Interactive Lecture
Understanding Stochastic Processes and Power Spectral Density
by Dr Arnaldo Delli Carri
Unlike deterministic vibrations where the response can be predicted exactly at any time, random vibrations are non-deterministic and can only be described statistically. Examples include earthquake ground motion, wind loads on structures, road roughness for vehicles, and acoustic noise.
Random vibrations require statistical methods — you can't describe them with a single amplitude and frequency. We characterise them using mean, variance, autocorrelation, and power spectral density (PSD), which shows how the signal's power is distributed across frequency.
A random process is stationary if its statistical properties do not change with time. For engineering applications, we often assume wide-sense stationary processes where the mean and autocorrelation are time-invariant.
Statistical Descriptors:
Mean: \(\mu_x = E[x(t)] = \text{constant}\)
Mean Square: \(E[x^2(t)] = \sigma_x^2 + \mu_x^2\)
Variance: \(\sigma_x^2 = E[(x(t) - \mu_x)^2]\)
Autocorrelation Function: \(R_{xx}(\tau) = E[x(t)x(t+\tau)]\)
The Power Spectral Density function \(S_{xx}(\omega)\) describes how the variance of a random process is distributed over frequency. It is the Fourier transform of the autocorrelation function.
Wiener-Khinchin Theorem:
\[S_{xx}(\omega) = \int_{-\infty}^{\infty} R_{xx}(\tau)e^{-i\omega\tau}d\tau\]
\[R_{xx}(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty} S_{xx}(\omega)e^{i\omega\tau}d\omega\]
Physical Interpretation: The area under the PSD curve equals the variance:
\[\sigma_x^2 = \int_{0}^{\infty} S_{xx}(\omega)d\omega\]
White noise has constant PSD across all frequencies (ideal but physically unrealizable):
\[S_{xx}(\omega) = S_0 \quad \text{for all } \omega\]
where \(S_0\) is a constant spectral density.
A practical approximation with constant PSD over a frequency range:
\[S_{xx}(\omega) = \begin{cases} S_0 & \omega_1 \leq \omega \leq \omega_2 \\ 0 & \text{otherwise} \end{cases}\]
Widely used for earthquake ground motion modelling:
\[S_{xx}(\omega) = S_0 \frac{\omega_g^4 + 4\zeta_g^2\omega_g^2\omega^2}{(\omega_g^2 - \omega^2)^2 + 4\zeta_g^2\omega_g^2\omega^2}\]
where \(\omega_g\) is the predominant ground frequency and \(\zeta_g\) is the damping ratio.
For a SDOF system with natural frequency \(\omega_n\) and damping ratio \(\zeta\) subjected to random base acceleration with PSD \(S_{uu}(\omega)\):
Frequency Response Function:
\[H(\omega) = \frac{1}{1 - r^2 + 2i\zeta r}\]
where \(r = \omega/\omega_n\) is the frequency ratio.
Response PSD:
\[S_{xx}(\omega) = |H(\omega)|^2 S_{uu}(\omega)\]
where \(|H(\omega)|^2\) is the squared magnitude of the frequency response function.
The mean square response equals the area under the response PSD — integrate \(S_{xx}(\omega)\) over all frequencies to recover the variance. For narrow-band excitation near the natural frequency, the response is sharply amplified, just as in harmonic resonance.
The root-mean-square (RMS) response is a key design parameter:
\[\sigma_x^2 = \int_{0}^{\infty} S_{xx}(\omega)d\omega = \int_{0}^{\infty} |H(\omega)|^2 S_{uu}(\omega)d\omega\]
\[x_{RMS} = \sigma_x = \sqrt{\sigma_x^2}\]
For white noise excitation that enters as an acceleration — either an applied force per unit mass (\(S_0\) = PSD of force/mass²) or a base acceleration (\(S_0\) = PSD of ground acceleration):
\[\sigma_x^2 = \frac{\pi S_0}{4\zeta\omega_n^3}\]
Both cases share this closed form because the forcing enters the equation of motion as an acceleration, so \(S_0\) has units of \((\text{m/s}^2)^2/(\text{rad/s})\) either way. Always confirm how \(S_0\) is defined before applying the formula.
Watch a SDOF mass-spring-damper system respond to random excitation in real time. The excitation signal is generated from the input PSD, and the system response is computed via numerical integration.
