Minimum Butterworth order (1 dB, 25 dB, ratio 1.5)

For a Butterworth response normalized to $\omega_p$ = 1, $\alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})$, with $\varepsilon^2 = 10^{0.1\alpha_p} - 1$. The stopband condition gives

$$n \ge \frac{\log_{10}\left[(10^{0.1\alpha_s} - 1)/(10^{0.1\alpha_p} - 1)\right]}{2\log_{10}(\omega_s/\omega_p)}$$ $$\begin{aligned} 10^{0.1\alpha_s} - 1 &= 10^{2.5} - 1 = 315.2278 \\ 10^{0.1\alpha_p} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} &= 1.5 \\ n &\ge \frac{\log_{10}(315.2278/0.25893)}{2\log_{10}(1.5)} = \frac{3.0854}{0.3522} = 8.761 \end{aligned}$$

The order must be an integer, so n = 9.

Check: with $n = 9$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 1.5^{18}) = 25.84$ dB $\ge 25$ dB.

Butterworth characteristic with minimum order:

$$|T(j\omega)|^2 = \frac{1}{1 + 0.2589\,\omega^{18}}, \qquad \varepsilon = 0.5088$$

($\omega$ normalized to $\omega_p$). It gives exactly 1 dB at $\omega$ = 1 and 25.84 dB at $\omega$ = 1.5. The half-power frequency is $\omega_{3dB} = \varepsilon^{-1/9} = 1.0780$, and the 9 poles lie on a circle of this radius, spaced 20° apart (one real pole at $-1.0780$).