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Why do these two voltage-to-torque methods behave so differently for my DC motor model?

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As my first real Simulink project, I'm making a reaction-wheel stabilizer. I'm using Simscape Multibody and Model Linearizer to control the motor with a linear quadratic regulator and the motor voltage as my actuation input. I'm ignoring inductance.

Here is a picture of the reaction-wheel rod model to give an idea of what I'm working with, along with the Simulink model I'm controlling it with. Note that Simscape Multibody is a simulation package which allows you to feed simulated measurements into a simulink model, and use Simulink outputs as torques and forces. The measurements on the left are taken directly from the physical model, and Simscape figures out how they should change based on the input torque shown on the right.

Wheel rod and simulink model

That section on the right with the manual switch decides which of two strategies are employed to turn voltage into torque. At first, I made my model assuming voltage instantaneously becomes torque using \$\tau = kv/R\$, but I know this was a gross simplification.

Seeking to see what a more realistic voltage to torque model would be, I first thought to find a Torque per Volt transfer function derived from \$v = iR + k\omega\$. I told the linearizer app to treat the TF as the gain \$kv/R\$, so I could see if the more realistic behavior would cause instability or if the controller would handle it:

$$v = iR + k\omega \\~\\kv - k^2\omega = \tau R \\~\\kV(s) - \frac {k^2} {Js} T(s) = T(s)R \\~\\kV(s) = T(s) \left[ R + \frac {k^2} {Js} \right] \\~\\\frac {T(s)} {V(s)} = \frac k {R + \frac {k^2} {Js}} = \frac J k \cdot \frac s {\left(\frac {JR} {k^2}\right)s + 1} \\~\\= \frac J k \left[ \frac s {\tau_ms + 1} \right]\tag 1$$

Where \$J\$ is the inertia of the rotor and all attached to it, \$k_e = k_t = k\$ is the torque constant in \$V/(rad/s)\$ or \$Nm/A\$, and \$\tau_m\$ is the mechanical time constant \$JR/k^2\$ in seconds. A datasheet described this constant as "the time it takes the shaft to reach 63% of equilibrium speed from rest due to a voltage step".

The following is the step response of this transfer function, and some simulation measurements with the manual switch in the up position, linearized assuming the TF is a gain \$kv/R\$ and with an appropriate LQR matrix computed for this linearization. In the simulation, the disturbance is a small torque CCW, then CW, then none:

Step response of the TF and some simulation curves

In the second strategy (manual switch down position), I attempted to forget the transfer function, and use the core law: \$v = iR + k\omega\$. Given the LQR's terminal voltage output, I compute \$\tau\$ as \$\tau = k(v - k\omega) / R\$ using the actuation voltage and angular velocity feedback from the motor. Again, I linearize the system, this time as-is, and apply the proper LQR.

Strategy two's curves

That's so much more reasonable. I expected both of these strategies to be equivalent, since they came from the same law.

What am I misunderstanding here?


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