Chapter 3: Spectral Response of Linear Systems
This chapter addresses the response of linear time-invariant systems to random excitations.
The focus is on calculating the mean, correlation, and spectral density of the response.
1. Problem Presentation
The typical problem considered in this chapter is a mechanical structure in small deformation submitted to random forces. The field of displacement is noted \(u(r,t)\) where \(r\) is a position on the system and \(t\) is time. The source positions are noted \(s_j\) for \(j=1,\dots,n\).
u (x,y, t )
f 1
f 2
f n
...
Figure 3: Schema of mechanical structure excited by \(n\) random forces.
The following assumptions are considered:
Sources:
The forces \(f_i(t)\) for \(i = 1, \dots, n\) are random and mutually stationary
Mean: \(\mu_{f_i}\)
Correlation: \(R_{f_i f_j}(\tau)\)
Spectral density: \(S_{f_i f_j}(\omega)\)
System:
The system is linear and time-invariant
Impulse response: \(h(r, s_j, t)\)
Frequency response function: \(H(r, s_j, \omega)\)
Modes: \(\psi_k(r)\), natural frequencies \(\omega_k\), and modal damping \(\xi_k\)
The response \(u(r, t)\) is given by the convolution of the impulse response and the input forces:
\[
u(r, t) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \tau) f_j(t - \tau) \, d\tau
\]
Remind: the convolution product is commutative.
\[
f*g(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) \, d\tau = \int_{-\infty}^{\infty} f(t-\tau) g(\tau) \, d\tau = g*f(t)
\]
2. Mean of Response
The mean of the response is:
\[
\mu_u(r, t) = \left\langle \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \tau) f_j(t - \tau) \, d\tau \right\rangle
\]
We can permute \(\sum\), \(\int\), and \(\langle \cdot \rangle\) (see chapter 1) and \(h\) is a deterministic function.
\[
\mu_u(r, t) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \tau) \left\langle f_j(t - \tau) \right\rangle \, d\tau
\]
Using the linearity of expectation and the stationarity of \(f_j(t)\):
\[
\mu_u(r, t) = \sum_{j=1}^n \mu_{f_j} \int_{-\infty}^{\infty} h(r, s_j, \tau) \, d\tau = \sum_{j=1}^n \mu_{f_j} H(r, s_j, 0)
\]
Remarks:
The right-hand side does not depend on \(t\), so \(\mu_\mu(r, t)\) is constant.
If the forces \(f_j\) are centered (\(\mu_{f_j} = 0\)), then \(\mu_u = 0\).
Modal development:
The modal development of the frequency response function at \(\omega=0\) reads
\[
H(r, s, 0) = \sum_{k \geq 0} H_k(0) \psi_k(s) \psi_k(r)
\]
where \(H_k(0) = \frac{1}{m_k \omega_k^2}\). Thus:
\[
\mu_u(r) = \sum_{j=1}^n \mu_{f_j} \sum_{k \geq 0} \frac{\psi_k(s_j) \psi_k(r)}{m_k \omega_k^2}
\]
Example: Mass-Spring System
For a mass-spring system with static force \(\mu_f \neq 0\):
\[
\mu_x = H(0) \mu_f = \frac{\mu_f}{k}
\]
This is equivalent to a static loading!
3. Correlation of Response
The correlation of the response at two points \(r_1\) and \(r_2\) is:
\[
R_{uu}(r_1, r_2, t, \tau) = \langle u(r_1, t) u(r_2, t + \tau) \rangle
\]
But
\[
u(r_1, t) = \sum_{i=1}^n \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) f_i(t - \theta_1) \, d\theta_1
\]
and
\[
u(r_2, t + \tau) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r_2, s_j, \theta_2) f_j(t +\tau- \theta_2) \, d\theta_2
\]
Substituting the expressions for \(u(r_i, t)\) gives
\[
R_{uu}(r_1, r_2, t, \tau) = \left\langle \sum_{i,j} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) h(r_2, s_j, \theta_2) f_i(t- \theta_1) f_j(t +\tau- \theta_2) \, d\theta_1 \, d\theta_2 \right\rangle
\]
By permuting the brackets and integrals
\[
R_{uu}(r_1, r_2, t, \tau) = \sum_{i,j} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) h(r_2, s_j, \theta_2) \left\langle f_i(t- \theta_1) f_j(t +\tau- \theta_2) \right\rangle \, d\theta_1 \, d\theta_2
\]
We recognize the expression of the correlation function
\[
R_{uu}(r_1, r_2, t, \tau) = \sum_{i,j} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) h(r_2, s_j, \theta_2) R_{f_i f_j}(\tau + \theta_1 - \theta_2) \, d\theta_1 \, d\theta_2
\]
Remark:
The right-hand side does not depend on \(t\), so \(u(r, t)\) is stationary.
