Shelving Filter (Low and High Shelf)

The same spring–mass resonator as in the bell filter, tapped differently: the position or inertial force of the mass plus the damper force are mixed into the direct signal. This boosts or cuts all low or all high frequencies – just like bass and treble controls.

Explanation Tasks Equivalent circuit

Oscilloscope: input and output

Frequency response

Listen (B key toggles)

filtered original

Speed

Input signal

Waveform

Sine Triangle Sawtooth Square Impulse

Frequency f Amplitude

Spring–mass system

Filter type

Low shelf (bass) High shelf (treble)

Spring stiffness k Mass m Gain G Quality factor Q

Presets

Transition frequency f₀
Resonance of the resonator
Effective mass
Damping ζ = 1/(2·Q)
Gain at f
Slow motion

Keyboard: Space = play/pause, F = presentation mode

What is happening?

What a shelving filter does

A shelving filter evenly boosts or cuts all frequencies below (low shelf) or above (high shelf) a transition frequency f₀. Unlike a high cut or low cut, the level does not keep falling but levels off on a plateau – the “shelf”. The bass and treble controls on amplifiers and mixing desks are shelving filters; in a parametric equaliser they are usually the lowest and the highest band.

This module shows that both variants are built from the same component as the bell filter: a spring–mass resonator from which signals are tapped, weighted and added to the original. The low and high shelf differ in just one tap – which is why you switch between them here with “Filter type”.

One resonator, three forces

In the animation the “Input u(t)” (blue; t stands for time) arrives from the left and splits at the branch point. One path runs dashed as the “direct signal” to the summing point Σ. The other moves the blue plate of the resonator: a spring with stiffness k, a mass m and a damper with damping coefficient c to the fixed frame. x is the displacement of the mass, its position.

The spring pulls with the force k·(u − x), the damper brakes with c·x′ (x′ is the velocity of the mass), and what is left accelerates the mass: m·x″ (x″ is its acceleration). Dividing everything by k gives u = x + (c/k)·x′ + (m/k)·x″. So the input signal splits into three parts that always add up exactly to u: the position x follows the low frequencies (a low-pass), the damper force is only large near resonance (a band-pass), and the inertial force carries the high frequencies (a high-pass).

Two weightings

Two of these parts are added to the direct signal. Each runs through a box with its factor, and the orange bar next to it shows its contribution live. Here A = 10^(G/40), where G is the gain in dB set with the “Gain G” control; A² is the desired gain on the plateau as a factor.

In the low shelf the position x is weighted with A² − 1. At low frequencies the mass follows the input completely, x = u, and the output is u + (A² − 1)·u = A²·u – exactly the set gain. At high frequencies the mass hardly moves, leaving the direct signal at 0 dB. The high shelf works the other way round: it weights the inertial force, which is only large at high frequencies.

The damper force with the factor A − 1 shapes the transition; without it a cut would not be the exact mirror image of a boost. Example for G = +9 dB: A = 10^(9/40) ≈ 1.68, so A² − 1 ≈ 1.82 and A − 1 ≈ 0.68. At 0 dB both factors are zero and the filter is neutral.

Transition frequency and effective mass

Spring stiffness k and mass m set the transition frequency f₀ = (1/2π) · √(k/m). With the default values (40 kN/m, 4 g) that is about 503 Hz, shown as “Transition frequency f₀”. There exactly half the gain is reached, i.e. +4.5 dB at +9 dB.

The resonator itself, however, does not resonate at f₀. A shelf reaches half its gain between its two corner points, and one of them is the resonance. So that f₀ stays put while you turn the gain – as on an equaliser in the studio – the model uses an effective mass: m·A for the low shelf, m/A for the high shelf. At +9 dB that is 6.7 g or 2.4 g, and the resonance is then at about 388 Hz or 652 Hz. Both are shown with the readouts. In the animation the mass block visibly grows or shrinks as you change the gain.

Quality factor, bump and dip

The quality factor Q sets the damping ζ (zeta; how strongly the damper brakes relative to spring and mass): ζ = 1/(2·Q). With Q = 0.707 the transition is as steep as possible without overshooting the plateau. Smaller values make it flatter and wider.

Larger values create a bump before and a dip after the transition. With “+12 dB, Q = 2” the low shelf boosts by 16.6 dB at about 322 Hz, 4.6 dB more than set, and cuts by 4.6 dB at about 787 Hz, although nothing is supposed to happen there. Some equalisers use this on purpose: the dip makes room above the bass, the bump emphasises the range right at the transition.

