2
$\begingroup$

I would like to write a 3 Band "EQ" with BiQuad Filters but for some reason, I get some phase distortion.

The problem already got discussed here.

I found out that I need to work with Allpass filters in order to compensate the phase response of my filters. Btw, the filters are Linkwitz-Riley 2nd order filters.

But when I try to copy the idea to my project, I still have some distortion.

// low pass calcuation for linkwitz riley filters

float lowCutoff; // the lower cutoff frequency
float highCutoff; // the higher cutoff frequency

double omega = M_PI * cutoffFrequency;  
double theta = omega / sampleRate;
double kappa = omega / tan(theta);
double delta = kappa * kappa + omega * omega + 2.0 * kappa * omega;

double a0 = (omega * omega) / delta;
double a1 = 2.0 * a0;
double a2 = a0;

double b1 = (-2.0 * kappa * kappa + 2.0 * omega * omega) / delta;
double b2 = (-2.0 * kappa * omega + kappa * kappa + omega * omega) / delta;

// high pass, only the calculation of a0 to b2 changes
// the intermediate values stay the same

double a0 = (kappa * kappa) / delta;
double a1 = -2.0 * a0;
double a2 = a0;

double b1 = (-2.0 * kappa * kappa + 2.0 * omega * omega) / delta;
double b2 = (-2.0 * kappa * omega + kappa * kappa + omega * omega) / delta;

// allpass filter to compensate phase response:

double q = 0.5;
double alpha = (tan(M_PI*q/sampleRate) - 1) / (tan(M_PI*0.5/sampleRate) + 1);
double beta =   -cos(lowCutoff  * M_PI / sampleRate);

double a0 = -alpha;
double a1 = beta*(1.0 - alpha);
double a2 = 1.0;
double b1 = beta*(1.0 - alpha);;
double b2 = -alpha;;

Since I use 2nd order filters, I also have to use a 2nd order all pass filter, right? I've read that for those filters and with this calculation, I hav to use q. So q should be the same as the q in the Linkwitz-Riley filter, which should be 0.5 right? (I've read this somewhere, not sure about it.)

Then, my actual processing looks something like this:

// filter names:
// loAPF, lowHPF, ... low: lower cutoffhigh: higher cutoff
// APF: allpass, HPF: highpass, LPF: lowpass

highSample =  lowAPF->tick(highHPF->tick(sample));
midSample =   lowHPF->tick(highLPF->tick(sample));
lowSample =   lowLPF->tick(highLPF->tick(sample));

Does somebody have any idea, why I might have problems with the phase response?

Thanks in advance!

|improve this question
$\endgroup$
  • $\begingroup$ I've just noticed that I used the high lowpass filtertwice, which isn't realy how you use dgital filters. :/ So this could be a problem but looking over, I also think that the phases are still off or not 360° shifted to each other. $\endgroup$ – ruhig brauner Nov 6 '14 at 20:48
  • $\begingroup$ a LRF shouldnt have a Q value at all... $\endgroup$ – bubu Aug 16 '19 at 6:08
  • $\begingroup$ That doesn't answer any of the points made in the question. Besides that, LRFs are put together usually with 2nd order IIR filters (cascaded) and thus have internal Q coefficients. They're however fixed by the design assumption that the sum of two filters (LP, HP) should have a flat cross over. So, A LRF doesn't have a parametric Q like a generic 2nd order lowpass. But it still has a Q parameter that is defined by the design assumption. $\endgroup$ – ruhig brauner Aug 16 '19 at 14:05
0
$\begingroup$

There is a reference at the Linkwitz website regarding the "The Duelund 3-way crossover filter function D3" that appears to be the 3-band that is not derived from splitting one band of a 2-band L-R crossover. I may have seen this before, but I am not familiar. I'll look at it now and see if I can grok it.

Here's another reference to the Duelund thingie.

Okay, it appears that this is the essential Duelund 3-way crossover:

$$ H_\text{LP}(s) = \frac{1}{s^4 + 2as^3 + (a^2+2)s^2 + 2as + 1} $$

$$ H_\text{BP}(s) = \frac{-(a^2+2)s^2}{s^4 + 2as^3 + (a^2+2)s^2 + 2as + 1} $$

$$ H_\text{HP}(s) = \frac{s^4}{s^4 + 2as^3 + (a^2+2)s^2 + 2as + 1} $$

and this all adds up to an All-pass filter. Looks like $a=2\sqrt2$ is recommended.

|improve this answer
$\endgroup$
0
$\begingroup$

A 2nd-order Linkwitz-Riley crossover needs 1st-order allpass phase compensation.

The Lowpass half of the LR crossover is:

$$\dfrac{1}{s^2+2s+1} $$

The Highpass half of the LR crossover is:

$$\dfrac{s^2}{s^2+2s+1} $$

With even-order LR crossovers, in order to sum-to-allpass one of the two sections has to have inverted polarity. Thus:

$$\dfrac{s^2}{s^2 + 2s +1} - \dfrac{1}{s^2 + 2s +1}= \dfrac{(s-1)(s+1)}{(s+1)(s+1)} = \dfrac{(s-1)}{(s+1)}$$

|improve this answer
$\endgroup$
  • $\begingroup$ Could you do me the favor and add some link which explaines what what 1/ (s²...) stands for? I sadly have no clue was s is and what the term means. ;D I've used 4th order and a 2nd order APF, which worked. But knowing how to solve this for 2nd order LR filters would be nice! :) $\endgroup$ – ruhig brauner Nov 12 '14 at 17:01
  • $\begingroup$ en.wikipedia.org/wiki/Laplace_transform $\endgroup$ – MackTuesday Dec 12 '14 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.