| Author |
Message |
Vladimir Vassilevsky
Guest
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 Posted:
Wed Dec 14, 2005 11:30 pm Post subject:
Re: Crossover networks. Can someone recall the name? |
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Greg Berchin wrote:
| Quote: | To have reasonable audio performance, those biquads should use 32x32 =
64 bit multiplications at least, as well as the noise shaping at
truncation of 64 to 32 bits. This relates to the straightforward 48kHz
implementation.
I respectfully disagree. I have implemented this in the DSP56K with 24
bit data, 24 bit coefficients, 56 bit accumulator, and no noise shaping.
The performance is quite "reasonable".
|
I have badly burned with that once and believe me I know the topic :(
The quantization noise in the biquad is amplified in the feedback path.
The noise gain is at the order of the square of (Q*Fc/Fs). Thus for the
100Hz cutoff at 48kHz sample rate the loss of precision will be over
100dB. Only 8 bits are left from your 24 bit data. This is very audible.
As for the coefficient accuracy, even 20 bit is enough however the data
should have 32bit width at the least.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com |
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Al Clark
Guest
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| Posted:
Wed Dec 14, 2005 11:39 pm Post subject:
Re: Crossover networks. Can someone recall the name? |
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Greg Berchin <76145.2455@compuswerve.com> wrote in
news:jfh0q1d2931d8ngbfd2rtu56k9hvf92m99@4ax.com:
| Quote: | On Wed, 14 Dec 2005 15:19:23 GMT, Al Clark <dsp@danvillesignal.com
wrote:
Imagine someone hits a drum. You can hear the drum before he hits it.
If the filters have a fairly regular passband ripple (kind of
sinusoidal), such as might be expected from a PM filter, then the pre-
echo will be concentrated. You can reduce this effect, by using a
filter
with a more "random" like passband ripple. BTW: (I learned all this
from
a discussion with rbj).
Peter Craven published an interesting article about this (in the
context
of antialias filters) last year: "Antialias Filters and System
Transient Response at High Sample Rates"; Journal of the Audio
Engineering Society, Volume 52, Number 3, March 2004.
Greg
|
This is an interesting paper. Thanks for the reference.
BTW: I'm not taking a stand on the best method of creating speaker
crossovers. I just remember a few of the issues.
In seems to me, that subwoofer crossovers are not nearly as critical as
the crossover between a tweeter and a midrange driver from an artifacts
point of view.
Obviously, one of the benefits of using a digital crossover is that you
can experiment by jusat changing the software.
--
Al Clark
Danville Signal Processing, Inc.
--------------------------------------------------------------------
Purveyors of Fine DSP Hardware and other Cool Stuff
Available at http://www.danvillesignal.com |
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Greg Berchin
Guest
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| Posted:
Wed Dec 14, 2005 11:53 pm Post subject:
Re: Crossover networks. Can someone recall the name? |
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On Wed, 14 Dec 2005 17:30:01 GMT, Vladimir Vassilevsky
<antispam_bogus@hotmail.com> wrote:
| Quote: | I have badly burned with that once and believe me I know the topic :(
The quantization noise in the biquad is amplified in the feedback path.
The noise gain is at the order of the square of (Q*Fc/Fs). Thus for the
100Hz cutoff at 48kHz sample rate the loss of precision will be over
100dB. Only 8 bits are left from your 24 bit data. This is very audible.
|
I do not disagree with your analysis. However there are creative ways
around the problem, and I used some of them. As a simple example,
consider that if one of the feedback coefficients is "4.19754", you can
simply shift the datum (hopefully you saved all 56 bits somewhere) two
bits left and add to that the upper 24 bits of the datum multiplied by
0.19754. It's not perfect, but I'm an engineer, not a mathematician.
Engineers enjoy the luxury of "good enough". :)
Greg |
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Greg Berchin
Guest
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| Posted:
Thu Dec 15, 2005 1:16 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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On Wed, 14 Dec 2005 23:27:57 +0100, Andrew Reilly
<andrew-newspost@areilly.bpc-users.org> wrote:
| Quote: | Surely not! Show me any linear phase FIR filter (digitally symmetrical
or not) and I'll show you a minimum phase filter with an identical
frequency response.
|
and
On 14 Dec 2005 14:47:02 -0800, "Ron N." <rhnlogic@yahoo.com> wrote:
| Quote: | Not true. To some degree you can vary the phase, given identical
(flat, brick, or whatever) frequency responses.
|
Gentlemen, while what you say is correct, you miss my point. My point
is that the impulse response of a rectangular magnitude response
(assuming a reasonably "smooth" accompanying phase response) decays at
rate of roughly 6 dB per doubling of time. That makes for a lot of
ringing in the time domain.
