Exercise 7.8.8: A magic tee in a monopulse radar

$[S]$ is symmetric (reciprocal), its diagonal is zero (matched), each column has two entries of magnitude $1/\sqrt{2}$ (unit length), and, for example, columns 3 and 4 give $\frac{1}{2}(1-1)=0$: lossless.

With $a_1=A$, $a_2=A{\mathrm{e}}^{\mathrm{j}\phi }$ and $|A{|}^2=1\mathrm{mW}$ (here the convention $P=|a{|}^2$),

$\phi =\frac{2\pi s}{\lambda }\sin \theta =\frac{2\pi \times 0.3}{0.03}\sin 0.5^{\circ }=0.548\mathrm{rad}=31.4^{\circ }$. Then $P_3=1.853\mathrm{mW}$, $P_4=0.147\mathrm{mW}$ and $P_4/P_3={\tan }^2(15.7^{\circ })=0.079$, that is $-11.0\mathrm{dB}$.

${\tan }^2(\phi /2)=0.01$ gives $\phi =2\arctan 0.1=0.199\mathrm{rad}$ and $\sin \theta =0.199\times 0.03/(2\pi \times 0.3)$, so $\theta =0.18^{\circ }$. The sign of $b_4/b_3$ (the phase $\mp 90^{\circ }$ of the difference relative to the sum) changes when the target crosses boresight: a phase detector comparing $\Delta$ with $\Sigma$ gives the side.

$P_3+P_4=2\mathrm{mW}$, the total input: the lossless tee only redistributes the power. On boresight ($\phi =0$) all of it goes to the sum port and $P_4=0$.