Transfer function from symbolic mass-spring-damper state model

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

with

$$A = \begin{bmatrix} 0 & 1 \\ -\dfrac{K}{M} & -\dfrac{B}{M} \end{bmatrix},\quad B_m = \begin{bmatrix} 0 \\ \dfrac{1}{M} \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}$$

(The input matrix is written $B_m$ to avoid confusion with the damping coefficient $B$.)

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

$$sI - A = \begin{bmatrix} s & -1 \\ \dfrac{K}{M} & s + \dfrac{B}{M} \end{bmatrix}$$ $$|sI - A| = s\left(s + \frac{B}{M}\right) + \frac{K}{M} = s^2 + \frac{B}{M}s + \frac{K}{M}$$

Step 2: Inverse

$$(sI - A)^{-1} = \frac{1}{s^2 + \frac{B}{M}s + \frac{K}{M}}\begin{bmatrix} s + \dfrac{B}{M} & 1 \\ -\dfrac{K}{M} & s \end{bmatrix}$$

Step 3: Transfer function

$$(sI-A)^{-1}B_m = \frac{1}{s^2 + \frac{B}{M}s + \frac{K}{M}}\begin{bmatrix} \dfrac{1}{M} \\ \dfrac{s}{M} \end{bmatrix}$$ $$\frac{Y(s)}{U(s)} = \begin{bmatrix} 1 & 0 \end{bmatrix}(sI-A)^{-1}B_m = \frac{1/M}{s^2 + \frac{B}{M}s + \frac{K}{M}} = \frac{1}{Ms^2 + Bs + K}$$

This is the transfer function of the mass–spring–damper system $M\ddot{y} + B\dot{y} + Ky = u$, with $\omega_n = \sqrt{K/M}$ and $\zeta = \dfrac{B}{2\sqrt{KM}}$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{Ms^2 + Bs + K}$.