--- title: "Rotational system θ1/T and electrical analogies" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-rot-j1-j2-k1k2k3-b1b2" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-rot-j1-j2-k1k2k3-b1b2 site: BCT Engineering --- # Rotational system θ1/T and electrical analogies Control Systems · Chapter 2: Component Modeling - Asked in: 2081 Baisakh, 10 marks - Video: [Mechanical Rotational System | problem 2 | Easy method | Torque Voltage and Torque Current circuits|](https://www.youtube.com/watch?v=TqeFbAXm7XY) — Witplum - 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-k1k2k3-b1b2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find the transfer function θ1(s)/T(s) for the given rotational mechanical system. Also draw analogous electrical networks. [Figure: fixed wall – torsional spring K1 – inertia J1 (torque T(t) applied, angle θ1); J1 connected to inertia J2 (angle θ2) by damper B1 in series with spring K2, in parallel with spring K3; J2 – damper B2 – fixed wall] ## Short answer Reduce series B1−K2 to B1K2s/(B1s+K2), parallel with K3 gives coupling Zc. θ1(s)/T(s) = (b+Zc)/[ab+Zc(a+b)] with a=J1s^2+K1, b=J2s^2+B2s; denominator is 5th order. T-V analogy: T→e, J→L, B→R, K→1/C; B1−K2 series maps to parallel R||C shared by loops, K3 adds series C. ## Answer Angles: $\theta_1$ ($J_1$, torque $T$), $\theta_2$ ($J_2$) and $\theta_3$ = junction of $B_1$ (on the $J_1$ side) and $K_2$ (on the $J_2$ side). $K_3$ couples $J_1$ and $J_2$ directly; $K_1$ and $B_2$ go to the frame. #### Differential equations (Laplace form) $$ \begin{aligned} J_1s^2\theta_1 + K_1\theta_1 + B_1s(\theta_1 - \theta_3) + K_3(\theta_1 - \theta_2) &= T \\ B_1s(\theta_3 - \theta_1) + K_2(\theta_3 - \theta_2) &= 0 \\ J_2s^2\theta_2 + B_2s\theta_2 + K_2(\theta_2 - \theta_3) + K_3(\theta_2 - \theta_1) &= 0 \end{aligned} $$ #### Reducing the series branch $B_1$ in series with $K_2$ is equivalent to $\dfrac{B_1sK_2}{B_1s + K_2}$; with $K_3$ in parallel the total coupling is $$ Z_c = K_3 + \frac{B_1K_2s}{B_1s + K_2} = \frac{Q}{B_1s + K_2}, \quad Q = K_3(B_1s + K_2) + B_1K_2s $$ With $a = J_1s^2 + K_1$ and $b = J_2s^2 + B_2s$: $$ \begin{aligned} (a + Z_c)\theta_1 - Z_c\theta_2 &= T \\ -Z_c\theta_1 + (b + Z_c)\theta_2 &= 0 \end{aligned} \qquad\Rightarrow\qquad \frac{\theta_1}{T} = \frac{b + Z_c}{ab + Z_c(a + b)} $$ #### Transfer function Multiplying by $(B_1s + K_2)$: $$ \frac{\theta_1(s)}{T(s)} = \frac{(B_1s + K_2)(J_2s^2 + B_2s) + Q}{(B_1s + K_2)(J_1s^2 + K_1)(J_2s^2 + B_2s) + Q\left[(J_1 + J_2)s^2 + B_2s + K_1\right]} $$ Expanded: $$ \frac{\theta_1}{T} = \frac{B_1J_2s^3 + (B_1B_2 + J_2K_2)s^2 + (B_1K_2 + B_1K_3 + B_2K_2)s + K_2K_3}{\Delta} $$ $$ \begin{aligned} \Delta =\; & B_1J_1J_2s^5 + (B_1B_2J_1 + J_1J_2K_2)s^4 \\ & + \left[B_1J_1(K_2 + K_3) + B_1J_2(K_1 + K_2 + K_3) + B_2J_1K_2\right]s^3 \\ & + \left[B_1B_2(K_1 + K_2 + K_3) + J_1K_2K_3 + J_2K_2(K_1 + K_3)\right]s^2 \\ & + \left[B_1K_1(K_2 + K_3) + B_2K_2(K_1 + K_3)\right]s + K_1K_2K_3 \end{aligned} $$ #### Analogous electrical networks | Mechanical | T-V (mesh) | T-I (node) | |---|---|---| | $T$ | $e$ | $i$ | | $J$ | $L$ | $C$ | | $B$ | $R$ | $1/R$ | | $K$ | $1/C$ | $1/L$ | | $\omega$ | $i$ | $v$ | **T-I network** (nodes $v_1, v_3, v_2$): ```text v1 RB1 v3 L2 v2 +----+----+----+--/\/\/--+--LLLL--+----+----+ | | | | | | | | | | +-------LLLL-------+ | | | | | L3 | | (i) C1 L1 C2 R2 | | | | | +----+----+----------------------------+----+ gnd ``` ($C_1 = J_1$, $L_1 = 1/K_1$, $R_{B1} = 1/B_1$, $L_2 = 1/K_2$, $L_3 = 1/K_3$, $C_2 = J_2$, $R_2 = 1/B_2$.) **T-V network** (mesh currents $i_1, i_2$ and inner loop $i_3$): ```text L1=J1 C1=1/K1 L2=J2 R2=B2 +-LLLL----||----+---------LLLL---/\/\-+ | | | (~) e C3=1/K3 | | i1 | i2 | | +---+---+ | | R1=B1 C2=1/K2 | | | i3 | | | +---+---+ | | | | +---------------+---------------------+ ``` $K_3$ (parallel coupling) becomes a series element $C_3$ in the shared branch, and the series pair $B_1$–$K_2$ becomes the parallel pair $R_1 \parallel C_2$. The mesh equations are exactly the three equations above with $\theta \to q$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the equation for junction θ3 state? A. B1 s(θ3 - θ1) + K2(θ3 - θ2) = 0 B. J1 s^2 θ3 = T C. B1 s θ3 = K2 θ2 D. K3(θ3 - θ1) = 0 Answer: A. The junction between B1 and K2 is massless so the two torques balance. (Key point: Junction) 2. B1 in series with K2 gives equivalent A. B1 K2 s / (B1 s + K2) B. B1 s + K2 C. B1 K2 D. (B1 s + K2)/(B1 K2 s) Answer: A. Series combinations add reciprocals of impedances. (Key point: Series branch) 3. How is K3 combined with the B1-K2 branch? A. In parallel, so impedances add B. In series C. It is ignored D. Multiplied Answer: A. K3 couples J1 and J2 directly, parallel to the series branch, so Zc = K3 + B1K2s/(B1s+K2). (Key point: Total coupling) 4. What is the leading term in the denominator Δ? A. J1 J2 s^4 B. B1 J1 J2 s^5 C. K1 K2 K3 D. B1 B2 s^2 Answer: B. The system is 5th order with leading term B1 J1 J2 s^5. 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