Butterworth order, poles and TF (1 kHz, 2.5 kHz)

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

$\alpha_{max}$ = 3 dB at $f_p$ = 1 kHz, $\alpha_{min}$ = 40 dB at $f_s$ = 2.5 kHz. Since only the ratio matters, $\omega_s/\omega_p = f_s/f_p = 2.5$.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{4} - 1 = 9999.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.3} - 1 = 0.99526 \\ \frac{\omega_s}{\omega_p} = \frac{2\pi(2500)}{2\pi(1000)} = 2.5 \\ n &\ge \frac{\log_{10}(9999.0000/0.99526)}{2\log_{10}(2.5)} = \frac{4.0020}{0.7959} = 5.028 \end{aligned}$$

The order must be an integer, so n = 6.

Check: with $n = 6$, $\varepsilon^2 = 0.99526$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.99526 \times 2.5^{12}) = 47.73$ dB $\ge 40$ dB.

Pole locations

Since $\alpha_{max}$ = 3 dB, $\varepsilon^2 = 0.995 \approx 1$, so 1 kHz is the half-power frequency and the poles lie on a circle of radius

$$\omega_c = 2\pi \times 1000 = 6283.2\ \text{rad/s}$$

For $n$ = 6, $\theta_k = (2k-1)\pi/12$ (15°, 45°, 75°), $p_k = -\omega_c\sin\theta_k \pm j\omega_c\cos\theta_k$:

Normalized poles: $-0.2588 \pm j0.9659$, $-0.7071 \pm j0.7071$, $-0.9659 \pm j0.2588$.

Transfer function

Normalized: $T_N(s) = \dfrac{1}{(s^2 + 0.5176s + 1)(s^2 + 1.4142s + 1)(s^2 + 1.9319s + 1)}$. Replacing $s$ by $s/\omega_c$:

$$T(s) = \frac{\omega_c^6}{(s^2 + 3252.4s + \omega_c^2)(s^2 + 8885.8s + \omega_c^2)(s^2 + 12138.2s + \omega_c^2)}$$

with $\omega_c^2 = 3.9478\times10^7$ and $\omega_c^6 = 6.153\times10^{22}$.