State equations of current-source RLC circuit

All branches are in parallel across the current source, so they share one voltage $v$ (the capacitor voltage).

State variables:

• $x_1 = i_1$: current in the 1 H inductor
• $x_2 = v$: voltage across the 1 F capacitor
• $x_3 = i_2$: current in the 1.5 Ω – 2 H branch

   +------+------+--------+
   |      |      |        |
  (I)^   1H     === 1F   1.5 ohm
   |      |i1    |        |
   |      |      |       2H  i2
   |      |      |        |
   +------+------+--------+

Equations

1 H inductor: $\;1\cdot\dfrac{di_1}{dt} = v \Rightarrow \dot{x}_1 = x_2$

KCL at the top node: $\;I = i_1 + 1\cdot\dfrac{dv}{dt} + i_2 \Rightarrow \dot{x}_2 = -x_1 - x_3 + I$

R–L branch: $\;v = 1.5i_2 + 2\dfrac{di_2}{dt} \Rightarrow \dot{x}_3 = 0.5x_2 - 0.75x_3$

State equation

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & -1 \\ 0 & 0.5 & -0.75 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}I$$

Output equation

Voltage across the 2 H inductor:

$$v_{L2} = 2\frac{di_2}{dt} = v - 1.5i_2 = x_2 - 1.5x_3$$ $$y = \begin{bmatrix} 0 & 1 & -1.5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + [0]\,I$$

Check: this model gives $\dfrac{V_{L2}(s)}{I(s)} = \dfrac{4s^2}{4s^3 + 3s^2 + 6s + 3}$, the same as the impedance method.