# LES filtering

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 June 17, 2005, 10:54 LES filtering #1 andy Guest   Posts: n/a We can decompose f into its filtered part [f] and subgrid part f": f = [f] + f" ------------------ 1 If we interpret [f] as the total-field filtered, can we then re-decompose it for a second time, i.e: f = [ [f] + f" ] + f" ------------------ 2 which is the same as: f = [ [f] ] + [f"] + f" ------------------ 3 Can some one please tell me if we can do this? Is equation 1 = equation 3? i.e. is [f] + f" = [ [f] ] + [f"] + f" for all types of filter? Thanks.

 June 17, 2005, 11:25 Re: LES filtering #2 agg Guest   Posts: n/a Yes, you can do that. In fact it is the base for scale similar models and I do not think it depends on the type of filter.

 June 17, 2005, 11:42 Re: LES filtering #3 Salvador Guest   Posts: n/a Yes but this decomposition doesn't help much. [f"] is NOT 0 so you do not go very far. Using spectral filters [f"]=0 and then [[f]]=[f] Check Pope's book on turbulent flows

 June 17, 2005, 11:48 Re: LES filtering #4 andy Guest   Posts: n/a You will have to say more about the properties of your filters. If [f"] = 0 and [[f]] = [f] then yes but this is trivial. Usually a second filter is different to the first and, hence, the subgrid components are different.

 June 17, 2005, 15:09 Re: LES filtering #5 noName Guest   Posts: n/a Your derivation is correct and [f] + f" = [ [f] ] + [f"] + f" is valid for all types of filters. The only thing you assumed is that the filter kernel is distributive, i.e. [ [f] + f'' ] = [ [f] ] + [f'']. In fact the filtering convolution integral is associative, commutative and distributive ... so the algebra is simple.

 June 17, 2005, 19:55 Re: LES filtering #6 andy Guest   Posts: n/a thanks guys this has been helpful.

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