the Bessel filter transfer function is defined via bessel polynomials. If we consider for example a 2nd order filter, the transfer function is: $$ H(s) = \frac{3}{s^2+3*s+3} $$ I wanted to build a simulation for such a filter with a Sallen-Key-Architecture. Therefore I consulted this design guide by TI. They define the transfer function of a 2nd order low pass as:

enter image description here

Ao is 1 since I want the gain to be unity. I looked at the table below in order to correctly calculate the C- and R-values.

enter image description here

Hence the transfer function becomes: $$ H(s) = \frac{1}{0.618*s^2+1.3617*s+1} $$

I ran the simulation and looked at the bode plot. It showed the desired result (the -3db cutoff frequency was as calculated).

However I do not understand why the transfer function looks so differently. Its definetely not a Bessel polynomial. I checked the step response and observed an overshoot of 0.4% as one would expect for a Bessel filter. Therefor I have 3 Questions:

  1. How come that the transfer function in the ti design guide is not a bessel polynomial.
  2. Should the pole location of a 2nd order Bessel filter be the same for any filter with a certain cutoff frequency?
  3. Can a second order bessel low pass have a different Q factor than 0.5773?


  • 1
    \$\begingroup\$ The constant in the denominator for the last equation H(s) must be "1" instead of "3". Last question: NO!. It is the Q factor only that determines the Bessel response. \$\endgroup\$
    – LvW
    Mar 13 '17 at 14:22
  • \$\begingroup\$ you are right, it was a typo. Still doesnt have to do much with a bessel polynomial. \$\endgroup\$
    – luis
    Mar 13 '17 at 14:25
  • \$\begingroup\$ Why not? It is a typical filter function with a frequency response called "Thomson-Bessel". What is your problem? Of course, the filter function is not identical to the "mathematical Bessel polynominal". \$\endgroup\$
    – LvW
    Mar 13 '17 at 15:57
  • \$\begingroup\$ I don't understand why you have this mathematical definition on the one side, and a transfer function that is completely different on the other side. Where do these coefficients come from? \$\endgroup\$
    – luis
    Mar 13 '17 at 16:01
  • 2
    \$\begingroup\$ The coefficients of the so-called Bessel filters are calculated on the requirement of a maximally flat group delay in the passband (to be compared with a maximally flat amplitude for Butterworth filtes). It can be shown that during calculation of the coefficients we make use of the known Bessel polynominals (this is a rather involved procedure) - but this does not mean that the magnitude of the transfer function has response which looks like Bessel functions. It is - as mentionmed - the mathematical procedure behind the finding of the coefficients. OK? \$\endgroup\$
    – LvW
    Mar 13 '17 at 16:07

How come that the transfer function in the TI design guide is not a Bessel polynomial.

Let's look at the transfer function you have written: -

\$H(s) = \dfrac{1}{0.618s^2+1.3617s+ 1}\$

Rearranging: -

\$H(s) = \dfrac{1.6181}{s^2+2.2034s+ 1.6181}\$

The equation is now in standard form : \$H(s) = \dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+ \omega_n^2}\$

And clearly \$\omega_n\$ = \$\sqrt{1.6181}\$ hence 2.2034/\$\sqrt{1.6181}\$ = 1.732. This bit is important because it is \$\sqrt3\$.

For a Bessel 2nd order low pass filter 2\$\zeta\$ = \$\sqrt3\$ hence zeta is 0.866. Test case: -

enter image description here

Picture source

In the picture I've manipulated R to give me a damping ratio (zeta) of precisely 1.732 - look at the peak in the step response - 1.00433 volts - exactly right for Bessel. Look at the phase delay plotted on the upper graph - maximally flat and gradually becoming 90 degrees at the natural resonant frequency. Fd (the damped frequency) is precisely 0.5 - also indicative of Bessel.

Can a second order Bessel low pass have a different Q factor than 0.5773?

0.5773 is the reciprocal of \$\sqrt3\$ and no it has to be that Q for a Bessel LPF.

  • 1
    \$\begingroup\$ The reason the polynomial is different is for the frequency scaling. A non-scaled Bessel (OP's first formula) will have \$\omega_0=\sqrt{3}\$, with an attenuation of ~-1.597dB@1Hz -- nothing unusual, Bessel is normally for flat group delay, not frequency -- so TI scaled it so that it's the classical -3dB. \$\endgroup\$ May 2 '18 at 16:25
  • \$\begingroup\$ @aconcernedcitizen why not make this an answer rather than pinning it to my answer unless, of course you are too-subtley pointing out an error in my answer that I'm too stupid to recognize? \$\endgroup\$
    – Andy aka
    May 2 '18 at 16:29
  • \$\begingroup\$ I had written this as a comment so that the answer to which I am making the comment can be updated, if necessary. But if you say it should be an answer, so be it. I don't know where did the "stupid" come from. \$\endgroup\$ May 2 '18 at 16:40

A Bessel filter has, as you correctly show in your first formula, \$\omega_0=\sqrt{3}\$. It's not unusual if you think that, normally, a Bessel filter is used for its flat group delay, rather than its frequency behaviour (as @LvW says in his comment). But implementing a filter with that transfer function will give a ~1.597dB@1Hz attenuation, which doesn't make the response a classical one. So, TI applied a frequency scaling so that the attenuation is -3dB@1Hz. As it so happens, the squared frequency (pulsation) is \$\phi\$=1.618..., after which they re-arranged the terms to fit their opamp topology.

  • \$\begingroup\$ That's great! And I just noticed that in the table at the top of this question, is 1/\$\phi\$=0.618... Maybe that's where the \$\phi\$ came from. \$\endgroup\$ Oct 9 '21 at 10:40
  • \$\begingroup\$ @MicroservicesOnDDD I hope you are not missing the point: just because there is a \$\varphi\$ in there it doesn't mean that the transfer function is based on it -- rather, it is a byproduct of the mathematics involved (as the last phrase in the aswer says: "As it so happens, [...]"). \$\endgroup\$ Oct 9 '21 at 11:39

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