Algebraic Properties of Lossless Junctions

Comparing (38) and (42) shows that the elements of the normalized scattering matrix ${\tilde {\mbox{\boldmath$A$}}}$ in (42) are

$$\begin{array}{l} {\tilde a }_{i,j} = a_{i,j} \frac{\sqrt{\Ga... ...amma_k}}\right. - \delta(i-j) = {\tilde a }_{j,i} \end{array}$$ (59)

where $\delta(n){\tiny\stackrel{\triangle}{=}}1$ for $n=0$ and $0$ for $n\neq 0$. Hence, the normalized scattering matrix is symmetric. As was shown in Theorem 1, it is easy to verify by direct computation that is also unitary (i.e., ), so that all its eigenvalues are on the unit circle. Since all eigenvalues of a real, symmetric matrix are real [66], ${\tilde {\mbox{\boldmath$A$}}}$ has eigenvalues only at $\pm 1$. Moreover, since by (32) and are related by a similarity transformation, also must have all eigenvalues at $+1$ or $-1$.

An elementary eigenvector analysis can be conducted using physical analogies. It is well known that a symmetric matrix has orthogonal eigenvectors. For the equal-impedance case, one eigenvector is always $e_0={\left[ \begin{array}{rrrr} 1 & 1 & \dots & 1\end{array} \right]}^T$ by symmetry: this corresponds to a collision of equal pressure waves at the junction, so the return scatter must be identical. This corresponds to the eigenvalue 1. For the -1 eigenvalues, a similar interpretation can be found: inject a unit pressure wave at all the branches but one, and ``pull out'' a pressure wave having magnitude $N-1$ at the remaining branch. In this case, the return scatter is inverted, since we have arranged that the pressure be zero at the junction and hence at each branch termination. In this way we can find $N$ eigenvectors analogous with $e_1={\left[ \begin{array}{rrrr} -(N-1) & 1 & \dots & 1\end{array} \right]}^T$ which span the ($N-1$)-dimensional subspace associated with the eigenvalue -1. Note that $e_0$ is orthogonal to this subspace, while the $e_i$ are not mutually orthogonal for $i>0$.

For unequal impedances, a similar physical interpretation can be found for the eigenvectors. If we supply equal pressure waves to all branches at the junction, the reflected waves must be equal by symmetry, since $p_i^-=p_J-p_i^+$, where $p_J$ is the junction pressure and all the $p_i^+$ are equal. Hence, $e_0={\left[ \begin{array}{rrrr} 1 & 1 & \dots & 1\end{array} \right]}^T$ remains an eigenvector corresponding to the eigenvalue 1. On the other hand, if we inject a unit pressure wave into all the branches but the $j$th and ``pull out'' a pressure wave having magnitude ${\sum_{i \neq j} \Gamma _i}/\Gamma_j$ at the $j$th branch, then the junction pressure $p_J$ is again forced to zero by construction and the return scatter at any branch is the negative of the incoming wave on that branch. In this way we can find $N$ eigenvectors analogous with $e_j= {\left[ \begin{array}{rrrrrrr} 1 & \dots & 1 & -{\sum_{i \neq j} \Gamma _i}/\Gamma_j & 1 & \dots & 1\end{array} \right]}^T$ spanning the ($N-1$)-dimensional subspace associated with the eigenvalue -1. In this case, none of the eigenvectors is necessarily orthogonal to the others.

The foregoing is an example of how physical intuition can help in finding algebraic properties of a given matrix in physical applications.

Another property of the scattering matrix ${\mbox{\boldmath$A$}}$ is that it is its own inverse: ${\mbox{\boldmath$A$}}^2 = {\mbox{\boldmath$I$}}$. This corresponds physically to the fact that if the results of a scattering operation are fed back to the same junction as incoming waves, the result must be the inverse of the original scattering. An implication of this is that lossless scattering networks can be run in reverse, i.e., by changing the directions of all the delay lines and computing the junctions as dictated by the wave impedances, the network will compute its own inverse. If there are inputs and outputs, they must be interchanged.