Determination of the hybrid modes Non-radiative dielectric waveguide



figure 4


with reference cross section of guide shown in fig. 1, tm , te fields can considered respect z longitudinal direction, along guide uniform. said, in nrd waveguide tm or (m≠0) te modes reference z direction cannot exist, because cannot satisfy conditions imposed presence of dielectric slab. yet, known propagation mode inside guiding structure can expressed superposition of tm field , te field reference z.


moreover, tm field can derived purely longitudinal lorentz vector potential





a
_




{\displaystyle {\underline {a}}}

. electromagnetic field can deduced general formulae:








h
_


=

×


a
_


 
 
 
 
 
,
 
 
 
 
 
 


e
_


=

j
ω
μ


a
_


+








a
_




j
ω
ε



 
 
 
 
 
 
(
8
)


{\displaystyle {\underline {h}}=\triangledown \times {\underline {a}}\ \ \ \ \ ,\ \ \ \ \ \ {\underline {e}}=-j\omega \mu {\underline {a}}+{\frac {\triangledown \triangledown \cdot {\underline {a}}}{j\omega \varepsilon }}\ \ \ \ \ \ (8)}


in dual manner, te field can derived purely longitudinal vector potential





f
_




{\displaystyle {\underline {f}}}

. electromagnetic field expressed by:








e
_


=


×


f
_


 
 
 
 
 
 
,
 
 
 
 
 
 


h
_


=

j
ω


f
_


+








f
_




j
ω
μ



 
 
 
 
 
 
 
(
9
)


{\displaystyle {\underline {e}}=-\triangledown \times {\underline {f}}\ \ \ \ \ \ ,\ \ \ \ \ \ {\underline {h}}=-j\omega {\underline {f}}+{\frac {\triangledown \triangledown \cdot {\underline {f}}}{j\omega \mu }}\ \ \ \ \ \ \ (9)}


due cylindrical symmetry of structure along z direction, can assume:








a
_


=

a

z





z

o


_


=

l

t
m


(
z
)

t

t
m


(
x
,
y
)



z

o


_


 
 
 
 
 
 
 
(
10
)


{\displaystyle {\underline {a}}=a_{z}{\underline {z_{o}}}=l^{tm}(z)t^{tm}(x,y){\underline {z_{o}}}\ \ \ \ \ \ \ (10)}








f
_


=

f

z





z

o


_


=

l

t
m


(
z
)

t

t
m


(
x
,
y
)



z

o


_


 
 
 
 
 
 
(
11
)


{\displaystyle {\underline {f}}=f_{z}{\underline {z_{o}}}=l^{tm}(z)t^{tm}(x,y){\underline {z_{o}}}\ \ \ \ \ \ (11)}


as known, in sourceless region, potential must satisfy homogeneous helmholtz equation:









2



a

z


+

k

2



a

z


=
0
 
 
 
 
 
 
(
12
)


{\displaystyle \triangledown ^{2}a_{z}+k^{2}a_{z}=0\ \ \ \ \ \ (12)}









2



f

z


+

k

2



f

z


=
0
 
 
 
 
 
 
(
13
)


{\displaystyle \triangledown ^{2}f_{z}+k^{2}f_{z}=0\ \ \ \ \ \ (13)}


from eqs. (10)-(13), obtain:










d

2


l


d

z

2





+

k

z


2


l
=
0
 
 
 
 
 
 
(
14
)


{\displaystyle {\frac {d^{2}l}{dz^{2}}}+k_{z}^{2}l=0\ \ \ \ \ \ (14)}









t


2


t
+

k

t


2


t
=
0
 
 
 
 
 
 
(
15
)


{\displaystyle \triangledown _{t}^{2}t+k_{t}^{2}t=0\ \ \ \ \ \ (15)}


where kz wave number in longitudinal direction,









t


2


=



2








2





z

2





 
 
 
a
n
d
 
 
 

k

t


2


=

k

2




k

z


2




{\displaystyle \triangledown _{t}^{2}=\triangledown ^{2}-{\frac {\partial ^{2}}{\partial z^{2}}}\ \ \ and\ \ \ k_{t}^{2}=k^{2}-k_{z}^{2}}

.


for case kz ≠ 0, general solution of eq. (14) given by:






l
(
z
)
=

l

o






e


j

k

z


z


+

l

o






e

j

k

z


z


 
 
