Power transfer

Due to the linearity and the symmetry of the device, it is sufficient to consider an excitation in only one of the external ports, say port $A$. Given input amplitudes $A = \sqrt{P_{\mbox{\scriptsize I}}}, \tilde{A} = 0$, one is interested in the transmitted power $P_{\mbox{\scriptsize T}} = \vert B\vert^2$ and the dropped power $P_{\mbox{\scriptsize D}} = \vert\tilde{B}\vert^2$. By solving equations (1.1) and (1.2) for amplitudes $B$ and $\tilde{B}$, one obtains

$$\tilde{B} = \left ( \frac{ {\sf {S}}_{\mbox{\scriptsize sb}} {\sf... ...{G}}^{2} } {1 - {\sf {S}}_{\mbox{\scriptsize bb}}^{2} {\sf {G}}^{2}} \right )A.$$ (1.3)

Let ${\sf {S}}_{\mbox{\scriptsize bb}} = \vert{\sf {S}}_{\mbox{\scriptsize bb}}\vert~\mbox{e}^{\displaystyle \mbox{i} \varphi}$, ${\sf {S}}_{\mbox{\scriptsize ss}} - {\sf {S}}_{\mbox{\scriptsize bs}} {\sf {S}... ...{\sf {S}}_{\mbox{\scriptsize bb}} = \rho~\mbox{e}^{\displaystyle \mbox{i} \psi}$ with $\varphi$, $\rho$ and $\psi$ real. Then the dropped power is given by

$$P_{\mbox{\scriptsize D}} = P_{\mbox{\scriptsize I}} \frac{ \vert{... ...ze bb}}\vert^2 \mbox{e}^{\displaystyle -\alpha L} \cos{(\beta L - 2 \varphi)}},$$ (1.4)

and the through power is given by

$$P_{\mbox{\scriptsize T}} = P_{\mbox{\scriptsize I}} \frac{ \vert{... ...ze bb}}\vert^2 \mbox{e}^{\displaystyle -\alpha L} \cos{(\beta L - 2 \varphi)}}.$$ (1.5)

Here $L$ is the total length of those parts of the cavity which are not already included in the couplers. According to equations (1.4) and (1.5), one can evaluate the throughput power and the dropped power, if the scattering matrix S and the cavity propagation constant $\gamma$ are available.