Random vibration analysis is standard practice in several industries:
| Concept | Equation | Description |
|---|---|---|
| Autocorrelation Function | \(R_{xx}(\tau) = E[x(t)x(t+\tau)]\) | Correlation between signal at different times |
| Power Spectral Density | \(S_{xx}(\omega) = \int_{-\infty}^{\infty} R_{xx}(\tau)e^{-i\omega\tau}d\tau\) | Fourier transform of autocorrelation |
| Variance from PSD | \(\sigma_x^2 = \int_{0}^{\infty} S_{xx}(\omega)d\omega\) | Area under PSD curve |
| White Noise PSD | \(S_{xx}(\omega) = S_0\) | Constant across all frequencies |
| Frequency Response Function | \(H(\omega) = \frac{1}{1 - r^2 + 2i\zeta r}\) | SDOF system transfer function, \(r = \omega/\omega_n\) |
| Response PSD | \(S_{xx}(\omega) = |H(\omega)|^2 S_{uu}(\omega)\) | Output PSD from input PSD |
| SDOF Response to White Noise | \(\sigma_x^2 = \frac{\pi S_0}{4\zeta\omega_n^3}\) | Same closed form for force/mass² or base-acceleration input — both enter as an acceleration; \(S_0\) is the PSD of that acceleration |
| Kanai-Tajimi Spectrum | \(S_{xx}(\omega) = S_0 \frac{\omega_g^4 + 4\zeta_g^2\omega_g^2\omega^2}{(\omega_g^2 - \omega^2)^2 + 4\zeta_g^2\omega_g^2\omega^2}\) | Earthquake ground motion model |
A SDOF system with natural frequency \(\omega_n = 20\) rad/s and damping ratio \(\zeta = 0.05\) is subjected to white noise base acceleration with PSD \(S_0 = 0.1\) (m/s²)²/(rad/s). Calculate the RMS displacement response.
Show Solution
Given: \(\omega_n = 20\) rad/s, \(\zeta = 0.05\), \(S_0 = 0.1\) (m/s²)²/(rad/s)
Step 1: Calculate variance of displacement response
For white noise base acceleration, the displacement variance is:
\[\sigma_x^2 = \frac{\pi S_0}{4\zeta\omega_n^3}\]
\[\sigma_x^2 = \frac{\pi \times 0.1}{4 \times 0.05 \times 20^3} = \frac{0.314}{1600} = 1.96 \times 10^{-4} \text{ m}^2\]
Step 2: Calculate RMS displacement
\[x_{RMS} = \sigma_x = \sqrt{1.96 \times 10^{-4}} = 0.014 \text{ m} = 14 \text{ mm}\]
Answer: The RMS displacement is 14 mm. Higher natural frequency or higher damping would reduce this response.
A system with \(\omega_n = 10\) rad/s and \(\zeta = 0.10\) experiences band-limited white noise from 5 to 15 rad/s with \(S_0 = 2.0\). Estimate the RMS response assuming the FRF is approximately constant within the band.
Show Solution
Given: \(\omega_n = 10\) rad/s, \(\zeta = 0.10\), \(S_0 = 2.0\), band: 5-15 rad/s
Step 1: Evaluate FRF at natural frequency
\[|H(\omega_n)|^2 = \frac{1}{(2\zeta)^2} = \frac{1}{(0.2)^2} = 25\]
Step 2: Calculate response variance (approximation)
Since the excitation band includes \(\omega_n\), we approximate:
\[\sigma_x^2 \approx |H(\omega_n)|^2 \times S_0 \times \Delta\omega\]
where \(\Delta\omega = 15 - 5 = 10\) rad/s (bandwidth)
\[\sigma_x^2 \approx 25 \times 2.0 \times 10 = 500\]
Step 3: RMS response
\[x_{RMS} = \sqrt{500} = 22.4\]
Answer: \(x_{RMS} \approx 22.4\), in the response units set by the definition of \(S_0\), using the constant-FRF band approximation.
Note: This is a rough approximation. For accurate results, integrate \(|H(\omega)|^2 S_0\) numerically over the frequency band.
For a system with \(\omega_n = 15\) rad/s subjected to white noise base acceleration with \(S_0 = 0.5\) (m/s²)²/(rad/s), calculate the RMS response for damping ratios of \(\zeta = 0.02\), \(0.05\), and \(0.10\). Discuss the trend.
Show Solution
Using: \(\sigma_x^2 = \frac{\pi S_0}{4\zeta\omega_n^3}\), with \(\omega_n^3 = 3375\) and \(\pi S_0 = 1.571\)
For ζ = 0.02:
\[\sigma_x^2 = \frac{\pi \times 0.5}{4 \times 0.02 \times 3375} = \frac{1.571}{270} = 5.82 \times 10^{-3}\,\text{m}^2 \quad\Rightarrow\quad x_{RMS} = 0.0763\,\text{m} = 76.3\,\text{mm}\]
For ζ = 0.05:
\[\sigma_x^2 = \frac{1.571}{675} = 2.33 \times 10^{-3}\,\text{m}^2 \quad\Rightarrow\quad x_{RMS} = 0.0482\,\text{m} = 48.2\,\text{mm}\]
For ζ = 0.10:
\[\sigma_x^2 = \frac{1.571}{1350} = 1.16 \times 10^{-3}\,\text{m}^2 \quad\Rightarrow\quad x_{RMS} = 0.0341\,\text{m} = 34.1\,\text{mm}\]
Answer: \(x_{RMS} = 76.3\,\text{mm}\) at \(\zeta = 0.02\), \(48.2\,\text{mm}\) at \(\zeta = 0.05\), and \(34.1\,\text{mm}\) at \(\zeta = 0.10\).
Conclusion: The variance scales as \(1/\zeta\), so the RMS scales as \(1/\sqrt{\zeta}\): doubling the damping from 0.05 to 0.10 cuts the RMS by \(\sqrt{2}\). Damping is the main lever in random vibration control.