Graphical design rule for Bessel-Thomson filter
\$\color{red}{\text{I made a mistake and I need to redo parts of the answer!}}\$ My apologies, however, the mistake is not that bad, since not all the answer is wrong. It bugged me why the group delay is not flat for a 13th order, or why the impulse response doesn't have a minor dip (since Bessel is not a Gaussian). I realized the mistake was already hinted at when I said I used the unsorted poles for the Bessel, which caused the transfer function to go wrong (though not by much). This means that I need to redo the part of the answer that deals with the comparison between the Bessel and the pole placement method; everything else is fine. Again, I am sorry for the mistake. Feel free to downvote, if you will.



Similar, though you'll not get away with just that, there is no closed form solution since they're found only based on the Bessel polynomials (i.e. root-finding). The poles are placed on an elipse, as Andy mentions, but with an offset in the right-hand side. Here's for N=13 for example (upper half):





Still, since the generating polynomial is fixed, i.e. only frequency scaling is needed, then the poles are also fixed and can be generated a priori, for a table (as an easier solution).



For clarification, here's the generating polynomial:

$$s^{13}+91*s^{12}+4095*s^{11}+120120*s^{10}+2552550*s^9+41351310*s^8+523783260*s^7+5237832600*s^6+41247931725*s^5+252070693875*s^4+1159525191825*s^3+3794809718700*s^2+7905853580625*s+7905853580625$$


\$\color{red}{\text{This is the initial mistake.}}\$ The transfer function was based on the poles, but the phase and group delay on the polynomial above, because both the atan() and the diff() blew up numerically.


and here are the poles (unsorted):

re=[-8.947709674391792,-8.470591771477185,-6.90037282614666,-8.830252084144904,-5.530680983344037,-7.844380277062596,-3.449867220628723];
im=[0.0,-3.483868450660993,-7.070644312152949,-1.736666400307631,-8.972247775155788,-5.254903406611962,-11.0739285522162];


for comparison, the poles of a Chebyshev with 0.01dB ripple:

re=[0.035061327,0.10314634,0.16523687,0.21772443,0.25755864,0.28242447,0.29087682];
im=[1.0338525,0.97376873,0.85709308,0.6906063,0.48398399,0.24923429,0.0];


Also, for comparison, the Bessel poles with a circle and the Chebyshev poles, both scaled for better comparison:



Note that the ellipse in the case of Chebyshev is aligned with the greater axis along the Y-axis, while the Bessel poles align themselves with the smaller axis on the Y-axis, while also having an offset.



I remember one book claiming that the poles lie on a circle, displaced to the right, and that they share the same angles as the Butterworth, but projected onto this circle. I used now an N=35 (odd for the extra, single real pole), with a circle, with proportional X and Y axis, but still scaled for better comparison:



The circle is scaled (both X and Y) by 37, and displaced to the right by 37-max(realpart(sBessel)). As you can see, the curves differ. One time I tried what you're asking now, by trying to approximate with a 90o rotated cosh() curve -- close, but no cigar, as they say. Here's a comparison:



I simply resigned and, years after, this question was asked on dsp.se (warning: lengthy post). I'm afraid that, sometimes, there just is no holy grail. In this case, you're stuck with the generating formula for the polynomial:

$$a_k=\frac{(2N-k)!}{2^{N-k}k!(N-k)!}$$

which can get "fluffy" for \$k\rightarrow 0\$, so the recursive one can get you a bit further, but with minor rounding issues:

$$\frac{a_{k+1}}{a_k}=\frac{2(N-k)}{(2N-k)(k+1)}$$

From then on it's the root-finding algorithm of your choice. Or, as I said, you can make tables, for example here's the complete roots of up to N=20, in double precision. Note: these are unscaled, i.e. calculated for delay, not frequency(!):

