The latest version of the biquad calculator. It also takes on the functionality of the frequency response grapher:

**bigger:** Yes.

**more filters:** I probably won’t go deep into allpass filters, but people ask about calculating their coefficients from time to time, so here it is. And added first order filters for comparison.

**phase plot:** In earlier versions of the calculator, phase wasn’t important, we’re interested in an amplitude response and live with the phase response. But in adding the allpass filter type, phase is everything. It’s also good to know for other filter types, and for plotting arbitrary coefficients.

**frequency response grapher:** Edit or paste in coefficients and complete the edit with tab or a click outside the editing field to plot it. You can change the plot controls, but if you change a filter control then the calculator will resume as a biquad calculator.

For instance, clear the *b* coefficients, and place this sequence into the a coefficients: 1,0,0,0,0,1. Then click on the graph or anywhere outside the edit field to graph it. That’s the response of summing a signal with a copy of it delayed by five samples, a simple FIR filter—a comb filter.

Because the calculator can also plot the response of arbitrary coefficients, the biquad calculator now displays the normalized b0 coefficient (1.0)—which you can ignore in a typical biquad implementation.

The coefficients fields accept values separated by almost anything—commas, spaces, new lines, for instance. And they ignore letters and values followed by “=”. You can use “a0 = 0.971, a1 = 0.215…”, for instance. Even scientific notation is accepted (“1.03e4”,).

Very cool. Thank you so much.

It doesn’t seem like the gain slider changes anything.

It’s subtle, but notice that the Gain field is dimmed for all but the filters where gain makes sense—peak and shelf filters.

Hai,

I liked your content. Thank you so much.

I have one small doubt.

We are doing audio Equalizer testing, where the default values we are using for the Equalizer are:

FC_lowShelf – 250

FC_highShelf – 1000

Sample rate – 44100

gain – 6

Q factor – 0.707

If I use the above values I am getting noise when I run my code.

What if I set “Gain= 0”? what is the impact?

And the values used for “Sample Rate”, “Qfactor”, and “Gain” are the default configurations?

I don’t understand the question. Are you using my biquad code, or other? For the shelves, Q is not used, so the setting doesn’t matter. Gain is the gain of the shelf. You should not get noise.

Hi Nigel,

thank you very much for the cool tool and webpage.

Having al look at your Biquad formulas I wonder, that some approaches for lowpass 2nd order coefficients use other formulas:

Q = 1/sqrt(2);

w0 = 2*pi*Fc/Fs;

alpha = sin(w0)/2*w0;

a0 = 1+alpha;

a1 =-2*cos(w0);

a2 = 1-alpha;

b1 = 1-cos(w0);

b0 = b1/2;

b2 = b0;

coeffs = [b0 b1 b2 a0 a1 a2]

coeffs_norm = coeffs / a0;

Do you have an idea what the connection is between these and your formulas ?

Best Regards

Dina

They yield the same coefficient values. The only differences are that Robert names his feedback coefficients a and I name mine b, and he uses the identity tan x = sin x / cos x—this can yield less error in for angle angles when using fixed point math for the calculation. They are both derived from the same analog prototypes in the s-domain, via the bilinear z transform, and produce the same values in floating point (with a and b swapped).

Hi Dina,

The alpha formula is not correct, you need to replace wo for Q at the division: alpha = sin(w0)/2*Q, then the coefficients are the same.

Hi Nigel,

thank you very much for your reply!

Why for the bandpass b1 (resp. a1) is always zero?

Can’t post more. Getting an Internal Server Error.

Hi Dina. It’s just due to the symmetry. A generic second-order bandpass places a zero at 1 and another at -1 on the horizontal axis of the z plane, to pull the response down from each end. The a coefficients dictate the positions of the zeros. The symmetry cancels out the a1 coefficient. You’ll see a different symmetry with the second-order lowpass and highpass, which place both zeros at -1 (Nyquist) for the lowpass, and 1 (0, DC) for the highpass—in both cases a0 and a2 will always be the same value. You can also play with this from the other direction with Pole-Zero placement widget.

Hi Nigel,

thank you very much for your reply!

1. Why for the bandpass b1 (resp. a1) is always zero?

(Whereas with matlab it seems to be composed of a low pass and a high pass filter

with b0-b1 always:

b0 b1 b2

1 2 1

1 -2 1)

2. And how can I calculate a 4th order Biquad with the Biquad formulas?

Dina

1. “1 2 1” looks like a lowpass filter. Are you sure you asked the matlab function for a bandpass?

