Page 354 - Electromagnetics

parameters to describe the cascaded system of two layers:
(n) (n) (n+1) (n+1)
T T T T
11 12 11 12 a n+2 = a n .
T (n) T (n) T (n+1) T (n+1) b n+2 b n
21 22 21 22
Since for a periodic layered medium the wave amplitudes should obey (4.422), we have

T 11 T 12 a n+2 a n jκL a n+2
= = e (4.424)
T 21 T 22 b n+2 b n b n+2
where L = n + n+1 is the period of the structure and
(n) (n) (n+1) (n+1)
T T T T
T 11 T 12 11 12 11 12
= (n) (n) (n+1) (n+1) .
T 21 T 22 T T T T
21 22 21 22
Equation (4.424) is an eigenvalue equation for κ and can be rewritten as
jκL
T 11 − e T 12 a n+2 = 0 .
T 21 T 22 − e jκL b n+2 0
This equation only has solutions when the determinant of the matrix vanishes. Expansion
of the determinant gives

T 11 T 22 − T 12 T 21 − e jκL (T 11 + T 22 ) + e j2κL = 0. (4.425)
The first two terms are merely
(n) (n) (n+1) (n+1)
T 11 T 12 11 12 11 12
T T T T
T 11 T 22 − T 12 T 21 = = (n) (n+1) .
T 21 T 22 T T T T
(n) (n+1)
21 22 21 22
Since we can show that
(n) (n)
T T Z n−1

11 12 = ,
T T Z n
(n) (n)
21 22
we have
Z n−1 Z n
T 11 T 22 − T 12 T 21 = = 1
Z n Z n+1
where we have used Z n−1 = Z n+1 because of the periodicity of the medium. With this,
(4.425) becomes
T 11 + T 22
cos κL = .
2
Finally, computing the matrix product and simplifying to find T 11 + T 22 , we have

cos κL = cos(k z,n n ) cos(k k,n−1 n−1 ) −
1 Z n−1 Z n
− + sin(k z,n n ) sin(k z,n−1 n−1 ) (4.426)
2 Z n Z n−1
or equivalently
1 (Z n−1 + Z n ) 2
cos κL = cos(k z,n n + k z,n−1 n−1 ) −
4 Z n Z n−1
1 (Z n−1 − Z n ) 2
− cos(k z,n n − k z,n−1 n−1 ). (4.427)
4 Z n Z n−1

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