Random vibration — PSD, RMS acceleration

What is Random Vibration?

WHY random, not deterministic?

• Launch acoustic noise: millions of pressure fluctuations from turbulent exhaust
• Aerodynamic buffeting: vortex shedding, boundary layer turbulence
• Engine roughness: combustion instabilities

These sources are too complex to predict instant-by-instant, but their statistical character is consistent and testable.

Power Spectral Density (PSD)

DERIVATION FROM FIRST PRINCIPLES:

Step 1: Frequency content of a signal Any time-domain signal $x(t)$ can be decomposed into frequency components via Fourier transform: $X(f) = \int_{-\infty}^{\infty} x(t) e^{-i 2\pi f t} \, dt$

WHY? Because $e^{i 2\pi f t}$ are the "pure tones" that form a complete basis for any function.

Step 2: Energy at each frequency The squared magnitude $|X(f)|^2$ represents energy density at frequency $f$. But for a random signal measured over finite time $T$, we need to normalize: $\text{Energy per unit time per unit frequency} = \frac{|X(f, T)|^2}{T}$

WHY divide by $T$? As we record longer, $|X(f)|^2$ grows with $T$ (more cycles accumulated). Dividing by $T$ gives a rate.

Step 3: Statistical average Since the signal is random, one realization isn't enough. We average over many realizations (ensemble average $\mathbb{E}[\cdot]$), then take the limit as $T \to \infty$ to get the steady-state power per frequency: $G_{xx}(f) = \lim_{T \to \infty} \frac{1}{T} \mathbb{E}[|X(f, T)|^2]$

WHAT DOES IT MEAN?

• $G_{xx}(f) = 0.01 \, \text{g}^2/\text{Hz}$ at $f = 100 \, \text{Hz}$: in a 1 Hz bandwidth around 100 Hz, the mean-square acceleration is $0.01 \, \text{g}^2$.

RMS Acceleration from PSD

DERIVATION:

Step 1: Mean-square acceleration PSD $G_{xx}(f)$ is power per Hz. To get total power (mean-square acceleration), integrate over all frequencies: $\overline{a^2} = \int_{f_1}^{f_2} G_{xx}(f) \, df$

WHY integration? Each frequency bin $df$ contributes $G_{xx}(f) \, df$ to the total. Suming (integrating) across bins gives the total.

Step 2: Root-mean-square RMS is the square root of mean-square: $a_{\text{RMS}} = \sqrt{\overline{a^2}} = \sqrt{\int_{f_1}^{f_2} G_{xx}(f) \, df}$

WHY take the square root? RMS has the same units as the original signal ($\text{g}$), making it physically interpretable as a "typical" acceleration magnitude. Mean-square has units $\text{g}^2$.

Worked Examples

Find: Overall RMS acceleration.

Solution:

Step 1: Apply the RMS formula: $a_{\text{RMS}} = \sqrt{\int_{20}^{2000} 0.04 \, df}$

WHY this works? PSD is constant, so the integral is just $G \times \Delta f$.

Step 2: Evaluate: $\int_{20}^{2000} 0.04 \, df = 0.04 \times (2000 - 20) = 0.04 \times 1980 = 79.2 \, \text{g}^2$

Step 3: Take square root: $a_{\text{RMS}} = \sqrt{79.2} \approx 8.9 \, \text{g}$

WHAT THIS MEANS: The structure experiences vibration equivalent to a steady $8.9 \, g$ acceleration, but spread across 20–2000 Hz.

Find: $a_{\text{RMS}}$.

Solution:

Step 1: Break the integral into regions: $a_{\text{RMS}} = \sqrt{\int_{20}^{100} 0.01 \, df + \int_{100}^{500} 0.04 \, df + \int_{500}^{2000} 0.01 \, df}$

WHY? PSD is piecewise constant, so integrate each segment separately.

Step 2: Evaluate each integral:

• Region 1: $0.01 \times (100 - 20) = 0.01 \times 80 = 0.8 \, \text{g}^2$
• Region 2: $0.04 \times (500 - 100) = 0.04 \times 400 = 16.0 \, \text{g}^2$
• Region 3: $0.01 \times (2000 - 500) = 0.01 \times 1500 = 15.0 \, \text{g}^2$

Step 3: Sum and take square root: $a_{\text{RMS}} = \sqrt{0.8 + 16.0 + 15.0} = \sqrt{31.8} \approx 5.64 \, \text{g}$

INSIGHT: Most energy comes from the 100–500 Hz "bump" (16 g² out of 31.8 g² total). This is where structural resonances often lie.

Structural Response to Random Vibration

For spacecraft, the excitation enters through the base (mounting interface) as an acceleration PSD $G_{\ddot{x}}(f)$. We want the response of the mass — either its absolute acceleration $\ddot{y}$ or the relative displacement $z = y - x$ (which drives stress).

DERIVATION SKETCH:

Step 1: The equation of motion for a base-excited mass, written in terms of relative displacement $z = y - x$, is: $\ddot{z} + 2\zeta\omega_n \dot{z} + \omega_n^2 z = -\ddot{x}(t)$ WHY this form? Substituting $y = x + z$ into $m\ddot{y} + c(\dot{y}-\dot{x}) + k(y-x) = 0$ and dividing by $m$ isolates the base acceleration $\ddot{x}$ as the forcing term.

