# Is it Possible to Derive a State Transition Matrix from an Unscented Transformation?

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?

• By state transition matrix, do you mean the Jacobian calculated from the nonlinear model of the system whose states you are trying to estimate? – Mr. Fegur Jul 10 '15 at 23:54
• Yes. To be explicit, if X is the vector of properties being estimated in the state, then the state transition matrix is: $\Phi = \frac{\partial X(t_{i+1})}{\partial X(t_i)}$ – davec Jul 13 '15 at 14:14

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.
• @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 – evan.oman Jan 3 '17 at 15:29