# Sound radiation of planar structure and nearfield acoustical holography : k-space analysis

Laurent Maxit

► To cite this version:

Laurent Maxit. Sound radiation of planar structure and nearfield acoustical holography: k-space analysis. Master. France. 2023. hal-05448826

HAL Id: hal-05448826

https://cel.hal.science/hal-05448826v1

Submitted on 8 Jan 2026

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons CC BY 4.0 - Attribution - International License
RECHERCHE

INSTITUT NATIONAL DES SCIENCES APPLIQUÉES DE LYON

<figure><img src="image_2.png" /></figure>

<figure><img src="image_3.png" /></figure>

<figure><img src="image_4.png" /></figure>

L. MAXIT

GRAN HORIZONTAL

Fig. 22

# Sound radiation of planar structure and nearfield acoustical holography:

## k-space analysis

Laboratoire Vibrations-Acoustique, INSA de Lyon
Bâtiment St. Exupéry, 25 bis av. Jean Capelle, 69621 Villeurbanne Cedex
laurent.maxit@insa-lyon.fr

<figure><img src="image_5.png" /></figure>

CeLyA

UNIVERSITE DE LYON

Centre Lyonnais
d'Acoustique

<figure><img src="image_6.png" /></figure>

lva

laboratoire
vibrations
acoustique

INSA

INSTITUT NATIONAL
DES SCIENCES
APPLIQUÉES
LYON
<figure><img src="image_7.png" /></figure>

"Fourier Acoustics: Sound Radiation and Nearfield Acoustical Holography" by Earl G. Williams, 1999.

k-space analysis ≈ spatial frequency analysis ≈ wavenumber analysis
# Illustration of k-space analysis for different applications

## - Ultrasonics

### Tendinopathy Discrimination by Use of Spatial Frequency Parameters in Ultrasound B-Mode Images

Gregory R. Bashford®, Senior Member, IEEE, Nicholas Tomsen, Shruti Arya, Judith M. Burnfield, and Kornelia Kulig

<figure><img src="image_8.png" /></figure>

Example image kernel and spatial frequency spectrum.

Left, typical 32 × 32 pixel kernel within the Achilles tendon of a control subject.

Right, spatial frequency spectrum obtained by taking the 2-D FFT of the left image.

## - Aeroacoustics

<figure><img src="image_9.png" /><figcaption>LES of spatially developing TBL at Mach 0.5
Gloerfelt and Margnat (2014).</figcaption></figure>

<figure><img src="image_10.png" /><figcaption>Wavenumber-frequency spectrum $10 \log_{10} \Phi_{pp}(k, \omega)$ of wall pressure
beneath a ZPG Mach 0.5 turbulent boundary layer</figcaption></figure>
- Structural acoustics (vibro-acoustics)

Examples of industrial structures for which the K-space analysis is of interest for understanding how these structures radiate

<figure><img src="image_11.png" /></figure>

<figure><img src="image_12.png" /></figure>

<figure><img src="image_13.png" /></figure>
- Structural acoustics (vibro-acoustics)

Modelling the acoustic radiation from infinite structures

<figure><img src="image_14.png" /><figcaption>Transversal velocity field of an
infinite periodically stiffened panel
excited by an harmonic point force</figcaption></figure>

<figure><img src="image_15.png" /><figcaption>Radial velocity field of an infinite
periodically stiffened shell excited
by an harmonic point force</figcaption></figure>

K-space analysis :

- A tool for understanding how structures
radiate in the far field

- Resolution of vibroacoustic problem
# Nearfield Acoustical Holography (NAH)

<figure><img src="image_16.png" /></figure>

Measurement of the
radiated pressure with a
microphone array

<figure><img src="image_17.png" /></figure>

Inverse problem: knowing the
measured pressures and having a
radiation model of the structure,
what is the velocity field of the
vibrating structure?
# OUTLINE

I – Helmholtz equation

II – Vibro-acoustic behavior of infinite thin panels

III – Vibro-acoustic behavior of finite thin panels

IV – Nearfield Acoustical Holography (NAH)
# I – Helmholtz equation

Assumptions on the acoustic domain:

- stagnant uniform fluid or gas

- perfect fluid (i.e. no viscosity, or heat conduction)

- No mass source or external volume force terms

Fluid equations with the previous assumptions:

- Mass conservation

$$ \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{u}) = 0 $$

- The momentum conservation law (second Newton's law)

$$ \rho \left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = - \nabla p $$

- The energy conservation law (adiabatic-reversible = isentropic processus)

$$ \frac{\partial s}{\partial t} + \mathbf{u} \cdot \nabla s = 0 $$

- An equation of state (thermodynamics)

$$ p = p(\rho, s) \quad \text{Isentropic flow} \quad \left( \frac{\partial p}{\partial s} \right)_{\rho} = 0 $$

$$ c^2 = \left( \frac{\partial p}{\partial \rho} \right)_s \quad \text{the square of the isentropic speed of sound} $$

<footnote>*S.W. Rienstra & A. Hirschberg, An Introduction to Acoustics, Eindhoven University of Technology, 2014</footnote>
# I – Helmholtz equation

Linearization and wave equation:

Small perturbations around the equilibrium position:

$p = p_0 + p', \rho = \rho_0 + \rho', s = s_0 + s', \mathbf{u} = \mathbf{u}_0 + \mathbf{u}'$ with $\mathbf{u}_0 = 0$ for stagnant fluid

<figure><img src="image_18.png" /></figure>

Linearized equations (taking the 1st order terms only):

$$ \frac{\partial \rho'}{\partial t} + \rho_0 \nabla \cdot \mathbf{u}' = 0 \quad (1) \qquad \rho_0 \frac{\partial \mathbf{u}'}{\partial t} + \nabla p' = 0 \quad (2) \qquad \mathbf{p}' = c_0^2 \rho' \quad (3) \qquad \frac{\partial s'}{\partial t} = 0 \quad (4) $$

Wave equation (introducing (3) in (1) and writing $\nabla \cdot (2) - \frac{\partial}{\partial t} (1)$)

$$ \nabla^2 p' - \frac{1}{c_0^2} \frac{\partial^2 p'}{\partial t^2} = 0 $$

Helmholtz equation (steady state harmonic variation $\mathbf{p}' = \bar{p} e^{i\omega t}$)

$$ \nabla^2 \bar{p} + \frac{\omega^2}{c_0^2} \bar{p} = 0 $$
# II – Vibro-acoustic behavior of infinite thin panels

## II.1 - Natural propagative waves

### II.1.a - For an acoustic medium

$$
\left\{
\begin{array}{ll}
\text{Helmholtz equation} & \nabla^2 p + \frac{\omega^2}{c_0^2} p = 0 \\
\text{Propagative plane wave} & p = Pe^{-ik_x x - ik_y y - ik_z z} \\
\text{Wavevector } \mathbf{k}: & \mathbf{k} = \begin{pmatrix} k_x \\ k_y \\ k_z \end{pmatrix}
\end{array}
\right.
\quad
\begin{array}{ll}
\text{Dispersion equation} & k = f(\omega) \\
\text{with the acoustic (natural) wavenumber} & k_0 = \frac{\omega}{c_0} \\
\text{Dispersion equation:} & \|\mathbf{k}\|_2 = k_0 \\
\text{an example with} & k_y = 0
\end{array}
$$

<figure><img src="image_19.png" /></figure>

<figure><img src="image_20.png" /></figure>
## II.1 - Natural propagative waves

### II.1.b - For a thin flexible plate

<figure><img src="image_21.png" /></figure>

Infinite thin plate made of an isotropic elastic material

Plate equation of motions with Love-Kirchhoff assumptions and time-harmonic dependence (pure bending, rotational inertia and shear stress not take into account, only adapted for "low" frequencies, use Mindlin assumptions for "high" frequencies):

$$D \left( \frac{\partial^4 w}{\partial x^4} + 2 \frac{\partial^4 w}{\partial x^2 \partial y^2} + \frac{\partial^4 w}{\partial y^4} \right) - \omega^2 \rho h w = 0 \quad \text{with the flexural rigidity } D = \frac{Eh^3}{12(1-\nu^2)}$$

Propagative plane wave $w = We^{-ik_x x - ik_y y}$

Dispersion equation $(k_x^2 + k_y^2)^2 = \frac{\omega^2 \rho h}{D}$

Propagative waves

$\Rightarrow$ For $k_y = 0, k_x^4 = \frac{\omega^2 \rho h}{D}$

$\Rightarrow$ 4 solutions: $k_x = \begin{cases} k_f \\ -k_f \\ ik_f \\ -ik_f \end{cases}$

Evanescente waves

$$\text{Flexural (natural) wavenumber } k_f = \sqrt{\omega \sqrt{\frac{\rho h}{D}}}$$
Forced response using the wave decomposition:

