Importance of constant delay filter; 3rd order TF

Importance of a constant delay filter

A constant delay (Bessel-Thomson) filter has a group delay $\tau(\omega) = -d\theta/d\omega$ that is maximally flat around $\omega = 0$, i.e. its phase is nearly linear in the passband.

• All frequency components of a signal are delayed by the same time, so the waveform shape is preserved (no phase distortion).
• Pulses and square waves pass with little overshoot and ringing. This matters in data transmission, video, radar, ECG and oscilloscope circuits.
• It is used as a delay line and in systems where the timing between signals matters.
• The magnitude response is less sharp than Butterworth, so it is chosen when phase matters more than selectivity.

Transfer function of the third-order constant delay 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 $s\tau_0$.