[0.8660254037844386i-1.5,-0.8660254037844386i-1.5]
[1.754380959783721i-1.838907322686957,-1.754380959783721i-1.838907322686957,-2.322185354626086]
[2.657418041856753i-2.103789397179628,-2.657418041856753i-2.103789397179628,0.8672341289345046i-2.896210602820372,-0.8672341289345046i-2.896210602820372]
[1.742661416183209i-3.351956399153524,-1.742661416183209i-3.351956399153524,3.57102292033797i-2.324674303181644,-3.57102292033797i-2.324674303181644,-3.646738595329665]
[0.8675096732313591i-4.248359395863367,-0.8675096732313591i-4.248359395863367,2.626272311447123i-3.735708356325813,-2.626272311447123i-3.735708356325813,4.492672953653945i-2.51593224781082,-4.492672953653945i-2.51593224781082]
[-4.971786858527892,1.73928606113053i-4.758290528154647,-1.73928606113053i-4.758290528154647,3.51717404770974i-4.070139163638142,-3.51717404770974i-4.070139163638142,5.420694130716758i-2.685676878943265,-5.420694130716758i-2.685676878943265]
[0.8676144453532826i-5.587886043262939,-0.8676144453532826i-5.587886043262939,4.414442500471611i-4.368289217202395,-4.414442500471611i-4.368289217202395,6.353911298604868i-2.838983948897615,-6.353911298604868i-2.838983948897615,2.616175152642267i-5.20484079063705,-2.616175152642267i-5.20484079063705]
[-6.29701918171626,1.737848383480994i-6.129367904273693,-1.737848383480994i-6.129367904273693,5.317271675435797i-4.638439887180668,-5.317271675435797i-4.638439887180668,7.291463688342168i-2.979260798180018,-7.291463688342168i-2.979260798180018,3.498156917885823i-5.604421819507492,-3.498156917885823i-5.604421819507492]
[0.8676651954556653i-6.92204490542646,-0.8676651954556653i-6.92204490542646,4.384947188943571i-5.967528328589314,-4.384947188943571i-5.967528328589314,6.224985482471234i-4.886219566858243,-6.224985482471234i-4.886219566858243,2.611567920796636i-6.61529096547683,-2.611567920796636i-6.61529096547683,8.232699459073597i-3.108916233649153,-8.232699459073597i-3.108916233649153]
[-7.622339845841585,3.489014503562782i-7.057892387669757,-3.489014503562782i-7.057892387669757,5.276191743697423i-6.301337454878748,-5.276191743697423i-6.301337454878748,1.737102820741282i-7.484229860704635,-1.737102820741282i-7.484229860704635,9.17711156870874i-3.229722089920541,-9.17711156870874i-3.229722089920541,7.13702075889222i-5.115648283905527,-7.13702075889222i-5.115648283905527]
[2.609066536949217i-7.997270599615764,-2.609066536949217i-7.997270599615764,0.8676935719771167i-8.253422011415825,-0.8676935719771167i-8.253422011415825,6.171534992991226i-6.61100424994881,-6.171534992991226i-6.61100424994881,8.052906864267905i-5.329708590886263,-8.052906864267905i-5.329708590886263,4.370169593404245i-7.465571240332478,-4.370169593404245i-7.465571240332478,10.12429680724084i-3.343023307800861,-10.12429680724084i-3.343023307800861]
[-8.94770967441898,3.483868450551646i-8.470591771510001,-3.483868450551646i-8.470591771510001,7.070644312151718i-6.900372826158152,-7.070644312151718i-6.900372826158152,1.736666400425321i-8.830252084116237,-1.736666400425321i-8.830252084116237,8.97224777515357i-5.530680983342347,-8.97224777515357i-5.530680983342347,5.254903406650159i-7.844380277035037,-5.254903406650159i-7.844380277035037,11.07392855221658i-3.449867220628742,-11.07392855221658i-3.449867220628742]
[2.607553324780497i-9.363145851070561,-2.607553324780497i-9.363145851070561,0.8677110294763433i-9.583171394019896,-0.8677110294763433i-9.583171394019896,6.143041071762656i-8.198846970087834,-6.143041071762656i-8.198846970087834,7.973217354159308i-7.172395962130479,-7.973217354159308i-7.172395962130479,4.361604177587814i-8.911000555481152,-4.361604177587814i-8.911000555481152,12.02573803225484i-3.551086883381187,-12.02573803225484i-3.551086883381187,9.894707597484578i-5.72035238382889,-9.894707597484578i-5.72035238382889]
