Random vibration — PSD, RMS acceleration: Visual walkthrough

Before any symbol appears, one promise: every letter is a picture first. Let us begin.

Step 1 — The wiggle: what a random signal even is

WHAT. We record acceleration over time. Call the recorded value at time $t$ the letter $x (t)$ — read it " $x$ of $t$ ", meaning "the number $x$ that depends on the moment $t$ ". Here $x$ is acceleration measured in g (multiples of Earth's gravity, $1 g \approx 9.81 m/s^{2}$ ), and $t$ is time in seconds.

WHY. Unlike a musical note (a smooth repeating wave), a launch signal jitters with no pattern you could predict a second ahead. But if you look at its spread — how far it typically swings from zero — that spread is stable. That stability is the only thing we can hold onto.

PICTURE. In the figure, the blue line is $x (t)$ : a jagged trace bouncing above and below the gray zero line. The orange band marks its typical swing. Notice the line never repeats, yet the width of the orange band stays the same across the whole record. That constant width is what "stationary random process" means.

Figure — Random vibration — PSD, RMS acceleration

Step 2 — Splitting the wiggle into pure tones

WHAT. We claim the jagged blue line is secretly a sum of smooth waves of many frequencies added together. A pure tone of frequency $f$ (in hertz, $Hz$ = cycles per second) is a wave that repeats $f$ times each second.

WHY this tool — the Fourier transform. We need a way to ask "how much of frequency $f$ is inside this mess?" The Fourier transform is exactly that measuring device. It answers one specific question: for each frequency $f$ , what is the size and timing of the wave at that frequency hiding inside $x (t)$ ? We choose it, and not simple averaging, because averaging destroys frequency information while the Fourier transform preserves it.

$X (f) = \int_{- \infty}^{\infty} the wiggle x (t) a probe tone at freq f e^{- i 2 π f t} d t$

$X (f)$ — the amount of frequency $f$ present (a complex number: its size is strength, its angle is timing).
$e^{- i 2 π f t}$ — the "probe": a spinning unit arrow that turns $f$ times per second. Multiplying by it and adding up (the $\int d t$ ) asks "does the wiggle spin in step with this tone?"
$\int_{- \infty}^{\infty} \dots d t$ — "add up over all time".

PICTURE. The figure shows the same jagged blue line on the left, and on the right three smooth colored sine waves (low, middle, high frequency) that add up to it. The arrow labelled "Fourier transform" points from the mess to its ingredients.

Prerequisite: this decomposition is the same idea used in Frequency Response Function (FRF) and Modal Analysis — split any motion into frequency ingredients.

Step 3 — From "amount" to "power": squaring it

WHAT. Take the size of $X (f)$ and square it: $∣ X (f) ∣^{2}$ . The bars $∣ \cdot ∣$ mean "length of the arrow" (ignore its timing/angle); squaring turns that length into a power-like quantity.

WHY square? Power in physics goes as amplitude squared (like kinetic energy $\propto$ velocity²). We do not care whether a wave pushes up or down — only how strongly it shakes. Squaring throws away sign and keeps strength. It also matches how "spread" is measured (variance is a mean of squares).

PICTURE. The figure plots $∣ X (f) ∣^{2}$ against $f$ : a bumpy curve, high where strong frequencies live, near zero elsewhere. Each vertical bar is the squared strength of one frequency.

Step 4 — Fixing the length: divide by $T$ , then average

WHAT. Our recording only lasted a finite time $T$ (seconds). Write $∣ X (f, T) ∣^{2}$ to remember the record length. We form $\frac{∣ X ( f , T ) ∣ ^{2}}{T} .$

WHY divide by $T$ ? Record longer and you accumulate more cycles, so $∣ X (f, T) ∣^{2}$ grows just because you watched longer — not because the physics got stronger. Dividing by $T$ converts a growing total into a stable rate: power per second of record.

WHY average, and take $T \to \infty$ ? One recording is one throw of a random dice. Different launches give slightly different curves. We average over many (the expected value $E [\cdot]$ , read "the mean over all possible recordings") and let $T \to \infty$ so the rate settles to its true steady value.

$G_{xx} (f) = lim_{T \to \infty} \frac{1}{T} E [∣ X (f, T) ∣^{2}]$

$G_{xx} (f)$ — the PSD: settled power per hertz, units $g^{2} / Hz$ .
$\frac{1}{T}$ — turns total into a rate (per second of record).
$E [\cdot]$ — average over many recordings (kills random luck).
$lim_{T \to \infty}$ — let the record grow forever so the rate stops wobbling.

PICTURE. The figure shows three thin jittery curves (three recordings) and one thick smooth curve — their average — which is $G_{xx} (f)$ . The jitter of individual runs washes out into a clean shape.

Step 5 — Area under the PSD = total mean-square shaking

WHAT. To get all the shaking power, add up every thin frequency slice. Each slice of width $df$ at frequency $f$ holds $G_{xx} (f) df$ of mean-square power. Summing slices is integration: $\overline{a^{2}} = \int_{f_{1}}^{f_{2}} G_{xx} (f) df .$

WHY integration — and why this tool? We have a quantity spread continuously over frequency (power density), and we want its total. Adding a continuum of tiny contributions is exactly what an integral does; a plain sum would only work if we had a finite list of frequencies. The bar over $a^{2}$ , $\overline{a^{2}}$ , means "mean square" — the average of acceleration squared.

