1
\$\begingroup\$

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

\$\endgroup\$
4
  • \$\begingroup\$ One might expect an error as group delay at resonance increases with Q \$\endgroup\$ Mar 1 at 0:28
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    Mar 1 at 8:28
  • 1
    \$\begingroup\$ If it helps, see this. \$\endgroup\$ Mar 1 at 15:11
  • \$\begingroup\$ @aconcernedcitizen it did, thank you! I upvoted your answer there too. \$\endgroup\$ Mar 1 at 22:40
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ You can select your own answer, it will help future searches to find an accepted solution to this kind of problem. \$\endgroup\$ Mar 2 at 10:09
  • \$\begingroup\$ @aconcernedcitizen yes I know but I have to wait 24 hours to do it :) 10 more to go! \$\endgroup\$ Mar 2 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.