Page 401 - Electromagnetics

Hence for waveguides filled with lossless media the total time-average power flow is given
by the superposition of the individual modal powers.
Simple formulas for the individual modal powers in a lossless guide may be obtained
by substituting the expressions for the fields. For TM modes we use (5.138) and (5.139)
to get
1 2 ∗ − j(k z −k z )
ˇ
ˇ
∗
∗
P av = Re |k z | Y e ˆ z · ∇ t ψ e × [ˆ z ×∇ t ψ ] dS
e
e
2 CS
1 2 − j(k z −k z )
∗
ˇ
ˇ
∗
= |k z | Re Y e ∗ e ∇ t ψ e ·∇ t ψ dS.
e
2 CS
ˇ
Here we have used (B.7) and ˆ z ·∇ t ψ e = 0. This expression can be simplified by using the
two-dimensional version of Green’s first identity (B.29):

∂b
2
(∇ t a ·∇ t b + a∇ b) dS = a dl.
t
S ∂n
ˇ
Using a = ψ e and b = ψ and integrating over the waveguide cross-section we have
ˇ ∗
e
ˇ ∗
∂ψ
ˇ ˇ ∗ ˇ 2 ˇ ∗ ˇ e
(∇ t ψ e ·∇ t ψ + ψ e ∇ ψ ) dS = dl.
e e ψ e
CS ∂n
ˇ
Substituting ∇ ψ =−k ψ and remembering that ψ e = 0 on we reduce this to
2 ˇ ∗
2 ˇ ∗
t
e
c
e

ˇ
ˇ ˇ
ˇ
∗
∗
∇ t ψ e ·∇ t ψ dS = k c 2 ψ e ψ dS. (5.153)
e
e
CS CS
Thus the power is
1 2 2 − j(k z −k z )z
ˇ ˇ
∗
∗
P av = Re Y e ∗ |k z | k e ψ e ψ dS.
c
e
2 CS
For modes above cutoff we have k z = β and Y e = ω /k z = ω /β. The power carried by
these modes is thus
1 2
ˇ ˇ
∗
P av = ω βk c ψ e ψ dS. (5.154)
e
2 CS
∗
For modes belowcutoff we have k z =− jα and Y e = jω /α.Thus Re{Y }= 0 and
e
P av = 0. For frequencies belowcutoff the fields are evanescent and do not carry power
in the manner of propagating waves.
For TE modes we may proceed similarly and show that
1 2
ˇ ˇ
∗
P av = ωµβk c ψ h ψ dS. (5.155)
h
2 CS
The details are left as an exercise.
Stored energy in a waveguide and the velocity of energy transport. Consider
a source-free section of lossless waveguide bounded on its two ends by the cross-sectional
ˇ c
ˇ i
surfaces CS 1 and CS 2 . Setting J = J = 0 in (4.156) we have
1
ˇ
(E × H ) · dS = 2 jω [ w e − w m ] dV,
ˇ ∗
2 S V
© 2001 by CRC Press LLC