Symmetry of Waveguide Junctions - Section 12 of Montgomery, Dicke, and Purcell

(Junctions are assumed to be lossless). Even w/o symmetry in a junction, the S parameters are constrained by conservation of energy and reciprocity: Sum |Sii |2 =1 and Sum j Sij Sjk* =0.

Symmetry, when present, lets us see by inspection that certain S parameters must be zero, or that a certain S parameter must be the negative of another. But MDP show that symmetry can produce other relations among the S parameters that are not apparent simply from instpection of the waveguide structure.

The elements of the S matrix are found to be functions of the eigenvalues of S. The eigenvalues can be found from symmetry; the symmetry operators are matrices R that commute with S. Because they commute, they have the same eigenvectors.

What exactly are eigenvectors of S? An eigenvector, A, is defined by SA = aA where a is the eigenvalue associated with the eigenvector. So A is a combination of inputs for which the output is proportional. A trivial example is a length of waveguide. A=1,1 is an eigen vector because SA = ej , ej = ej (1,1) , where theta is is the electrical length of the guide. Note that the eigenvalue is ej and that the magnitude of the eigenvalue is unity. All the S matrix eigenvalues for lossless waveguide junctions have unit magnitude. Another eigenvector of the length of waveguide is 1, -1. (For this simplest device, any vector is an eigenvector since this S is diagonal).

What are the eigenvalues of the S matrix? They are just the (constant) phase difference between the input voltages and output voltages. If, for a given eigenvector, Vi is the input voltage at the ith port, then the output voltge from that port will be ej Vi (same theta for all i). If we are considering the impedance matrix rather than the S matrix, an eigenvalue is the impedance at the ith port, i.e. Vi /Ii . For a given eigenvector, every port presents the same impdeance.

Next simplest device: waveguide with a thin iris at lambda/2 from each end. Thinking transmission line, we imagine terminating the far end. That puts Z0 in parallel with the susceptance, Y, of the iris. The input half-wave line doesn't change the impedance, so the input admittance is Y || Z0 . So S11= (zin-1)/(zin+1) = (1-yin)/(1+yin) =

(1-[y+1])/(1+[y+1]) = y/(2+y) = jb/(2+jb). The other element of S, S12, is just the voltage at the matched load: S12 = 2 z/(1+z) = 2/(1+ y+1) = 2/(2+y) = 2/(2+jb).

The sum of the squares of the magnitudes of S11 and S12 is b2 /(b2 +4) + 4/(b2 +4) = 1.

Formal solution: