Multimodal scattering matrix

In sections 6.3 and 6.4 we performed multimodal scattering from a discontinuity for an example in which there was no backward going wave on the right hand side. In order to create the multimodal version of our scattering equation (6.17), we must drop this assumption and express the pressure and volume velocity as the sum of forward and backward going waves on both sides. The aim then is to start with the known forward going and backward going waves at the input to a discontinuity (ie. from an experimentally measured input impulse response), and calculate the forward and backward going waves on the other side of the discontinuity. This method was suggested by van Walstijn [60].

We will denote the forward going pressures on surface 1 and surface 2 respectively as ${\mathbf P}_+^{(1)}$ and ${\mathbf P}_+^{(2)}$. The backward going pressure on surface 1 and surface 2 respectively will be ${\mathbf P}_-^{(1)}$ and ${\mathbf P}_-^{(2)}$. Again the surfaces are as shown in figure 5.1. Using the pressure projection equation (2.79) we get:

$$\left( {\mathbf P}_{+}^{(1)} + {\mathbf P}_{-}^{(1)} \right)... ...\mathbf P}_{-}^{(2)} \right), \mbox{\hspace{1cm}} S_1 < S_2.$$ (6.20)

Putting equations (6.3) and (6.4) into the volume velocity projection equation (2.83) gives:

$$\left(Z_c^{(2)}\right)^{-1} \left( {\mathbf P}_{+}^{(2)} - {... ...{\mathbf P}_{-}^{(1)} \right), \mbox{\hspace{1cm}} S_1 < S_2$$ (6.21)

These two simultaneous equations can be solved by eliminating ${\mathbf P}_{-}^{(2)}$ to give

$${{\mathbf P}_{+}^{(2)} =}$$

$$\frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} + F^{-1} \right) {\mathbf P}_{+}^{(1)}$$

$$+ \frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} - F^{-1} \right) {\mathbf P}_{-}^{(1)}.$$ (6.22)

Eliminating ${\mathbf P}_{+}^{(2)}$ in the simultaneous equations gives

$${{\mathbf P}_{-}^{(2)} =}$$

$$- \frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} - F^{-1} \right) {\mathbf P}_{+}^{(1)}$$

$$+ \frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} + F^{-1} \right) {\mathbf P}_{-}^{(1)}.$$ (6.23)

It is possible to express both equations (6.22) and (6.23) in one equation. This is done by making column vectors consisting of the forward going and backward going pressure vectors end to end. If the vectors ${\mathbf P}_{+}^{(2)}$ and ${\mathbf P}_{-}^{(2)}$ are each truncated to have $N$ elements, the resulting end to end vector will have a length of $2N$ elements. This is then set equal to a single scattering matrix with $2N \times 2N$ elements multiplied by a vector of length $2N$ consisting of the pressure vectors ${\mathbf P}_{+}^{(1)}$ and ${\mathbf P}_{-}^{(1)}$ placed end to end. The first $N$ columns in our scattering matrix then come from the coefficients of ${\mathbf P}_{+}^{(1)}$ in the previous equations and the second $N$ columns from the coefficients for ${\mathbf P}_{-}^{(1)}$. The resulting equation is

$$\left( \begin{array}{c} {\mathbf P}_+^{(2)} \\ {\mathbf P}_... ...{\mathbf P}_+^{(1)} \\ {\mathbf P}_-^{(1)} \end{array} \right)$$ (6.24)

EDIT: Equation 6.24 this is now assumed to be erroneous. Please download Bjørn Kolbrek's corrections.
where ${\mathcal E}$ is an $N \times N$ matrix given by

$${\mathcal E} = \frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} + F^{-1} \right)$$ (6.25)

and ${\mathcal F}$ is an $N \times N$ matrix given by

$${\mathcal F} = \frac{1}{2} \left( Z_c^{(2)} F^{\mathrm{T}} \left(Z_c^{(1)}\right)^{-1} - F^{-1} \right).$$ (6.26)

Notice that if there is no backward going wave in section 2, then ${\mathbf P}_-^{(2)} = 0$, and from equation (6.24), ${\mathbf P}_-^{(1)} = {\mathcal E}^{-1} {\mathcal F} {\mathbf P}_+^{(1)}$ which agrees with the results of section 6.3.

Each of the elements in equation (6.24) is frequency dependent so we must compute the scattering matrix for each frequency in the Fourier transform of the pressure signals and perform the matrix multiplication in the frequency domain. This method means a significant increase in the computational load. Only a small number of modes could be added for this method of bore reconstruction to be feasible.

In addition to scattering across discontinuities, scattering along cylindrical sections is also necessary. We will therefore require multimodal versions of equations (6.18) and (6.19) which give the pressure value at surface 3 from the pressure value at surface 2 in figure 5.1. The modes propagate independently between these surfaces because the cross-section remains constant. For a given frequency of sound, the wavelength along the axis of the duct depends on the mode under consideration, with higher order modes having longer wavelengths untill at high enough mode numbers the waves are exponentially damped. This means the pressure is not just delayed by travel along the pipe, it also experiences dispersion.

The dispersion is described in the time domain by Morse and Ingard [22] p.498. Firstly, below cut-off the $n \ge 1$ modes are exponentially damped meaning that the low frequencies are filtered out. Above cut-off, the phase velocity is larger than $c$ and the group velocity is smaller than $c$, with both converging on $c$ in the high frequency limit. The time domain impulse transfer function for a higher mode pressure distribution is calculated by Fourier transform to give an impulse travelling at the speed of sound $c$ with a wake trailing behind. Physically this is realistic since an impulse contains all frequency components, with the very high frequencies responsible for the sharp impulse travelling with a group velocity of $c$ and the lower frequency components following behind.

Using this impulse transfer function for time domain convolution with the forward and backward going pressure vectors is a possible solution for scattering down a cylindrical section. Alternatively, frequency domain multiplication can be used. The pressure amplitude of the $n$th higher mode in a uniform pipe is described in equation (2.33). Projecting the forward going part gives

$$P_+^{(3)} = e^{- i k_n d} P_+^{(2)}$$ (6.27)

and the backward going part gives

$$P_-^{(3)} = e^{i k_n d} P_-^{(2)}.$$ (6.28)