Frequency domain decomposition

Assumptions we use

The derivation is exact under A0–A6 and A8. A7 is an estimation assumption; A9 and A10 are the FDD and EFDD approximations. Each is used where noted and revisited in §9.

Model

A0

Output-only. Only $y(t)$ is measured; $x(t)$ is unknown and modeled as random.

A1

Linear, time-invariant, finite-dimensional. $n$ coordinates, $r$ inputs, $p$ sensors; small displacements about equilibrium; $M,C,K$ constant.

A2

Viscous damping. Constant symmetric $C$. General damping makes the modes complex; the residue derivation still applies.

A3

Linear input and output maps. Forces enter as $Bx$, sensors read $Lq$; no sensor dynamics, delays, or nonlinearities.

A4

Matrix properties. $M=M^T\succ0$, $K=K^T\succeq0$, $C=C^T\succeq0$; $K\succ0$ if restrained. Derived from energy in §2.4.

A5

Simple, stable eigenvalues. The $2n$ eigenvalues are distinct, stable, and in conjugate pairs; every pole is simple. Repeated or defective roots are excluded.

Excitation and measurement

A6

Stationary random excitation. $x$ is zero-mean, wide-sense stationary, with finite variance and a spectral density matrix; ergodicity lets time averages estimate ensemble averages.

A7

Accurate estimate. $\widehat G_y$ is close to $G_y$; finite-record errors are treated in Appendix D.

A8

Clean, synchronized measurements. Noise is zero or uncorrelated with the response; channels are synchronized and calibrated.

A9

One dominant mode in the band. $G_x(f)\approx Q_x$, and the residual $E_m$ is small relative to the selected term. Made precise in §6.

A10

One mode in the bell. The EFDD bell contains one mode and its correlation decays as a single exponential.

Conventions

A11

Fourier and spectral conventions. $j^2=-1$; Fourier transform $e^{-j2\pi ft}$; $G_y$ two-sided unless marked $^+$; one-sided spectra double interior bins but not DC or Nyquist; $f$ in hertz, $\omega=2\pi f$; displacement PSD in $\text{unit}^2/\text{Hz}$.

A12

Scale and phase are not identified. The shape is fixed only up to a complex scale; modal mass is not identifiable.

Notation

A dot is a time derivative, an overbar complex conjugation, and $^H$ the conjugate transpose. Scalar examples use $u$ for displacement and $F$ for force. $H(s)$ is the transfer matrix; $^H$ is an operation.

$t$

Time; seconds.

$f,\ \omega$

Frequency in hertz; $\omega=2\pi f$ in radians per second.

$j$

Imaginary unit, $j^2=-1$.

$s=\sigma+j\omega$

Complex Laplace variable.

$q(t)\in\mathbb R^n$

Displacements from equilibrium; metres.

$x(t)\in\mathbb R^r$

Force inputs; newtons.

$y(t)\in\mathbb R^p$

Measured outputs; sensor units.

$n,\ r,\ p$

Coordinates, inputs, sensors.

$M,\ C,\ K$

Mass, damping, stiffness; $n\times n$.

$B,\ L$

Input $n\times r$ and output $p\times n$ matrices.

$Z(s)$

Dynamic stiffness $s^2M+sC+K$.

$\mathcal T(s)$

Force-to-coordinate transfer $Z(s)^{-1}B$; $H=L\mathcal T$.

$H(s)$

Transfer matrix $LZ(s)^{-1}B$; $p\times r$. $H_{ik}$ maps input $k$ to sensor $i$.

$\lambda_m$

Eigenvalue (pole); inverse seconds.

$\phi_m$

Mode shape at the $n$ coordinates; complex in general.

$v_m=L\phi_m$

Measured shape; $p\times1$.

$\gamma_m$

Input-coupling vector; $r\times1$.

$R_m=v_m\gamma_m^T$

Residue matrix; $p\times r$, rank one.

$\alpha_m,\ \omega_{d,m},\ \omega_{n,m},\ \zeta_m$

Decay rate, damped frequency, natural frequency, damping ratio.

$G_y(f),\ G_x(f)$

Output and input PSD matrices; $p\times p$, $r\times r$.

$Y_T(f)$

Fourier transform of a finite record of $y$.

$U,\ S,\ s_k,\ u_k$

Eigendecomposition or SVD: vectors, values, value $k$, vector $k$.

$\operatorname{MAC}(u,v)$

Modal assurance criterion, in $[0,1]$.

