Apr 2, 2001
Notes for Lecture
Notes for Lecture #30
Electromagnetic wave guides
In order to understand the operation of a wave guide, we must
first learn how electromagnetic waves behave in a dissipative
medium. A plane wave solution to Maxwell's equations of the
form:
|
E = E0eik [^(k)] ·r - i wt and B = |
k
w
|
|
^
k
|
×E0 eik [^(k)] ·r - i wt |
| (1) |
for the electric and magnetic fields, with the wave vector k
satisfying the relation:
We can determine the complex wavevector kr + i ki according to
|
kr = |
æ
ç
ç
ç
è
|
2
|
|
ö
÷
÷
÷
ø
|
1/2
|
and ki = |
æ
ç
ç
ç
è
|
2
|
|
ö
÷
÷
÷
ø
|
1/2
|
|
| (3) |
The form of the frequency dependent constants R and
I depend on the materials. For the Drude model at
low frequency (Eq. 7.56), R = w2 meb and I = wms, for
example. The value of ki determines the rate of decay of the
field amplitudes in the vicinity of the surface, with the skin
depth given by d º 1/ki. In the limit that
I >> R, as in the case of a good
conductor at low frequency, d » (2/(wms) )1/2.
For an ïdeal" conductor I ® ¥, so
that the fields are confined to the surface. Because of the
field continuity conditions at the surface of the conductor, this
means that, Btangential ¹ 0 (because there can
be a surface current), Enormal ¹ 0 (because
there can be a surface charge), and Bnormal = 0.
Suppose we construct a wave guide from an ïdeal" conductor,
designating [^(z)] as the propagation direction. We
will assume that the fields take the form:
|
E = E(x,y) eikz-iwt and B = B(x,y) eikz-iwt |
| (4) |
inside the pipe, where now k and e are assumed to be
real. Assuming that there are no sources inside the pipe, the
fields there must satisfy Maxwell's equations (8.16) which expand
to the following :
|
|
¶Bx
¶x
|
+ |
¶By
¶ y
|
+ i k Bz = 0. |
| (5) |
|
|
¶Ex
¶x
|
+ |
¶Ey
¶ y
|
+ i k Ez = 0. |
| (6) |
|
|
¶Ey
¶x
|
- |
¶Ex
¶ y
|
= i wBz. |
| (9) |
|
|
¶Bz
¶y
|
- i k By = -i mew Ex. |
| (10) |
|
i k Bx - |
¶Bz
¶x
|
= -i mew Ey. |
| (11) |
|
|
¶By
¶x
|
- |
¶Bx
¶ y
|
= -i mewEz. |
| (12) |
Combining Faraday's Law and Ampere's Law, we find that each field
component must satisfy a two-dimensional Helmholz equation:
|
|
æ
ç
è
|
|
¶2
¶x2
|
+ |
¶2
¶ y2
|
- k2 + mew2 |
ö
÷
ø
|
Ex(x,y) = 0, |
| (13) |
with similar expressions for each of the other field components.
For the rectangular wave guide discussed in Section 8.4 of your
text a solution for a TE mode can have:
|
Ez(x,y) º 0 and Bz(x,y) = B0 cos |
æ
ç
è
|
m px
a
|
ö
÷
ø
|
cos |
æ
ç
è
|
n p y
b
|
ö
÷
ø
|
, |
| (14) |
with k2 º k2mn = mew2 -[([(m p)/ a])2 + ([(n p)/ b])2] . From this result and Maxwell's
equations, we can determine the other field components. For
example Eqs. (7-8) simplify to
|
Bx = - |
k
w
|
Ey and By = |
k
w
|
Ex. |
| (15) |
These results can be used in Eqs. (10-11) to solve for
the fields Ex and Ey and Bx and By:
|
Ex = |
w
k
|
By = |
-i w
k2-mew2
|
|
¶Bz
¶y
|
= |
-iw
|
|
é
ê
ë
|
|
æ
ç
è
|
m p
a
|
ö
÷
ø
|
2
|
+ |
æ
ç
è
|
n p
b
|
ö
÷
ø
|
2
|
|
ù
ú
û
|
|
|
|
n p
b
|
B0 cos |
æ
ç
è
|
m px
a
|
ö
÷
ø
|
sin |
æ
ç
è
|
n p y
b
|
ö
÷
ø
|
, |
| (16) |
and
|
Ey = - |
w
k
|
Bx = |
i w
k2-mew2
|
|
¶Bz
¶x
|
= |
i w
|
|
é
ê
ë
|
æ
ç
è
|
m p
a
|
ö
÷
ø
|
2
|
+ |
æ
ç
è
|
n p
b
|
ö
÷
ø
|
2
|
|
ù
ú
û
|
|
|
|
n p
a
|
B0 sin |
æ
ç
è
|
m px
a
|
ö
÷
ø
|
cos |
æ
ç
è
|
n p y
b
|
ö
÷
ø
|
. |
| (17) |
One can check this result to show that these results satisfy the
boundary conditions. For example, Etangential = 0 is satisfied since Ex(x,0) = Ex(x,b) = 0 and Ey(0,y) = Ey(a,y) = 0. This was made possible choosing ÑBzûsurface·[^(n)] = 0, where
[^(n)] denotes a unit normal vector pointing out of the
wave guide surface.
File translated from TEX by TTH, version 2.20.
On 2 Apr 2001, 10:30.