Random vibration — PSD, RMS acceleration

3.6.11 · Physics › Spacecraft Structures & Systems Engineering

Random Vibration Kya Hai?

Random kyun, deterministic kyun nahi?

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

Ye sources itne complex hain ki instant-by-instant predict nahi ho sakte, lekin inका statistical character consistent aur testable hota hai.

Power Spectral Density (PSD)

FIRST PRINCIPLES SE DERIVATION:

Step 1: Signal ka frequency content Koi bhi time-domain signal $x(t)$ ko Fourier transform ke zariye frequency components mein decompose kiya ja sakta hai: $X(f)={\int }_{-\infty }^{\infty }x(t)e^{-i2\pi ft}dt$

Kyun? Kyunki $e^{i2\pi ft}$ woh "pure tones" hain jo kisi bhi function ke liye ek complete basis banate hain.

Step 2: Har frequency par energy Squared magnitude $|X(f){|}^2$ frequency $f$ par energy density represent karta hai. Lekin finite time $T$ mein measure ki gayi random signal ke liye, humein normalize karna padta hai: $\text{Energy per unit time per unit frequency}=\frac{|X(f,T){|}^2}{T}$

$T$ se divide kyun karte hain? Jab hum zyada der record karte hain, toh $|X(f){|}^2$ $T$ ke saath badhta hai (zyada cycles accumulate hote hain). $T$ se divide karne par ek rate milta hai.

Step 3: Statistical average Kyunki signal random hai, ek realization kaafi nahi hai. Hum kai realizations par average lete hain (ensemble average $\mathbb{E}[\cdot ]$), phir $T\to \infty$ ka limit lete hain taaki steady-state power per frequency mile: $G_{xx}(f)={\lim }_{T\to \infty }\frac{1}{T}\mathbb{E}[|X(f,T){|}^2]$

ISKA MATLAB KYA HAI?

• $G_{xx}(f)=0.01{\text{g}}^2/\text{Hz}$ at $f=100\text{Hz}$: 100 Hz ke aaspaas 1 Hz bandwidth mein, mean-square acceleration $0.01{\text{g}}^2$ hai.

PSD Se RMS Acceleration

DERIVATION:

Step 1: Mean-square acceleration PSD $G_{xx}(f)$ power per Hz hai. Total power (mean-square acceleration) paane ke liye, saari frequencies par integrate karo: $\overline{a^2}={\int }_{f_1}^{f_2}{G}_{xx}(f)df$

Integration kyun? Har frequency bin $df$, total mein $G_{xx}(f)df$ contribute karta hai. Bins par suming (integrating) se total milta hai.

Step 2: Root-mean-square RMS, mean-square ka square root hai: $a_{\text{RMS}}=\sqrt{\overline{a^2}}=\sqrt{{\int }_{f_1}^{f_2}{G}_{xx}(f)df}$

Square root kyun lete hain? RMS ke units original signal ($\text{g}$) jaisi hoti hain, jisse yeh ek "typical" acceleration magnitude ki tarah physically interpretable ho jaata hai. Mean-square ke units ${\text{g}}^2$ hote hain.

Worked Examples

Dhundhna hai: Overall RMS acceleration.

Solution:

Step 1: RMS formula apply karo: $a_{\text{RMS}}=\sqrt{{\int }_{20}^{2000}0.04df}$

Ye kyun kaam karta hai? PSD constant hai, isliye integral sirf $G\times \Delta f$ hai.

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

Step 3: Square root lo: $a_{\text{RMS}}=\sqrt{79.2}\approx 8.9\text{g}$

ISKA MATLAB: Structure $8.9g$ steady acceleration ke equivalent vibration experience karta hai, lekin 20–2000 Hz mein spread hua.

Dhundhna hai: $a_{\text{RMS}}$.

