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How deep should a subwoofer play?

Extension to 20 Hz is not automatically better. Useful bandwidth depends on programme material, room gain, maximum SPL, placement and integration.

Depth needs an operating level

A claim of 20 Hz is incomplete without the reference level and conditions. A system that reaches 20 Hz only quietly may be less useful than one with more clean output at 30 Hz. Consider frequency extent, desired level, available space and the listening area together.

Frequency is not loudness

20 Hz means twenty cycles per second. At an assumed sound speed of 343 m/s its wavelength is about 17.2 m, while 40 Hz is about 8.6 m. A room does not need to contain a full wavelength to reproduce that frequency. Room boundaries instead influence how pressure is distributed and how sound builds up.

Music and film ask different questions

Some instruments and electronic recordings contain useful energy below 30 Hz; others benefit more from balanced output between roughly 40 and 100 Hz. Film soundtracks may contain very deep effects, but not every release does. Bass management can also send low-frequency content from the main channels to the subwoofer, beyond the dedicated effects channel.

An extra octave can be expensive

For an idealised small direct-radiating piston at fixed sound pressure, halving frequency requires about four times the excursion. More cone area or multiple drivers can share this task. A resonator can help within a limited region. More amplifier power cannot remove mechanical excursion limits, and deep boost below vented or PR tuning can be particularly demanding.

The room can hide or exaggerate extension

A strong peak around 55 Hz may sound impressive even without deep extension. A cancellation around 45 Hz can make an otherwise capable system seem thin. Placement is therefore part of the design, not cosmetic finishing. Compare source and listener positions and assess all important seats rather than assuming a different subwoofer alone will solve the problem.

Integrate the crossover

Around the crossover, mains and subwoofer operate together. Their acoustic phases, levels and delays must combine appropriately. The lowest possible crossover is not automatically best: small mains may gain useful headroom from earlier relief. Very high crossing, unwanted harmonics or placement can make the subwoofer easier to localise. Measure the individual and combined responses.

Choose a practical target

A compact music system can benefit greatly from clean output to 35–40 Hz. A more demanding music and film system may target 25–30 Hz, while deliberate reproduction of the lowest effects octave can justify 20 Hz or below. These are orientation points, not quality grades. Judge clean level, integration, unwanted rattles and repeatable measurements alongside familiar recordings.

Explore the relationship
Interactive experiment

One octave lower: what happens to excursion?

At the same radiating area and sound pressure, move from 60 to 30 Hz. The simple low-frequency model needs four times the excursion.

Model limits: Relative low-frequency direct-piston model. No port or passive-radiator contribution, room or absolute SPL prediction.

A closer look · model and equation

For a small direct-radiating piston at fixed distance, approximately p ∝ f²·Sd·x. At fixed area and sound pressure, required excursion therefore scales as 1/f².

Workshop takeaway

Decide depth, level and space together. The lowest octave is useful when the system can reproduce it cleanly.

Electrical foundationsFiltersMeasurement

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

04 · Sound speed 343 m/s, frequency 1 kHz. Find wavelength.

Show worked answer
  1. Wavelength = speed/frequency.
  2. 343/1000.
  3. Wavelength = 0.343 m = 34.3 cm.
Calculation
\lambda=\frac{c}{f}=0.343\,m

17 · A midwoofer needs 1 mm excursion at 100 Hz. Estimate excursion at 50 Hz for the same SPL in the simple piston model.

Show worked answer
  1. Excursion is approximately inversely proportional to frequency squared.
  2. Half frequency requires four times excursion.
  3. Result: 4 mm; enclosure effects excluded here.
Calculation
x_2=x_1\left(\frac{f_1}{f_2}\right)^2=4\,mm

18 · Two identical 8 Ω drivers are wired in parallel. Find nominal load and effect on amplifier current at the same voltage.

Show worked answer
  1. Nominal equivalent: 4 Ω.
  2. At constant voltage total current doubles in the resistive model.
  3. Check amplifier capability and actual impedance curve.
Calculation
R_p=\frac{8\cdot8}{8+8}=4\,\Omega

20 · Put a two-way design workflow in order: level, driver data, branches, sum, angles and loading.

Show worked answer
  1. Measure installed drivers with impedance and shared timing.
  2. Check overlap, choose targets, filter branches and match levels.
  3. Verify complex sum; check angles, impedance and loading, then iterate.
20 exercises · PDF ↓Teacher solutions · PDF ↓

Match drivers together →

Further technical reading

Colloms & Darlington (2005): High Performance Loudspeakers, 6th ed. Section 4.8 (Sub-woofers and Extended Low Frequencies), starting on printed page 190. Source information (COLLOMS-2005-6) →