Exercise 7.8.7: Two three-ports

Both matrices are symmetric: reciprocal. $[S{]}_A$: each column has $\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1$, and, for example, columns 1 and 2 give $\frac{1}{9}(-2-2+4)=0$: lossless, but not matched ($S_{ii}=-\frac{1}{3}$). $[S{]}_B$: matched ($S_{ii}=0$), but each column has squared length $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$: lossy.

A: $\frac{1}{9}=11.1\%$ reflected ($RL=9.54\mathrm{dB}$), $\frac{4}{9}=44.4\%$ to each of ports 2 and 3 ($IL=3.52\mathrm{dB}$), nothing absorbed. B: nothing reflected, $\frac{1}{4}\mathrm{W}$ to each output ($IL=6.02\mathrm{dB}$), and $0.5\mathrm{W}$ absorbed in the resistors.

Let $S_{ii}=0$ and $S_{ij}=S_{ji}$. Unit columns give $|S_{12}{|}^2+|S_{13}{|}^2=1$, $|S_{12}{|}^2+|S_{23}{|}^2=1$ and $|S_{13}{|}^2+|S_{23}{|}^2=1$, so all three equal $\frac{1}{2}$. Orthogonality of columns 1 and 2 requires $S_{31}{S}_{32}^{\ast }=0$, so $S_{13}$ or $S_{23}$ vanishes, a contradiction. A lossless matched three-port must be non-reciprocal (the circulator) or contain loss (splitter B).

A is three equal lines joined at one point. A wave on line 1 meets the two others in parallel, $Z_0/2$, so $\Gamma =\frac{Z_0/2-Z_0}{Z_0/2+Z_0}=-\frac{1}{3}$ and the voltage $1+\Gamma =\frac{2}{3}$ continues on each. B is the three-resistor splitter with $Z_0/3$ in each arm.

Splitter B: port 1 sees $Z_0/3+(Z_0/3+Z_0)/2=Z_0$, matched. Each output receives $\frac{1}{2}$ of the input voltage, a quarter of the power.