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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)

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