<figure><img src="image_22.png" /></figure>

Infinite thin plate excited by a normal line force

Wave decomp. for each part (with non
reflective conditions at infinite):

$$
\left\{
w_1(x) = A_1 e^{ik_f x} + B_1 e^{k_f x}, x \in [-\infty, 0]
\right.
$$

$$
\left\{
w_2(x) = A_2 e^{-ik_f x} + B_2 e^{-k_f x}, x \in [0, \infty]
\right\}
$$

Continuity relations at the junction:

$$
w_1(0) = w_2(0)
$$

$$
\frac{\partial w_1}{\partial x}(0) = \frac{\partial w_2}{\partial x}(0)
$$

$$
D \frac{\partial^2 w_1}{\partial x^2}(0) = D \frac{\partial^2 w_2}{\partial x^2}(0)
$$

$$
D \frac{\partial^3 w_1}{\partial x^3}(0) - F = D \frac{\partial^3 w_2}{\partial x^3}(0)
$$

<figure><img src="image_23.png" /></figure>

$$
A_1 + B_1 = A_2 + B_2
$$

$$
iA_1 + B_1 = -iA_2 - B_2
$$

$$
-A_1 + B_1 = -A_2 + B_2
$$

$$
-ik_f^3 A_1 + k_f^3 B_1 + \frac{F}{D} = ik_f^3 A_2 - k_f^3 B_2
$$

<figure><img src="image_24.png" /></figure>

$$
A_1 = A_2
$$

$$
B_2 = -iA_2
$$

$$
B_1 = -iA_2
$$

$$
A_2 = \frac{F}{4iDk_f^3}
$$

Analytical expression of the plate displacement:

$$
w_2(x) = \frac{F}{4iDk_f^3} e^{-ik_f x} - \frac{F}{4Dk_f^3} e^{-k_f x}, x \in [0, \infty]
$$
## II.1 - Natural propagative waves

### II.1.c - The critical frequency (or acoustic coincidence frequency)

<figure><img src="image_25.png" /></figure>

Critical frequency (flexural celerity = acoustic celerity): $k_0 = k_f$

$$f_c = \frac{c_0^2}{2\pi} \sqrt{\frac{\rho h}{D}} = \frac{c_0^2}{2\pi} \sqrt{\frac{12\rho(1-\nu^2)}{Eh^2}} \rightarrow 5.8 \text{ kHz for a 2mm-thick steel plate}$$
## II.2 – Mathematical formulation and resolution in the k-space

<figure><img src="image_26.png" /></figure>

Resolution using the spatial Fourier transform:

$$f(x, y) \rightarrow \tilde{f}(k_x, k_y) = \int \int_{-\infty}^{+\infty} f(x, y)e^{-ik_x x} e^{-ik_y y} dx dy$$

$$\tilde{f}(k_x, k_y) \rightarrow f(x, y) = \frac{1}{4\pi^2} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(k_x, k_y) e^{ik_x x} e^{ik_y y} dk_x dk_y$$
Property of the Fourier transform on the space derivative:

$$ \frac{\partial^p \partial^q f}{\partial x^p \partial y^q} (x, y) \rightarrow (ik_x)^p (ik_y)^q \tilde{f}(k_x, k_y) $$

...

Vibroacoustic problem in the wavenumber space (i.e. k-space):

$$
\left\{
\begin{array}{l}
\tilde{v}_z(k_x, k_y) = i\omega\tilde{w}(k_x, k_y) \\
\\
Z_p \tilde{v}(k_x, k_y) = \tilde{F}(k_x, k_y) - \tilde{p}(k_x, k_y, 0) \quad \text{with} \quad Z_p = \left[ \frac{D(1+i\eta)(k_x^2+k_y^2)^2 - \omega^2 \rho h}{i\omega} \right] \\
\\
\frac{\partial^2 \tilde{p}}{\partial z^2}(k_x, k_y, z) + (k_0^2 - k_x^2 - k_y^2)p(k_x, k_y, z) = 0 \\
\\
\frac{\partial \tilde{p}}{\partial z}(k_x, k_y, 0) = -\rho_0(i\omega)\tilde{v}(k_x, k_y) + \text{Sommerfeld conditions}
\end{array}
\right.
$$
Resolution of the fluid equations:

Homogeneous 2nd order differential equation about z:

$$ \frac{\partial^2 \tilde{p}}{\partial z^2}(k_x, k_y, z) + (k_0^2 - k_x^2 - k_y^2) \tilde{p}(k_x, k_y, z) = 0 $$

Solution on the following form: $\tilde{p}(k_x, k_y, z) = \tilde{p}(k_x, k_y, 0)e^{ik_z z}$ with $k_z$ the vertical wavenumber is unknown

→ Characteristic equation of the differential equation: $k_z^2 = k_0^2 - k_x^2 - k_y^2$

Solutions satisfying the Sommerfeld conditions:

$$ k_z = \begin{cases} -\sqrt{k_0^2 - k_x^2 - k_y^2} & \text{if } \sqrt{k_x^2 + k_y^2} < k_0 \\ i\sqrt{k_x^2 + k_y^2 - k_0^2} & \text{if } \sqrt{k_x^2 + k_y^2} > k_0 \end{cases} \rightarrow \begin{array}{l} \text{Propagative waves} \\ \text{Evanescent waves} \end{array} $$

Euler equation:

$$ \tilde{p}(k_x, k_y, 0) = Z_a \tilde{v}(k_x, k_y) \quad \text{with} \quad Z_a = -\frac{\rho_0 \omega}{k_z} $$

$$ \text{then } \tilde{p}(k_x, k_y, z) = Z_a \tilde{v}(k_x, k_y) e^{ik_z z} $$
Vibroacoustic problem in the wavenumber space (i.e. k-space):

$$
\left\{
\begin{array}{ll}
Z_p \tilde{v}(k_x, k_y) = \tilde{F} - \tilde{p}(k_x, k_y, 0) & \text{with } Z_p = \left[ D(1+i\eta)(k_x^2+k_y^2)^2 - \omega^2 \rho h \right] / i\omega \\[1.5em]
\tilde{p}(k_x, k_y, z) = Z_a \tilde{v}(k_x, k_y) e^{ik_z z} & \text{with } Z_a = -\frac{\rho_0 \omega}{k_z}, \quad k_z = \begin{cases} -\sqrt{k_0^2 - k_x^2 - k_y^2} & \text{if } \sqrt{k_x^2 + k_y^2} < k_0 \\ i\sqrt{k_x^2 + k_y^2 - k_0^2} & \text{if } \sqrt{k_x^2 + k_y^2} > k_0 \end{cases}
\end{array}
\right.
$$

Spectral velocity (i.e. velocity in the wavenumber space):

$$
\tilde{v}(k_x, k_y) = \frac{\tilde{F}(k_x, k_y)}{Z_p(k_x, k_y) + Z_a(k_x, k_y)}
$$

Spectral pressure at z (i.e. pressure in the wavenumber space for z given):

$$
\tilde{p}(k_x, k_y, z) = \Gamma(k_x, k_y, z) \tilde{v}(k_x, k_y)
$$

with the spectral propagator $\Gamma(k_x, k_y, z) = Z_a(k_x, k_y)e^{ik_z z}$

RK: Spectral force for a normal point force at $(x_0, y_0)$

$$
\tilde{F}(k_x, k_y) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \delta(x-x_0) \delta(y-y_0) e^{-ik_x x} e^{-ik_y y} dx dy = e^{-ik_x x_0} e^{-ik_y y_0}
$$
## II.3 – Estimation of the plate velocity in the physical space

Inverse space - Fourier transform:

$$ \tilde{f}(k_x, k_y) \xrightarrow{\text{I.F.T.}} f(x, y) = \frac{1}{4\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \tilde{f}(k_x, k_y) e^{j(k_x x + k_y y)} dk_x dk_y $$

→ No analytical expressions for the Fourier transform of the spectral displacement/pressure in general

→ Alternative: use an inverse Discrete Fourier Transform (iDFT) for approximating the solution

Truncation and sampling of the wavenumber domain $D$:

$$ \tilde{F}_{n_{k_x}, n_{k_y}} = \tilde{f}(n_{k_x} \delta k_x, n_{k_y} \delta k_y), \quad \forall (n_x, n_y) \in \left( \left[ \left( -\frac{\bar{k}_x}{\delta k_x}, \frac{\bar{k}_x}{\delta k_x} \right] \times \left( -\frac{\bar{k}_y}{\delta k_y}, \frac{\bar{k}_y}{\delta k_y} \right) \right) \right) $$

where $\delta k_x, \delta k_y$ are the wavenumber resolution, and, $\bar{k}_x, \bar{k}_y$ the cut-off wavenumbers of $D$.
<figure><img src="image_27.png" /></figure>

Truncated and discretized
wavenumber space

<figure><img src="image_28.png" /></figure>

Truncated and discretized
physical space

$$
\text{Properties of the iDFT: } \delta x = \frac{2\pi}{\Delta k_x} = \frac{2\pi}{k_x^{\max} - k_x^{\min}} \rightarrow \delta x = \frac{\pi}{k_x}, \delta y = \frac{\pi}{k_y} \\
\Delta X = X^{\max} - X^{\min} = \frac{2\pi}{\delta k_x} \rightarrow X^{\max} = \frac{\pi}{\delta k_x}, Y^{\max} = \frac{\pi}{\delta k_y}
$$

