电磁场与电磁波第25讲矩形波导的传播特性
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Electromagnetic Field and Wave 电磁场与电磁波
2011. 5.21
Review
1. General Wave Behaviors along Uniform Guiding Structures
e ze jt e( jt z) e ze j(t z)
E(x, y, z;t) Re[E0 (x, y)e( jt z) ]
0
Ez
yb
0 kyb n
ky
n
b
(n 1, 2,3...)
Ez Ez0 (x, y)e z
(C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
m
C2 C4 sin( a
x) sin( n
b
y)e z
E0
b
,
x a z
the component Ez should first be solved, and from which the other components can be derived.
The z-component of the electric field intensity can be written as
variables is used. Let
Ez0( x,y ) X ( x )Y( y )
We obtain
X Y h2 XY
where X" denotes the second derivative of X with respect to x, and Y" denotes the second derivative of Y with respect to y.
Ez ( x, y, z ) Ez0( x, y )e z
5
It satisfies the following scalar Helmholtz equation, i.e.
(
2 x2
2 y 2
h2
)Ez0(
x, y
)
0
In order to solve the above equation, the method of separation of
2xyE ( 2 k2 )E 0 and 2xyH ( 2 k2 )H 0
Ex0
1 h2
Ez0 x
j
H z 0 y
Ey0
1 h2
Ez0 y
Baidu Nhomakorabea
j
H
0 z
x
H
0 x
1 h2
H z 0 x
j
Ez0 y
H
0 y
1 h2
H z 0 y
j
Ez0 x
2
2xyE ( 2 k2 )E 0
X X
kx2
Y Y
k
2 y
where k x and k y are called the separation constants, and they can be found by using the boundary conditions.
Obviously
h2 kx2 ky2
X X
Ez Ez0 (x, y)e z (C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
Boundary conditions: aˆn2 (E1 E2 ) 0
E
0 y
Ez0
x0
0
Et
0
E
0 y
Ex0
Ez0 Ez0
xa y0
0 0
2xyE ( 2 k2 )E 0 and 2xyH ( 2 k2 )H 0
6
X Y h2 XY
The only way the equation can be satisfied is that both terms on the left
side are constants. Now let
kx2
X
k
2 x
X
0
2 kx2 0 jkx
X C1 cos kx x C2 sin kx x
7
The two equations are second order ordinary differential equations,
and the general solutions, are respectively
X C1 cos kx x C2 sin kx x
Y C3 cos ky y C4 sin ky y
where all the constants C1 , C2 , C3 , C4 , and k x , k y , depend on the boundary conditions.
The z-component of the electric field intensity can be written as
()
ZTE
1
fc f
2
()
3
Main topic
1. Rectangular Waveguides
4
1. Rectangular Waveguides
1.1 TM waves Select the rectangular coordinate system and let the broad side
be placed along the x-axis, the narrow side along the y-axis, and the propagating direction be along the z-axis.
For TM waves, Hz = 0 , and according
y
to the method of longitudinal fields,
Ex0 Ez0 yb 0
y
b
,
x a z
8
Ez Ez0 (x, y)e z (C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
Ez
x0
0 C1
0
Ez
xa
0 kxa
m
kx
m
a
(m 1, 2,3...)
Ez
y0
0 C3
2 h2 k2
and 2xyH ( 2 k2 )H 0
h2 c2
or
fc
2
h
k
1
fc f
2
(rad/m).
g
2
k
2
2
1
fc f
2
1
fc f
up
u
u
1
fc f
2
1
ug d / d u
1
fc f
2
g
u.
ugup u2.
2
ZTM
1
fc f
2011. 5.21
Review
1. General Wave Behaviors along Uniform Guiding Structures
e ze jt e( jt z) e ze j(t z)
E(x, y, z;t) Re[E0 (x, y)e( jt z) ]
0
Ez
yb
0 kyb n
ky
n
b
(n 1, 2,3...)
Ez Ez0 (x, y)e z
(C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
m
C2 C4 sin( a
x) sin( n
b
y)e z
E0
b
,
x a z
the component Ez should first be solved, and from which the other components can be derived.
The z-component of the electric field intensity can be written as
variables is used. Let
Ez0( x,y ) X ( x )Y( y )
We obtain
X Y h2 XY
where X" denotes the second derivative of X with respect to x, and Y" denotes the second derivative of Y with respect to y.
Ez ( x, y, z ) Ez0( x, y )e z
5
It satisfies the following scalar Helmholtz equation, i.e.
(
2 x2
2 y 2
h2
)Ez0(
x, y
)
0
In order to solve the above equation, the method of separation of
2xyE ( 2 k2 )E 0 and 2xyH ( 2 k2 )H 0
Ex0
1 h2
Ez0 x
j
H z 0 y
Ey0
1 h2
Ez0 y
Baidu Nhomakorabea
j
H
0 z
x
H
0 x
1 h2
H z 0 x
j
Ez0 y
H
0 y
1 h2
H z 0 y
j
Ez0 x
2
2xyE ( 2 k2 )E 0
X X
kx2
Y Y
k
2 y
where k x and k y are called the separation constants, and they can be found by using the boundary conditions.
Obviously
h2 kx2 ky2
X X
Ez Ez0 (x, y)e z (C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
Boundary conditions: aˆn2 (E1 E2 ) 0
E
0 y
Ez0
x0
0
Et
0
E
0 y
Ex0
Ez0 Ez0
xa y0
0 0
2xyE ( 2 k2 )E 0 and 2xyH ( 2 k2 )H 0
6
X Y h2 XY
The only way the equation can be satisfied is that both terms on the left
side are constants. Now let
kx2
X
k
2 x
X
0
2 kx2 0 jkx
X C1 cos kx x C2 sin kx x
7
The two equations are second order ordinary differential equations,
and the general solutions, are respectively
X C1 cos kx x C2 sin kx x
Y C3 cos ky y C4 sin ky y
where all the constants C1 , C2 , C3 , C4 , and k x , k y , depend on the boundary conditions.
The z-component of the electric field intensity can be written as
()
ZTE
1
fc f
2
()
3
Main topic
1. Rectangular Waveguides
4
1. Rectangular Waveguides
1.1 TM waves Select the rectangular coordinate system and let the broad side
be placed along the x-axis, the narrow side along the y-axis, and the propagating direction be along the z-axis.
For TM waves, Hz = 0 , and according
y
to the method of longitudinal fields,
Ex0 Ez0 yb 0
y
b
,
x a z
8
Ez Ez0 (x, y)e z (C1 cos kx x C2 sin kx x)(C3 cos ky y C4 sin ky y)e z
Ez
x0
0 C1
0
Ez
xa
0 kxa
m
kx
m
a
(m 1, 2,3...)
Ez
y0
0 C3
2 h2 k2
and 2xyH ( 2 k2 )H 0
h2 c2
or
fc
2
h
k
1
fc f
2
(rad/m).
g
2
k
2
2
1
fc f
2
1
fc f
up
u
u
1
fc f
2
1
ug d / d u
1
fc f
2
g
u.
ugup u2.
2
ZTM
1
fc f