ISEGORIA / MATH ENCYCLOPEDIA
Filters and crossovers: splitting sound between drivers
A speaker with two drivers has to send the bass to one and the treble to the other, and the two halves have to add back up to the original in the air in front of it. Resistors, capacitors and inductors make filters whose behaviour is read off a Bode plot; crossover filters are designed so that their outputs sum, as complex numbers, to something flat; and the physical distance between the drivers turns that sum into a function of where the listener sits, which is the problem a coaxial driver solves.
Before you begin: AC circuits, complex numbers and Fourier analysis
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. RC and RLC filters and the Bode plot
A capacitor’s impedance 1/jωC falls with frequency and an inductor’s jωL rises, so a voltage divider built from them lets some frequencies through and blocks others. On the right, the Bode plot: gain in decibels and phase against frequency on a log scale, where each pole bends the gain down by 20 dB per decade and turns the phase by 90°. On the left, what that means for an actual sine wave: drag the test frequency and watch the output shrink and lag behind the input. The series RLC low-pass has a resonance at f₀ = 1/2π√(LC) with quality factor Q = √(L/C)/R: at high Q the gain peaks at f₀ before it falls, and the phase swings through −90° exactly there.
Worked example. With \(R=10\ \Omega\) and \(C=10\ \mu\mathrm F\), the RC low-pass has \(f_c=1/(2\pi\times10^{-4})=1592\ \mathrm{Hz}\), where the gain is \(1/\sqrt2\), \(-3\ \mathrm{dB}\), and the phase \(-45^\circ\). Ten times higher, at 15.9 kHz, the gain is \(-20\ \mathrm{dB}\). Adding \(L=1\ \mathrm{mH}\) in series gives \(f_0=1592\ \mathrm{Hz}\) and \(Q=\sqrt{10^{-3}/10^{-5}}/10=1\): at \(f_0\) the gain is exactly \(Q\), 0 dB, and a decade above it has fallen by 40 dB.
Watch out. These are ideal components driving no load. A passive crossover in a speaker drives a driver, whose impedance is anything but a resistor, as the loudspeaker topic shows, so real passive networks include impedance-compensation branches or are replaced by active filters before the amplifier. The time trace shows the steady state; switching a filter on also produces a transient, not drawn.
Why does each pole cost 20 dB per decade?
Far above a pole at \(\omega_p\), its factor \(1/(1+j\omega/\omega_p)\) behaves like \(\omega_p/(j\omega)\): the magnitude is inversely proportional to frequency. A factor of 10 in frequency is a factor of 10 in amplitude, and \(20\log_{10}10=20\ \mathrm{dB}\). An \(n\)-th order filter has \(n\) poles, so its skirt falls at \(20n\) dB per decade, 6n dB per octave, and its phase ultimately turns by \(90n\) degrees.
2. Adding the halves back together
A crossover splits the signal into a low-pass for the woofer and a high-pass for the tweeter, and the listener hears their sum. What matters is not only how loud each half is at the crossover frequency but its phase, because the two add as complex numbers. On the left, at the frequency chosen on the right, the two outputs are drawn as arrows in the complex plane, head to tail, with their sum. A first-order pair sums perfectly flat. A second-order Butterworth pair is 180° apart at the crossover and cancels into a deep notch, or, with the tweeter wired in reverse, bumps up 3 dB. Linkwitz–Riley filters are Butterworth filters squared: each half is 6 dB down at the crossover and in phase with the other, so they sum to exactly unity gain, with a smooth phase rotation that is the price of steep slopes.
Worked example. At the crossover frequency \(s=j\). For the second-order Butterworth pair, \(H_{\rm lp}=1/(\sqrt2 j)\) and \(H_{\rm hp}=-1/(\sqrt2j)\): equal and opposite, sum zero. Invert the tweeter and they add to \(\sqrt2/j\), magnitude \(\sqrt2\), +3 dB. For LR4, each is \(1/(\sqrt2j)^2=-\tfrac12\): both \(-6\ \mathrm{dB}\), in phase, summing to \(-1\), magnitude exactly 1 but 180° of phase shift; far below the crossover the phase is 0 and far above it is \(-360^\circ\).
