--- title: "Two-inertia rotational system transfer function and T-I analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-rot-j1-j2-k" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-rot-j1-j2-k site: BCT Engineering --- # Two-inertia rotational system transfer function and T-I analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2081 Bhadra, 6+2 marks - Video: [Problem on Mechanical Rotational System](https://www.youtube.com/watch?v=qHbUeuVFwLE) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-rot-j1-j2-k - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Obtain transfer function for the following system considering angular position of object second (θ2) as output. Also develop Torque-Current Analogous Circuit. [Figure: torque T(t) applied to inertia J1 (angle θ1(t)) which has viscous friction D1 to the fixed frame; J1 is connected through a shaft of torsional stiffness K to inertia J2 (angle θ2(t)), which has viscous friction D2 to the fixed frame] ## Short answer θ2(s)/T(s) = K / [s{ J1 J2 s^3 + (J1 D2 + J2 D1)s^2 + (K(J1+J2) + D1 D2)s + K(D1+D2) }]. In torque-current analogy, torque → current source, inertias → capacitances, friction → conductances, stiffness → inductance. ## Answer **Assumptions:** zero initial conditions; $D_1$, $D_2$ are viscous friction coefficients to the fixed frame. #### Equations (Laplace form) Inertia $J_1$: $$ T(s) = (J_1s^2 + D_1s + K)\theta_1 - K\theta_2 $$ Inertia $J_2$: $$ 0 = -K\theta_1 + (J_2s^2 + D_2s + K)\theta_2 $$ #### Solving (Cramer's rule) $$ \Delta = (J_1s^2 + D_1s + K)(J_2s^2 + D_2s + K) - K^2 $$ $$ \theta_2 = \frac{K\,T(s)}{\Delta} $$ Expanding $\Delta$: $$ \Delta = J_1J_2s^4 + (J_1D_2 + J_2D_1)s^3 + \left[K(J_1+J_2) + D_1D_2\right]s^2 + K(D_1+D_2)s $$ $$ \frac{\theta_2(s)}{T(s)} = \frac{K}{s\left[J_1J_2s^3 + (J_1D_2 + J_2D_1)s^2 + \{K(J_1+J_2) + D_1D_2\}s + K(D_1+D_2)\right]} $$ The factor $s$ in the denominator shows that a constant torque makes the whole system rotate continuously (angle grows with time). #### Torque–current (T-I) analogy Torque → current source, inertia $J$ → capacitance $C = J$, viscous friction $D$ → conductance ($R = 1/D$), stiffness $K$ → inductance $L = 1/K$, angular velocity $\omega$ → node voltage. Each inertia is a node. ```text ω1 ω2 +----o----+---[L]---+----o | | | | | (T) C1 R1 C2 R2 | | | | | +----+----+---------+----+ ground C1=J1, R1=1/D1, L=1/K, C2=J2, R2=1/D2 ``` Node equations: $$ \begin{aligned} T &= C_1\frac{d\omega_1}{dt} + \frac{\omega_1}{R_1} + \frac{1}{L}\int(\omega_1 - \omega_2)dt \\ 0 &= C_2\frac{d\omega_2}{dt} + \frac{\omega_2}{R_2} + \frac{1}{L}\int(\omega_2 - \omega_1)dt \end{aligned} $$ which match the mechanical equations with $\omega = \dot\theta$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The equation for J2 is A. T = (J2 s^2 + D2 s) theta2 B. 0 = (J2 s^2 + D2 s - K) theta2 C. 0 = K theta1 + (J2 s^2 + K) theta2 D. 0 = -K theta1 + (J2 s^2 + D2 s + K) theta2 Answer: D. J2 has inertia, friction and shaft torque. (Key point: Equation for J2) 2. The factor s in the denominator means A. constant torque makes the system rotate continuously B. it is stable with zero error C. it has no inertia D. it oscillates with constant amplitude Answer: A. Integrator-like behaviour. (Key point: Free rotation) 3. In the T-I analogy, K becomes A. a capacitor C = K B. an inductor L = 1/K between nodes C. a resistor R = K D. a source Answer: B. J->C, D->1/R, K->1/L. 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