Lab 6

Equations
We start with two equations, one electrical and the other mechanical. Summing the loop voltages
gives:
dia
(t)
(1) La + R ai
a (t) = ea
(t) − eb
(t)
dt
Summing the torques present at the motor’s output shaft gives
(2) J ω&
(t) + bω(
t)
= T(t)
Equations (1) and (2) are two first-order ODE’s. They are related by the motor’s torque-current
relationship and by the back-emf produced by the motor rotation:
(3) T(t) = kmia(t)
(4) eb(t) = kmω(t)
Note that the same constant (km) appears in both equations (3) and (4). This is only true if we use
SI units. Two constants (called the motor torque constant and the back-emf (electromotive force)
coefficient) are required when English units are used.
The motor’s time response has two time constants (τ) – we’ll call these an electrical and a
mechanical time constant. The electrical time constant is almost always a lot faster than the
mechanical time constant, and is often neglected in modeling the motor dynamics. This is called a
dominant-pole approximation, and is a useful approach to simplifying the model.
To test this assumption, the electrical time constant can be measured by locking the motor shaft in
place. Then eb = kmω(t) = 0, and equation (1) simplifies to:
dia
(t)
(5) La + R ai
a (t) = ea
(t)
dt
If we apply a voltage step to the motor (ea(t) = ea for t > 0), we expect to see an exponential
La
ea
response with a time constant given by τ = and a steady-state current of i ss = . We can
R a
R a
use measurements of τ and iss to determine Ra and La.
When the motor shaft is unlocked, we can use steady-state measurements of the input current,
motor speed, and applied voltage to determine the torque constant (km) and the viscous damping
coefficient (b). In steady-state, all the derivative terms in equations (1) and (2) are zero, so we
have:
(6) k mωss
= ea
− R ai
a, ss
k mi
a, ss
(7) ωss
=
b
The final parameter to be measured is the motor’s moment of inertia, J. If the electrical time
constant is very fast, the dynamic model for the motor is simplified by neglecting the first term in
equation (1):
(8) i (t) =
e (t) − e (t)
R a a a b
8