PHY.5:5.1 - Keep a motor’s torque while eliminating fast current dynamics

Consider an ideal linear motor over a range in which resistance R, inductance L, inertia J, damping b and conversion constant k are positive and constant. In consistent SI units, use the same k for torque per current and back voltage per angular speed:

L*i' = V - R*i - k*omega

J*omega' = k*i - b*omega

The electrical and mechanical balances describe the assumed device, including its load in J and b. Saturation, variable load or a different drive would require the corresponding physical relations.

Suppose the work needs the slow speed response. The electrical relaxation time is tau=L/R. The current toward which the electrical part relaxes is q(t)=(V(t)-k*omega(t))/R. Replacing i by q gives

J*omega' = (k/R)*V - (b+k^2/R)*omega.

The eliminated current still supplies driving torque and additional damping. The slow response time of this candidate is J/(b+k^2/R). Compare tau with that time and the drive’s variation time.

To examine the neglected response, set e=i-q. The full electrical equation gives tau*e'=-e-tau*q'. If |q'|<=K on the interval, integration yields

|e(t)| <= |e(0)|*exp(-t/tau) + tau*K*(1-exp(-t/tau)).

The bound on q’ can come from |q'|<=(|V'|+k*|omega'|)/R and the allowed drive and acceleration, using the full physical account where needed. A bound inferred only by assuming the proposed approximation would leave that assumption unresolved.

For example, let R=2 ohms, L=0.02 henry, k=0.1 in the stated SI convention, J=0.02 kg m² and b=0.01 N m s per radian. Then tau=0.01 s and the candidate slow time is about 1.33 s. With |e(0)|<=0.5 ampere and K=1 ampere per second, the current departure at 0.05 s is at most 0.01331 ampere, giving a torque departure at that instant of at most 0.001331 N m relative to kq.

The slow speed error is a different output. With the same initial speed and drive, let delta be full speed minus reduced speed. It satisfies

J*delta' + (b+k^2/R)*delta = k*e.

Let B(s) denote the current-departure bound above. Since the initial speed difference is zero and the response kernel is positive, integration gives

|delta(t)| <= (k/J)*integral_0^t exp(-(b+k^2/R)*(t-s)/J)*B(s) ds.

For the stated values, this speed-departure bound at 0.05 s is about 0.026064 rad/s, hence less than 0.02607 rad/s. An allowed error of 0.03 rad/s therefore permits the reduced calculation for that speed reading. The integral carries the earlier transient into the answer; the small current departure at the final instant alone would not give this bound.

Changed work. If the next question concerns torque immediately after switching, the bound includes the initial current departure. Use the electrical transient. If the drive varies on the electrical relaxation time, recompute its departure instead of extending the slow-drive approximation. These returns change which physical response is retained.