Hi, Does anyone know of a parameteric description for the transfer function (in Laplace domain) of a 4th order Bessel Thompson LPF, in terms of its 3dB bandwidth? If not, is there an empirical way to scale the filter coeffs to get a specific 3dB cut off point? Thanks, Venugopal

# Parametric description of Transfter function for 4th order Bessel filter

Started by ●April 6, 2005

Reply by ●April 7, 20052005-04-07

in article 1112838492.022063.167610@f14g2000cwb.googlegroups.com, balasubv@hotmail.com at balasubv@hotmail.com wrote on 04/06/2005 21:48:> Does anyone know of a parameteric description for the transfer function > (in Laplace domain) of a 4th order Bessel Thompson LPF, in terms of its > 3dB bandwidth? If not, is there an empirical way to scale the filter > coeffs to get a specific 3dB cut off point?i've never done a Bessel filter (never sharp enough for me), but my trusty Lindquist "Active Network Design" says that the 3 dB corner frequency relative to the normalized frequency is (approximately) omega_3dB ~= sqrt((2*order - 1)*ln(2)) how it's derived, i do not know. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Reply by ●April 7, 20052005-04-07

On Thu, 07 Apr 2005 01:18:49 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:>>Lindquist "Active Network Design" says that the 3 dB corner frequency >>relative to the normalized frequency is (approximately) >> >> omega_3dB ~= sqrt((2*order - 1)*ln(2)) >> >>how it's derived, i do not know.I have in my notes the following for Bessel filters: 1st order Denominator Polynomial: s + 1 w0: 1.00000000000000 w3dB: 1.00000000000000 2nd order Denominator Polynomial: s^2 + 3s + 3 w0: 1.73205080756888 w3dB: 1.36165412871613 3rd order Denominator Polynomial: s^3 + 6s^2 + 15s + 15 w0: 2.46621207433047 w3dB: 1.75567236868121 4th order Denominator Polynomial: s^4 + 10s^3 + 45s^2 + 105s + 105 w0: 3.20108587294368 w3dB: 2.11391767490422 I have up through 10th order, but you only asked for 4th. I found w0 by calculating the "nth-root" of the final term in each of the polynomials. I found w3dB by solving the transfer functions for magnitude equal to 1/sqrt(2), using Matlab. Interesting that the values included above do not match Lindquist's predictions very well. Another useful relationship, albeit a bit off-topic: Bessel LPF approximates Gaussian LPF; the higher the order, the better the approximation. In a Gaussian, the relationship between the half-amplitude (-6 dB) frequency and the half-power (-3 dB) frequency is: f6/f3 = sqrt(2) You can prove this with the defining equation for the Gaussian characteristic {exp[-(x^2)/(2*(sigma^2))]}. Greg Berchin

Reply by ●April 7, 20052005-04-07

in article 6pga51t9d93vqa6s68rv5ur3ro6tuj7gpj@4ax.com, Greg Berchin at 76145.2455@compuswerve.com wrote on 04/07/2005 11:09:> On Thu, 07 Apr 2005 01:18:49 -0400, robert bristow-johnson > <rbj@audioimagination.com> wrote: > >>> Lindquist "Active Network Design" says that the 3 dB corner frequency >>> relative to the normalized frequency is (approximately) >>> >>> omega_3dB ~= sqrt((2*order - 1)*ln(2)) >>> >>> how it's derived, i do not know. > > I have in my notes the following for Bessel filters: > > 1st order > Denominator Polynomial: > s + 1 > w0: > 1.00000000000000 > w3dB: > 1.00000000000000 > > 2nd order > Denominator Polynomial: > s^2 + 3s + 3 > w0: > 1.73205080756888 > w3dB: > 1.36165412871613 > > 3rd order > Denominator Polynomial: > s^3 + 6s^2 + 15s + 15 > w0: > 2.46621207433047 > w3dB: > 1.75567236868121 > > 4th order > Denominator Polynomial: > s^4 + 10s^3 + 45s^2 + 105s + 105 > w0: > 3.20108587294368 > w3dB: > 2.11391767490422your polynomials seem to agree with Lindquist (at least at 3rd order).> I have up through 10th order, but you only asked for 4th. > > I found w0 by calculating the "nth-root" of the final term in each > of the polynomials.i don't understand. isn't w0 always 1 with normalized s? i don't get what w0 is.> I found w3dB by solving the transfer > functions for magnitude equal to 1/sqrt(2), using Matlab.fine. that's the true w3dB. how does that compare to sqrt((2*order - 1)*ln(2)) ? (being too lazy to do it myself.)> Interesting that the values included above do not match > Lindquist's predictions very well.i wanna see the comparison. or is that what we see above?> Another useful relationship, albeit a bit off-topic: Bessel LPF > approximates Gaussian LPF; the higher the order, the better the > approximation. In a Gaussian, the relationship between the > half-amplitude (-6 dB) frequency and the half-power (-3 dB) > frequency is: > f6/f3 = sqrt(2) > You can prove this with the defining equation for the Gaussian > characteristic {exp[-(x^2)/(2*(sigma^2))]}.ya. Gaussian filters are easier to figger out than Bessel. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."

Reply by ●April 7, 20052005-04-07

On Thu, 07 Apr 2005 15:39:01 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:>>i don't understand. isn't w0 always 1 with normalized s?Not with Bessel. The denominator polynomials are of the form: s^n + ... + w0^n Thus w0 is the nth-root of the final term in each polynomial.>>fine. that's the true w3dB. how does that compare to >>sqrt((2*order - 1)*ln(2)) ? (being too lazy to do it myself.) >> >>> Interesting that the values included above do not match >>> Lindquist's predictions very well. >> >>i wanna see the comparison. or is that what we see above?I believe that is what we see above. Greg

Reply by ●April 7, 20052005-04-07

On Thu, 07 Apr 2005 18:18:33 -0500, Greg Berchin <76145.2455@compuswerve.com> wrote:>>>>i don't understand. isn't w0 always 1 with normalized s? >> >>Not with Bessel. The denominator polynomials are of the form: >> >> s^n + ... + w0^nI guess I should say, "Not with any of the Bessel derivations I've ever seen in any references." It would be possible, of course, to manipulate the polynomials into the s^n + ... + 1 form, but for some reason I've never seen it like that. Probably related to the formulation of the Bessel Functions themselves. Greg