--- title: "State equations of two-source circuit with 2 F" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-two-source-1-2h-2f" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-two-source-1-2h-2f site: BCT Engineering --- # State equations of two-source circuit with 2 F Control Systems · Chapter 8: State Space Analysis - Asked in: 2074 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-two-source-1-2h-2f - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question Write the state equation for the circuit shown below. Also write output equation. [Figure: source V1 on the left in series with 1 Ω and 2 H inductor to node A; 2 F capacitor from node A to the bottom (common) line; node A through 1 H inductor and 2 Ω resistor to source V2 on the right; both sources return to the common bottom line] ## Short answer Define states x1=i1 (2H inductor), x2=i2 (1H inductor), x3=vC (capacitor). The state equation is x' = [ -0.5 0 -0.5; 0 -2 1; 0.5 -0.5 0 ] x + [0.5 0; 0 -1; 0 0] [V1; V2], output y = [0 0 1]x. ## Answer 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) ```text 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. $C = [0 \;\; 2 \;\; 0]$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which states are used? A. V1, V2 and vC B. i1, i2 and vC C. i1 only D. Resistor voltages Answer: B. Two inductors and one capacitor. (Key point: State variables) 2. What is x1'? A. -x1 - x3 + V1 B. -0.5 x1 + 0.5 x3 + 0.5 V1 C. -0.5 x1 - 0.5 x3 + 0.5 V1 D. -2 x1 - 0.5 x3 + V1 Answer: C. 2 di1/dt = V1 - i1 - vC, divided by 2. (Key point: Left loop equation) 3. What is x2'? A. -2 x2 + x3 + V2 B. 2 x2 - x3 - V2 C. -x2 + x3 - V2 D. -2 x2 + x3 - V2 Answer: D. 1 di2/dt = vC - 2 i2 - V2. (Key point: Right loop) 4. If the output is the voltage across the 2 ohm resistor, what is C? A. [0 2 0] B. [0 0 1] C. [2 0 0] D. [0 1 0] Answer: A. Voltage across 2 ohm = 2 i2 = 2 x2. 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