Transfer function from state equations, output x1

Step 1: Write the state model

$$\dot{x}_1 = -3x_1 + x_2, \qquad \dot{x}_2 = -2x_1 + u, \qquad y = x_1$$ $$A = \begin{bmatrix} -3 & 1 \\ -2 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix},\quad D = 0$$

Step 2: Transfer function $C(sI-A)^{-1}B$

$$sI - A = \begin{bmatrix} s+3 & -1 \\ 2 & s \end{bmatrix},\qquad |sI - A| = s(s+3) + 2 = s^2 + 3s + 2$$ $$(sI-A)^{-1} = \frac{1}{s^2+3s+2}\begin{bmatrix} s & 1 \\ -2 & s+3 \end{bmatrix}$$ $$(sI-A)^{-1}B = \frac{1}{s^2+3s+2}\begin{bmatrix} 1 \\ s+3 \end{bmatrix}, \qquad \frac{Y(s)}{U(s)} = \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+3s+2}\begin{bmatrix} 1 \\ s+3 \end{bmatrix}$$ $$\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 3s + 2} = \frac{1}{(s+1)(s+2)}$$

Check by direct Laplace transform (zero initial conditions)

$$\begin{aligned} sX_1 &= -3X_1 + X_2 \;\Rightarrow\; X_2 = (s+3)X_1 \\ sX_2 &= -2X_1 + U \;\Rightarrow\; s(s+3)X_1 + 2X_1 = U \\ \frac{Y}{U} &= \frac{X_1}{U} = \frac{1}{s^2+3s+2} \end{aligned}$$

Poles at $s = -1, -2$: the system is stable and overdamped.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s^2+3s+2}$.