I am studying the book "B. Sicilliano -Robotics. Modelling, Planning and Control" chapter 11, Input/output linearization topic. Where does the formula for theta in 11.75 come from?

In the case of the unicycle, consider the following outputs:

$ y_1 = x + b cos θ; $

$ y_2 = y + b sin θ $

with $ b \neq 0 $. They represent the Cartesian coordinates of a point B located along the sagittal axis of the unicycle at a distance |b| from the contact point of the wheel with the ground. How to find the following input-output linearization?

$ \dot{y_1} = u_1; $

$ \dot{y_2} = u_2; $

$ \dot{\theta} = \frac{u_2 \cos\theta − u_1\sin\theta}{b} $

I am not very good at robotics, I would like to get an answer simple to understand.

Thank you for all the help.

  • $\begingroup$ For those who don't have the book, it would be great that you mention what do the variables (b, u_1, u_2) stand for. $\endgroup$ – Joe Feb 8 at 9:58
  • $\begingroup$ academia.edu/23785978/… $\endgroup$ – jsotola Feb 8 at 16:44

If you observe that $\omega = \dot{\theta}$, you can readily derive the equation at hand from:

$ \left[ \begin{array}{c} v \\ \omega \end{array} \right] = \ \left[ \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta/b & \cos\theta/b \ \end{array} \right] \ \left[ \begin{array}{c} u_1 \\ u_2 \end{array} \right]. $

  • 1
    $\begingroup$ Thank you. Can't believe I missed that! $\endgroup$ – Leon Rai Feb 8 at 11:00

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