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I think I understand the discrete Kalman Filter very well. But I don't exactly follow the continuous time KF.

In discrete Kalman Filter, we assume the measurement noise is "white". The error of our measurement in this moment is not correlated to the error we observe the next moment. Can we realistically assume that in continuous filter? When the time between measurements is so small, there have to be some correlation?

I'm also not sure why $R_t = \frac{R}{\Delta t}$ in the MIT 2.160 | Spring 2006 - System Identification, Estimation, and Learning - Lecture 007. Why the smaller the $\Delta t$ is, the bigger the measurement noise is?

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The noise energy, in the AWGN model, depends on the LPF of the sampling system.
Basically determined by the bandwidth defined by the Sampling Frequency.

See

The notes make the implicit assumption the filter matches the time interval.
Hence smaller time interval means larger bandwidth which means larger noise energy.

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  • $\begingroup$ Thanks. Can you elaborate how $R$ in $R_t = R/\Delta t$ is defined? what "time interval" does the filter match? The filter operates continuously so there isn't an "interval"? It's also not clear to me why smaller time intervals means larger bandwidth means larger noise... Sorry for the basic questions, I have been using discrete KF in a statistical context so the EE aspects are actually new to me. $\endgroup$ Commented Nov 19 at 20:02
  • $\begingroup$ Also, how does continuous Kalman filter handle the issue when measurements are really close to each other, the measurements mostly likely have correlation? $\endgroup$ Commented Nov 19 at 23:18
  • $\begingroup$ @HelenSmith, In the 1st page, at the definition the notes states that $\Delta t$ is the sampling interval hence the analysis. Regarding the bandwidth and noise, read the links I posted. $\endgroup$
    – Royi
    Commented Nov 20 at 5:14
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In discrete Kalman Filter, we assume the measurement noise is "white". ... Can we realistically assume that in continuous filter? When the time between measurements is so small, there have to be some correlation?

The noise doesn't have to be completely white -- it just needs to be flat enough and wideband enough that there's nothing to be gained by modeling it as colored.

For a steady-state Kalman, as long as the noise bandwidth ends up being at least four times faster than the dynamics of the filter. If you need to know whether it matters in those first few moments as the filter is spinning up, that's a separate question*.

Why the smaller the $\Delta t$ is, the bigger the measurement noise is?

I asked just that question the first time I saw that in a textbook (Optimal State Estimation by Simon, in my case).

If the textbook that the lecture refers to uses the same technique, it's because the author assumes an integrating system, and then finds the measurement noise that has the same result.

Note that this result probably isn't true for a different starting system -- I'm not sure why the authors don't start by just saying "c'mon, it's different because it's continuous-time". But they don't. Personally, if I were doing a continuous-time Kalman, I'd use the (hopefully) known spectral density of the process and measurement noise for $\mathbf Q$ and $\mathbf R$, rather than messing around with discrete-time counterparts.


* And one I'd have to work to answer, either by doing some math or finding some papers. I think the first approach I'd take would be to model the system with colored noise, then compare the performance of a Kalman that also models that colored noise vs. one that pretends that the noise is white. That may not give a general result, but it would at least answer the question for a specific system.

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    $\begingroup$ What kind of noise would be "flat enough and wideband enough that there's nothing to be gained by modeling it as colored." $\endgroup$ Commented Nov 20 at 12:41
  • $\begingroup$ Edited! I meant to include that in my answer the first time around -- thanks! $\endgroup$
    – TimWescott
    Commented Nov 20 at 15:35

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