Acoustic loads — SPL, octave band analysis: Visual walkthrough

This page rebuilds the whole acoustic-loads story from nothing but a wiggling air molecule. We will earn every symbol before using it: pressure, RMS, the decibel, the reference number $20 μ Pa$ , and finally the rule that stitches many octave bands into one Overall SPL. Follow the pictures — they carry the derivation.

Parent: Acoustic Loads — SPL, Octave Band Analysis.

Step 1 — What a sound "pressure" actually is

WHAT. Air normally pushes on every surface with the steady atmospheric pressure, about $p_{0} = 1 0^{5} Pa$ (pascals — newtons of push per square metre). A sound wave is a tiny extra push-and-pull riding on top of that steady value. We call this wobble $p (t)$ : it goes slightly positive (compression), then slightly negative (rarefaction), over and over.

WHY. Before we can measure "how loud," we must name the thing we measure. Loudness is not the total pressure $p_{0} + p (t)$ — a microphone ignores the constant $p_{0}$ and reports only the wobble $p (t)$ . So $p (t)$ is our raw ingredient.

PICTURE. In the figure the flat line is $p_{0}$ ; the wavy line is the true pressure. The shaded gap between them — swinging above and below — is $p (t)$ , the sound.

Figure — Acoustic loads — SPL, octave band analysis

Step 2 — Why we can't just average $p (t)$ : enter RMS

WHAT. We want one number for "how big is the wobble." The obvious idea — average $p (t)$ over time — fails: the wave spends equal time positive and negative, so its average is exactly zero. Instead we square first (making every value positive), average that, then take the square root. This is the root-mean-square, written $p_{rms}$ .

WHY square? Two reasons that are really one. (1) Squaring kills the sign, so cancellation can't hide the wave. (2) The energy a wave carries is proportional to pressure squared — so the mean of $p^{2}$ is the physically meaningful average, not the mean of $p$ .

PICTURE. The figure shows the raw wave (crossing zero), then the squared wave $p^{2}$ (all above zero, a hump train), then the flat dashed line at the mean of $p^{2}$ ; $p_{rms}$ is the square root of that height.

Step 3 — From pressure to energy flow (intensity)

WHAT. A wave doesn't just squeeze air in place; it also nudges molecules along, giving them a velocity $v (t)$ . The power per square metre carried past a surface is the intensity $I$ . For a plane wave, pressure and velocity are locked together by one number — the acoustic impedance $ρ c$ .

WHY this tool? We need intensity because energy is what adds up when many sources or many frequency bands combine (coming in Steps 6–8). Pressure alone doesn't add cleanly; intensity does. The link $p = ρ c v$ is what lets us trade pressure for energy — see 2.5.8 Acoustic Impedance and Transmission for why $ρ c$ is the natural "stiffness × speed" of the medium.

PICTURE. The figure shows a slab of air: a pressure arrow pushing right, a velocity arrow moving right, and the product $p \cdot v$ shaded as the energy streaming through the face. The slope of the $p$ -vs- $v$ line is exactly $ρ c$ .

Step 4 — Why loudness needs a logarithm

WHAT. The quietest audible sound has $p_{rms} \approx 20 μ Pa$ (that's $20 \times 1 0^{- 6}$ Pa). A jet at full blast is around $200$ Pa. That is a ten-million-to-one span. Writing "0.00002 Pa … 200 Pa" is hopeless. So we compress it with $lo g_{10}$ : the logarithm asks "ten to what power gives this ratio?" and turns multiplication into addition.

WHY the log and not, say, a square root? A square root shrinks the range only a little ( $1 0^{7} \approx 3162$ , still huge). The logarithm shrinks any number of zeros to a small count: $lo g_{10} (1 0^{7}) = 7$ . Because our ears themselves respond roughly to ratios (doubling the pressure feels like a fixed step up whether you start soft or loud), the log matches perception too.

PICTURE. Left panel: a linear axis where the whisper is invisibly close to zero and the jet runs off the page. Right panel: the same points on a log axis, now evenly spread from $- 6$ to $+ 2$ in exponent — readable.

