Random vibration — PSD, RMS acceleration: Question bank

Below, a quick vocabulary anchor so every reveal reads cleanly. Notation note: to avoid symbol clashes, this page writes the record length as $T_{rec}$ (seconds of recording) and reserves $T (f)$ for transmissibility; relative displacement is written $z_{rel}$ so it is never mistaken for the damping ratio $ζ$ .

Recall Symbol refresher (open if any notation feels unfamiliar)

$x (t)$ ::: the raw acceleration signal in time — a wiggly, unpredictable curve.
$X (f, T_{rec})$ ::: the Fourier transform of $x (t)$ computed over a record of finite length $T_{rec}$ (in seconds); it reports how much of frequency $f$ lives in that record. $T_{rec}$ is record length only — never transmissibility.
$G_{xx} (f)$ ::: Power Spectral Density — how much mean-square "stuff" ( $g^{2}$ ) sits in each 1 Hz slice of frequency, so its units are $g^{2} / Hz$ . Assumed one-sided here (defined only for $f \geq 0$ ).
$a_{RMS}$ ::: the square root of the area under the one-sided PSD curve; a single number in $g$ describing overall intensity.
$E [\cdot]$ ::: ensemble average — the average over many independent recordings of the same random environment, as opposed to a time average over one recording.
$r = f / f_{n}$ ::: the dimensionless frequency ratio; $r = 1$ means "at resonance." Terms like $(f / f_{n})^{2}$ are just $r^{2}$ .
$f_{n}$ , $ζ$ , $Q$ ::: natural frequency, damping ratio, and quality factor $Q = 1/ (2 ζ)$ — the peakiness of a resonance.
$z_{rel} = y - x$ , $H_{z}$ ::: relative displacement (mass motion $y$ minus base motion $x$ ) and its frequency-domain filter $∣ H_{z} (f) ∣^{2}$ ; this is what bends the mounts and drives stress. ( $z_{rel}$ is a displacement, $ζ$ is a damping ratio — do not confuse them.)
$T (f)$ ::: transmissibility — output acceleration divided by input base acceleration, frequency by frequency. It is a complex quantity with magnitude $∣ T ∣$ and a phase; random-vibration RMS uses only $∣ T ∣^{2}$ .

Before the reveals, three pictures anchor the whole topic. Refer back to them as you argue each item.

Figure 1 — a PSD is an area, not a height. The flat and piecewise PSDs from the parent note; RMS is $shaded area$ .

Figure — Random vibration — PSD, RMS acceleration

Figure 2 — transmissibility magnitude, all three regimes. Low frequency rides at 1, resonance spikes to $\approx Q$ , and high frequency rolls off — with the damping term keeping it from ever hitting zero.

Figure 3 — the base-excited mass. Base moves $x$ , mass moves $y$ ; the mounts stretch by the relative displacement $z_{rel} = y - x$ , which is what actually causes stress.

True or false — justify

PSD units are g per Hz, so RMS is just PSD times bandwidth

False — PSD units are

g^{2} / Hz

(mean-square), so its integral gives

g^{2}

; you must take a square root at the end to get

g

Doubling the flat PSD level doubles the RMS acceleration

False — RMS scales with the square root of PSD, so doubling the level multiplies RMS by

2 \approx 1.41

, not 2.

A flat PSD means the structure feels equal acceleration at every frequency

True in the sense of equal

g^{2} / Hz

per bin, but the total energy is dominated by the wide high-frequency span because more Hz sit up there (Figure 1) — a flat PSD is not a flat energy contribution.

If you halve the frequency band of a flat PSD, you halve the RMS

False — you halve the area (mean-square), so RMS drops by only

2

, to about

71%

of its original value.

For a stationary random process, the mean and variance are constant in time

True — that constancy of statistics is the definition of stationary; individual samples still jump around unpredictably.

Random vibration and sine vibration test the same failure mechanism

False — sine excites one frequency at a time to hunt resonances, while random excites all frequencies at once, closer to real launch and better for cumulative fatigue.

At resonance, the response PSD equals the input PSD

False — the response PSD is

∣ T (f) ∣^{2}

times the input, and near

f_{n}

that factor is roughly

Q^{2}

, so a flat input becomes a sharp spike (Figure 2).

Increasing damping

ζ

always lowers the response RMS

Mostly true near resonance (it flattens the peak), but it raises transmissibility above resonance because the damper transmits force, so the net effect depends on where the input energy lies.

A one-sided PSD and a two-sided PSD give the same RMS if used correctly

True — the two-sided PSD spreads the same energy over

\pm f

at half the height, so its integral over all

f

equals the one-sided integral over

f \geq 0

; mixing conventions (one-sided height, two-sided limits) is what introduces a spurious factor of 2.

Spot the error

a_{RMS} = \int G_{xx} (f) df

" — what's missing?

The square root — that integral is mean-square in

g^{2}

;

a_{RMS} = \int G_{xx} df

"I integrated a one-sided PSD from

- f_{2}

f_{2}

Wrong limits — a one-sided PSD is only defined for

f \geq 0

, so you integrate

0

f_{2}

; extending to negatives double-counts and inflates RMS by

2

"The resonance factor

1/ [(1 - r^{2})^{2} + (2 ζ r)^{2}]

is the transmissibility."

That is only the denominator (dynamic amplification) shape; true base-acceleration transmissibility multiplies it by the numerator

1 + (2 ζ r)^{2}

, which forces

∣ T ∣ \to 1

at low frequency (where

r \to 0

"For a piecewise PSD, sum the RMS of each band:

a = 0.8 + 16 + 15

Wrong — you add the mean-squares (areas) first, then take one square root:

0.8 + 16 + 15

. RMS values don't add linearly.

