smoothing and filter frequency

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Hallo,

is there a way to relate the "num" parameter in binomial smoothing to the filter frequency of a corresponding Gauss-filter (or experimental Bessel-Filter)?

Essentialy I want to filter/smooth the data in processing and want to relate this to a possible experimental filtering.

Thanks in advance

Schmaud
I believe the 'num' parameter is equivalent to the number of iterative passes of the binomial filter. That is, using a filter with 'num' is the same as sequentially smoothing with num=1, 'num' times. This means the Gaussian variance is proportional to 'num', the standard deviation is proportional to sqrt(num), and the frequency bandwidth is inversely proportional to the standard deviation. You can do the math for the particular definition of filter frequency you wish to apply.

May 24, 2017 at 02:46 am - Permalink

Hallo,

just an update: I did some test and for "smooth" the following seems to apply approximately:

Cuttoff frequency = sampling frequency / 5 / root(par)
( with cutoff frequency is the frequency where only 1/root(2)of the amplitude is left after filtering)

This is of course not very analytical, for the more theoretically inclined:

Some more information than in the online help is here:
https://www.wavemetrics.com/products/igorpro/dataanalysis/signalprocess…

The mentioned algorithm Marchand and Marmet (1983) is here:
http://aip.scitation.org/doi/pdf/10.1063/1.1137498


Best

schmaud

August 16, 2017 at 07:13 am - Permalink

Ralf,

In the limit of large smoothing parameters (which is the Gaussian filter), your approximate numerical factor (1/5) is ideally sqrt(2 ln2)/(2*pi) = 0.1874

Also, for n>50, to achieve highest accuracy use Variable/G root:V_doOrigBinomSmooth=1

Even for low filter orders, a further improvement in finding the closest corresponding Gaussian filter (and its corresponding band-width) is to use (for unit sampling interval) a Gaussian variance sigma^2 = n/2 + 1/8 .

EDIT UPDATE: the exact result for the binomial filter cutoff frequency is actually buried in the above Marchand reference. After rewriting their equation (10) for unit-time sampling with my IP analysis, and taking its inverse,  Fcutoff = (1/pi) * acos( 0.5^(1/(4*n)) . This is easily derivable from the starting kernel {1/4, 1/2, 1/4} but is not as transparent as using the Gaussian approximation.

August 29, 2017 at 05:08 am - Permalink

This was very helpfull - thanks everyone!
Filtering is quite somewhat more complicated than initially thought...

s.r.chinn wrote:
your approximate numerical factor (1/5) is ideally sqrt(2 ln2)/(2*pi) = 0.1874

Thank you - I prefer a rational explanation over some fudge factor any day.

johnweeks wrote:
File->Example Experiments->

I had not used them in years - I completely forgot how useful they are.
Thanks a lot for the great resource!

August 17, 2017 at 06:58 am - Permalink