Page 404 - Electromagnetics

where v = 1/(µ ) 1/2 . Thus the TM 11 mode has the lowest cutoff frequency of any TM
mode. There is a range of frequencies for which this is the only propagating TM mode.
For TE modes the solution is subject to
˜
∂ψ h (ρ,ω)
˜
ˆ n ·∇ t ψ h (ρ,ω) = = 0, ρ ∈ .
∂n
At x = 0 we have
˜
∂ψ h
= 0
∂x
leading to A x = 0.At y = 0 we have
˜
∂ψ h
= 0
∂y
leading to A y = 0.At x = a we require sin k x a = 0 and thus
nπ
k x = , n = 0, 1, 2,....
a
Similarly, from the condition at y = b we find
mπ
k y = , m = 0, 1, 2,....
b
The case n = m = 0 is not allowed since it produces the trivial solution. Thus
nπx mπy

˜
ψ h (x, y,ω) = B nm cos cos , m, n = 0, 1, 2,..., m + n > 0.
a b
From (5.140)–(5.142) we find that the fields are
nπx mπy ∓ jk z z

˜
2
H z = k B nm cos cos e ,
c nm
a b
nπ nπx mπy mπ nπx mπy
˜
H t =± jk z B nm ˆ x sin cos + ˆ y cos sin e ∓ jk z z ,
a a b b a b
mπ nπx mπy nπ nπx mπy
˜
E t = jk z Z h B nm ˆ x cos sin − ˆ y sin cos e ∓ jk z z .
b a b a a b
Here
η
.
Z h = !
1 − ω 2 /ω 2
c nm
In this case the modes are designated TE nm . The cutoff wavenumber of the TE nm mode
is
%
nπ mπ
2 2
= + , m, n = 0, 1, 2,..., m + n > 0
k c nm
a b
and the cutoff frequency is
%
nπ mπ
2 2
= v , m, n = 0, 1, 2,..., m + n > 0
ω c nm +
a b
where v = 1/(µ ) 1/2 . Modes having the same cutoff frequency are said to be degenerate.
This is the case with the TE and TM modes. However, the field distributions differ and
thus the modes are distinct. Note that we may also have degeneracy among the TE
© 2001 by CRC Press LLC