The transverse resonance method Non-radiative dielectric waveguide

figure 3

in order obtain dispersion relation possible proceed in 2 different ways. first one, simpler analytic point of view, consists of applying transverse resonance method obtain transverse equivalent network. according method apply resonance condition along transverse direction. condition brings transcendental equation that, numerically solved, gives possible values transverse wavenumbers. exploiting well-known relation of separability links wavenumbers in various directions , frequency, possible obtain values of longitudinal propagation constant kz various modes.

it supposed radiation losses, because metallic plates have finite width, negligible. in fact, supposing field evanescent in outside air-regions negligible @ aperture, can assume situation substantially coincides ideal case of metallic plates having infinite width. thus, can assume transverse equivalent network shown in fig. 2. in kxε , kx0 wavenumbers in x transverse direction, in dielectric , in air, respectively; yε , y0 associated characteristic admittances of equivalent transmission line. presence of metallic plates, considered conductive, imposes possible values wavenumber in y vertical direction:




k

y


=



m
π

a




{\displaystyle k_{y}={\frac {m\pi }{a}}}

, m = 0, 1, 2, ... these values same in air in dielectric regions. above mentioned, wavenumbers must satisfy separability relations. in air region, assimilated vacuum, have:

k

o


2


=


(



2
π


λ

o




)


2


=

k

x
o


2


+

k

y


2


+

k

z


2


=
−


|

k

x
o


|


2


+

k

y


2


+

β

2


 
 
 
 
(
1
)


{\displaystyle 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 , λo wavenumber , wavelength in vacuum, respectively. have assumed kz = β, because structure non-radiating , lossless, , kxo= – j | kxo | , because field has evanescent in air regions. in dielectric region, instead, have:

k

2


=

k

o


2



ε

r


=


(



2
π

λ


)


2


=

k

x
ε


2


+

k

y


2


+

k

z


2


=

k

x
ε


2


+

k

y


2


+

β

2


 
 
 
 
 
 
(
2
)


{\displaystyle 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 , λ wavenumber , wavelength, respectively in dielectric region ,




ε

r




{\displaystyle \varepsilon _{r}}

relative dielectric constant.

unlikely kxo, kxε real, corresponding configuration of standing waves inside dielectric region. wavenumbers ky , kz equal in regions. fact due continuity conditions of tangential components of electric , magnetic fields, @ interface. consequence, have continuity of voltage , current in equivalent transmission line. transverse resonance method automatically takes account boundary conditions @ metallic walls , continuity conditions @ air-dielectric interface.

analyzing possible transverse modes, in air regions (being



a
<



λ

o


2




{\displaystyle a<{\frac {\lambda _{o}}{2}}}

) mode m=0 can propagate along x; mode tem mode traveling obliquely in xz plane, non-zero field components ey,hx, hz. mode results above cutoff, no matter small is, not excited if symmetry of structure reference middle plane y = a/2 preserved. in fact, in symmetrical structures, modes different polarizations of exciting field not excited. in dielectric region, instead, have



λ
=



λ

o




ε

r







{\displaystyle \lambda ={\frac {\lambda _{o}}{\sqrt {\varepsilon _{r}}}}}

. mode index m above cutoff if a/λ > m/2. example, if εr = 2.56, (polystyrene), f = 50  ghz , = 2.7  mm, have a/λo = 0.45 , a/λ = 0.72. therefore in dielectric region modes m=1 above cutoff, while modes m=2 below cutoff (1/2 < 0.72 < 1).

in nrd guide, in h guide, due presence of dielectric strip boundary conditions cannot satisfied tem, tm or (m≠0) te modes reference longitudinal z direction. modes of structure hybrid, both longitudinal field components different zero. fortunately, desired mode tm mode reference horizontal x direction, along equivalent transmission line has been adopted. therefore, according known expressions of characteristic admittances of tm modes, have:

y

o


=



ω

ε

o




k

x
o




 
 
 
 

y

ε


=



ω

ε

o



ε

r




k

x
ε




 
 
 
 
 
(
3
)


{\displaystyle y_{o}={\frac {\omega \varepsilon _{o}}{k_{xo}}}\ \ \ \ y_{\varepsilon }={\frac {\omega \varepsilon _{o}\varepsilon _{r}}{k_{x\varepsilon }}}\ \ \ \ \ (3)}

where

k

x
o


=
−
j

|

k

x
o


|



{\displaystyle k_{xo}=-j\left|k_{xo}\right|}

the transverse equivalent network of fig. 2 further simplified using geometrical symmetry of structure reference middle plane x=0 , considering polarization of electric field required mode, orthogonal middle plane. in case, possible bisect structure vertical metallic plane without changing boundary conditions , internal configuration of electromagnetic field. corresponds short circuit bisection in equivalent transmission line, simplified network shows in fig. 3.

then possible apply transverse resonance condition along horizontal x direction expressed relation:

y
←


+


y
→


=
0
 
 
 
 
 
 


{\displaystyle {\overleftarrow {y}}+{\overrightarrow {y}}=0\ \ \ \ \ \ }

where

 
 
 
 
 
 
