All-pass in delay equalization; 4th order Bessel-Thomson

Importance of all-pass filters in delay equalization

An all-pass filter has $|T(j\omega)| = 1$ at all frequencies but a frequency-dependent phase. A first-order section is $T(s) = \dfrac{s-a}{s+a}$ and a second-order section is $T(s) = \dfrac{s^2 - \frac{\omega_0}{Q}s + \omega_0^2}{s^2 + \frac{\omega_0}{Q}s + \omega_0^2}$: zeros are mirror images of the poles.

• Sharp filters (Chebyshev, elliptic) and transmission channels have non-constant group delay, largest near the band edge. This causes phase distortion of pulses and data.
• An all-pass filter cascaded with such a filter adds delay without changing the magnitude.
• Its delay is designed to be large where the original delay is small, so that the total delay is nearly flat: $\tau_{total}(\omega) = \tau_{filter}(\omega) + \tau_{AP}(\omega) \approx$ constant.
• Used in telephone/modem lines, video and data channels, and after sharp anti-aliasing filters.

 tau    filter delay   +  all-pass delay  = total
  |      _/\              \_    _/         ______
  |   __/   \               \__/
  +--------- w        +-------- w    +-------- w

Fourth-order Bessel-Thomson low-pass filter

Ideal 1 s delay: $T(s) = e^{-s} = 1/(\cosh s + \sinh s)$. Expand $\coth s$ as a continued fraction and truncate after four terms:

$$\coth s \approx \frac{1}{s} + \cfrac{1}{\cfrac{3}{s} + \cfrac{1}{\cfrac{5}{s} + \cfrac{1}{\cfrac{7}{s}}}}$$

Working from the bottom:

$$\begin{aligned} \frac{5}{s} + \frac{s}{7} &= \frac{s^2+35}{7s} \\ \frac{3}{s} + \frac{7s}{s^2+35} &= \frac{10s^2 + 105}{s(s^2+35)} \\ \frac{1}{s} + \frac{s(s^2+35)}{10s^2+105} &= \frac{s^4 + 45s^2 + 105}{10s^3 + 105s} \end{aligned}$$

So $m(s) = s^4 + 45s^2 + 105$ (approximates $\cosh s$) and $n(s) = 10s^3 + 105s$ (approximates $\sinh s$):

$$B_4(s) = s^4 + 10s^3 + 45s^2 + 105s + 105$$

(Check by recursion: $B_4 = 7B_3 + s^2B_2 = 7(s^3+6s^2+15s+15) + s^2(s^2+3s+3)$, which gives the same polynomial.)

$$T(s) = \frac{105}{s^4 + 10s^3 + 45s^2 + 105s + 105}$$

This gives unity DC gain and a delay of $105/105 = 1$ s at low frequency; replace $s$ by $s\tau_0$ for a delay of $\tau_0$.