10.2 Passive Optical Couplers
407
M ϕ =
exp( jkL/2)
0
0
e xp(− jkL/2)
(10.33)
The optical output fields E out,1 and E out,2 from the two central arms can be related
to the input fields E in,1 and E in,2 by
E out,1
E out,2
= M
E in,1
E in,2
(10.34)
where M is the product of the matrices M coupler , M ϕ , and M coupler ,
M = M coupler · M ϕ · M coupler =
M 11 M 21
M 12 M 22
= j
sin(kL/2) cos(kL/2)
cos(kL/2) − sin(kL/2)
(10.35)
Because the device of interest is a multiplexer, it is necessary to have the inputs
to the MZI be at different wavelengths; that is, E in,1 is at λ 1 and E in,2 is at λ 2 . Then,
from Eq. (10.34), the output field E out,1 and E out,2 are each the sum of the individual
contributions from the two input fields:
E out,1 = j
E in,1 (λ 1 ) sin(k 1 L/2) + E in,2 (λ 2 ) cos(k 2 L/2)
(10.36)
E out,2 = j
E in,1 (λ 1 ) cos(k 1 L/2) − E in,2 (λ 2 ) sin(k 2 L/2)
(10.37)
where k j = 2πn eff /λ j . The output powers are then found from the light intensity,
which is the square of the field strengths. Thus,
P out,1 = E out,1 E
∗
out,1 = sin
2
(k 1 L/2)P in,1 + cos
2
(k 2 L/2)P in,2
(10.38)
P out,2 = E out,2 E
∗
out,2 = cos
2
(k 1 L/2)P in,1 + sin
2
(k 2 L/2)P in,2
(10.39)
where P in,j = |E in, j |
2
= E in,j ·E
*
in,j . In Eqs. (10.38) and (10.39), the cross terms
are dropped because their frequency is twice the optical carrier frequency, which is
beyond the response capability of the photodetector.
From Eqs. (10.38) and (10.39), it can be seen that if one wants all the power from
both inputs to leave the same output port (e.g., port 2), then it is necessary to have
k 1 L/2 = π and k 2 L/2 = π/2, or
(k 1 − k 2 )L = 2π n e f f
1
λ 1
−
1
λ 2
L = π
(10.40)
407
M ϕ =
exp( jkL/2)
0
0
e xp(− jkL/2)
(10.33)
The optical output fields E out,1 and E out,2 from the two central arms can be related
to the input fields E in,1 and E in,2 by
E out,1
E out,2
= M
E in,1
E in,2
(10.34)
where M is the product of the matrices M coupler , M ϕ , and M coupler ,
M = M coupler · M ϕ · M coupler =
M 11 M 21
M 12 M 22
= j
sin(kL/2) cos(kL/2)
cos(kL/2) − sin(kL/2)
(10.35)
Because the device of interest is a multiplexer, it is necessary to have the inputs
to the MZI be at different wavelengths; that is, E in,1 is at λ 1 and E in,2 is at λ 2 . Then,
from Eq. (10.34), the output field E out,1 and E out,2 are each the sum of the individual
contributions from the two input fields:
E out,1 = j
E in,1 (λ 1 ) sin(k 1 L/2) + E in,2 (λ 2 ) cos(k 2 L/2)
(10.36)
E out,2 = j
E in,1 (λ 1 ) cos(k 1 L/2) − E in,2 (λ 2 ) sin(k 2 L/2)
(10.37)
where k j = 2πn eff /λ j . The output powers are then found from the light intensity,
which is the square of the field strengths. Thus,
P out,1 = E out,1 E
∗
out,1 = sin
2
(k 1 L/2)P in,1 + cos
2
(k 2 L/2)P in,2
(10.38)
P out,2 = E out,2 E
∗
out,2 = cos
2
(k 1 L/2)P in,1 + sin
2
(k 2 L/2)P in,2
(10.39)
where P in,j = |E in, j |
2
= E in,j ·E
*
in,j . In Eqs. (10.38) and (10.39), the cross terms
are dropped because their frequency is twice the optical carrier frequency, which is
beyond the response capability of the photodetector.
From Eqs. (10.38) and (10.39), it can be seen that if one wants all the power from
both inputs to leave the same output port (e.g., port 2), then it is necessary to have
k 1 L/2 = π and k 2 L/2 = π/2, or
(k 1 − k 2 )L = 2π n e f f
1
λ 1
−
1
λ 2
L = π
(10.40)