--- title: "State model of two-source RLC circuit" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-model-two-source-circuit" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-model-two-source-circuit site: BCT Engineering --- # State model of two-source RLC circuit Control Systems · Chapter 8: State Space Analysis - Asked in: 2075 Chaitra, 6 marks - Video: [#217 Problem- 1 state space model of electrical circuit || EC Academy](https://www.youtube.com/watch?v=eGD40W4SKWI) — EC Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-model-two-source-circuit - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question Develop state space model for circuit below. [Figure: two-source circuit; source V1 on the left and V2 on the right; top branch: V1(+) – 2 Ω – node A – 1 H – V2(+); bottom branch: V1(−) – 2 H – node B – 3 Ω – V2(−); 2 F capacitor connected between node A and node B] ## Short answer Choose states x1=i1 (left inductor), x2=i2 (right inductor), x3=vC. Then x' = [ -1 0 -0.5; 0 -3 1; 0.5 -0.5 0 ] x + [0.5 0; 0 -1; 0 0] [V1; V2], output y=vC = [0 0 1]x. ## Answer 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 ```text 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 $C = [0 \;\; 1 \;\; 0]$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the states? A. Two capacitor voltages and one current B. Two inductor currents and the capacitor voltage C. Both source voltages D. Only the capacitor voltage Answer: B. Two inductors and one capacitor give three states. (Key point: State choice) 2. What is x1' (left loop, 2 ohm and 2 H)? A. -2 x1 - x3 + V1 B. -x1 + 0.5 x3 - 0.5 V1 C. -x1 - 0.5 x3 + 0.5 V1 D. -0.5 x1 - 0.5 x3 + 0.5 V1 Answer: C. 2 di1/dt = V1 - 2 i1 - vC, so divide by 2. (Key point: Left loop) 3. How is the capacitor voltage derived? A. 2 dvC/dt = i1 + i2 B. dvC/dt = i1 - i2 C. 2 dvC/dt = vC - V1 D. 2 dvC/dt = i1 - i2 Answer: D. The capacitor carries the difference of the two loop currents; C = 2 F. 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