Try it: the 100 Hz sawtooth contains all harmonics – 100, 200, 300 Hz and so on. In the frequency response they appear as pale blue bars (“Harmonics at input”), the orange bars (“Harmonics at output”) show the result. As the scale only spans ±24 dB, they are drawn relative to the strongest harmonic at the input. Switch between Q = 0.707 and Q = 2 and watch the 3rd and the 8th harmonic.

Low and high, boost and cut

Switch from low to high shelf with the same settings: the frequency response is mirrored at f₀. What the low shelf does one octave below f₀, the high shelf does one octave above it. With the default values and +9 dB the low shelf boosts 200 Hz by 8.7 dB; the high shelf reaches the same 8.7 dB only at 503²/200 Hz ≈ 1267 Hz.

Boost and cut are exact mirror images too: at 300 Hz +10.4 dB (G = +12 dB) becomes exactly −10.4 dB (G = −12 dB). Two such filters in series give a neutral signal again.

With the impulse you see the impulse response. The kick passes straight through via the direct signal. In the low shelf a broad hump follows, because the position of the mass follows the kick only sluggishly. In the high shelf the kick itself is amplified, because the inertial force reacts immediately. With Q = 2 the resonator visibly rings.

In the studio, on the amp, in mastering

The low shelf gives a kick drum or a whole mix more foundation, or takes out muddiness, without cutting the low end off entirely like a low cut. The high shelf is the classic control for “air” and sheen on vocals and cymbals, or it tames harsh highs. In mastering you usually work with small gains of 1 to 2 dB because every change affects the whole mix.

Combined, the two make a tilt EQ: a low shelf at −3 dB and a high shelf at +3 dB with the same transition frequency tilt the spectrum like a seesaw around f₀. Like every analogue filter, a shelf shifts the phase around the transition: with the default values (low shelf, +9 dB, Q = 0.707) it is about −41° at f₀, and no more at any other frequency.

What the model simplifies

Note on time scale: all values are real physical quantities – the frequencies lie in the audible range. As that would be far too fast for the eye, the animation automatically runs in slow motion; the factor is shown with the readouts.

The slow motion follows the signal frequency: the input signal always appears with one period every two seconds, multiplied by the “Speed” control. At 100 Hz the readout “Slow motion” therefore shows 1 : 200, at 20 Hz 1 : 40.

The model is ideally linear: the spring obeys Hooke's law exactly, the damper brakes exactly in proportion to velocity, and the weighting and summing point work without delay or noise. This makes it mathematically identical to the analogue prototype of the widely used shelving filters by Robert Bristow-Johnson (RBJ). Some equalisers set the slope S instead of the quality factor Q; S = 1 corresponds to Q = 0.707 here. The effective mass that moves with the gain is a calculation step of the model, not real mechanics. Digital equalisers compute the filter from samples; near half the sample rate the transition is squeezed there, which this continuous-time model does not show. The “Amplitude” control only changes the size of the signals, not the gain.

Tasks

Practice questions

Questions from studio, stage and mastering – to think about, discuss and check with the module.

1. Why do you often boost the low end of a mix with a low shelf rather than with a bell filter at 60 Hz?

Show answer

A low shelf boosts everything below the transition frequency evenly: the fundamentals of bass and kick drum as well as the range below them. The balance within the low end is preserved. A bell filter only emphasises a band around 60 Hz; if an instrument's fundamental lies beside it, it hardly gains. However, the shelf also boosts low-frequency rumble, which is why it is often combined with a low cut. If only one particular range is booming, the bell filter is the right tool.
2. A low shelf boosts by 6 dB with a transition frequency f₀ = 200 Hz (Q = 0.707). By how much is 100 Hz boosted, and by how much 400 Hz?

Show answer

At f₀ it is exactly half, i.e. +3 dB. One octave below, at 100 Hz, it is about +5.6 dB; one octave above, at 400 Hz, only about +0.4 dB. The two add up to the full gain: 5.6 + 0.4 = 6 dB. This holds for every pair of frequencies equally far from f₀ on the logarithmic axis. At 50 Hz the plateau of 6.0 dB is practically reached.
3. What do you get from a low shelf at −3 dB combined with a high shelf at +3 dB, both with f₀ = 1 kHz?

Show answer

A tilt EQ: the spectrum tilts like a seesaw around 1 kHz. There −1.5 dB and +1.5 dB cancel out; at 50 and 200 Hz −3 dB remain, at 5 and 15 kHz +3 dB. The sound becomes brighter without the overall loudness changing much. With the signs reversed it becomes warmer.

Tasks

Predict first, then try it out, then open the solution. “Load starting point” sets the matching initial values.

1. Which harmonics?

The sawtooth has 100 Hz, the low shelf boosts by 9 dB, the transition frequency f₀ is about 503 Hz. Which harmonics gain more than 6 dB? Which one gains about half the gain? Look at the bars in the frequency response.