"Peak transmissibility magnitude is

∣ T ∣^{2} = Q

The squared magnitude is

\approx Q^{2}

(here 100); the transmissibility magnitude itself is

\approx Q = 10

. Don't confuse the "

∣ T ∣ \approx Q

" approximation with the "

∣ T ∣^{2} \approx Q^{2} + 1

" statement.

"Since

Q = 1/ (2 ζ)

and

ζ = 0.05

, then

Q = 0.1

Sign/reciprocal slip —

Q = 1/ (2 \times 0.05) = 10

, not

0.1

; small damping means large

Q

T (f)

is a real number, so I can just add response contributions arithmetically."

Wrong —

T (f)

is complex, with a phase that swings through

- 9 0^{\circ}

at resonance; only its squared magnitude

∣ T ∣^{2}

enters the PSD relation, so you multiply PSDs, never add signed transmissibilities.

"The relative-displacement filter

∣ H_{z} ∣^{2}

and the transmissibility

∣ T ∣^{2}

peak at the same value."

They peak at nearly the same frequency but not the same value;

∣ H_{z} ∣^{2}

carries the

1/ ω_{n}^{4}

scaling for displacement, while

∣ T ∣^{2}

is dimensionless — different quantities entirely.

"The stress in the structure is set by the absolute acceleration

\overset{y}{¨}

Stress is driven by relative displacement

z_{rel} = y - x

(how much the mass bends its mounts, Figure 3), which uses

∣ H_{z} ∣^{2}

, not the absolute-acceleration transmissibility.

Why questions

Why divide

∣ X (f, T_{rec}) ∣^{2}

by the record length

T_{rec}

in the PSD definition?

Because a longer recording accumulates more cycles, so

∣ X (f, T_{rec}) ∣^{2}

grows with

T_{rec}

; dividing gives a rate (power per unit time) that converges to a steady value.

Why do we use an ensemble average

E [\cdot]

(average over many recordings) instead of one measurement?

A single random realization is noisy and non-repeatable; averaging over the ensemble extracts the stable statistical content — and for an ergodic process a long time average of one record converges to that same ensemble average, which is why one long test can stand in for many.

Why take RMS at all instead of reporting the peak acceleration?

A random signal has no fixed peak — extreme values occur rarely and unpredictably, so the RMS gives a repeatable, physically meaningful "typical" magnitude for design.

Why does the absolute-acceleration transmissibility approach 1 (not 0) at very low frequency?

Well below

f_{n}

(where

r \to 0

) the mass rides rigidly with the base, so output equals input (left edge of Figure 2); the

+ 1

in the numerator encodes exactly this rigid-body following.

Why does even a flat input PSD produce a peaked response PSD?

Because the structure acts as a resonant filter — transmissibility

∣ T (f) ∣^{2}

amplifies the band near

f_{n}

(Figure 2), so uniform input energy gets concentrated there.

Why is the 100–500 Hz "bump" in a launch spectrum the region engineers worry about?

It carries the most energy (largest area), and structural resonances often sit there, so amplification lands right where the input is already strong.

Why does the Fourier transform use complex exponentials

e^{- i 2 π f t}

as its building blocks?

They are the pure single-frequency tones that form a complete basis, so any signal

x (t)

can be written as a weighted sum of them — that's what lets us talk about "energy at frequency

f

Why do the SDOF formulas always appear in the ratio

r = f / f_{n}

rather than

f

alone?

Because the physics only cares whether you drive above or below the natural frequency, not the absolute Hz; normalizing by

f_{n}

makes the response shape universal and dimensionless.

Edge cases

What is the RMS of a PSD that is zero everywhere?

Zero — no energy in any band means the integral is zero and

a_{RMS} = 0

; a silent input yields a silent response.

What happens to peak transmissibility as damping

ζ \to 0

Q = 1/ (2 ζ) \to \infty

, so at resonance

∣ T ∣^{2} \approx 1 + 1/ (4 ζ^{2})

blows up. The

+ 1

is just the low-frequency (rigid-body) asymptote riding along; the resonant growth comes from the

1/ (4 ζ^{2}) \approx Q^{2}

term, so for small

ζ

the exact peak is

Q^{2} + 1

, not literally

Q^{2}

. Real damping always keeps it finite.

What is the high-frequency limit of the absolute-acceleration transmissibility (

f ≫ f_{n}

, i.e.

r \to \infty

∣ T ∣^{2} \to (2 ζ r)^{2} / r^{4} = (2 ζ)^{2} / r^{2} \to 0

, so it rolls off like

1/ r

in magnitude (the right tail of Figure 2). Crucially the roll-off is set by the damper term, so heavier damping means a slower fall-off above resonance — the mass is increasingly isolated from the base, but never perfectly, because the damper keeps passing force.

If the input PSD has zero content exactly at

f_{n}

, is there still amplification?

No response energy can appear where there is no input energy to amplify; transmissibility multiplies input, and anything times zero is zero at that frequency.

What is transmissibility at

f = 0

(a static offset,

r = 0

Exactly 1 — the mass moves with the base as a rigid body, no relative motion and no dynamic amplification (left edge of Figure 2).

Can two very different PSD shapes give the same

a_{RMS}

Yes — RMS only measures total area, so a tall narrow PSD and a low wide one can match in RMS while causing very different fatigue because the damaging frequencies differ.

What if the frequency band

[f_{1}, f_{2}]

excludes the resonance

f_{n}

You miss the amplified band entirely, so the computed RMS understates the true structural response — always integrate over a range that contains the resonances.

Does adding a second band with the same PSD level far above the first change RMS a lot?

If that band is wide it adds substantial area and can dominate RMS, since a flat level over more Hz contributes more mean-square even when the level looks "small."