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Greg Berchin
Guest

Posted: Thu Dec 15, 2005 9:15 am Post subject: Re: Crossover networks. Can someone recall the name?

On 14 Dec 2005 19:14:31 -0800, "Ron N." <rhnlogic@yahoo.com> wrote:

Quote:

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?

Yes! Exactly!

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.

Greg

Greg Berchin
Guest

Posted: Thu Dec 15, 2005 9:16 am Post subject: Re: Crossover networks. Can someone recall the name?

On Thu, 15 Dec 2005 02:23:25 GMT, Vladimir Vassilevsky
<antispam_bogus@hotmail.com> wrote:

Quote:

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.

20Hz? I thought that we were talking about 100Hz.

Quote:

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.

The techniques that I mentioned ARE ways to increase numeric precision,
but they don't require construction of traditional double precision
operations.

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

Okay, so I pulled a number out of thin air, without really thinking
about the resulting locations of the poles. It was a poor example of a
good idea.

Quote:

The 24dB/oct Xovers are "good enough" for most practical purposes.

That's a sweeping generalization about something that requires
consideration on a case-by-case basis.

Quote:

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.

The transition band slope is only half the problem. The other half is
the flatness in the passband. A 4th order Bessel will have essentially
no ringing at all, while a 4th order Chebyshev might ring excessively.

Greg

robert bristow-johnson
Guest

Posted: Thu Dec 15, 2005 9:16 am Post subject: Re: Crossover networks. Can someone recall the name?

Al Clark wrote:

Quote:

I think the main criticism with Linear Phase FIR filters is that you can
often hear pre-echos caused by the fact that the impulse response is
symmetrical around the center.

but it's not so much the ripples ramping up to the main lobe of the
impulse response, Al. it's a consequence of using the Parks-McClellan
algorithm to design an "equi-ripple" filter. those ripples in the
pass-band are like multiplying by a cosine which has the effect of
translation in the other domain.

Quote:

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).

this is what i meant, Al - you can see a picture of it at:

http://www.circuitcellar.com/library/eq/154/4.htm

below, at bottom, is a simple MATLAB program that will compare a LPF
design using Parks-McClellan (otherwize known as "remez" in MATLAB
land), Least Squares, and a Kaiser-windowed sinc methods. if the
design method is Parks-McClellan, it is likely that the ripples *will*
look like a cosine with a pretty constant frequency (and P-McC will
make sure the amplitude is constant) over the passband. this is
equivalent to taking a smoother LPF frequency response and multiplying
it by a small cosine biased up by the constant 1. multiplying in the
frequency-domain is the same as convoluting in the time-domain.
multiplying by the constant 1 is like convolving with a delta function
(which changes nothing) but multiplying by the small cosine is like
convolving by a pair of offset deltas that are small in amplitude.
that means the main LPF impulse will be translated to the edges before
and after the main impulse (a pre-echo and post-echo). temporal
masking might hide the post-echo, but not the pre-echo.

run the following MATLAB code if you can to see. (you will have to fix
the word-wrapping that Google Groups puts in.) you can visually see
the pre-echo and post-echo that will go away if you use a different
design method.

r b-j

% lpf.m

% Basic Parks-McClellan (or Least Squares or windowed sinc) Low-Pass
Filter Design

if exist('filename') ~= 1,
filename = 'lpfcoefs.txt'; % filename for FIR coefs output
end;

% the input parameters are not changed if they already exist

if exist('SR') ~= 1,
SR = 96; % sampling rate
end;

if exist('PB') ~= 1,
PB = 20; % top of passband region (where filter should be flat
and 0 dB)
end;

if exist('PBR') ~= 1,
PBR = 0.5; % max passband ripple in dB which is twice the max
error in passband
end;

if exist('SB') ~= 1,
% SB = SR/2 - PB;
SB = 20.5; % bottom of stopband region (where filter should be
flat and -inf dB)
end;

if exist('SBG') ~= 1,
SBG = -84; % max stop band gain in dB
end;

R = (PB + SB)/SR;

% Number of required FIR taps, from a formula in Oppenhiem & Schafer
p. 480
% the more complex and accurate MATLAB function remlpord() could be
used instead.
% for nicety, we always round up to the nearest odd number of FIR
taps.
% when N is odd, the delay (N-1)/2 is always an integer number of
samples

N = 2*ceil(0.0342416*SR*(-0.5*SBG - 10*log10(10^(0.5*PBR/20)-1) -
13)/(SB-PB) - 0.5) + 1

if 0 % set this to 1 if you wanna get coefficients
for windowed sinc LPF (for testing)

h = zeros(1, N+1); % zero the whole thang

t = linspace(-R*(N-1)/2, R*(N-1)/2, N);
h = [0 (R*sinc(t)) .* kaiser(N, 0.1102*(-SBG-8.7))'] ; %
kaiser windowed sinc()
clear t;

else

W = ( (10^(0.5*PBR/20) - 1) )/( 10^(SBG/20) ) % linear
stopband/passband weighting ratio

f = [0 PB/(0.5*SR) SB/(0.5*SR) 1]; % Define passband and
stopband regions
m = [1 1 0 0]; % Define passband and
stopband magnitudes
w = [1 W ]; % Define passband and
stopband error weights

h = zeros(1, N+1);

