7
$\begingroup$

I would like to run a Kalman filter over a set of measurements which may (will) be correlated. Essentially, each new measurement contains (say) 90% of the same information from the previous measurement. The measurements are made by integrating over time, and subsequent measurements are taken by overlapping integrations.

For various reasons, I can't necessarily decimate the measurements so that they are independent of each other, but if I apply each measurement at its full weighting, the filter will become over confident and its covariance will shrink too quickly.

My first thought is that there must be someway to increase the covariance from each measurement so that it doesn't overly affect the filter state, but I would like a rigorously correct solution, and I welcome any pointers.

A least squares approach which allowed expressing the correlated measurements would also be welcome, but I kind of prefer the Kalman filter because it gives me a running covariance as well as the filtered result.

Thank you in advance for any pointers! Is there a better stack exchange for this question?

$\endgroup$
1
$\begingroup$

The measurements are made by integrating over time, and subsequent measurements are taken by overlapping integrations.

Have you considered creating a "rate" state and taking measurements in the "rate" state.

For example suppose your state was position and your sensor was a velocity sensor and the "position measurement" is the integral of the velocity sensor measurements.

Instead reformulate your problem with 2 states, a position state and a rate state. Take the measurements in rate. For example \begin{equation} \left[ \begin{array}{c} x \\ \dot{x} \end{array} \right]_{t+1} = \left[ \begin{array}{c} 1 & \Delta t \\ 0 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ \dot{x} \end{array} \right]_{t} + q_t \end{equation} \begin{equation} y_t = \left[ \begin{array}{c} 0 & 1 \\ \end{array} \right] \left[ \begin{array}{c} x \\ \dot{x} \end{array} \right] + r_t \end{equation}

$\endgroup$

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.