# Sensor fusion - Kalman for two identical position sensors

I require a little help in the implementation of my filter. I am currently working with two Leapmotion devices for the physiological study of hand vibrations. For this we call each of the devices "L leader" and "L support". I have problems with the C matrix, which is responsible for modeling the operation of the sensors. In my case, Leapmotion devices deliver data on position (x, y, z) and of course with this it is possible to detach the speed and use it in the filter, leaving L leader (Px, Py, Pz, Vx, Vy, Vz) and L support (Px, Py, Pz, Vx, Vy, Vz). This would then be as follows for the merger?

What data should be put in Xk*?

How do I model matrix C?

Should the H matrix be a 6x6 identity matrix?

Thanks for read, excuse my bad english.

• If you have two identical sensors, and they're measuring the same thing, why not just average the readings? – Chuck Dec 11 '18 at 15:37
• Two reasons, first, L support goes through a process of rotation and translational through a homogeneous matrix, generating data with a certain level of noise, very different from the original noise. Second, I read that Kalman solves the problem of period differences between the acquired data, which would be of great help to me. – gondsieg Dec 11 '18 at 19:30

The matrix $$C$$ usually is denoted the output matrix of your system. So e.g. if you are interested in the 3 positions $$x,y,z$$ your matrix would look like $$C=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$
The matrix $$H$$ on the other hand, is the matrix describing uncertainties in your system (which could also be measurement noise) and has the dimension $$3\times r$$ for the case of a 3-dimensional output. $$r$$ is the dimension of your assumed uncertainty (could be 6 for the noise of your $$2\times3$$ dimensional sensors). In practice, it is feasible to assume the uncertainties are not affecting each other, in this case the matrix $$H$$ simplifies to a diagonal matrix.