Load starting point

Show solution

The 1st to 4th harmonics (100 to 400 Hz) gain more than 6 dB: about 9.0, 8.7, 7.9 and 6.3 dB. The 5th harmonic (500 Hz) lies almost exactly at f₀ and gains about 4.5 dB, i.e. half. Above that the boost falls off quickly; at 1 kHz only about 0.6 dB is left. In the oscilloscope the sawtooth becomes more rounded because the low harmonics gain relative to the high ones.
2. Mirror image

A 300 Hz sine runs through the low shelf at +12 dB. Read the “Gain at f”. Then set −12 dB. What do you notice? How do the effective mass and the resonance of the resonator change?

Load starting point

Show solution

+10.4 dB becomes exactly −10.4 dB: boost and cut are mirror images, two such filters in series cancel out. The effective mass drops from about 8.0 g to about 2.0 g, i.e. by a factor of 10^(24/40) ≈ 4. The resonance rises from about 356 Hz to about 711 Hz, while the transition frequency f₀ stays at 503 Hz.
3. Bump and dip

The low shelf boosts by 12 dB with Q = 0.707. Watch the 3rd harmonic (300 Hz) and the 8th harmonic (800 Hz). Then set Q = 2. What happens to the two?

Load starting point

Show solution

With Q = 0.707 the 3rd harmonic gains about 10.4 dB, the 8th about 1.9 dB. With Q = 2 the 3rd harmonic is boosted by about 16.4 dB – more than the set gain – while the 8th is cut by about 4.6 dB. The lightly damped resonator creates a bump before the transition and a dip after it. The two plateaus stay at +12 dB and 0 dB.
4. From low to high

A 200 Hz sine runs through the low shelf at +9 dB. Note the gain. Then switch to the high shelf. How much is left? At which frequency does the high shelf show the same value as the low shelf did before?

Load starting point

Show solution

The low shelf boosts 200 Hz by about 8.7 dB, the high shelf only by about 0.3 dB. The high shelf reaches the same 8.7 dB at f₀²/200 Hz ≈ 1267 Hz: the frequency response is mirrored at f₀, and on the logarithmic axis 200 Hz and 1267 Hz are equally far from 503 Hz (a factor of 2.5 each). In the animation the inertial force is now weighted instead of the position, and the effective mass drops from 6.7 g to 2.4 g.
5. Impulse response

An impulse runs through the low shelf at +12 dB. What does the output look like in the oscilloscope? Then switch to the high shelf. After that, go back to the low shelf and set Q = 2.

Load starting point

Show solution

With the low shelf the impulse passes straight through via the direct signal, followed by a broad, flat hump: the weighted contribution of the position x, which follows the kick only sluggishly. With the high shelf the impulse itself is amplified because the inertial force reacts immediately; it is followed by a flat tail in the opposite direction. With Q = 2 the resonator in the low shelf rings for several periods, close to its resonance of about 356 Hz. The ringing belongs to the bump in the frequency response.

Electrical equivalent circuit

The resonator of the shelving filter also contains the series resonant circuit of inductor L, resistor R and capacitor C. The input voltage Uₑ divides among the three components, and the partial voltages always add up exactly to Uₑ: U_L + U_R + U_C = Uₑ. For the low shelf U_C is weighted with (A² − 1), for the high shelf U_L. In both cases U_R is added with (A − 1).

U_C corresponds to the spring force k·x, i.e. the position, U_R to the damper force c·v and U_L to the inertial force m·a. In electronics, weighting and sum are op-amp circuits. The values follow the controls; L is the effective mass m·A or m/A.

Mechanics	Electrics
Mass m	Inductor L = m
Spring stiffness k	Capacitor C = 1/k
Damping c	Resistor R = c
Force F	Voltage U
Velocity v	Current I
Displacement x	Charge q
f₀ = 1/(2π√(m/k))	f₀ = 1/(2π√(LC))
ζ = c/(2√(km))	ζ = (R/2)·√(C/L)

Three partial voltages, three filters

At low frequencies the reactance of the inductor (ωL) is small and that of the capacitor (1/(ωC)) large: almost the whole voltage is across the capacitor, so U_C is a low-pass. At high frequencies it is the other way round, U_L is a high-pass. Near resonance the two reactances cancel and the voltage is across the resistor: U_R is a band-pass – the same one the bell filter uses.

Because the three partial voltages always add up to Uₑ, all basic filters can be mixed from them. The state-variable filter, a common circuit in synthesisers that provides low-pass, band-pass and high-pass at the same time, uses the same principle.

