--- title: "State equations of current-source RLC circuit" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-current-source-1h-1f-2h" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-current-source-1h-1f-2h site: BCT Engineering --- # State equations of current-source RLC circuit Control Systems · Chapter 8: State Space Analysis - Asked in: 2072 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-current-source-1h-1f-2h - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question Develop state space equations for the following circuit considering voltage of 2H inductor as output. I is input to the system. [Figure: current source I in parallel with a 1 H inductor, a 1 F capacitor, and a branch of 1.5 Ω resistor in series with a 2 H inductor] ## Short answer States: x1=i1 (1H), x2=vC, x3=i2 (2H branch). State equation: x' = [0 1 0; -1 0 -1; 0 0.5 -0.75] x + [0; 1; 0] I; output y = v_{L2} = [0 1 -1.5]x. ## Answer 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 ```text +------+------+--------+ | | | | (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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do all branches share the voltage v? A. They are in series B. They are in parallel across the current source C. The inductors cancel D. The capacitor is ideal Answer: B. Parallel branches have the same voltage. (Key point: Common voltage) 2. Which are chosen as states? A. I, v and i1 B. vL1 and vL2 C. i1 (1 H), v (1 F) and i2 (R-L branch) D. i1 and i2 only Answer: C. One state per energy store. (Key point: States) 3. Which equation gives x2'? A. x2' = x1 + x3 - I B. x2' = -x1 + x3 + I C. x2' = x2 - x3 D. x2' = -x1 - x3 + I Answer: D. I = i1 + C dv/dt + i2 with C = 1. (Key point: KCL) 4. What is the output equation for the voltage across the 2 H? A. y = x2 - 1.5 x3 B. y = x2 C. y = 2 x3 D. y = x2 + 1.5 x3 Answer: A. vL2 = v - 1.5 i2. 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