$\delta,\ T_d,\ \mathcal R_m(\tau)$

Log decrement, damped period, autocorrelation at lag $\tau$.

$\mathbb E[\cdot]$

Expectation over records.

$\|\cdot\|,\ \|\cdot\|_2$

Euclidean norm; matrix spectral norm.

The argument in one page

Each link uses only the link before it and the assumptions above.

Energy gives the equations of motion. Free vibration gives poles and mode shapes. Forcing makes each pole a rank-one residue. Random forcing makes the response PSD a sum of those rank-one terms. When one term dominates, the leading eigenvector of the PSD is the measured shape. The width of that term gives the damping.

1The measurement problem

Sensors record the response $y(t)$; the forces $x(t)$ are unknown. We want the frequencies at which the structure vibrates, the relative motion of the sensors at those frequencies, and the decay rate. A mode shape is a fixed pattern of relative amplitude and phase: with one mode, every record is a scalar multiple of one vector, so the ratio between two sensors does not change with time. With several modes the ratio varies, and the pattern is visible only in a narrow band around a resonance.

FDD estimates the cross-spectral matrix $G_y(f)$ and decomposes it at each frequency; the dominant direction at a peak is the shape. EFDD selects a band around a peak, inverse-transforms it, and fits the correlation decay to get damping. Because the inputs are not measured, the shape is determined only up to a complex scale; §6 shows why.

2Energy and the equation of motion

FDD relies on symmetry and positive semidefiniteness, which come from energy. We derive the equation of motion from energy and check it with Lagrange’s equations.

2.1Energy of a vibrating structure

Let $q(t)\in\mathbb R^n$ collect displacements from static equilibrium. For a linear structure,

$M,K,C$ are the mass, stiffness, and viscous damping matrices. External forces do virtual work

where $B$ is $n\times r$; $Bx$ is the generalized force. Sensors read

with $L$ of size $p\times n$. The quadratic forms follow from a Taylor expansion about equilibrium: the linear term vanishes at a stationary point, the constant can be taken as zero, and the quadratic term dominates for small displacements (A1). The matrices are second derivatives of the energy, hence symmetric.

2.2One degree of freedom

Take a mass $\mu>0$, spring $k>0$, damper $b\ge0$, force $F(t)$, and displacement $u(t)$. The energies and virtual work are

Energy balance equates the rate of stored energy to the supplied power minus the dissipated power:

Differentiate and factor:

When $\dot u\ne0$, divide. When $\dot u=0$, continuity of $\ddot u$ and $F$ makes the bracket vanish as a limit. Hence

Each term is a force. Dividing by $\mu$ gives the normalized form

Lagrange’s equations give the same result and generalize to matrices. With Lagrangian $T-V$, dissipation $D$, and generalized force $Q$,

Here $T-V=\tfrac12\mu\dot u^2-\tfrac12ku^2$, $D=\tfrac12b\dot u^2$, and $Q=F$. The derivatives are

Substitution returns (2.4).

2.3Many degrees of freedom

For the full structure, energy balance is

With $M,K$ constant and symmetric,

Collecting terms gives

This holds for every $\dot q$. Taking $\dot q=e_i$ for each $i$ makes every component of the bracket zero, so

2.4What the matrices mean

Symmetry. $q^TAq=q^TA^Tq$, so only the symmetric part contributes; take $M,K,C$ symmetric. For $M,K$ this follows from energy; for $C$ it is A2.

Definiteness. $A$ is positive semidefinite if $x^TAx\ge0$, positive definite if strict for $x\ne0$. $T\ge0$, with equality only at $\dot q=0$, so $M\succ0$ and invertible. $V\ge0$; a restrained structure has $K\succ0$, a free-free structure has $K\succeq0$ with rigid-body $K\phi=0$. $D\ge0$, so $C\succeq0$, possibly singular.

Units. $q$ in metres and $x$ in newtons give $M$ in kilograms, $C$ in newton-seconds per metre, $K$ in newtons per metre; each term of (2.7) is a force. $B$ absorbs input units and $L$ sensor units.

Two masses. Masses $m_1,m_2$ are joined by springs $k_1,k_2,k_3$ and dampers $c_1,c_2,c_3$, with $k_1,c_1$ from the wall to $m_1$, $k_2,c_2$ between the masses, and $k_3,c_3$ from $m_2$ to the wall. With $q=[q_1,q_2]^T$,

The off-diagonal entries are negative because the middle spring and damper pull the masses together. The energies are

Each bracket is a sum of squares with positive coefficients, so $V,D\ge0$. $K\succ0$ when $k_1+k_2>0$ and $k_1k_2+k_1k_3+k_2k_3>0$, which holds for positive springs. A force on $m_1$ gives $B=[1,0]^T$; measuring both displacements gives $L=I$.

