Random vibration — PSD, RMS acceleration: Worked examples

Before the examples, three tiny building blocks we'll reuse everywhere.

The figure below is your master key: it shows the three shapes a PSD segment can take, and the area recipe each demands. Come back to it before every example — each worked problem is just one of these shapes.

Figure — Random vibration — PSD, RMS acceleration

Trace the three coloured regions with your eye. The magenta rectangle (flat PSD) has area = height $\times$ width — that is Examples 1, 2, 4, 7-region-3, 8. The violet ramp (a straight line on log–log paper) bulges into a curved area needing the power-law integral — that is Examples 3 and 7-region-2. The orange spike (resonance) has almost all its area packed into a narrow band — that is Examples 5 and 6. Three shapes → three area recipes → the whole topic.

The scenario matrix

Cell	Case class	What makes it tricky	Example
A	Flat PSD, single band	baseline area = $G Δ f$	Ex 1
B	Piecewise-flat (bands stacked)	sum of rectangles	Ex 2
C	Sloping segment (ramp, dB/oct)	log–log ramp needs a special integral	Ex 3
D	Degenerate: zero-width band ( $f_{1} = f_{2}$ )	area collapses to zero	Ex 4
E	Notched / dip spectrum	subtract, don't add	Ex 4
F	Resonance amplification (transmissibility)	flat input $\to$ peaked output	Ex 5
G	Limiting case: $ζ \to 0$ (undamped)	RMS $\to \infty$ , why testing needs damping	Ex 6
H	Real-world word problem (launch qual)	pick $f_{1}, f_{2}$ , convert to $g$	Ex 7
I	Exam twist: work backwards (given RMS, find level)	invert the area formula	Ex 8

Nine cells, eight examples (some hit two cells). Each is labelled.

Example 1 — Cell A: Flat PSD, single band

Forecast: Guess the answer before reading on. Rectangle area, then square-root. Is it closer to $2 g$ or $20 g$ ?

Write the area as a rectangle. $\overline{a^{2}} = G \times Δ f = G \times (f_{2} - f_{1}) = 0.04 \times (2000 - 20)$ Why this step? A flat PSD is a constant height, so the "area under the curve" is just height $\times$ width, where the width is the bandwidth $Δ f = f_{2} - f_{1}$ — no integral needed.
Compute the area. $\overline{a^{2}} = 0.04 \times 1980 = 79.2 g^{2}$ Why this step? This is the mean-square (units $g^{2}$ ). It is not the answer yet — the units are wrong for a "typical acceleration."
Square-root to get RMS. $a_{RMS} = 79.2 \approx 8.90 g$ Why this step? RMS restores the units to $g$ , giving a number you can compare to a steady acceleration.

Verify: Units: $(g^{2} / Hz) \times Hz = g^{2}$ ; $g^{2} = g$ . ✓ Sanity: $8.9 g$ over a very wide band is typical of real launch bracketry.

Example 2 — Cell B: Piecewise-flat spectrum

Forecast: Three rectangles. Which one holds the most area — the wide low bands, or the tall middle band?

One rectangle per band. $A_{1} = 0.01 (100 - 20) = 0.8, A_{2} = 0.04 (500 - 100) = 16.0, A_{3} = 0.01 (2000 - 500) = 15.0$ (all in $g^{2}$ ) Why this step? Areas simply add — mean-square is additive across disjoint frequency bands because energy in separate bands doesn't interfere.
Sum the areas. $\overline{a^{2}} = 0.8 + 16.0 + 15.0 = 31.8 g^{2}$ Why this step? Total mean-square = total area under the whole (stepped) curve.
Square-root. $a_{RMS} = 31.8 \approx 5.64 g$ Why this step? The RMS is the square root of the mean-square; it restores the units to $g$ so the answer reads as a typical acceleration magnitude rather than a power.

Verify: The middle band alone ( $16 g^{2}$ ) is half the total energy despite being narrow — because its height is 4× the others. This is why the 100–500 Hz "bump" matters for Fatigue Analysis. ✓

Example 3 — Cell C: Sloping segment (dB/octave ramp)

Forecast: A ramp on log–log paper is a curve on linear paper. Will the answer be bigger or smaller than a plain trapezoid $\frac{1}{2} (G_{1} + G_{2}) Δ f$ ?

The violet region in the figure below is exactly this shape — a log–log straight line that, seen on ordinary linear axes, bulges above the naive straight chord (dashed magenta). We compute the true bulged area.

