State model of two-source RLC circuit

Choose the inductor currents and the capacitor voltage as states. The left loop current flows through 2 Ω and 2 H (in series); the right loop current flows through 1 H and 3 Ω (in series); the capacitor carries the difference.

• $x_1 = i_1$: current in the 2 H inductor (left loop, clockwise)
• $x_2 = i_2$: current in the 1 H inductor (right loop, from A towards $V_2$)
• $x_3 = v_C = v_A - v_B$: voltage on the 2 F capacitor

     2 ohm   A    1 H
  +--/\/\/--+---UUUU---+
  |   i1 -> |  i2 ->   |
 (V1)      === 2F     (V2)
  |         |          |
  +--UUUU---+--/\/\/---+
      2 H   B   3 ohm

Left loop (KVL)

$$V_1 = 2i_1 + v_C + 2\frac{di_1}{dt} \;\Rightarrow\; \dot{x}_1 = -x_1 - 0.5x_3 + 0.5V_1$$

Right loop (KVL)

From A through 1 H, $V_2$ and 3 Ω back to B:

$$v_C = 1\cdot\frac{di_2}{dt} + V_2 + 3i_2 \;\Rightarrow\; \dot{x}_2 = -3x_2 + x_3 - V_2$$

Capacitor (KCL at A)

$$2\frac{dv_C}{dt} = i_1 - i_2 \;\Rightarrow\; \dot{x}_3 = 0.5x_1 - 0.5x_2$$

State model

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & -0.5 \\ 0 & -3 & 1 \\ 0.5 & -0.5 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0.5 & 0 \\ 0 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

Output: no output is marked, so the capacitor voltage is taken as output:

$$y = v_C = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

(For the current in $V_2$ as output, use $C = [0 \;\; 1 \;\; 0]$.)