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TUTORIALS10 min read

Composite STC Calculation: Wall + Door + Window — Full Worked Example

Calculate composite STC for an office wall with a door and window. See how an STC 28 door drags down an STC 55 wall — with full area-weighted transmission loss at every 1/3 octave band.

AcousPlan Editorial · March 18, 2026

Every opening in a wall — a door, a window, a duct penetration, a gap at the skirting board — degrades its sound insulation. The composite STC calculation quantifies exactly how much. This article calculates composite STC for an office wall with a door and an interior window (sidelight), showing every step of the area-weighted transmission loss method per ASTM E413.

The Assembly

A private office wall separating a manager's office from an open-plan area:

  • Total wall area (including door and window): 4.0 m wide × 2.7 m high = 10.8 m²
  • Door: single solid-core door with perimeter seals — 0.9 m × 2.1 m = 1.89 m² — STC 38
  • Interior sidelight window: fixed single glazing — 0.6 m × 1.5 m = 0.90 m² — STC 28
  • Wall assembly: double-layer 13 mm gypsum board each side, 90 mm steel studs at 400 mm, R15 insulation — remaining area = 10.8 − 1.89 − 0.90 = 7.95 m² (net wall area) — STC 55

The Method

Per ASTM E413-22:

  1. Express each element's transmission loss as a transmission coefficient: τ = 10^(−TL/10)
  2. Compute the area-weighted composite: τ_c = Σ(Si × τi) / S_total
  3. Convert back: TL_c = −10 × log10(τ_c)
  4. Fit the composite TL spectrum to the STC contour

Step 1 — Octave-Band TL Values

The STC contour requires 1/3 octave bands from 125 to 4000 Hz (16 bands). For this example we use representative 1/3 octave values for each assembly. In practice, these come from ASTM E90 laboratory tests.

Wall assembly (STC 55) — 1/3 octave TL:

Freq (Hz)TL_wall (dB)Freq (Hz)TL_wall (dB)Freq (Hz)TL_wall (dB)
1253963057200062
1604380058250062
20046100060315061
25049125061400060
31552160062
40054
50056

Solid-core door with seals (STC 38) — 1/3 octave TL:

Freq (Hz)TL_door (dB)Freq (Hz)TL_door (dB)Freq (Hz)TL_door (dB)
1252463040200043
1602680041250043
20029100042315042
25032125042400041
31534160043
40037
50039

Fixed single glazing (STC 28) — 1/3 octave TL:

Freq (Hz)TL_glass (dB)Freq (Hz)TL_glass (dB)Freq (Hz)TL_glass (dB)
1251763027200035
1601880028250037
20020100029315036
25022125030400034
31524160032
40025
50026

Step 2 — Compute τ for Each Element

Convert TL to transmission coefficient: τ = 10^(−TL/10)

At 500 Hz as a demonstration:

  • Wall: τ_wall = 10^(−56/10) = 10^(−5.6) = 2.51 × 10⁻⁶
  • Door: τ_door = 10^(−39/10) = 10^(−3.9) = 1.26 × 10⁻⁴
  • Glass: τ_glass = 10^(−26/10) = 10^(−2.6) = 2.51 × 10⁻³
Note the enormous range: glass transmits 2510/2.51 = 1000× more energy than the wall at 500 Hz.

Step 3 — Area-Weighted Composite TL

At 500 Hz:

ElementArea (m²)τSi × τi
Wall7.952.51 × 10⁻⁶1.995 × 10⁻⁵
Door1.891.26 × 10⁻⁴2.381 × 10⁻⁴
Glass0.902.51 × 10⁻³2.259 × 10⁻³
Total10.82.519 × 10⁻³

τ_composite = 2.519 × 10⁻³ / 10.8 = 2.332 × 10⁻⁴

TL_composite = −10 × log10(2.332 × 10⁻⁴) = −10 × (−3.632) = 36.3 dB

The wall contributes 0.795% of the composite τ. The door contributes 9.5%. The glass contributes 89.7% of the composite transmission at 500 Hz, despite covering only 8.3% of the area.

Step 4 — Full 1/3 Octave Composite TL Table

Freq (Hz)TL_wallTL_doorTL_glassτ_wall × 7.95τ_door × 1.89τ_glass × 0.90Σ(Si×τi)τ_cTL_c (dB)
1253924171.00×10⁻³2.99×10⁻²1.80×10⁻¹2.10×10⁻¹1.94×10⁻²17.1
1604326183.98×10⁻⁴1.89×10⁻²1.43×10⁻¹1.63×10⁻¹1.51×10⁻²18.2
2004629201.99×10⁻⁴9.48×10⁻³9.00×10⁻²9.97×10⁻²9.23×10⁻³20.3
2504932221.00×10⁻⁴4.75×10⁻³5.68×10⁻²6.17×10⁻²5.71×10⁻³22.4
3155234245.01×10⁻⁵2.99×10⁻³3.58×10⁻²3.88×10⁻²3.60×10⁻³24.4
4005437253.16×10⁻⁵1.50×10⁻³2.85×10⁻²3.00×10⁻²2.78×10⁻³25.6
5005639262.00×10⁻⁵2.38×10⁻⁴2.26×10⁻³2.52×10⁻³2.33×10⁻⁴36.3
6305740271.59×10⁻⁵1.89×10⁻⁴1.79×10⁻³2.00×10⁻³1.85×10⁻⁴37.3
8005841281.26×10⁻⁵1.50×10⁻⁴1.43×10⁻³1.59×10⁻³1.47×10⁻⁴38.3
10006042297.95×10⁻⁶1.19×10⁻⁴1.14×10⁻³1.27×10⁻³1.17×10⁻⁴39.3
12506142306.31×10⁻⁶1.19×10⁻⁴9.00×10⁻⁴1.03×10⁻³9.50×10⁻⁵40.2
16006243325.01×10⁻⁶9.48×10⁻⁵5.68×10⁻⁴6.68×10⁻⁴6.19×10⁻⁵42.1
20006243355.01×10⁻⁶9.48×10⁻⁵2.85×10⁻⁴3.85×10⁻⁴3.56×10⁻⁵44.5
25006243375.01×10⁻⁶9.48×10⁻⁵1.79×10⁻⁴2.79×10⁻⁴2.59×10⁻⁵45.9
31506142366.31×10⁻⁶1.19×10⁻⁴2.26×10⁻⁴3.52×10⁻⁴3.26×10⁻⁵44.9
40006041347.95×10⁻⁶1.50×10⁻⁴3.58×10⁻⁴5.16×10⁻⁴4.78×10⁻⁵43.2