Find the exponent $n$ of the power law. A straight line on log–log axes means $G (f) = G_{1} (f / f_{1})^{n}$ . Reading the two endpoints, the exponent is $n = \frac{l n ( G _{2} / G _{1} )}{l n ( f _{2} / f _{1} )} = \frac{l n ( 0.08/0.01 )}{l n ( 400/100 )} = \frac{l n 8}{l n 4} = \frac{2.0794}{1.3863} = 1.5$ Why this step? On log–log paper the slope is $\frac{Δ ( lo g G )}{Δ ( lo g f )}$ , which is exactly the exponent $n$ . We need $n$ to pick the right area integral. (This is not $lo g_{2} 8 = 3$ ; that would be the exponent only if the frequency ratio were also a single doubling, which it isn't — the frequency spans 2 octaves.)
Use the log–log ramp area formula. $A = \int_{f_{1}}^{f_{2}} G_{1} (\frac{f}{f _{1}})^{n} df = \frac{G _{1} f _{1}}{n + 1} [(\frac{f _{2}}{f _{1}})^{n + 1} - 1]$ Why this step? A power law $f^{n}$ integrates by the power rule (valid because $n = 1.5 \neq = - 1$ ); factoring $G_{1} f_{1} / (n + 1)$ keeps it tidy.
Plug in $n = 1.5$ , $G_{1} = 0.01$ , $f_{1} = 100$ , $f_{2} / f_{1} = 4$ . $A = \frac{0.01 \times 100}{2.5} [4^{2.5} - 1] = \frac{1}{2.5} \times (32 - 1) = \frac{31}{2.5} = 12.4 g^{2}$ Why this step? $4^{2.5} = 4^{2} \cdot 4^{0.5} = 16 \times 2 = 32$ ; minus 1 gives 31; the prefactor $G_{1} f_{1} / (n + 1) = 1/2.5 = 0.4$ .
(Optional) RMS of just this segment: $12.4 \approx 3.52 g$ .

Verify: The naive linear trapezoid gives $\frac{1}{2} (0.01 + 0.08) (300) = 13.5 g^{2}$ . The correct power-law area $12.4 g^{2}$ is close but smaller here (for a gentle $n = 1.5$ ramp the curve bows only slightly below the chord). ✓ Units: $(g^{2} / Hz) \cdot Hz = g^{2}$ . ✓

Example 4 — Cells D & E: Zero-width band and a notch

Forecast: (a) is a trap — what is the area of a rectangle with zero width? (b) Does the notch help much?

Part (a): degenerate band. $A = G \times (f_{2} - f_{1}) = G \times 0 = 0 g^{2}$ Why this step? A single frequency has zero bandwidth ( $Δ f = 0$ ). A true pure tone would be a Dirac spike (infinite height, finite area) — that belongs to Sine Vibration Testing, not random vibration. Random PSD always needs a finite width to carry energy.
Part (b): full area if there were no notch. $A_{full} = 0.05 \times (2000 - 20) = 0.05 \times 1980 = 99.0 g^{2}$ Why this step? Start from the un-notched rectangle, then correct for the dip.
Subtract what the notch removes. The notch replaces $0.05$ with $0.005$ over a $40$ Hz window, i.e. removes $(0.05 - 0.005) \times 40$ . $Δ A = (0.05 - 0.005) \times (220 - 180) = 0.045 \times 40 = 1.8 g^{2}$ Why this step? A dip subtracts area. Compute the missing rectangle and take it away — do not re-add it.
Net area and RMS. $\overline{a^{2}} = 99.0 - 1.8 = 97.2 g^{2}, a_{RMS} = 97.2 \approx 9.86 g$ Why this step? Once the corrected total area (mean-square) is known, taking its square root gives the RMS in $g$ — the single final operation that turns a $g^{2}$ power into a physical acceleration.

Verify: Without the notch, $a_{RMS} = 99 = 9.95 g$ . The notch barely helps (9.95 → 9.86 g) because it's only 40 Hz wide out of 1980 — a good reminder that a narrow Transmissibility dip does little for broadband RMS, even if it saves a specific resonant part. ✓

Example 5 — Cell F: Resonance amplification (transmissibility)

Before this example, two pieces of notation to earn.

Formula Transmissibility

$∣ T (f) ∣^{2}$ — the full frequency-dependent form For a single-degree-of-freedom (SDOF) oscillator of natural frequency $f_{n}$ and damping ratio $ζ$ , driven through its base, write the frequency ratio $r \equiv f / f_{n}$ . The Frequency Response Function (FRF) that maps input base-acceleration PSD to output absolute-acceleration PSD is the transmissibility: $∣ T (f) ∣^{2} = \frac{1 + ( 2 ζ r ) ^{2}}{[ 1 - r ^{2} ] ^{2} + ( 2 ζ r ) ^{2}}, r = \frac{f}{f _{n}}$ Read it off the three regimes:

Low frequency $r \to 0$ : numerator $\to 1$ , denominator $\to 1$ , so $∣ T ∣^{2} \to 1$ — the mass just rides along with the base (no amplification).
At resonance $r = 1$ : the $[1 - r^{2}]^{2}$ term vanishes, leaving $∣ T ∣^{2} = \frac{1 + ( 2 ζ ) ^{2}}{( 2 ζ ) ^{2}} = 1 + \frac{1}{4 ζ ^{2}}$ — a tall peak controlled by $ζ$ .
High frequency $r \to \infty$ : numerator $\sim (2 ζ r)^{2}$ , denominator $\sim r^{4}$ , so $∣ T ∣^{2} \to (2 ζ / r)^{2} \to 0$ — the mass is isolated, its motion dies away.