[-10.27310955148198,3.480671268214976i-9.859567223419484,-3.480671268214976i-9.859567223419484,7.034393625952233i-8.532459059160995,-7.034393625952233i-8.532459059160995,1.736388856012094i-10.17091406847279,-1.736388856012094i-10.17091406847279,8.878982621996924i-7.429396992165036,-8.878982621996924i-7.429396992165036,5.242258876713885i-9.323599304919446,-5.242258876713885i-9.323599304919446,12.97950107076231i-3.647356862491653,-12.97950107076231i-3.647356862491653,10.81999913763804i-5.900151713629612,-10.81999913763804i-5.900151713629612]
[2.606567011382309i-10.71898582131243,-2.606567011382309i-10.71898582131243,4.356163385056269i-10.32511960145284,-4.356163385056269i-10.32511960145284,6.125760887225088i-9.712326332501009,-6.125760887225088i-9.712326332501009,0.8677225109985072i-10.91188607722687,-0.8677225109985072i-10.91188607722687,9.787697438361704i-7.673240790885078,-9.787697438361704i-7.673240790885078,11.74787493845505i-6.07124138290424,-11.74787493845505i-6.07124138290424,13.93502847581496i-3.739231797160583,-13.93502847581496i-3.739231797160583,7.928772856867366i-8.84796819655695,-7.928772856867366i-8.84796819655695]
[-11.59852952544957,1.736201495207083i-11.50807674884866,-1.736201495207083i-11.50807674884866,5.23407489400232i-10.76413417397734,-5.23407489400232i-10.76413417397734,8.825998303451005i-9.147588677578124,-8.825998303451005i-9.147588677578124,7.012009979228726i-10.08029444442791,-7.012009979228726i-10.08029444442791,10.69914507525592i-7.9054495961617,-10.69914507525592i-7.9054495961617,12.67812022904479i-6.234580978311283,-12.67812022904479i-6.234580978311283,3.478543926896344i-11.23343683286985,-3.478543926896344i-11.23343683286985,14.89215892466672i-3.82717378510033,-14.89215892466672i-3.82717378510033]
[0.8677305796056393i-12.23990211013843,-0.8677305796056393i-12.23990211013843,4.352480023166813i-11.71894899465382,-4.352480023166813i-11.71894899465382,7.9008930883336i-10.43001303090171,-7.9008930883336i-10.43001303090171,6.114394005840858i-11.18003883474541,-6.114394005840858i-11.18003883474541,9.725900329506054i-9.433132214976286,-9.725900329506054i-9.433132214976286,2.605887611187429i-12.06813579593398,-2.605887611187429i-12.06813579593398,13.61054734922753i-6.390972783893709,-13.61054734922753i-6.390972783893709,15.85075359693817i-3.911572291156902,-15.85075359693817i-3.911572291156902,11.61313174828707i-8.127283943599762,-11.61313174828707i-8.127283943599762]
[-12.92396298726643,3.477057739745347i-12.59706211536081,-3.477057739745347i-12.59706211536081,6.997077172796814i-11.57560275196964,-6.997077172796814i-11.57560275196964,10.62832089711397i-9.70610250233404,-10.62832089711397i-9.70610250233404,1.73606799805627i-12.84282859307907,-1.73606799805627i-12.84282859307907,5.228447830672733i-12.17923016627348,-5.228447830672733i-12.17923016627348,12.52948385810331i-8.339800733603411,-12.52948385810331i-8.339800733603411,14.54499130235651i-6.541095058909744,-14.54499130235651i-6.541095058909744,16.81069206004072i-3.992758917950229,-16.81069206004072i-3.992758917950229,8.792293285710413i-10.76353766688636,-8.792293285710413i-10.76353766688636]
[0.8677350518003363i-13.56742501366895,-0.8677350518003363i-13.56742501366895,4.349859625596589i-13.09881927110951,-4.349859625596589i-13.09881927110951,7.882058991191003i-11.95308937929071,-7.882058991191003i-11.95308937929071,9.686092710205683i-11.08258050261995,-9.686092710205683i-11.08258050261995,2.605405205905522i-13.41259743649624,-2.605405205905522i-13.41259743649624,13.44804526520383i-8.543895716554248,-13.44804526520383i-8.543895716554248,11.53311485564302i-9.967762520706822,-11.53311485564302i-9.967762520706822,6.106481551595795i-12.61728471920278,-6.106481551595795i-12.61728471920278,17.77186906891292i-4.071018561839362,-17.77186906891292i-4.071018561839362,15.48130618749379i-6.685526878511447,-15.48130618749379i-6.685526878511447]


