State space model of circuit with current source

For an electric circuit the natural state variables are the capacitor voltages and inductor currents, since they describe the stored energy. Here the states are $v_{C1}$, $i_L$ and $v_{C2} = v_o$; the inputs are $V_i$ and $I_s$.

Circuit

Vi --[4 ohm]--+--[2 H, iL ->]--+-------+
              |                |       |
           0.25 F            0.5 F   1 ohm   Is (up into
              |                |       |     node Vc2)
GND ----------+----------------+-------+

State variables and inputs

$$x_1 = v_{C1},\quad x_2 = i_L,\quad x_3 = v_{C2} = v_o, \qquad u_1 = V_i,\quad u_2 = I_s$$

KCL at node $V_{C1}$

Current in through 4 Ω = capacitor current + inductor current:

$$\frac{V_i - x_1}{4} = 0.25\,\dot{x}_1 + x_2 \;\Rightarrow\; \dot{x}_1 = -x_1 - 4x_2 + V_i$$

KVL around the inductor

$$2\,\dot{x}_2 = x_1 - x_3 \;\Rightarrow\; \dot{x}_2 = 0.5x_1 - 0.5x_3$$

KCL at node $V_{C2}$

Inductor current + source current = capacitor current + resistor current:

$$x_2 + I_s = 0.5\,\dot{x}_3 + \frac{x_3}{1} \;\Rightarrow\; \dot{x}_3 = 2x_2 - 2x_3 + 2I_s$$

State equation

$$\begin{bmatrix}\dot{v}_{C1}\\ \dot{i}_L\\ \dot{v}_{C2}\end{bmatrix} = \begin{bmatrix}-1 & -4 & 0\\ 0.5 & 0 & -0.5\\ 0 & 2 & -2\end{bmatrix} \begin{bmatrix}v_{C1}\\ i_L\\ v_{C2}\end{bmatrix} + \begin{bmatrix}1 & 0\\ 0 & 0\\ 0 & 2\end{bmatrix} \begin{bmatrix}V_i\\ I_s\end{bmatrix}$$

Output equation

$$v_o = \begin{bmatrix}0 & 0 & 1\end{bmatrix}\begin{bmatrix}v_{C1}\\ i_L\\ v_{C2}\end{bmatrix} + \begin{bmatrix}0 & 0\end{bmatrix}\begin{bmatrix}V_i\\ I_s\end{bmatrix}$$

Check

• Characteristic equation: $|sI - A| = s^3 + 3s^2 + 5s + 5$.
• $V_o/V_i = \dfrac{1}{s^3 + 3s^2 + 5s + 5}$. At DC this is $1/5$, which matches the resistive divider $1/(4+1)$.
• $V_o/I_s = \dfrac{2(s^2 + s + 2)}{s^3 + 3s^2 + 5s + 5}$. At DC this is $0.8\ \Omega$, which matches $1 \parallel 4 = 0.8\ \Omega$.

Answer: $\dot{x} = Ax + Bu$, $v_o = Cx$ with $A = \begin{bmatrix}-1 & -4 & 0\\ 0.5 & 0 & -0.5\\ 0 & 2 & -2\end{bmatrix}$, $B = \begin{bmatrix}1 & 0\\ 0 & 0\\ 0 & 2\end{bmatrix}$, $C = [0\;\;0\;\;1]$, $D = 0$. (If the current source points the other way, the sign of the $I_s$ column in $B$ changes.)