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Conditions for Losslessness

The scattering matrices for lossless physical waveguide junctions give an apparently unexplored class of lossless FDN prototypes. However, this is just a subset of all possible lossless feedback matrices. We are therefore interested in the most general conditions for losslessness of an FDN feedback matrix.

Consider the general case in which ${\bf A}$ is allowed to be any scattering matrix, i.e., it is associated with a not-necessarily-physical junction of $N$ physical waveguides. Following the definition of losslessness in classical network theory, we may say that a waveguide scattering matrix $A$ is said to be lossless if the total complex power [1] at the junction is scattering invariant, i.e.,

$\displaystyle {\bf {p}^+}^\ast {\bf\Gamma}\bf {p}^+$ $\textstyle =$ $\displaystyle {\bf {p}^-}^\ast {\bf\Gamma}\bf {p}^-$  
$\displaystyle \Rightarrow \quad {\bf A}^\ast {\bf\Gamma}{\bf A}$ $\textstyle =$ $\displaystyle {\bf\Gamma}$ (27)

where ${\bf\Gamma}$ is any Hermitian, positive-definite matrix (which has an interpretation as a generalized junction admittance). The form $x^\ast {\bf\Gamma}
x$ is by definition the square of the elliptic norm of $x$ induced by ${\bf\Gamma}$, or $ \vert\vert x\vert\vert _{\bf\Gamma}^2 = x^\ast {\bf\Gamma}x$. Setting ${\bf\Gamma}={\bf I}$, we obtain that ${\bf A}$ must be unitary. This is the case commonly used in current FDN practice.

The following theorem gives a general characterization of lossless scattering:
Theorem 1: A scattering matrix (FDN feedback matrix) ${\bf A}$ is lossless if and only if its eigenvalues lie on the unit circle and it admits a basis of linearly independent eigenvectors.
In general, the Cholesky factorization ${\bf\Gamma}= {\bf U}^\ast {\bf U}$ gives an upper triangular matrix ${\bf U}$ which converts ${\bf A}$ to a unitary matrix via similarity transformation: ${\bf A}^\ast {\bf\Gamma}{\bf A}= {\bf\Gamma}\Rightarrow
{\bf A}^\ast {\bf U}^\...
... {\bf U}^\ast {\bf U}\Rightarrow {\tilde {\bf A}}^\ast {\tilde {\bf A}}={\bf I}$, where ${\tilde {\bf A}}={\bf U}{\bf A}{\bf U}^{-1}$. Hence, the eigenvalues of every lossless scattering matrix lie on the unit circle. It readily follows from similarity to ${\tilde {\bf A}}$ that ${\bf A}$ admits $N$ linearly independent eigenvectors. In fact, ${\tilde {\bf A}}$ is a normal matrix (since it is unitary), and normal matrices admit a basis of linearly independent eigenvectors [21].

Conversely, assume $\vert\lambda\vert = 1$ for each eigenvalue of ${\bf A}$, and that there exists a matrix ${\bf T}$ of linearly independent eigenvectors of ${\bf A}$. Then the matrix ${\bf T}$ diagonalizes ${\bf A}$ to give ${\bf T}^{-1}{\bf A}{\bf T}= {\bf D}\Rightarrow
{\bf T}^\ast {\bf A}^\ast {{\bf T}^{-1}}^\ast = {\bf D}^\ast $, where ${\bf D}= \mbox{diag}(\lambda_1,\dots,\lambda_N)$. Multiplying, we obtain ${\bf T}^\ast {\bf A}^\ast {{\bf T}^{-1}}^\ast {\bf T}^{-1}{\bf A}{\bf T}={\bf D...
...}^\ast {{\bf T}^{-1}}^\ast {\bf T}^{-1}{\bf A}={{\bf T}^{-1}}^\ast {\bf T}^{-1}$. Thus, (27) is satisfied for ${\bf\Gamma}={{\bf T}^{-1}}^\ast {\bf T}^{-1}$ which is Hermitian and positive definite. $\Box$

Thus, lossless scattering matrices may be fully parametrized as ${\bf A}=
{\bf T}^{-1}{\bf D}{\bf T}$, where ${\bf D}$ is any unit-modulus diagonal matrix, and ${\bf T}$ is any invertible matrix. In the real case, we have ${\bf D}= \mbox{diag}(\pm 1)$ and ${\bf T}\in\Re^{N\times N}$.

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``Circulant and Elliptic Feedback Delay Networks for Artificial Reverberation'', by Davide Rocchesso and Julius O. Smith III, preprint of version in IEEE Transactions on Speech and Audio, vol. 5, no. 1, pp. 51-60, Jan. 1996.

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Copyright © 2005-03-10 by Davide Rocchesso and Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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