Welcome, Guest
Username: Password: Remember me
NOTE: This forum is community powered. Please be mindful that long time community members are here to help as part of a community effort. If you have a specific issue (e.g. hardware, failure), please use our tech support portal (Support menu - > Contact Us). Thanks a lot of your help in making a better community. :-)

TOPIC: How calculate biquad coefficients for 6dB filter

How calculate biquad coefficients for 6dB filter 1 year 3 months ago #33035

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi

I own 4 pieces of PWR-ICE125, and I am in the process of building a 3-way speakersystem. I want to create speakers with Duelund Syncron Filters. These filters are different from normal textbook filters. The individual speaker-units need acoustically to follow certain target-curves.

A friend of mine does passive Duelund Syncron Filters for 3-way speakers. He speaks of cascading several 6 dB filters to create hybrid crossovers with several cut-off- frequencies, thereby approaching the desired filter-slopes.

We want to try to translate that approach to the digital domain by using the "Advanced" mode in the crossover-section of the PWR-ICE125 plug-in. But how can I cascade to, three or four 6 dB filters?

I have played with the all-digital-coefs v1.2, and I have not succeeded in finding a way to create a 6dB Butterworth filter. The program only seems to create 2. order filters!? Am I right in that? It would be nice with some explanation to that program, i.e. what does it create, or where to find what Q-value to choose!

In my old 2x4 Plug-in there is a biquad-calculater that can produce coefficients for 6dB filters, though for a different sampling-rate. In the 2x4 Plug-in I managed to create a hybrid crossover with more than one cut-off-frequency.

BUT how do I do that in the PWR-ICE125- Plug-in?

Is there an alternative biquad-calculater somewhere that can be used? Or could the biquad-calculater from the 2x4 be made available for 96kHz in some way?

Loooong story. I hope someone can help.

Regards from Denmark

Steffen
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 1 year 3 months ago #33041

  • dreite
  • dreite's Avatar
  • Offline
  • Platinum Boarder
  • Posts: 1179
  • Thank you received: 594
  • Karma: 82
Your post is somewhat confusing.
6db Butterworth filters all have a Q=0.707.

If I understand your objective correctly, I think the easiest approach would be to use the "basic" function of the GUI to select the a 6db/BW filter at your desired frequency then switch to the advanced mode and copy the coefficients. Repeat that process with your array of desired cutoff frequencies. You can then take the results (up to 8) and paste into the "advanced" portion of the Xover section of your PWR-ICE plugin.

You might also consider purchasing one of the 2x4 software plugins since they have a handy biquad calculator built in.

Dave.
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 1 year 3 months ago #33043

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi Dave

Thank You so far. And sorry for the confusion.

The ideer of purchasing a 2x4 plug-in sounds Ok. As I understand it, I would need the Plug-in for the 2x4 HD version with 96khz samplingrate. This is probably the best choice at the moment.


I tried going from Basic to Advanced in the PWR-ICE 2x2 GUI, and that is also an option, though i really have to maintain focus not to get confused! And it seems, that for simple 6 dB BW filters, the biquad 1 is used for the basic`s Lowpass and the biquad 5 is used for the Highpass. So I need to be very awake here what biquad to copy!

By the way, I compared the coefficients from the GUI and the All-digital-spreadsheet for a High Pass BW 12 dB/oct filter, and they where not exactly the same!? I guess that can be due to the accuracy of decimals in the Q settings?

When I but 1000 Hz and Q=0,707 in the spreadsheet i got coefficients similar to the GUI`s 12dB/oct BW-filter! I think the spreadsheet can only produce 12dB/oct filters with different Q´s, because biquads inherently are 2. order! I produced coefficients in the GUI Basic for a 6dB BW High pass filter, switched to Advanced and noticed that b2 and a2 are zero, only 1.order components. Well and how do I make them?

Well, that got complicated again!!! I keep on looking and learning.

Bedtime for me now.

Steffen
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 1 year 3 months ago #33049

  • dreite
  • dreite's Avatar
  • Offline
  • Platinum Boarder
  • Posts: 1179
  • Thank you received: 594
  • Karma: 82
What you're attempting to do is not uncomplicated and could get tedious. Take your time.