Watch out. The sum is computed for ideal filters and for two drivers that are themselves flat and occupy the same point in space. Real drivers have their own roll-offs, which become part of the acoustic filter, so a real design targets acoustic Linkwitz–Riley slopes and compensates the drivers’ contribution. A flat magnitude with a rotating phase is an all-pass response: the waveform of a square wave is changed, though its spectrum is not, and whether that is audible at typical crossover frequencies is debated.
Why do Linkwitz–Riley filters sum flat while Butterworth filters of the same order do not?
A Butterworth low-pass and high-pass are power complementary: \(|H_{\rm lp}|^2+|H_{\rm hp}|^2=1\), so their powers add to one, but their voltages, which is what the air adds, are not in phase at the crossover. Squaring a Butterworth filter makes the pair amplitude complementary instead: with \(B(s)\) the Butterworth denominator, \(1/B(s)^2+s^{2n}/B(s)^2\) has numerator \(1+s^{2n}\), which for even order factors as \(B(s)B(-s)\) up to sign, leaving the all-pass \(B(-s)/B(s)\), whose magnitude is 1 at every frequency.
3. When the drivers are not in the same place
The crossover sums perfectly only for a listener equidistant from both drivers. A tweeter mounted 15 cm above the woofer is closer to a listener above the axis and farther from one below, and near the crossover frequency, where both drivers play equally, that path difference turns the in-phase sum into partial cancellation: the vertical radiation pattern grows lobes and nulls, and if the tweeter’s acoustic centre also sits behind the woofer’s, the main lobe tilts. Change the spacing and the depth offset, then drag the listener round the polar plot and see the response at that seat on the right. Set the spacing to zero, a coaxial driver with the tweeter at the woofer’s centre, and the lobes disappear.
Worked example. With an LR4 crossover at 2.5 kHz and the drivers 15 cm apart vertically, a listener \(20^\circ\) above the axis hears the tweeter \(0.15\sin20^\circ=5.1\ \mathrm{cm}\) earlier, \(0.15\ \mathrm{ms}\), which at 2.5 kHz is \(0.37\) of a cycle, \(134^\circ\): the two halves, each \(-6\ \mathrm{dB}\), add to \(|\cos67^\circ|=0.39\), \(-8\ \mathrm{dB}\). The first null is where the delay is half a cycle, \(\sin\theta=343/(2\times2500\times0.15)\), \(\theta=27^\circ\). A 2 cm depth offset shifts the whole pattern, tilting the main lobe \(\arctan(0.02/0.15)=7.6^\circ\) upwards, towards the tweeter, where its extra depth is paid back by its height.
Watch out. The drivers are modelled as point sources in the far field, with ideal flat responses and no directivity of their own; the previous topic’s beaming would further narrow the woofer’s contribution near the crossover. A digital delay can remove a depth offset exactly on axis, but no delay can remove the vertical path difference for every listening angle at once; only putting the acoustic centres in the same place does that. In a coaxial driver the tweeter still sits slightly behind the woofer’s cone plane, which is corrected by delay, and the woofer cone shapes the tweeter’s dispersion.
Why can a delay fix the depth offset but not the vertical spacing?
The depth offset adds nearly the same delay \(d_z\cos\theta/c\) at every angle close to the axis, so delaying the woofer by \(d_z/c\) cancels it for all listeners in front. The vertical spacing adds a delay \(-h\sin\theta/c\) that changes sign from above to below the axis: any fixed electronic delay makes it right at exactly one angle and wrong everywhere else. Only \(h=0\) makes the delay zero for every angle, which is the geometric argument for coincident drivers.
In practice
Where this mathematics and physics is at work, in explainers that take the real thing apart.