posted 1 years ago

\dot{\mathbf{x}}=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -12 & -20 & -9 \end{array}\right] \mathbf{x}+\left[\begin{array}{c} 0 \\ 0 \\ 0.4 \end{array}\right] u \quad y=\left[\begin{array}{lll} 1 & 0 & 0 \end{array}\right] \mathbf{x}

a. Use MATLAB to determine the eigenvalues.

b. Describe the free response of the output y(r) given an arbitrary initial state x(0).

c. Use MATLAB or Simulink to verify your answer in part (b). The initial state vector is

\mathbf{x}(0)=\left[\begin{array}{lll} x_{1}(0) & x_{2}(0) & x_{3}(0) \end{array}\right]^{T}=\left[\begin{array}{lll} 2 & -0.5 & 0 \end{array}\right]^{T}

posted 1 years ago

\dot{\mathbf{x}}=\left[\begin{array}{cc} -0.2 & -0.6 \\ 2 & -4 \end{array}\right] \mathbf{x}+\left[\begin{array}{c} 0 \\ 1.5 \end{array}\right] u \quad y=\left[\begin{array}{ll} 1 & 0 \end{array}\right] \mathbf{x}

a.Compute the eigenvalues "by hand."

b. Use MATLAB to verify your answer in part (a).

c. Describe the free response of the output y(t) given an arbitrary initial state x(0).

d. Use MATLAB or Simulink to verify your answer in part (c). The initial state vector is

\mathbf{x}(0)=\left[x_{1}(0) \quad x_{2}(0)\right]^{T}=\left[\begin{array}{ll} -2 & -1 \end{array}\right]^{T}