→ The wavenumber resolutions and the cut-off wavenumbers should be chosen carefully to avoid the loss of information and the aliasing effect.
## II.4 – Analysis of the plate velocity in the k-space WITHOUT fluid

Spectral velocity response of an in-vacuo plate excited by a point force at (0,0)

$$ \tilde{v}_{in-vacuo}(k_x, k_y) = \frac{F_0}{Z_p(k_x, k_y)} $$

$$ Z_p = \left[ \frac{D(1+i\eta)(k_x^2 + k_y^2)^2 - \omega^2 \rho h}{i\omega} \right] $$

Steel plate - Thickness= 2 mm

<figure><img src="image_29.png" /></figure>

<figure><img src="image_30.png" /></figure>

Dash line, flexural wavenumber (kf)

→ $k_y=0$: line force along y at x=0
Spectral velocity response of an **in-vacuo** plate excited by a point force at (0,0)

$$ \tilde{v}_{in-vacuo}(k_x, 0) = \frac{F_0}{Z_p(k_x, 0)} $$

$$ Z_p(k_x, 0) = \frac{[D(1+i\eta)k_x^4 - \omega^2\rho h]}{i\omega} $$

$$ Z_p(k_x, 0) = D \frac{[1+i\eta]k_x^4 - k_f^4}{i\omega} $$

<figure><img src="image_31.png" /></figure>

$$ k_f = \sqrt{\omega \sqrt{\frac{\rho h}{D}}} $$

<figure><img src="image_32.png" /></figure>

$$ Z_p(k_x, 0) \approx \frac{[-\omega^2 \rho h]}{i\omega} \quad Z_p(k_f, 0) \approx \frac{iD\eta k_f^4}{i\omega} \quad Z_p(k_x, 0) \approx \frac{Dk_x^4}{i\omega} $$

Mass controlled

Damping controlled

Stiffness controled
Criteria for choosing the wavenumber resolutions and the cut-off wavenumbers:

<figure><img src="image_33.png" /></figure>
Comparison with the analytical result (Matlab code: `plateEast.m`)

$$ 
\begin{aligned}
\kappa_x &= 2 \\
K_x &= 1.5
\end{aligned}
 $$

<figure><img src="image_34.png" /></figure>

<figure><img src="image_35.png" /></figure>

$$ 
\begin{aligned}
\kappa_x &= 4 \\
K_x &= 4
\end{aligned}
 $$

<figure><img src="image_36.png" /></figure>

<figure><img src="image_37.png" /></figure>
## II.5 – Analysis of the plate velocity in the k-space WITH fluid

Spectral velocity response of an **air** loaded plate excited by a point force at (0,0)

Steel plate in air - Thickness= 2 mm - Critical frequency= 5.8kHz

$$ \tilde{v}(k_x, 0) = \frac{F_0}{Z_p(k_x, 0) + Z_a(k_x, 0)} $$

Matlab code: spectral_fluid_loaded hjg_plate velocity.m

<figure><img src="image_38.png" /></figure>

! Below the
critical
frequency !
Spectral velocity response of an **air** loaded plate excited by a point force at (0,0)

Steel plate in air - Thickness= 2 mm - Critical frequency= 5.8kHz

**Below** the critical frequency

$$ \tilde{v}(k_x, 0) = \frac{F_0}{Z_p(k_x, 0) + Z_a(k_x, 0)} $$

$$ \lim_{k_x \to k_0} Z_a(k_x, 0) = \infty $$

<figure><img src="image_39.png" /></figure>

Plate velocity - 3000 Hz

<figure><img src="image_40.png" /></figure>

<figure><img src="image_41.png" /></figure>

<figure><img src="image_42.png" /></figure>

**! Below** the
critical
frequency !

for $f < f_c$

$Z_a(k_f, 0) \in \square^-$

$|Z_p(k_f, 0)| \gg Z_a(k_f, 0)$

for light fluid (like air)

<figure><img src="image_43.png" /></figure>

Plate velocity - 3000 Hz
Spectral velocity response of a **water** loaded plate excited by a point force at (0,0)

Steel plate in water - Thickness= 40 mm - Critical frequency= 5.7kHz

<figure><img src="image_44.png" /></figure>

! Below the
critical
frequency !

Dash line, flexural wavenumber (kf)
Solid line, acoustic wavenumber (k₀)
Why the peak has shift?

As $f < f_c \rightarrow k_f > k_0$

$$ \text{For } k_x \approx k_f \rightarrow \left| \begin{array}{l} k_z = i\sqrt{k_x^2 - k_0^2} \\ \hat{Z}_p(k_x, 0) + \hat{Z}_a(k_x, 0) = \left[ Dk_x^4 - \omega^2 \left( \rho h + \frac{\rho_0}{\sqrt{k_x^2 - k_0^2}} \right) \right] / i\omega \end{array} \right. $$

Fluid added mass

(inertial effect of the fluid on the shell)

For an air loaded plate, $\frac{\rho_0}{\sqrt{k_x^2 - k_0^2}}$ is generally negligible compared to $\rho h$.

Estimation of an equivalent mass density for the panel:

Considering the fluid added mass term with, $k_x = k_f$, one deduces $\rho_{eq} \approx \rho + \frac{\rho_0}{h\sqrt{k_f^2 - k_0^2}}$

Estimation of the flexural wavenumber taking the fluid added mass effect:

$$ k_f^{FL} = \sqrt{\omega \sqrt{\frac{\rho_{eq} h}{D}}} \quad \text{At } 3 \text{ kHz, } k_f^{FL} = 18.4 \text{ m}^{-1} \quad k_f = 17.3 \text{ m}^{-1} $$
Spectral velocity response of an **air** loaded plate excited by a point force at (0,0)

Steel plate in air - Thickness= 2 mm - Critical frequency= 5.7kHz

<figure><img src="image_45.png" /></figure>
Spectral velocity response of a **air** loaded plate excited by a point force at (0,0)

Steel plate in air - Thickness= 2 mm - Critical frequency= 5.7kHz

Focus on wavenumbers close to the acoustic wavenumber (i.e. zoom)

<figure><img src="image_46.png" /></figure>

Scholte-Stoneley wave:

- Surface wave (appears at the fluid-structure interface)

- Fluid-borne wave

- Evanescente in the normal direction of the panel

- Fewly damped (as the fluid is no damped) → Very narrow peak

$$k_{Scholte} = 147.839685 \text{ m}^{-1}$$
Spectral velocity response of a **water** loaded plate excited by a point force at (0,0)

Steel plate in water - Thickness= 40 mm - Critical frequency= 5.7kHz

<figure><img src="image_47.png" /></figure>

→ Contrary to the frequencies below the critical frequency, the flexural peak is not shifted, but it is less narrow.
Why the peak does not shift and is larger?

As $f > f_c \Rightarrow k_f < k_0$

$$ \text{For } k_x \approx k_f \Rightarrow \left| k_z = -\sqrt{k_0^2 - k_x^2} \quad Z_p(k_x, 0) + Z_a(k_x, 0) = \left[ D(1+i\eta)k_x^4 + i \frac{\rho_0\omega^2}{\sqrt{k_0^2-k_x^2}} - \omega^2\rho h \right] / i\omega \right. \\ \left. \underbrace{\text{Fluid added damping}} \right\} $$

Fluid added damping

(panel radiation into the fluid)

Estimation of an equivalent damping for the panel:

Considering the fluid added damping term with, $k_x = k_f$, one deduces

$$ \eta_{eq} \approx \eta + \frac{\rho_0 \omega^2}{D k_f^4 \sqrt{k_0^2 - k_f^2}} $$

At 8 kHz, with $\eta = 0.02 \Rightarrow \eta_{eq} \approx 0.18 \Rightarrow$ The internal panel damping is negligible compared to the radiation damping.

RK: for an air loaded plate, $\eta_{eq} \approx 0.0211 \Rightarrow$ The radiation damping is negligible compared to the internal panel damping.
## II.6 – Analysis of the plate velocity in the physical space

Matlab code: `fluid_loadedPlateVelocity.m`

Steel plate in water - Thickness= 40 mm - Critical frequency= 5.7kHz

3 kHz

<figure><img src="image_48.png" /><figcaption>Velocity - (dB, ref. 5d-8 m³/s)</figcaption></figure>

<figure><img src="image_49.png" /></figure>

8 kHz

<figure><img src="image_50.png" /><figcaption>Velocity - (dB, ref. 5d-8 m³/s)</figcaption></figure>

<figure><img src="image_51.png" /></figure>
## II.7 – Analysis of the radiated pressure in the k-space

Spectral pressure response of a water loaded plate excited by a point force at (0,0)

Steel plate in water - Thickness= 40 mm - Critical frequency = 5.7kHz

<figure><img src="image_52.png" /></figure>

Matlab code:

fluid_loadedzelatepressure.m

! Below the
critical frequency !

