Butterworth order, poles and transfer function

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

$\alpha_p$ = 0.5 dB, $\alpha_s$ = 20 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 2000 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\ \frac{\omega_s}{\omega_p} = \frac{2000}{1000} = 2 \\ n &\ge \frac{\log_{10}(99.0000/0.12202)}{2\log_{10}(2)} = \frac{2.9092}{0.6021} = 4.832 \end{aligned}$$

The order must be an integer, so n = 5.

Check: with $n = 5$, $\varepsilon^2 = 0.12202$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.12202 \times 2^{10}) = 21.00$ dB $\ge 20$ dB.

Pole locations

$\varepsilon = \sqrt{0.12202} = 0.34931$. Choosing the passband edge exactly at $\omega_p$ (this leaves a margin in the stopband), the poles lie on a circle of radius

$$\omega_o = \frac{\omega_p}{\varepsilon^{1/n}} = \frac{1000}{0.34931^{1/5}} = 1234.12\ \text{rad/s}$$

at angles $\theta_k = \dfrac{(2k-1)\pi}{2n}$ measured from the $j\omega$ axis: $p_k = -\omega_o\sin\theta_k \pm j\omega_o\cos\theta_k$.

           jw
       x   |       radius 1234 rad/s
    x      |       poles 36 deg apart
 x---------+------ sigma
    x      |
       x   |

Transfer function

Normalized 5th-order Butterworth: $(s+1)(s^2+0.618s+1)(s^2+1.618s+1)$. Replacing $s$ by $s/\omega_o$:

$$T(s) = \frac{\omega_o^5}{(s + 1234.12)(s^2 + 762.73s + 1.5231\times10^6)(s^2 + 1996.85s + 1.5231\times10^6)}$$

with $\omega_o^5 = 2.8628\times10^{15}$, $\omega_o^2 = 1.5231\times10^6$, so $T(0) = 1$.