Calculating Bessel Filter poles?

I've seen a number of references to calculating the filter poles by the following method:

The poles of the Bessel filter can be determined by locating all of the poles on a circle and separating their imaginary parts by:

2/n

where n is the number of poles.

e.g. page 8.23 on the link below:

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However using this method I can't anything like the pole locations I see in published tables.

E.g. take a 5th order filter, n = 5

2/n = 0.4 so their should be poles with imaginary parts of +/- 0.4j but tabulated values give poles with imaginary components:

0

+/- 0.7201 +/- 1.4756

What have I missed?

I wrote a python script which solves the reverse Bessel polynomials and this matches the values in published tables, and if I actually wanted to design a filter I would probably use a free online calculator, but I'm curious to know why I can't get the 2/n method to work.

Thanks for any help,

Gareth.

Reply to
Gareth
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With me cracking a book, my first question is have you checked if the tables are designed for 1 second time delay rather than 1 rad/sec filtering? That is, maybe there is just frequency scaling between your solution paths.

Reply to
miso

Make that "without me cracking a book"

Reply to
miso

lter.pdf

I'm not sure (yet). But I think the 2/n is an increment, not the explicit imaginary point of the poles (that would not make sense, as *all* imaginary portions of each pole would be at 0.4, and then no way could they be on a circumference).

I played with it, and could not get it to quite work either. I would inter pret the 2/n "separation" as something like {-4/5, -2/5, 0, 2/5, 4/5} for t he imaginary part of the 5 poles. Furthermore, there has to at least be a scale factor difference given the published table in the same document.

They mention that the poles are on a circle. The don't say where the cente r of the circle is. The only thing is that the poles are apparently on the circumference, and the imaginary parts have the specified separation, in t his 2/n ratiometric proportion.

This was an interesting image:

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Now in that one, the center would not be at , but could only be at , where x0 is positive real. Naturally, I don't know if the image is correct, but it gets one thinking, in any case.

But if it is true that the points lie on a circle, and you know the general ized equation for a circle, you may be able to find out the center of the c ircle and get a toe-hold on the problem that way. After all, you have 5 poi nts that lie on the circle from the table. Perhaps this is a geometry prob lem. Then you can see how the 2/n factor works into it.

Since the vector length from the origin to the poles are not equal amongst the poles (aside from the conjugates), it suggests the center of the circle cannot be the origin. The questions are: (1) are the poles on a circle, a nd (2) where is the circle's center?

It is too bad it is so poorly explained by Analog. In fact, portions of it must be wrong. If such a fact is so important, they should make it clear or say nothing.

Reply to
Simon S Aysdie

ositive non-zero real component. In addition, the radius of the circle is not 1 for a 1 rad/second cutoff frequency, as it is for the Butterworth. I 'll dig into my notes to see if I can find the equation for the center of t he circle and the radius. In any case, using the method in the referenced article yields approximate, not exact pole locations.> 2/n

lter.pdf

Reply to
Jon Lark

positive non-zero real component. In addition, the radius of the circle i s not 1 for a 1 rad/second cutoff frequency, as it is for the Butterworth. I'll dig into my notes to see if I can find the equation for the center of the circle and the radius. In any case, using the method in the reference d article yields approximate, not exact pole locations.> 2/n

The article 'mumbles' that the 'highest' poles are only 1/n below where the circle crosses the imaginary axis. I think that might fix the origin, but I'm too lazy to figure it out. (I look up numbers in a table.) I like Bessel filters... Butterworth's ring too much in the time domain.

George H.

Reply to
George Herold

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