Regarding the difference in coefficients, I'm not sure what you're referring to. Maybe you could elaborate further?
Either utility, if programming the same filter at the same Fs, should generate the same set of coefficients.

Yes, b2 and a2 coefficients would be zero in a first-order filter.

Dave.
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 1 year 3 months ago #33062

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi again

I did purchase the 2x4 HD Plug-IN. It does NOT contain a biquad-calculater as does the old 48 kHz 2x4 Plug-IN. Hmm.

BUT I CAN use the 2x4 HD Plug-IN as a biquad-calculater for 6 dB/oct BW-filters by choosing a filter in the BASIC mode and switch to the ADVANCED mode. The Plug-IN will automatically compute biquad-coefficients for Low-pass filters in the biquads 1-4 and biquad-coefficients for High-pass filters in the biquads 5-8. So again I need to keep focus, and copy the right biquads!

The advantage by using the 2x4 HD Plug-IN to calculate biquad-coefficients for the PWR-ICE-Plug-IN is, that I keep things sepparated and minimize chances for error end confusion! And yes, it still is a complicated affair. But on the other hand, I did by the PWR-ICE125 exactly because it can be programmed so flexible. I just need to know how.


And now to some "Research":

I did evaluate a 12dB/oct BW High-Pass-filter in the 2x4 HD Plug-IN, the PWR-ICE-Plug-IN and the All-digital-coefs v1.2 and compared the b0-coefficient. I evaluated the b0-coefficient in the spreadsheed for Q-values with increased amount of decimals.

2x4 HD
b0=0,9547740306487544

PWR-ICE
b0=0,9547740306487544

All-Digital- Coefs
b0=0,954767656510722 @ Q=0,707
b0=0,945773625922957 @ Q=0,7071
b0=0,954774043721377 @ Q=0,707107
b0=0,954774031784389 @ Q=0,7071068
b0=0,954774030650375 @ Q=0,707106781
b0=0,954774030662312 @ Q=0,7071067812
b0=0,954774030661715 @ Q=0,70710378119
b0=0,954774030661536 @ Q=0,707106781187

2x4 HD
b0=0,9547740306487544

PWR-ICE
b0=0,9547740306487544

So, as You can se: The two 96 KHz Plug-IN´s compute the same b0-coefficients. But the All-digital-coefs does compute different value, but approximating the Plug-IN-values with increasing amount of decimals in the Q-setting!!!!!

I guess, it does not really matter, as long as the coefficients produce a stable biquad!? But it was interesting and educational.

Regards from Denmark

Steffen
Last Edit: 1 year 3 months ago by supersteff.
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 8 months 3 weeks ago #35778

  • JJ2106
  • JJ2106's Avatar
  • Offline
  • Expert Boarder
  • Posts: 106
  • Thank you received: 8
  • Karma: 1
Sorry to restart this thread after such a long time.
The value computed by All-digital coeffs is the correct one.
I assume you took Fc=500Hz, Fs=48Khz.
If you enter Q=1/2^0.5, you will obtain b0 = 0.954774030661509 which is he final value.
For a quick check , I enclose a model which I wrote, and will give you the exact same value, even though the method is different.
This model also gives you the coefficients for first-order (6dB) filters, except first-order all-pass, wich i haven't included yet.
Please rename the file to ..xls, as this isn't a zipped file.
Attachments:
The administrator has disabled public write access.
The following user(s) said Thank You: supersteff

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35892

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi.

Nice with a little response.

I did evaluate a 1000Hz, 12dB/oct BW High-Pass-filter with a 96000 Hz samplingfrequency, as that is what the PWR-ICE is working with.

I guess you get the "same" result for b0 if you use 500 Hz and 48000Hz, two values that are half as much as mine!?

I am going to have a look at your spreadsheet.

I don't know much about filters and what distinguish them, but i know a bit of math, so that I can improvise a bit and play with things, to find out how it works. But I can not find a way to ask "All-digital-coefs v1.2 " for a 6 dB filter!? "All-digital-coefs v1.2 " only asks for corner frequency, Q and samplingfrequency in the inputsection, and then computes a 2. order filter.

