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One pole LP and HP

References : Posted by Bram
Code :
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



Comments


Added on : 23/03/06 by skyphos_atw[ AT ]hotmail[ DOT ]com
Comment :
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 ]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((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 No
Comment :
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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