State equations of DC motor circuit

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 $\dfrac{\Omega_m(s)}{E_a(s)} = \dfrac{K_t}{(L_as+R_a)(J_ms+B_m) + K_tK_b}$.