Simply Supported Edge (1)

Conditions (5.44b) are somewhat more complicated to implement than the others, because none of the variables calculated on the boundary can be zeroed out by short- or open-circuiting. From Figure 5.20(b), we can see that in addition to the gyrator coupling between the two meshes that must be maintained, we also must keep the waveguides which lie along the boundary.

In the problem interior, parallel and series junctions in the five-variable mesh are connected, through waveguides, to four neighboring junctions; on the southern boundary, however, each is connected to three--two to the east and west) and one to the north. Due to this asymmetry, we might suspect that it will be necessary to adjust the boundary waveguide impedances away from the values that they would take on the interior (which is $r_{3}$). In fact, it is possible to show that by introducing transformers, with turns ratios of $n=2$ in these waveguides, we indeed have a lossless termination which satisfies conditions (5.44b). We must set the boundary immittances to

$$\tilde{Y}_{x^{-},i+\frac{1}{2},0} = \tilde{Y}_{x^{+},i+\frac{1}{2},0}= \frac{1}{2r_{3}}\hspace{1.0in}Z_{x^{-},i,0} =Z_{x^{+},i,0} = \frac{r_{3}}{2}$$

Notice that each waveguide now includes scaling factors (of 2 and 1/2), and that the impedances at either end are no longer identical, due to the transformer impedance matching. The self-loop immittances should be set according to

$$\tilde{Y}_{c,i+\frac{1}{2},0} = \frac{2v_{0}}{(D(1-\nu))_{i+\frac{1}{2},0}}-\frac{2}{r_{3}}$$

$$\tilde{Z}_{c,i,0} = \frac{v_{0}(\rho h^{3})_{i,0}}{12}-r_{2}-r_{3}$$

$$Z_{c,i,0} = = \frac{v_{0}}{(\kappa^{2}Gh)_{i,0}}-r_{1}$$

which are precisely half the values they would take at interior junctions, from (5.45) (and thus the positivity condition on these immittances is no different from the condition over interior self-loop immittances). We also mention that the gyrator coefficient, which takes on a value of $R_{G} = \Delta$ over the interior, should also be halved to $R_{G} = \Delta/2$ at the boundary junctions.