--- title: "State space model of circuit with current source" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-circuit-4ohm-2h-is" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-circuit-4ohm-2h-is site: BCT Engineering --- # State space model of circuit with current source Control Systems · Chapter 8: State Space Analysis - Asked in: 2078 Kartik, 6 marks - Video: [State Space Analysis for Electrical Systems](https://www.youtube.com/watch?v=Lm86H82z7pE) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-circuit-4ohm-2h-is - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question Find state space representation of the system. [Figure: voltage source Vi in series with 4 Ω to node Vc1; 0.25 F capacitor from Vc1 to ground; 2 H inductor (current iL) from Vc1 to node Vc2; 0.5 F capacitor (voltage Vo) from Vc2 to ground; 1 Ω resistor from Vc2 to ground; current source is injecting into node Vc2 from ground] ## Short answer Choose states: vC1, iL, vC2=vo. Then state equation: dot{x} = A x + B u, with A = [[-1,-4,0],[0.5,0,-0.5],[0,2,-2]], B = [[1,0],[0,0],[0,2]], C = [0,0,1], D=0. Inputs: Vi and Is. ## Answer 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 ```text 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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the natural state variables for this circuit? A. Resistor voltages and source currents B. Capacitor voltages and inductor current C. Only the output voltage D. Source voltage and current Answer: B. Capacitor voltages and inductor currents describe stored energy, so they are the states: vC1, iL, vC2. (Key point: Choice of state variables) 2. Which KCL relation at node Vc1 gives x1'? A. (Vi - x1)/4 = 0.25 x1' + x2 B. (Vi - x1)/4 = x2 - 0.25 x1' C. Vi/4 = 0.25 x1' + x2 D. (Vi - x1)/4 = 0.25 x1' Answer: A. Current through 4 ohm splits into the capacitor and the inductor, giving x1' = -x1 - 4x2 + Vi. (Key point: KCL at node Vc1) 3. What is the state equation for the inductor current x2 = iL? A. x2' = 0.5 x1 - 0.5 x3 B. x2' = 2 x1 - 2 x3 C. x2' = -0.5 x1 + 0.5 x3 D. x2' = 0.5 x3 Answer: A. L diL/dt = vC1 - vC2 with L = 2 H, so x2' = 0.5x1 - 0.5x3. (Key point: Inductor equation) 4. Which is the equation for x3 = vC2? A. x3' = 2x2 - 2x3 + 2Is B. x3' = 0.5x2 - x3 + Is C. x3' = -2x2 - 2x3 + 2Is D. x3' = 2x1 - 2x3 Answer: A. x2 + Is = 0.5 x3' + x3/1, so x3' = 2x2 - 2x3 + 2Is. (Key point: Third state equation) 5. How is the result checked at DC? A. Vo/Vi = 1/5 matches the divider 1/(4+1) B. Vo/Vi = 1/4 matches the 4 ohm resistor C. Vo/Vi = 0 because capacitors short at DC D. Vo/Vi = 1 because the inductor shorts the capacitors Answer: A. At DC capacitors are open and the inductor is a short, so 4 ohm and 1 ohm form a divider giving 1/5. 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