<figure><img src="image_53.png" /></figure>
Spectral pressure response of a water loaded plate excited by a point force at (0,0)

Steel plate in water - Thickness= 40 mm - Critical frequency = 5.7kHz

<figure><img src="image_54.png" /></figure>

! Above the
critical frequency !

$$k_{Scholte} = 147.839685 \text{ m}^{-1}$$

$$k_0 = 147.839654 \text{ m}^{-1}$$

<figure><img src="image_55.png" /></figure>

→ Above fc, propagative flexural waves
induced propagative waves into the fluid
Spectral pressure response of a water loaded plate excited by a point force at (0,0)

Steel plate in water - Thickness= 40 mm - Critical frequency = 5.7kHz

Focus on wavenumbers close to the acoustic wavenumber (i.e. zoom)

<figure><img src="image_56.png" /></figure>

Above the critical frequency !

$$k_{Scholte} = 147.839685 \text{ m}^{-1}$$

$$k_0 = 147.839654 \text{ m}^{-1}$$

<figure><img src="image_57.png" /></figure>

Propagative radiation

Evanescent radiation

→ Scholte-Stoneley waves induced evanescent waves into the fluid, but only very slightly decreasing

→ Effect of the damping/viscosity of the fluid? Open question?...
**Case of a monochromatic space excitation:**

Pressure induced by a wall plane wave of wavevector $(\bar{k}_x, \bar{k}_y) : F(x, y) = e^{i\bar{k}_x x + i\bar{k}_y y}$

Spectral force:

$$ \tilde{F}(k_x, k_y) = 4\pi^2 \delta(k_x - \bar{k}_x) \delta(k_y - \bar{k}_y) \quad \text{because} \quad F(x, y) = \frac{1}{4\pi^2} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{F}(k_x, k_y) e^{ik_x x} e^{ik_y y} dk_x dk_y $$

Spectral pressure:

$$ \tilde{p}(k_x, k_y, z) = \frac{4\pi^2 Z_a(k_x, k_y)}{Z_p(k_x, k_y) + Z_a(k_x, k_y)} e^{ik_z z} \delta(k_x - \bar{k}_x) \delta(k_y - \bar{k}_y) $$

Pressure in the physical space:

$$ p(x, y, z) = \frac{Z_a(\bar{k}_x, \bar{k}_y)}{Z_p(\bar{k}_x, \bar{k}_y) + Z_a(\bar{k}_x, \bar{k}_y)} e^{i\bar{k}_z z} e^{i\bar{k}_x x + i\bar{k}_y y} $$
Examples of the different modes of radiation ($\bar{k}_y = 0$)

$$0 < \bar{k}_x < k_0$$

$$p(x, 0, z)$$

$$\bar{k}_x = 0$$

<figure><img src="image_58.png" /></figure>

Direction of wave

$$k_x = k_0 \sin \theta$$

$$k_z = k_0 \cos \theta$$

<figure><img src="image_59.png" /></figure>

<figure><img src="image_60.png" /></figure>

<figure><img src="image_61.png" /><figcaption>Grazing plane wave</figcaption></figure>

<figure><img src="image_62.png" /></figure>

$$\bar{k}_x > k_0$$

<figure><img src="image_63.png" /></figure>

$\lambda > \lambda_x$ Direction of exponential decay

<figure><img src="image_64.png" /><figcaption>Evanescent wave</figcaption></figure>
## II.8 – Pressure field in the physical space

<figure><img src="image_65.png" /></figure>

Discrete domain definitions: D, wavenumber space; R, physical space.

$$ \delta x = \frac{2\pi}{\bar{k}_x}, \quad \delta y = \frac{2\pi}{\bar{k}_y}. $$

$$ \Delta X = \frac{2\pi}{\delta k_x}, \quad \Delta Y = \frac{2\pi}{\delta k_y}. $$

MATLAB codes: - `fluid_loadedPlatePressure.m` (pressure for a given Z in function of X) → Study the influence of the wavenumber resolution

- `fluid_loadedPlatePressure_map.m`
(pressure map in function of X and Z)
Line excitation (ky=0)
1000 Hz

<figure><img src="image_66.png" /></figure>

<figure><img src="image_67.png" /></figure>

<figure><img src="image_68.png" /></figure>

<figure><img src="image_69.png" /></figure>
Line excitation (ky=0)

1000 Hz

<figure><img src="image_70.png" /><figcaption>Subsonic components</figcaption></figure>

<figure><img src="image_71.png" /><figcaption>Total</figcaption></figure>

<figure><img src="image_72.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_73.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_74.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_75.png" /></figure>
Line excitation (ky=0)
16000 Hz

<figure><img src="image_76.png" /></figure>

<figure><img src="image_77.png" /></figure>

<figure><img src="image_78.png" /></figure>

<figure><img src="image_79.png" /></figure>
Line excitation (ky=0)
6000 Hz

<figure><img src="image_80.png" /></figure>

<figure><img src="image_81.png" /></figure>

<figure><img src="image_82.png" /></figure>

<figure><img src="image_83.png" /></figure>
Line excitation (ky=0)
6000 Hz

<figure><img src="image_84.png" /><figcaption>Subsonic components</figcaption></figure>

<figure><img src="image_85.png" /><figcaption>Total</figcaption></figure>

<figure><img src="image_86.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_87.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_88.png" /><figcaption>Pressure - (dB, ref. 1d-6 Pa)</figcaption></figure>

<figure><img src="image_89.png" /></figure>
## II.9 – Radiation power and radiation efficiency

### Reminder: Parseval-Plancherel formula

$$ \int_{-\infty}^{\infty} f(x) g^*(x) dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{f}(k_x) \tilde{g}^*(k_x) dx $$

$$ \int_{-\infty}^{\infty} |f(x)|^2 dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} |\tilde{f}(k_x)|^2 dx $$

### Demo:

$$ \begin{align*} \int_{-\infty}^{\infty} f(x) g^*(x) dx &= \int_{-\infty}^{\infty} f(x) \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{g}(k_x) dk_x dx \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x) \tilde{g}(k_x) dk_x dx \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x) \tilde{g}(k_x) dx dk_x \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{g}(k_x) \int_{-\infty}^{\infty} f(x) dx dk_x \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{g}(k_x) \tilde{f}(k_x) dk_x \end{align*} $$
Radiation efficiency:

$$ \sigma = \frac{\Pi_{rad}}{\rho_0 c_0 SV^2} \quad \text{with} $$

$$ \text{- Sum of the quadratic velocity:} \quad SV^2 = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \bar{v}^2(x, y, t) dx dy dt $$

$$ \text{- Radiated power:} \quad \Pi_{rad} = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \bar{p}(x, y, 0, t) \bar{v}(x, y, t) dx dy dt $$

<figure><img src="image_90.png" /></figure>

Harmonic excitation, stationary response and complex notation

$$ \bar{v}(x, y, t) = \text{Re}\{v(x, y)e^{j\omega t}\} $$

$$ \bar{p}(x, y, z, t) = \text{Re}\{p(x, y, z)e^{j\omega t}\} $$

$$ SV^2 = \frac{1}{2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |v(x, y)|^2 dx dy $$

$$ \Pi_{rad} = \frac{1}{2} \text{Re} \left\{ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} p(x, y, 0) v^*(x, y) dx dy \right\} $$
Using the Parseval-Plancherel formula

$$SV^2 = \frac{1}{2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |v(x, y)|^2 dxdy = \frac{1}{8\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |\tilde{v}(k_x, k_y)|^2 dk_x dk_y$$

$$\Pi_{rad} = \frac{1}{8\pi^2} \text{Re} \left\{ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{p}(k_x, k_y, 0) \tilde{\nu}^*(k_x, k_y) dk_x dk_y \right\}$$

$$\Pi_{rad} = \frac{1}{8\pi^2} \text{Re} \left\{ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} Z_a(k_x, k_y) |\tilde{v}(k_x, k_y)|^2 dk_x dk_y \right\}$$

$$\Pi_{rad} = \frac{1}{8\pi^2} \text{Re} \left\{ \iint_{\Omega_a} Z_a(k_x, k_y) |\tilde{v}(k_x, k_y)|^2 dk_x dk_y \right\}$$
with $$\Omega_a = \{(k_x, k_y) \in \Box^2, \sqrt{k_x^2 + k_y^2} \le k_0\}$$
Numerical evaluation of the previous quantities:

- Truncation of the wavenumber space with a cut-off wavenumber (only $\bar{k}$ for velocity)

$$k_x \in [-\bar{k}, \bar{k}], k_y \in [-\bar{k}, \bar{k}] \quad \bar{k} = \kappa \max[k_0, k_f] \quad \kappa > 2$$

- Discretisation of the wavenumber space with a wavenumber resolution $\delta k$

Remark: Other interpretation of the radiated power

No energy dissipation into the fluid

Integrating the radial component of the intensity vector over a large sphere of radius $R$ concentric with the sound source

