Transfer Function of Bessel Filter

Question

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:

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.

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?

Thanks!

Answer 1

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: -

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.

Answer 2

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.