--- title: "Three-inertia rotational system θ3/T and T-V analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:de-tf-tv-three-inertia" source: https://www.bctengineer.com/q/semester_3/control_systems/de-tf-tv-three-inertia site: BCT Engineering --- # Three-inertia rotational system θ3/T and T-V analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2068 Baisakh (old course), 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/de-tf-tv-three-inertia - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Write differential equation and obtain transfer function of the mechanical system as shown below considering θ3 as output. Also draw torque voltage analogy network. [Figure: torque T applied to inertia J1 (angle θ1, friction B1); J1 – torsional spring K1 – inertia J2 (angle θ2, friction B2); J2 – torsional spring K2 – inertia J3 (angle θ3); J3 – torsional spring K3 – fixed wall] ## Short answer Torque equations: J1θ1''+B1θ1'+K1(θ1-θ2)=T, J2θ2''+B2θ2'+K1(θ2-θ1)+K2(θ2-θ3)=0, J3θ3''+K2(θ3-θ2)+K3θ3=0. Transfer function θ3(s)/T(s)=K1K2/Δ, steady-state gain 1/K3. Torque-voltage analogy: L=J, R=B, C=1/K, mesh currents as velocities. ## Answer Angles $\theta_1, \theta_2, \theta_3$ of $J_1, J_2, J_3$; torque $T$ on $J_1$. $J_3$ has no friction (none shown); $K_3$ ties $J_3$ to the wall. #### Differential equations $$ \begin{aligned} J_1\ddot\theta_1 + B_1\dot\theta_1 + K_1(\theta_1 - \theta_2) &= T \\ J_2\ddot\theta_2 + B_2\dot\theta_2 + K_1(\theta_2 - \theta_1) + K_2(\theta_2 - \theta_3) &= 0 \\ J_3\ddot\theta_3 + K_2(\theta_3 - \theta_2) + K_3\theta_3 &= 0 \end{aligned} $$ #### Laplace form (matrix) $$ \begin{bmatrix} J_1s^2 + B_1s + K_1 & -K_1 & 0 \\ -K_1 & J_2s^2 + B_2s + K_1 + K_2 & -K_2 \\ 0 & -K_2 & J_3s^2 + K_2 + K_3 \end{bmatrix} \begin{bmatrix}\theta_1\\ \theta_2\\ \theta_3\end{bmatrix} = \begin{bmatrix} T\\0\\0\end{bmatrix} $$ #### Transfer function By Cramer's rule, the numerator is the product of the couplings $K_1K_2$: $$ \frac{\theta_3(s)}{T(s)} = \frac{K_1K_2}{\Delta} $$ $$ \begin{aligned} \Delta =\; & (J_1s^2 + B_1s + K_1)\left[(J_2s^2 + B_2s + K_1 + K_2)(J_3s^2 + K_2 + K_3) - K_2^2\right] \\ & - K_1^2(J_3s^2 + K_2 + K_3) \end{aligned} $$ $\Delta$ is 6th order, with leading term $J_1J_2J_3s^6$ and constant term $K_1K_2K_3$, so the steady-state gain for a constant torque is $1/K_3$ (all of $T$ is finally held by $K_3$). #### Torque-voltage analogy $T\to e$, $J\to L$, $B\to R$, $K\to 1/C$, $\dot\theta\to i$ (mesh currents $i_1, i_2, i_3$). $$ \begin{aligned} L_1\frac{di_1}{dt} + R_1i_1 + \frac{1}{C_1}\int(i_1 - i_2)dt &= e \\ L_2\frac{di_2}{dt} + R_2i_2 + \frac{1}{C_1}\int(i_2 - i_1)dt + \frac{1}{C_2}\int(i_2 - i_3)dt &= 0 \\ L_3\frac{di_3}{dt} + \frac{1}{C_2}\int(i_3 - i_2)dt + \frac{1}{C_3}\int i_3dt &= 0 \end{aligned} $$ ```text L1 R1 L2 R2 L3 C3 +LLLL/\/\-+----LLLL/\/\-+-----LLLL----||--+ | | | | (~) e C1=1/K1 C2=1/K2 | | i1 | i2 | i3 | +---------+-------------+-----------------+ ``` ($L_k = J_k$, $R_k = B_k$, $C_k = 1/K_k$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation describes J3? A. J3 θ3'' + K2(θ3 - θ2) + K3 θ3 = 0 B. J3 θ3'' + B3 θ3' + K3 θ3 = T C. J3 θ3'' + K2 θ2 = 0 D. J3 θ3'' + K3(θ3 - θ2) = 0 Answer: A. J3 has no friction or applied torque, and is held by K2 from J2 and by K3 to the wall. (Key point: Equation for J3) 2. Why is the numerator of θ3/T equal to K1 K2? A. Torque passes through springs K1 and K2 in series to reach J3 B. It is the sum of the spring constants C. K3 transmits the torque D. It is the determinant Answer: A. T acts on J1 and reaches J3 through the couplings K1 and K2, whose product forms the Cramer numerator. (Key point: Numerator) 3. What is the steady-state value of θ3/T? A. 1/K1 B. 1/K2 C. 1/K3 D. K1K2K3 Answer: C. Finally all of T is held by K3 so θ3 = T/K3. (Key point: Steady-state gain) 4. In the torque-voltage analogy, what do J, B and K become? A. L, R, 1/C B. C, 1/R, 1/L C. R, L, C D. L, 1/R, C Answer: A. T to e, J to L, B to R, K to 1/C, and angular velocity to mesh current. (Key point: T-V analogy) ## Related in Control Systems - [Two-inertia rotational system transfer function and T-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-rot-j1-j2-k) — Control Systems, chapter 2 - [Rotational system θ/T and T-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-ti-rot-j1-j2-k1k2) — Control Systems, chapter 2 - [Gear-train rotational system transfer function](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-gear-train-rotational) — Control Systems, chapter 2 - [Two-mass system transfer function and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-two-mass-k1k2b1b2) — Control Systems, chapter 2 - [Rotational system θ2/T with T-V and T-I analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-tv-ti-j1-j2-d1) — Control Systems, chapter 2, asked 2× - [Rotational system θ1/T and electrical analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-rot-j1-j2-k1k2k3-b1b2) — Control Systems, chapter 2 ## Related in other subjects - [Voltage ratio transfer function of two-section LC network](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Avoltage-ratio-tf-two-1h-shunt) — Electrical Circuit Theory, chapter 5 - [Voltage ratio transfer function, port 2 terminated by 2 H](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Avoltage-ratio-tf-two-port-terminated) — Electrical Circuit Theory, chapter 5