---
title: "State equations of DC motor circuit"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-dc-motor-armature-field"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-dc-motor-armature-field
site: BCT Engineering
---

# State equations of DC motor circuit

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2071 Chaitra, 5 marks
- Video: [State space control of armature controlled dc motor](https://www.youtube.com/watch?v=VBny1gap860) — Karthikeyan K
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-dc-motor-armature-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

Develop state equation for motor circuit below. [Figure: DC motor; armature circuit with supply Ea, resistance Ra, inductance La, armature current Ia and back emf Eb; separate field circuit with supply ef, resistance Rf, inductance Lf and field current If; motor torque Tm, speed ωm, angle θm; load inertia Jm and friction Bm]

## Short answer

Define state vector [i_a, i_f, ω_m, θ_m]^T. For linear analysis with constant field flux, the state equations are ̇i_a = -(R_a/L_a)i_a - (K_b/L_a)ω_m + (1/L_a)E_a; ̇i_f = -(R_f/L_f)i_f + (1/L_f)e_f; ̇ω_m = (K_t/J_m)i_a - (B_m/J_m)ω_m; ̇θ_m = ω_m.

## Answer

Take the currents in the two inductances and the mechanical speed and angle as state variables:

$$
x_1 = i_a,\quad x_2 = i_f,\quad x_3 = \omega_m,\quad x_4 = \theta_m
$$

#### Basic equations
- Armature circuit: $\;L_a\dfrac{di_a}{dt} + R_ai_a + e_b = E_a$
- Field circuit: $\;L_f\dfrac{di_f}{dt} + R_fi_f = e_f$
- Back emf: $\;e_b = K_b\,\omega_m$ (with $K_b \propto$ field flux)
- Motor torque: $\;T_m = K_t\,i_a$ (with $K_t \propto$ field flux)
- Load: $\;J_m\dfrac{d\omega_m}{dt} + B_m\omega_m = T_m$, and $\dfrac{d\theta_m}{dt} = \omega_m$

Strictly, flux is proportional to $i_f$, so $T_m = K i_f i_a$ and $e_b = K i_f\omega_m$, which are non-linear. For a linear model the field flux is taken as constant at its operating value (armature control), so $K_t$ and $K_b$ are constants.

#### State equations

$$
\begin{aligned}
\dot{x}_1 &= -\frac{R_a}{L_a}x_1 - \frac{K_b}{L_a}x_3 + \frac{1}{L_a}E_a \\
\dot{x}_2 &= -\frac{R_f}{L_f}x_2 + \frac{1}{L_f}e_f \\
\dot{x}_3 &= \frac{K_t}{J_m}x_1 - \frac{B_m}{J_m}x_3 \\
\dot{x}_4 &= x_3
\end{aligned}
$$

#### Matrix form

$$
\begin{bmatrix} \dot{i}_a \\ \dot{i}_f \\ \dot{\omega}_m \\ \dot{\theta}_m \end{bmatrix} = \begin{bmatrix} -\dfrac{R_a}{L_a} & 0 & -\dfrac{K_b}{L_a} & 0 \\ 0 & -\dfrac{R_f}{L_f} & 0 & 0 \\ \dfrac{K_t}{J_m} & 0 & -\dfrac{B_m}{J_m} & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} i_a \\ i_f \\ \omega_m \\ \theta_m \end{bmatrix} + \begin{bmatrix} \dfrac{1}{L_a} & 0 \\ 0 & \dfrac{1}{L_f} \\ 0 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} E_a \\ e_f \end{bmatrix}
$$

Output (shaft position): $y = \theta_m = [0\;\;0\;\;0\;\;1]\,x$ (or $[0\;\;0\;\;1\;\;0]\,x$ for speed).

With constant field the armature-controlled motor uses only $i_a$, $\omega_m$, $\theta_m$, giving $\dfrac{\Omega_m(s)}{E_a(s)} = \dfrac{K_t}{(L_as+R_a)(J_ms+B_m) + K_tK_b}$.

## Check yourself

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

1. Which is the armature circuit equation?
   A. La dia/dt + Ra ia = eb
   B. La dia/dt + Ra ia + eb = Ea
   C. Ra ia = Ea + eb
   D. La dia/dt - Ra ia = Ea
   Answer: B. KVL in armature includes back emf eb. (Key point: Armature equation)

2. How are eb and Tm related to variables?
   A. eb = Kb ia and Tm = Kt wm
   B. eb = Kb theta
   C. eb = Kb wm and Tm = Kt ia
   D. Tm = Kt eb
   Answer: C. Back emf is proportional to speed, torque to armature current. (Key point: Back emf and torque)

3. Why are Kt and Kb taken constant?
   A. Because the field current is zero
   B. Because armature resistance is zero
   C. Because inertia is constant
   D. Field flux is assumed constant, avoiding the non-linear if*ia term
   Answer: D. Torque is actually proportional to if*ia, which is non-linear. (Key point: Linearisation)

4. How many states in this model?
   A. 4: ia, if, wm, theta
   B. 2
   C. 3
   D. 5
   Answer: A. Two inductances plus speed and angle. (Key point: Number of states)

## Related in Control Systems

- [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
- [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
- [DC motor θm/Va via block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-dc-motor-bd-thetam-va) — 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
- [State equations of current-source RLC circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-current-source-1h-1f-2h) — Control Systems, chapter 8

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