Solution:

Step 1: Integral ko regions mein tod do: $a_{\text{RMS}}=\sqrt{{\int }_{20}^{100}0.01df+{\int }_{100}^{500}0.04df+{\int }_{500}^{2000}0.01df}$

Kyun? PSD piecewise constant hai, isliye har segment alag integrate karo.

Step 2: Har integral evaluate karo:

• 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 karo aur square root lo: $a_{\text{RMS}}=\sqrt{0.8+16.0+15.0}=\sqrt{31.8}\approx 5.64\text{g}$

INSIGHT: Zyaadatar energy 100–500 Hz "bump" se aati hai (31.8 g² total mein se 16 g²). Yahan structural resonances aksar hoti hain.

Random Vibration Par Structural Response

Spacecraft ke liye, excitation base (mounting interface) ke through acceleration PSD $G_{\ddot{x}}(f)$ ke roop mein enter karta hai. Hum mass ka response chahte hain — ya toh uski absolute acceleration $\ddot{y}$ ya relative displacement $z=y-x$ (jo stress drive karta hai).

DERIVATION SKETCH:

Step 1: Base-excited mass ke liye equation of motion, relative displacement $z=y-x$ ke terms mein likha gaya, kuch aisa hai: $\ddot{z}+2\zeta {\omega }_n\dot{z}+{\omega }_n^{2}z=-\ddot{x}(t)$ Ye form kyun? $m\ddot{y}+c(\dot{y}-\dot{x})+k(y-x)=0$ mein $y=x+z$ substitute karne par aur $m$ se divide karne par base acceleration $\ddot{x}$ forcing term ki tarah isolate ho jaata hai.

Step 2: Frequency domain mein, relative-displacement transfer function hai: $H_z(f)=\frac{Z}{\ddot{X}}=\frac{-1}{{\omega }_n^2\left[1-(f/f_n{)}^2+i2\zeta (f/f_n)\right]}$ toh $|H_z(f){|}^2=\frac{1/{\omega }_n^4}{{\left[1-(f/f_n{)}^2\right]}^2+{\left[2\zeta \frac{f}{f_n}\right]}^2}$ Ye stress-producing relative motion ke liye sahi filter hai.

Step 3: Absolute acceleration output mein extra numerator term $1+(2\zeta f/f_n{)}^2$ aata hai (damper resonance ke upar bhi force transmit karta hai), jo upar wala transmissibility $|T(f){|}^2$ deta hai.

YE KYUN MATTER KARTA HAI: $f\approx f_n$ ke paas, $|T{|}^2$ aur $|H_z{|}^2$ dono sharply peak karte hain (low $\zeta$ ke liye), us band ko amplify karte hain. Even flat input PSD ke saath, response PSD resonance par spike karta hai.

Dhundhna hai: Resonance par peak response acceleration PSD.

Solution:

Step 1: Resonance par ($f=f_n$), denominator ka pehla term vanish ho jaata hai aur $f/f_n=1$, toh: $|T(f_n){|}^2=\frac{1+(2\zeta {)}^2}{(2\zeta {)}^2}=1+\frac{1}{4{\zeta }^2}$

Kyun? $1-(f/f_n{)}^2=0$ set karo; numerator aur denominator dono apne $(2\zeta {)}^2$ terms rakhte hain.

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

Note: Quality factor $Q=1/(2\zeta )=10$ hai, toh peak transmissibility magnitude $\approx Q=10$ hai (yaani $|T{|}^2\approx Q^2=100$); extra $+1$ ek small correction hai.

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

INTERPRETATION: Structure 200 Hz par vibration ko roughly $Q^2=100$ se amplify karta hai. Isliye random vibration environments mein low damping dangerous hai.

Common Mistakes

Ye sahi kyun lagta hai: Number chhota hai, toh ek direct measurement jaisa lagta hai.