Yes, the changes in phase that you mention WILL change specific
characteristics of the impulse response -- for example, going from zero
phase to minimum phase will change it from noncausal (starting at
t=-infinity, peaking at t=0, and extending to t=+infinity) to causal
(starting at t=0, peaking soon thereafter, and extending to
t=+infinity). But Parseval's relation tells us that the integral, over
all time, of the squared magnitude in the time domain is proportional to
the integral, over all frequencies, of the squared magnitude in the
frequency domain. So changing the phase response does NOT change the
amount of energy in the impulse response; it just "moves it around".
Greg |
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Ron N.
Guest
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| Posted:
Thu Dec 15, 2005 1:16 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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Greg Berchin wrote:
| Quote: | On Wed, 14 Dec 2005 22:22:28 +0100, Andrew Reilly
andrew-newspost@areilly.bpc-users.org> wrote:
a perfect (for whatever
definition of perfect actually works in practice) brick-wall crossover
ought to improve the off-axis behaviour by ensuring that there's no
interference (no frequencies in common to multiple drivers).
... which brings us full-circle, back to the original issue of transient
response. The closer the filters are to true brickwall in the frequency
domain, the closer to sin(x)/x they are in the time domain.
|
Not true. To some degree you can vary the phase, given identical
(flat, brick, or whatever) frequency responses. Minimum/maximum
phase versus linear-phase for instance, merely by moving some
zeros inside/outside the unit circle (or left/right half plane in the
analog case).
IMHO. YMMV.
--
rhn A.T nicholson d.O.t C-o-M |
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Andrew Reilly
Guest
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| Posted:
Thu Dec 15, 2005 1:16 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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On Wed, 14 Dec 2005 15:50:46 -0600, Greg Berchin wrote:
| Quote: | On Wed, 14 Dec 2005 22:22:28 +0100, Andrew Reilly
andrew-newspost@areilly.bpc-users.org> wrote:
a perfect (for whatever
definition of perfect actually works in practice) brick-wall crossover
ought to improve the off-axis behaviour by ensuring that there's no
interference (no frequencies in common to multiple drivers).
... which brings us full-circle, back to the original issue of transient
[Sorry for missing the opening of the thread.]
response. The closer the filters are to true brickwall in the frequency
domain, the closer to sin(x)/x they are in the time domain.
|
Surely not! Show me any linear phase FIR filter (digitally symmetrical
or not) and I'll show you a minimum phase filter with an identical
frequency response. Got my trusty Hilbert transform here in my back
pocket...
Of course, you don't have to live only at the three usual points:
minimum, linear, maximum phase: life gets more interesting in between.
Cheers,
--
Andrew |
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Greg Berchin
Guest
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| Posted:
Thu Dec 15, 2005 1:16 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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On Wed, 14 Dec 2005 22:22:28 +0100, Andrew Reilly
<andrew-newspost@areilly.bpc-users.org> wrote:
| Quote: | a perfect (for whatever
definition of perfect actually works in practice) brick-wall crossover
ought to improve the off-axis behaviour by ensuring that there's no
interference (no frequencies in common to multiple drivers).
|
.... which brings us full-circle, back to the original issue of transient
response. The closer the filters are to true brickwall in the frequency
domain, the closer to sin(x)/x they are in the time domain.
Greg |
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Andrew Reilly
Guest
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| Posted:
Thu Dec 15, 2005 1:16 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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On Tue, 13 Dec 2005 21:44:58 -0600, Greg Berchin wrote:
| Quote: | On Tue, 13 Dec 2005 22:29:32 -0500, Jerry Avins <jya@ieee.org> wrote:
On axis. Their outstanding property is superior (but hardly perfect)
performance off axis.
