Subwoofer tools.
ENCLOSURES & PORTS / THE TOOLKIT

Sealed box.

Enter a net volume to predict the ideal response, or choose a target system Q to find a starting box volume. Every result uses a stated, small-signal model.

01 / THE INPUTS
Driver parameters
Hz
Enclosure design
Electrical damping
02 / THE RESULTS
Net enclosure volume
47.08L
System resonance (fc)
53.03Hz
System Q (Qtc)
0.707
-3 dB cutoff (f3)
53.03Hz
Response peak above reference
0.00dB
No peak above reference

Normalized sealed response

LOG FREQUENCY
Normalized sealed response: Relative level (dB) against frequency in hertz.-43.2-33.92-24.65-15.37-6.13.18102050100200Relative level (dB)Frequency (Hz)

Read this result with its assumptions.

Effective driver Qts: 0.4000. The -3 dB reference is the ideal upper passband.

Ideal airtight, lossless enclosure. Stuffing, leakage, absorption, and changes in air loading are not included.

Units & assumptions

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03 / UNDER THE HOOD

The math, in plain sight.

A simpler box. A clearer starting point.

Volume, system resonance, and damping

A sealed cabinet adds an air spring to the suspension. A smaller box raises resonance and changes damping. This model assumes an airtight, lossless box with negligible changes in air loading.

α=VasVb,fc=fs1+α,Qtc=Qts,eff1+α\alpha=\frac{V_{as}}{V_b},\qquad f_c=f_s\sqrt{1+\alpha},\qquad Q_{tc}=Q_{ts,\mathrm{eff}}\sqrt{1+\alpha}
Vb=Vas(Qtc/Qts,eff)2−1for Qtc>Qts,effV_b=\frac{V_{as}}{(Q_{tc}/Q_{ts,\mathrm{eff}})^2-1}\quad\text{for }Q_{tc}>Q_{ts,\mathrm{eff}}

Use Qts,eff = Qts when added series resistance is negligible. If resistance is included, the entered Qes and Qms determine the effective damping instead.

Normalized response and cutoff

The -3 dB point is relative to the ideal high-frequency reference, not the response peak. At Qtc = 1/sqrt(2), f3 equals fc. A high-Q system can peak above its reference; a low f3 alone does not mean flat bass extension.

H(s)=s2s2+(ωc/Qtc)s+ωc2,ωc=2πfcH(s)=\frac{s^2}{s^2+(\omega_c/Q_{tc})s+\omega_c^2},\qquad\omega_c=2\pi f_c
∣H∣=x2(1−x2)2+(x/Qtc)2,x=ffc|H|=\frac{x^2}{\sqrt{(1-x^2)^2+(x/Q_{tc})^2}},\qquad x=\frac f{f_c}
f3=fcQtc−2−2+(2−Qtc−2)2+42f_3=f_c\sqrt{\frac{Q_{tc}^{-2}-2+\sqrt{(2-Q_{tc}^{-2})^2+4}}{2}}

Worked example and limitations

For fs = 30 Hz, Vas = 100 L, Qts = 0.40, and target Qtc = 0.707, the estimated volume is 47.1 L, with resonance and cutoff both approximately 53.0 Hz.

Stuffing, leaks, and absorption can change the measured values. In particular, leakage does not leave an exact two-pole system characterized by one Qtc.