4.1.4.1 Rigid Boundaries

Here, we are looking for the mode shapes of our system with given boundary conditions, eg. rigid conditions such as ${\mathit{v}}_x(0)={\mathit{v}}_x(L)=0$. So, the solution (4.2) must fulfil these conditions:

  (4.28)

  (4.29)

The first equation gives $\mathit{A}=\mathit{B}$. Inserting this in the second equation leads to the following solutions for pressure and velocity

  (4.30)

  (4.31)

The choice $n=0$ corresponds to a static pressure. These wave numbers correspond to modal frequencies and wave lengths

  (4.32)

4.1.4.2 Modal Coordinates and Matrix Representation

Mode shapes are natural shapes of the system. That means the system tends to vibrate in this form at the specific frequency. We know that the mode shapes can be used for coordinate transformations such as when the state of the system is given in modal coordinates ${\mathit{p}}_n^{\prime }$

  (4.33)

Entering this into the one-dimensional inhomogeneous Helmholtz Equation (2.123) gives

  (4.34)

For the derivation of the modal coordinates ${\mathit{p}}_n^{\prime }$ we need the orthogonality relations according to equations (1.111) and (1.112). As we don’t have a discrete system the vector product is replaced by a scalar product of the continuous mode

  (4.35)

The complex conjugate does not have any effect, as the mode shape is supposed to be real. But in later application we may have complex modes that require the scalar product in complex form. For practical reasons we normalize the modes with

  (4.36)

so that

  (4.37)

We make use of the orthogonality by multiplying Equation (4.34) from the left with ${\mathrm{\Phi }}_{p,m}^{*}$, and keeping in mind that the source term ${\mathit{f}}_q$ contains the volume source strength density ${\mathit{q}}_s$, so that we have to multiply with the tube cross-section $A$ to get the dimensions right

  (4.38)

providing an expression similar to (1.28) and (1.126)

  (4.39)

An equivalent relationship can be found with the free boundary modes.

4.1.4.3 Modal Density

When systems are excited by broadband signals, it is relevant how many modes are in the frequency range of interest. The more modes are available, the higher the probability for the exciting force to excite resonators. We define the number of modes until frequency $\omega$ as

  (4.40)

$\mathrm{\Delta }{\omega }_n$ is the distance between each modal frequency as shown in Figure 4.4. The modal density is the number of modes per band given by

  (4.41)

These quantities are indicators of the dynamic complexity. A high number of modes in a band may denote high dynamic complexity. In addition they are criteria for the probability to excite a mode when excited by signals with a specific bandwidth.

4.1.4.4 Damping

Damping is considered with Equation (2.59) by using a complex wavenumber

  (4.42)

With this assumption we end up with a modal damping according to (1.63) and (1.127). So, the damping in the wave propagation is in line with the modal damping expressions from Chapter 1.

4.1.4.5 Modal Frequency Response

The first example with a piston excitation at $x=0$ is taken as first application of the modal method. According to Equation (4.39) the response is

  (4.43)

We assume a piston vibrating with ${\hat{v}}_x$ at $x=0$, so the volumetric source strength density would be ${\mathit{q}}_s(x)=\hat{Q}\delta (x-x_0)=\delta (x-x_0)A{\hat{v}}_{x1}$. Using the sifting property of the delta function the result is

  (4.44)

and transforming back into real space with (4.33) gives

  (4.45)

Obviously, in numerical implementation the infinite series must be replaced by a certain number $N$ of modes. As we now the exact solution from Equation (4.26), we can calculate the error

  (4.46)

A modal solution may require a large number of modes to synthesize the exact solution similar to the Fourier transform of a rectangular signal. If the solution is similar to a mode, saying that $k$ is chosen as $k=k_n$, the approximation is naturally exact if the damping is not too high. In Figure 4.5 six modes ($N=6$) in the sum of Equation (4.45) are shown. On the left hand side the system is excited at resonance frequencies, the upper two are well represented, but the case $n=8$ shows that six modes are not sufficient. The required shape to represent the solution is simply not available in the used mode set. When the system is excited out of resonance as shown on the right hand side, even for low frequencies the representation is not perfect, the different colors show increasing numbers of modes. Investigating the absolute value of the modal coordinates as shown in Figure 4.6 the above considerations are confirmed. The out-of-resonance excitation requires a reasonable number of modes, with decreasing contribution, the more the modal frequency is separated from the frequency of excitation.

Choosing a high value of $n=50$ we see that due to large damping even the result of resonance frequencies $N=6$ leads to visible deviations, but for $N=100$ both curves coincide. Consequently for out-of-resonance frequencies 100 modes are required for reasonable precision. In other words, for systems large compared to the wavelength and with damping, a larger number of modes is required to describe the dynamics correctly.

4.2.1.1 Modal Density

The number of modes in a three-dimensional system must be naturally higher than for one-dimensional systems. We can count the possible modes below a specific frequency using Equation (4.62). As the mode count becomes important when the system gets complex and thus many modes exist, we need a more appropriate way to estimate the number of modes that occur until frequency $\omega$ or $k=\omega /c_0$.

A geometrical approach is used to estimate the mode count with some corrections as proposed by Maa (1939). In Figure 4.9 the plane of modal wavenumbers for the tangential modes $n_z=0$ is shown. Each oblique mode occupies a full rectangle of area $k_x{k}_y={\pi }^2/(L_x{L}_y)$ in the quarter circle of radius $k=\omega /c_0$. The number of modes would be the quarter circle area divided by the rectangle area. However, the axial modes with additionally $n_x=0$ or $n_y=0$ are not correctly considered. Only half of their occupied area is taken into account by the quarter circle. This is corrected by adding half-rectangles of area $k\pi /(2L_x)$ and $k\pi /(2L_y)$. The static mode is covered by one quarter through the quarter circle, but two quarters are already compensated by rectangular area. So, only a correction by ${\pi }^2/(4L_x{L}_y)$ is necessary.

