Two-inertia rotational system transfer function and T-I analogy

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.

      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 $\omega = \dot\theta$.