 
 
 
 
 
(
16
)


{\displaystyle l(z)=l_{o}^{\dotplus }e^{-jk_{z}z}+l_{o}^{-}e^{jk_{z}z}\ \ \ \ \ \ \ (16)}


in following suppose direct traveling wave present (lo = 0). wavenumbers ky , kz must same in dielectric in air regions in order satisfy continuity condition of tangential field components. moreover, kz must same both in tm in te fields.


eq. (15) can solved separation of variables. letting t (x,y) = x(x) y(y), obtain:










d

2


x


d

x

2





+

k

x


2


x
=
0
 
 
 
 
 
 
 
(
17
)


{\displaystyle {\frac {d^{2}x}{dx^{2}}}+k_{x}^{2}x=0\ \ \ \ \ \ \ (17)}










d

2


y


d

y

2





+

k

y


2


y
=
0
 
 
 
 
 
 
 
 
(
18
)


{\displaystyle {\frac {d^{2}y}{dy^{2}}}+k_{y}^{2}y=0\ \ \ \ \ \ \ \ (18)}


where



y
(
y
)
=

c

1


cos

(

k

y


y
)
+

c

2


s
i
n
(

k

y


y
)


{\displaystyle y(y)=c_{1}\cos(k_{y}y)+c_{2}sin(k_{y}y)}


for tm field, solution of eq. (18), taking account boundary conditions @ y = 0 , y = a, given by:






s
i
n
(



m
π

a


y
)
 
 
 
 
 
(
m
=
1
,
2
,
3
,
.
.
.
)


{\displaystyle sin({\frac {m\pi }{a}}y)\ \ \ \ \ (m=1,2,3,...)}

.


for te field, have analogously:






c
o
s
(



m
π

a


y
)
 
 
 
 
 
 
(
m
=
1
,
2
,
3
,
.
.
.
)


{\displaystyle cos({\frac {m\pi }{a}}y)\ \ \ \ \ \ (m=1,2,3,...)}

.


as far eq. (17) concerned, choose form general solution:






x
(
x
)
=

c

3



e


j

k

x


x


+

c

4



e

j

k

x


x




{\displaystyle x(x)=c_{3}e^{-jk_{x}x}+c_{4}e^{jk_{x}x}}


therefore, various regions assume:


dielectric region (-w < x < w)







t

ε


t
m


=
s
i
n
(



m
π

a


y
)

(
a

e


j

k

x
ε


x


+
b

e

j

k

x
ε


x


)
 
 
 
 
 
m
=
0
,
1
,
2
,
.
.
.
 
 
 
 
 
 
(
19
)


{\displaystyle t_{\varepsilon }^{tm}=sin({\frac {m\pi }{a}}y)\cdot (ae^{-jk_{x\varepsilon }x}+be^{jk_{x\varepsilon }x})\ \ \ \ \ m=0,1,2,...\ \ \ \ \ \ (19)}







t

ε


t
e


=
c
o
s
(



m
π

a


y
)

(
c

e


j

k

x
ε


x


+
d

e

j

k

x
ε


x


)
 
 
 
 
 
 
m
=
1
,
2
,
3
,
.
.
.
 
 
 
 
(
20
)


{\displaystyle t_{\varepsilon }^{te}=cos({\frac {m\pi }{a}}y)\cdot (ce^{-jk_{x\varepsilon }x}+de^{jk_{x\varepsilon }x})\ \ \ \ \ \ m=1,2,3,...\ \ \ \ (20)}


where







k

x
ε


=

k

o





ε

r



(



m
π



k

o


a




)

2



(



k

z



k

o





)

2




 
 
 
 
(
21
)


{\displaystyle k_{x\varepsilon }=k_{o}{\sqrt {\varepsilon _{r}-({\frac {m\pi }{k_{o}a}})^{2}-({\frac {k_{z}}{k_{o}}})^{2}}}\ \ \ \ (21)}


air region on right (x > w)







t

o
+


t
m


=
e
s
i
n
(



m
π

a


y
)

e


j

k

x
o


(
x

w
)


 
 
 
 
(
22
)