2.5Lagrange’s equations for the matrix case

With $q$ as the generalized coordinate and $Q=Bx$,

Since $T=\tfrac12\dot q^TM\dot q$ and $V$ does not depend on $\dot q$,

because $M$ is constant. Since $V=\tfrac12q^TKq$, component $i$ of its gradient is

using $K_{ji}=K_{ij}$ and renaming the index. Also $\partial D/\partial\dot q=C\dot q$. Substituting into (2.9) with $\partial(T-V)/\partial q=-\partial V/\partial q$ gives

which is (2.7). Lagrange’s form holds for any generalized coordinates and allows $M$ to be non-diagonal.

3Modal analysis: free vibration

Free vibration is the motion with no force. Its solutions are the modes: fixed spatial patterns that evolve at a single complex rate.

3.1The exponential ansatz and the quadratic eigenproblem

Set $x=0$ in (2.7):

This linear constant-coefficient equation has exponential solutions; polynomial factors occur only for repeated eigenvalues, excluded by A5. Seek

Here $c$ marks the complex auxiliary solution, $\phi\ne0$ is the spatial pattern, and $\lambda$ is the complex rate. Substitution gives

Cancel the exponential and define

Nonzero $\phi$ makes $Z(\lambda)$ singular; a zero determinant gives a null vector. The scale of $\phi$ is arbitrary; only its direction matters. This is a quadratic eigenvalue problem. $\det Z(s)$ has degree $2n$ because its leading term is $\det(M)s^{2n}$, so there are $2n$ roots, and real matrices give conjugate pairs. A numerical solver uses the $2n\times2n$ state form $z=[q;\dot q]$, whose eigenvectors stack $\phi_m$ on $\lambda_m\phi_m$ (Appendix A).

3.2Real motion from complex modes

$q_c$ is complex, but the physical $q$ is real. Real matrices give $\bar\lambda$ and $\bar\phi$; combine the pair:

by $z+\bar z=2\operatorname{Re}z$. Each coordinate is real, and the $2n$ real constants in the $n$ coefficients $c_m$ match the initial conditions. Write $\lambda=\sigma+j\omega_d$:

Stability means $\sigma<0$. Choose $\omega_d>0$ and write $\lambda=-\alpha+j\omega_d$ with $\alpha>0$: $\alpha$ is the decay rate, $\omega_d$ the damped angular frequency. For $C=0$ or proportional damping, $\phi$ can be real; for general $C$ it is complex and records relative phases, not a complex displacement.

3.3The equivalent single-degree-of-freedom oscillator

Natural frequency and damping ratio come from a scalar oscillator. With mass $\mu>0$, spring $k>0$, damper $b\ge0$, set $F=0$ in (2.4) and divide by $\mu$:

With $b=0$, $\ddot u+(k/\mu)u=0$, so $\omega_n^2=k/\mu$ and $f_n=\omega_n/(2\pi)$. Restoring the damper and substituting $u=e^{st}$:

The discriminant decides: $b^2<4\mu k$ gives complex roots and oscillation, $b^2>4\mu k$ gives real negative roots and no oscillation, and equality gives a double root. The boundary is $b_{\mathrm{crit}}=2\sqrt{\mu k}$, so

The dimensionless $\zeta$ is $<1$ underdamped, $=1$ critical, $>1$ overdamped; the factor 2 comes from (3.6). Since $b/\mu=2\zeta\omega_n$, dividing (2.4) by $\mu$ gives

For $\zeta<1$, $s=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}$. Comparing with $\lambda=-\alpha\pm j\omega_d$ gives $\alpha=\zeta\omega_n$ and $\omega_d=\omega_n\sqrt{1-\zeta^2}$. Matching the full polynomial,

so $\zeta_m\omega_{n,m}=\alpha_m$, $\omega_{n,m}^2=\alpha_m^2+\omega_{d,m}^2$, and

These are the modal parameters. $\omega_{n,m}=|\lambda_m|$ is the equivalent-oscillator parameter, not necessarily an undamped structural frequency (§3.5).

3.4Interactive: damping regimes

Release the oscillator from $u(0)=1$, $\dot u(0)=0$. For $\zeta<1$, $u=e^{-\alpha t}(C_1\cos\omega_dt+C_2\sin\omega_dt)$; the conditions give $C_1=1$ and $C_2=\alpha/\omega_d$. For $\zeta=1$, $u=(C_1+C_2t)e^{-\omega_nt}$ gives $C_1=1$, $C_2=\omega_n$. For $\zeta>1$, $\beta=\omega_n\sqrt{\zeta^2-1}$, $\lambda_{1,2}=-\alpha\pm\beta$, and the two conditions give the coefficients below:

The critical case is the limit of the underdamped case; the overdamped case reduces to it as $\beta\to0$. The figure shows the poles and the response.

3.5Two meanings of natural frequency

“Remove the damping” has two meanings. For the equivalent oscillator, set its $\zeta$ to zero while keeping $\omega_{n,m}=|\lambda_m|$. For the structure, set $C=0$ in (2.7) and solve

For one mass and spring both give $\sqrt{k/\mu}$. Under proportional damping $C=aM+bK$, the damped modes are the undamped modes with an added decay; Appendix A shows $\omega_{n,r}=\Omega_r$ and $\zeta_r=(a/\Omega_r+b\Omega_r)/2$. For general damping, modes can mix undamped shapes and $|\lambda_m|$ need not equal any $\Omega_r$. Here $\omega_{n,m}=|\lambda_m|$.

The next part connects these modes to forced measurements through the transfer matrix.

4The transfer matrix and its residues

Free vibration gives the poles; the transfer matrix gives their residues. It maps forces to measurements in the frequency domain.

4.1The Laplace transform

For a signal $q(t)$ on $t\ge0$,

This is a Fourier transform of $q(t)e^{-\sigma t}$; choosing $\sigma$ large enough makes it converge. The derivative rule follows from integration by parts with $u=e^{-st}$, $dv=\dot q\,dt$:

The upper limit vanishes by the choice of $\sigma$; the lower limit gives $-q(0)$. Applying the rule again,

With zero initial conditions each derivative is multiplication by $s$; the initial-condition response can be added separately. On the imaginary axis, $s=j\omega=j2\pi f$, this is the stationary frequency response.

4.2From the equation of motion to $H(s)$

Transform (2.7) with zero initial conditions:

Where $Z(s)$ is invertible,

so

$H$ is $p\times r$; $H_{ik}$ maps input $k$ to sensor $i$. Its entries share the denominator $\det Z(s)$. $Z$ is singular exactly at the eigenvalues, so the poles of $H$ are eigenvalues unless cancelled by $L$ or $B$.

4.3Partial fractions at a simple pole

For one entry with a simple pole at $\lambda_m$,

where $g_{ik,m}$ is finite at $\lambda_m$. This is exact. Multiply by $s-\lambda_m$ and take the limit:

The second term vanishes, so the limit isolates the residue. Entrywise this gives a matrix $R_m$ and a remainder:

$R_m$ is $p\times r$; $H_{\mathrm{rest},m}$ contains the conjugate pole and all other modes. A simple pole (A5) makes the principal part one term.

4.4Why the residue is rank one

Separate the force-to-coordinate transfer from the sensor map:

Let $W_m$ be the residue of $\mathcal T$ at $\lambda_m$:

Multiply $Z(s)\mathcal T(s)=B$ by $s-\lambda_m$ and take the limit:

Each column of $W_m$ lies in the nullspace of $Z(\lambda_m)$, the set of vectors mapped to zero. At a simple eigenvalue that nullspace is one-dimensional, so every column is a multiple of $\phi_m$. If input $k$ has scalar $\gamma_{m,k}$,

Apply $L$, which is constant and passes through the limit:

The residue is rank one: every column is a multiple of the measured shape $v_m$. If $v_m=0$ the sensors do not see the mode; if $\gamma_m=0$ the inputs do not excite it.

For two inputs with coefficients $2$ and $-3$, and $v_m=[1,-2]^T$,

Both columns keep the ratio $1:-2$; an input vector combines them into $v_m(2X_1-3X_2)$. The full $H$ need not be rank one, because it contains the other modes. The rank-one statement is about one residue.

4.5The full modal sum

For real matrices the conjugate pole has residue $\bar R_m$. Summing the $n$ pairs,

This is exact under A5, not a one-mode assumption. A zero $v_m$, zero $\gamma_m$, or cancellation removes a term from $H$; the mode still exists.

