Electrical – Third order filter – I need to verify calculations

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.

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 tto be that Q for a Bessel LPF.

First I don't understand why you are refering to page 450 of the book;

I found your circuit on page 456. Figure 11.23 "High-pass equal-component" (VCVS).

In the book they describe two types of Sallen-Key Topology filters, one is the "unit-gain"-version which as the name suggests has A_v=1. and the other is the "equal-component"-version (the one you have), this one also has a specific/fixed gain associated with it which is A=3−α this is what the book sayes about the "equal-component"-version on page 449:

"We see that the gain and damping of the filter are linked together. Indeed, for a certain damping factor, only one specific gain will work properly: A=3−α"

Since we know that for a butterworth-filter α must be sqrt(2) that determines our gain. So to answer the question;

How can I alter the above equations to provide a gain of 2 Av while using the Butterworth coefficients?

You can't without changing the basic circuit because the gain is determined by the topology and the choise of α for a butterworth-filter.

Now to answer the broader question of

How to design a 2nd Order High Pass Butterworth filter with a gain of 6 dB?

You can easily make the gain of your circuit almost anything you want by just adding a single resistor and fiddleing with the values of the existing like this;

The circuit you have can be turned into this one:

^{simulate this circuit – Schematic created using CircuitLab}

By replacing R2 and Rf with voltage-dividers

Now the gain of your new circuit is going to be (3-α)(R3+R4)/R4

To make this work the following have to be true:

R3//R4=R2 <- The thevinin equivalent of R3//R4 has to be equal to the original R2

R5//R6=Rf <- The thevinin equivalent of R5//R6 has to be equal to the original Rf

R3/R4=R5/R6 <- The two voltage-dividers have to divide the output by the same amount.

Now R6 and Ri can of course be combined, but for the sake of understanding the circuit I left them seperate.

If I was you though I would go for the "unit gain"-type and then do as I have described using R3=R4 to amplify the output by 2 to get A_v=2

EDIT:

I followed the example in the book for a unit-gain type, I chose 1kHz cutoff and simulated it in LT-spice with the results I got for the resistors and caps. here is a screenshot of the simulation in LT-spice showing cutoff at 1kHz, 0dB in-band gain and butterworth responce;

I then replaced the feedback resistors with voltage dividers as per my suggestion and simulated the results, below is a screenshot of the simulation in LT-spice, showing 6dB in-band gain, cutoff at 1kHz and butterworth-responce.

Sorry I know the pictures are hard to make out.