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Old   June 17, 2005, 11:54
Default LES filtering
  #1
andy
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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.
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Old   June 17, 2005, 12:25
Default Re: LES filtering
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agg
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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.
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Old   June 17, 2005, 12:42
Default Re: LES filtering
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Salvador
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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
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Old   June 17, 2005, 12:48
Default Re: LES filtering
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andy
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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.
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Old   June 17, 2005, 16:09
Default Re: LES filtering
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noName
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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.
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Old   June 17, 2005, 20:55
Default Re: LES filtering
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andy
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thanks guys this has been helpful.

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