5The response spectral matrix

Operational forces are unknown and vary between records. We average products of Fourier coefficients, not the response itself; that average is the spectral matrix.

5.1From records to a spectral matrix

Let $T$ be the record duration and $y(t)$ the $p$-channel measurement. Define

$Y_T$ is random; $\mathbb E$ averages over records. The $(i,k)$ entry of $Y_TY_T^H$ is $Y_i\bar Y_k$. The factor $1/T$ converts squared transform (amplitude squared times time squared) to a density (amplitude squared per hertz). Integrating a diagonal entry over all frequencies gives that channel’s mean square, equal to its variance because the mean is zero.

The matrix is two-sided. For real signals the negative frequencies mirror the positive ones; a one-sided spectrum, marked $^+$, doubles interior positive bins and keeps DC and Nyquist unchanged. The definition assumes wide-sense stationarity (A6): expected products depend only on time separation, so the limit exists. Ergodicity lets a time average estimate the ensemble average; estimation details are in Appendix D.

5.2Hermitian positive semidefiniteness

For any $Y$,

A matrix equal to its conjugate transpose is Hermitian; expectation preserves this, so $G_y$ is Hermitian. Its diagonal entries are averages of $|Y_i|^2$, hence real and nonnegative. For any complex weight vector $a$, $|a^HY_T|^2\ge0$, and averaging gives

So $G_y$ is Hermitian positive semidefinite. The spectral theorem gives an orthonormal eigenbasis and real eigenvalues; positive semidefiniteness makes them nonnegative. With unitary $U$ and diagonal $\Lambda$,

For such a matrix the singular values are the eigenvalues and the left singular vectors are the eigenvectors. The rank is the number of nonzero eigenvalues; a rank-one matrix has one eigenvector and a one-dimensional nullspace.

5.3The input–output relation

At a fixed frequency, the stationary output transform is the transfer matrix on the imaginary axis times the input transform:

This is (4.4) at $s=j2\pi f$, for the forced part after the transient has decayed. Substitute into the outer product:

At fixed frequency $H$ is constant across realizations, so it passes through the expectation. Applying (5.1),

$G_x$ is $r\times r$ and Hermitian positive semidefinite; $G_y$ is $p\times p$. The force spectrum describes the excitation, $H$ the response and the sensors.

5.4Conventions and estimation

This page defines entry $(i,k)$ as $Y_i\bar Y_k$. Some software uses $\bar Y_iY_k$, the transpose and conjugate; the estimated shape is then conjugated. Do not mix conventions. A one-sided spectrum keeps $f\ge0$ and doubles interior bins so that integrating over $f\ge0$ matches the two-sided mean square; DC and Nyquist are not doubled. The Welch estimator, its normalization, and the resolution–variance tradeoff are in Appendix D.

6When one mode dominates

The response matrix is an exact sum of rank-one terms. FDD uses the band where one term dominates and the others are a perturbation. This section isolates that term and bounds the remainder.

6.1Isolating one pole

On the imaginary axis, $s=j\omega$ with $\omega=2\pi f$, separate mode $m$:

$E_m$ contains the conjugate pole and every other mode and is finite at $\omega_{d,m}$ (A5). The selected denominator is

At $\omega=\omega_{d,m}$ its magnitude is $\alpha_m$, so the selected term grows like $1/\alpha_m$. The remainder has no such denominator.

6.2The exact four-term expansion

With $P=R_m/(j\omega-\lambda_m)$ and $E=E_m$, substitute into (5.3):

This is exact. The first term has the selected pole twice; the middle terms mix it with the remainder; the last has the remainder twice. Since $P^H=R_m^H/(-j\omega-\bar\lambda_m)$,

With $R_m^H=\bar\gamma_mv_m^H$,

Define the scalar weight

The numerator is $(\bar\gamma_m)^HG_x(\bar\gamma_m)\ge0$ because $G_x$ is positive semidefinite. The direction $v_m$ is fixed; only $d_m$ varies with frequency. For constant $G_x\approx Q_x$,

a Lorentzian peaked at $\omega_{d,m}$, with height set by the numerator and width by $\alpha_m$.

6.3The residual bound

Dropping the last three terms of (6.2) is legitimate when they are small. Weight selected term and remainder by the excitation; let $G_x^{1/2}$ be the PSD square root and

Then $P_wP_w^H=PG_xP^H$, $P_wE_w^H=PG_xE^H$, $E_wP_w^H=EG_xP^H$, $E_wE_w^H=EG_xE^H$, so