Wave Propagation in Linear Media | re-examined

3.2 Excursion: a delay line
sections of the transmission line do not in uence each other. As the voltage and current in the
line stand for the electric and magnetic elds, respectively, this is tantamount to the exclusive
occurence of TEM waves [93]. A slightly less restrictive de nition based on the alternative
description of the elds by a scalar and a vector potential was given by Paschke [92]. The use
of the equivalent circuit is then justi ed when the contribution of the vector potential to the
electric eld is negligible compared with that of the scalar potential, so that the electric eld
is almost curl-free (r E ' 0). Apart from TEM waves, where the requirement is identically
satis ed, it is a good approximation also for slow waves characterised by vp c.
3.2 Excursion: a delay line
As a second example, we review now the model of a delay line shown in g. 3.2 , which was
already analysed by Borgnis [38] as well as Kleen and Poschl [39]. The walls of the wave
guide consist of conducting combs with narrowly spaced teeth. For the sake of analytical
simplicity, we assume that the extention in z-direction is in nite and all eld components in
this direction are constant. In addition, since the boundaries are perfectly conducting, there
is no component of the electric eld in z-direction. The magnetic eld, on the contrary, does
exist in this direction only. Hence we have
E z=0; H x=H y=0;
@
@z
=0: (3.16)
With these restrictions and the assumption of a linear and isotropic medium, the Maxwell
equations reduce to (from r H =@D=@t)
and (from r E=,@B=@t)
@Hz @y = " @Ex @t
;
@Hz @x = ," @Ey @t
@Ey @x , @Ex @y = , @Hz @t
(3.17)
: (3.18)
Note that the other two Maxwell equations are also satis ed. It follows directly from our
assumptions (3.16) that rH = 0, and rE = 0 can be shown by di erentiating this equation
with respect to t and those of (3.17) with respect to x and y.
The di erential equations can be combined to yield a wave equation for the transverse magnetic
component
@2H z
@x2 + @2Hz @y2 = " @2H z
@t2 : (3.19)
Like before, we set H z = Hze j(!t,kx) , and with c 2 =1=( ") and = p k 2 , (!=c) 2 we nally
obtain the two possible solutions for the amplitude of the magnetic eld, an antisymmetric
and a symmetric mode,
Hz = C sinh y ; Hz = C cosh y (3.20)
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