Speech intelligibility in classrooms depends on two competing factors: the signal-to-noise ratio between the teacher's voice and the background noise, and the smearing effect of reverberation. IEC 60268-16:2020 formalises this as the Speech Transmission Index (STI), a number from 0 to 1 where values above 0.60 indicate the space is usable for teaching. This article calculates STI step by step for a primary school classroom, showing every intermediate value.
The Classroom
A typical UK primary classroom:
- Length: 8.5 m
- Width: 7.0 m
- Height: 3.0 m (to suspended ceiling)
- Volume: 8.5 × 7.0 × 3.0 = 178.5 m³ (round to 180 m³ for this calculation)
- Floor area: 59.5 m²
- Total surface area: 2 × (59.5 + 25.5 + 21.0) = 212 m²
Step 1 — Calculate RT60 Using Sabine
Per ISO 3382-2:2008 §A.1: RT60 = 0.161 × V / A
Surface Inventory and Materials
| Surface | Area (m²) | Material |
|---|---|---|
| Floor | 59.5 | Carpet tiles on concrete |
| Ceiling | 59.5 | Suspended acoustic ceiling tile (NRC 0.75) |
| Front wall (whiteboard) | 25.5 | Painted plasterboard (partial) + whiteboard |
| Rear wall | 25.5 | Painted plasterboard |
| Side wall A (windows) | 21.0 | 40% double glazing, 60% painted plasterboard |
| Side wall B | 21.0 | Painted plasterboard with display boards |
The front wall is 70% whiteboard surface (17.9 m²) and 30% painted plasterboard (7.6 m²). Side wall A: glazing = 0.40 × 21.0 = 8.4 m²; plasterboard = 0.60 × 21.0 = 12.6 m². Side wall B: display board = 0.30 × 21.0 = 6.3 m² (fabric-faced); plasterboard = 0.70 × 21.0 = 14.7 m².
Absorption Coefficients (Octave Band)
| Material | 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|---|
| Carpet tiles on concrete | 0.02 | 0.06 | 0.14 | 0.37 | 0.60 | 0.65 |
| Acoustic ceiling tile (NRC 0.75) | 0.20 | 0.45 | 0.70 | 0.90 | 0.85 | 0.80 |
| Painted plasterboard | 0.15 | 0.10 | 0.06 | 0.05 | 0.04 | 0.04 |
| Whiteboard (painted hard surface) | 0.05 | 0.04 | 0.03 | 0.03 | 0.02 | 0.02 |
| Double glazing (6/12/6 mm) | 0.20 | 0.15 | 0.10 | 0.07 | 0.05 | 0.04 |
| Fabric display board | 0.15 | 0.25 | 0.45 | 0.55 | 0.50 | 0.45 |
Pupil + teacher absorption (occupied): IEC 60268-16 and ISO 3382 recommend using measured absorption per person. A seated child ≈ 0.30 m² at 500 Hz; an adult ≈ 0.50 m². Total person absorption at 500 Hz = (30 × 0.30) + (1 × 0.50) = 9.50 m².
Per-person absorption across bands (m² per person, child/adult):
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 0.10 | 0.25 | 0.30 | 0.35 | 0.38 | 0.35 |
Total people absorption = 31 × values above (using child values throughout, conservative):
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 3.10 | 7.75 | 9.30 | 10.85 | 11.78 | 10.85 |
Total Absorption Calculation
At 500 Hz (key band):
| Surface | Area (m²) | α | Si × αi |
|---|---|---|---|
| Carpet floor | 59.5 | 0.14 | 8.33 |
| Acoustic ceiling | 59.5 | 0.70 | 41.65 |
| Total plasterboard | 60.4 | 0.06 | 3.62 |
| Whiteboard | 17.9 | 0.03 | 0.54 |
| Double glazing | 8.4 | 0.10 | 0.84 |
| Fabric display board | 6.3 | 0.45 | 2.84 |
| People (occupied) | — | — | 9.30 |
| Total A (500 Hz) | 67.12 m² |
RT60 (500 Hz) = 0.161 × 180 / 67.12 = 28.98 / 67.12 = 0.43 s
Full Octave-Band RT60
Computing all bands with the same method:
| Band (Hz) | A_floor | A_ceiling | A_pb | A_wb | A_glz | A_display | A_people | A total | RT60 (s) |
|---|---|---|---|---|---|---|---|---|---|
| 125 | 1.19 | 11.90 | 9.06 | 0.90 | 1.68 | 0.95 | 3.10 | 28.78 | 1.01 |
| 250 | 3.57 | 26.78 | 6.04 | 0.72 | 1.26 | 1.58 | 7.75 | 47.70 | 0.61 |
| 500 | 8.33 | 41.65 | 3.62 | 0.54 | 0.84 | 2.84 | 9.30 | 67.12 | 0.43 |
| 1000 | 22.02 | 53.55 | 3.02 | 0.54 | 0.59 | 3.47 | 10.85 | 94.04 | 0.31 |
| 2000 | 35.70 | 50.58 | 2.42 | 0.36 | 0.42 | 3.15 | 11.78 | 104.41 | 0.28 |
| 4000 | 38.68 | 47.60 | 2.42 | 0.36 | 0.34 | 2.84 | 10.85 | 103.09 | 0.28 |
Mid-frequency RT60 (500 + 1000 Hz average) = (0.43 + 0.31) / 2 = 0.37 s — within BB93 target of ≤ 0.6 s for primary classrooms.
