DC motor θm/Va via block diagram

With $I_f$ constant the motor is armature controlled; flux is constant, so torque $\propto I_a$ and back emf $\propto$ speed.

Equations

$$\begin{aligned} V_a(s) &= (r_a + sL_a)I_a(s) + E_b(s) && \text{(armature KVL)} \\ \tau(s) &= K_m I_a(s) && \text{(torque)} \\ \tau(s) &= (Js^2 + fs)\,\theta_m(s) && \text{(load)} \\ E_b(s) &= K_b\,s\,\theta_m(s) && \text{(back emf)} \end{aligned}$$

($K_b$ = back-emf constant; in SI units $K_b = K_m$.)

Constructing the block diagram

Each equation is written as output = block × input:

• $I_a = \dfrac{1}{r_a + sL_a}(V_a - E_b)$
• $\tau = K_m I_a$
• $\omega_m = \dfrac{1}{Js + f}\tau$, and $\theta_m = \dfrac{1}{s}\omega_m$
• $E_b = K_b\,\omega_m$ (feedback)

 Va  +     ┌────────┐ Ia ┌────┐ τ ┌────────┐ ωm  ┌───┐ θm
 ───►(Σ)──►│   1    ├───►│ Km ├──►│   1    ├──┬─►│1/s├──►
      ▲ -  │ ra+sLa │    └────┘   │ Js + f │  │  └───┘
      │    └────────┘             └────────┘  │
      │ Eb           ┌────┐                   │
      └──────────────┤ Kb │◄──────────────────┘
                     └────┘

Reduction

1. Combine the three forward blocks in cascade:

$$G(s) = \frac{K_m}{(r_a + sL_a)(Js + f)}$$

2. Eliminate the back-emf feedback loop ($H = K_b$):

$$\frac{\omega_m(s)}{V_a(s)} = \frac{G}{1 + GK_b} = \frac{K_m}{(r_a + sL_a)(Js + f) + K_mK_b}$$

3. Multiply by the integrator $1/s$:

$$\frac{\theta_m(s)}{V_a(s)} = \frac{K_m}{s\left[(r_a + sL_a)(Js + f) + K_mK_b\right]}$$

Expanded:

$$\frac{\theta_m(s)}{V_a(s)} = \frac{K_m}{s\left[L_aJs^2 + (r_aJ + L_af)s + r_af + K_mK_b\right]}$$

Simplified form

Neglecting $L_a$ (electrical time constant much smaller than mechanical):

$$\frac{\theta_m(s)}{V_a(s)} = \frac{K}{s(\tau_m s + 1)}, \quad K = \frac{K_m}{r_af + K_mK_b}, \quad \tau_m = \frac{r_aJ}{r_af + K_mK_b}$$

The back emf acts like extra viscous friction $K_mK_b/r_a$, which is why the effective time constant is reduced.