---
title: "Rotational system θ2/T with T-V and T-I analogies"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-tv-ti-j1-j2-d1"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-tv-ti-j1-j2-d1
site: BCT Engineering
---

# Rotational system θ2/T with T-V and T-I analogies

Control Systems · Chapter 2: Component Modeling

- Asked in: 2079 Bhadra, 8 marks; 2076 Chaitra, 8 marks (asked 2 times)
- 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-tv-ti-j1-j2-d1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find the transfer function θ2(s)/T(s) for the mechanical rotational system of figure below. Also draw the T-V and T-I analogy circuit of the system. [Figure: fixed wall – viscous damper D1 – inertia J1 (torque T(t) applied, angle θ1); J1 – damper D – torsional spring K – inertia J2 (angle θ2); J2 – damper D2 – fixed wall]

## Short answer

θ2(s)/T(s) = DK / [s{ (Ds+K)(J1s+D1)(J2s+D2) + DK((J1+J2)s + D1+D2) }]. In torque-voltage analogy, torque maps to voltage, inertias to inductors, dampers to resistors, springs to capacitors; in torque-current analogy, torque maps to current, inertias to capacitors, dampers to resistors, springs to inductors.

## Answer

**Assumptions:** zero initial conditions; $\theta_3$ is the angle of the junction between damper $D$ and spring $K$ (D on the $J_1$ side). The series $D$–$K$ coupling is massless.

#### Equations (Laplace form)
Inertia $J_1$:

$$
T(s) = (J_1s^2 + D_1s)\theta_1 + Ds(\theta_1 - \theta_3)
$$

Junction: $\;Ds(\theta_3 - \theta_1) + K(\theta_3 - \theta_2) = 0$

Inertia $J_2$: $\;(J_2s^2 + D_2s)\theta_2 + K(\theta_2 - \theta_3) = 0$

#### Solving
Damper and spring in series transmit the same torque, so they act as one element with "stiffness"

$$
K_e(s) = \frac{DsK}{Ds + K}
$$

Then $(J_1s^2+D_1s+K_e)\theta_1 - K_e\theta_2 = T$ and $-K_e\theta_1 + (J_2s^2+D_2s+K_e)\theta_2 = 0$, giving

$$
\frac{\theta_2}{T} = \frac{K_e}{(J_1s^2+D_1s)(J_2s^2+D_2s) + K_e\left[(J_1+J_2)s^2 + (D_1+D_2)s\right]}
$$

Multiply top and bottom by $(Ds+K)$ and cancel the common $s$:

$$
\frac{\theta_2(s)}{T(s)} = \frac{DK}{s\left[(Ds+K)(J_1s+D_1)(J_2s+D_2) + DK\{(J_1+J_2)s + D_1 + D_2\}\right]}
$$

Expanded:

$$
\frac{\theta_2(s)}{T(s)} = \frac{DK}{s\left[a_3s^3 + a_2s^2 + a_1s + a_0\right]}
$$

$$
\begin{aligned}
a_3 &= DJ_1J_2, \quad a_2 = D(D_1J_2 + D_2J_1) + KJ_1J_2 \\
a_1 &= DD_1D_2 + K\left[D(J_1+J_2) + D_1J_2 + D_2J_1\right] \\
a_0 &= K\left[D(D_1+D_2) + D_1D_2\right]
\end{aligned}
$$

#### Torque–voltage (T-V) analogy
Torque → voltage, $J$ → $L$, $D$ → $R$, $K$ → $C = 1/K$, angular velocity → loop current. The series $D$–$K$ pair becomes $R \parallel C$ shared by the two loops.

```text
  +--[L1]--[R1]--+-----+--[L2]--[R2]--+
  |              |     |              |
 (T)  i1 ->      R     C     <- i2    |
  |              |     |              |
  +--------------+-----+--------------+
 L1=J1, R1=D1, R=D, C=1/K, L2=J2, R2=D2
```

#### Torque–current (T-I) analogy
Torque → current source, $J$ → $C$, $D$ → conductance ($R = 1/D$), $K$ → $L = 1/K$, angular velocity → node voltage.

```text
      ω1          ω3           ω2
  +----o----+--[R]--o--[L]--+----o
  |    |    |               |    |
 (T)   C1   R1              C2   R2
  |    |    |               |    |
  +----+----+---------------+----+
              ground
 C1=J1, R1=1/D1, R=1/D, L=1/K, C2=J2, R2=1/D2
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Damper D and spring K in series act as
   A. Ke(s) = D s K/(Ds + K)
   B. D s + K
   C. D K s
   D. K/(Ds)
   Answer: A. Series elements carry the same torque. (Key point: Series element)

2. theta2/T has a factor s in the denominator because
   A. it is unstable
   B. D and K have no ground connection, so the system can rotate freely
   C. K is zero
   D. J2 is zero
   Answer: B. No restoring torque to ground. (Key point: Pole at origin)

3. In the T-V analogy, J, D and K map to
   A. C=J, G=D, L=1/K
   B. R=J, L=D, C=K
   C. L=J, R=D, C=1/K
   D. L=1/J, R=1/D, C=K
   Answer: C. Torque-voltage. (Key point: T-V analogy)

4. In the T-I analogy, torque becomes
   A. a voltage source and velocity a loop current
   B. a resistor and velocity a node voltage
   C. a capacitor and velocity a charge
   D. a current source and angular velocity a node voltage
   Answer: D. Torque-current: T->I, w->V. (Key point: T-I 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
- [Three-inertia rotational system θ3/T and T-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ade-tf-tv-three-inertia) — 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
- [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
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