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Butterworth Lowpass Filter

Theory

The magnitude-squared response of an analog lowpass Butterworth filter is given by

$$
|H_a(j\Omega) |^2 = \frac{1}{1+(\Omega/\Omega_c)^{2N}}
$$

Passband ripple: $\delta_s$
Stopband ripple: $\delta_s$

$$
|H_a(j\Omega_p) |^2 = \frac{1}{1+(\Omega_p/\Omega_c)^{2N}} = (1-\delta_p)^2 = \frac{1}{1+\epsilon^2}
$$

$$
|H_a(j\Omega_s) |^2 = \frac{1}{1+(\Omega_s/\Omega_c)^{2N}} = \delta_s^2 = \frac{1}{A^2}
$$

Solving above

$$
N = \frac{1}{2} \frac{log{10}[(A^2-1)/\epsilon^2]}{log{10}(\Omega_s/\Omega_p)}
$$

Example

Passband edge frequency : $\Omega_p = 1$
Stopband edge frequency : $\Omega_s = 2.56$
Passband ripple: $\delta_p = 0.01$
Stopband ripple: $\delta_s = 0.005$

calculating $\epsilon$ and $A$

$$
(1-\delta_p)^2 = \frac{1}{1+\epsilon^2}
$$

$$
\delta_s^2 = \frac{1}{A^2}
$$

$\epsilon = 0.1425$ and $A = 200$ , Then calculating N

$$
N = \frac{1}{2} \frac{log{10}[(A^2-1)/\epsilon^2]}{log{10}(\Omega_s/\Omega_p)} = 7.7092
$$

$N$ should be an integer, so $N = 8$

Then calculating $\Omega_c$

$$
(\Omega_p/\Omega_c)^{2N} = \epsilon^2
$$

or

$$
\frac{1}{1+(\Omega_s/\Omega_c)^{2N}} = \delta_s^2
$$

It can gives the minimum of $\Omega_c = 1.2785 $ and maximum of $\Omega_c = 1.3201$

Here shows the diffrent $\Omega_c $ of the magnitude response