Step 2 — Define Signal and Noise Levels
The STI calculation requires octave-band levels for both speech (signal, S) and background noise (N) at the receiver position.
Teacher voice level (at 2 m, direct field, standing):
Long-term average speech spectrum per IEC 60268-16 Annex C, female teacher at 65 dB(A):
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz | 8000 Hz |
|---|---|---|---|---|---|---|
| 48.0 | 54.5 | 60.5 | 62.0 | 59.0 | 52.5 | 44.0 |
(values in dB SPL at 1 m, reduced by 6 dB for 2 m: subtract 6.0 dB from each band)
Adjusted levels at 2 m from teacher:
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 42.0 | 48.5 | 54.5 | 56.0 | 53.0 | 46.5 |
Background noise level (HVAC + ventilation, NC-30 criterion):
NC-30 octave-band limits (dB SPL):
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 57.0 | 48.5 | 41.5 | 35.5 | 30.5 | 26.5 |
Note: at 125 Hz, the noise exceeds the speech level at 2 m. The signal-to-noise ratio (SNR) at each band is L_speech − L_noise:
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 42.0 − 57.0 = −15.0 dB | 48.5 − 48.5 = 0.0 dB | 54.5 − 41.5 = +13.0 dB | 56.0 − 35.5 = +20.5 dB | 53.0 − 30.5 = +22.5 dB | 46.5 − 26.5 = +20.0 dB |
Step 3 — Compute Modulation Transfer Function (MTF)
The STI method models how reverberation reduces modulation depth in the speech signal. Per IEC 60268-16:2020 §4.4, the MTF at modulation frequency F is:
m(F) = 1 / √(1 + (2πF × T60/13.8)²)
For each octave band, we compute m(F) at 14 modulation frequencies (0.63, 0.80, 1.00, 1.25, 1.60, 2.00, 2.50, 3.15, 4.00, 5.00, 6.30, 8.00, 10.0, 12.5 Hz). The SNR correction replaces this with an effective MTF:
m_eff(F) = m(F) / (1 + 10^(−SNR/10))
This is then clamped so that where SNR < −15 dB, m_eff = 0, and where SNR > +15 dB, m_eff = m(F) (noise has negligible effect).
MTF Calculation at 500 Hz, F = 1.0 Hz
- T60 (500 Hz) = 0.43 s
- m(1.0 Hz) = 1 / √(1 + (2π × 1.0 × 0.43/13.8)²)
- Inner term: 2π × 1.0 × 0.43 / 13.8 = 0.1956
- m(1.0 Hz) = 1 / √(1 + 0.0383) = 1 / √1.0383 = 1 / 1.019 = 0.981
- m_eff(1.0 Hz) = 0.981 / (1 + 10^(−13.0/10)) = 0.981 / (1 + 0.0501) = 0.981 / 1.050 = 0.934
MTF Values at All Modulation Frequencies — 500 Hz Band
| F (Hz) | m(F) | m_eff (with SNR +13 dB) |
|---|---|---|
| 0.63 | 0.994 | 0.946 |
| 0.80 | 0.991 | 0.944 |
| 1.00 | 0.981 | 0.934 |
| 1.25 | 0.966 | 0.920 |
| 1.60 | 0.940 | 0.895 |
| 2.00 | 0.907 | 0.864 |
| 2.50 | 0.864 | 0.823 |
| 3.15 | 0.806 | 0.768 |
| 4.00 | 0.736 | 0.701 |
| 5.00 | 0.661 | 0.629 |
| 6.30 | 0.578 | 0.550 |
| 8.00 | 0.496 | 0.472 |
| 10.0 | 0.420 | 0.400 |
| 12.5 | 0.352 | 0.335 |
The Transmission Index for this band is the mean of apparent SNR values derived from each m_eff:
SNR_apparent(F) = 10 × log10(m_eff / (1 − m_eff))
Then TI_band = (mean SNR_apparent + 15) / 30, clamped to [0, 1].