 


y
←


 
 
 
a
n
d
 
 
 


y
→


 
 


{\displaystyle \ \ \ \ \ \ \ {\overleftarrow {y}}\ \ \ and\ \ \ {\overrightarrow {y}}\ \ }

are admittances looking toward left , right respectively, reference arbitrary section t.

selecting reference section shown in fig. 3, have





y
→


=

y

o




{\displaystyle {\overrightarrow {y}}=y_{o}}

, because line infinite toward right. looking toward left have:

y
←


=
−
j

y

ε


c
o
t
(

k

x
ε


w
)


{\displaystyle {\overleftarrow {y}}=-jy_{\varepsilon }cot(k_{x\varepsilon }w)}

then introducing expression of characteristic admittances resonance condition:

−
j

y

ε


c
o
t
(

k

x
ε


w
)
+

y

o


=
0


{\displaystyle -jy_{\varepsilon }cot(k_{x\varepsilon }w)+y_{o}=0}

the dispersion equation derived:

−
j

ε

r



k

x
o


c
o
t
(

k

x
ε


w
)
+

k

x
ε


=
0
 
 
 
 
(
4
)


{\displaystyle -j\varepsilon _{r}k_{xo}cot(k_{x\varepsilon }w)+k_{x\varepsilon }=0\ \ \ \ (4)}

moreover, (1) , (2) have:

k

x
o


=




k

o


2



−
(



m
π

a



)

2


−

β

2




=

k

o




1
−
(



m
π


a

k

o






)

2


−



β

2



k

o








{\displaystyle 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

x
o


=




k

o


2



ε

r



−
(



m
π

a



)

2


−

β

2




=

k

o





ε

r


−
(



m
π


a

k

o






)

2


−



β

2



k

o








{\displaystyle 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 can assume normalized unknown





β

k

o




=



ε

e
f
f






{\displaystyle {\frac {\beta }{k_{o}}}={\sqrt {\varepsilon _{eff}}}}

,




ε

e
f
f




{\displaystyle \varepsilon _{eff}}

so-called effective relative dielectric constant of guide.

the cutoff frequency fc obtained solving dispersion equation β =0.

it important notice that, due presence of 2 dielectrics, solution depends on frequency, is, value of β frequency cannot obtained cutoff frequency, 1 dielectric only, which:



β
=



k

2


−

k

t


2




=



ω

2


μ
ε
−

ω

c


2


μ
ε




{\displaystyle \beta ={\sqrt {k^{2}-k_{t}^{2}}}={\sqrt {\omega ^{2}\mu \varepsilon -\omega _{c}^{2}\mu \varepsilon }}}

. in our case, instead, necessary solve dispersion equation, each value of frequency. in dual manner, te modes reference x can considered. expressions characteristic admittances in case (μ=μo):

y

o


=



k

x
o




μ

o


ω



 
 
 
,
 
 
 
 

y

ε


=



k

x
ε




μ

o


ω



 
 
 
 
 
 
 
(
5
)


{\displaystyle y_{o}={\frac {k_{xo}}{\mu _{o}\omega }}\ \ \ ,\ \ \ \ y_{\varepsilon }={\frac {k_{x\varepsilon }}{\mu _{o}\omega }}\ \ \ \ \ \ \ (5)}

moreover, in case magnetic field orthogonal middle plane x=0. therefore, possible bisect structure perfect magnetic wall, corresponds bisection open circuit, obtaining circuit shown in fig. 4. then, reference t plane, be:





y
→


=

y

o


 
 
 
,
 
 
 


y
←


=
j

y

ε


t
a
n
(

k

x
ε


w
)


{\displaystyle {\overrightarrow {y}}=y_{o}\ \ \ ,\ \ \ {\overleftarrow {y}}=jy_{\varepsilon }tan(k_{x\varepsilon }w)}

, dispersion equation obtained:

j

k

x
ε


t
a
n
(

k

x
ε


w
)
+

k

x
o


=
0
 
 
 
(
6
)


{\displaystyle jk_{x\varepsilon }tan(k_{x\varepsilon }w)+k_{xo}=0\ \ \ (6)}

obviously, results, here obtained dispersive behavior, obtained complete transverse equivalent network, without bisections, shown in fig. 2. in case, reference t plane, have:

y
→


=

y

o


 
 
 
 
 
 
 
 


y
←


=

y

ε






y

o


+
j

y

ε


t
a
n
(

k

x
ε


b
)



y

ε


+
j

y

o


t
a
n
(

k

x
ε


b
)





{\displaystyle {\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

ε






y

o


+
j

y

ε


t
a
n
(

k

x
ε


b
)



y

ε


+
j

y

o


t
a
n
(

k

x
ε


b
)



=
0
 
 
 
 
 
 
 
 
 
(
7
)


{\displaystyle 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)}

we must specify if tm or te modes considered reference x direction, eqs. (3) or (5) can used relevant characteristic admittances.

then, shown, transverse resonance method allows obtain dispersion equation nrd waveguide.

yet, electromagnetic field configuration in 3 regions has not been considered in details. further information can obtained method of modal expansion.