2te−t(cos⁡2t−12sin⁡2t)]\begin{bmatrix} e^{-t}(\cos 2t + \dfrac{1}{2} \sin 2t ) & -\dfrac{5}{2}e^{-t} \sin 2t \\ \dfrac{1}{2}e^{-t} \sin 2t & e^{-t} ( \cos 2t - \dfrac{1}{2} \sin 2t ) \end{bmatrix}

4

[e−t(cos⁡2t+12sin⁡2t)12e−tsin⁡2t−52e−tsin⁡2te−t(cos⁡2t−12sin⁡2t)]\begin{bmatrix} e^{-t}(\cos 2t + \dfrac{1}{2} \sin 2t ) & \dfrac{1}{2}e^{-t} \sin 2t \\ - \dfrac{5}{2}e^{-t} \sin 2t & e^{-t} ( \cos 2t - \dfrac{1}{2} \sin 2t ) \end{bmatrix}

previous article
next article