The response PSD is then, band by band, $G_{\overset{y}{¨} \overset{y}{¨}} (f) = ∣ T (f) ∣^{2} G_{\overset{x}{¨} \overset{x}{¨}} (f)$ .

Forecast: The input is flat and calm. How tall does the output spike get at $f = f_{n}$ ?

The figure shows the calm flat magenta input line and the violet response that spikes 100-fold at $f_{n}$ — the orange dotted line marks the natural frequency. Almost all the response energy lives under that narrow violet peak.

Transmissibility at resonance. Set $r = 1$ in the full form above; the $[1 - r^{2}]^{2}$ term vanishes: $∣ T (f_{n}) ∣^{2} = \frac{1 + ( 2 ζ ) ^{2}}{( 2 ζ ) ^{2}} = 1 + \frac{1}{4 ζ ^{2}} = 1 + \frac{1}{4 ( 0.05 ) ^{2}} = 101$ Why this step? The FRF denominator loses its big term at resonance, leaving only the small $(2 ζ)^{2}$ — so the ratio blows up. This is the resonance peak.
Peak response PSD. $G_{\overset{y}{¨} \overset{y}{¨}} (f_{n}) = ∣ T (f_{n}) ∣^{2} G_{\overset{x}{¨} \overset{x}{¨}} = 101 \times 0.02 = 2.02 g^{2} / Hz$ Why this step? Response PSD = transmissibility $^{2}$ × input PSD, band by band. The output spikes 101× above the flat input at $f_{n}$ .
Response RMS via Miles' equation (area under the peak). The classic Modal Analysis result for a flat input through one mode is $a_{RMS,resp} = \frac{π}{2} f_{n} Q G_{\overset{x}{¨} \overset{x}{¨}}, Q = \frac{1}{2 ζ} = 10$ $= \frac{π}{2} \times 200 \times 10 \times 0.02 = 62.83 \approx 7.93 g$ Why this step? The response area is dominated by the narrow resonant peak; Miles' equation is the closed-form area of that peak (bandwidth $\sim \frac{π}{2} f_{n} / Q$ times peak height). It saves integrating the whole $∣ T ∣^{2}$ curve.

Verify: Peak PSD $2.02 g^{2} / Hz$ vs input $0.02$ → exactly 101× ✓. Miles' $g$ : input RMS over the band would be tiny, but resonance pumps it up to $\approx 7.9 g$ . The Structural Damping $ζ$ controls both the peak (101) and the RMS ( $\propto Q$ ). ✓

Example 6 — Cell G: Limiting case $ζ \to 0$ (undamped)

Forecast: With no friction, does the answer stay finite or run off to infinity?

Peak transmissibility limit. $∣ T (f_{n}) ∣^{2} = 1 + \frac{1}{4 ζ ^{2}} ζ \to 0 \infty$ Why this step? Zero damping means nothing removes energy at resonance, so the amplification grows without bound as $ζ \to 0$ .
Response RMS limit. $a_{RMS,resp} = \frac{π}{2} f_{n} Q G_{\overset{x}{¨} \overset{x}{¨}}, Q = \frac{1}{2 ζ} ζ \to 0 \infty$ Why this step? Since the quality factor $Q = 1/ (2 ζ)$ appears inside the Miles square root, and $Q \to \infty$ as $ζ \to 0$ , the RMS $\propto 1/ ζ$ also diverges. Physically this says: with no energy loss, each resonant cycle adds to the last and the response builds without limit — the undamped model predicts an infinite RMS, which no real structure can reach.
Physical resolution. Why this step? Real structures always carry some Structural Damping ( $ζ$ typically $0.01$ – $0.05$ ), and nonlinear effects (joint friction, material hysteresis, yielding) cap the response before it runs away. The divergent limit is therefore a warning, not a prediction: it tells us damping is exactly what keeps random-vibration response finite, which is why every qualification test is run with a realistic, non-zero $ζ$ and never assumes an undamped mode.

Verify: Plug a small $ζ = 0.01$ : $∣ T ∣^{2} = 1 + 1/ (4 \cdot 1 0^{- 4}) = 2501$ ; $Q = 50$ ; RMS $= \frac{π}{2} \cdot 200 \cdot 50 \cdot 0.02 = 314.2 \approx 17.7 g$ . As $ζ$ shrinks 5× (0.05→0.01), RMS grows by $5 \approx 2.24$ × (7.93→17.7). ✓ Confirms $\propto 1/ ζ$ and that the trend heads to infinity.