Step 5 — Assembling the SPL formula

WHAT. We now define the Sound Pressure Level (SPL). We compare our sound's intensity to the hearing-threshold intensity $I_{ref} = 1 0^{- 12} W/m^{2}$ , take the log, and scale by 10 to make decibels. Because $I \propto p^{2}$ , the pressure form carries a 20 instead of a 10.

WHY the 10, and why does it become 20? The "deci" in decibel means we use tenths of a bel, so intensity level is $10 lo g_{10} (I / I_{ref})$ . Swapping intensity for pressure uses $I \propto p^{2}$ , and $lo g_{10} (x^{2}) = 2 lo g_{10} (x)$ — that factor of 2 rides along with the 10 to give 20. The reference $p_{ref} = 20 μ Pa$ is not arbitrary: it is exactly the pressure matching $I_{ref}$ through Step 3, $p_{ref} = I_{ref} ρ c$ .

PICTURE. A flow arrow: raw $p_{rms}$ → divide by $p_{ref}$ → take $lo g_{10}$ → multiply by 20 → dB. Each box is annotated with what leaves it. A side ruler shows landmark levels: 0 dB threshold, 60 dB talk, 120 dB pain, 150 dB fairing.

Step 6 — Why decibels never add like ordinary numbers

WHAT. Two engines each at 145 dB do not make 290 dB. Decibels are logs of energy, and to add energies you must first climb out of the log, add the real intensities, then climb back in.

WHY do intensities add and not pressures? Two uncorrelated sources have random relative phase. When you square the summed pressure, $⟨(p_{1} + p_{2})^{2} ⟩ = ⟨ p_{1}^{2} ⟩ + ⟨ p_{2}^{2} ⟩ + 2 ⟨ p_{1} p_{2} ⟩$ , and that last cross-term averages to zero because the phases wander independently. What survives is $⟨ p_{1}^{2} ⟩ + ⟨ p_{2}^{2} ⟩$ — the two intensities, added. So energy is the currency that sums.

PICTURE. Left: two random waves and their sum; the cross-term area is drawn half positive, half negative, cancelling. Right: two intensity bars stacking to one taller bar, with the dB tick showing the stack rises only $10 lo g_{10} 2 \approx 3$ dB.

Step 7 — Slicing the spectrum into octave bands

WHAT. Real launch noise isn't one tone; it's energy spread across frequencies. We chop that continuous spectrum into octave bands — each band a 2:1 frequency ratio, centred on standard values $31.5, 63, 125, 250, 500, 1000, 2000, 4000, 8000$ Hz.

WHY $2$ edges? We want the band to be logarithmically centred, because frequency perception (and structural resonances) are ratio-based. Demanding $lo g f_{c} = \frac{1}{2} (lo g f_{low} + lo g f_{high})$ forces $f_{c}^{2} = f_{low} f_{high}$ . Combine with the octave rule $f_{high} = 2 f_{low}$ and you get $f_{low} = f_{c} / 2$ , $f_{high} = 2 f_{c}$ . That is where the $2$ comes from — no magic, just "put $f_{c}$ in the geometric middle."

PICTURE. A log-frequency axis with tick marks at the standard centres; over each centre a bracket runs from $f_{c} / 2$ to $2 f_{c}$ , and every bracket has the same width on the log axis (that's what "octave" means visually).

Step 8 — Stitching the bands back into one number (OASPL)

WHAT. Given each band's level $L_{i}$ (dB), the Overall SPL is not the loudest band and not the sum of dB values. Since the bands cover non-overlapping frequencies, their energies simply add — exactly the logic of Step 6, now for $N$ pieces.

WHY sum $1 0^{L_{i} /10}$ ? Each band's intensity is $I_{i} = I_{ref} 1 0^{L_{i} /10}$ (undo the log). Non-overlapping bands can't interfere, so total energy is $\sum I_{i}$ . Put that back through $10 lo g_{10}$ and the $I_{ref}$ cancels, leaving a clean sum-of-tens formula.

