Band-pass level loss: what a filter chain costs you

Almost every driver in a real system is band-passed. A midbass unit has a high-pass below it to keep it out of the subwoofer's range and a low-pass above it to hand over to the tweeter. Set both to the values the design calls for, check the levels, and the channel will not reach the output the rest of the system does. Nothing is broken, and the reason is not in any preset.

The two filters are in series, so the signal passes through one and then the other, and their responses multiply. Multiplication is the whole story: two responses that are each below unity multiply to something below both of them, and that is exactly what a band-pass is. Set the corners to 63 Hz and 300 Hz with 4th-order Linkwitz-Riley filters and the highest point of the band is 0.75 dB down. Move the band to the subwoofer's 25–63 Hz and it is 2.54 dB down.

The calculator below reports the peak of the band, the closed form for it, and how much of the level a correction would cost you in headroom.

Two filters, two ways to connect them

The first thing to get straight is that the same two filters give two different answers depending on what they are connected to, and only one of them is the familiar one.

Feed a low-pass and a high-pass to two separate drivers — a woofer and a tweeter, a sub and a main — and the two acoustic outputs add. Addition is complex: the result depends on the phase as much as the level, which is why crossovers are designed around alignments that put the two branches in phase at the corner. A matched 4th-order Linkwitz-Riley pair does exactly that, and the sum is unity across the whole overlap. Put the same pair of filters in series on one channel, as on every band-passed driver, and the responses multiply. No phase question arises, because one signal simply passes through both: the answer is the product of two curves that are each below unity.

The two answers are not close to each other. With both corners at 80 Hz:

A 4th-order Linkwitz-Riley low-pass and high-pass at the same 80 Hz corner, in dB

Frequency	Two drivers, summed	One channel, in series
40 Hz	0.00	−25.14
60 Hz	0.00	−14.77
80 Hz	0.00	−12.04
100 Hz	0.00	−13.72
160 Hz	0.00	−25.14

Zero on the left is not an approximation: a matched Linkwitz-Riley pair sums to unity at every one of those frequencies, which is the property the alignment is chosen for. The right column is the same filters doing the same thing to the same signal, and the band is 12 dB down at its own centre. Nothing is misaligned, nothing is broken, and no amount of delay or polarity changes it, because there is only one source.

Where the loss comes from

A crossed-over pair loses nothing at the corner because each branch is 6.02 dB down there and the two of them are in phase, so an amplitude of 0.5 added to an amplitude of 0.5 gives 1. That is the design. A band-pass inherits the same 6.02 dB at each of its two corners, but the two corners are at different frequencies and the band between them peaks somewhere in the middle, where each filter is less than 6.02 dB down but neither is at unity.

For a Linkwitz-Riley pair the peak sits at the geometric mean of the two corners, and that is exact rather than approximate. Set r as the low corner divided by the high corner; at the geometric mean both filters sit at the same normalised frequency, and the product works out to a simple expression in r alone. For 63–300 Hz, r is 0.21, the peak is at 137.5 Hz, and the loss is 0.75 dB. For 25–63 Hz, r is 0.40, the peak is at 39.7 Hz, and the loss is 2.54 dB.

The loss is bigger when the two corners are closer together, because r moves toward 1 and each filter is closer to its own corner at the peak. A subwoofer band, with its corners less than an octave and a half apart, therefore loses more than a midrange band four octaves wide.

The closed form, and two things it says that nobody expects

Writing the two Butterworth-family responses at the geometric mean and multiplying them gives the loss in closed form. For a Linkwitz-Riley band-pass of order M, and for a Butterworth band-pass of order N, with r the ratio of the corners:

Two consequences fall straight out of those expressions, and both are the opposite of what most people assume.

A steeper slope loses less. More attenuation in the stopband sounds like more damage to the signal, but a steeper transition also leaves the passband flatter, so the peak sits closer to unity. Over the same 63–300 Hz band:

Loss at the passband peak, 63–300 Hz, corners matched

Shape	Slope	Loss
Linkwitz-Riley 2nd order	12 dB/octave	3.31 dB
Linkwitz-Riley 4th order	24 dB/octave	0.75 dB
Linkwitz-Riley 8th order	48 dB/octave	0.03 dB
Butterworth 2nd order	12 dB/octave	0.37 dB
Butterworth 4th order	24 dB/octave	0.02 dB

Two coincident corners lose a fixed amount, whatever the order. Let r go to 1 and the expressions collapse: every Linkwitz-Riley band-pass loses 12.04 dB, and every Butterworth band-pass loses 6.02 dB. The order disappears from the answer entirely. That is because the corner level is a definition rather than a property — a Linkwitz-Riley filter is 6.02 dB down at its corner at 2nd, 4th and 8th order alike, and a Butterworth is 3.01 dB down — so two coincident corners always multiply the same two numbers.

When the two slopes differ

The closed form above assumes both branches are described by the same exponent, which means the same family and the same order. Real systems often do not: a band-pass built with a 4th-order high-pass and a 2nd-order low-pass is a perfectly reasonable thing to want, and many processors make it easy to build by accident.

