DC motor differential equation

Question

This question involves finding the transfer function for the system, but I first need to get the differential equations correct. Have I set up the gearbox correctly?

Answer 1

$$T_m-T_L-J\dfrac{d\Omega_m}{dt}=0$$

$$T_m-D_L\dfrac{N_1}{N_2}-(J_a+J_L(\dfrac{N_1}{N_2})^2)\dfrac{d\Omega_m}{dt}=0$$

This is for mechanical part of the equation. The electrical is little bit tricky, not so simple, because you have a feedback of back EMF voltage and armature inductivity.

$$e_a(t)=R_a\cdot i_a+L_a\frac{di_a}{dt}+K_\Phi\Omega$$

As said the total moment of inertia seen from motor is: \$J_a + J_L(\dfrac{N_1}{N_2})^2\$ and the torque at motor is: \$D_L\dfrac{N_1}{N_2}\$, not the \$D_a + D_L(\dfrac{N_1}{N_2})^2\$, what is \$D_a\$ anyway? Further, the solution of the differential equation should show the transient response, but without taking into account the armature inductivity is useless, it would be like omitting the moment of inertia. Thus, it would be simpler to compute the steady state without differentiating or you may have to expand the equation to full problem.

Have a look on my previous answer

EDIT: I got \$D_L\$ and \$D_a\$ meaning from JonRB answer it's damping or friction of the load and rotor respectively. Now, from one of my previous answers:

$$ \dfrac{\Omega_m(s)}{u_q(s)} =\dfrac{\dfrac{k_\Phi}{L_qJ}}{s^2 + s\dfrac{R_qJ+L_qF}{L_qJ}+\dfrac{R_qF+k_\Phi^2}{L_qJ}} $$ You can replace \$F\$ with \$D_a + D_L(\dfrac{N_1}{N_2})\$, but you have the inductivity. Rearranging the equation, we get:

$$ \dfrac{\Omega_m(s)}{u_q(s)} =\dfrac{{k_\Phi}}{{L_qJ}s^2 + s{R_qJ+L_qF}+{R_qF+k_\Phi^2}} $$ Omitting the inductance: $$ \dfrac{\Omega_m(s)}{u_q(s)} =\dfrac{{k_\Phi}}{s{R_qJ}+{R_qF+k_\Phi^2}} $$ $$ \dfrac{\Omega_L(s)}{e_a(s)} =\dfrac{N_1}{N_2}\dfrac{{k_\Phi}}{s{R_a(J_a + J_L(\dfrac{N_1}{N_2})^2)}+{R_a(D_a + D_L(\dfrac{N_1}{N_2}))+k_\Phi^2}} $$

Finally, let's integrate the angular velocity to get the angular displacement: $$ \dfrac{\Theta_L(s)}{e_a(s)} =\dfrac{N_1}{N_2}\cdot\dfrac{1}{s}\cdot\dfrac{{k_\Phi}}{s{R_a(J_a + J_L(\dfrac{N_1}{N_2})^2)}+{R_a(D_a + D_L(\dfrac{N_1}{N_2}))+k_\Phi^2}} $$

\$k_\Phi [V\cdot s/rad]= k_i[Nm/A]\$, so replace the constants accordingly, but if you have both and unequal, then replace \$k_\Phi^2=k_\Phi\cdot k_i\$. The transfer function is in the s-domain, as an engineer would understand. You can still transform it in the less understandable time domain.

Answer 2

Consider an electrical motor on its own:

A voltage source exciting the stator resistance & inductance. The resultant current produces torque which produces an acceleration against an inertia, with some damping.

We know:

**MOTOR: Electromagnetic & mechanical **

\$T = K_t i \$

torque equals constant multiplied by current

\$V = K_e \omega = K_e \dot{\Theta_a}\$

We know the backEMF voltage is proportional to velocity

\$T = J \ddot{\Theta_a} + D\dot{\Theta_a} \$

The Motor torque is that needed to accelerate an inertia with the damping torque subtracted.

\$ L_a\frac{di}{dt} + R_ai_a = e_a - K_e \dot{\Theta_a}\$

The voltage across the stator resistance and inductance equals the supply voltage minus the backEMF voltage (velocity dependant). With the stator current being that which flows through this network.

S-Domain (for clarity)

\$s(Js + D)\Theta_a(s) = K_tI(s) \$

\$(L_as + R_a)I(s) = E_a(s) - Ke\Theta_a(s) \$

Exam Question

The question authors query has additional complication via a load inertia & load damping all coupled via gearing.

Equally it does not include stator inductance (???) yet expects the final equation to include inertia and acceleration (???). One could argue in steady-state the inductor term drops out of the final equation (speed has settled, load has settled, backEMF has settled -> no change on the electrical side). HOWEVER, if steady-state is assumed then inertia & acceleration drops out.

Personally I think the question has been stripped back too much, but if that is what has been shown... lets drop the \$ L_a\frac{di}{dt}\$ term.

By reflecting the inertia and the damping to the primary side of the gearbox:

\$ J = J_a + (\frac{N_1}{N_2})^2J_L \$

\$ D = D_a + (\frac{N_1}{N_2}) D_L \$

The question is using the subscript m to denote the total inertia and load as seen by the motor. The load specific has a subscript L and the "motor specific" has a subscript a (to denote armature?)

Combining:

\$e_a = \Theta_a s (\frac{R(Js + D)}{K_t} + K_e )\$

\$e_a = \Theta_a s (\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

\$e_a = \dot{\Theta_a}(\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

Finally as it is needed to be in terms of \$\Theta_L\$

\$e_a = \dot{\Theta_a}\frac{N_1}{N_2}(\frac{R((J_a + (\frac{N_1}{N_2})^2J_L)s + (D_a + (\frac{N_1}{N_2}) D_L))}{K_t} + K_e )\$

since \$\Theta_m = \frac{N_2}{N_1}\Theta_L \Rightarrow \Theta_L = \frac{N_1}{N_2}\Theta_m\$