# One pole LP and HP¶

**Author or source:**Bram**Created:**2002-08-26 23:33:27

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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¶

**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
```