# Kalman Filter: Do we need to measure each entry of the output-sequence, or can we derive them?

A Kalman-Filter gets applied to a state-space model in order to obtain estimates of the state-vector.

Example: Assume a car moving on a straight line in x-direction (this is a 1-dimensional problem). The State-space model could be described by:

$$\begin{bmatrix} x_k \\ \dot{x}_k \\ \ddot{x}_k \end{bmatrix} = \begin{bmatrix} 1 & T & \frac12\,T^2 \\ 0 & 1 & T \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{k-1} \\ \dot{x}_{k-1} \\ \ddot{x}_{k-1} \end{bmatrix} + w_{k-1}$$

$$\begin{bmatrix} y_k \\ \dot{y}_k \\ \ddot{y}_k \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_k \\ \dot{x}_k \\ \ddot{x}_k \end{bmatrix} + v_k$$

The y-vector is our measurement-sequence. And here comes my question: For the Kalman-Filter to work, is it necessary to measure each of the entries of the y-vector?

Coming back to the car: I can measure the number of rotations of the driving-shaft using a wheel-encoder. From this I can infer the distance travelled, which is my x-coordinate assuming a movement on a straight line in x-direction.

I cannot measure the velocity directly. But I can infer the velocity from the number of edges detected within a certain time-gap, using the wheel encoder. And I can derivate velocity in order to obtain acceleration.

So, what I am saying is: In order to obtain each entry of the y-vector, I am using but one sensor: The wheel encoder. Is this valid when applying the Kalman-Filter equations?

• Is the first question about if KF can have missing data? The second question is asking the adequacy of encoder as sensor? May 6, 2019 at 8:53
• I think I am just asking one question here: For the given state-space model, do I need to have sensors directly measuring y, d/dt y and d^2/dt^2 y ? May 7, 2019 at 9:04

If you formulate your Kalman filter with the output matrix as eye(3) - $$\left[\begin{array} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$$
then you need to provide all those measurements, but you could alternatively define the output matrix as eye(1,3) -
$$\left[\begin{array} \\ 1 & 0 & 0 \\ \end{array}\right]$$
• @Chunk: How did you find out that the system is still observable? - the observation matrix is [C, CA]^T and that would be [1 0 0, 1 T 0.5T^2] and its determinant is 0, isn't it? Jun 9, 2019 at 19:15
• whops, I made a mistake. The observation matrix would be C, CA, CA^2]^T and that's 1 0 0, 1 T 0.5T^2, 1 2T 2T^2]^T and its determinant is not zero Jun 9, 2019 at 19:21