>>>>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

I 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

Reply by Greg Berchin●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 robert bristow-johnson●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:
> 2.11391767490422

your 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 Greg Berchin●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 robert bristow-johnson●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 6, 20052005-04-06

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