Intensity estimated from plane wave approximation:

$$Z = \frac{p}{\nu} = \rho_0 c_0 \qquad I = \frac{1}{2} p \nu^* = \frac{1}{2 \rho_0 c_0} |p|^2$$

$$\Pi_{\text{rad}} = \frac{1}{2\rho_0 c_0} \iint_{\Sigma^{\infty}} |p|^2 dS \quad \sum^{\infty} \text{Hemi sphere of large radius}$$
# Radiation power and radiation efficiency – Air loaded panel

Steel plate in air - Thickness= 2 mm - Critical frequency= 5.8kHz

<figure><img src="image_91.png" /></figure>

<figure><img src="image_92.png" /></figure>

<figure><img src="image_93.png" /></figure>

<figure><img src="image_94.png" /></figure>

<figure><img src="image_95.png" /></figure>
# Radiation power and radiation efficiency – Water loaded panel

Steel plate in water - Thickness= 40 mm - Critical frequency = 5.4kHz

<figure><img src="image_96.png" /></figure>

<figure><img src="image_97.png" /></figure>

<figure><img src="image_98.png" /></figure>

<figure><img src="image_99.png" /></figure>
ON THE RADIATION EFFICIENCY OF INFINITE
PLATES SUBJECT TO A POINT LOAD IN WATER

J. HANNSEN SU AND R. VASUDEVAN

Signature Control Technology, Naval Surface Warfare Center, Carderock Division,
9500 MacArthur Boulevard, West Bethesda, Maryland 20817-5700, U.S.A.

(Received 17 July 1995, and in final form 7 July 1997)

**Journal of Sound and Vibration (1997) 208, 441-445.**

In this paper, an approach to calculate the radiation efficiency for infinite plates has been developed and validated. The comparison between the results from the thin plate theory and those from the thick plate theory indicates the thin plate theory is adequate for radiated power calculation up to 0.7 times the coincidence frequency but starts to underestimate significantly for frequencies larger than twice the coincidence frequency. As far as the radiation efficiency is concerned, the thin plate theory is adequate up to 0.2 times the coincidence frequency, and for higher frequencies, it overestimates the efficiency (see Figure 2). Consequently, it predicts a lower frequency at which the radiation efficiency approaches unity when compared to the thick plate theory. It is concluded that the heavier fluid loading causes the radiation efficiency to approach unity at a higher frequency.

The results obtained are valid for the radiation efficiency of finite plates of real material excited by a point load in water for frequencies near and above the coincidence frequency.
# Results for a brass 12.7mm thick plate excited by a point excitation

Critical frequency in water: 26 420 Hz

<figure><img src="image_100.png" /></figure>

The calculated (a) radiation efficiency and (b) radiated power of an infinite plate loaded with water on one side and subjected to a point load. ——, By the "thick plate" theory; ---, by the "thin plate" theory.

→ As shown on this example, the thin plate (love-kirchhoff) model may not be adapted around and above the critical frequency

→ Calculation in the k-space with the thick plate (mindlin) model can be performed
Influence of the type of excitation on the radiation efficiency

<figure><img src="image_101.png" /></figure>

Radiation efficiency: (a), acoustic excitation; (b), mechanical point force.
Macadam 1977

→ Radiation efficiency depend on the type of excitation below the critical frequency
## II.10 – Evaluation of the radiated pressure in the far field

**Stationary Phase theorem:**

Let $\varphi(k_x)$ a continuous function and $\psi(k_x)$ a continuous with a single extremum $\bar{k}_x$ such that $\frac{\partial^2 \psi}{\partial k_x^2}(\bar{k}_x) \neq 0$.

The integral $I = \int_{-\infty}^{+\infty} \phi(k_x) e^{j\psi(k_x)} dk_x$ can be approximated by:

$$ egin{cases} I \approx \frac{\sqrt{2\pi}\varphi(\bar{k}_x)e^{\frac{j\pi}{4}+j\psi(\bar{k}_x)}}{\sqrt{\left|\frac{\partial^2\psi}{\partial k_x^2}(\bar{k}_x)\right|}}, & \text{si } \frac{\partial^2\psi}{\partial k_x^2}(\bar{k}_x) > 0, \\ I \approx \frac{\sqrt{2\pi}\varphi(\bar{k}_x)e^{-\frac{j\pi}{4}+j\psi(\bar{k}_x)}}{\sqrt{\left|\frac{\partial^2\psi}{\partial k_x^2}(\bar{k}_x)\right|}}, & \text{si } \frac{\partial^2\psi}{\partial k_x^2}(\bar{k}_x) < 0. \end{cases} $$

**Principle:** - $\varphi$ varies slowly than $e^{j\psi(k_x)}$;

- when the phase of $\psi$ varies of $2\pi$, the integral is null;

- only the « points » when the phase is stationary (null derivative) contribute to the integral.
$$p(x, y, z) = \frac{1}{8\pi^2} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} Z_a(k_x, k_y) \tilde{v}(k_x, k_y) e^{ik_x x + ik_y y + ik_z z} dk_x dk_y$$

$$p(x, y, z) = \frac{1}{8\pi^2} \int_{-k_0}^{+k_0} \int_{-k_0}^{+k_0} Z_a(k_x, k_y) \tilde{v}(k_x, k_y) e^{ik_x x + ik_y y - i\sqrt{k_0^2 - k_x^2 - k_y^2} z} dk_x dk_y$$

$$\psi(k_x, k_y) = k_x x + k_y y - \sqrt{k_0^2 - k_x^2 - k_y^2} z$$

<figure><img src="image_102.png" /></figure>

$$\begin{cases} x = R \sin \theta \cos \phi \\ y = R \sin \theta \sin \phi \\ z = R \cos \theta \end{cases}$$

Radiated pressure in the far field (i.e. $k_0 R \ll 1$):

$$p_{\infty}(R, \theta, \phi) = \frac{i\omega\rho_0}{2\pi R} \tilde{v}(\bar{k}_x, \bar{k}_y) e^{-ik_0 R} \quad \text{with} \quad \begin{cases} \bar{k}_x = k_0 \sin \theta \cos \phi \\ \bar{k}_y = k_0 \sin \theta \sin \phi \end{cases}$$
# Directivity of the far-field radiated noise– Water loaded panel

Steel plate in water - Thickness= 40 mm - Critical frequency = 5.4kHz – R=10 m –
Line excitation (ky=0)

3000 Hz

8000 Hz

16000 Hz

<figure><img src="image_103.png" /></figure>

<figure><img src="image_104.png" /></figure>

<figure><img src="image_105.png" /></figure>

<figure><img src="image_106.png" /></figure>

<figure><img src="image_107.png" /></figure>

<figure><img src="image_108.png" /></figure>
## II.11 - Transmission loss of an infinite plate

<figure><img src="image_109.png" /></figure>

How can we introduce the plane wave excitation in the VA problem?

- With a source term $q$ in the right member of the Helmholtz equation:

$$ \nabla^2 p - \frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} = q \quad \Rightarrow \text{Definition of } q? $$

$$ n \cdot \nabla p = \rho_0 \omega^2 w $$

- Using the « blocked » pressure induced by the external source
Description of an external source by its « blocked » pressure

<figure><img src="image_110.png" /></figure>

Problem with rigid wall:

$$ \nabla^2 p_b - \frac{1}{c_0^2} \frac{\partial^2 p_b}{\partial t^2} = q $$

$$ n \cdot \nabla p_b = 0 $$

Blocked pressure on the wall:
(source-image theory)

$$ p_b(M') = 2 p_0 e^{-i \vec{k}_i \cdot \vec{OM}'} = 2 p_0 e^{-ik_x x} $$

with $k_x = k_0 \sin \theta$

$$ n \cdot \nabla p_b = 0 $$

Pressure decomposition: $p = p_b + p_{rad}$

<figure><img src="image_111.png" /></figure>

Linear operator

$$ \nabla^2 p_{rad} - \frac{1}{c_0^2} \frac{\partial^2 p_{rad}}{\partial t^2} = 0 \qquad n \cdot \nabla p_{rad} = \rho_0 \omega^2 w $$

$p_{rad}$, the radiated pressure respects the homogeneous Helmholtz eq. and the Euler eq.

→ If the blocked pressure is known, it is not necessary to define the source term $q$
Transmission loss of an infinite plate for an incident acoustic wave

Resolution in the wavenumber space with $\tilde{p}_b(k_x, k_y) = 2p_0\delta(k_x - \sin(\theta)k_0)$

Transmission loss: $R = 10\log_{10}\left(\frac{W_t}{W_i}\right) = 10\log_{10}\left(\frac{p_t^2}{p_o^2}\right)$

<figure><img src="image_112.png" /></figure>

Incident angle from the normal

<figure><img src="image_113.png" /></figure>
Transmission loss of an infinite plate under a diffuse field

Diffuse field = an infinite set of uncorrelated acoustic plane waves with uniformly distributed incidence angles

<figure><img src="image_114.png" /></figure>

Random diffuse field incidence TL for a 25 mm thick steel panel in air
blue line, TL for a random diffuse field; grey line, TL for various angles of incidence.

