Gear-train rotational system transfer function

Reflect everything to one shaft. Mechanical impedances move across a gear pair by $(N_{dest}/N_{source})^2$, and torques by $N_{dest}/N_{source}$.

Step 1: Reflect the input torque to the $N_2$ shaft

$$T_2 = T\,\frac{N_2}{N_1} = T\cdot\frac{33}{11} = 3T$$

Step 2: Reflect the load damping to the $N_3$ shaft

$$D_{eq} = 0.04\left(\frac{N_3}{N_4}\right)^2 = 0.04\left(\frac{50}{10}\right)^2 = 1\ \text{N-m-s/rad}$$

Step 3: Equivalent system

Let $\theta_2$ = angle of the $N_2$ shaft (with $J = 1$, $D = 2$) and $\theta_3$ = angle of gear $N_3$ (end of the spring, $K = 3$, carrying the reflected damping 1):

 3T ─► [J=1, D=2] ──spring K=3── (θ3) ── D=1 ─► frame
        θ2

Equations (Laplace):

$$\begin{aligned} (s^2 + 2s + 3)\,\theta_2 - 3\,\theta_3 &= 3T \\ -3\,\theta_2 + (s + 3)\,\theta_3 &= 0 \end{aligned}$$

Step 4: Solve

From the second equation $\theta_2 = \dfrac{(s+3)}{3}\theta_3$. Substituting:

$$\begin{aligned} \left[\frac{(s^2+2s+3)(s+3)}{3} - 3\right]\theta_3 &= 3T \\ \left[(s^3 + 5s^2 + 9s + 9) - 9\right]\theta_3 &= 9T \\ \frac{\theta_3(s)}{T(s)} &= \frac{9}{s(s^2 + 5s + 9)} \end{aligned}$$

Step 5: Output shaft

$$\theta_L = \theta_3\,\frac{N_3}{N_4} = 5\,\theta_3$$

Answer:

$$\frac{\theta_L(s)}{T(s)} = \frac{45}{s(s^2 + 5s + 9)}$$

The pole at $s=0$ appears because no spring ties the system to the frame, so a constant torque gives a steadily increasing angle.