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Assume we have a linear state-space model: $$ z_{k} = Hx_{k} + v_{k}\\ x_{k} = F x_{k-1}+ w_{k}. $$

We are interested in filtering, i.e. we aim to estimate $E[x_{n}|z_{0}, \dots, z_{n}]$. If the measurement noise goes to zero, i.e. $var(v_{k}) \to 0$, then filtering does not give much to us and the filtered data is almost the same as original.

What is the opposite situation, when $var(w_{k}) \approx 0$ and $var(v_{k})$ is fixed? What would be the outcome of Kalman filter?

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  • $\begingroup$ I think you have a mistake in your question. v_k is measurement noise. w_k is process noise. $\endgroup$
    – Anton
    Jan 23, 2020 at 13:14
  • $\begingroup$ Dear @Anton, you are right! Thank you! I have edited. $\endgroup$
    – ABK
    Jan 23, 2020 at 19:27

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Then your filter strictly follows the process model and does not give attention to the measurements. You see very smooth output which seems nice, however it is not correct. Your filter will fail when something happens not fitting to the process model.

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  • $\begingroup$ Dear @deathiscertainlifeisnot, could you, please, clarify the first sentence? Actually, I tried to conclude something from predict/update equations, but I did not succeed. $\endgroup$
    – ABK
    Jan 23, 2020 at 12:47
  • $\begingroup$ Considering the Bayesian filtering, your tuning parameters are process and measurement noises. As you stated, if you make process noise goes to zero, then your filter will weight the measurements moret than the process model. Thus, your filter will follow your measurements dynamically and your result will be almost the same with the measurements. On the other hand, if you make your process noise goes to zero, you say to filter that I trust my process model. So your filter directly takes its results from process model. That is why it does not give attention to the measurments. $\endgroup$
    – deathiscertainlifeisnot
    Jan 23, 2020 at 13:00

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