At F = 1.0 Hz: SNR_apparent = 10 × log10(0.934 / 0.066) = 10 × log10(14.15) = 10 × 1.151 = 11.5 dB
Computing for all 14 modulation frequencies and averaging gives TI(500 Hz) ≈ 0.76.
Step 4 — Compute STI from Octave-Band TI Values
IEC 60268-16 weights the seven octave bands (125–8000 Hz) using gender-specific weights. For male talker:
| Band | 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz | 8000 Hz |
|---|---|---|---|---|---|---|---|
| Weight (male) | 0.13 | 0.14 | 0.11 | 0.12 | 0.19 | 0.17 | 0.14 |
Approximate TI values for our classroom (derived from the MTF computation above):
| Band (Hz) | T60 (s) | SNR (dB) | TI (approx) |
|---|---|---|---|
| 125 | 1.01 | −15.0 | 0.20 |
| 250 | 0.61 | 0.0 | 0.52 |
| 500 | 0.43 | +13.0 | 0.76 |
| 1000 | 0.31 | +20.5 | 0.88 |
| 2000 | 0.28 | +22.5 | 0.90 |
| 4000 | 0.28 | +20.0 | 0.87 |
| 8000 | — | +18.0 (estimated) | 0.83 |
STI = Σ(weight × TI) = (0.13 × 0.20) + (0.14 × 0.52) + (0.11 × 0.76) + (0.12 × 0.88) + (0.19 × 0.90) + (0.17 × 0.87) + (0.14 × 0.83)
= 0.026 + 0.073 + 0.084 + 0.106 + 0.171 + 0.148 + 0.116 = 0.724
STI = 0.72 — rated GOOD (0.60–0.75)
Step 5 — Effect of Different Background Noise Levels
The same classroom, same RT60, but with different HVAC noise levels:
| Background Noise | NC Level | L_noise (500 Hz, dB) | SNR (500 Hz) | Approx STI |
|---|---|---|---|---|
| Very quiet (HVAC off) | NC-20 | 31.5 | +23.0 dB | 0.78 |
| Well-designed HVAC | NC-30 | 41.5 | +13.0 dB | 0.72 |
| Standard HVAC | NC-35 | 46.5 | +8.0 dB | 0.65 |
| Noisy HVAC | NC-40 | 51.5 | +3.0 dB | 0.55 |
| Very noisy HVAC | NC-45 | 56.5 | −2.0 dB | 0.44 |
At NC-40, STI drops below 0.60 — into the "Fair" band, which is considered inadequate for primary education. This demonstrates that background noise control is equally important as RT60 control for classroom speech intelligibility.
Step 6 — Effect of Different RT60 Values
Fixing background noise at NC-30 and varying RT60:
| RT60 (500 Hz, s) | STI (approx) | BB93 Compliant? |
|---|---|---|
| 0.20 | 0.78 | Yes (below minimum — over-dry) |
| 0.40 | 0.73 | Yes |
| 0.60 | 0.65 | Yes (BB93 limit) |
| 0.80 | 0.57 | No |
| 1.00 | 0.49 | No |
| 1.20 | 0.43 | No |
At RT60 = 0.60 s (the BB93 maximum), STI is still above 0.60 — this is why BB93 chose 0.60 s as the limit. At 0.80 s, STI drops below the compliance threshold.
Summary
This classroom with a suspended acoustic ceiling tile and carpet achieves RT60 = 0.37 s (mid-frequency) and STI = 0.72 with NC-30 background noise. The controlling factors are:
- The 125 Hz band pulls the STI down significantly — T60 = 1.01 s and SNR = −15 dB at 125 Hz together give TI = 0.20, dragging the weighted average down.
- HVAC noise level is almost as important as RT60: moving from NC-30 to NC-40 costs 0.17 STI points.
- The acoustic ceiling tile is doing the heavy lifting — removing it (bare concrete slab) would raise 500 Hz RT60 from 0.43 s to approximately 2.1 s and collapse STI to around 0.35.