Computed

100 Hz peaking eq

Boosts or cuts a band around f0 and leaves both ends of the spectrum at unity. Solved at 8 sample rates, with the poles, the real −3 dB point and the word length it stops working at.

Open the interactive designer →All peaking eq frequencies

Magnitude response at 48 kHz

Every vertex is 20·log₁₀|H(e^jω)| evaluated on the unit circle — not a sketch of the filter's shape. The faint traces behind it are the same corner at every Q in the sweep below.

Gain at f0 (100 Hz)

+6.00 dB

exact at every sample rate — the bilinear transform maps this value, not just the frequency

Pole radius at 48 kHz

0.9953773

conjugate pair at ±0.70°, 0.00462 from the circle

−3 dB point

61.7 Hz

0.617× f0 at Q = 1.0000

16-bit fixed point

holds

largest pole 0.9953812 in Q1.14

Coefficients, at every sample rate

The cookbook computes w₀ = 2πf₀/F_s, so the same filter is a different set of numbers at every rate. Rates whose Nyquist limit is at or below 100 Hz are absent because the filter does not exist there — equivalently, the f₀/Nyquist column never reaches 100%. That is why this table is 8 rows and a 20 Hz page is 8.

Sample rate	b0	b1	b2	a1	a2	Pole r	f₀/Nyquist
8 kHz	1.026894	-1.939957	0.919062	-1.939957	0.945956	0.972603	2.5%
16 kHz	1.013641	-1.971066	0.958946	-1.971066	0.972587	0.986198	1.3%
22.1 kHz	1.009937	-1.979227	0.970094	-1.979227	0.980031	0.989965	0.9%
32 kHz	1.006869	-1.985813	0.979327	-1.985813	0.986196	0.993074	0.6%
44.1 kHz	1.004994	-1.989762	0.984970	-1.989762	0.989964	0.994970	0.5%
48 kHz	1.004590	-1.990605	0.986186	-1.990605	0.990776	0.995377	0.4%
96 kHz	1.002300	-1.995335	0.993077	-1.995335	0.995377	0.997686	0.2%
192 kHz	1.001152	-1.997675	0.996534	-1.997675	0.997686	0.998842	0.1%

const float b0 = 1.00459014f, b1 = -1.99060546f, b2 = 0.98618587f;
const float a1 = -1.99060546f, a2 = 0.99077601f;   // a0 == 1

What word length this filter survives

The fixed-point rows round all five coefficients to one shared scale, which is what a q15/q31 biquad section does with its post-shift.Pole radius is solved from the quadratic, not taken as √|a₂| — once rounding pushes the poles onto the real axis those two disagree, and the convenient one reports a comfortable margin on a filter that has already left the unit circle. Error is the worst deviation from float64 across frequencies where the response is within 40 dB of its own peak; below that it is measuring the −200 dB floor.

Word format	Q format	Largest pole	√\|a₂\| says	Stable	Worst error in band
float64	—	0.9953773	0.9953773	yes	reference
float32	—	0.9953773	0.9953773	yes	0.0008 dB
32-bit fixed	Q1.30	0.9953773	0.9953773	yes	0.0000 dB
24-bit fixed	Q1.22	0.9953773	0.9953773	yes	0.0121 dB
16-bit fixed	Q1.14	0.9953812	0.9953812	yes	0.3819 dB

What Q does at 100 Hz

f₀ does not move with Q — the −3 dB point does. They are the same frequency only at Q = 1/√2, which is why a Butterworth corner is the one people quote and why every other Q surprises somebody.

Q	At f₀	Peak	−3 dB	Pole r	Group delay at f₀
0.5	+6.00 dB	+6.00 dB @ 101 Hz	242.1 Hz	0.990776	1.121 ms
0.7071	+6.00 dB	+6.00 dB @ 101 Hz	193.6 Hz	0.993469	1.586 ms
1	+6.00 dB	+6.00 dB @ 101 Hz	162.1 Hz	0.995377	2.243 ms
2	+6.00 dB	+5.99 dB @ 101 Hz	128.3 Hz	0.997686	4.484 ms
4	+6.00 dB	+5.96 dB @ 101 Hz	113.5 Hz	0.998842	8.955 ms
10	+6.00 dB	+5.77 dB @ 101 Hz	105.6 Hz	0.999537	22.171 ms

What gain does at 100 Hz

A peaking filter puts its full gain at f₀ and returns to unity at both ends of the spectrum.

Gain	At f₀	b0	a1	a2	Pole r
-12 dB	−12.00 dB	0.990348	-1.974050	0.974219	0.987026
-6 dB	−6.00 dB	0.995431	-1.981510	0.981680	0.990798
-3 dB	−3.00 dB	0.997746	-1.984393	0.984563	0.992252
+3 dB	+3.00 dB	1.002259	-1.988876	0.989047	0.994508
+6 dB	+6.00 dB	1.004590	-1.990605	0.990776	0.995377
+12 dB	+12.00 dB	1.009746	-1.993290	0.993461	0.996725

Questions this filter answers

What are the biquad coefficients for a 100 Hz peaking EQ filter at 48 kHz?

b0 = 1.004590, b1 = -1.990605, b2 = 0.986186, a1 = -1.990605, a2 = 0.990776, with a0 normalised to 1 — the RBJ Audio EQ Cookbook form at Q = 1.0000 and +6 dB of gain. Every other sample rate in the table above gives different numbers for the same filter, because w0 = 2πf0/Fs and every cosine and sine downstream of it moves.

Is a 100 Hz peaking EQ filter stable in 16-bit fixed point?

Yes. Rounding the five coefficients to a shared 1.14 scale leaves the largest pole at 0.9953812, against 0.9953773 exact, and the response drifts by at most 0.382 dB inside the band. 24-bit takes that to 0.0121 dB.

Where is the real −3 dB point of a 100 Hz peaking EQ filter?

61.7 Hz, which is 0.617× the 100 Hz corner. f0 and the −3 dB point are the same frequency only at Q = 1/√2; this page is designed at Q = 1.0000, and the Q sweep above shows the point moving from 242.1 Hz to 105.6 Hz across the sweep while f0 never moves.

How close to the unit circle are the poles of a 100 Hz peaking EQ filter?

0.9953773 at 48 kHz, as a conjugate pair at ±0.70°. Pole radius rises toward 1 as the corner frequency falls relative to the sample rate — the same filter at 192 kHz sits at 0.9988423 and at 8 kHz at 0.9726028. That distance is the whole story of the fixed-point table: a pole a few parts in 10⁵ from the circle has nowhere to be rounded to.

The neighbouring corners

One third-octave either side, and the same 100 Hz corner as every other filter type.

Method and limits. Coefficients follow the RBJ Audio EQ Cookbook, the bilinear transform of the analog prototype prewarped so the corner lands exactly on f0 — which is why +6.00 dB at 100 Hz holds at every sample rate in the table rather than only at low f0/Fs. Shelves fix the slope at S = 1, matching Web Audio's BiquadFilterNode, so Q is not read for those two types. Pole and zero radii are the roots of the quadratic, not sqrt(|a2|). The fixed-point rows model a single shared coefficient scale and no other quantisation: they say nothing about signal-path headroom, limit cycles or the accumulator width your implementation uses, all of which can make a filter that passes this table still misbehave. Nothing on this page is fetched or estimated — it is solved from the type and the frequency in the URL.