Question 10.21: Stability Analysis in State Space a. Compute the poles of th...
Stability Analysis in State Space
a. Compute the poles of the system described by
\dot{\mathbf{x}}= \begin{bmatrix} 0 & 1 \\ 0 & -16.883 \end{bmatrix} \mathbf{x}+\begin{bmatrix} 0 \\ 3.778 \end{bmatrix}u
y=\begin{bmatrix} 1 & 0 \end{bmatrix}\mathbf{x}
b. Verify the results by converting the state-space representation to a transfer function and then identifying the poles of the transfer function.
a. The characteristic equation is
\left| s\mathbf{I} – \mathbf{A} \right| = \begin{vmatrix} s & -1 \\ 0 & s+16.883 \end{vmatrix} = s^2 + 16.883s = 0
which yields the poles s_1 = 0 and s_2 = −16.883.
b. As presented in Section 4.4, state-space equations for a single-input–singleoutput system can be converted to a transfer function using
G(s) =\mathbf{C}(s\mathbf{I} − \mathbf{A})^{−1}\mathbf{B} +D
Substituting the system matrices \mathbf{A, B, C}, and D, which has the value of 0 in this example, gives
G(s) = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} s & -1 \\ 0 & s+16.883 \end{bmatrix}^{-1}\begin{bmatrix} 0 \\ 3.778 \end{bmatrix} + 0 = \begin{bmatrix} 1 & 0 \end{bmatrix} \frac{\begin{bmatrix} s+16.883 & 1 \\ 0 & s \end{bmatrix}}{s^2 +16.883s}\begin{bmatrix} 0 \\ 3.778 \end{bmatrix}
= \frac{3.778}{s^2 +16.883s}
The characteristic equation is s^2 + 16.883s = 0, which yields the poles at 0 and −16.883. The results agree with the poles obtained in Part (a).