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I'm a student that recently started taking a course on cognitive robotics. The book I use is Probabilistic Robotics by Thrun Burgard and Fox.

In the EKF algorithm, we linearized the action model in the following way

$$g(u_t,\:x_{t-1})\:=\:g(u_t,\:\mu _{t-1})+G_t\cdot (x_{t-1}-\mu \:_{t-1})$$

$g(u_t,\:x_{t-1})$ is the action model and $G_t$ is its Jacobian matrix with respect to the state $x_{t-1}$.

I don't see how this guarantees linearity because $g$ could be nonlinear in $u_t$. The authors don't mention anything about why this is the case.

In other words, I imagined that the multivariate taylor expansion for this where we get a linear function in both $u_t$ and $x_{t-1}$

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  • $\begingroup$ "The authors don't mention anything about why this is the case" don't accuse the authors if you didn't read the book deeply enough. This is not linear rather is linearized model (i.e. big difference). The authors extensively explained it and you are using Extended Kalman filter not Kalman filter. $\endgroup$
    – CroCo
    Commented Oct 20, 2022 at 19:57
  • $\begingroup$ Thank for your nice comment. Mentioning that the author's didn't mention why this is the case doesn't mean I'm accusing them for insufficiently explaining the concept. It only means that there might be some fact that I'm not aware of that justifies approximating this way which is what brought me here. Furthermore, authors mention "Linearization approximates the nonlinear function g by a linear function that is tangent to g" which doesn't tally with your statement that this is "not linear rather is linearized". You can also check the mathematical definition of "linearization" to check. $\endgroup$
    – Essam
    Commented Oct 21, 2022 at 6:09
  • $\begingroup$ I believe you might be misunderstanding my question. Because $g$ is multivariate, I expected linearization to take the following form described here web.ma.utexas.edu/users/m408m/Display14-4-3.shtml However, the term responsible for linearizing $g$ in terms of the action $u_{t}$ seems to be missing in the authors' formulation. $\endgroup$
    – Essam
    Commented Oct 21, 2022 at 6:12

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