In order to obtain the dispersion relation it is possible to proceed in two different ways. The first one, which is simpler from the analytic point of view, consists of applying the transverse resonance method to obtain a transverse equivalent network. According to this method, the resonance condition along a
transverse direction will be applied. This condition brings to a transcendental equation that, numerically solved, gives possible values for the
transverse wavenumbers. Exploiting the well-known relation of
separability which links the
wavenumbers in the various directions and the frequency, it is possible to obtain the values of the longitudinal propagation constant kz for the various modes. It is supposed that the radiation losses, because actually the metallic plates have a finite width, are negligible. In fact, supposing that the field evanescent in the outside air-regions is negligible at the
aperture, one can assume that the situation substantially coincides with the ideal case of the metallic plates having infinite width. Thus, one can assume the transverse equivalent network shown in Fig. 2. In it kxε and kx0 are the wavenumbers in the x transverse direction, in the dielectric and in the air, respectively; Yε and Y0 are the associated characteristic admittances of the equivalent
transmission line. The presence of the metallic plates, considered perfectly conductive, imposes the possible values for the wavenumber in the y vertical direction: k_{y}=\frac{m\pi }{a}, with m = 0, 1, 2, ... These values are the same in the air as in the dielectric regions. As above mentioned, the wavenumbers must satisfy the separability relations. In the air region, assimilated to a vacuum, one obtains k_{o}^{2}=\left ( \frac{2\pi }{\lambda_{o}} \right )^{2}=k_{xo}^{2}+k_{y}^{2}+k_{z}^{2}=- \left | k_{xo} \right | ^{2} +k_{y}^{2}+\beta ^{2} \ \ \ \ (1) being ko and λo the wavenumber and the wavelength in a vacuum, respectively. It is assumed that kz = β, because the structure is non-radiating and lossless, and moreover kxo= – j | kxo | , because the field has to be
evanescent in the air regions. In the dielectric region, instead, it is k^{2}=k_{o}^{2}\varepsilon _{r}=\left ( \frac{2\pi }{\lambda} \right )^{2}=k_{x\varepsilon }^{2}+k_{y}^{2}+k_{z}^{2}= k_{x\varepsilon }^{2}+k_{y}^{2}+\beta ^{2} \ \ \ \ \ \ (2) where k and λ are the wavenumber and the wavelength, respectively in the dielectric region and \varepsilon _{r} is the relative
dielectric constant. Unlikely kxo, kxε is real, corresponding to a configuration of
standing waves inside the dielectric region. The wavenumbers ky and kz are equal in all the regions. This fact is due to the continuity conditions of the tangential components of the electric and
magnetic fields, at the interface. As a consequence, one obtains the continuity of voltage and current in the equivalent transmission line. Thus the transverse resonance method automatically takes into account the boundary conditions at the metallic walls and continuity conditions at the air-dielectric interface. Analyzing the possible transverse modes, in the air regions (being a) only the mode with m=0 can propagate along x; this mode is a TEM mode traveling obliquely in the xz-plane, with the
non-zero field components Ey, Hx, Hz. This mode always results above cutoff, no matter small
a is, but it is not excited if the symmetry of the structure with reference to the middle plane y = a/2 is preserved. In fact, in symmetrical structures, modes with different polarizations from that of the exciting field are not excited. In the dielectric region, instead, one has \lambda =\frac{\lambda _{o}}{\sqrt{\varepsilon _{r}}} . The mode with index m is above cutoff if a/λ > m/2. For example, if εr = 2.56, (
polystyrene), f = 50 GHz and a = 2.7 mm, it is a/λo = 0.45 and a/λ = 0.72. Therefore in the dielectric region the modes with m=1 are above cutoff, while the modes with m=2 are below cutoff (1/2 Y_{o}=\frac{\omega \varepsilon _{o}}{k_{xo }} \ \ \ \ Y_{\varepsilon }=\frac{\omega \varepsilon _{o}\varepsilon _{r}}{k_{x\varepsilon }} \ \ \ \ \ (3) where k_{xo}=-j\left | k_{xo} \right | The transverse equivalent network of Fig. 2 is further simplified using the geometrical symmetry of the structure with reference to the middle plane x=0 and considering the polarization of the electric field for the required mode, which is
