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I have an application where I am using an unscented Kalman filter to process data.

While the unscented transformation eliminates the linearization assumption used with the typical state-transition matrix in the extended Kalman filter, there are still areas where the state-transition matrix is useful.

Is there a way to derive a state-transition matrix from a set of propagated sigma points from an unscented transformation?

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  • $\begingroup$ By state transition matrix, do you mean the Jacobian calculated from the nonlinear model of the system whose states you are trying to estimate? $\endgroup$ – Mr. Fegur Jul 10 '15 at 23:54
  • $\begingroup$ Yes. To be explicit, if X is the vector of properties being estimated in the state, then the state transition matrix is: $\Phi = \frac{\partial X(t_{i+1})}{\partial X(t_i)}$ $\endgroup$ – davec Jul 13 '15 at 14:14
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May I ask why you want to obtain the state transition matrix? The unscented filter is used to bypass an inaccurate linearization assumption, which may not reflect what is actually going on and thus throws off linearization techniques like the EKF.

Anyway, one way I would do it is to use the posterior densities $P_{k}$ and $P_{k+1}$ (which you already have thanks to the unscented transform) to weight your sigma points $x_{k}$ and $x_{k+1}$. Next, let $A \in \mathbb{R}^{n \times n}$ be the matrix of unknown coefficients, and define $$ x_{k+1} = Ax_{k}. $$ A least squares fit should then give you $A$, the state transition matrix.

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  • $\begingroup$ As to why: I have two different algorithms and existing implementations that operate over the measurement space and state space respectively to provide additional statistics from the filtering process. I could probably reformulate them for the unscented approach, but I have a fairly linear problem, and so I was hoping to use the existing implementations for some quick analysis. $\endgroup$ – davec Jul 13 '15 at 19:54
  • $\begingroup$ As for you recommendation: I think that makes sense. I will give it a try, see how it works, and report back. Thanks! $\endgroup$ – davec Jul 13 '15 at 19:56
  • $\begingroup$ @Mr.Fegur "A least squares fit should then give you A, the state transition matrix." What would this look like? I haven't seen anything where A is being solved for, rather x_k $\endgroup$ – evan.oman Jan 3 '17 at 15:29

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