2
$\begingroup$

For the non-additive noise case, \begin{equation} x_k = f(x_{k-1}, u_{k-1}, \xi_{k-1}) \\ y_k = h(x_k, \nu_k) \end{equation}

the EKF takes into account the jacobian wrt to the noise terms $ L_{k-1} = \frac{\partial f}{\partial \xi} |_{\hat{x}_{k-1|k-1}, u_{k-1}} $ and $ M_{k} = \frac{\partial h}{\partial \nu} |_{\hat{x}_{k|k-1}} $

I know that jacobian wrt to state: $ A = \frac{\partial f}{\partial x}$ and $ H = \frac{\partial h}{\partial x}$ are evaluated at means of noises i.e. at $\xi = 0, \nu = 0$ resp.

But I'm not clear where to evaluate $L, M$ ? The above expression for $L_{k-1} , M_k$ tells only the $x, u$ but not the noise terms. Should $\xi = 0, \nu = 0$ or some random sample?

$\endgroup$
0
$\begingroup$

Yes, you should use 0 for both. As the EKF assumes that noises have 0 mean, I would evaluate them there. If you look up the Wikipedia page, it also suggests the same.

If you would use a single sample, then

  1. You would need a random number generator (if you use the algorithm on an embedded system, you might want to avoid this).
  2. Most importantly, this would introduce high variance - you can use multiple samples, but then using the mean could be a compromise.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.