Three-inertia rotational system θ3/T and T-V analogy

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 $i_1, i_2, i_3$).

$$\begin{aligned} L_1\frac{di_1}{dt} + R_1i_1 + \frac{1}{C_1}\int(i_1 - i_2)dt &= e \\ L_2\frac{di_2}{dt} + R_2i_2 + \frac{1}{C_1}\int(i_2 - i_1)dt + \frac{1}{C_2}\int(i_2 - i_3)dt &= 0 \\ L_3\frac{di_3}{dt} + \frac{1}{C_2}\int(i_3 - i_2)dt + \frac{1}{C_3}\int i_3dt &= 0 \end{aligned}$$

  L1  R1         L2  R2         L3      C3
 +LLLL/\/\-+----LLLL/\/\-+-----LLLL----||--+
 |         |             |                 |
(~) e    C1=1/K1       C2=1/K2             |
 |   i1    |     i2      |       i3        |
 +---------+-------------+-----------------+

($L_k = J_k$, $R_k = B_k$, $C_k = 1/K_k$.)