Relation between group delay, pole frequency and quality factor

Question

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 — Tony Stewart EE75, Mar 01 '21 at 00:28
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 01 '21 at 08:28
If it helps, see [this](https://electronics.stackexchange.com/q/521118/95619). — a concerned citizen, Mar 01 '21 at 15:11
@aconcernedcitizen it did, thank you! I upvoted your answer there too. — Granger Obliviate, Mar 01 '21 at 22:40

score 3 · Accepted Answer · answered Mar 01 '21 at 13:41

3

Ok, I figured it out! It was a mistake on my part because I assumed that

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

While in fact

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

Now it makes sense.

answered Mar 01 '21 at 13:41

Granger Obliviate

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You can select your own answer, it will help future searches to find an accepted solution to this kind of problem. – a concerned citizen Mar 02 '21 at 10:09
@aconcernedcitizen yes I know but I have to wait 24 hours to do it :) 10 more to go! – Granger Obliviate Mar 02 '21 at 12:20

Relation between group delay, pole frequency and quality factor

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