Correct. For perfect off-axis response, the low frequency driver and
the high frequency driver would have to be coincident. I think of that
not as a limitation of the crossover, but of the loudspeaker.
|
Although I own several pairs of very nice speakers with coincident drivers
(SEAS, I think), that isn't the only option: a perfect (for whatever
definition of perfect actually works in practice) brick-wall crossover
ought to improve the off-axis behaviour by ensuring that there's no
interference (no frequencies in common to multiple drivers).
I'm afraid that I am yet to get around to experimenting with this in my
own systems, but my coleagues who build speaker controller boxes which
have the ability to do this sort of crossover (amongst others) report that
they can work very well in practice.
--
Andrew |
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Guest
|
| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: questions raised by reading and thinking with possibly m |
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Randy Yates wrote:
....
| Quote: | Consider the filter coefficients determined as
function y = test(x)
%function y = test(x)
n = [-25 : 25];
Fs = 1;
Ts = 1/Fs;
t = n*Ts;
plot(sinc(t+1/7));
These are neither symmetric nor antisymmetric in the sense you defined,
and yet this is a linear phase filter, is it not?
No, it's not.
True, but this is:
b[n] = sinc(n + 1/7)
|
Since Fs = Ts = 1, I don't see a difference between sinc(t+1/7) and
sinc(n+1/7) ? |
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Vladimir Vassilevsky
Guest
|
| Posted:
Thu Dec 15, 2005 8:23 am Post subject:
Re: Crossover networks. Can someone recall the name? |
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|
Greg Berchin wrote:
| Quote: | The quantization noise in the biquad is amplified in the feedback path.
The noise gain is at the order of the square of (Q*Fc/Fs). Thus for the
100Hz cutoff at 48kHz sample rate the loss of precision will be over
100dB. Only 8 bits are left from your 24 bit data. This is very audible.
I do not disagree with your analysis. However there are creative ways
around the problem, and I used some of them.
|
Perhaps we did not quite understand each other.
1. Âuild a biquad HPF with Fc = 20Hz and Fs = 48kHz.
2. Apply a small signal to the input.
3. Observe the amount of trash at the output.
This is a problem I am talking about. Not much can be done about it
unless noise shaping is used or the numeric precision is increased.
| Quote: | As a simple example,
consider that if one of the feedback coefficients is "4.19754",
|
How could it be?
For the stability of the biquad it is required that -2 < A1 < +2,
-1 < A2 < +1
| Quote: | you can
simply shift the datum (hopefully you saved all 56 bits somewhere) two
bits left and add to that the upper 24 bits of the datum multiplied by
0.19754. It's not perfect, but I'm an engineer, not a mathematician.
Engineers enjoy the luxury of "good enough". :)
|
The 24dB/oct Xovers are "good enough" for most practical purposes. It
does not really matter if it is FIR or IIR and what is the particular
type of the response. The 24dB slope is not going to produce much ripple
in the transient anyway.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com |
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Ron N.
Guest
|
| Posted:
Thu Dec 15, 2005 9:14 am Post subject:
Re: Crossover networks. Can someone recall the name? |
|
|
Greg Berchin wrote:
| Quote: | On Wed, 14 Dec 2005 22:22:28 +0100, Andrew Reilly
andrew-newspost@areilly.bpc-users.org> wrote:
a perfect (for whatever
definition of perfect actually works in practice) brick-wall crossover
ought to improve the off-axis behaviour by ensuring that there's no
interference (no frequencies in common to multiple drivers).
... which brings us full-circle, back to the original issue of transient
response. The closer the filters are to true brickwall in the frequency
domain, the closer to sin(x)/x they are in the time domain.
.... |
and
....
| Quote: | On 14 Dec 2005 14:47:02 -0800, "Ron N." <rhnlogic@yahoo.com> wrote:
Not true. To some degree you can vary the phase, given identical
(flat, brick, or whatever) frequency responses.
Gentlemen, while what you say is correct, you miss my point. My point
is that the impulse response of a rectangular magnitude response
(assuming a reasonably "smooth" accompanying phase response) decays at
rate of roughly 6 dB per doubling of time. That makes for a lot of
ringing in the time domain.
