Armature-controlled DC motor with gear and load

The printed question is incomplete. It is read here as the standard problem: find the load-angle transfer function $\theta_L(s)/E_a(s)$ of an armature-controlled DC motor driving a load through a gear pair $N_1{:}N_2$. Motor data: inertia $J_m$, friction $D_m$; load data: inertia $J_L$, friction $D_L$; shafts are taken as rigid.

Assumptions

• Field current $I_f$ (through $L_f$, $R_f$) is constant, so flux is constant.
• Torque $T_m = K_tI_a$; back emf $e_b = K_b\dot\theta_m$.
• Gears are ideal (no loss, no backlash): $\theta_L = \dfrac{N_1}{N_2}\theta_m$.

Step 1: Electrical equation

$$e_a = R_ai_a + L_a\frac{di_a}{dt} + e_b \;\Rightarrow\; I_a(s) = \frac{E_a(s) - K_bs\theta_m(s)}{R_a + sL_a}$$

Step 2: Reflect the load to the motor shaft

Load impedances move to the motor side multiplied by $(N_1/N_2)^2$:

$$J_{eq} = J_m + J_L\left(\frac{N_1}{N_2}\right)^2, \qquad D_{eq} = D_m + D_L\left(\frac{N_1}{N_2}\right)^2$$

Step 3: Mechanical equation at the motor shaft

$$T_m = K_tI_a = J_{eq}\ddot\theta_m + D_{eq}\dot\theta_m \;\Rightarrow\; K_tI_a(s) = s(J_{eq}s + D_{eq})\theta_m(s)$$

Step 4: Motor-angle transfer function

Substituting $I_a$:

$$K_t\left[E_a - K_bs\theta_m\right] = s(R_a + sL_a)(J_{eq}s + D_{eq})\theta_m$$ $$\frac{\theta_m(s)}{E_a(s)} = \frac{K_t}{s\left[(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b\right]}$$

Step 5: Load-angle transfer function

$$\frac{\theta_L(s)}{E_a(s)} = \frac{N_1}{N_2}\cdot\frac{K_t}{s\left[(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b\right]}$$

Expanded denominator: $s\left[L_aJ_{eq}s^2 + (R_aJ_{eq} + L_aD_{eq})s + R_aD_{eq} + K_tK_b\right]$.

The load speed is $\Omega_L(s) = s\theta_L(s)$, so $\dfrac{\Omega_L(s)}{E_a(s)} = \dfrac{(N_1/N_2)K_t}{(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b}$.

Block diagram

 Ea +   ┌──────┐Ia┌──┐Tm┌───────┐ωm┌───┐θm┌─────┐ θL
 ──►(Σ)►│  1   ├─►│Kt├─►│   1   ├┬►│1/s├─►│N1/N2├──►
    ▲-  │Ra+sLa│  └──┘  │Jeq s+ ││ └───┘  └─────┘
    │   └──────┘        │  Deq  ││
    │ Eb      ┌────┐    └───────┘│
    └─────────┤ Kb │◄────────────┘
              └────┘

Simplified form ($L_a \approx 0$)

$$\frac{\theta_L(s)}{E_a(s)} = \frac{K}{s(\tau s + 1)}, \quad K = \frac{(N_1/N_2)K_t}{R_aD_{eq} + K_tK_b}, \quad \tau = \frac{R_aJ_{eq}}{R_aD_{eq} + K_tK_b}$$

A large reduction ratio ($N_2 \gg N_1$) makes the load inertia and friction look much smaller to the motor, which is why gears are used for matching.