Poles and zeros from state model

Find the transfer function $C(sI-A)^{-1}B$; its denominator roots are the poles and numerator roots are the zeros.

$$A = \begin{bmatrix} -5 & -6 \\ 1 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 2 \end{bmatrix}$$

Step 1

$$sI - A = \begin{bmatrix} s+5 & 6 \\ -1 & s \end{bmatrix},\qquad |sI-A| = s(s+5) + 6 = s^2 + 5s + 6 = (s+2)(s+3)$$ $$\operatorname{adj}(sI-A) = \begin{bmatrix} s & -6 \\ 1 & s+5 \end{bmatrix},\qquad \operatorname{adj}(sI-A)B = \begin{bmatrix} s \\ 1 \end{bmatrix}$$

Step 2

$$C\,\operatorname{adj}(sI-A)B = s + 2$$ $$\frac{Y(s)}{U(s)} = \frac{s+2}{(s+2)(s+3)}$$

Poles and zeros

• Poles (eigenvalues of $A$): $s = -2$ and $s = -3$
• Zero: $s = -2$

The zero at $-2$ cancels the pole at $-2$, so the input–output transfer function reduces to $\dfrac{1}{s+3}$. The mode $e^{-2t}$ exists inside the system but does not appear at the output: the system is controllable ($|[B\;\; AB]| = 1 \ne 0$) but not observable ($\left|\begin{smallmatrix} C \\ CA \end{smallmatrix}\right| = 0$). Both poles are in the left half-plane, so the system is stable.