Butterworth characteristics and order (1 dB, 12 dB)

Characteristics of the Butterworth response

The Butterworth (maximally flat) low-pass response has magnitude

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2\omega^{2n}}$$

(normalized so that $\omega_p$ = 1 rad/s; with $\varepsilon = 1$ the half-power frequency is 1 rad/s). Its characteristics are:

• Maximally flat at $\omega = 0$: the first $2n-1$ derivatives of $|T(j\omega)|^2$ are zero at $\omega = 0$, so the passband is as flat as possible.
• Monotonic: no ripple in the passband or the stopband; the gain falls continuously with frequency.
• $|T(j0)| = 1$; at $\omega = 1$ (when $\varepsilon = 1$) $|T| = 1/\sqrt{2}$ (−3 dB) for every order.
• Roll-off of $20n$ dB/decade far into the stopband; the transition is wider than Chebyshev or elliptic for the same $n$.
• All-pole function: all transmission zeros are at $\omega = \infty$.
• Poles lie on a circle of radius $\omega_o = \varepsilon^{-1/n}$ in the left-half s-plane, equally spaced by $\pi/n$, symmetric about the real axis.
• Phase is fairly linear (better than Chebyshev), with a moderate delay peak near the cut-off.
• As $n \to \infty$ the response approaches the ideal brick-wall.

 |T|
  1 |----.__
    |       `.   n = 2
0.707 - - - - :\.  n = 4
    |         : \ \.
    |         :  `. `-._
    +---------+---------`----> w
              1

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, $\alpha_{max}$ = 1 dB, $\omega_s$ = 3000 rad/s, $\alpha_{min}$ = 12 dB.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{1.2} - 1 = 14.8489 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} = \frac{3000}{2000} = 1.5 \\ n &\ge \frac{\log_{10}(14.8489/0.25893)}{2\log_{10}(1.5)} = \frac{1.7585}{0.3522} = 4.993 \end{aligned}$$

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

Check: with $n = 5$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 1.5^{10}) = 12.02$ dB $\ge 12$ dB.

Answer: order $n = 5$.

The calculated value 4.993 is very close to 5, so a 5th-order filter only just meets the stopband requirement (12.02 dB).