<figure><img src="image_115.png" /></figure>

Panel damping loss factor: 0.005

<figure><img src="image_116.png" /></figure>

Panel damping loss factor: 0.05

Stiffness law
(9 dB/oct)
# III – Vibroacoustic response of a finite panel excited by a mechanical force

## III.1 – Plate velocity

<figure><img src="image_117.png" /></figure>

Simply supported plate excited by a normal mechanical point force

Finite structure → Modal behavior

Modal information for the S-S plate:

- Modal angular frequency:

$$ \omega_{pq} = \sqrt{\frac{D}{\rho h} \left[ \left( \frac{p\pi}{a} \right)^2 + \left( \frac{q\pi}{b} \right)^2 \right]} $$

- Mode shape:

$$ \phi_{pq}(x, y) = \sin(k_p x)\sin(k_q y) \text{ with } k_p = \frac{p\pi}{a}, k_q = \frac{q\pi}{b} $$

- Generalized mass: $M_{pq} = \rho h a / 4$

<figure><img src="image_118.png" /></figure>

<figure><img src="image_119.png" /></figure>

<figure><img src="image_120.png" /></figure>
Plate displacement response (by modal expansion method):

$$v(x, y, t) = \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} \frac{i\omega F_{pq} \varphi_{pq}(x, y)}{M_{pq} (-\omega^2 + i\omega\omega_{pq}\eta_{pq} + \omega_{pq}^2)} e^{i\omega t}$$

with the generalized forces $F_{pq} = \int_{0}^{a} \int_{0}^{b} p(x, y) \phi_{pq}(x, y) dx dy$

<figure><img src="image_121.png" /><figcaption>Illustration of the modal response</figcaption></figure>
Example for a harmonic mechanical point force at $(x_0, y_0)$

<figure><img src="image_122.png" /></figure>

$$v(x, y, \omega) = \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} \frac{i \omega F_0 \varphi_{pq}(x_0, y_0) \varphi_{pq}(x, y)}{M_{pq} (-\omega^2 + i\omega\omega_{pq}\eta_{pq} + \omega_{pq}^2)}$$

with $F_0$ the magnitude of the force.

Displacement in the wavenumber space:

$$\tilde{v}(k_x, k_y, \omega) = \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} \frac{i \omega F_0 \varphi_{pq}(x_0, y_0) \tilde{\varphi}_{pq}(k_x, k_y)}{M_{pq} (-\omega^2 + i\omega\omega_{pq}\eta_{pq} + \omega_{pq}^2)}$$

$$\int_{0}^{a} \sin(k_p x) e^{-ik_x x} dx = \int_{0}^{a} \frac{e^{ik_p x} - e^{-ik_p x}}{2i} e^{-ik_x x} dx = \dots = \frac{p\pi}{a} \frac{(-1)^p e^{-ik_x a} - 1}{k_x^2 - (p\pi/a)^2}$$

$$\tilde{\varphi}_{pq}(k_x, k_y) = k_p k_q \left[ \frac{(-1)^p e^{-ik_x a} - 1}{k_x^2 - k_p^2} \right] \left[ \frac{(-1)^q e^{-ik_y b} - 1}{k_y^2 - k_q^2} \right] \quad \text{with} \quad k_q = \frac{q\pi}{b}, \quad k_p = \frac{p\pi}{a}$$
the modal wavenumbers
Plate response for a mechanical point force at (-0.3m, -0.2m) – 500 Hz

<figure><img src="image_123.png" /><figcaption>Velocity in the physical space (dB, ref. 1 m/s) - Finite plate</figcaption></figure>

<figure><img src="image_124.png" /><figcaption>Velocity in the wavenumber space (dB, ref. 1 m3/s) - Infinite plate</figcaption></figure>

<figure><img src="image_125.png" /></figure>

Infinite plate

<figure><img src="image_126.png" /><figcaption>Velocity in the wavenumber space (dB, ref. 1 m3/s) - Finite plate</figcaption></figure>

Simply supported steel plate 1.2m x 1m x 2mm
$$ \tilde{\varphi}_{pq}(k_x, k_y) = k_p k_q \left[ \frac{(-1)^p e^{-ik_x a} - 1}{k_x^2 - k_p^2} \right] \left[ \frac{(-1)^q e^{-ik_y b} - 1}{k_y^2 - k_q^2} \right] $$

<figure><img src="image_127.png" /></figure>

<figure><img src="image_128.png" /><figcaption>Modal wavenumbers of resonant modes, flexural wavenumber and acoustic wavenumber versus frequency
Fahy, F. – Sound and structural vibration, 1985.</figcaption></figure>

<figure><img src="image_129.png" /></figure>
$$A_{pq} = \frac{i\omega F_{pq}}{M_{pq}(-\omega^2 + i\omega\omega_{pq}\eta_{pq} + \omega_{pq}^2)}$$

<figure><img src="image_130.png" /></figure>

$$\tilde{v}(k_x, k_y, \omega) = \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} \frac{i\omega F_{pq} \tilde{\varphi}_{pq}(k_x, k_y)}{M_{pq}(-\omega^2 + i\omega\omega_{pq}\eta_{pq} + \omega_{pq}^2)}$$
### III.2 – Radiated pressure

<figure><img src="image_131.png" /></figure>

$$p(k_x, k_y, z) \approx \Gamma(k_x, k_y) \tilde{v}(k_x, k_y)$$

with the spectral propagator:

$$\Gamma(k_x, k_y) \approx -\frac{\rho_0 \omega}{k_z} e^{ik_z z}$$

$$k_z = \begin{cases} -\sqrt{k_0^2 - k_x^2 - k_y^2} & \text{if } \sqrt{k_x^2 + k_y^2} < k_0 \\ i\sqrt{k_x^2 + k_y^2 - k_0^2} & \text{if } \sqrt{k_x^2 + k_y^2} > k_0 \end{cases}$$

→!! Light fluid assumption !!

## Spectral radiated pressure at different distances from the panel

<figure><img src="image_132.png" /></figure>

<figure><img src="image_133.png" /></figure>

<figure><img src="image_134.png" /></figure>

<figure><img src="image_135.png" /></figure>

<figure><img src="image_136.png" /></figure>

<figure><img src="image_137.png" /></figure>
$$p(k_x, k_y, z) = Z_a \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} A_{pq} \tilde{\varphi}_{pq}(k_x, k_y) e^{ik_z z}$$

$\tilde{\varphi}_{pq}(k_x, k_y)$

<figure><img src="image_138.png" /><figcaption>Spectral modeshape - 498Hz</figcaption></figure>

<figure><img src="image_139.png" /></figure>

$$A_{pq} = \frac{i\omega F_0 \varphi_{pq}(x_0, y_0)}{M_{pq}(-\omega^2 + i\omega\omega_{pq}\eta_{pq} + (\omega_{pq})^2)}$$

$\tilde{\varphi}_{pq}(k_x, k_y)e^{ik_z z}$

<figure><img src="image_140.png" /><figcaption>Spectral mode radiation - 498Hz - Z=1 m</figcaption></figure>

<figure><img src="image_141.png" /></figure>
<figure><img src="image_142.png" /><figcaption>Edge modes</figcaption></figure>

<figure><img src="image_143.png" /><figcaption>Corner modes</figcaption></figure>

<figure><img src="image_144.png" /></figure>

<figure><img src="image_145.png" /></figure>

Area radiating effectively in the far field
<figure><img src="image_146.png" /></figure>

<figure><img src="image_147.png" /></figure>

<figure><img src="image_148.png" /></figure>

<figure><img src="image_149.png" /></figure>

<figure><img src="image_150.png" /></figure>

<figure><img src="image_151.png" /></figure>
<figure><img src="image_152.png" /><figcaption>Radiating modeshape - 419Hz</figcaption></figure>

<figure><img src="image_153.png" /></figure>

<figure><img src="image_154.png" /><figcaption>Radiating modeshape - 421Hz</figcaption></figure>

<figure><img src="image_155.png" /></figure>

<figure><img src="image_156.png" /><figcaption>Radiating modeshape - 455Hz</figcaption></figure>

<figure><img src="image_157.png" /></figure>

→ Compare the amplitude between the standard modeshape and the radiating modeshape

(see in the literature the comparison between active intensity and non-negative intensity/supersonic intensity (Williams))
# IV – Nearfield Acoustical Holography

## IV.1 – Principle

<figure><img src="image_158.png" /></figure>

→ Williams and Maynard, 1980, Holographic Imaging without the wavelength resolution limit, Phys. Rev. Lett., 45, pp.554-557.

