Page 304 - Electromagnetics

in certain layers are nonuniform. For the case of perpendicular polarization we can write
˜ i
˜
˜ r
the electric field in region n, 0 ≤ n ≤ N − 1,as E ⊥n = E + E where
⊥n ⊥n
˜ i
e
E = ˆ ya n+1 e − jk x,n x − jk z,n (z−z n+1 ) ,
⊥n
˜ r
E = ˆ yb n+1 e − jk x,n x + jk z,n (z−z n+1 ) ,
e
⊥n
˜
˜ i
and the magnetic field as H ⊥n = H ⊥n + H r ⊥n where
˜ i
H ⊥n = −ˆ xk z,n + ˆ zk x,n a n+1 e − jk x,n x − jk z,n (z−z n+1 ) ,
e
k n η n
˜ r
H = +ˆ xk z,n + ˆ zk x,n b n+1 e − jk x,n x + jk z,n (z−z n+1 ) .
e
⊥n
k n η n
When n = N there is no reflected wave; in this region we write
˜
e
E ⊥N = ˆ ya N+1 e − jk x,N x − jk z,N (z−z N ) ,
˜ −ˆ xk z,N + ˆ zk x,N − jk x,N x − jk z,N (z−z N )
H ⊥N = a N+1 e e .
k N η N
Since a 1 is the known amplitude of the incident wave, there are 2N unknown wave am-
plitudes. We obtain the necessary 2N simultaneous equations byapplying the boundary
conditions at each of the interfaces. At interface n located at z = z n , 1 ≤ n ≤ N − 1,we
have from the continuityof tangential electric field
a n + b n = a n+1 e − jk z,n (z n −z n+1 ) + b n+1 e + jk z,n (z n −z n+1 )
while from the continuityof magnetic field
k z,n−1 k z,n−1 k z,n − jk z,n (z n −z n+1 ) k z,n + jk z,n (z n −z n+1 )
−a n + b n =−a n+1 e + b n+1 e .
k n−1 η n−1 k n−1 η n−1 k n η n k n η n
Noting that the wave impedance of region n is
k n η n
Z ⊥n =
k z,n
and defining the region n propagation factor as

˜
P n = e − jk z,n n
where n = z n+1 − z n , we can write

˜ 2
˜
˜
a n P n + b n P n = a n+1 + b n+1 P , (4.305)
n
˜ ˜ Z ⊥n−1 Z ⊥n−1 ˜ 2
−a n P n + b n P n =−a n+1 + b n+1 P . (4.306)
n
Z ⊥n Z ⊥n
We must still applythe boundaryconditions at z = z N . Proceeding as above, we find
˜
that (4.305) and (4.306) hold for n = N if we set b N+1 = 0 and P N = 1.
The 2N simultaneous equations (4.305)–(4.306) maybe solved using standard matrix
methods. However, through a little manipulation we can put the equations into a form
easilysolved byrecursion, providing a verynice physical picture of the multiple reflections
that occur within the layered medium. We begin by eliminating b n bysubtracting (4.306)
from (4.305):

˜ Z ⊥n−1 ˜ 2 Z ⊥n−1
2a n P n = a n+1 1 + + b n+1 P 1 − . (4.307)
n
Z ⊥n Z ⊥n

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