4. Spectral Density of Response
Using the Wiener-Khinchin theorem, the spectral density of the response is:
\[
S_{uu}(r_1, r_2, \omega) = \int_{-\infty}^{\infty} R_{uu}(r_1, r_2, \vartheta) e^{-i \omega \tau} \, d\tau
\]
Substituting the expression for \(R_{uu}\) and permuting the integrals:
\[
S_{uu}(r_1, r_2, \omega) = \sum_{i,j} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) h(r_2, s_j, \theta_2) \int_{-\infty}^{\infty} R_{f_i f_j}(\tau + \theta_1 - \theta_2) e^{-i \omega \tau} \, d\tau d\theta_1 \, d\theta_2
\]
By the change of variables (\(\tau' = \tau + \theta_1 - \theta_2\)) and re-ordering the integrals
\[
S_{uu}(r_1, r_2, \omega) = \sum_{i,j} \int_{-\infty}^{\infty} h(r_1, s_i, \theta_1) e^{i \omega \theta_1} \, d\theta_1 \int_{-\infty}^{\infty} h(r_2, s_j, \theta_2) e^{-i \omega \theta_2} \, d\theta_2 \int_{-\infty}^{\infty} R_{f_i f_j}(\tau') e^{-i \omega \tau} \, d\tau
\]
We recognize the frequency response functions \(H(r_i, s_j, \omega)\) and the power spectral density \(S_{f_i f_j}(\omega)\)
\[
S_{uu}(r_1, r_2, \omega) = \sum_{i,j} S_{f_i f_j}(\omega) \overline{H}(r_1, s_i, \omega) H(r_2, s_j, \omega)
\]
Modal development:
Since
\[
H(r_1, s_i, \omega) = \sum_{k \geq 0} H_k(\omega) \psi_k(r_1) \psi_k(s_i), \quad H(r_2, s_j, \omega) = \sum_{l \geq 0} H_l(\omega) \psi_l(r_2) \psi_l(s_j)
\]
the modal development of \(S_{uu}\) is
\[
S_{uu}(r_1, r_2, \omega) = \sum_{i,j} \sum_{k,l \geq 0} \overline{H}_k(\omega) H_l(\omega) \psi_k(r_1) \psi_k(r_2) \psi_k(s_i) \psi_k(s_j) S_{f_i f_j}(\omega)
\]
Assuming light damping (\(\xi_k \ll 1\)), the cross terms \(H_k(\omega) H_l(\omega)\) are negligible for \(k \neq l\). Thus:
\[
S_{uu}(r_1, r_2, \omega) = \sum_{i,j} \sum_{k \geq 0} |H_k(\omega)|^2 \psi_k(r_1) \psi_k(r_2) \psi_k(s_i) \psi_k(s_j) S_{f_i f_j}(\omega)
\]
In the special case of a unique source \(s\) and a unique receiver point \(r\)
\[
S_{uu}(r, \omega) = \sum_{k \geq 0} |H_k(\omega)|^2 \psi_k(r)^2 \psi_k(s)^2 S_{f f}(\omega)
\]
5. Root-Mean-Square of Response
The root-mean-square of response \(u(r,t)\) is:
\[
u_\text{RMS}^2(r) = \langle u(r,t)^2 \rangle = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_{u u}(r, \omega) \, d\omega
\]
\[
u_\text{RMS}^2(r) = \sum_{i,j=1}^n \int_{-\infty}^{\infty} \overline{H}(r, s_i, \omega) H(r, s_j, \omega) S_{f_i f_j}(\omega) \, \frac{d\omega}{2\pi}
\]
In the case of a single source \(s\) of constant power spectral density \(S_0\) (white noise), this reduces to
\[
u_\text{RMS}^2(r) = S_0 \int_{-\infty}^{\infty} \left|H(r, s, \omega)\right|^2 \, \frac{d\omega}{2\pi}
\]
Example: White-Noise Force
Consider a single resonator excited by a white-noise force \(f(t)\) with power spectral density \(S_0\).
