Transfer function, poles and zeros from 3×3 state model

For $\dot{x} = Ax + Bu$, $y = Cx + Du$:

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 0 & 1 & 0 \\ -1 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$$

Step 1: $sI - A$ and determinant

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 1 & s+1 & 0 \\ -1 & 0 & s \end{bmatrix}$$

Expanding along the third column:

$$|sI - A| = s\begin{vmatrix} s & -1 \\ 1 & s+1 \end{vmatrix} = s\left[s(s+1) + 1\right] = s^3 + s^2 + s$$

Step 2: Needed adjoint elements

$C$ picks row 3 and $B$ picks column 2, so only $\operatorname{adj}(sI-A)_{32}$ is needed. It is the cofactor of element (2,3):

$$\operatorname{adj}_{32} = (-1)^{2+3}\begin{vmatrix} s & -1 \\ -1 & 0 \end{vmatrix} = -\left[(s)(0) - (-1)(-1)\right] = 1$$

(The full column $\operatorname{adj}(sI-A)B = [\,s,\; s^2,\; 1\,]^T$.)

Step 3: Transfer function

$$\frac{Y(s)}{U(s)} = \frac{1}{s^3 + s^2 + s} = \frac{1}{s(s^2 + s + 1)}$$

Poles and zeros

• Poles: $s(s^2 + s + 1) = 0$:

$$s = 0, \qquad s = \frac{-1 \pm \sqrt{1-4}}{2} = -0.5 \pm j0.866$$

• Zeros: the numerator is a constant, so there are no finite zeros (three zeros at infinity).

The pole at the origin makes the system marginally stable (type 1); the complex pair has $\omega_n = 1$ rad/s and $\zeta = 0.5$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s(s^2+s+1)}$; poles $0,\ -0.5 \pm j0.866$; no finite zeros.