{\displaystyle t_{o+}^{tm}=esin({\frac {m\pi }{a}}y)e^{-jk_{xo}(x-w)}\ \ \ \ (22)}







t

o
+


t
e


=
f
c
o
s
(



m
π

a


y
)

e


j

k

x
o


(
x

w
)


 
 
 
 
 
 
 
 
 
(
23
)


{\displaystyle t_{o+}^{te}=fcos({\frac {m\pi }{a}}y)e^{-jk_{xo}(x-w)}\ \ \ \ \ \ \ \ \ (23)}


air region on left (x < w)







t

o



t
m


=
g
s
i
n
(



m
π

a


y
)

e

j

k

x
o


(
x
+
w
)


 
 
 
 
 
 
 
(
24
)


{\displaystyle t_{o-}^{tm}=gsin({\frac {m\pi }{a}}y)e^{jk_{xo}(x+w)}\ \ \ \ \ \ \ (24)}







t

o



t
e


=
h
c
o
s
(



m
π

a


y
)

e

j

k

x
o


(
x

w
)


 
 
 
 
 
 
 
 
(
25
)


{\displaystyle t_{o-}^{te}=hcos({\frac {m\pi }{a}}y)e^{jk_{xo}(x-w)}\ \ \ \ \ \ \ \ (25)}


in air regions have:







k

x
o


=

k

o




1

(



m
π



k

o


a




)

2



(



k

z



k

o





)

2




 
 
 
 
 
 
 
(
26
)


{\displaystyle k_{xo}=k_{o}{\sqrt {1-({\frac {m\pi }{k_{o}a}})^{2}-({\frac {k_{z}}{k_{o}}})^{2}}}\ \ \ \ \ \ \ (26)}


the 8 constants a, b, c, d, e, f, g, h determined imposing 8 continuity conditions tangential components ey, ez, hy, hz of electromagnetic field @ x = w , @ x = – w.


the various field components given by:







e

x


=


1

j
ω
ε






d
l



d

z








t



x




t
m



l





t



y




t
e


=





k

z




ω
ε



l





t



x




t
m



l





t



y




t
e


 
 
 
 
 
 
 
(
27
)


{\displaystyle e_{x}={\frac {1}{j\omega \varepsilon }}{\frac {\mathrm {dl} }{\mathrm {d} z}}{\frac {\partial t}{\partial x}}^{tm}-l{\frac {\partial t}{\partial y}}^{te}={\frac {-k_{z}}{\omega \varepsilon }}l{\frac {\partial t}{\partial x}}^{tm}-l{\frac {\partial t}{\partial y}}^{te}\ \ \ \ \ \ \ (27)}







e

y


=


1

j
ω
ε






d
l



d

z








t



y




t
m



l





t



x




t
e


=





k

z




ω
ε



l





t



y




t
m



l





t



x




t
e


 
 
 
 
 
 
(
28
)


{\displaystyle e_{y}={\frac {1}{j\omega \varepsilon }}{\frac {\mathrm {dl} }{\mathrm {d} z}}{\frac {\partial t}{\partial y}}^{tm}-l{\frac {\partial t}{\partial x}}^{te}={\frac {-k_{z}}{\omega \varepsilon }}l{\frac {\partial t}{\partial y}}^{tm}-l{\frac {\partial t}{\partial x}}^{te}\ \ \ \ \ \ (28)}







e

z


=



k

t


2



j
ω
ε



l
 

t

t
m


=




k

2




k

z


2




j
ω
ε



l
 

t

t
m


 
 
 
 
 
 
 
(
29
)


{\displaystyle e_{z}={\frac {k_{t}^{2}}{j\omega \varepsilon }}l\ t^{tm}={\frac {k^{2}-k_{z}^{2}}{j\omega \varepsilon }}l\ t^{tm}\ \ \ \ \ \ \ (29)}







h

x


=
l





t



y




t
m


+


1

j
ω
μ






d
l



d

z








t



x




t
e


=
l





t



y




t
m






k

z



ω
μ



l





t



x




t
e


 
 
 
 
 
(
30
)


{\displaystyle h_{x}=l{\frac {\partial t}{\partial y}}^{tm}+{\frac {1}{j\omega \mu }}{\frac {\mathrm {dl} }{\mathrm {d} z}}{\frac {\partial t}{\partial x}}^{te}=l{\frac {\partial t}{\partial y}}^{tm}-{\frac {k_{z}}{\omega \mu }}l{\frac {\partial t}{\partial x}}^{te}\ \ \ \ \ (30)}







h

y


=

l





t



x




t
m


+


1

j
ω
μ






d
l



d

z








t



y




t
e


=

l





t



x




t
m






k

z



ω
μ



l





t



y




t
e


 
 
