# Relation between group delay, pole frequency and quality factor

So I'm solving some problems on filters and came up with an apparent contradiction that I would like to get clear. Considering a low-pass biquadratic section:

$$T(s)=\frac{V_o(s)}{V_i(s)}=\frac{K_0\omega_0^2}{s^2+\frac{\omega_0}{Q_0}s+\omega_0^2}$$

So in the first problem that this appeared I was expected to calculate the group-delay at pole frequency.

I considered the phase being given by (because in this exercise the gain is negative)

$$\phi(\omega)=\pi-\arctan(\frac{\frac{\omega\omega_0}{Q_0}}{\omega_0^2-\omega^2})$$

and the group delay calculated with $$\tau_g(\omega)=-\frac{d\phi}{d\omega}$$

I come up with $$\tau_g(\omega)=\frac{Q_0\omega_0(\omega_0^2+\omega^2)}{Q_0^2(\omega_0^2-\omega^2)^2+\omega_0^2\omega^2}$$

Finally at $$\ \omega_0 \$$: $$\tau_g(\omega_0)=\frac{2Q_0}{\omega_0}$$

Done!

Now off with the second exercise, I have to consider the same biquadratic section but now it is a Bessel filter with constant group delay $$\\tau_0 \$$. Now the author calculates the group delay as a function of the pole frequency and the quality factor and comes up with.

$$\tau_0=\frac{1}{Q_0\omega_0}$$

I have no idea where this expression comes from and why it contradicts my answer of the previous exercise. Is this because this a Bessel filter now? How do we arrive with this new expression? What are the math steps? Thank you in advance

• One might expect an error as group delay at resonance increases with Q Mar 1 at 0:28
• When the pole-Q is high, the phase function has a steep slope at the pole frequency - that means: Group delay is proportional to the pole-Q. The last equation must be wrong.
– LvW
Mar 1 at 8:28
• If it helps, see this. Mar 1 at 15:11
• @aconcernedcitizen it did, thank you! I upvoted your answer there too. Mar 1 at 22:40

$$\tau_0=\tau_g(\omega_0)$$
$$\tau_0=\tau_g(0)$$