orthogonal vector to the middle plane. In this case, it is possible to bisect the structure with a vertical metallic plane without changing the boundary conditions and thus the internal
configuration of the electromagnetic field. This corresponds to a
short circuit bisection in the equivalent transmission line, as the simplified network shows in Fig. 3. Then it is possible to apply the transverse resonance condition along the horizontal x direction expressed by the relation: \overleftarrow{Y}+\overrightarrow{Y}=0 \ \ \ \ \ \ where \ \ \ \ \ \ \ \overleftarrow{Y} \ \ \ and \ \ \ \overrightarrow{Y} \ \ are the admittances looking toward left and right respectively, with reference to an arbitrary section T. Selecting the reference section as shown in Fig. 3, \overrightarrow{Y}=Y_{o} , because the line is infinite toward right. Looking toward left it is \overleftarrow{Y}=-jY_{\varepsilon } cot(k_{x\varepsilon }w) Then introducing the expression of the characteristic admittances into the resonance condition: -jY_{\varepsilon } cot(k_{x\varepsilon }w)+Y_{o}=0 the dispersion equation is derived: -j\varepsilon _{r}k_{xo}cot(k_{x\varepsilon }w )+k_{x\varepsilon }=0 \ \ \ \ (4) Moreover, from (1) and (2) one obtains k_{xo}=\sqrt{{k_{o}^{2}}-(\frac{m\pi }{a})^{2}-\beta ^{2}}= k_{o}\sqrt{1-(\frac{m\pi }{ak_{o}})^{2}-\frac{\beta ^{2}}{k_{o}}} k_{xo}=\sqrt{{k_{o}^{2}\varepsilon _{r}}-(\frac{m\pi }{a})^{2}-\beta ^{2}}= k_{o}\sqrt{\varepsilon _{r}-(\frac{m\pi }{ak_{o}})^{2}-\frac{\beta ^{2}}{k_{o}}} Therefore one can assume the normalized unknown \frac{\beta }{k_{o}}=\sqrt{\varepsilon _{eff}}, where \varepsilon _{eff} is the so-called effective relative dielectric constant of the guide. The
cutoff frequency fc is obtained by solving the dispersion equation for β =0. It is important to notice that, due to the presence of two dielectrics, the solution depends on frequency, that is, the value of β for any frequency cannot be simply obtained from the cutoff frequency, as it would be for one dielectric only, for which: \beta =\sqrt{k^{2}-k_{t}^{2}}=\sqrt{\omega ^{2}\mu \varepsilon -\omega _{c}^{2}\mu \varepsilon } . In our case, instead, it is necessary to solve the dispersion equation, for each value of frequency. In dual manner, TE modes with reference to x can be considered. The expressions for the characteristic admittances are in this case (μ=μo): Y_{o}=\frac{k_{xo}}{\mu _{o}\omega } \ \ \ , \ \ \ \ Y_{\varepsilon }=\frac{k_{x\varepsilon }} {\mu _{o}\omega } \ \ \ \ \ \ \ (5) Moreover, in this case the magnetic field is orthogonal to the middle plane x=0. Therefore, it is possible to bisect the structure with a perfect magnetic wall, which corresponds to a bisection with an open circuit, obtaining the circuit shown in Fig. 4. Then, with reference to the T plane, it will be: \overrightarrow{Y}=Y_{o} \ \ \ , \ \ \ \overleftarrow{Y}=jY_{\varepsilon }tan(k_{x\varepsilon }w), from which the
dispersion equation is obtained: jk_{x\varepsilon }tan(k_{x\varepsilon }w)+k_{xo}=0 \ \ \ (6) Obviously, the results, here obtained for the dispersive behavior, could be obtained from the complete transverse equivalent network, without bisections, shown in Fig. 2. In this case, with reference to the T plane, one obtains \overrightarrow{Y}=Y_{o} \ \ \ \ \ \ \ \ \overleftarrow{Y}=Y_{\varepsilon }\frac{Y_{o}+jY_{\varepsilon }tan(k_{x\varepsilon }b)}{Y_{\varepsilon }+jY_{o }tan(k_{x\varepsilon }b)} and then Y_{o}+Y_{\varepsilon }\frac{Y_{o}+jY_{\varepsilon }tan(k_{x\varepsilon }b)}{Y_{\varepsilon }+jY_{o}tan(k_{x\varepsilon }b)}=0 \ \ \ \ \ \ \ \ \ (7) This depends on whether TM or TE modes are considered with reference to the x direction, so that Eqs. (3) or (5) can be used for the relevant characteristic admittances. Then, as previously shown, the transverse resonance method allows us to easily obtain the dispersion equation for the NRD waveguide. Yet, the electromagnetic field configuration in the three regions has not been considered in details. Further information can be obtained with the method of modal expansion. ==Determination of the hybrid modes==