I doubt you'll need more, but, if you do, I can copy-paste. The tables are very useful when you have memory to spare, as opposed to cycles. All you need from now on is frequency scaling, as these will get you nice 2nd order stages.



Update: Andy's post remimded me that once I concocted up a frequency scaling formula (from the then zunzun.com., now defunct, sadly), but it works decently. For example, for -3dB point, a sweep from N=2 to 32 gives the difference between the first and the last trace of ~0.31dB, and ~0.0125dB between adjacent traces. It's not perfect, but it works:

$$\omega_{scale}(A_{sc})=8091309.68544832\exp\left[-0.5\left(0.09397449321551755(\ln{N}-8.03901973218457)^2+0.009140987415805315\left(\ln{A_{sc}}-54.61336204495193\right)^2\right)\right]+0.02602784079436049$$

where Asc is the attenuation, in dB, at fc and N the order. As a small example, for the same 13th order and 3dB, the scaling would be |H(j4.13082549938354)|, while the formula says |H(j4.125564879197584)|, which gives -3.0025dB (0.7077408150981647). It's not limited to only 3dB: if you want 1.57dB, then \$\omega_{scale}\$ should be 2.99434327282329, while the formula says \$\omega_{scale}\$=3.001850652953856, which results in -1.577946667040319dB (0.8338782890589183). I say it's not bad.




\$\color{red}{\text{This part is redone.}}\$ I will keep the mistakes and pictures only as links, for shaming.


Update: I just tried what you propose, that is, to compare the Bessel with the pole placement on the circle, as the document from analog.com, and wherever elsewhere I read, say. First, since I already have N=13 above, I made the example for N=13. Second, I scaled the Bessel poles to match the X-axis.

Since the imaginary part of the poles are separated by 2/n and placed on a circle (non-displaced), then all you have to do is generate a list based on that, and the realpart is simply \$\Re=\sqrt{1-\Im^2}\$:

im=[0,0.1538461538461539,0.3076923076923077,0.4615384615384616,0.6153846153846154,0.7692307692307693,0.9230769230769231];
re=[1.0,0.9880948137434714,0.9514859136040755,0.8871201995900613,0.7882269819968921,0.6389710663783135,0.3846153846153845];


And this is how both poles look like compared to the unit circle. From here on, Bessel is blue.



Next, make the transfer functions and compare them. I applied frequency scaling to both so they have -3dB@1Hz: Bessel= 3.277105084487313, p.p.=0.3193551457708009. Interesting enough, the reverse of each is close to the other (less so on lower orders). Notice that Bessel has a steeper rolloff. In addition, the magnitude according to the long polynomial above is also plotted as the dashed green line; since it overlaps with the blue one, the blue one is kept as reference from now on.



(wrong: https://i.stack.imgur.com/7PtRa.png)

and the difference between them (it shows zero towards the end because of numerical inaccuracies, given the huge numbers in the original Bessel polynomial -- no longer the case, the transfer function is made up of 2nd order sections, made up of sorted out poles.):



(wrong: https://i.stack.imgur.com/VEbgY.png)

Then, the phases. Not surprisingly, differences due to the different rolloff:



(wrong: https://i.stack.imgur.com/EH5z1.png)

and the difference:



(wrong: https://i.stack.imgur.com/VQIA8.png)

And the most important, the group delay Bessel is truncated for the same reasons as above). Note that the p.p. method has lower delay, due to the slower rolloff, but also that it is not as flat as Bessel:



(wrong: https://i.stack.imgur.com/yWv7g.png)

and the difference (both normalized to 1, for easier comparison):



(wrong: https://i.stack.imgur.com/lVjQI.png)

Update: The flatness of the group delay can be verified with the derivative:



Conclusion: the pole placement is not a Bessel response, but it comes very reasonably close, so if you don't mind the minor differences, this is a very convenient and, perhaps most importantly, cheap way to generate the poles by avoiding the expensive root-finding algorithms. Note, however, that I only used N=13 for this, so, for an attempt at completion's sake, here's what the differences for N=5 look like, in order: magnitude, phase, group delay, update and flatness of group delay:






(wrong: https://i.stack.imgur.com/YVA1j.png, https://i.stack.imgur.com/gNKCc.png, https://i.stack.imgur.com/gqegm.png)



As a minor addendum, here are the impulse responses of the two 5th order, with the same Bessel=blue (using the -3dB frequency scaled versions):

(wrong: https://i.stack.imgur.com/BG4MF.png)

[I added this part at the end]


\$\color{red}{\text{End of redone, part 1.}}\$




Well, you've opened up an old wound, congratulations. I thought about modifying the Bessel poles (blue) by projecting them onto the unit circle along the X-axis, so they will lose whatever curve they normally have and convert them, forcefully (red). In addition, by comparison, the black squares are the p.p. method.




