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 One pole LP and HPReferences : Posted by BramCode : LP: recursion: tmp = (1-p)*in + p*tmp with output = tmp coefficient: p = (2-cos(x)) - sqrt((2-cos(x))^2 - 1) with x = 2*pi*cutoff/samplerate coeficient approximation: p = (1 - 2*cutoff/samplerate)^2 HP: recursion: tmp = (p-1)*in - p*tmp with output = tmp coefficient: p = (2+cos(x)) - sqrt((2+cos(x))^2 - 1) with x = 2*pi*cutoff/samplerate coeficient approximation: p = (2*cutoff/samplerate)^2

 CommentsAdded on : 23/03/06 by skyphos_atw[ AT ]hotmail[ DOT ]comComment : coefficient: p = (2-cos(x)) - sqrt((2-cos(x))^2 - 1) with x = 2*pi*cutoff/samplerate p is always -1 using the formula above. The square eliminates the squareroot and (2-cos(x)) - (2-cos(x)) is 0.Added on : 24/03/06 by q[ AT ]qComment : Look again. The -1 is inside the sqrt.Added on : 11/08/08 by batlord[[ DOT ]A[ DOT ]T[ DOT ]]o2[[ DOT ]D[ DOT ]O[ DOT ]T[ DOT ]]plComment : skyphos: sqrt((2-cos(x))^2 - 1) doesn't equal sqrt((2-cos(x))^2) + sqrt(- 1) so -1 can be inside the sqrt, because (2-cos(x))^2 will be always >= one. Added on : 12/07/09 by NoComment : HP is wrong! Or at least it does not work here. It acts like a lofi low-shelf. However this works: HP: recursion: tmp = (1-p)*in + p*tmp with output = in-tmp coefficient: p = (2-cos(x)) - sqrt((2-cos(x))^2 - 1) with x = 2*pi*cutoff/samplerate coeficient approximation: p = (1 - 2*cutoff/samplerate)^2

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