Yes, the changes in phase that you mention WILL change specific
characteristics of the impulse response -- for example, going from zero
phase to minimum phase will change it from noncausal (starting at
t=-infinity, peaking at t=0, and extending to t=+infinity) to causal
(starting at t=0, peaking soon thereafter, and extending to
t=+infinity). But Parseval's relation tells us that the integral, over
all time, of the squared magnitude in the time domain is proportional to
the integral, over all frequencies, of the squared magnitude in the
frequency domain. So changing the phase response does NOT change the
amount of energy in the impulse response; it just "moves it around".
|
I see. So what you are saying is that even if a brick-wall filter
of some sort can keep the frequency bands from interfering, in
trade, the time domain energy can then interfere, since it is split
into the tails of two different impulse responses which then need
to cancel?
So if you don't have perfectly matched co-axial transducers, you are
stuck with either off-axis frequency response or transient response
distortion?
I think this makes sense. Something about not being able to
reproduce a 2D/3D field from a small finite number of point
sources (unless identical to the original field generators)...
IMHO. YMMV.
--
rhn A.T nicholson d.O.t C-o-M |
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Randy Yates
Guest
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| Posted:
Thu Dec 15, 2005 9:15 am Post subject:
Re: questions raised by reading and thinking with possibly m |
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abariska@student.ethz.ch writes:
| Quote: | Randy Yates wrote:
...
Consider the filter coefficients determined as
function y = test(x)
%function y = test(x)
n = [-25 : 25];
Fs = 1;
Ts = 1/Fs;
t = n*Ts;
plot(sinc(t+1/7));
These are neither symmetric nor antisymmetric in the sense you defined,
and yet this is a linear phase filter, is it not?
No, it's not.
True, but this is:
b[n] = sinc(n + 1/7)
Since Fs = Ts = 1, I don't see a difference between sinc(t+1/7) and
sinc(n+1/7) ?
|
There isn't a difference in that respect. The difference is that n is not
constrained to be between -25 and +25.
--
% Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC % She love the way Puccini lays down a tune, and
%%% 919-577-9882 % Verdi's always creepin' from her room."
%%%% <yates@ieee.org> % "Rockaria", *A New World Record*, ELO
http://home.earthlink.net/~yatescr |
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Jerry Avins
Guest
|
| Posted:
Thu Dec 15, 2005 9:15 am Post subject:
Re: Crossover networks. Can someone recall the name? |
|
|
Greg Berchin wrote:
| Quote: | On 14 Dec 2005 19:14:31 -0800, "Ron N." <rhnlogic@yahoo.com> wrote:
|
...
| Quote: | There are conditions under which those "two different impulse responses"
actually DO cancel ... mathematically. But getting them to cancel
acoustically is a very difficult matter.
|
Especially if they originate from different locations.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Guest
|
| Posted:
Thu Dec 15, 2005 9:15 am Post subject:
Re: questions raised by reading and thinking with possibly m |
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|
Randy Yates wrote:
| Quote: | abariska@student.ethz.ch writes:
Randy Yates wrote:
...
Consider the filter coefficients determined as
function y = test(x)
%function y = test(x)
n = [-25 : 25];
Fs = 1;
Ts = 1/Fs;
t = n*Ts;
plot(sinc(t+1/7));
These are neither symmetric nor antisymmetric in the sense you defined,
and yet this is a linear phase filter, is it not?
No, it's not.
True, but this is:
b[n] = sinc(n + 1/7)
Since Fs = Ts = 1, I don't see a difference between sinc(t+1/7) and
sinc(n+1/7) ?
There isn't a difference in that respect. The difference is that n is not
constrained to be between -25 and +25.
|
Ah, you didn't mention that. You claim that the infinitely long
sequence
b[n] = sinc(n + 1/7),
for all n, is linear-phase. That could be true. Do you have an idea for
a proof? |
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Randy Yates
Guest
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| Posted:
Thu Dec 15, 2005 9:15 am Post subject:
Re: questions raised by reading and thinking with possibly m |
|
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abariska@student.ethz.ch writes:
| Quote: | [...]
You claim that the infinitely long sequence
b[n] = sinc(n + 1/7),
for all n, is linear-phase. That could be true. Do you have an idea
for a proof?
|
Not a complete one, but a possible outline could be this: A digital
filter is linear-phase if and only if its continuous-time counterpart
(i.e., the continuous-time signal interpolated from the digital sequence)
is linear-phase. The interpolation of the given sequence is an offset
sinc(), which is linear phase.
--
% Randy Yates % "The dreamer, the unwoken fool -
%% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..."
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO
http://home.earthlink.net/~yatescr |
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