Step 5 — STC Rating from the Composite TL Curve

Per ASTM E413, fit the standard STC contour to the composite TL values such that:

  • No single deficiency (STC contour value minus measured TL) exceeds 8 dB
  • The sum of all deficiencies does not exceed 32 dB
The STC contour is shifted downward until both rules are satisfied. Reviewing the TL_c values:

At 500 Hz, TL_c = 36.3 dB. The STC 37 contour passes through 40 dB at 500 Hz — giving a deficiency of 3.7 dB. Checking all bands against the STC 28 contour (which passes through approximately 28 dB at 500 Hz and 31 dB at 1000 Hz):

The composite TL curve has a pronounced dip in the 125–400 Hz range (17–26 dB) and rises steeply at 500 Hz (36 dB) due to the different low-frequency behaviour of the elements. Fitting the STC contour:

Composite STC = 28 — controlled by the glazed sidelight.

The STC 55 wall contributes almost nothing to the composite rating because the sidelight dominates transmission across all frequencies.

Step 6 — Improvement Scenarios

Scenario A: Replace Single Glazing with Acoustic Double Glazing (STC 38)

Replacing the STC 28 sidelight with acoustic double glazing (6/12/6 mm, STC 38):

At 500 Hz: τ_glass_new = 10^(−38/10) = 1.58 × 10⁻⁴ τ_c = (7.95 × 2.51×10⁻⁶ + 1.89 × 1.26×10⁻⁴ + 0.90 × 1.58×10⁻⁴) / 10.8 = (1.995×10⁻⁵ + 2.381×10⁻⁴ + 1.422×10⁻⁴) / 10.8 = 4.004×10⁻⁴ / 10.8 = 3.71×10⁻⁵

TL_c = −10 × log10(3.71×10⁻⁵) = 44.3 dB at 500 Hz (was 36.3 dB)

Composite STC improves to approximately 37 — still controlled by the door.

Scenario B: Upgrade Both Door (STC 45) and Glazing (STC 38)

Upgraded door at 500 Hz: τ_door_new = 10^(−45/10) = 3.16 × 10⁻⁵

τ_c = (7.95 × 2.51×10⁻⁶ + 1.89 × 3.16×10⁻⁵ + 0.90 × 1.58×10⁻⁴) / 10.8 = (1.995×10⁻⁵ + 5.97×10⁻⁵ + 1.422×10⁻⁴) / 10.8 = 2.22×10⁻⁴ / 10.8 = 2.06×10⁻⁵

TL_c = −10 × log10(2.06×10⁻⁵) = 46.9 dB at 500 Hz

Composite STC ≈ 42 — a 14-point improvement over the original assembly.

Scenario C: Remove Sidelight (Wall Only + Upgraded Door STC 45)

With no glazing, using upgraded door only:

At 500 Hz:

  • S_wall = 10.8 − 1.89 = 8.91 m²
  • S_total = 10.8 m²
  • τ_c = (8.91 × 2.51×10⁻⁶ + 1.89 × 3.16×10⁻⁵) / 10.8
  • = (2.237×10⁻⁵ + 5.97×10⁻⁵) / 10.8 = 8.21×10⁻⁵ / 10.8 = 7.60×10⁻⁶
  • TL_c = −10 × log10(7.60×10⁻⁶) = 51.2 dB
Composite STC ≈ 47 — close to the base wall's capability.

Improvement Summary

ScenarioDoor STCGlazing STCComposite STCSpeech Privacy
Original382828Fair
A — Better glass383837Good
B — Better door + glass453842Good
C — No glass + better door4547Very Good
D — No glass, no opening55Excellent

Key Lessons

1. The weakest element governs composite STC. An STC 28 sidelight (0.9 m²) completely dominates a wall system with an STC 55 wall (7.95 m²) and STC 38 door (1.89 m²).

2. Area fraction multiplied by linear τ gives the true weighting. The glass has τ = 2.51×10⁻³ at 500 Hz. The wall has τ = 2.51×10⁻⁶ — a 1000× ratio. The glass is 8.3% of the area, but at 500 Hz it contributes 89.7% of the total transmitted energy.

3. Improving the dominant element has most impact. Upgrading the STC 55 wall to STC 60 would have essentially no measurable effect on composite STC. Upgrading the STC 28 sidelight to STC 38 raises composite STC by 9 points.

4. Log-linear mismatch is the core trap. When specifying partitions, designers often specify elements by STC rating. The gap between STC 28 and STC 55 looks like 27 points — but in terms of transmission coefficient, the glass is 10^((55−28)/10) = 501× more transmissive than the wall. One single-pane sidelight destroys the investment in a high-STC wall.

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