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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Cookbook formulae for audio EQ biquad filter coefficients</title>
<script>
MathJax = {
tex: {
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<script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js">
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<style media="screen">
body {
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margin-left: auto;
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dl {
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</head>
<body>
<h1>Cookbook formulae for audio equalizer biquad filter coefficients</h1>
<dl>
<dt>Author</dt>
<dd>
<a href="mailto:rbj@audioimagination.com">Robert Bristow-Johnson</a>
</dd>
<dt>Editor</dt>
<dd>
<a href="mailto:schepers@w3.org">Doug Schepers (W3C)</a>
</dd>
</dl>
<p>Adapted from <a href=
"https://webaudio.github.io/Audio-EQ-Cookbook/Audio-EQ-Cookbook.txt">Audio-EQ-Cookbook.txt</a>,
by Robert Bristow, with permission.</p>
<section>
<h2>Biquad Filter Formulae</h2>
<p>All filter transfer functions were derived from analog prototypes (that
are shown below for each equalizer (EQ) filter type) and had been digitized
using the Bilinear Transform (BLT). BLT frequency warping has been taken
into account for both significant frequency relocation (this is the normal
"prewarping" that is necessary when using the BLT) and for bandwidth
readjustment (since the bandwidth is compressed when mapped from analog to
digital using the BLT).</p>
<p>First, given a biquad transfer function defined as:</p>
<!-- displayed-eqn is just a random tag to keep html tidy from indenting the
contents which could mangle the equations. It's not used for anything else, and
it's not a custom element.
-->
<displayed-eqn> $$
\begin{equation}
H(z) = \frac{b_0 + b_1z^{-1}+b_2z^{-2}}{a_0 + a_1z^{-1}+a_2z^{-2}}
\end{equation}
$$
</displayed-eqn>
<p>This shows 6 coefficients instead of 5 so, depending on your
architecture, you will likely normalize \( a_0 \) to be 1 and perhaps also
\(b_0\) to 1 (and collect that into an overall gain coefficient). Then your
transfer function would look like:</p>
<displayed-eqn> $$
\begin{equation}
H(z) = \frac{\left(\displaystyle\frac{b_0}{a_0}\right) + \left(\displaystyle\frac{b_1}{a_0}\right)z^{-1}+\left(\displaystyle\frac{b_2}{a_0}\right)z^{-2}}{1 + \left(\displaystyle\frac{a_1}{a_0}\right)z^{-1}+\left(\displaystyle\frac{a_2}{a_0}\right)z^{-2}}
\label{direct-form-1}
\end{equation}
$$
</displayed-eqn>
<p>or:</p>
<displayed-eqn> $$
\begin{equation}
H(z) = \left(\frac{b_0}{a_0}\right) \frac{1 + \left(\displaystyle\frac{b_1}{b_0}\right)z^{-1}+\left(\displaystyle\frac{b_2}{b_0}\right)z^{-2}}{1 + \left(\displaystyle\frac{a_1}{a_0}\right)z^{-1}+\left(\displaystyle\frac{a_2}{a_0}\right)z^{-2}}
\end{equation}
$$
</displayed-eqn>
<p>The most straight forward implementation would be the "Direct Form 1"
(\(eq. \ref{direct-form-1}\)):</p>
<displayed-eqn> $$
\begin{align}
y[n] = \left(\frac{b_0}{a_0}\right)x[n] & +
\left(\frac{b_1}{a_0}\right)x[n-1] + \left(\frac{b_2}{a_0}\right)x[n-2]
\nonumber \\
&-\left(\frac{a_1}{a_0}\right)y[n-1] - \left(\frac{a_2}{a_0}\right)y[n-2]
\end{align}
$$
</displayed-eqn>
<p>This is probably both the best and the easiest method to implement in