Well, thanks for now

Steffen
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35893

  • dreite
  • dreite's Avatar
  • Offline
  • Platinum Boarder
  • Posts: 1179
  • Thank you received: 594
  • Karma: 82
As mentioned above, all the functionality you need to create a set of first-order coefficients is built right into your PWR-ICE software plugin. You'll just have to "cascade" them correctly when pasting into the advanced portion of your plugin to create the multiple 6db/octave turnover points for your objective.

I'm not sure why you're using the "All-digital...." spreadsheet, since that is based around generating bi-quad coefficients.......which you're not interested in here.

Dave.
The administrator has disabled public write access.
The following user(s) said Thank You: devteam

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35895

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi Dave

Well, I actually figured out a way to make my filters by copying the bi-quad coefficients from the 2x4 HD plug in (also 96 kHz) and paste them in the PWR-ICE plug in. It is more safe to produce coefficients in a separate plug in, because switching between Basic to Advanced mode in the PWR-ICE
Plug in created problems for me.

But initially I thought I'd have to use "All digital...." to use the advanced mode in the PWR-ICE plug in.

I will still use "All-digital...." to produce bi-quad coefficients for a Linkwitz transform.

Steffen
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35909

  • JJ2106
  • JJ2106's Avatar
  • Offline
  • Expert Boarder
  • Posts: 106
  • Thank you received: 8
  • Karma: 1
Hi Steffen,
Here is an htm file which explains how to calculate first order coefficients for an allpass filter.
Transposing this for lopass and hipass filters is trivial.
JJ
Pls: pls rename file xxx.htm before launching it
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<!-- saved from url=(0053)www.earlevel.com/Digital%20Audio/Bilinear.html -->
<HTML>
<HEAD>
<TITLE>The Bilinear Transform</TITLE>
<META http-equiv=Content-Type content="text/html; charset=iso-8859-1">
<META content="MSHTML 6.00.2900.2963" name=GENERATOR>
</HEAD>
<BODY text=#333333 vLink=#66099 aLink=#ff0000 link=#990000 bgColor=#ffffff
leftMargin=8 topMargin=8>

<H3>The bilinear z transform</H3>

<P>The bilinear transform is the most popular method of converting analog filter
prototypes in the s domain to the z domain so we can implement them as digital
filters. The reason we are interested in these s domain filters is that analog
filter theory has been around longer than digital filter theory, so we have
available to us a number of popular and useful filters in the s domain. The
standard biquads used heavily in audio processing are direct implementations
from the s domain using the bilinear transform.

<P>The theory of the bilinear transform is well documented in DSP texts and on
the web, so rather than spend time going into the theory, we'll cut to the chase
and show how to do the transform, using a first order all-pass filter as an
example.

<P>The s domain transfer function of a first order all-pass filter is:

<BLOCKQUOTE><P><PRE>
H(s) = (1 - s) / (1 + s)
</PRE>
</BLOCKQUOTE>

<P>To move to the z domain, we need to substitute for s in terms of z. At the
same time, we "pre-warp" the response using the TAN function in order to wrap
the infinite s plane onto the circular z plane. The combined substitution is:

<BLOCKQUOTE><P><PRE>
s = (1 - z^-1) / (K*(1 + z^-1)) </PRE>where:
<P><Font face="Courier New" size=-1>
K = TAN (</FONT><Font face=Symbol>p</FONT><Font face="Courier New" size=-1> * Fc / Fs)</FONT>

<P>Fc is the desired corner frequency, and Fs is the sampling frequency.
</BLOCKQUOTE>

<P>Here's our z domain transfer function after substituting for s:

<BLOCKQUOTE><P><PRE>
1 - (1 - z^-1) / (K*(1 + z^-1))
H(z) =
1 + (1 - z^-1) / (K*(1 + z^-1))
</PRE>
</BLOCKQUOTE>

<P>In order to solve for our filter coefficients, we need to get the equation
into the form of a first order polynomial in the numerator (the zero) and
another first order polynomial in the denominator (the pole). That is, we need
it to look something like this:

<P><BLOCKQUOTE>
<PRE>
b0 + b1*z^-1
H(z) =
a0 + a1*z^-1
<PRE>
</BLOCKQUOTE>

<P> We multiply the nominator and the denominator of the substituted transfer function by:

<P><BLOCKQUOTE>
<PRE>
K*(1 + z^-1)
</PRE>
</BLOCKQUOTE>

<P> We obtain the following expression:

<P><BLOCKQUOTE>
<PRE>
K*(1 + z^-1) - (1 - z^-1)
H(z) =
K*(1 + z^-1) + (1 - z^-1)
</PRE>
</BLOCKQUOTE>

<P> We rearrange the numerator and the denominator in descending powers of z to obtain:

<P><BLOCKQUOTE>
<PRE>
(K - 1) + (K + 1)*z^-1
H(z) =
(K + 1) + (K - 1)*z^-1
</PRE>
</BLOCKQUOTE>

<P>Our coefficients are now evident:

<P><BLOCKQUOTE>
<PRE>
b0 = K - 1 b1 = K + 1 b2 = 0

a0 = K + 1 a1 = K - 1 a2 = 0
</PRE>
</BLOCKQUOTE>

<P>Because the (a0) coefficient is associated with y[n], the filter output, we
divide everything by (a0) to normalize the filter:

<P><BLOCKQUOTE>
<PRE>
K - 1
b0 =
b1 = 1 b2 = 0
K + 1
K - 1
a0 = 1 a1 =
a2 = 0
K + 1
</PRE>
</BLOCKQUOTE>

<P>Original article: <A href="www.earlevel.com/Digital%20Audio/Bilinear.html" target=_blank>www.earlevel.com/Digital%20audio/Bilinear.html</A>

</BODY>
</HTML>
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35910

  • JJ2106
  • JJ2106's Avatar
  • Offline
  • Expert Boarder
  • Posts: 106
  • Thank you received: 8
  • Karma: 1
I resubmit because the previous version was not readable. Hope it will be better.

Hi Steffen,
Here is an htm file which explains how to calculate first order coefficients for an allpass filter.
Transposing this for lopass and hipass filters is trivial.
JJ
Pls: pls rename file xxx.htm before launching it

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<!-- saved from url=(0053)www.earlevel.com/Digital%20Audio/Bilinear.html -->
<HTML>
<HEAD>
<TITLE>The Bilinear Transform</TITLE>
<META http-equiv=Content-Type content="text/html; charset=iso-8859-1">
<META content="MSHTML 6.00.2900.2963" name=GENERATOR>
</HEAD>
<BODY text=#333333 vLink=#66099 aLink=#ff0000 link=#990000 bgColor=#ffffff
leftMargin=8 topMargin=8>

<H3>The bilinear z transform</H3>

<P>The bilinear transform is the most popular method of converting analog filter
prototypes in the s domain to the z domain so we can implement them as digital
filters. The reason we are interested in these s domain filters is that analog
filter theory has been around longer than digital filter theory, so we have
available to us a number of popular and useful filters in the s domain. The
standard biquads used heavily in audio processing are direct implementations
from the s domain using the bilinear transform.

<P>The theory of the bilinear transform is well documented in DSP texts and on
the web, so rather than spend time going into the theory, we'll cut to the chase
and show how to do the transform, using a first order all-pass filter as an
example.

<P>The s domain transfer function of a first order all-pass filter is:

<BLOCKQUOTE><P><PRE>
H(s) = (1 - s) / (1 + s)
</PRE>
</BLOCKQUOTE>

<P>To move to the z domain, we need to substitute for s in terms of z. At the
same time, we "pre-warp" the response using the TAN function in order to wrap
the infinite s plane onto the circular z plane. The combined substitution is:

<BLOCKQUOTE><P><PRE>
s = (1 - z^-1) / (K*(1 + z^-1)) </PRE>where:
<P><Font face="Courier New" size=-1>
K = TAN (</FONT><Font face=Symbol>p</FONT><Font face="Courier New" size=-1> * Fc / Fs)</FONT>

<P>Fc is the desired corner frequency, and Fs is the sampling frequency.
</BLOCKQUOTE>