When they differ the closed form stops being the peak, and it is worth being precise about how it fails, because it fails in one direction only. The peak moves up in frequency away from the geometric mean, and the closed form then overstates the loss. Over every combination of shapes and corners tested it never understated it. Two examples at 63–300 Hz:

Closed form against the measured peak for mismatched pairs, 63–300 Hz

Low corner	High corner	Closed form	Actual peak loss	Peak moves to
LR4	LR4	0.7497 dB	0.7495 dB	137.5 Hz (the geometric mean)
LR4	BW2	0.5623 dB	0.5293 dB	150.5 Hz
LR2	BW6	1.6561 dB	0.7880 dB	221.1 Hz

The last row is the worst case found. The peak moves from 137.5 Hz to 221.1 Hz, a little over eight semitones, and the closed form overstates the loss by 0.87 dB. That is far too large to wave away as a rounding difference, so the calculator reports both figures whenever the pair is not matched and lets you see the gap. The practical reading: for a matched pair the closed form is exact and can be trusted; for a mismatched one, use the measured number, and treat the mismatch as a reason to expect the band to be less predictable than the arithmetic suggests.

It is not always a loss

Everything above concerns the Butterworth and Linkwitz-Riley families, whose second-order sections sit at or below Q = 0.7071. Raise the Q above that and the band resonates, the peak goes above unity, and the correction is a cut instead of a boost. Cascading four second-order sections — two high-pass and two low-pass, all at the same Q — between 63 Hz and 300 Hz:

What a higher section Q does to the same band, 63–300 Hz

Section Q	Passband peak
0.5	−6.62 dB
0.7071	−0.75 dB
1	+3.28 dB
2	+13.39 dB

So the direction of the correction is not fixed by the shape of the band, only by the Q of the sections in it. A resonant band is not a defect — it is how a band-pass is built when you want one — but it does mean the level budget has to be read off the actual curve rather than assumed from the word “band-pass”.

Putting the level back, and what it costs

Correcting the loss means adding gain, and gain is not free: it comes out of the headroom between where the signal peaks and the ceiling. If the chain loses 2.54 dB and the signal already peaks at 0 dBFS, the correction lifts it to +2.54 dBFS and the channel is over the ceiling.

Headroom left after correcting a 2.54 dB loss

Level-matching reference	Peak after correction	Headroom left
0 dBFS	+2.54 dBFS	−2.54 dB
−6 dBFS	−3.46 dBFS	3.46 dB
−10 dBFS	−7.46 dBFS	7.46 dB
−18 dBFS	−15.46 dBFS	15.46 dB

The arithmetic is exactly one for one: matching levels 10 dB lower buys exactly 10 dB of room for the correction. This is the whole reason the reference level used during level matching matters more than it looks like it should. Set up the system with a full-scale sine and the correction has nowhere to go; set it up with a −10 or −18 dBFS sine and the same correction fits easily. The cost is only that the measurement is made 10 or 18 dB further from the noise floor, which on any modern converter is not a limitation.

One consequence is worth stating plainly, because it is easy to get backwards: a processor that reserves a few decibels of gain above its nominal maximum output is not a design flourish, it is making room for exactly this. So is a level-matching procedure that starts from a low reference. Both are the same allowance, priced in different places.

Where the arithmetic and the processor part company

The closed form above is an analogue identity. The filters a processor runs are digital, made from the bilinear transform, whose corner lands below where the analogue prototype asked for it by an amount that grows with frequency. The two figures therefore do not agree exactly, and the disagreement is a property of the realisation rather than an error in either one.

At 48 kHz, with the corners four times apart, the difference between the closed form and the measured peak is 0.0004 dB when the high corner is at 300 Hz, 0.005 dB at 1 kHz, 0.019 dB at 2 kHz, 0.075 dB at 4 kHz and 0.31 dB at 8 kHz. Below about 2% of the sample rate the closed form is good to a hundredth of a decibel and can be used as it stands; above that it drifts, in one direction, and the measured peak is the figure to quote. This is why the calculator has a sample-rate field at all: the loss itself is a property of the band, but the agreement between the two ways of predicting it is a property of the rate.

Five mistakes that survive review

1. Treating the band-pass loss as a crossover problem. No delay, no polarity and no re-timing changes it, because there is only one source and the loss is a product rather than a sum. If a band-passed channel is short of level, the fix is in the level budget, not in the alignment.
2. Expecting a steeper filter to cost more level. It costs less, by a lot: 0.03 dB at 8th order against 3.31 dB at 2nd over the same band. What a steeper filter costs is group delay and a more abrupt transition, not level.
3. Correcting the corners instead of the peak. The loss is not at either corner — it is in between, at the geometric mean, and at the corners themselves the level is 6.02 dB down by definition. Compensating at a corner applies the correction where the response is already correct.
4. Using the closed form on a mismatched pair. It overstates the loss, by up to 0.87 dB in the cases checked, and it never understates it. The peak also moves nearly seven semitones, so a correction placed at the geometric mean is in the wrong place.
5. Correcting the loss without knowing what it costs. Adding 2.5 dB at a 0 dBFS reference puts the channel over the ceiling. Decide the reference level first, then decide how much of the loss to correct — or correct it in the analogue stage where the headroom actually is.

Summary

Two filters in series multiply, so a band-passed channel's passband peak sits below unity, and the amount depends only on the ratio of the two corners and the slopes. For a matched pair the loss is 40 log₁₀(1 + r^M/2) dB for Linkwitz-Riley and 20 log₁₀(1 + r^N) dB for Butterworth, at the geometric mean of the corners. Steeper slopes lose less. Higher Q can turn the loss into a gain. And correcting any of it costs the same number of decibels of headroom, which is why the reference level chosen during level matching decides what the rest of the chain can afford.

The calculator is an engineering aid rather than a measurement; see the disclaimer, and the tool index for the rest of the series.