2. Biquads are inherently second order (the “bi”). But you can make higher order filters out of biquads. See Cascading filters.

Hi Nigel,

1. Yes, I’m sure. Therefore I thought, it seems to be composed of a low pass and a high pass filter. Post the matlab code:

n = 2;

ftype = ‘bandpass’

fs = 100e6;

Wn = [1e6 3e6] / fs * 0.5

[z,p,k] = butter(n, Wn, ftype);

[sos, g] = zp2sos(z, p, k);

sos =

1.0000 2.0000 1.0000 1.0000 -1.9676 0.9693

1.0000 -2.0000 1.0000 1.0000 -1.9865 0.9868

g =

2.4136e-004

https://de.mathworks.com/help/signal/ref/butter.html

– see the bottom example.

2. Thank you for the link. I will have a look at. Does this also work for bandpass filters?

Dina

1. OK, but that’s a high pass filter and a lowpass filter. Taken together, you get bandpass. But that’s a completely different thing than a second order bandpass derived by bilinear transform. So, I don’t understand the question. Matlab is giving you a lowpass filter, and the same thing transposed to highpass (their b1 flipped in sign). If you run through both, you get a fourth order bandpass.

2. A similar point to the one above. Yes, it works, but will it give you the bandpass filter you want? There are limited absolutes in filters—you’ll only get exactly the same thing if you specify exactly the same constraints—this is why some 12 dB/oct synth filters don’t sound like other designs of the same thing. (Which one is “correct”?) A first order highpass or lowpass will have the expected rolloff and corner, in general, but even then it depends on where the zero is (compare the “one-pole lp” lowpass with the first order lowpass). The Cascading filters calculator produces the exact expected results for a Butterworth (which means maximally flat, in which case Q= 0.7071) second order lowpass derived by bilinear z transform. If you want a Butterworth filter of a higher order than 2, the tool will give you the info to build exactly that. But if you want an arbitrary Q, it won’t give you what you want. In the case of bandpass, Q is an important setting—whether the tool give you the bandpass you want is up to your needs.

Hi Nigel,

thank you very much for the helpful reply.

1. It was just a matter of uncertainty, whether I understood the matlab approach correctly.

2. No, a lowpass- highpass composition is certainly not what I want.

3. Related to the higher order cascaded filter: Does the filter with the first Q have to be the first? In other words: is the order relevant?

4. Is there a way to construct a bandpass filter with a frequency range instead of a single frequency?

Dina

3. Related to the higher order cascaded filter: Does the filter with the first Q have to be the first?

It’s not critical. Basically, if limited computing precision is available (rarely the case these days), then you want to have the stages with more gain (includes high Q) first, just so they aren’t boosting accumulated errors from previous stages.

4. Is there a way to construct a bandpass filter with a frequency range instead of a single frequency?

No. There is just a single complex conjugate pair of poles in a second order filter, a single frequency that gets pushed up (a single pole in a tent). In the Pole-Zero placement v2, uncheck “Pair” for the zeros. Slide one zero all the ways left (-1), the other all the way right. That’s a bandpass filter. Set “Pole mag” higher for greater Q, slide “Pole angle” to visit different frequencies. Those are the two degrees of control you have with bandpass.

Hi Nigel,

3. O.K. that makes sense.

4. I see. But if there would be another second order bandpass filter with the same zeros like the first and the pair of poles with a certain slightly different angle and magnitude, the “tent” seems to get a “wider roof”.

Isn’t there a formula to calculate the poles and zeros for a given -3dB “roof beginning” to a given -3DB “end of the roof”?

Thank you very much and sorry for insisting.

Dina

“and the pair of poles with a certain slightly different angle and magnitude”—That’s what sets the frequency and Q of the bandpass. You’ll get a better understanding of poles and zeros if you experiment with the Pole-Zero placement widget. Remember that for the poles to move off the horizontal axis, they have to be a conjugate pair. In other word, there may be two poles, but you can only set them to one frequency. It’s not like two poles in a tent, it’s like one, the other is a mirror image. The top and bottom half of the pole-zero plot are always mirror images in any realizable filter. It’s for the same reason that the upper-half any sampled signal mirrors the lower half.Hello Nigel,

thank you very much for the update.

Would it be possible to add Q to the shelf filters, such that their steepness can be controlled? I’ve seen this with another EQ editor of which I’m rather sure it uses biquads, and found use for it.

