K Constant for the rotation

The transfer function of the open-loop system is:

\[BO(p) = \frac{L_e(p)}{E(p)} = \frac{k_p k_v e^{-p T_s / 2}}{(1 + p \tau)p}\]

With \(T_s = \frac{1}{100} s\), \(k_v = 17.5 rad / s\), and \(\tau = 0.159 s\)

Using the gain margin criteria:

\[\phi (BO(jw_1)) = - \pi\] \[-w_1 \frac{T_s}{2} - arctan(w_1 \tau) - \frac{\pi}{2} = - \pi\] \[-w_1 \frac{T_s}{2} - arctan(w_1 \tau) = -\frac{\pi}{2}\] \[w_1 \frac{T_s}{2} + arctan(w_1 \tau) = \frac{\pi}{2}\] \[w_1 = 35.282 rad / s\]

Using -6 dB as target value : \(\| BO(jw_1) \| = -6 dB \simeq 0.5\)

\[\frac{k_{pi1} k_v}{w_1 \sqrt{1 + (w_1 \tau)^2}} = 0.5\] \[k_{p1} = \frac{w_1 \sqrt{1 + (w_1 \tau)^2}}{2 k_v} = 5.74 rad^{-1}\]

Using the phase margin criteria:

\[M\phi(BO(jw_2)) = \frac{-5\pi}{6}\] \[-w_2 \frac{T_s}{2} - arctan(w_2 \tau) - \frac{\pi}{2} = \frac{-5\pi}{6}\] \[-w_2 \frac{T_s}{2} - arctan(w_2 \tau) = \frac{-2\pi}{6}\] \[w_2 \frac{T_s}{2} + arctan(w_2 \tau) = \frac{2\pi}{6}\] \[w_2 = 9.761\]

As \(\| Bo(jw_2) \| = 1\) :

\[\frac{k_{p_2} k_v}{w_2 \sqrt{1 + (w_2 \tau)^2}} = 1\] \[k_{p_2} = \frac{w_2 \sqrt{1 + (w_2 \tau)^2}}{k_v} = 1. 0298 rad^{-1}\]

The gain is equal to the lowest value, giving a gain for the rotation of \(1. 0298 rad^{-1}\).

Criticism

The practical results showed that with this rotation gain, the robot didn’t always finish the rotation. Near the end of the rotation, the error was so small that the voltage applied to the motors wasn’t enough to overcome the friction forces. 3 solutions were considered :

Adding a minimal value to the voltage that would ensure it is always high enough to overcome the friction
Adding a small constant to the voltage to compensate the friction’s effect or to increase the gain.
Multiplying the gain

In the end, the gain was multiplied by a factor of 3 (found by trial and error) and the results were sufficiently adequate to retain this solution.