Delay in filter and 3rd order Bessel-Thomson

Delay in a filter

Delay is the time by which a signal component is shifted while passing through the filter. If the phase response is $\theta(\omega)$, the group delay is $\tau(\omega) = -\dfrac{d\theta(\omega)}{d\omega}$. A filter with constant $\tau$ (linear phase) delays all frequency components equally, so the waveform shape is preserved.

Third-order Bessel-Thomson filter

Approach (Storey/Thomson): the ideal delay of 1 s is $T(s) = e^{-s} = \dfrac{1}{\sinh s + \cosh s}$. Write

$$\coth s = \frac{\cosh s}{\sinh s} = \frac{1}{s} + \cfrac{1}{\cfrac{3}{s} + \cfrac{1}{\cfrac{5}{s} + \cfrac{1}{\cfrac{7}{s}+\cdots}}}$$

Truncating after $n$ terms gives $\coth s \approx m(s)/n(s)$, where $m$ (even) approximates $\cosh s$ and $n$ (odd) approximates $\sinh s$. Then $T(s) \approx K/[m(s)+n(s)]$.

For order 3 keep the first three terms ($1/s$, $3/s$, $5/s$). First simplify the inner part:

$$\begin{aligned} \frac{3}{s} + \frac{1}{5/s} &= \frac{3}{s} + \frac{s}{5} = \frac{s^2+15}{5s} \\ \coth s &\approx \frac{1}{s} + \frac{5s}{s^2+15} = \frac{s^2 + 15 + 5s^2}{s(s^2+15)} = \frac{6s^2+15}{s^3+15s} \end{aligned}$$

So $m(s) = 6s^2 + 15$ and $n(s) = s^3 + 15s$, and

$$B_3(s) = m(s)+n(s) = s^3 + 6s^2 + 15s + 15$$

Choosing $K = 15$ for unity DC gain:

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

Check: the delay at DC is $b_1/b_0 = 15/15 = 1$ s. The poles are $s = -2.3222$ and $s = -1.8389 \pm j1.7544$ (normalized to 1 s delay). For a delay of $\tau_0$, replace $s$ by $s\tau_0$.

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