The expectation of elastic energy in the system is:
\[
V = \frac{1}{2} k \langle x^2 \rangle
\]
and the expectation of kinetic energy is
\[
K = \frac{1}{2} m \langle \dot{x}^2 \rangle
\]
But the power spectral densities of \(x\) and \(\dot{x}\) are:
\[
S_{xx}(\omega) = S_0 \left| H(\omega)\right|^2, \quad S_{\dot{x}\dot{x}}(\omega) = S_0 \omega^2 \left| H(\omega)\right|^2
\]
Thus:
\[
V = \frac{1}{2} k \times \frac{S_0}{2\pi} \int_{-\infty}^{\infty} \frac{1}{\left| m\omega_0^2 \left( 1+ 2 i \xi \frac{\omega}{\omega_0} - \frac{\omega^2}{\omega_0^2} \right) \right|^2} \, d\omega
\]
And:
\[
K = \frac{1}{2} m \times \frac{S_0}{2\pi} \int_{-\infty}^{\infty} \frac{\omega^2}{\left| m\omega_0^2 \left( 1+ 2 i \xi \frac{\omega}{\omega_0} - \frac{\omega^2}{\omega_0^2} \right) \right|^2} \, d\omega
\]
But the above integral can be calculated exactly with the residue theorem. We get
\[
V = K
\]
6. Force-Response Cross-Correlation
The cross-correlation between the force \(f_i(t)\) and the response \(u(r, t)\) is:
\[
R_{f_i u}(r, t, \tau) = \langle f_i(t) u(r, t + \tau) \rangle
\]
But
\[
u(r, t+\tau) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \theta) f_j(t+\tau - \theta) \, d\theta
\]
Substituting the expression for \(u(r, t+\tau)\) and permuting the brackets and integral:
\[
R_{f_i u}(r, t, \tau) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \theta) \langle f_i(t) f_j( t + \tau -\theta) \rangle \, d\theta
\]
We recognize the correlation function \(R_{f_i f_j}\)
\[
R_{f_i u}(r, t, \tau) = \sum_{j=1}^n \int_{-\infty}^{\infty} h(r, s_j, \theta) R_{f_i f_j}(\tau - \theta) \, d\theta
\]
Remarks:
The right-hand side does not depend on \(t\), so \(f_i\) and \(u\) are mutually stationary.
7. Force-Response Cross-Spectrum
By Wiener-Khinchin
\[
S_{f_i u}(r, \omega) = \int_{-\infty}^\infty R_{f_i u}(r,\tau) e^{-i\omega \tau} \, d\tau
\]
Substituting the expression for \(R_{f_i u}(\tau)\) and permuting the integrals
\[
S_{f_i u}(r, \omega) = \sum_{j=1}^n \int_{-\infty}^\infty h(r, s_j, \theta) \int_{-\infty}^\infty R_{f_i f_j}(\tau-\theta) e^{-i\omega \tau} \, d\tau d\theta
\]
With the change of variables \(\tau'=\tau-\theta\) and re-ordering the terms
\[
S_{f_i u}(r, \omega) = \sum_{j=1}^n \int_{-\infty}^\infty h(r, s_j, \theta) e^{-i\omega \theta} \, d\theta \int_{-\infty}^\infty R_{f_i f_j}(\tau') e^{-i\omega \tau'} \, d\tau'
\]
Finally, the cross-spectral density is:
\[
S_{f_i u}(r, \omega) = \sum_{j=1}^n H(r, s_j, \omega) S_{f_i f_j}(\omega)
\]
8. Root-Mean-Square Value of Products
The cross-product force-response root-mean-square value is:
\[
\langle f_i u \rangle = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_{f_i u}(r, \omega) \, d\omega = \sum_{j=1}^n \int_{-\infty}^{\infty} H(r, s_j, \omega) S_{f_i f_j}(\omega) \, \frac{d\omega}{2\pi}
\]
Example: White-Noise Force
For a white-noise force \(f(t)\) with power spectral density \(S_0\), the expectation of mechanical power injected into the system is:
\[
P_{\text{inj}} = \langle f \dot{x} \rangle
\]
But the force-velocity cross-spectral density is:
\[
S_{f \dot{x}}(\omega) = i \omega H(\omega) S_0
\]
Thus:
\[
P_{\text{inj}} = \frac{S_0}{2\pi} \int_{-\infty}^{\infty} \frac{i\omega}{m\omega_0^2 \left( 1+ 2 i \xi \frac{\omega}{\omega_0} - \frac{\omega^2}{\omega_0^2} \right)} \, d\omega
\]
But the above integral can be calculated exactly with the residue theorem. Finally
\[
P_{\text{inj}} = \frac{S_0}{2m}
\]