Transfer function from state model [0 1; −5 −6]

For a state model $\dot{x} = Ax + Bu$, $y = Cx + Du$, the transfer function is

$$T(s) = \frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D$$

Here

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

Step 1: $(sI - A)$ and its determinant

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

Step 2: Inverse

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

Step 3: Transfer function

$$\begin{aligned} (sI-A)^{-1}B &= \frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} \\ T(s) &= \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} = \frac{1}{s^2+6s+5} \end{aligned}$$ $$\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 6s + 5} = \frac{1}{(s+1)(s+5)}$$

Step 4: Stability

Characteristic equation: $|sI - A| = s^2 + 6s + 5 = 0$, so the eigenvalues of $A$ (poles) are

$$s = -1, \qquad s = -5$$

Both are real and negative (in the left half of the s-plane), so the system is stable. It is overdamped: comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$, $\omega_n = \sqrt5 = 2.24$ rad/s and $\zeta = 6/(2\sqrt5) = 1.34$. Its unit step response settles to $1/5 = 0.2$ without oscillation.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s^2+6s+5}$; poles at $-1$ and $-5$, so the system is stable.