Fix: PSD power density hai, amplitude nahi. Acceleration paane ke liye, tumhe:

1. Mean-square paane ke liye PSD ko frequency par integrate karo: $\overline{a^2}=\int G(f)df$
2. Square root lo: $a_{\text{RMS}}=\sqrt{\overline{a^2}}$

Units poori kahani batate hain: ${\text{g}}^2/\text{Hz}$ ko integration (removing /Hz) aur square root (removing the square) chahiye taaki $\text{g}$ bane.

Ye sahi kyun lagta hai: Combine karne ke liye hum cheezein add karte hain, na?

Fix: RMS values root-sum-square ke zariye combine hoti hain kyunki vibrations uncorrelated hain: $a_{\text{total}}=\sqrt{a_A^2+a_B^2}=\sqrt{3^2+4^2}=5\text{g}$

Kyun? Mean-square accelerations add hote hain (kyunki independent signals ke liye $\overline{(a_A+a_B{)}^2}=\overline{a_A^2}+\overline{a_B^2}$ hota hai), isliye square root lene se pehle add karo.

Ye sahi kyun lagta hai: Mathematical definitions aksar infinite limits use karti hain.

Fix: Real systems ki finite bandwidth hoti hai. Sensors DC (0 Hz) ya ultrasonic frequencies (>20 kHz) measure nahi kar sakte. Test specifications hamesha $f_1$ aur $f_2$ dete hain: $a_{\text{RMS}}=\sqrt{{\int }_{f_1}^{f_2}G(f)df}$

Is range ke bahar integrate karne se woh noise ya energy include ho jaati hai jo structure tak transmit nahi hoti.

Testing and Qualification

Random vibration testing kyun?

• Launches minutes tak chalte hain; testing equivalent fatigue ke hours recreate karti hai.
• Miles' Equation ek SDOF resonator ke response acceleration ka RMS deta hai jab usse ek flat input acceleration PSD $G_0$ (units g²/Hz) ke saath subject kiya jaaye. Iska standard, correctly-derived form hai: $a_{\text{RMS}}=\sqrt{\frac{\pi }{2}{f}_{n}Q{G}_0}$ jahaan $Q=1/(2\zeta )$ quality factor hai.

Ye kyun kaam karta hai: Sharp resonance ek narrow band-pass filter ki tarah kaam karta hai jiska effective bandwidth $\approx \frac{\pi }{2}\frac{f_n}{Q}$ hai lekin $Q^2$ se amplified hai; product $\frac{\pi }{2}{f}_{n}Q$ pick out karta hai. Miles' result directly fatigue estimates mein feed hota hai (equivalent-static loads ke liye $3\sigma$ acceleration ke zariye).

Test setup:

1. Spacecraft ko ek shaker table par mount karo.
2. Launch vehicle spec (jaise Atlas V, Falcon 9) ke hisaab se controlled PSD input apply karo.
3. Critical components par response accelerometers monitor karo.
4. Har axis ke liye 1–2 minutes run karo (3 axes: X, Y, Z).

Connections

• Frequency Response Function (FRF) – $H(f)$ frequency domain mein input ko output se relate karta hai
• Modal Analysis – Real structures mein multiple resonances hoti hain; har mode PSD par respond karta hai
• Fatigue Analysis – Random vibration stress cycles ke zariye cumulative damage karta hai
• Sine Vibration Testing – Deterministic single-frequency test; random se alag
• Acoustic Loading – Sound pressure, pressure fluctuations ke zariye random vibration create karta hai
• Shock Response Spectrum (SRS) – Transient events (pyro shocks) ke liye alag analysis
• Structural Damping – $\zeta$ resonance amplification control karta hai; PSD response ke liye critical
• Transmissibility – Base-excitation ratio $T(f)$ output to input acceleration ka

#flashcards/physics

Power Spectral Density (PSD) ka physical meaning kya hai? :: PSD $G_{xx}(f)$ frequency $f$ par mean-square acceleration per unit frequency hai, jo batata hai ki vibration energy frequencies mein kaise distribute hoti hai. Units: g²/Hz.

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps