Consider a very simple, discrete-time constant position-type model for state updating in a Kalman filter:

$$ x_{k+1} = x_k + w_k $$

The Kalman filter will be run with update interval $T_s$ such that $x_k$ represents position at time $t=kT_s$.

Typically, the process noise $w_k$ is modeled as a zero-mean, white stationary random process of variance $\sigma^2_w$.

Now, assume that the position time series is strictly band-limited with bandwidth far lower than the Nyquist frequency $1/(2T_s)$. This corresponds to a high degree of oversampling.

In principle (say, with zero measurement noise), the process noise could be recovered by applying a simple differencing high pass filter with transfer function:

$$ H(z) = 1 - z^{-1} $$

to the position time series. However, if the position time series is strictly bandlimited, then the recovered process noise is strictly bandlimited, which violates the white process noise model.

Does this mean that the Kalman filter needs to use a correlated process noise model for highly oversampled systems?

In my case, the state space model (which, in fact, is constant velocity not constant position) describes the time evolution of a certain biological quantity. The discrete-time state space model has not been chosen from a precise description of the dynamics of the underlying physical (physiological) phenomena, (This is because such descriptions are complex and require access to additional biological quantities which are rarely knowable in practice). Rather, the constant velocity model is a common choice in the literature motivated primarily by expedience, simplicity, and good tracking results.

In the literature, all results seem to be reported for near-Nyqvist sampling. Some studies in the literature analyzed real biological data (again, sampled near Nyqvist) and claimed that the process noise for this model was white. In my case, for historical reasons, I am oversampling by a good factor of 15. When I analyzed a very clean sample signal, I found that at the oversampled rate, the process noise was highly correlated. A decimated-to-near-Nyqvist version of the signal gave rise to a significantly lower amount of temporal correlation in the process noise. This is what motivated my question.

  • $\begingroup$ I'm sorry but if you have a constant signal model (exact model), then your process noise should be zero... If you want to assume a non zero process noise $w_k$, it means you do not have an exactly constant signal model...? $\endgroup$ – Fat32 Sep 7 '17 at 19:53
  • $\begingroup$ This is a question of terminology. In target tracking, there are "constant position", "constant velocity," and "constant acceleration" models with process noise. See e.g.,(ai.mit.edu/courses/6.801/Fall2002/lect/lect21.pdf) slides 19-20 $\endgroup$ – rhz Sep 8 '17 at 0:28

The answer to your question is : maybe. When one is at the discrete time model, it is generally assumed that you solved the state transition equations from the continuous time differential equations where the physics have already been appropriately modeled, which is where your answer lies. Time discretization is based on sampling the continuous time state transition matrix, thus the sampling intervals aren't related to Nyquist, they correspond to the fidelity of the modeling. Shorter intervals tend to be more accurate approximations. The time intervals don't have to be uniform, but uniform intervals provide a less complicated implementation.

While shorter time intervals correspond to a heuristic where shorter has better approximation, you raise a valid concern, sampling too closely, may indeed introduce correlation in the process noise. This can be solved with an augmented state.

These are all considerations. Kalman Filters need to be tuned. The measurement noise has been neglected in this discussion.

So, the answer to your question is that it really depends.

  • $\begingroup$ Yes. additional info added to body of question. I think you're right. Another issue for me is distinguishing between measurement noise outliers and the equivalent of a “maneuver” in the state. See: dsp.stackexchange.com/questions/43560/… $\endgroup$ – rhz Sep 11 '17 at 1:25
  • $\begingroup$ if you use the Innovations form of the Kalman Filter, the independence (whiteness) of the innovations provides a diagnostic of the fidelity (order) of the filter $\endgroup$ – Stanley Pawlukiewicz Sep 11 '17 at 1:35

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