% h = [0 firls(N-1, f, m, w)]; % Compute impulse response from
Least Squares

h = [0 remez(N-1, f, m, w)]; % Compute impulse response from
Parks-McClellan

end

% h(1) = 0 and h((N+1)/2) = max
% symmetry: h((N+1)/2 + k) = h((N+1)/2 - k)
% this should result in zero phase
% in the frequency response.

plot(h/R, '.'); % plot scaled impulse response pause;

H = fft([h([(N+1)/2+1:N+1]) zeros(1, 3*(N+1)) h([1:(N+1)/2])]); %
fft after zero-padding to extend length by factor of 4

freq = linspace(0, 1-2/(N+1), 2*N+1); % set of frequencies
from DC to just under Nyquist

plot(freq, 20*log10(abs(H(1:2*N+1))+1.0e-7)); % dB by linear
freq plot
pause;

axis([0 0.5 -1 1]); % look closely at passband ripple
pause;

semilogx(freq(2:2*N+1), 20*log10(abs(H(2:2*N+1))+1.0e-7), 'g'); % dB
by log freq plot, skip DC
pause;

axis([0.05 0.5 -2 2]);

h = h/R;

outfile=fopen(filename,'wt+');
fprintf(outfile, '%1.8e\n', h(2:N+1));
fclose(outfile);

Vladimir Vassilevsky
Guest

Posted: Thu Dec 15, 2005 9:16 am Post subject: Re: Crossover networks. Can someone recall the name?

Greg Berchin wrote:

Quote:

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.

20Hz? I thought that we were talking about 100Hz.

I suggested to try lower frequency just to make the artifacts more
pronounced. It will be also very noticeable at 100Hz. The 24bit
precision at Fs = 48kHz without noise shaping allows to build biquads
with Fc no lower then about 500Hz. If the Fc is lower the performance is
likely to be insufficient for the quality audio.

Quote:

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.

The techniques that I mentioned ARE ways to increase numeric precision,
but they don't require construction of traditional double precision
operations.

The problem is that you have to increase the precision significantly. At
the 100Hz cutoff the loss of precision can be as high as 16 bits. Small
tricks can save one or two bits however it is not going to help much.
Noise shaping in biquad is good and simple solution.

Quote:

The 24dB/oct Xovers are "good enough" for most practical purposes.
That's a sweeping generalization about something that requires
consideration on a case-by-case basis.

This is the engineering approach and also my observation from the
practice. In most cases, steep slopes do not really make noticeable
improvement.

Quote:

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.

The transition band slope is only half the problem. The other half is
the flatness in the passband. A 4th order Bessel will have essentially
no ringing at all, while a 4th order Chebyshev might ring excessively.

The reasonably designed 24dB IIR filter will have relatively mild
ringing. High Q stages are very sensitive to the problems with numeric
precision and the overflows, therefore I wouldn't recommend Chebyshev.
Old good Butterworth will work just fine.

Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

Guest

Posted: Thu Dec 15, 2005 3:39 pm Post subject: Re: questions raised by reading and thinking with possibly m

Randy Yates wrote:

Quote:

abariska@student.ethz.ch writes:
[...]
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.

That won't work. For an FIR with real coefficients, the if and only if
condition for linear phase response is that the impulse response be
symmetric or antisymmetric. This would contradict your statement.

I found the correct condition in that link that Peter K posted
(http://0xdc.com/paper.pdf):

"
Theorem: A real, discrete signal x(n) has linear phase response iff

x_a(t) := sum_n x(n) sinc( pi (t-n) )

is symmetric about some point d.
"

That excludes your truncated asymmetrically sampled symmetric
continuous signal (because the finite number of asymmetric samples
result in an asymmetric reconstructed signal x(t)).

Regards,
Andor

Guest

Posted: Thu Dec 15, 2005 3:42 pm Post subject: Re: questions raised by reading and thinking with possibly m

Quote:

Randy Yates wrote:

abariska@student.ethz.ch writes:
[...]
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.

That won't work.

Rereading your statement, it seems to be exactly the stated theorem, so
I retract that.

:-)

Regards,
Andor

Guest

Posted: Thu Dec 15, 2005 3:50 pm Post subject: Re: questions raised by reading and thinking with possibly m

I wrote:

Quote:

"
Theorem: A real, discrete signal x(n) has linear phase response iff

x_a(t) := sum_n x(n) sinc( pi (t-n) )

is symmetric about some point d.
"

Hmm. What about antisymmetric impulse responses?

Jerry Avins
Guest