Step 2: In the frequency domain, the relative-displacement transfer function is: $H_z(f) = \frac{Z}{\ddot{X}} = \frac{-1}{\omega_n^2\left[1 - (f/f_n)^2 + i\,2\zeta (f/f_n)\right]}$ so $|H_z(f)|^2 = \frac{1/\omega_n^4}{\left[1 - (f/f_n)^2\right]^2 + \left[2\zeta \frac{f}{f_n}\right]^2}$ This is the correct filter for stress-producing relative motion.

Step 3: The absolute acceleration output introduces the extra numerator term $1 + (2\zeta f/f_n)^2$ (the damper transmits force even above resonance), giving the transmissibility $|T(f)|^2$ above.

WHY THIS MATTERS: Near $f \approx f_n$, both $|T|^2$ and $|H_z|^2$ peak sharply (for low $\zeta$), amplifying that band. Even with a flat input PSD, the response PSD spikes at resonance.

Find: Peak response acceleration PSD at resonance.

Solution:

Step 1: At resonance ($f = f_n$), the denominator's first term vanishes and $f/f_n = 1$, so: $|T(f_n)|^2 = \frac{1 + (2\zeta)^2}{(2\zeta)^2} = 1 + \frac{1}{4\zeta^2}$

WHY? Set $1-(f/f_n)^2 = 0$; both numerator and denominator keep their $(2\zeta)^2$ terms.

Step 2: Calculate: $|T(200)|^2 = 1 + \frac{1}{4 \times (0.05)^2} = 1 + \frac{1}{0.01} = 1 + 100 = 101$

Note: The quality factor $Q = 1/(2\zeta) = 10$, so the peak transmissibility magnitude is $\approx Q = 10$ (i.e. $|T|^2 \approx Q^2 = 100$); the extra $+1$ is a small correction.

Step 3: Response PSD: $G_{\ddot{y}\ddot{y}}(200) = 101 \times 0.02 = 2.02 \, \text{g}^2/\text{Hz}$

INTERPRETATION: The structure amplifies vibration at 200 Hz by roughly $Q^2 = 100$. This is why low damping is dangerous in random vibration environments.

Common Mistakes

Why it feels right: The number is small, so it seems like a direct measurement.

The fix: PSD is power density, not amplitude. To get acceleration, you must:

1. Integrate PSD over frequency to get mean-square: $\overline{a^2} = \int G(f) \, df$
2. Take square root: $a_{\text{RMS}} = \sqrt{\overline{a^2}}$

Units tell the story: $\text{g}^2/\text{Hz}$ needs integration (removing /Hz) and square root (removing the square) to become $\text{g}$.

Why it feels right: We add things to combine them, right?

The fix: RMS values combine via root-sum-square because vibrations are uncorrelated: $a_{\text{total}} = \sqrt{a_A^2 + a_B^2} = \sqrt{3^2 + 4^2} = 5 \, \text{g}$

WHY? Mean-square accelerations add (since $\overline{(a_A + a_B)^2} = \overline{a_A^2} + \overline{a_B^2}$ for independent signals), so you add before taking the square root.

Why it feels right: Mathematical definitions often use infinite limits.

The fix: Real systems have finite bandwidth. Sensors can't measure DC (0 Hz) or ultrasonic frequencies (>20 kHz). Test specifications always give $f_1$ and $f_2$: $a_{\text{RMS}} = \sqrt{\int_{f_1}^{f_2} G(f) \, df}$

Integrating outside this range includes noise or energy not transmitted to the structure.

Testing and Qualification

WHY random vibration testing?

• Launches last minutes; testing recreates hours of equivalent fatigue.
• Miles' Equation gives the RMS of the response acceleration of an SDOF resonator subjected to a flat input acceleration PSD $G_0$ (units g²/Hz). Its standard, correctly-derived form is: $a_{\text{RMS}} = \sqrt{\frac{\pi}{2}\, f_n \, Q \, G_0}$ where $Q = 1/(2\zeta)$ is the quality factor.

WHY it works: The sharp resonance acts as a narrow band-pass filter of effective bandwidth $\approx \frac{\pi}{2}\frac{f_n}{Q}$ but amplified by $Q^2$; the product picks out $\frac{\pi}{2} f_n Q$. Miles' result feeds directly into fatigue estimates (via the $3\sigma$ acceleration for equivalent-static loads).

Test setup:

1. Mount spacecraft on a shaker table.
2. Apply controlled PSD input per launch vehicle spec (e.g., Atlas V, Falcon 9).
3. Monitor response accelerometers on critical components.
4. Run for 1–2 minutes per axis (3 axes: X, Y, Z).

Connections

• Frequency Response Function (FRF) – $H(f)$ relates input to output in frequency domain
• Modal Analysis – Real structures have multiple resonances; each mode responds to PSD
• Fatigue Analysis – Random vibration causes cumulative damage via stress cycles
• Sine Vibration Testing – Deterministic single-frequency test; different from random
• Acoustic Loading – Sound pressure creates random vibration via pressure fluctuations
• Shock Response Spectrum (SRS) – Different analysis for transient events (pyro shocks)
• Structural Damping – $\zeta$ controls resonance amplification; critical for PSD response
• Transmissibility – Base-excitation ratio $T(f)$ of output to input acceleration

#flashcards/physics

What is the physical meaning of Power Spectral Density (PSD)? :: PSD $G_{xx}(f)$ is the mean-square acceleration per unit frequency at frequency $f$, describing how vibration energy is distributed across frequencies. Units: g²/Hz.

Concept Map

Hinglish (regional understanding)