State equations of two-source circuit with 2 F

Take the two inductor currents and the capacitor voltage as states (assumed directions shown):

• $x_1 = i_1$: current in the 2 H inductor, from $V_1$ towards node A
• $x_2 = i_2$: current in the 1 H inductor, from node A towards $V_2$
• $x_3 = v_C$: voltage on the 2 F capacitor (node A to common line)

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

Left loop (KVL)

$$V_1 = 1\cdot i_1 + 2\frac{di_1}{dt} + v_C \;\Rightarrow\; \dot{x}_1 = -0.5x_1 - 0.5x_3 + 0.5V_1$$

Right loop (KVL)

$$v_C = 1\cdot\frac{di_2}{dt} + 2i_2 + V_2 \;\Rightarrow\; \dot{x}_2 = -2x_2 + x_3 - V_2$$

KCL at node A

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

State equation

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} -0.5 & 0 & -0.5 \\ 0 & -2 & 1 \\ 0.5 & -0.5 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0.5 & 0 \\ 0 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

Output equation

No output is marked in the figure, so the capacitor voltage (node A voltage) is taken as the output:

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

If, for example, the voltage across the 2 Ω resistor is the output, then $y = 2x_2$, i.e. $C = [0 \;\; 2 \;\; 0]$.