 
 
(
31
)


{\displaystyle h_{y}=-l{\frac {\partial t}{\partial x}}^{tm}+{\frac {1}{j\omega \mu }}{\frac {\mathrm {dl} }{\mathrm {d} z}}{\frac {\partial t}{\partial y}}^{te}=-l{\frac {\partial t}{\partial x}}^{tm}-{\frac {k_{z}}{\omega \mu }}l{\frac {\partial t}{\partial y}}^{te}\ \ \ \ (31)}







h

z


=



k

t


2



j
ω
μ



l
 

t

t
e


=




k

2




k

z


2




j
ω
μ



l
 

t

t
e


 
 
 
 
(
32
)


{\displaystyle h_{z}={\frac {k_{t}^{2}}{j\omega \mu }}l\ t^{te}={\frac {k^{2}-k_{z}^{2}}{j\omega \mu }}l\ t^{te}\ \ \ \ (32)}


imposing continuity conditions @ each interface, have:










k

z



ω

ε

o











t

o





y




t
m


+






t

o





x




t
e


=




k

z



ω

ε

o



ε

r











t

ε





y




t
m


+






t

ε





x




t
e




{\displaystyle -{\frac {k_{z}}{\omega \varepsilon _{o}}}{\frac {\partial t_{o}}{\partial y}}^{tm}+{\frac {\partial t_{o}}{\partial x}}^{te}=-{\frac {k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}{\frac {\partial t_{\varepsilon }}{\partial y}}^{tm}+{\frac {\partial t_{\varepsilon }}{\partial x}}^{te}}










k

o


2




k

z


2




j
ω

ε

o





 

t

o


t
m


=




k

o


2



ε

r




k

z


2




j
ω

ε

r



ε

o






t

ε


t
m




{\displaystyle {\frac {k_{o}^{2}-k_{z}^{2}}{j\omega \varepsilon _{o}}}\ t_{o}^{tm}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \varepsilon _{r}\varepsilon _{o}}}t_{\varepsilon }^{tm}}













t

o





x




t
m






k

z



ω
μ









t

o





y




t
e


=







t

ε





x




t
m






k

z



ω
μ









t

ε





y




t
e




{\displaystyle -{\frac {\partial t_{o}}{\partial x}}^{tm}-{\frac {k_{z}}{\omega \mu }}{\frac {\partial t_{o}}{\partial y}}^{te}=-{\frac {\partial t_{\varepsilon }}{\partial x}}^{tm}-{\frac {k_{z}}{\omega \mu }}{\frac {\partial t_{\varepsilon }}{\partial y}}^{te}}










k

o


2




k

z


2




j
ω
μ



 

t

o


t
e


=





k

o


2



ε

r




k

z


2




j
ω
μ




t
e




{\displaystyle {\frac {k_{o}^{2}-k_{z}^{2}}{j\omega \mu }}\ t_{o}^{te}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \mu }}^{te}}


where first members referred air-regions, , second members dielectric-region.


introducing eqs. (19), (20), , (22)-(25) in 4 continuity conditions @ x = w, e , f constants can expressed in terms of a, b, c, d, linked 2 relations.


similarly @ interface x = -w, g , h constants can expressed in terms of a, b, c, d. expressions of electromagnetic field components become:


dielectric region (-w < x < w)







e

x


=

[
j




k

x
ε



k

z




ω

ε

o



ε

r





(
a

e


j

k

x
ε


x



b

e

j

k

x
ε


x


)
+



m
π

a


(
c

e


j

k

x
ε


x


+
d

e

j

k

x
ε


x


)
]

s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
33
)


{\displaystyle e_{x}=\left[j{\frac {k_{x\varepsilon }k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}(ae^{-jk_{x\varepsilon }x}-be^{jk_{x\varepsilon }x})+{\frac {m\pi }{a}}(ce^{-jk_{x\varepsilon }x}+de^{jk_{x\varepsilon }x})\right]sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (33)}







e

y


=

[




k

z



ω

ε

o



ε

r








m
π

a


(
a

e


j

k

x
ε


x


+
b

e

j

k

x
ε


x


)

j

k

x
ε


(
c

e


j

k

x
ε


x



d

e

j

k

x
ε


x


)
]

c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
(
34
)


