Acoustic loads — SPL, octave band analysis: Worked examples

This page is the drill hall for the parent topic. The parent built the formulas; here we hit every case those formulas can produce — every sign of the number, every degenerate input, every limiting value, plus a real-world word problem and an exam trap.

Before we start, let us pin down every symbol we use, so nothing appears unexplained:

Recall The formulas we lean on

SPL (forward): $L = 20 lo g_{10} (\frac{p _{rms}}{p _{ref}})$ , with $p_{ref} = 20 \times 1 0^{- 6}$ Pa. SPL (inverse): $p_{rms} = p_{ref} \times 1 0^{L /20}$ . Combining bands/sources (energy sum): $L_{tot} = 10 lo g_{10} (\sum_{i} 1 0^{L_{i} /10})$ , because each term $1 0^{L_{i} /10} = I_{i} / I_{ref}$ is that source's ground-level intensity. Here $lo g_{10} (x)$ asks "10 to what power gives $x$ ?"; $1 0^{a}$ undoes it. dB is a ratio on a log ruler, never an absolute pressure.

The scenario matrix

Every problem this topic throws at you falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it tricky	Example
A. Forward dB → pressure → force	plug the inverse formula	Ex 1
B. Backward pressure → dB	apply the log	Ex 2
C. Equal sources add	log of a sum, both equal	Ex 3
D. Unequal sources add	one term dominates	Ex 4
E. Degenerate: one source is ZERO / silence	$lo g_{10} (0) = - \infty$	Ex 5
F. Limiting: many equal bands	$+ 10 lo g_{10} N$ growth	Ex 6
G. Real-world word problem	choose the right tool	Ex 7
H. Exam twist: subtraction / negative dB	"remove" a source, negative $Δ$	Ex 8

Figure — Acoustic loads — SPL, octave band analysis

Figure 1 — dB levels as heights on a log ladder (left) vs. their intensities as ground-level bars that may be summed (right).

Example 1 — Cell A: dB → pressure → force

Worked example Fairing SPL to panel force

Given: A fairing interior measures 148 dB. A flat antenna of area $A = 1.5 m^{2}$ faces the wave. Find: RMS pressure, and the RMS force felt by the antenna.

Forecast: Guess — is the pressure a few Pa, a few hundred, or thousands? Write your guess.

Step 1. Invert the SPL formula: $p_{rms} = p_{ref} \times 1 0^{L /20}$ . Why this step? We are given a dB number and want a physical pressure, so we must undo the log — that is exactly what $1 0^{(\cdot)}$ does. $p_{rms} = 20 \times 1 0^{- 6} \times 1 0^{148/20} = 20 \times 1 0^{- 6} \times 1 0^{7.4}$

Step 2. Evaluate $1 0^{7.4} = 1 0^{7} \times 1 0^{0.4} = 1 0^{7} \times 2.512 = 2.512 \times 1 0^{7}$ . Why this step? Splitting the exponent into a whole part and a decimal part lets you read the magnitude by eye ( $1 0^{7}$ ) and the mantissa separately ( $1 0^{0.4} \approx 2.51$ ). $p_{rms} = 20 \times 1 0^{- 6} \times 2.512 \times 1 0^{7} = 502.4 Pa$

Step 3. Force $= p_{rms} \times A = 502.4 \times 1.5 = 753.6 N$ . Why this step? Pressure is force per area; multiplying by the exposed area gives the actual push. This is the oscillating force that fatigues the structure.

Verify: Units: $Pa \times m^{2} = \frac{N}{m ^{2}} \times m^{2} = N$ ✓. Sanity: 148 dB is 2 dB below the 150 dB / 632 Pa case in the parent note; 2 dB less ≈ factor $1 0^{- 2/20} = 0.794$ , and $632 \times 0.794 \approx 502$ Pa ✓.

Example 2 — Cell B: pressure → dB

Example 3 — Cell C: two EQUAL uncorrelated sources

Example 4 — Cell D: two UNEQUAL sources

Worked example Loud source plus a quiet one

Given: Source 1 = 150 dB, Source 2 = 138 dB, uncorrelated. Find: Combined SPL.

Forecast: Will the 138 dB source add a lot, a little, or nothing? Guess.

