# One pole LP and HP¶

• Author or source: Bram
• Created: 2002-08-26 23:33:27
code
 ```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 ```

```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
```