Mathematical meaning of window averaging in fieldAverage functionObject

Tobermory · November 15, 2020, 12:47

I have been taking a look at the fieldAverage functionObject, for time averaging fields during a simulation. The basic implementation is cunningly coded to use the minimum amount of storage, by calculating the mean and variance at the new time step using the latest data and the mean from the old time step. You can show through some simple algebra that this provides an exact estimation of the mean and variance, based on the complete sample set.

Now, my question is, what do you get when you use the "window" function?

In an ideal world, for a window size of w samples, it would be the average over the last w samples, but this is clearly impossible to calculate without storing w samples, since you can't simply discard the (n-w)'th sample. But does the window function give something that approximates this? Let's take a look.

The algorithm for the mean value of

samples (see fieldAverageTemplates.C) is: $\overline{U}_n = (1-\beta)\overline{U}_{n-1} + \beta U_n$ , where $\beta = \delta t/T=1/n$ (assuming no windowing and equal spaced samples) and $\overline{U}_{n-1}$ is the average of the first

samples. Expanding out $\overline{U}_{n-1}$ , you can show that this reduces to $\overline{U}_n = (1/n) \sum_{i=1}^n{U_i}$ , as desired.

Now, if windowing is active with a window size of

, then when the number of samples exceeds the window size the following limit is applied: $\beta = \delta t/w = 1/m$ , where

is now a constant. Let's take an example with a window size of 2 samples (

here):

$\overline{U}_1 =U_1$
$\overline{U}_2 =\frac{1}{2}(U_1 + U_2)$
$\overline{U}_3 =\frac{1}{4}(U_1 + U_2) + \frac{1}{2}U_3$
$\overline{U}_4 =\frac{1}{8}(U_1 + U_2) + \frac{1}{4}U_3 + \frac{1}{2}U_4$
$\overline{U}_5 =\frac{1}{16}(U_1 + U_2) + \frac{1}{8}U_3 + \frac{1}{4}U_4 + \frac{1}{2}U_5$

etc. - you can see the pattern. My question is - what the heck is this quantity? It clearly slowly diminuishes the effect of the earlier samples, but the average is slightly biased to the most recent samples (that's exaggerated in the above example; for larger values of m, then $1 -\beta \approx 1$ and the biasing is less strong). It's certainly not a balanced average of the recent

samples.

So what is the physical meaning of the windowed average, and when might you want to use it?

navidmt · December 31, 2022, 05:38

Very good thread! Have you got the answer? If so, would you mind sharing it here? Thanks

Tobermory · December 31, 2022, 10:44

Given the silence from the community on this one, I think it's safe to assume that there probably is no clear meaning to the quantity and I'm not sure why/when you would ever want to use it.

Here's my interpretation of the quantity: start off with the initial definition:
$\overline{U}_n = (1-\beta)\overline{U}_{n-1} + \beta U_n$
and rearrange:
$(\Delta \overline{U})_n = \overline{U}_n - \overline{U}_{n-1} = \beta (U_n - \overline{U}_{n-1} )$
and now you see that the change in the average value is the weighted difference between the new instantaneous value and the old mean, with the weight being $\beta = 1/m$ .

This can be visualised in the following chart, for a sequence of n=20 and window values of m = 2, 5 and 10. The chart shows the factors on the

values in the arithmetic sequence (refer back to my original post) ... as can be seen, there's no real meaning to the result - it's not an average of the last m values, which was what I originally thought it might be aiming to provide. It's just a complex mash of all of the previous values, with higher weighting to the more recent values.

So, all in all - unless someone can provide a justification, my advice is to ignore it.

November 15, 2020, 12:47	Mathematical meaning of window averaging in fieldAverage functionObject	#1
Tobermory Senior Member Join Date: Apr 2020 Location: UK Posts: 736 Rep Power: 14	I have been taking a look at the fieldAverage functionObject, for time averaging fields during a simulation. The basic implementation is cunningly coded to use the minimum amount of storage, by calculating the mean and variance at the new time step using the latest data and the mean from the old time step. You can show through some simple algebra that this provides an exact estimation of the mean and variance, based on the complete sample set. Now, my question is, what do you get when you use the "window" function? In an ideal world, for a window size of w samples, it would be the average over the last w samples, but this is clearly impossible to calculate without storing w samples, since you can't simply discard the (n-w)'th sample. But does the window function give something that approximates this? Let's take a look. The algorithm for the mean value of samples (see fieldAverageTemplates.C) is: $\overline{U}_n = (1-\beta)\overline{U}_{n-1} + \beta U_n$ , where $\beta = \delta t/T=1/n$ (assuming no windowing and equal spaced samples) and $\overline{U}_{n-1}$ is the average of the first samples. Expanding out $\overline{U}_{n-1}$ , you can show that this reduces to $\overline{U}_n = (1/n) \sum_{i=1}^n{U_i}$ , as desired. Now, if windowing is active with a window size of , then when the number of samples exceeds the window size the following limit is applied: $\beta = \delta t/w = 1/m$ , where is now a constant. Let's take an example with a window size of 2 samples ( here): $\overline{U}_1 =U_1$ $\overline{U}_2 =\frac{1}{2}(U_1 + U_2)$ $\overline{U}_3 =\frac{1}{4}(U_1 + U_2) + \frac{1}{2}U_3$ $\overline{U}_4 =\frac{1}{8}(U_1 + U_2) + \frac{1}{4}U_3 + \frac{1}{2}U_4$ $\overline{U}_5 =\frac{1}{16}(U_1 + U_2) + \frac{1}{8}U_3 + \frac{1}{4}U_4 + \frac{1}{2}U_5$ etc. - you can see the pattern. My question is - what the heck is this quantity? It clearly slowly diminuishes the effect of the earlier samples, but the average is slightly biased to the most recent samples (that's exaggerated in the above example; for larger values of m, then $1 -\beta \approx 1$ and the biasing is less strong). It's certainly not a balanced average of the recent samples. So what is the physical meaning of the windowed average, and when might you want to use it?

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December 31, 2022, 05:38		#2
navidmt New Member navid toussi Join Date: Nov 2015 Posts: 20 Rep Power: 10	Very good thread! Have you got the answer? If so, would you mind sharing it here? Thanks