Wave Propagation in Linear Media | re-examined
Wave Propagation in Linear Media | re-examined
Wave Propagation in Linear Media | re-examined
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3<strong>Wave</strong> propagation <strong>in</strong> electromagnetic transmission l<strong>in</strong>es<br />
What we want to study is the evolution of the elds if we switch on the transverse component<br />
of the magnetic eld at t = 0 <strong>in</strong> such a way that only the TE01 mode is excited. This<br />
assumption is quite <strong>re</strong>asonable s<strong>in</strong>ce this mode has the lowest cuto f<strong>re</strong>quency and is thus<br />
the dom<strong>in</strong>at<strong>in</strong>g mode <strong>in</strong> the wave guide [15, 93]. So the boundary condition <strong>re</strong>ads<br />
Hy(0;t)= (t)H0cos !0t s<strong>in</strong> y<br />
b<br />
: (3.91)<br />
With the substitution H0 = j 0<br />
! E and tak<strong>in</strong>g the <strong>re</strong>al parts of the complex functions, we<br />
nally obta<strong>in</strong> the steady-state elds<br />
H0<br />
Ez;s<br />
Hx;s<br />
H0<br />
Hy;s<br />
H0<br />
=<br />
p =" =<br />
r<br />
1<br />
1 , !0<br />
!p<br />
= s<strong>in</strong> y<br />
b e, 0x cos !0t<br />
r !p<br />
!0<br />
1<br />
2<br />
2<br />
, 1<br />
cos y<br />
b e, 0x cos !0t<br />
s<strong>in</strong> y<br />
b e, 0x s<strong>in</strong> !0t ;<br />
(3.92)<br />
whe<strong>re</strong> 0 = 1p<br />
!p c<br />
2 , !0 2 is the attenuation constant at the signal f<strong>re</strong>quency. Compar<strong>in</strong>g<br />
these solutions with the ones we obta<strong>in</strong>ed for the cur<strong>re</strong>nt (3.69) and voltage (3.71) <strong>in</strong> the<br />
plasma transmission l<strong>in</strong>e, we notice a strik<strong>in</strong>g similarity: the transverse component of the<br />
magnetic eld, Hx, cor<strong>re</strong>sponds to the cur<strong>re</strong>nt <strong>in</strong> the transmission l<strong>in</strong>e, and the electric eld<br />
to the negative voltage. As the <strong>in</strong>itial conditions a<strong>re</strong> also the same, we can apply the same<br />
method we used successfully <strong>in</strong> the last section to obta<strong>in</strong> the transient parts of the<br />
With the familiar scaled variables<br />
elds.<br />
= c<br />
!p<br />
; T = !pt; X= !px<br />
c<br />
; = !0<br />
!p<br />
; Y = y<br />
b<br />
we can immediately write down the complete solutions for the transverse elds,<br />
H0<br />
Hy<br />
H0<br />
= s<strong>in</strong> Y<br />
Ez p = s<strong>in</strong> Y<br />
="<br />
h e ,X p 1, 2<br />
, 2 Z 1<br />
0<br />
cos T ,<br />
s<strong>in</strong> ( X) cos T p 1+ 2<br />
1+ 2 , 2<br />
h<br />
,X<br />
e p 1, 2 1<br />
p s<strong>in</strong> T ,<br />
1= 2 , 1<br />
, 2 Z 1<br />
0<br />
p 1+ 2 cos ( X) s<strong>in</strong> T p 1+ 2<br />
62<br />
1+ 2 , 2<br />
d<br />
i<br />
; (3.93)<br />
d<br />
i :<br />
(3.94)<br />
(3.95)