\$\color{red}{\text{Redone, part 2.}}\$ 


and the magnitudes of the Bessel (blue) near the converted Bessel (red dashed magenta) and p.p. (black red) -- for some reason, the rolloff is slower for p.p., I must have some typo somewhere, I won't find it today:.



(wrong: https://i.stack.imgur.com/vOmao.png)

All for N=13, and the results are consistent for 5, 9, 25, so on it seems. The conclusion remains: not Bessel, but darn close. Pick your choice.



This should be the last edit (before I go over the rabbit hole's event horizon) to address the issue of the displacement of the underlying circle. Doubts crept up so I wanted to clear this out. From previous pictures, it's clear it's not a circle, it's not some cosh(), but something else, but numbers are clearer than pictures, so I started a reductio ad absurdum: what if it is a circle? Then it should be scaled and displaced. Here's the basic idea:



The circle of radius OM (grey, dashed) is the unit circle, and the blue one of radius MQ would be the underlying circle. At point C is a pole whose coordinates are known. OM is also known, so \$AM=1-\Re(C)\$ amd \$AC=\Im(C)\$ => the red angle (ignore the reading), \$\alpha=\arctan\frac{AM}{AC}\$, while the green angle, \$\beta=\frac{\pi}{2}-\alpha=\arctan\frac{AQ}{AC}\$ => \$MQ=AM+AC\tan\left(\frac{\pi}{2}-\arctan\frac{AM}{AC}\right)\$, with the displacement being a simple subtraction.

But, for N=13, calculating the radius of the underlying circle should be the same for all the poles, yet it isn't. Here's the values for all the radii for the positive imaginary parts of the poles that are not on the X-axis or Y-axis (5 poles, see 1st picture):


\$\color{red}{\text{Redone, part 3.}}\$ Since the poles were mingled, the results came mingled, plus, I calculated the diameter. Not a biggie this one, since the result is the same: no underlying circle.


[2.882849152139202,2.896382080602158,2.920436970785266,2.957894588441385,3.014834976457822]
[1.448191040223894,1.460218485342188,1.478947294206138,1.507417488214167,1.553654588674449]

And here is a graphical representation of what the underlying circles would be for each of these poles. Note that, not only they don't overlap, but only one of each pole lies on a circle at a time (of course, the one after the circle has been calculated):



This should be the proof that, however flat the group delay is from the pole placement approximation, it's just an approximation, not a Bessel in the true sense. IMHO, this should have been specified by both analog.com and whatever other sources mention it -- that it's an approximation, a very good good one, but not Bessel(-Thompson).



And, since the wound goes deep, here's some more. It strikes me that, the larger the order, the more the p.p. method converges towards a Gaussian filter, and, sure enough, here's a plot of a reference gaussian function, \$\exp\left(-\frac{\ln2}{2}x^2\right)\$ (black), an approximated transfer function "a la Bessel" (blue, frequency scaling needed), and a free, non-frequency scaled version of the p.p. method (red):



If you can't see any difference it's because of the small things in life:



Does this mean that the p.p. method converges towards a Gaussian filter? No. Here are the poles of the Gaussian (blue) and the Bessel (red), scaled for comparison with the unit circle:



They're even more spread out. But are they, at least, placed on a circle, displaced or not? Here are the results of the radii of the circles, in a method similar to one for Bessel, above:

[2.216482009751927,2.067108140129843,1.988005879596608,1.939764970045811,1.910091446017072]

and the graphical representation, after another forced attempt at projecting them onto the unit circle (again, same way as Bessel, above):



For completeness, here's the impulse reponse of the Gaussian (blue), compared to the Bessel (red), and the p.p. method (green):



Given this final proof, the p.p. method is even better in terms of time-domain compared to Gaussian, but that also means that, frequency-wise, it's a mess. However, I am thinking of another pproach, but that will be for another day.