Order and Cutt off
The Transfer function for Butterworth Low pass Filter is given as
\[
|T_n(j\omega)|^2 = \frac{1}{1 + \left( \frac{\omega}{\omega_0} \right)^{2n}} \tag{1}
\]
Where, ω(0) is cut off frequency
Given specification, we know
\[
\text{At } \omega = \omega_p, \quad \alpha = \alpha_p = \alpha_{\text{max}}
\]
\[
\text{At } \omega = \omega_s, \quad \alpha = \alpha_s = \alpha_{\text{min}}
\]
The attenuation for equation 1 can be represented as
\[
\alpha = -20 \log_{10} \left( |T(j\omega)| \right)
\]
\[
= -20 \log_{10} \left( \frac{1}{\sqrt{1 + \left( \frac{\omega}{\omega_0} \right)^{2n}}} \right)
\]
\[
= 10 \log_{10} \left( 1 + \left( \frac{\omega}{\omega_0} \right)^{2n} \right)
\]
\[
\alpha = 10 \log_{10} \left( 1 + \left( \frac{\omega}{\omega_0} \right)^{2n} \right)
\]
\[
\Rightarrow 10^{\frac{\alpha}{10}} = 1 + \left( \frac{\omega}{\omega_0} \right)^{2n}
\]
\[
\Rightarrow \left( \frac{\omega}{\omega_0} \right)^{2n} = 10^{\frac{\alpha}{10}} - 1
\]
\[
\Rightarrow \frac{\omega}{\omega_0} = \left( 10^{\frac{\alpha}{10}} - 1 \right)^{\frac{1}{2n}}
\]
\[
\text{At } \omega = \omega_p, \quad \alpha = \alpha_p = \alpha_{\text{max}}
\]
\[
\Rightarrow \frac{\omega_p}{\omega_0} = \left( 10^{\frac{\alpha_{\text{max}}}{10}} - 1 \right)^{\frac{1}{2n}} \tag{2}
\]
\[
\text{Similarly, at } \omega = \omega_s,\quad \alpha = \alpha_s = \alpha_{\text{min}}
\]
\[
\Rightarrow \frac{\omega_s}{\omega_0} = \left(10^{\frac{\alpha_{\text{min}}}{10}} - 1\right)^{\frac{1}{2n}} \tag{3}
\]
\[
\text{From equations (2) and (3):}
\]
\
\[
\frac{\omega_p}{\omega_0} = \left(10^{\frac{\alpha_{\text{max}}}{10}} - 1\right)^{\frac{1}{2n}}, \quad
\frac{\omega_s}{\omega_0} = \left(10^{\frac{\alpha_{\text{min}}}{10}} - 1\right)^{\frac{1}{2n}}
\]
\[
\Rightarrow \frac{\omega_s / \omega_0}{\omega_p / \omega_0} =
\frac{
\left(10^{\frac{\alpha_{\text{min}}}{10}} - 1\right)^{\frac{1}{2n}}
}{
\left(10^{\frac{\alpha_{\text{max}}}{10}} - 1\right)^{\frac{1}{2n}}
} = \left(
\frac{
10^{\frac{\alpha_{\text{min}}}{10}} - 1
}{
10^{\frac{\alpha_{\text{max}}}{10}} - 1
}
\right)^{\frac{1}{2n}}
\]
\[
\Rightarrow \left(\frac{\omega_s}{\omega_p}\right)^{2n} = \frac{
10^{\frac{\alpha_{\text{min}}}{10}} - 1
}{
10^{\frac{\alpha_{\text{max}}}{10}} - 1
}
\]
\[
\text{Taking } \log_{10} \text{ on both sides:}
\]
\[
2n \cdot \log_{10} \left(\frac{\omega_s}{\omega_p}\right) =
\log_{10} \left(
\frac{
10^{\frac{\alpha_{\text{min}}}{10}} - 1
}{
10^{\frac{\alpha_{\text{max}}}{10}} - 1
}
\right)
\]
\[
\Rightarrow n = \frac{1}{2} \cdot \frac{
\log_{10} \left(
\frac{
10^{\frac{\alpha_{\text{min}}}{10}} - 1
}{
10^{\frac{\alpha_{\text{max}}}{10}} - 1
}
\right)
}{
\log_{10} \left(\frac{\omega_s}{\omega_p}\right)
}
\]
The above equation represents the order of Butterworth low pass approximations.
It can also be wrriten as
\[
n = \frac{
\log_{10} \left( \frac{10^{\frac{\alpha_{\max}}{10}} - 1}{10^{\frac{\alpha_{\min}}{10}} - 1} \right)
}{
2 \cdot \log_{10} \left(\frac{\omega_p}{\omega_s}\right)
}
\]