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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

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in article 1...@f14g2000cwb.googlegroups.com,
b...@hotmail.com at b...@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                  r...@audioimagination.com

"Imagination is more important than knowledge."

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On Thu, 07 Apr 2005 01:18:49 -0400, robert bristow-johnson
<r...@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

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in article 6...@4ax.com, Greg Berchin at
7...@compuswerve.com wrote on 04/07/2005 11:09:

> On Thu, 07 Apr 2005 01:18:49 -0400, robert bristow-johnson
> <r...@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                  r...@audioimagination.com

"Imagination is more important than knowledge."

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On Thu, 07 Apr 2005 15:39:01 -0400, robert bristow-johnson
<r...@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

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On Thu, 07 Apr 2005 18:18:33 -0500, Greg Berchin
<7...@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^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

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