Example 7 — Cell H: Real-world launch qualification

Forecast: Three shapes — rectangle, ramp, rectangle. Predict which contributes most.

Region 1 (flat, 20–80 Hz). $A_{1} = 0.026 \times (80 - 20) = 0.026 \times 60 = 1.56 g^{2}$ Why this step? Rectangle rule for a flat band.
Region 2 (log–log ramp, 80–350 Hz). Get the exponent from the two endpoints exactly as in Example 3: $n = \frac{l n ( 0.11375/0.026 )}{l n ( 350/80 )} = \frac{l n 4.375}{l n 4.375} = 1$ so $G (f) = 0.026 (f /80)^{1}$ . Apply the ramp area formula with $n = 1$ : $A_{2} = \frac{G _{1} f _{1}}{n + 1} [(f_{2} / f_{1})^{n + 1} - 1] = \frac{0.026 \times 80}{2} [(4.375)^{2} - 1]$ $= 1.04 \times (19.1406 - 1) = 1.04 \times 18.1406 = 18.866 g^{2}$ Why this step? A ramp is a power law $f^{n}$ ; integrate it with the $(n + 1)$ rule, not a trapezoid. Here the numbers happen to give a clean $n = 1$ (level rises in the same ratio as frequency).
Region 3 (flat, 350–2000 Hz). $A_{3} = 0.16 \times (2000 - 350) = 0.16 \times 1650 = 264.0 g^{2}$ Why this step? Rectangle rule again; note this tall, wide band dominates.
Total and RMS. $\overline{a^{2}} = 1.56 + 18.866 + 264.0 = 284.43 g^{2}, a_{RMS} = 284.43 \approx 16.9 g$ Why this step? Sum the three sub-areas into the total mean-square, then square-root once to get the overall RMS in $g$ .

Verify: Region 3 holds $264/284 \approx 93%$ of the energy — makes sense, it's the highest level over the widest span. A realistic small-sat qual RMS of $\sim 17 g$ . ✓ Units all $g^{2}$ before the root. ✓ This feeds directly into Fatigue Analysis and Acoustic Loading margins.

Example 8 — Cell I: Exam twist (work backwards)

Forecast: We know the answer (RMS) and must recover the input (level). Invert the area formula.

Write the forward relation. $a_{RMS}^{2} = G Δ f = G (f_{2} - f_{1}) \Rightarrow 1 0^{2} = G \times (2000 - 20)$ Why this step? For a flat spectrum, mean-square = level × bandwidth. We just solve for the unknown $G$ .
Solve for $G$ . $G = \frac{100}{1980} = 0.05051 g^{2} / Hz$ Why this step? Dividing the required mean-square ( $100 g^{2}$ ) by the bandwidth ( $1980$ Hz) gives the per-Hz level.
Round for the test spec. $G \approx 0.0505 g^{2} / Hz$ Why this step? Test consoles want the PSD level, not the RMS; this is the number you type in.

Verify: Forward-check: $0.05051 \times 1980 = 100.0 = 10.0 g$ ✓. This inversion is exactly what a Sine Vibration Testing engineer does when translating a "10 g RMS" requirement into a shaker command. ✓

Recall Self-test (reveal after guessing)

What is $Δ f$ ? ::: The bandwidth $f_{2} - f_{1}$ , the width in Hz of the band you integrate over. What does the overline in $\overline{a^{2}}$ mean? ::: The average (ensemble / time mean) of the squared acceleration — the mean-square, in $g^{2}$ . Ramp area on log–log uses which exponent? ::: $n = ln (G_{2} / G_{1}) / ln (f_{2} / f_{1})$ , then the $(n + 1)$ power-law area formula — NOT the linear trapezoid. Zero-width band contributes how much mean-square? ::: Zero — random PSD needs finite bandwidth to carry energy. A notch (dip) does what to the area? ::: Subtracts the removed rectangle; never re-add it. Why integrate only over positive frequencies? ::: The one-sided PSD folds the symmetric negative-frequency half onto $f \geq 0$ ; negative $f$ is meaningless in a test spec. Full transmissibility form at resonance $r = 1$ ? ::: $∣ T ∣^{2} = 1 + 1/ (4 ζ^{2})$ ; low- $f$ it $\to 1$ , high- $f$ it $\to 0$ . Response RMS scales with damping how? ::: $\propto 1/ ζ$ (via $Q = 1/2 ζ$ ), diverging as $ζ \to 0$ . Given a target RMS on a flat band, find the level how? ::: $G = a_{RMS}^{2} /Δ f$ .