$G_{xx} (f) df$ — power in one thin slice (height $\times$ width).
$\int_{f_{1}}^{f_{2}}$ — glue all slices from $f_{1}$ to $f_{2}$ together.
$\overline{a^{2}}$ — total mean-square acceleration, units $g^{2}$ .

PICTURE. The figure shades the whole area under a PSD curve between $f_{1}$ and $f_{2}$ . One thin slice is highlighted in orange with its height $G_{xx} (f)$ and width $df$ labelled, showing it is a tiny rectangle. The shaded total is $\overline{a^{2}}$ .

Step 6 — The square root: back to honest units

WHAT. The area is in $g^{2}$ . Take its square root to return to $g$ : $a_{RMS} = \overline{a^{2}} = \int_{f_{1}}^{f_{2}} G_{xx} (f) df .$

WHY. "RMS" = Root of the Mean of the Squares. Squaring (Step 3) and averaging (Step 5) left us in $g^{2}$ ; the root undoes the squaring so the answer is a real, feelable acceleration in $g$ — a "typical" shake magnitude.

PICTURE. The figure is a simple bar: a tall $g^{2}$ box (the area) with an arrow "" collapsing it into a shorter $g$ bar (the RMS). The words " $g^{2} \to g$ " sit on the arrow.

Step 7 — Edge and degenerate cases

WHAT & WHY. A formula you can only use on friendly inputs is a trap. Walk the corners:

Zero PSD everywhere ( $G_{xx} = 0$ ): area $= 0$ , so $a_{RMS} = 0$ . No shaking in, no shaking out. Correct.
A spike (single tone): if all energy sits at one exact frequency, the density $G_{xx}$ there is infinite while its width is zero — a PSD cannot describe a pure tone. That is why single tones belong to Sine Vibration Testing, not random analysis.
Very narrow band: if $f_{2} \to f_{1}$ , area $\to 0$ , so $a_{RMS} \to 0$ . A whisker-thin band carries almost no power.
Piecewise input: split the area into pieces and add — nothing new, just Step 5 applied per segment. Region areas add as $g^{2}$ , and only the final total gets rooted.

PICTURE. The figure shows the launch spectrum of parent Example 2 as three colored rectangles ( $0.8$ , $16.0$ , $15.0 g^{2}$ ). Their heights are the PSD levels, their widths the bands, and a bracket shows the areas summing to $31.8 g^{2}$ before one final root gives $5.64 g$ .

Step 8 — What resonance does to the picture

WHAT. Real structures do not pass shaking through untouched. A part with natural frequency $f_{n}$ (its favourite ringing frequency) and damping ratio $ζ$ (how quickly it settles) amplifies the input near $f_{n}$ . The response PSD is $G_{\overset{y}{¨} \overset{y}{¨}} (f) = ∣ T (f) ∣^{2} G_{\overset{x}{¨} \overset{x}{¨}} (f),$ where $∣ T (f) ∣^{2}$ is the transmissibility — the frequency-by-frequency gain.

WHY it changes RMS. Even a flat input becomes a peaked output: the area under the response curve is dominated by a tall spike at $f_{n}$ . So $a_{RMS}$ of the response is larger than of the input, concentrated near resonance.

PICTURE. The figure overlays a flat blue input PSD and the orange response PSD with a sharp peak at $f_{n} = 200 Hz$ . At resonance $∣ T ∣^{2} = 1 + 1/ (4 ζ^{2}) = 101$ for $ζ = 0.05$ , so the peak response density is $0.02 \times 101 = 2.02 g^{2} / Hz$ — a red dot marks it.

See Transmissibility, Structural Damping, and Fatigue Analysis for where this spike does its damage.

The one-picture summary

Everything on one canvas: wiggle → tones → square → average (PSD) → area → root → RMS, with resonance reshaping the curve.

Recall Feynman retelling — say it plain

I record a shaky line. I can't predict it, but its spread is steady. I ask "how much of each pitch is in there?" — that's the Fourier transform. I square the answer to get strength (I don't care up or down), divide by how long I watched so it's a fair rate, and average over many tries so luck cancels. That gives me the PSD: a curve of shaking-power at every frequency. To get one number, I find the area under that curve — total mean-square shaking in $g^{2}$ — and take a square root to get back to plain $g$ . That number, $a_{RMS}$ , is my "typical shake". If the part likes to ring at some frequency, it swells the curve there, so its own RMS is bigger than what came in. That's the whole story, from wiggle to one honest number.

Recall Quick self-test

Why divide $∣ X (f, T) ∣^{2}$ by $T$ ? ::: Longer records inflate the total just by watching longer; dividing gives a stable rate (power per second). Why square before integrating? ::: To measure strength regardless of sign, and to match how power/variance is defined. Why one final square root, not per segment? ::: Powers ( $g^{2}$ ) add; the root restores honest $g$ units and must be applied to the total only. Flat input, resonant part — is output RMS bigger or smaller? ::: Bigger; $∣ T ∣^{2}$ spikes at $f_{n}$ and swells the area under the response PSD.