{\displaystyle e_{y}=\left[-{\frac {k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}{\frac {m\pi }{a}}(ae^{-jk_{x\varepsilon }x}+be^{jk_{x\varepsilon }x})-jk_{x\varepsilon }(ce^{-jk_{x\varepsilon }x}-de^{jk_{x\varepsilon }x})\right]cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ (34)}







e

z


=




k

o


2



ε

r




k

z


2




j
ω

ε

o



ε

r





(
a

e


j

k

x
ε


x



b

e

j

k

x
ε


x


)
s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
35
)


{\displaystyle e_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \varepsilon _{o}\varepsilon _{r}}}(ae^{-jk_{x\varepsilon }x}-be^{jk_{x\varepsilon }x})sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (35)}







h

x


=

[



m
π

a


(
a

e


j

k

x
ε


x


+
b

e

j

k

x
ε


x


)
+
j




k

x
ε



k

z




ω
μ



(
c

e


j

k

x
ε


x



d

e

j

k

x
ε


x


)
]

c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
(
36
)


{\displaystyle h_{x}=\left[{\frac {m\pi }{a}}(ae^{-jk_{x\varepsilon }x}+be^{jk_{x\varepsilon }x})+j{\frac {k_{x\varepsilon }k_{z}}{\omega \mu }}(ce^{-jk_{x\varepsilon }x}-de^{jk_{x\varepsilon }x})\right]cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ (36)}







h

y


=

[
j

k

x
ε


(
a

e


j

k

x
ε


x



b

e

j

k

x
ε


x


)
+



k

z



ω
μ






m
π

a


(
c

e


j

k

x
ε


x


+
d

e

j

k

x
ε


x


)
]

s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
(
37
)


{\displaystyle h_{y}=\left[jk_{x\varepsilon }(ae^{-jk_{x\varepsilon }x}-be^{jk_{x\varepsilon }x})+{\frac {k_{z}}{\omega \mu }}{\frac {m\pi }{a}}(ce^{-jk_{x\varepsilon }x}+de^{jk_{x\varepsilon }x})\right]sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ (37)}







h

z


=




k

o


2



ε

r




k

z


2




j
ω
μ



(
c

e


j

k

x
ε


x


+
d

e

j

k

x
ε


x


)
c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
 
 
 
 
(
38
)


{\displaystyle h_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \mu }}(ce^{-jk_{x\varepsilon }x}+de^{jk_{x\varepsilon }x})cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ \ \ \ \ (38)}


air region on right (x > w)







e

x


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[



j

k

x
o



k

z




ω

ε

o



ε

r





(
a
 

e


j

k

x
ε


w


+
b
 

e

j

k

x
ε


w


)
+



m
π

a


(
c
 

e


j

k

x
ε


w


+
d
 

e

j

k

x
ε


w


)
]

e


j

k

x
o


(
x

w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
 
(
39
)


{\displaystyle e_{x}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[{\frac {jk_{xo}k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}(a\ e^{-jk_{x\varepsilon }w}+b\ e^{jk_{x\varepsilon }w})+{\frac {m\pi }{a}}(c\ e^{-jk_{x\varepsilon }w}+d\ e^{jk_{x\varepsilon }w})]e^{-jk_{xo}(x-w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ \ (39)}







e

y


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[




k

z



ω

ε

o



ε

r








m
π

a


(
a
 

e


j

k

x
ε


w


+
b
 

e

j

k

x
ε


w


)

j

k

x
o


(
c
 

e


j

k

x
ε


w


+
d
 

e

j

k

x
ε


w


)
]

e


j

k

x
o


(
x

w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
40
)


{\displaystyle e_{y}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[-{\frac {k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}{\frac {m\pi }{a}}(a\ e^{-jk_{x\varepsilon }w}+b\ e^{jk_{x\varepsilon }w})-jk_{xo}(c\ e^{-jk_{x\varepsilon }w}+d\ e^{jk_{x\varepsilon }w})]e^{-jk_{xo}(x-w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (40)}







e

z


=




k

o


2



ε

r




k

z


2




j
ω

ε

o



ε

r





(
a
 

e


j

k

x
ε


w


+
b
 

e

j

k

x
ε


w


)

e


j

k

x
o


(
x

w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
(
41
)


