# Lowpass Butterworth Filter

#### schmitt trigger

Joined Jul 12, 2010
510
N is the filter order.
Since there are no “non-integer” filter orders, you round UP to the next integer.

#### pinkyponky

Joined Nov 28, 2019
318
N is the filter order.
Since there are no “non-integer” filter orders, you round UP to the next integer.
I knew that, it is a filter order, I would like consider this like a example and I have to derive filter order for my circuit. So that's why I asked how to derive that.

#### LvW

Joined Jun 13, 2013
1,586
The formula you have shown (magnitude of H_lp) applies for Butterworth filters only.
The general expression for the magnitude of a Butterworth response is
|H(jw)| = 1/[SQRT(1+Ω to the power of 2n)].
The quantity Ω is the frequency parameter - normalized to the defined cut-off.
So - when the required magnitude is given you can solve for another unknown (in ths case "n").

#### pinkyponky

Joined Nov 28, 2019
318
The formula you have shown (magnitude of H_lp) applies for Butterworth filters only.
The general expression for the magnitude of a Butterworth response is
|H(jw)| = 1/[SQRT(1+Ω to the power of 2n)].
The quantity Ω is the frequency parameter - normalized to the defined cut-off.
So - when the required magnitude is given you can solve for another unknown (in ths case "n").
Hi LvW,

Here, the problem is I don't know how to solve the 'n' value. I'm asking you to help me how to solve the 'n' value and if explain step by step it's grateful.
Thank you.

#### LvW

Joined Jun 13, 2013
1,586
Hi LvW,

Here, the problem is I don't know how to solve the 'n' value. I'm asking you to help me how to solve the 'n' value and if explain step by step it's grateful.
Thank you.
Applying the antilog function we have an equation:
0.0707=1/SQRT (....).
After squaring both sides it should be not a problem to solve for "n".