For a 2nd order Bessel low pass, I have a pole-zero map which has a pole couple of the form: \$p_{1/2} = a \pm jb\$.

Can I deduce the cutoff frequency from the pole location? Does it contain any more information?

EDIT: Since the calculation of the cutoff frequency in the answer depends on the filter type I am wondering, if it is possible to determine the cutoff frequency for any kind of 2nd order low pass solely by the pole location (under the assumption they are complex conjugate)?


1 Answer 1


Yes - you can. There are filter tables for all the well-known lowpass approximations - including the Thomson-Bessel response. These tables list the relavant pole parameters for the various filter orders.

For second order Bessel response the relevant parameters are

$$Q_p=0.5773,\space \Omega_p=\frac{\omega_p}{\omega_0}=1.732$$

Explanation: The pole quality factor \$Q_p\$ is a measure for the angle \$\beta\$ between the pole position and the negative-real axis. We have \$Q_p=\frac12\cos\beta\$ . This angle can also be expressed using the given values for the real part (\$a\$) and the img. part (\$b\$) of the pole.

The normalized angular frequency \$\Omega_p=\frac{\omega_p}{\omega_0}\$ is the ratio of the actual angular pole frequency \$\omega_p\$ divided by the actual cut-off frequency \$\omega_0\$. The actual frequency \$\omega_p\$ is identical to the magnitude of the line which connects the origin and the pole position. This magnitude can also be expressed using the known values for \$a\$ and \$b\$.

In this context, you should note that the above mentioned value \$\Omega_p=1.7320\$ is based on the following definition of cut-off: The cut-off angular frequency \$\omega_0\$ is defined as the inverse of the group delay, hence: \$\omega_0=\frac{1}{\tau}\$.

This commonly used definition makes sense because - in most cases - the Bessel response is applied because of its time behaviour (delay filter). This is in contrast to other responses (Butterworth, Chebyshev, ...) where the frequency response matters primarily.

If the cut-off frequency \$\omega_0\$ is defined using the 3dB criterion for the magnitude the following normalized value applies: \$\Omega_p=\frac{\omega_p}{\omega_0}=1.2723\$.

  • \$\begingroup\$ Since there are two definitions for cutoff frequency. Is there a relation which expresses wo for one definition in terms of wo of the other definition, i.e. something like $$ w_{o-3dB} = \alpha \cdot w_{o-tau}$$ ? \$\endgroup\$
    – luis
    Mar 17, 2017 at 15:00
  • \$\begingroup\$ Thank you very mch for editing ..... \$\endgroup\$
    – LvW
    Apr 1, 2020 at 18:56
  • 1
    \$\begingroup\$ Answer to the first comment (question): No - because there is no singular definition for the cut-off in the time domain....it depends on the application. Example for specification: A Bessel lowpass shall have a group delay ripple (deviation from the basic value) of max. 10% at the frequency f1=1.5kHz. At the frequency f2=2kHz this lowpass shall have an attenuation of at least 10 dB. Both requirements have to be met at the same time....and it turns out that only a 4th order Bessel filter can fulfill these requirements. \$\endgroup\$
    – LvW
    Apr 1, 2020 at 19:09

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