Butterworth order (0.5 dB, 22 dB)

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for the given specifications

$\omega_p$ = 2000 rad/s, $\alpha_{max}$ = 0.5 dB, $\omega_s$ = 3000 rad/s, $\alpha_{min}$ = 22 dB.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2.2} - 1 = 157.4893 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\ \frac{\omega_s}{\omega_p} = \frac{3000}{2000} = 1.5 \\ n &\ge \frac{\log_{10}(157.4893/0.12202)}{2\log_{10}(1.5)} = \frac{3.1108}{0.3522} = 8.833 \end{aligned}$$

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

Check: with $n = 9$, $\varepsilon^2 = 0.12202$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.12202 \times 1.5^{18}) = 22.58$ dB $\ge 22$ dB.

Answer: a 9th-order Butterworth filter (the narrow transition ratio of 1.5 and small passband loss make the order high).