The total area is given by sum of

1. Quarter circle of area $\pi k^2/4$.
2. Two rectangles of size $\pi k/(2L_x)$ and $\pi k/(2L_y)$.
3. A quarter rectangle of size ${\pi }^2/(4L_x{L}_y)$.

The total area of all these components divided by the area of one oblique mode area provides an estimate for two-dimensional systems.

For the three-dimensional wavenumber grid in $k$-space a similar compensation is required. See Figure 4.10 for details. The octant of radius $k$ is divided by the cube space any oblique mode is occupying, i.e. $\mathrm{\Delta }V={\pi }^3/(L_x{L}_y{L}_z)$ to estimate the mode count. The oblique modes with all $n_i\ne 0$ are considered correctly.

The cubes corresponding to the tangential modes are cut in half; this is compensated by three quarter slabs. Only the quarter of the axial modes is covered by the octant, but one half is already considered by the two adjacent quarter slabs. So, we compensate by the columns with length $k$ and quarter area. To be fully consequent the static mode is considered by one-eighth from the octant and two times the contribution of the three slabs and columns. Finally, a missing one-eighth must be added to the full volume.

The summarized total mode count volume has the following components:

1. An octant of a sphere of radius $k=\omega /c_0$ with $V_{\text{obl}}=\frac{\pi k^3}{6}$ .
2. The sum of the volume of three quarter slabs of radius $k$ and with thicknesses $\pi /(2L_x),\pi /(2L_y)$ and $\pi /(2L_z)$ respectively.
3. The sum of the volume of three rectangular columns of length $k$ and cross-sectional area ${\pi }^2/(4L_x{L}_y)$, ${\pi }^2/(4L_y{L}_z)$ and ${\pi }^2/(4L_x{L}_z)$.
4. A volume ${\pi }^3/(8L_x{L}_y{L}_z)$ for the static mode.

Thus, the total volume divided by the volume on one mode ${\pi }^3/(L_x{L}_y{L}_z)={\pi }^3/V$ is

  (4.69)

Introducing the quantities perimeter $P=4(L_x+L_y+L_z)$ and surface $S=2(L_x{L}_y+L_y{L}_z+L_x{L}_z)$ we get the approximate expression for the mode count formulated for basic geometric quantities $V$, $S$, and $P$

  (4.70)

and for the modal density

  (4.71)

In Figure 4.11 the exact curves derived by counting all modes with ${\omega }_n\le \omega$ and the estimated curves from function (4.70) are shown. At high mode count the estimation is precise enough. The more irregular the cavity the worse the estimation becomes. Irregular cavities mean having a large perimeter and surface compared to the volume, hence flat and column shaped cavities.

Figure 4.12 shows precise and estimated modal density for the same rooms. Here, the derivative leads to more sensitivity to small errors. Irregular cavities lead to inaccurate modal density estimations.

4.2.3.1 Point Sources

A point source at the wall is represented by a small piston with $R\ll \lambda$ as shown in Figure 4.14. In this case the velocity at the wall leads to the following volume source

  (4.85)

Together with (4.84) the response can be calculated. This expression can also be derived by introducing the vibration as a boundary condition applying $\mathrm{\nabla }\mathit{p}({\mathbf{r}}_0)=j\omega {\rho }_0\mathit{v}$. As we need the radiation impedance for later applications the mechanical impedance is derived by using the source area $A$.

  (4.86)

This impedance can be used when vibrating surfaces are described by discrete meshes. Every vibrating node is related to a specific area and thus experiences a mechanical radiation impedance from the fluid. In Figure 4.13 the mechanical impedance of a point source in the room wall is shown and compared to the impedance of the circular piston. In the range of validity for the point source, i.e. $\lambda \ll R$ or $kR<1$ the real part of the room point impedance equals the piston result.

2.3.2 Radiation Impedance and Power

With the above defined impedance the radiated power is given by

  (4.87)

From the considerations in section 2.4.1 we know that the radiation impedance is more appropriate to describe radiation of point sources. For the rectangular room this reads as

  (4.88)

for the power radiated into the room we get

  (4.89)

4.2.3.3 Rectangular Piston in the Wall

The room itself is assumed to be source free, so $f({\mathbf{r}}_0)=0$. Therefore, there is no contribution from the volume term in the Kirchhoff integral (2.137). We assume a rectangular piston in the wall at $x=0$; the dimensions are as shown in Figure 4.14.

  (4.90)

The surface integral over the room surface is zero except at the wall at $x_0=0$ and the limit of the rectangular surface

  (4.91)

This integral can be analytically solved for the modal Green’s function from Equation (4.82)

  (4.92)

Assuming $x_0=0$ the solution for $A_m=A_{m_x,m_y,m_z}$ is

  (4.93)

With Equation (4.92) the pressure field due to the vibrating piston can be derived. But, we are interested in the radiation impedance to the piston, in order to get the power radiated into the room.

For this we have to integrate the pressure over the piston as in Equation (2.146)

  (4.94)

because we have to validate the same integral again. In Figure 4.15 the impedance is shown. Please note that the radiation impedance approaches ${\mathit{z}}_a={\rho }_0{c}_0$ in the free field limit.