---
title: "DC motor θm/Va via block diagram"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-dc-motor-bd-thetam-va"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-dc-motor-bd-thetam-va
site: BCT Engineering
---

# DC motor θm/Va via block diagram

Control Systems · Chapter 2: Component Modeling

- Asked in: 2074 Chaitra, 8 marks
- Video: [Example: Motor Transfer Function](https://www.youtube.com/watch?v=VlRpvyG-Tz8) — The Ryder Project
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-dc-motor-bd-thetam-va
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find the transfer function θm(s)/Va(s) of the system below by constructing the block diagram. [Figure: armature-controlled DC motor; armature voltage Va drives armature current Ia through resistance ra and inductance La; field current If = constant; back emf Eb; motor torque constant Km; motor shaft (angle θm, torque τ) drives a load of inertia J with viscous friction f to the fixed frame]

## Short answer

Armature-controlled DC motor: from KVL and dynamics, block diagram has forward path Km/[(ra+sLa)(Js+f)] and 1/s, with feedback Kb. θm(s)/Va(s) = Km / {s[(ra+sLa)(Js+f) + KmKb]}. Expanded: Km / [s(LaJ s^2 + (raJ+La f)s + ra f + KmKb)]. Neglecting La gives K/(s(τm s+1)).

## Answer

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)

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is torque proportional to Ia?
   A. Ra is zero
   B. Field current is constant, so flux is constant
   C. Armature is open
   D. Va is constant
   Answer: B. With I_f constant the flux is constant, so torque = Km Ia and back emf is proportional to speed. (Key point: Armature control)

2. Eliminating the back-emf loop gives omega/Va equal to:
   A. 1 / (Js + f)
   B. Km / ((ra + sLa)(Js + f) + Km Kb)
   C. Km Kb / (...)
   D. Km / ((ra + sLa)(Js + f))
   Answer: B. G/(1 + G Kb) with G = Km/((ra+sLa)(Js+f)). (Key point: Loop reduction)

3. Neglecting La, the back emf acts like:
   A. Extra inertia
   B. A spring
   C. Extra viscous friction Km Kb/ra, reducing the time constant
   D. No effect
   Answer: C. The denominator becomes ra f + Km Kb, so effective friction rises and tau_m falls. (Key point: Effective friction)

## Related in Control Systems

- [Armature-controlled DC motor transfer function and block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-armature-dc-motor) — Control Systems, chapter 2
- [Armature-controlled DC motor with gear and load](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Adc-motor-gear-load) — Control Systems, chapter 2
- [Dynamic model of armature-controlled DC motor](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Adc-motor-de-dynamic-model) — Control Systems, chapter 2