{\displaystyle e_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \varepsilon _{o}\varepsilon _{r}}}(a\ e^{-jk_{x\varepsilon }w}+b\ e^{jk_{x\varepsilon }w})e^{-jk_{xo}(x-w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ (41)}







h

x


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[



m
π

a




1

ε

r




(
a
 

e


j

k

x
ε


w


+
b
 

e

j

k

x
ε


w


)
+



j

k

x
o



k

z




ω
μ



(
c
 

e


j

k

x
ε


w


+
d
 

e

j

k

x
ε


w


)
]

e


j

k

x
o


(
x

w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
(
42
)


{\displaystyle h_{x}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[{\frac {m\pi }{a}}{\frac {1}{\varepsilon _{r}}}(a\ e^{-jk_{x\varepsilon }w}+b\ e^{jk_{x\varepsilon }w})+{\frac {jk_{xo}k_{z}}{\omega \mu }}(c\ e^{-jk_{x\varepsilon }w}+d\ e^{jk_{x\varepsilon }w})]e^{-jk_{xo}(x-w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ (42)}







h

y


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[
j



k

x
o



ε

r




(
a
 

e


j

k

x
ε


w


+
b
 

e

j

k

x
ε


w


)
+



k

z



ω
μ






m
π

a


(
c
 

e


j

k

x
ε


w


+
d
 

e

j

k

x
ε


w


)
]

e


j

k

x
0


(
x

w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
43
)


{\displaystyle h_{y}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[j{\frac {k_{xo}}{\varepsilon _{r}}}(a\ e^{-jk_{x\varepsilon }w}+b\ e^{jk_{x\varepsilon }w})+{\frac {k_{z}}{\omega \mu }}{\frac {m\pi }{a}}(c\ e^{-jk_{x\varepsilon }w}+d\ e^{jk_{x\varepsilon }w})]e^{-jk_{x0}(x-w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (43)}







h

z


=




k

o


2



ε

r




k

z


2




j
ω
μ



(
c
 

e


j

k

x
ε


w


+
d
 

e

j

k

x
ε


w


)

e


j

k

x
o


(
x

w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
(
44
)


{\displaystyle h_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \mu }}(c\ e^{-jk_{x\varepsilon }w}+d\ e^{jk_{x\varepsilon }w})e^{-jk_{xo}(x-w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ (44)}


air region on left (x < -w)







e

x


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[




j

k

x
o



k

z




ω

ε

o



ε

r





(
a
 

e

j

k

x
ε


w


+
b
 

e


j

k

x
ε


w


)
+



m
π

a


(
c
 

e

j

k

x
ε


w


+
d
 

e


j

k

x
ε


w


)
]

e

j

k

x
o


(
x
+
w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
 
(
45
)


{\displaystyle e_{x}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[{\frac {-jk_{xo}k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}(a\ e^{jk_{x\varepsilon }w}+b\ e^{-jk_{x\varepsilon }w})+{\frac {m\pi }{a}}(c\ e^{jk_{x\varepsilon }w}+d\ e^{-jk_{x\varepsilon }w})]e^{jk_{xo}(x+w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ \ (45)}







e

y


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[




k

z



ω

ε

o



ε

r








m
π

a


(
a
 

e

j

k

x
ε


w


+
b
 

e


j

k

x
ε


w


)

j

k

x
o


(
c
 

e

j

k

x
ε


w


+
d
 

e


j

k

x
ε


w


)
]

e

j

k

x
o


(
x
+
w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
46
)


{\displaystyle e_{y}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[-{\frac {k_{z}}{\omega \varepsilon _{o}\varepsilon _{r}}}{\frac {m\pi }{a}}(a\ e^{jk_{x\varepsilon }w}+b\ e^{-jk_{x\varepsilon }w})-jk_{xo}(c\ e^{jk_{x\varepsilon }w}+d\ e^{-jk_{x\varepsilon }w})]e^{jk_{xo}(x+w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (46)}







e

z


=




k

o


2



ε

r




k

z


2




j
ω

ε

o



ε

r





(
a
 

e

j

k

x
ε


w


+
b
 

e


j

k

x
ε


w


)

e


j

k

x
o


(
x
+
w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
(
47
)