→ Inverse problem: knowing the measured pressures in a plane at a given distance of the radiated structure, what is the velocity field of the structure?
Spectral pressure from the spectral velocity at $z=0$ :

$$ \tilde{p}(k_x, k_y, z) = \Gamma(k_x, k_y, z) \tilde{v}(k_x, k_y), \forall z \ge 0 $$

with the spectral propagator: $\Gamma(k_x, k_y, z) = Z_a(k_x, k_y) e^{ik_x z}$

$$ \text{with: } k_z = \begin{cases} -\sqrt{k_0^2 - k_x^2 - k_y^2} & \text{if } \sqrt{k_x^2 + k_y^2} < k_0 \\ i\sqrt{k_x^2 + k_y^2 - k_0^2} & \text{if } \sqrt{k_x^2 + k_y^2} > k_0 \end{cases} $$

Spectral pressure at $z_h$ from the spectral velocity at $z=0$

$$ \tilde{p}(k_x, k_y, z_h) = \Gamma(k_x, k_y, z_h) \tilde{v}(k_x, k_y) $$

Spectral velocity at $z=0$ from the spectral pressure at $z_h$

$$ \tilde{v}_z(k_x, k_y) = G(k_x, k_y, z_h) \tilde{p}(k_x, k_y, z_h) $$

with the inverse spectral propagator:

$$ G(k_x, k_y, z_h) = \frac{1}{\Gamma(k_x, k_y, z_h)} = \frac{e^{-ik_z z_h}}{Z_a(k_x, k_y)} $$
The process can be decomposed in 4 steps:

(1) Measure the pressure field in a plane (called hologram) at $z_h$ distance from the panel $p(k_x, k_y, z_h)$

(2) Compute the spectral pressure $\tilde{p}(k_x, k_y, z_h)$

(3) Multiply by the inverse propagator $G(k_x, k_y, z_h) \tilde{p}(k_x, k_y, z_h) = \tilde{v}_z(k_x, k_y)$

(4) Compute the inverse Fourier transforms $\tilde{v}_z(k_x, k_y) \rightarrow v_z(x, y)$

Remark: possibility to reconstruct also the wall pressure and the normal acoustic intensity

Because NAH solves an inverse problem, the mathematical solution must be approached with some caution to assure that the solution is unique and stable.⁴ That is, straightforward implementation of Eq. (3.4) will lead to disaster! How this equation must be modified to avoid disaster is the springboard for the NAH technique.

E.G. Williams, FOURIER Acoustics: Sound Radiation and Neafield Acoustical Holography, Academic press, 1999.
IV.2 – k-space filter

Arrays of microphone → Measurement at discrete points $\delta x$ → Use of Discrete
Fourier Transform for estimating the spectral pressure → introduction of a sampling
wavenumber (i.e. Cutoff wavenumber) $\bar{k}_x = \frac{\pi}{\delta x}$

$$1\text{D example: } v(x) = \sin(3.5\pi x) \Pi_{[-1,1]}(x)$$

<figure><img src="image_159.png" /></figure>

Velocity to reconstruct (target)

<figure><img src="image_160.png" /></figure>

Spectral velocity: sinc function

<figure><img src="image_161.png" /></figure>

<figure><img src="image_162.png" /></figure>

Large errors
Modification of the filter shape (for 'reducing' the edge reconstruction errors)

$$ \bar{\Pi}(k_x/(2k_c)) = \begin{cases} 1 - \frac{1}{2}e^{-(1-|k_x|/k_c)/\alpha} & |k_x| < k_c \\ \frac{1}{2}e^{(1-|k_x|/k_c)/\alpha} & |k_x| > k_c \end{cases} $$

<figure><img src="image_163.png" /></figure>

<figure><img src="image_164.png" /><figcaption>With rectangular filter</figcaption></figure>

<figure><img src="image_165.png" /><figcaption>With tapered filter</figcaption></figure>

→ Very slight improvement with the tapered filter

→ Use greater kc for better reconstruction at the edges
IV.3 – Influence of measurement noise

<figure><img src="image_166.png" /></figure>

<figure><img src="image_167.png" /></figure>

$z_h=0.1$ m, 500 Hz, Air

<figure><img src="image_168.png" /></figure>

<figure><img src="image_169.png" /></figure>
Influence of measurement noise: use of k-filter with cutoff wavenumber $k_c=40 \text{ 1/m}$.

<figure><img src="image_170.png" /></figure>

<figure><img src="image_171.png" /></figure>

<figure><img src="image_172.png" /></figure>

<figure><img src="image_173.png" /></figure>
IV.4 – Influence of the standoff distance $z_h$

Inverse propagator

<figure><img src="image_174.png" /></figure>

$$G(k_x, k_y, z_h) = \frac{e^{-ik_z z_h}}{Z_a(k_x, k_z)}$$

<figure><img src="image_175.png" /></figure>

<figure><img src="image_176.png" /></figure>
IV.5 – Finite Measurement Aperture Effect

Measurement on an infinite plane is not possible in practice

→ Measurement on a rectangular area (called measurement aperture)

$$p_{mes}(x, y) = p(x, y) \Pi(x / L_x) \Pi(y / L_y)$$

$$\tilde{p}_{mes}(k_x, k_y) = \tilde{p}(k_x, k_y) * \tilde{\Pi}_{Aperture}(k_x, k_y)$$

$$\tilde{\Pi}_{Aperture}(k_x, k_y) = \text{sinc}(k_x L_x / 2) \text{sinc}(k_y L_y / 2)$$

The convolution with the sinc functions spread the spectrum of $\tilde{p}$.

Non-physical contribution may appear and amplified by the inverse propagator.

→ In practice, the measurement aperture is always larger than the actual source.

The size of the aperture should be larger than twice the actual source. This may guarantees that the measured pressure field drops off significantly towards the edges of the aperture.
IV.6 – Discretization and Aliasing

Microphone spacing should be chosen such that the aliasing effect
may be negligible.

Conclusion

Because NAH solves an inverse problem, the mathematical solution must be approached with some caution to assure that the solution is unique and stable.⁴ That is, straightforward implementation of Eq. (3.4) will lead to disaster! How this equation must be modified to avoid disaster is the springboard for the NAH technique.

For more details, see E.G. Williams, FOURIER Acoustics: Sound Radiation and Neafield Acoustical Holography, Academic press, 1999.

Quentin Leclere, quentin.leclere@insa-lyon.fr

Bernard Laulagnet, bernard.laulagnet@insa-lyon.fr

Laboratoire Vibrations Acoustique, INSA-Lyon

<figure><img src="image_177.png" /><figcaption>RMS velocity maps in the near-field region of a gasoline engine: left, 600 Hz; center, 1100 Hz; right, 4200 Hz.</figcaption></figure>

<figure><img src="image_178.png" /><figcaption>Left: velocity of the membrane at 4 cm distance from the acoustic source at 2700 Hz. Right: back-propagated velocity field in the source plane using NAH. The chromatic scale represents negative, zero and positive acoustic values.</figcaption></figure>
# Appendix

## SOUND RADIATION OF A PERIODICALLY STIFFENED
PLATE EXCITED BY A POINT FORCE
# I - INTRODUCTION

Stiffened structures are very common for practical applications

**Example:**

- Fuselage/wing of an airplane

- Railway car structure

- Submarine pressure hull

<figure><img src="image_179.png" /></figure>

Need of « basic models » for understanding how stiffened structures radiated in the near field/far field

- effects of stiffener dynamics, of periodicity

- Influence of stop band/pass band (Block-Floquet wave)
## II – FLEXURAL BLOCH-FLOQUET WAVES

One-dimensional periodic structure: example of the simply supported beam

<figure><img src="image_180.png" /></figure>

<figure><img src="image_181.png" /></figure>

D.J. Mead, Wave propagation in continuous periodic structures: research contributions from Southampton 1964-1995, JSV (1996) 190, 495-524.
## II – FLEXURAL BLOCH-FLOQUET WAVES

One-dimensional periodic structure: example of the simply supported beam

<figure><img src="image_182.png" /></figure>

<figure><img src="image_183.png" /></figure>

D.J. Mead, Wave propagation in continuous periodic structures: research contributions from Southampton 1964-1995, JSV (1996) 190, 495-524.

