Designing a constant delay filter (2nd order example)

Constant delay filter

A constant delay filter (Bessel-Thomson filter) is a low-pass filter whose group delay $\tau(\omega) = -d\theta/d\omega$ is (nearly) constant over the passband, i.e. its phase is linear, $\theta(\omega) \approx -\omega\tau_0$. The ideal is $T(s) = e^{-s\tau_0}$: every frequency is delayed by the same $\tau_0$, so the output is an undistorted, delayed copy of the input. Since $e^{-s\tau_0}$ is not rational, it is approximated by an all-pole function whose delay is maximally flat at $\omega = 0$.

Design procedure

1. Specify the required delay $\tau_0$ at DC and the allowed delay error (e.g. delay within 1% up to a frequency $\omega_1$), plus any magnitude requirement.
2. Normalize to $\tau_0 = 1$ s: the normalized frequency is $\omega\tau_0$.
3. Choose the order $n$ from Bessel delay/magnitude curves so that the delay and loss at $\omega_1\tau_0$ are within limits.
4. Take the Bessel polynomial $B_n(s)$ (from the recursion $B_n = (2n-1)B_{n-1} + s^2B_{n-2}$, or derived by the maximally-flat-delay method below): $T(s) = B_n(0)/B_n(s)$.
5. Denormalize with $s \to s\tau_0$.
6. Realize the factors with active biquads (Sallen-Key, MFB) or a passive ladder, then impedance-scale to practical values.

Example: second-order filter

Take a second-order all-pole low-pass function normalized for a delay of 1 s at DC:

$$T(s) = \frac{b_0}{s^2 + b_1 s + b_0}$$

Phase and delay. With $s = j\omega$:

$$\theta(\omega) = -\tan^{-1}\frac{b_1\omega}{b_0 - \omega^2}, \qquad \tau(\omega) = -\frac{d\theta}{d\omega} = \frac{b_1(b_0 + \omega^2)}{(b_0-\omega^2)^2 + b_1^2\omega^2}$$

Expanding the denominator:

$$\tau(\omega) = \frac{b_1 b_0 + b_1\omega^2}{b_0^2 + (b_1^2 - 2b_0)\omega^2 + \omega^4}$$

Conditions for maximally flat delay equal to 1. $\tau(\omega)$ stays at 1 for as many powers of $\omega$ as possible if the numerator matches the denominator term by term:

$$\begin{aligned} \omega^0:&\quad b_1 b_0 = b_0^2 \;\Rightarrow\; b_1 = b_0 \\ \omega^2:&\quad b_1 = b_1^2 - 2b_0 \;\Rightarrow\; b_0 = b_0^2 - 2b_0 \;\Rightarrow\; b_0 = 3 \end{aligned}$$

So $b_0 = b_1 = 3$ and

$$T(s) = \frac{3}{s^2 + 3s + 3}, \qquad \tau(\omega) = \frac{9 + 3\omega^2}{9 + 3\omega^2 + \omega^4}$$

The delay is $1$ s at $\omega = 0$ and deviates only by the $\omega^4$ term, i.e. it is maximally flat. The poles are $s = -1.5 \pm j0.866$ ($\omega_0 = \sqrt{3} = 1.732$, $Q = 1/\sqrt{3} = 0.577$). The denominator $s^2+3s+3$ is the second-order Bessel polynomial. For a delay of $\tau_0$ seconds, replace $s$ by $s\tau_0$:

$$T(s) = \frac{3/\tau_0^2}{s^2 + (3/\tau_0)s + 3/\tau_0^2}$$

Numerical example: design a second-order constant delay filter with $\tau_0 = 1$ ms.

$$T(s) = \frac{3\times10^6}{s^2 + 3000s + 3\times10^6},\qquad \omega_0 = \sqrt{3\times10^6} = 1732.05\ \text{rad/s},\quad Q = \frac{1732.05}{3000} = 0.5774$$

Realize it with a unity-gain Sallen-Key low-pass section with $R_1 = R_2 = R$, feedback capacitor $C_1$ and grounded capacitor $C_2$. For this circuit $Q = \frac{1}{2}\sqrt{C_1/C_2}$ and $\omega_0 = 1/(R\sqrt{C_1C_2})$:

$$\begin{aligned} C_1/C_2 &= 4Q^2 = 4/3; \quad \text{choose } C_2 = 0.1\ \mu\text{F} \Rightarrow C_1 = 0.1333\ \mu\text{F} \\ R &= \frac{1}{\omega_0\sqrt{C_1C_2}} = \frac{1}{1732.05 \times 1.1547\times10^{-7}} = 5\ \text{k}\Omega \end{aligned}$$

Answer: $R_1 = R_2 = 5$ kΩ, $C_1 = 0.1333\ \mu$F, $C_2 = 0.1\ \mu$F, unity-gain buffer; delay $\approx 1$ ms over the passband.