the 56K and other fixed-point or floating-point architectures with a double
wide accumulator.</p>
<p>Begin with these user defined parameters:</p>
<dl>
<dt>\(F_s\)</dt>
<dd>the sampling frequency</dd>
<dt>\(f_0\)</dt>
<dd>Center Frequency or Corner Frequency, or shelf midpoint frequency,
depending on which filter type. The "significant frequency". "wherever
it's happenin', man."</dd>
<dt>\(\mathrm{dBgain}\)</dt>
<dd>used only for peaking and shelving filters</dd>
<dt>\(Q\) or \(\mathrm{BW}\) or \(S\)</dt>
<dd>
<dl>
<dt>\(Q\)</dt>
<dd>the EE kind of definition, except for peakingEQ in which A*Q is
the classic EE Q. That adjustment in definition was made so that a
boost of N dB followed by a cut of N dB for identical Q and f0/Fs
results in a precisely flat unity gain filter or "wire".</dd>
<dt>\(\mathrm{BW}\)</dt>
<dd>the bandwidth in octaves (between -3 dB frequencies for BPF and
notch or between midpoint (dBgain/2) gain frequencies for peaking
EQ)</dd>
<dt>\(S\)</dt>
<dd>a "shelf slope" parameter (for shelving EQ only). When S = 1, the
shelf slope is as steep as it can be and remain monotonically
increasing or decreasing gain with frequency. The shelf slope, in
dB/octave, remains proportional to S for all other values for a fixed
f0/Fs and dBgain</dd>
</dl>
</dd>
</dl>
<p>Then compute a few intermediate variables:</p>
<ol>
<li>
<displayed-eqn> $$
\begin{align*}
A &= \sqrt{10^{\mathrm{dBgain}/20}} & \\
& = 10^{\mathrm{dBgain}/40} & \textrm{(for peaking and shelving EQ filters only)}
\end{align*}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
\omega_0 = 2 \pi \frac{f_0}{F_s}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
\begin{align*}
& \cos \omega_0 \\
& \sin \omega_0
\end{align*}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
\begin{align*}
\alpha &= \frac{\sin \omega_0}{2 Q} & \textrm{(case: Q)} \\
&= \sin \omega_0 \, \sinh\left(\frac{\log 2}{2} \cdot
\mathrm{BW} \cdot
\frac{\omega_0}{\sin \omega_0}\right) & \textrm{(case: BW)} \\
&= \frac{\sin \omega_0}{2} \sqrt{\left(A +
\frac{1}{A}\right) \left(\frac{1}{S} - 1\right) + 2} & \textrm{(case: S)}
\end{align*}
$$
</displayed-eqn>
</li>
</ol>
<p>FYI: The relationship between bandwidth and Q is</p>
<p class="label">digital filter with BLT</p>
<displayed-eqn> $$
\frac{1}{Q} = 2\sinh\left(\frac{\log 2}{2} \cdot \mathrm{BW} \cdot
\frac{\omega_0}{\sin \omega_0}\right)
$$
</displayed-eqn>
<p>or</p>
<p class="label">analog filter prototype</p>
<displayed-eqn> $$
\frac{1}{Q} = 2 \sinh\left(\frac{\log 2}{2} \cdot \mathrm{BW}\right)
$$
</displayed-eqn>
<p>The relationship between shelf slope and Q is</p>
<displayed-eqn> $$
\frac{1}{Q} = \sqrt{\left(A + \frac{1}{A}\right) \left(\frac{1}{S} -
1\right) + 2}
$$
</displayed-eqn>
<displayed-eqn> $$
2\sqrt{A}\,\alpha = (\sin \omega_0)\, \sqrt{\left(A^2 +
1\right)\left(\frac{1}{S}-1\right) + 2A}
$$
</displayed-eqn>
<p>is a handy intermediate variable for shelving EQ filters.</p>
<p>Finally, compute the coefficients for whichever filter type you
want:</p>
<p>(The analog prototypes, H(s), are shown for each filter type for
normalized frequency.)</p>
<dl>
<dt>LPF</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{1}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= \frac{1-\cos\omega_0}{2} \\
b_1 &= 1-\cos\omega_0 \\
