In discussing smoothing filters, Numerical Recipes p. 772 says

... irregularly sampled data, where the values $f_i$ are not uniformly spaced ...
one can simply pretend that the data points are equally spaced ... a rough criterion:
If the change in $f$ across the full width of the $N$ point window is less than $\sqrt{N/2}$ times the measurement noise on a single point, then the cheap method can be used.

Where does this come from ? It looks odd: for $f(x) = \text{sin} ( 2 \pi f x )$ on [$-N \dots N]$,
change in $f = 2 < \sqrt{N/2}\ \sigma$ would require bigger $N$ for smaller $\sigma$ . Or have I misunderstood ?

Also, does this work for smoothing data on non-uniformly spaced 2d squares, 3d cubes ... ?

Added: https://gist.github.com/denis-bz/5957279 is a short Python program that tries to elucidate this, for quantization noise (sample at integers +- random-uniform( -1/2, 1/2 )).

  • $\begingroup$ I can only provide my scientific hunch on where to look... If the error in the x direction of the data is greater than the magnitude to the spacing, then shifting the grid doesn't really alter the shape of the data very much. I imaging by looking at the standard error (which is defined as $\sqrt(\sigma/N)$ by the way, similar function to the one you mention), that you can derive the condition yourself. $\endgroup$
    – boyfarrell
    Jul 1 '13 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.