PICTURE. Five vertical bars, one per band, heights = $1 0^{L_{i} /10}$ (relative intensity). They stack into one tall bar; a dB scale on the right shows the stack lands at 143.4 dB — barely above the tallest single bar (140 dB) because the 500 Hz band dominates.

Step 9 — Edge and degenerate cases (never leave the reader stranded)

WHAT & WHY & PICTURE (three quick cases):

Silence, $p_{rms} \to 0$ . Then $p_{rms} / p_{ref} \to 0$ and $lo g_{10} (0) \to - \infty$ : SPL dives to $- \infty$ dB. Physically: perfect silence is "infinitely below" threshold. The picture shows the SPL curve plunging off the bottom as pressure approaches zero.
Exactly at threshold, $p_{rms} = p_{ref}$ . Ratio $= 1$ , $lo g_{10} 1 = 0$ , so SPL $= 0$ dB. Zero dB is not silence — it's the faintest audible tone. The curve crosses the axis here.
One band dominates in OASPL. If band $k$ is $\geq 10$ dB above all others, $1 0^{L_{k} /10}$ swamps the sum and $L_{OASPL} \approx L_{k}$ . The picture: one giant bar and negligible others — the overall level "clamps" to the peak. This is why finding the loudest band (and whether it hits a resonance) is the whole game.

The one-picture summary

One diagram, the whole chain: a wiggling molecule → sound pressure $p (t)$ → RMS → intensity $\propto p^{2}$ → log-compress → SPL in dB → slice into octave bands → energy-sum back to OASPL.

Recall Feynman retelling — say it like a story

Air is normally pushing on everything with a steady weight. A sound is a tiny extra jiggle on top of that push. If I try to average the jiggle I get zero, because it goes up as much as down — so I square it first (which also happens to be the energy), average, and square-root back; that's the RMS jiggle. For a pure tone the squared wave hovers around half its peak (because $sin^{2}$ averages to a half), so the RMS comes out to the peak over $2$ . That jiggle carries energy, and how much energy streams past a wall depends on the air's own stiffness-times-speed, a number called $ρ c$ ; the energy goes as jiggle-squared. But quiet and loud sounds differ by ten million times, so I can't write them on one ruler — I take a logarithm, which just counts zeros, and I compare against the faintest sound a person can hear. Scale by twenty (ten for the decibel, two because energy is pressure-squared) and I have the decibel, the SPL. Decibels are logs, so I can never add them directly: to combine sounds I climb out of the log to real energy, add the energies, climb back. Launch noise is spread over many pitches, so I cut the spectrum into octaves — each a doubling of frequency, centred in the middle on a log ruler, which is why the edges have that $2$ . Finally, to get one grand total I add up every band's energy and take the log once more: that's the OASPL. And the punchline — because it's energy-adding, the loudest band almost always wins; the overall level barely rises above the biggest single voice in the choir.

Recall Quick self-check

Why 20 and not 10 in SPL? ::: Intensity level uses 10; pressure enters squared, and $lo g (p^{2}) = 2 lo g p$ , so 10×2 = 20. Why does a sine give $p_{rms} = \overset{p}{^} / 2$ ? ::: Because $sin^{2}$ time-averages to $\frac{1}{2}$ (using $sin^{2} = \frac{1}{2} (1 - cos 2 t)$ , whose $cos$ part averages to 0), so $p_{rms} = \frac{1}{2} \overset{p}{^}$ . Why do intensities add but not pressures for uncorrelated sources? ::: The pressure cross-term $2 ⟨ p_{1} p_{2} ⟩$ averages to zero at random phase, leaving $⟨ p_{1}^{2} ⟩ + ⟨ p_{2}^{2} ⟩$ — i.e. intensities. What is the OASPL when one band is 10 dB above the rest? ::: Approximately that band's own level; it dominates the energy sum. What SPL corresponds to $p_{rms} = p_{ref}$ ? ::: Exactly 0 dB (the ratio is 1, $lo g_{10} 1 = 0$ ).

Related environments to keep in view: 3.6.13 Shock Loads and SRS, 3.6.14 Combined Environmental Testing.