One pole LP and HP¶
- Author or source: Bram
- Created: 2002-08-26 23:33:27
1 2 3 4 5 6 7 8 9 | 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
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Comments¶
- Date: 2006-03-23 15:39:07
- By: moc.liamtoh@wta_sohpyks
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.
- Date: 2006-03-24 09:37:19
- By: q@q
Look again. The -1 is inside the sqrt.
- Date: 2008-08-11 09:34:07
- By: batlord[.A.T.]o2[.D.O.T.]pl
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.
- Date: 2009-07-13 01:22:43
- By: No
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