All-pass filters in delay equalization; 3rd order Bessel-Thomson

Importance of all-pass filters (delay equalization)

An all-pass filter has $|T(j\omega)| = 1$ at all frequencies and changes only the phase. Its zeros are mirror images of its poles about the $j\omega$ axis:

$$T_1(s) = \frac{a - s}{a + s}, \qquad T_2(s) = \frac{s^2 - \dfrac{\omega_o}{Q}s + \omega_o^2}{s^2 + \dfrac{\omega_o}{Q}s + \omega_o^2}$$

The first-order section has phase $\theta = -2\tan^{-1}(\omega/a)$ and delay $\tau = \dfrac{2a}{a^2 + \omega^2}$.

     jw
  x  |  o      x pole, o zero
     |
  x  |  o      |T(jw)| = 1 for all w
 ----+-------> sigma

Importance:

• Sharp filters (Chebyshev, elliptic) have a delay that peaks near the band edge. Cascading an all-pass network adds delay where the filter's delay is small, making the total delay flat without changing the magnitude response. This is delay equalization.
• Used in data transmission, telephone lines, video and radar where pulse shape must be kept.
• Used as phase shifters (e.g. 90° networks for SSB modulation) and as pure delay lines.

Third-order Bessel–Thomson low-pass filter

The ideal constant-delay function (delay normalized to 1 s) is

$$T(s) = e^{-s} = \frac{1}{\cosh s + \sinh s}$$

Write $M = \cosh s$ (even part) and $N = \sinh s$ (odd part). Their ratio has the continued-fraction expansion

$$\coth s = \frac{M}{N} = \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \cfrac{1}{\dfrac{5}{s} + \cfrac{1}{\dfrac{7}{s} + \cdots}}}$$

For an $n$th-order approximation, keep the first $n$ terms, write the result as $M/N$ with polynomials, and use $T(s) = K/(M + N)$, with $K$ chosen so that $T(0) = 1$.

Third order (first three terms):

$$\begin{aligned} \frac{M}{N} &= \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \dfrac{s}{5}} = \frac{1}{s} + \frac{5s}{15 + s^2} \\ &= \frac{15 + s^2 + 5s^2}{s(s^2 + 15)} = \frac{6s^2 + 15}{s^3 + 15s} \end{aligned}$$

So $M = 6s^2 + 15$, $N = s^3 + 15s$ and

$$T(s) = \frac{15}{s^3 + 6s^2 + 15s + 15}$$

The same polynomial follows from the Bessel recursion $B_n = (2n-1)B_{n-1} + s^2B_{n-2}$ with $B_0 = 1$, $B_1 = s + 1$: $B_2 = s^2 + 3s + 3$ and $B_3 = 5B_2 + s^2B_1 = s^3 + 6s^2 + 15s + 15$.

Poles: $s = -2.322$ and $s = -1.839 \pm j1.754$. The delay is $\tau(0) = 15/15 = 1$ s and stays at 0.996 s at $\omega$ = 1 rad/s and 0.887 s at 2 rad/s.

For an actual delay $\tau_o$, replace $s$ by $s\tau_o$ (frequency scaling by $k_f = 1/\tau_o$).