Butterworth characteristics and 5th order TF

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

Fifth-order Butterworth transfer function

For the normalized ($\varepsilon$ = 1, $\omega_c$ = 1 rad/s) case the poles lie on the unit circle at

$$p_k = -\sin\theta_k \pm j\cos\theta_k, \qquad \theta_k = \frac{(2k-1)\pi}{2n} = 18°, 54°, 90°$$

(Each pair gives $s^2 + 2\sin\theta_k\, s + 1$.)

$$\begin{aligned} T(s) &= \frac{1}{(s+1)(s^2 + 0.618s + 1)(s^2 + 1.618s + 1)} \\ &= \frac{1}{s^5 + 3.236s^4 + 5.236s^3 + 5.236s^2 + 3.236s + 1} \end{aligned}$$