<P>Here's our z domain transfer function after substituting for s:

<BLOCKQUOTE><P><PRE>

H(z) = [ 1 - (1 - z^-1) / (K*(1 + z^-1))] / [1 + (1 - z^-1) / (K*(1 + z^-1))]
</PRE>
</BLOCKQUOTE>

<P>In order to solve for our filter coefficients, we need to get the equation
into the form of a first order polynomial in the numerator (the zero) and
another first order polynomial in the denominator (the pole). That is, we need
it to look something like this:

<P><BLOCKQUOTE>
<PRE>
H(z) = (b0 + b1*z^-1) / (a0 + a1*z^-1)
<PRE>
</BLOCKQUOTE>

<P> We multiply the nominator and the denominator of the substituted transfer function by:

<P><BLOCKQUOTE>
<PRE>
K*(1 + z^-1)
</PRE>
</BLOCKQUOTE>

<P> We obtain the following expression:

<P><BLOCKQUOTE>
<PRE>
H(z) = [K*(1 + z^-1) - (1 - z^-1)] / [ K*(1 + z^-1) + (1 - z^-1)]
</PRE>
</BLOCKQUOTE>

<P> We rearrange the numerator and the denominator in descending powers of z to obtain:

<P><BLOCKQUOTE>
<PRE>
H(z) = [ (K - 1) + (K + 1)*z^-1] / [(K + 1) + (K - 1)*z^-1]
</PRE>
</BLOCKQUOTE>

<P>Our coefficients are now evident:

<P><BLOCKQUOTE>
<PRE>
b0 = K - 1
b1 = K + 1
b2 = 0

a0 = K + 1
a1 = K - 1
a2 = 0
</PRE>
</BLOCKQUOTE>

<P>Because the (a0) coefficient is associated with y[n], the filter output, we
divide everything by (a0) to normalize the filter:

<P><BLOCKQUOTE>
<PRE>
b0= (K - 1) / (K + 1)
b1 = 1
b2 = 0

a0 = 1
a1 = (K - 1) / (K + 1)
a2 = 0

</PRE>
</BLOCKQUOTE>

<P>Original article: <A href="www.earlevel.com/Digital%20Audio/Bilinear.html" target=_blank>www.earlevel.com/Digital%20audio/Bilinear.html</A>

</BODY>
</HTML>
The administrator has disabled public write access.

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35912

  • JJ2106
  • JJ2106's Avatar
  • Offline
  • Expert Boarder
  • Posts: 106
  • Thank you received: 8
  • Karma: 1
And now, an htm page for a second order lopass. Again, generalizing to other filter types is trivial, assuming the transfer function in s is known.
JJ

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<!-- saved from url=(0053)www.earlevel.com/Digital%20Audio/Bilinear.html -->
<HTML>
<HEAD>
<TITLE>The Bilinear Transform</TITLE>
<META http-equiv=Content-Type content="text/html; charset=iso-8859-1">
<META content="MSHTML 6.00.2900.2963" name=GENERATOR>
</HEAD>
<BODY text=#333333 vLink=#66099 aLink=#ff0000 link=#990000 bgColor=#ffffff
leftMargin=8 topMargin=8>

<H3>The bilinear z transform</H3>

<P>The bilinear transform is the most popular method of converting analog filter
prototypes in the s domain to the z domain so we can implement them as digital
filters. The reason we are interested in these s domain filters is that analog
filter theory has been around longer than digital filter theory, so we have
available to us a number of popular and useful filters in the s domain. The
standard biquads used heavily in audio processing are direct implementations
from the s domain using the bilinear transform.

<P>The theory of the bilinear transform is well documented in DSP texts and on
the web, so rather than spend time going into the theory, we'll cut to the chase
and show how to do the transform, using a second order lowpass filter as an
example.

<P>The s domain transfer function of a second order lowpass filter is:

<BLOCKQUOTE><P><PRE>
H(s) = 1 / (s^2 + A*s + 1)
</PRE>
</BLOCKQUOTE>

<P>where (A) is the damping coefficient, i.e. the inverse of the filter's Q.