b_2 &= \frac{1-\cos\omega_0}{2} \\
a_0 &= 1 + \alpha \\
a_1 &= -2\cos\omega_0 \\
a_2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>HPF</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{s^2}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= \frac{1 + \cos \omega_0}{2} \\
b_1 &= -(1 + \cos \omega_0) \\
b_2 &= \frac{1 + \cos \omega_0}{2} \\
a_0 &= 1 + \alpha \\
a_1 &= -2\cos \omega_0 \\
a_2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>BPF<br>
(constant skirt gain,<br>
peak gain = Q)</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{s}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b0 &= \frac{\sin\omega_0}{2} = Q \alpha \\
b1 &= 0 \\
b2 &= -\frac{\sin\omega_0}{2} = -Q \alpha \\
a0 &= 1 + \alpha \\
a1 &= -2\cos\omega_0 \\
a2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>BPF<br>
(constant 0 dB peak gain)</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{\displaystyle\frac{s}{Q}}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= \alpha \\
b_1 &= 0 \\
b_2 &= -\alpha \\
a_0 &= 1 + \alpha \\
a_1 &= -2\cos \omega_0 \\
a_2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>notch</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{s^2 + 1}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= 1 \\
b_1 &= -2\cos \omega_0 \\
b_2 &= 1 \\
a_0 &= 1 + \alpha \\
a_1 &= -2\cos \omega_0 \\
a_2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>APF</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{s^2 - \displaystyle\frac{s}{Q} + 1}{s^2 + \displaystyle\frac{s}{Q} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= 1 - \alpha \\
b_1 &= -2\cos\omega_0 \\
b_2 &= 1 + \alpha \\
a_0 &= 1 + \alpha \\
a_1 &= -2\cos\omega_0 \\
a_2 &= 1 - \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>peakingEQ</dt>
<dd>
<displayed-eqn> $$
H(s) = \frac{s^2 + s\displaystyle\frac{A}{Q} + 1}{s^2 + \displaystyle\frac{s}{AQ} + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= 1 + \alpha A \\
b_1 &= -2\cos\omega_0 \\
b_2 &= 1 - \alpha A \\
a_0 &= 1 + \frac{\alpha}{A} \\
a_1 &= -2\cos\omega_0 \\
a_2 &= 1 - \frac{\alpha}{A}
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>lowShelf</dt>
<dd>
<displayed-eqn> $$
H(s) = A \frac{s^2 + \displaystyle\frac{\sqrt{A}}{Q} s + A}
{A*s^2 + \displaystyle\frac{\sqrt{A}}{Q} s + 1}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= A\left( (A+1) - (A-1)\cos\omega_0 + 2\sqrt{A}\, \alpha \right) \\
b_1 &= 2A\left( (A-1) - (A+1)\cos\omega_0 \right) \\
b_2 &= A\left( (A+1) - (A-1)\cos\omega_0 - 2\sqrt{A}\, \alpha \right) \\
a_0 &= (A+1) + (A-1)\cos\omega_0 + 2\sqrt{A}\, \alpha \\
a_1 &= -2\left( (A-1) + (A+1)\cos\omega_0 \right) \\
a_2 &= (A+1) + (A-1)\cos\omega_0 - 2\sqrt{A}\, \alpha
\end{align*}
$$
</displayed-eqn>
</dd>
<dt>highShelf</dt>
<dd>
<displayed-eqn> $$
H(s) = A \frac{As^2 + \displaystyle\frac{\sqrt{A}}{Q}s + 1}
{s^2 + \displaystyle\frac{\sqrt{A}}{Q} s + A}
$$
</displayed-eqn>
<displayed-eqn> $$
\begin{align*}
b_0 &= A\left( (A+1) + (A-1)\cos\omega_0 + 2\sqrt{A}\alpha \right) \\
b_1 &= -2A\left( (A-1) + (A+1)\cos\omega_0 \right) \\
b_2 &= A\left( (A+1) + (A-1)\cos\omega_0 - 2\sqrt{A}\alpha \right) \\
a_0 &= (A+1) - (A-1)\cos\omega_0 + 2\sqrt{A}\alpha \\
a_1 &= 2\left( (A-1) - (A+1)\cos\omega_0 \right) \\
a_2 &= (A+1) - (A-1)\cos\omega_0 - 2\sqrt{A}\alpha
\end{align*}
$$
</displayed-eqn>
