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Coupled Horizontal and Vertical Waves

No vibrating string in musical acoustics is truly rigidly terminated, because such a string would produce no sound through the body of the instrument.5.17Yielding terminations result in coupling of the horizontal and vertical planes of vibration. In typical acoustic stringed instruments, nearly all of this coupling takes place at the bridge of the instrument.

Figure 4.26: Digital waveguide model of a string in which vertical and horizontal planes of vibration are coupled linearly at the bridge.
\begin{figure}\input fig/fdlscoupled.pstex_t
\end{figure}

Figure 4.26 illustrates the more realistic case of two planes of vibration which are linearly coupled at one end of the string (the ``bridge''). Denoting the traveling force waves entering the bridge from the vertical and horizontal vibration components by $ F_v^+(z)$ and $ F_h^+(z)$, respectively, the outgoing waves in each plane are given by

$\displaystyle \left[\begin{array}{c} F_v^-(z) \\ [2pt] F_h^-(z) \end{array}\rig...
...] \left[\begin{array}{c} F_v^+(z) \\ [2pt] F_h^+(z) \end{array}\right] \protect$ (5.24)

as shown in the figure.

In physically symmetric situations, we expect $ H_{vh}(z) = H_{hv}(z)$. That is, the transfer function from horizontal to vertical waves is normally the same as the transfer function from vertical to horizontal waves.

If we consider a single frequency $ \omega $, then the coupling matrix with $ z = e^{j\omega T}$ is a constant (generally complex) matrix (where $ T$ denotes the sampling interval as usual). An eigenanalysis of this matrix gives information about the modes of the coupled system and the damping and tuning of these modes [559].

As a simple example, suppose the coupling matrix $ \mathbf{H}(e^{j\omega T})$ at some frequency has the form

$\displaystyle \mathbf{H}(e^{j\omega T}) = \left[\begin{array}{cc} A & B \\ [2pt] B & A \end{array}\right]
$

where $ A$ and $ B$ are any