Despite the numerous studies of wave motion in continuous periodic systems over the past 40 years, a simple physical explanation has yet to be presented for the very existence of frequency-propagation zones and attenuation zones. However, even if there is no simple answer to the question "Why does wave motion of one frequency propagate freely while motion of another frequency is attenuated?", reliable prediction methods do exist for the properties of free motion and the magnitudes of forced motion.
## II – PERIODICALLY STIFFENED PLATE

<figure><img src="image_184.png" /></figure>

Fluid load panel periodically stiffened by line beams

Stiffened panel: 2-D system, periodic in one direction (i.e. x axis)

<figure><img src="image_185.png" /></figure>

Space-Fourier transform about y-axis

For each wavenumber $k_y$, one-dimensional periodic system

(see L. MAXIT - Wavenumber space and physical space responses of a periodically ribbed plate to a point drive: A discrete approach. *Applied Acoustics*, 70 (2009) 563-578.) [10.1016/j.apacoust.2008.06.012](https://doi.org/10.1016/j.apacoust.2008.06.012)
## Some elements for the spectral displacement calculation

<figure><img src="image_186.png" /></figure>

### Formulation of vibro-acoustic interaction (1):

Love – Kirchhoff assumptions

<figure><img src="image_187.png" /></figure>
Formulation of vibro-acoustic interaction (2):

<figure><img src="image_188.png" /></figure>

Beam assumptions (without warping)

- Bending motions:

$$E_p^* I_x \frac{\partial^4 W}{\partial y^4}(x, y) - \omega^2 \rho_p S_p W(x, y) = F(x, y)$$

<figure><img src="image_189.png" /></figure>

- Torsion motions:

$$-G_p^* J_p \frac{\partial^3 W}{\partial x \partial y^2}(x, y) - \omega^2 \rho_p I_0 \frac{\partial W}{\partial x}(x, y) = M(x, y)$$

<figure><img src="image_190.png" /></figure>
Formulation of vibro-acoustic interaction (3):

Fluid domain (water)

<figure><img src="image_191.png" /></figure>

- Helmholtz equation in the fluid domain:

$$ \Delta p + \frac{\omega^2}{c_0^2} p = 0 $$

- Euler equation and normal velocity continuity at z=0:

$$ - \rho_0 \omega^2 W(x, y) = - \frac{\partial p}{\partial z}(x, y, 0) $$

- Sommerfeld radiation conditions on a fictive infinite surface, $\Sigma_{inf}$ (no reflection)
Resolution using the Space-Fourier transform:

$$f(x, y) \xrightarrow{T.F.} \tilde{f}(k_x, k_y) = \int_{\Box^2} f(x, y) e^{-j(k_x x + k_y y)} dxdy$$

$$\text{Properties: } \frac{\partial^p \partial^q W}{\partial x^p \partial y^q} (x, y) \rightarrow (ik_x)^p (ik_y)^q \tilde{W}(k_x, k_y)$$

$$\sum_{n \in \mathbb{Z}} \delta(x - nd) \xrightarrow{\text{F.T.}} \frac{1}{d} \sum_{n \in \mathbb{Z}} \delta\left(k_x + \frac{2\pi n}{d}\right) \quad (\text{Poisson formula})$$

Results: Plate displacements in the wave number domain $(k_x, k_y)$

$$\tilde{W}(k_x, k_y) = \tilde{W}_0(k_x, k_y) \Gamma(k_x, k_y)$$

Displacement of the bare plate
$$\tilde{W}_0(k_x, k_y) = \frac{F_0 e^{-jk_x x_0}}{Z(k_x, k_y)}$$

Rib effect:

$$\Gamma = \frac{A_p Z_p S_1 (S_1 - \hat{S}_1) - (1 + A_p S_2) [1 + Z_p (S_0 - \hat{S}_0)] - k_x A_p [Z_p (\hat{S}_0 S_1 - S_0 \hat{S}_1) - \hat{S}_1]}{A_p Z_p S_1^2 - (1 + A_p S_2) (1 + Z_p S_0)}$$
# DEFINITION OF CRITERIONS

## 1 - CRITERION ON THE MAXIMUM WAVENUMBERS

<figure><img src="image_192.png" /><figcaption>Level of the displacement in the wavenumber domain $(k_x, k_y)$</figcaption></figure>
# DEFINITION OF CRITERIONS

## 1 - CRITERION ON THE MAXIMUM WAVENUMBERS

Term 20\*log(S₀) in the wavenumber domain (kₓ, kᵧ)

<figure><img src="image_193.png" /></figure>

Maximums for: $\sqrt{\left(k_{x}+\frac{2\pi n}{d}\right)^{2}+k_{y}^{2}} \approx k_{f,water}^{plate}$
# DEFINITION OF CRITERIONS

## 1 - CRITERION ON THE MAXIMUM WAVENUMBERS

<figure><img src="image_194.png" /></figure>

Criterions defined with these observations:

$$ \bar{k}_x = \kappa_x \max \left[ k_0, k_{f,water}^{plate} + \frac{2\pi\chi}{d} \right] \quad \bar{k}_y = \kappa_y \max \left[ k_0, k_{f,water}^{plate}, k_f^{beam}, k_t^{beam} \right] \\ \text{with } \kappa_x = 1,5 \quad \kappa_y = 1,5 \quad \chi = 2 $$
# DEFINITION OF CRITERIONS

## 2 - DISCRETE SPACE CRITERION $\delta k_x, \delta k_y$

Half power bandwidths for a 4th order admittance
(plate/beam flexural motion)

<figure><img src="image_195.png" /></figure>

Half power bandwidths for a 2nd order admittance
(beam torsion motion)

$$ \Delta_{-3dB}^{(2)} = \eta \bar{k} $$

Discretisation space criterion:

$$ \delta k_x = \frac{1}{K} \Delta_{-3dB}^{(4), plate} \qquad \delta k_y = \frac{1}{K} \min ( [ \Delta_{-3dB}^{(2), beam}, \Delta_{-3dB}^{(4), plate}, \Delta_{-3dB}^{(4), beam} ] ) $$
with K greater than 3.

Only valid for the displacements and the pressure in the near field ($k_0 z << 1$)

$$ \tilde{p}_0(k_x, k_y, z) = \frac{\rho_0 \omega^2 F_0 e^{-j k_x x_0} e^{-\sqrt{k_x^2 + k_y^2 - k_0^2} z}}{Z(k_x, k_y) \sqrt{k_x^2 + k_y^2 - k_0^2}} $$
# DEFINITION OF CRITERIONS

## 3 - ALIASING

Discrete Fourier transform ➔ Aliasing phenomena

<figure><img src="image_196.png" /></figure>

Illustration of the superposition of an original field $W_{plate}^{\infty}$ with its aliasing $\hat{W}_{plate}^{\infty}$

Asymptotic formulations of the spatial decrease of the displacement field for an infinite plate and an infinite beam (flexural, torsion)

➔ Definition of criterions describing where the aliasing may be neglected
NUMERICAL VALIDATION

Test case in the literature:

Mace B.R., JSV n°73 (4), p.
487-504, 1980

<figure><img src="image_197.png" /></figure>

<figure><img src="image_198.png" /></figure>

<figure><img src="image_199.png" /></figure>

Input mobility versus frequency
Solid, results of Mace, dash, present approach.
EXAMPLES OF RESULTS

FFT⁻¹ (~30 seconds)

<figure><img src="image_200.png" /><figcaption>(a) in the wavenumber space ($k_x, k_y$)</figcaption></figure>

<caption>(b) in the physical space (x,y)</caption>

Level of the displacement at 1000 Hz – x₀=0.3m

Stiffened plate (naval test case)
# EXAMPLES OF RESULTS

## Comparison with and without stiffeners

<figure><img src="image_201.png" /><figcaption>(a) Un-stiffened plate</figcaption></figure>

<figure><img src="image_202.png" /><figcaption>(b) Stiffened plate</figcaption></figure>

Level of the displacement at 1000 Hz – $x_0=0.3m$

Results in the physical space (x,y)
EXAMPLES OF RESULTS

Comparison with different excitation points

<figure><img src="image_203.png" /><figcaption>(a) $x_0=0.3\text{m}$ (excitation between 2 ribs)</figcaption></figure>

<figure><img src="image_204.png" /><figcaption>(b) $x0=0\text{m}$ (excitation on a rib)</figcaption></figure>

Level of the displacement at 3000 Hz – $x_0=0.3\text{m}$

Results in the physical space (x,y)
EXAMPLES OF RESULTS

<figure><img src="image_205.png" /></figure>
<figure><img src="image_206.png" /><figcaption>Plate accelerations in the physical space (x,y) -400 Hz - dB ref. $1 \mu\text{m/s}^{-2}$</figcaption></figure>

<figure><img src="image_207.png" /></figure>
<figure><img src="image_208.png" /><figcaption>Radiated plate accelerations in the physical space (x,y) -400 Hz - dB ref. 1 µm/s$^{-2}$</figcaption></figure>

<figure><img src="image_209.png" /><figcaption>Radiated pressure ($R=20\text{m}$) -400 Hz - dB ref. 1 µPa</figcaption></figure>

<figure><img src="image_210.png" /></figure>
# EXAMPLES OF RESULTS

## Sound radiated by a stiffened plate

<figure><img src="image_211.png" /></figure>

Spectrum of the radiated pressure in the
normal direction to the plate (R=20m)
Displacements in (kx,ky) space

<figure><img src="image_212.png" /></figure>

Displacements in (x,y) space

<figure><img src="image_213.png" /></figure>

FF radiating displacements in (x,y) space

<figure><img src="image_214.png" /></figure>

Radiated pressure in the far field (dB)

<figure><img src="image_215.png" /></figure>
1000 Hz

<figure><img src="image_216.png" /></figure>

<figure><img src="image_217.png" /></figure>

<figure><img src="image_218.png" /></figure>

<figure><img src="image_219.png" /></figure>