Components that move with the gain

To keep the transition frequency f₀ fixed, the model shifts the resonance of the circuit: L becomes m·A in the low shelf and m/A in the high shelf, while C = 1/k stays. The damping ζ = 1/(2·Q), on the other hand, depends only on Q; hence R = √(L/C)/Q.

With the default values (k = 40 kN/m, m = 4 g, low shelf, G = +9 dB, Q = 0.707) C = 25 µF, L ≈ 6.7 mH and R ≈ 23 Ω. The circuit resonates at about 388 Hz, while the transition frequency is 503 Hz.

Try it

Change the gain G: C stays put, L and R move with it, as do the two factors. Switch between low and high shelf: the weighting (A² − 1) moves from the capacitor to the inductor, and L jumps from m·A to m/A. At 0 dB both factors are zero and Uₐ = Uₑ.

Why this works

Mechanics and electrics are described here by the same second-order differential equation. For the mass, m·x″ + c·x′ + k·x = F; for the series resonant circuit, L·q″ + R·q′ + q/C = U. Comparing term by term, m corresponds to L, c to R and k to 1/C – as well as displacement x to charge q, velocity v to current I and force F to voltage U.

This mapping is called the force–voltage analogy or impedance analogy. In this module it is converted with a factor of 1: 1 kg corresponds to 1 H, 1 N·s/m to 1 Ω and 1 m/N of compliance to 1 F. The starting values give very real components: 1 mH, 25 µF and just under 9 Ω.

Mass ↔ inductor

A mass resists any change in its velocity (F = m·a), an inductor resists any change in its current (U = L·dI/dt). Both store kinetic energy: ½·m·v² in the mass, ½·L·I² in the magnetic field of the inductor.

Their opposition grows with frequency (ω·m and ω·L). That is why a heavy mass can no longer keep up at high frequencies – and why an inductor blocks high frequencies.

Spring ↔ capacitor

A tensioned spring stores potential energy ½·k·x², a charged capacitor stores ½·q²/C in its electric field. The capacitance corresponds to the compliance 1/k of the spring: a soft spring is a large capacitor, a stiff spring a small one.

Their opposition is large at low frequencies (k/ω and 1/(ω·C)) and falls as the frequency rises – the exact opposite of mass and inductor. Where both are equal lies the resonant frequency f₀.

Damper ↔ resistor

The damper produces a force proportional to velocity (F = c·v), the resistor a voltage proportional to current (U = R·I). They are the only components that do not store energy but turn it into heat. They therefore determine how quickly an oscillation dies away and how high the resonance peak becomes.

The formulas carry over directly: f₀ = 1/(2π·√(L·C)) and ζ = (R/2)·√(C/L); the quality factor is Q = 1/(2ζ). With 1 mH and 25 µF, f₀ = 1/(2π·√(2.5·10⁻⁸ s²)) ≈ 1007 Hz. For Butterworth damping (ζ ≈ 0.707) you need R = 2ζ·√(L/C) ≈ 1.414 · 6.32 Ω ≈ 8.94 Ω.

The second analogy

You can also map the other way round: force ↔ current and velocity ↔ voltage (force–current or mobility analogy). Then the mass becomes a capacitor, the spring an inductor with L = 1/k and the damper a resistor with R = 1/c. The advantage: the circuit looks like the mechanics. Parts that move at the same velocity are in parallel electrically.

You know this form from the equivalent circuit of loudspeakers (Thiele–Small). There, the moving mass of the diaphragm appears on the electrical side as a capacitor, the compliance of the suspension as an inductor and the mechanical losses as a resistor – together a parallel resonant circuit. That is why the impedance curve of a loudspeaker has a peak at its resonant frequency.

Limits of the analogy

The schematic contains ideal components: real inductors have wire resistance, real capacitors have losses, real springs have a mass of their own. The source has no internal resistance, and there is no load at the output. In a real circuit, the internal resistance of the source adds to R because it lies in the same loop, and thus changes the damping. A load at the output, on the other hand, cannot simply be added to R: it is in parallel with the component across which the output is taken – the capacitor in the low-pass, the inductor in the high-pass. It then changes not only the damping but also the gain in the passband; in the low-pass, low frequencies no longer come through at full level. The circuit is then no longer a pure series RLC loop.

That the values are so handy with a factor of 1 is due to the small masses and stiff springs of this module. A mass of 1 kg would be an inductor of 1 H – a large, heavy component. That is why active filters for low frequencies are usually built without inductors, for example as Sallen–Key filters made of an op-amp, resistors and capacitors. The transfer function stays the same.

Settings

Sound

Level calibration (dB SPL)

MIDI keyboard

Shelving Filter (Low and High Shelf)