The “cookbook formulae” (http://shepazu.github.io/Audio-EQ-Cookbook/audio-eq-cookbook.html) have it as well. I fail to transfer that form into yours, looks so different…

Yes, that’s a bit of a can of worms, I found. Robert’s cookbook implementation is certainly one way, but in my mind it has a feature that’s less useful from a musical perspective—on shelf boost, it produces a peak above the shelf level. Many analog shelf filter implementations have a dip before the shelf. My guess is that it was not a targeted design feature, but simply because they made the shelf by summing a lowpass or highpass with variable gain with the direct signal; the inevitable phase difference will cause that dip. I did try to emulate that at one time by manipulating the pole and zero motion in the s-plane prototype, but ultimately there is no single “correct” way to do it, I gave up because I’m not designing a product and have no reason to go down the road of implementing a certain behavior. And trying to make a more general design that can have different behaviors was too much work. If you check both analog and digital shelf filters from many sources, their behaviors vary widely.

I did some more comparisons between the cookbook formulae and yours. The only filters where both implementations give identical results are lowpass and highpass. For the others, there seems to be a different interpretation of the Q value. The result can be matched when changing Q. Except the shelf filters, here Fc seems off to me. The cookbook shelfs yield 50% of the step amplitude at the center frequency, are point symmetric around it. Yours rather have a 3db point there, the symmetry is further away.

A more detailled breakdown of the 2nd order filters:

Peak:

The gain setting has a dependency to the relative “spread” of the hump. Example: 10 dB boost gives a remaining 58% of the boost level at 2*Fc, versus 1 dB boost has only 34% at 2*Fc. (With the cookbook formulae, this ratio stays approximately constant. However, with very high boost values the hump gets “pointy”, changes its shape.) By changing Q, the results can be matched. The correction is no constant factor, it depends on the boost level. It does not depend on the center frequency.

Low pass, high pass:

Identical results, cheers!

Band pass, all pass, notch:

Different Q interpretation. The results can be matched by changing Q, the “correction factor” is however not constant, but depends on Fc.

Low shelf, high shelf:

Here, it’s rather a different Fc interpretation! The results can be matched by changing Fc to achieve symmetry. (And selecting Q=1 on the cookbook side.) It’s approximately a factor of 0.75 for low shelf and 1/0.75 for high shelf.

I did not see the dip or peak before the shelf which you mentioned, with Q=1.

I can give you some code for comparison, if you like, get in touch.

On bandpass, Robert has two interpretations of bandwidth, IIRC, my implementation matches one of them. His other doesn’t have the peak constant at 0 dB, it’s more of an engineering choice, probably not one you’d implement for an EQ plug-in.

On shelves, we differ, as you noted. I prefer the frequency to be the edge of the step. In other words, if I raise a low shelf at 100 Hz, I want the response to be mostly flat at 100 Hz. But, it’s all a musical choice. iZotope puts theirs in the middle. For Waves, PEQ middle and Q does nothing; REQ edge and Q is very limited but gives the dip all the time. Motu MW EQ is edge, full Q but only puts the dip at the baseline (no edge peaking), unlike Robert’s. Motu’s older PEQ is edge, no Q available. None of it’s wrong, just choices. Though I think Robert’s Q peaking at the edge of a shelf boost is not the right thing, musically. If you’re already boosting a shelf, I don’t think it’s often that you want an additional peak above that at the edge. (And if you do, add a peaking filter.)

Notch: Are you sure? I thought I implemented s^2 / (s^2 + s/Q + 1), looks like that’s what Robert’s doing too. I’ll have to check the APF.

Lowpass and highpass definition is pretty universal, at least for BLT-derived, because the zeros are always at the extremes, the poles at the frequency. For notch, the frequency is obvious (where the zeros sits on the unit circle, complex conjugate), and you’re just sliding the poles toward the zeros. For something like shelf, you have to decide what frequency and Q even mean. For bandpass and peaking filters, you have to decide what bandwidth means (and Q is related to bandwidth). But that’s OK, because even analog implementations are different.

Hi Nigel,

Great work have been done here.

I was trying to use this calculations in Sigma Studio, however it seems that some the values are stocked, from what i found i need to rearrange results to this:

b2: a2,

b1: a1,

b0: a0,

a2: inverted b2

a1: inverted b1

Why is this? Same goes with miniDSP coefs

Stian

Thanks, Stian. See here about conventions.

Hello, and thanks for this site.

but: i use minidsp!

your biquads of the a site is b! and the b´s are the a´s, and there turns – to+ und + to-. so it´s right!

grüße aus münchen

David