{\displaystyle e_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \varepsilon _{o}\varepsilon _{r}}}(a\ e^{jk_{x\varepsilon }w}+b\ e^{-jk_{x\varepsilon }w})e^{-jk_{xo}(x+w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ (47)}







h

x


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[



m
π

a




1

ε

r




(
a
 

e

j

k

x
ε


w


+
b
 

e


j

k

x
ε


w


)




j

k

x
o



k

z




ω
μ



(
c
 

e

j

k

x
ε


w


+
d
 

e


j

k

x
ε


w


)
]

e

j

k

x
o


(
x
+
w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
 
(
48
)


{\displaystyle h_{x}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[{\frac {m\pi }{a}}{\frac {1}{\varepsilon _{r}}}(a\ e^{jk_{x\varepsilon }w}+b\ e^{-jk_{x\varepsilon }w})-{\frac {jk_{xo}k_{z}}{\omega \mu }}(c\ e^{jk_{x\varepsilon }w}+d\ e^{-jk_{x\varepsilon }w})]e^{jk_{xo}(x+w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ \ (48)}







h

y


=




k

o


2



ε

r




k

z


2





k

o


2




k

z


2





[

j



k

x
o



ε

r




(
a
 

e

j

k

x
ε


w


+
b
 

e


j

k

x
ε


w


)
+



k

z



ω
μ






m
π

a


(
c
 

e

j

k

x
ε


w


+
d
 

e


j

k

x
ε


w


)
]

e

j

k

x
o


(
x
+
w
)


s
i
n
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
 
(
49
)


{\displaystyle h_{y}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{k_{o}^{2}-k_{z}^{2}}}[-j{\frac {k_{xo}}{\varepsilon _{r}}}(a\ e^{jk_{x\varepsilon }w}+b\ e^{-jk_{x\varepsilon }w})+{\frac {k_{z}}{\omega \mu }}{\frac {m\pi }{a}}(c\ e^{jk_{x\varepsilon }w}+d\ e^{-jk_{x\varepsilon }w})]e^{jk_{xo}(x+w)}sin({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ \ (49)}







h

z


=




k

o


2



ε

r




k

z


2




j
ω
μ



(
c
 

e

j

k

x
ε


w


+
d
 

e


j

k

x
ε


w


)

e

j

k

x
o


(
x
+
w
)


c
o
s
(



m
π

a


y
)

e


j

k

z


z


 
 
 
 
 
(
50
)


{\displaystyle h_{z}={\frac {k_{o}^{2}\varepsilon _{r}-k_{z}^{2}}{j\omega \mu }}(c\ e^{jk_{x\varepsilon }w}+d\ e^{-jk_{x\varepsilon }w})e^{jk_{xo}(x+w)}cos({\frac {m\pi }{a}}y)e^{-jk_{z}z}\ \ \ \ \ (50)}


these expressions not directly provided transverse resonance method.


finally, remaining continuity conditions homogeneous system of 4 equations in 4 unknowns a, b, c, d, obtained. non-trivial solutions found imposing determinant of coefficients vanishes. in way, using eqs. (21) , (26) dispersion equation, gives possible value longitudinal propagation constant kz various modes, obtained.


then, unknowns a, b, c, d can found, apart arbitrary factor.


in order obtain cutoff frequencies of various modes sufficient set kz=0 in determinant , solve equation, simplified, reference frequency. similar simplification not occur when using transverse resonance method since kz implicitly appears; equations solved in order obtain cutoff frequencies formally same.


a simpler analysis, expanding again field superposition of modes, can obtained taking account electric field orientation required mode , bisecting structure conducting wall, has been done in fig. 3. in case, there 2 regions, 6 unknowns have determined , continuity conditions 6 (continuity of ey, ez, hy, hz x = w , vanishing of ey, ez x=0).


finally important notice resulting dispersion equation factorizable in product of 2 expressions, coincide dispersion equation te , tm modes reference x direction, respectively. solutions belong these 2 classes of modes.







Comments

Popular posts from this blog

United Kingdom List of armoured fighting vehicles by country

Episodes List of 2 Broke Girls episodes

Advert synopses Orson Welles Paul Masson adverts