Step 1. Energy sum: $L_{tot} = 10 lo g_{10} (1 0^{150/10} + 1 0^{138/10})$ . Why this step? Only energies (ground values $I / I_{ref}$ ) may be added — climb down, add, climb up.

Step 2. $1 0^{15.0} = 1.000 \times 1 0^{15}$ , $1 0^{13.8} = 6.310 \times 1 0^{13}$ . Sum $= 1.0631 \times 1 0^{15}$ . Why keep both? To see the dominance: the quiet source is only $6.3%$ of the loud one's energy.

Step 3. $L_{tot} = 10 lo g_{10} (1.0631 \times 1 0^{15}) = 10 (15.0266) = 150.27 dB$ . Why this step? Having summed the energies on the ground, we climb back up the log ladder with $10 lo g_{10}$ to report the answer in dB — the final leg of the down–add–up cycle.

Verify: Difference is 12 dB. Rule of thumb: a source 10 dB down adds ≈ 0.4 dB; 12 dB down should add slightly less. We got $+ 0.27$ dB ✓.

Figure 2 plots exactly this rule: the horizontal axis is the gap between the two sources, the curve is the dB you add on top of the louder one. Trace the orange dot at a 12 dB gap down to $+ 0.27$ dB (this example) and the green dot at a 0 dB gap up to $+ 3.0$ dB (Ex 3). The steady fall of the curve is why a distant, quieter source barely matters.

Figure 2 — dB added on top of the louder source as a function of the level gap between two uncorrelated sources.

Example 5 — Cell E: degenerate input, one source is SILENCE

Example 6 — Cell F: limiting behaviour, many equal bands

Worked example Ten equal octave bands

Given: A spectrum has 10 octave bands, each at 130 dB. Find: The OASPL (Overall SPL), and the general rule.

Forecast: Is it near 130, near 140, or well above 200 dB? Guess.

Step 1. $N$ equal bands ⇒ $L_{OASPL} = L + 10 lo g_{10} N = 130 + 10 lo g_{10} 10$ . Why this step? Non-overlapping bands hold independent energy $I$ , so their ground-values just sum — identical maths to equal sources.

Step 2. $10 lo g_{10} 10 = 10$ dB exactly. $L_{OASPL} = 130 + 10 = 140 dB$

Step 3 — the limit. In general $N$ equal bands add $10 lo g_{10} N$ dB. This grows logarithmically: $N = 2 \to + 3$ , $N = 10 \to + 10$ , $N = 100 \to + 20$ . Doubling $N$ always adds exactly $+ 3$ dB, forever. Why does it never blow up fast? Because the log ladder compresses even huge $N$ — you cannot get a runaway dB total by stacking bands.

Verify: $10 lo g_{10} (10 \times 1 0^{13}) = 10 lo g_{10} (1 0^{14}) = 140$ dB ✓.

Figure 3 draws this ceiling: the curve is $10 lo g_{10} N$ against $N$ on a log axis, with dots at $N = 2, 10, 100$ landing on $+ 3, + 10, + 20$ dB. Notice how each tenfold jump in the number of bands buys only another flat $+ 10$ dB — that visual flattening is the whole safety argument.

Figure 3 — the added level $10 lo g_{10} N$ grows only logarithmically with the number of equal bands $N$ .

Example 7 — Cell G: real-world word problem

Worked example Will the solar array survive?

Given: A solar array (area $A = 3 m^{2}$ ) sits in a fairing whose octave-band SPLs are: 125 Hz → 134, 250 Hz → 139, 500 Hz → 141, 1000 Hz → 137 dB. The array's structural qualification limit is a peak oscillating force of 3000 N. (Take peak $\approx 2 p_{rms}$ , and use the OASPL as the effective broadband pressure.) Find: Does the array pass?

Forecast: Pass with margin, just pass, or fail? Guess.

