General 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. The results below are adapted from [465,388].

Consider the general case in which $\mathbf{A}$ 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 $\mathbf{A}$ is said to be lossless if the total complex power [35] at the junction is scattering invariant, i.e.,

$${\mathbf{p}^+}^\ast {\bm \Gamma}\mathbf{p}^+ = {\mathbf{p}^-}^\ast {\bm \Gamma}\mathbf{p}^-$$

$$\implies \quad \mathbf{A}^\ast {\bm \Gamma}\mathbf{A} = {\bm \Gamma} \protect$$ (C.145)

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

The following theorem gives a general characterization of lossless scattering:

Theorem: A scattering matrix (FDN feedback matrix) $\mathbf{A}$ is lossless if and only if its eigenvalues lie on the unit circle and its eigenvectors are linearly independent.

Proof: Since ${\bm \Gamma}$ is positive definite, it can be factored (by the Cholesky factorization) into the form ${\bm \Gamma}= \mathbf{U}^\ast \mathbf{U}$ , where $\mathbf{U}$ is an upper triangular matrix, and $\mathbf{U}^\ast$ denotes the Hermitian transpose of $\mathbf{U}$ , i.e., $\mathbf{U}^\ast \isdef \overline{\mathbf{U}}^T$ . Since ${\bm \Gamma}$ is positive definite, $\mathbf{U}$ is nonsingular and can be used as a similarity transformation matrix. Applying the Cholesky decomposition ${\bm \Gamma}= \mathbf{U}^\ast \mathbf{U}$ in Eq.(C.146) yields

$$\begin{eqnarray*} & & \mathbf{A}^\ast {\bm \Gamma}\mathbf{A}= {\bm \Gamma}\\ &\implies& \mathbf{A}^\ast \mathbf{U}^\ast \mathbf{U}\mathbf{A}= \mathbf{U}^\ast \mathbf{U}\\ &\implies& (\mathbf{U}^{-\ast}\mathbf{A}^\ast \mathbf{U}^\ast )(\mathbf{U}\mathbf{A}\mathbf{U}^{-1}) = \mathbf{I}\\ &\implies& \tilde{\mathbf{A}}^\ast \tilde{\mathbf{A}}= \mathbf{I} \end{eqnarray*}$$

where $\mathbf{U}^{-\ast}\isdef (\mathbf{U}^{-1})^\ast$ , and

$$\tilde{\mathbf{A}}\isdef \mathbf{U}\mathbf{A}\mathbf{U}^{-1}$$

is similar to $\mathbf{A}$ using $\mathbf{U}^{-1}$ as the similarity transform matrix. Since $\tilde{\mathbf{A}}$ is unitary, its eigenvalues have modulus 1. Hence, the eigenvalues of every lossless scattering matrix lie on the unit circle in the $z$ plane. It readily follows from similarity to $\tilde{\mathbf{A}}$ that $\mathbf{A}$ admits $N$ linearly independent eigenvectors. In fact, $\tilde{\mathbf{A}}$ is a normal matrix ( $\mathbf{A}\tilde{\mathbf{A}}= \tilde{\mathbf{A}}\mathbf{A}$ ), since every unitary matrix is normal, and normal matrices admit a basis of linearly independent eigenvectors [349].

Conversely, assume $\vert\lambda\vert = 1$ for each eigenvalue of $\mathbf{A}$ , and that there exists a matrix $\mathbf{E}$ of linearly independent eigenvectors of $\mathbf{A}$ . The matrix $\mathbf{E}$ diagonalizes $\mathbf{A}$ to give $\mathbf{E}^{-1}\mathbf{A}\mathbf{E}= \mathbf{D}$ , where $\mathbf{D}=$   diag$(\lambda_1,\dots,\lambda_N)$ . Taking the Hermitian transform of this equation gives $\mathbf{E}^\ast \mathbf{A}^\ast \mathbf{E}^{-\ast}= \mathbf{D}^\ast$ . Multiplying, we obtain $\mathbf{E}^\ast \mathbf{A}^\ast \mathbf{E}^{-\ast}\mathbf{E}^{-1}\mathbf{A}\mathbf{E}=\mathbf{D}^\ast \mathbf{D}= \mathbf{I} \implies \mathbf{A}^\ast \mathbf{E}^{-\ast}\mathbf{E}^{-1}\mathbf{A}=\mathbf{E}^{-\ast}\mathbf{E}^{-1}$ . Thus, (C.146) is satisfied for ${\bm \Gamma}=\mathbf{E}^{-\ast}\mathbf{E}^{-1}$ which is Hermitian and positive definite. $\Box$

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

Note that not all lossless scattering matrices have a simple physical interpretation as a scattering matrix for an intersection of $N$ lossless reflectively terminated waveguides. In addition to these cases (generated by all non-negative branch impedances), there are additional cases corresponding to sign flips and branch permutations at the junction. In terms of classical network theory [35], such additional cases can be seen as arising from the use of ``gyrators'' and/or ``circulators'' at the scattering junction [437]).