1 、+。 (3 2 28 。+61 Fig.1.The resonance frequencies of the symmetric (lower curve) and antisymmetric(upper curve)mode of a coupled resonator as a function of detuning 1. acquire the character of the mode to whose frequency the eigenvalue is the closest. It is of interest to ask for the implications of this simple coupled mode formalism.Consider first the case when the two resonators have equal frequencies.Then,if resonator 1 is excited at t=0,a=1,and resonator 2 is unexcited,a2 =0,the initial conditions are matched by a superposition of equal amounts of the symmetric and antisymmetric solutions.The phases of the two solutions evolve at different rates and after the time t=/2 all of the excitation will have been transferred to the other resonator.The excitation oscillates back and forth between the two resonators.When the frequencies of the Fig.2.Coupled metallic rectangular waveguide resonators that permit exact analysis. two uncoupled resonators are not equal,and initially only one resonator is excited,the transfer is not complete. by setting B.A Simple Example 1 Figure 2 shows a simple example that permits both a rigorous analysis and yields easily to the coupled mode for- malism.In this way one may compare the exact answer with We use energy arguments to derive the coupling.Consider the approximate solution.The two resonators are metallic the time rate of change of the energy in mode (1): rectangular waveguides partially filled with a dielectric so d that the empty waveguides are below cutoff at the resonance a1P=jN12aia2-jxza1o时. (2.13) frequency.The uncoupled waveguides are each terminated in infinite air-filled waveguides.The question is as to how This energy change must be equal to the power fed into one may find the coupling coefficient for this structure,and mode (1)by mode(2).Mode(2)finds the perturbation of thus the value of the beat frequency of the two eigenmodes. dielectric constant e within resonator (1)and drives a Denote the spatial distribution of the dielectric constant that polarization current density through that perturbation that forms cavity (1)by is equal to Eo +6E1 2.9) jwP12 jwde1a2e2 (2.14) and the one that forms resonator (2)by The power fed into mode (1)by this current density is Eo +662. (2.10) equal to The actual distribution is 1 4 juwP12·ajeidv+c.c. e=e。+6e1+6e2, (2.11) =jk12aja2-jki2a1a2 Denote the electric field patterns of modes(1)and(2)by 1 er and e2,respectively.The energy in mode (1)is 4 jwbeie2ejdvaja2 +c.c.(2.15) cleil2dv a12 (2.12) and must be responsible for the rate of growth in time of the energy in resonator (1)due to the presence of mode (2) 1508 PROCEEDINGS OF THE IEEE,VOL 79,NO.10.OCTOBER 1991