</dd>
</dl>
<p>FYI: The bilinear transform (with compensation for frequency warping)
substitutes:</p>
<p class="label">(normalized)</p>
<displayed-eqn> $$
s \leftarrow \frac{1}{\tan\displaystyle\frac{\omega_0}{2}} \frac{1-z^{-1}}{1+z^{-1}}
$$
</displayed-eqn>
<p>and makes use of these trig identities:</p>
<ol>
<li>
<displayed-eqn> $$
\tan\frac{\omega_0}{2} = \frac{\sin\omega_0}{1+\cos\omega_0}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
\left(\tan\frac{\omega_0}{2}\right)^2 = \frac{1-\cos\omega_0}{1+\cos\omega_0}
$$
</displayed-eqn>
</li>
</ol>
<p>resulting in these substitutions:</p>
<ol>
<li>
<displayed-eqn> $$
1 \leftarrow \frac{1+\cos\omega_0}{1+\cos\omega_0}\frac{1+2z^{-1}+z^{-2}}{1+2z^{-1}+z^{-2}}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
\begin{align*}
s \leftarrow & \frac{1+\cos\omega_0}{\sin\omega_0}
\frac{1-z^{-1}}{1+z^{-1}} \\
& = \frac{1+\cos\omega_0}{\sin\omega_0} \frac{1-z^{-2}}{1+2z^{-1}+z^{-2}}
\end{align*}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
s^2 \leftarrow \frac{1+\cos\omega_0}{1-\cos\omega_0} \frac{1-2z^{-1}+z^{-2}}{1+2z^{-1}+z^{-2}}
$$
</displayed-eqn>
</li>
</ol>
<p>The factor:</p>
<displayed-eqn> $$
\frac{1+\cos\omega_0}{1+2z^{-1}+z^{-2}}
$$
</displayed-eqn>
<p>is common to all terms in both numerator and denominator, can be
factored out, and thus be left out in the substitutions above resulting
in:</p>
<ol>
<li>
<displayed-eqn> $$
1 \leftarrow \frac{1+2z^{-1}+z^{-2}}{1+\cos\omega_0}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
s \leftarrow \frac{1-z^{-2}}{\sin\omega_0}
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
s^2 \leftarrow \frac{1-2z^{-1}+z^{-2}}{1-\cos\omega_0}
$$
</displayed-eqn>
</li>
</ol>
<p>In addition, all terms, numerator and denominator, can be multiplied by
a common</p>
<displayed-eqn> $$
\sin^2\omega_0
$$
</displayed-eqn>factor, finally resulting in these substitutions:
<ol>
<li>
<displayed-eqn> $$
1 \leftarrow (1 + 2z^{-1} + z^{-2}) (1 - \cos\omega_0)
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
s \leftarrow (1-z^{-2})\sin\omega_0
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
s^2 \leftarrow (1 - 2z^{-1} + z^{-2}) (1 + \cos\omega_0)
$$
</displayed-eqn>
</li>
<li>
<displayed-eqn> $$
1 + s^2 \leftarrow 2\, (1 - 2\cos\omega_0\,z^{-1} + z^{-2})
$$
</displayed-eqn>
</li>
</ol>
<p>The biquad coefficient formulae above come out after a little
simplification.</p>
</section>
<section id="acknowledgements">
<h2>Acknowledgements</h2>
<p>Special thanks to <a href="mailto:rbj@audioimagination.com">Robert
Bristow-Johnson</a> for creating the Audio EQ Cookbook and permitting its
adaption and use for the Web Audio API.</p>
<p>Thanks to <a href="mailto:peter.krautzberger@mathjax.org">Peter
Krautzberger</a> for help in adapting these mathematical formulae to
MathML, and to the whole <a href="https://www.mathjax.org/">MathJax</a>
team for making the JavaScript extension that makes the use of math on the
web possible.</p>
<p>Thanks to Peter Jipsen, creator of <a href=
"http://www1.chapman.edu/~jipsen/mathml/asciimath.html">ASCIIMathML</a>,
for making it easier to convert ASCII math notation to MathML.</p>
<p>Converted to using TeX formulas instead of MathML by <a href=
"mailto:rtoy@chromium.org">Raymond Toy</a>.</p>
</section>
</body>
</html>