Page 306 - Electromagnetics

interface n + 1 with amplitude a n+1 as consisting of two terms. The first term is the
wave transmitted through interface n (at z = z n ). This wave must propagate through a
˜ ˜
distance n to reach interface n +1 and thus has an amplitude a n T n P n . The second term
is the reflection at interface n of the wave traveling in the −z direction within region n.
˜
˜
The amplitude of the wave before reflection is merely b n+1 P n , where the term P n results
from the propagation of the negatively-traveling wave from interface n + 1 to interface
n. Now, since the interfacial reflection coefficient at interface n for a wave incident from
region n is the negative of that for a wave incident from region n − 1 (since the wave
is traveling in the reverse direction), and since the reflected wave must travel through a
distance n from interface n back to interface n + 1, the amplitude of the second term is
˜
˜
˜
b n+1 P n (− n )P n . Finally, remembering that b n+1 = R n+1 a n+1 , we can write
˜
˜
˜ ˜
˜
˜
a n+1 = a n T n P n + a n+1 R n+1 P n (− n )P n .
This equation is exactlythe same as (4.309) which was found using the boundarycon-
ditions. Bysimilar reasoning, we maysaythat the wave traveling in the −z direction
in region n − 1 consists of a term reflected from the interface and a term transmitted
˜
through the interface. The amplitude of the reflected term is merely a n n . The amplitude
˜
of the transmitted term is found byconsidering b n+1 = R n+1 a n+1 propagated through a
distance n and then transmitted backwards through interface n. Since the transmission
˜
coefficient for a wave going from region n to region n − 1 is 1 + (− n ), the amplitude of
˜
˜
˜
the transmitted term is R n+1 P n (1 − n )a n+1 . Thus we have
˜
˜
˜
˜
b n = n a n + R n+1 P n (1 − n )a n+1 ,
which is identical to (4.311).
We are still left with the task of solving for the various field amplitudes. This can be
˜
˜
done using a simple recursive technique. Using T n = 1 + n we find from (4.309) that
˜ ˜
(1 + n )P n
a n+1 = a n . (4.313)
˜ ˜
1 + n R n+1 P ˜ 2
n
Substituting this into (4.311) we find
˜
˜
n + R n+1 P ˜ 2
n
b n = a n . (4.314)
˜ ˜
˜ 2
1 + n R n+1 P n
Using this expression we find a recursive relationship for the global reflection coefficient:
˜
˜
n + R n+1 P ˜ 2
b n
˜ n
R n = = . (4.315)
˜ ˜
˜ 2
1 + n R n+1 P
a n n
The procedure is now as follows. The global reflection coefficient for interface N is, from
(4.312),
˜
˜
R N = b N /a N = N . (4.316)
˜
˜
This is also obtained from (4.315) with R N+1 = 0. We next use (4.315) to find R N−1 :
˜
˜
˜ 2
N−1 + R N P
˜ N−1
R N−1 = .
˜
˜
˜ 2
1 + N−1 R N P
N−1
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