<P>To move to the z domain, we need to substitute for s in terms of z. At the
same time, we "pre-warp" the response using the TAN function in order to wrap
the infinite s plane onto the circular z plane. The combined substitution is:

<BLOCKQUOTE><P><PRE>
s = (1 - z^-1) / (K*(1 + z^-1)) </PRE>where:
<P><Font face="Courier New" size=-1>
K = TAN (</FONT><Font face=Symbol>p</FONT><Font face="Courier New" size=-1> * Fc / Fs)</FONT>

<P>Fc is the desired corner frequency, and Fs is the sampling frequency.
</BLOCKQUOTE>

<P>Here's our z domain transfer function after substituting for s:

<BLOCKQUOTE><P><PRE>
H(z) = 1 /[(1 - z^-1)^2 / K^2*(1 +z^-1)^2) + A*(1 - z^-1) / K*(1 + z^-1) + 1]
</PRE>
</BLOCKQUOTE>

<P>In order to solve for our filter coefficients, we need to get the equation
into the form of a second order polynomial in the numerator (the zeroes) and
another second order polynomial in the denominator (the poles). That is, we need
it to look something like this:

<P><BLOCKQUOTE>
<PRE>
b0 + b1*z^-1 + b2*z^-2
H(z) =
a0 + a1*z^-1 + a2*z^-2
</PRE>
</BLOCKQUOTE>

<P> We multiply the nominator and the denominator of the substituted transfer function by:

<P><BLOCKQUOTE>
<PRE>
K^2*(1 + z^-1)^2
</PRE>
</BLOCKQUOTE>

<P> We obtain the following expression:

<P><BLOCKQUOTE>
<PRE>
K^2*(1 + z^-1)^2
H(z) =
(1 - z^-1)^2 + A*K*(1 - z^-1)*(1 + z^-1) + K^2*(1 + z^-1)^2
</PRE>
</BLOCKQUOTE>

<P> We expand the numerator and the denominator, and rearrange them in descending powers of z to obtain:

<P><BLOCKQUOTE>
<PRE>
K^2 + 2*K^2*z^-1 + K^2*z^-2
H(z) =
(K^2 + A*K + 1) + 2*(K^2 - 1)*z^-1 + (K^2 -A*K + 1)*z^-2
</PRE>
</BLOCKQUOTE>

<P>Our coefficients are now evident:

<P><BLOCKQUOTE>
<PRE>
b0 = K^2 b1 = 2*K^2 b2 = K^2

a0 = K^2 + A*K + 1 a1 = 2*(K^2 - 1) a2 = K^2 - A*K + 1
</PRE>
</BLOCKQUOTE>

<P>Because the (a0) coefficient is associated with y[n], the filter output, we
divide everything by (a0) to normalize the filter. Here we also note the
relationships between the (b) coefficients:

<P><BLOCKQUOTE>
<PRE>
K^2
b0 =
b1 = 2*b0 b2 = b0
K^2 + A*K + 1
2*(K^2 - 1) K^2 - A*K + 1
a0 = 1 a1 =
a2 =
K^2 + A*K + 1 K^2 + A*K + 1
</PRE>
</BLOCKQUOTE>

<P>Original article: <A href="www.earlevel.com/Digital%20Audio/Bilinear.html" target=_blank>www.earlevel.com/Digital%20audio/Bilinear.html</A>

</BODY>
</HTML>
The administrator has disabled public write access.
The following user(s) said Thank You: devteam

How calculate biquad coefficients for 6dB filter 8 months 2 weeks ago #35913

  • supersteff
  • supersteff's Avatar
  • Offline
  • Fresh Boarder
  • Posts: 14
  • Thank you received: 1
  • Karma: 0
Hi JJ2206

Thank you for all that theory and where to find it. I see, my math is a bit rusty by now, so it's a bit over my head. But I think your contribution is a valuable resource for others too.

My project is stuck right now, although I have got my digital signal chain up running. At the moment I'm only using one out of 4 channels on each side to drive my Visaton 200 b full range speakers.

Steffen
The administrator has disabled public write access.
Moderators: devteam
Time to create page: 0.176 seconds