Step 1. OASPL $= 10 lo g_{10} (1 0^{13.4} + 1 0^{13.9} + 1 0^{14.1} + 1 0^{13.7})$ . Why this step? We must combine the bands into a single broadband level before converting to pressure. Terms: $2.512 \times 1 0^{13}, 7.943 \times 1 0^{13}, 1.259 \times 1 0^{14}, 5.012 \times 1 0^{13}$ . Sum $= 3.0165 \times 1 0^{14}$ . $OASPL = 10 lo g_{10} (3.0165 \times 1 0^{14}) = 144.80 dB$

Step 2. Convert to $p_{rms}$ : $p_{rms} = 20 \times 1 0^{- 6} \times 1 0^{144.80/20} = 20 \times 1 0^{- 6} \times 1 0^{7.2398}$ . Why this step? Force lives in the physical world, so leave dB behind. $1 0^{7.2398} = 1.738 \times 1 0^{7}$ , giving $p_{rms} = 347.5 Pa$ .

Step 3. Peak pressure $\approx 2 \times 347.5 = 491.4 Pa$ . Peak force $= 491.4 \times 3 = 1474.3 N$ . Why $2$ ? A sinusoid's peak is $2$ times its rms — we design against the peak, not the average.

Step 4. Compare: $1474 N < 3000 N$ ⇒ PASS, with a margin of $3000/1474 = 2.03 \times$ . Why this step? Qualification is a pass/fail gate: the predicted peak load must sit below the tested limit, and the ratio of the two is the safety margin engineers report to sign off the hardware.

Verify: Units $Pa \cdot m^{2} = N$ ✓. Cross-check the dominant band: the 500 Hz band alone is 141 dB; OASPL of 144.8 dB is only $+ 3.8$ dB above it — consistent because one band carries ~42% of the energy ✓. This is the same energy-summing logic as random vibration PSD integration.

Example 8 — Cell H: exam twist, source subtraction (negative dB)

Worked example Removing background noise

Given: A test rig measures 147 dB with the shaker running. With the shaker off (background only), the mic reads 141 dB. An examiner asks: what is the shaker-only SPL? Find: The shaker's contribution alone.

Forecast: Is the shaker-only level 6 dB, 141 dB, or ~146 dB? Guess.

Step 1. Energies subtract on the ground: $I_{shaker} = I_{tot} - I_{bg}$ . Why this step? Total energy $I$ = shaker + background (uncorrelated), so isolating the shaker means removing the background energy — a subtraction, the mirror image of Ex 4.

Step 2. $L_{shaker} = 10 lo g_{10} (1 0^{147/10} - 1 0^{141/10})$ . $1 0^{14.7} = 5.012 \times 1 0^{14}$ , $1 0^{14.1} = 1.259 \times 1 0^{14}$ . Difference $= 3.753 \times 1 0^{14}$ . Why can this be done? Only because $I_{tot} > I_{bg}$ ; if background ever exceeded total, the measurement is impossible (you'd take a log of a negative — the exam's hidden degeneracy check). $L_{shaker} = 10 lo g_{10} (3.753 \times 1 0^{14}) = 145.74 dB$

Step 3. Interpret the "negative dB" idea: the correction is $145.74 - 147 = - 1.26$ dB. Because the background is only 6 dB down, it still steals a non-trivial $1.26$ dB from the reading. Why care? If background is <10 dB below total, you must subtract it or you overstate the shaker.

Verify: Re-add: $10 lo g_{10} (1 0^{14.574} + 1 0^{14.1}) = 10 lo g_{10} (3.753 \times 1 0^{14} + 1.259 \times 1 0^{14}) = 10 lo g_{10} (5.012 \times 1 0^{14}) = 147.0$ dB ✓. We recovered the original total.

Recall check

Recall Why do dB never add arithmetically?

Because dB is a height on a log ladder; only the ground-level energies ( $1 0^{L /10} = I / I_{ref}$ ) may be summed. Add on the ground, then climb back up with $10 lo g_{10}$ to return to dB. ::: Log-scale ⇒ convert to intensity, sum intensities, convert back.

Doubling equal sources adds how many dB?

Exactly

10 lo g_{10} 2 \approx 3.01

dB.

Ten equal bands raise the level by how much?

10 lo g_{10} 10 = 10

dB.

What is the SPL of true silence?

- \infty

dB (not 0 dB).

To remove a known background from a measurement, you…

subtract intensities, then convert back to dB.

See also: 3.6.10 Structural Natural Frequencies and Mode Shapes · 3.6.13 Shock Loads and SRS · 3.6.14 Combined Environmental Testing · 2.5.8 Acoustic Impedance and Transmission.