
One pole LP and HP
References : Posted by Bram Code : LP:
recursion: tmp = (1p)*in + p*tmp with output = tmp
coefficient: p = (2cos(x))  sqrt((2cos(x))^2  1) with x = 2*pi*cutoff/samplerate
coeficient approximation: p = (1  2*cutoff/samplerate)^2
HP:
recursion: tmp = (p1)*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 
Comments
Added on : 23/03/06 by skyphos_atw[ AT ]hotmail[ DOT ]com Comment : coefficient: p = (2cos(x))  sqrt((2cos(x))^2  1) with x = 2*pi*cutoff/samplerate
p is always 1 using the formula above. The square eliminates the squareroot and (2cos(x))  (2cos(x)) is 0.
Added on : 24/03/06 by q[ AT ]q Comment : 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 ]]pl Comment : skyphos:
sqrt((2cos(x))^2  1) doesn't equal
sqrt((2cos(x))^2) + sqrt( 1)
so
1 can be inside the sqrt, because (2cos(x))^2 will be always >= one.
Added on : 12/07/09 by No Comment : HP is wrong!
Or at least it does not work here. It acts like a lofi lowshelf. However this works:
HP:
recursion: tmp = (1p)*in + p*tmp with output = intmp
coefficient: p = (2cos(x))  sqrt((2cos(x))^2  1) with x = 2*pi*cutoff/samplerate
coeficient approximation: p = (1  2*cutoff/samplerate)^2

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