Page 257 - Fiber Optic Communications Fund

238 Fiber Optic Communications

1
out
q = [−iq + q ], (5.132)
2 r LO
2
[ out ]
q
out 1
q = . (5.133)
q out
2
The photocurrents I Q+ and I Q− can be calculated as in Section 5.6.2. Proceeding as before, the quadrature
current I is given by
Q
I = I Q+ − I Q− (5.134)
Q
=−RA A Im {s(t) exp [−i( t +Δ)]}. (5.135)
IF
r LO
∘
For the I-branch, the LO output is phase-shifted by 90 . The in-phase component of the current is
I = I I+ − I I− (5.136)
I
= RA A Re {s(t) exp [−i( t +Δ)]}. (5.137)
r LO IF
For homodyne receivers, = 0. When Δ = 0, from Eqs. (5.136) and (5.134), we find
IF
I = RA A Re {s(t)}, (5.138)
I r LO
I =−RA A Im {s(t)}, (5.139)
Q r LO
I = I − iI = RA A s(t). (5.140)
I Q r LO
∘
The components inside the rectangular line constitute a 2 × 490 hybrid. The transfer matrix of the 2 × 490 ∘
hybrid can be written as
⎡ 1 −i⎤
⎢−i 1 ⎥
T = . (5.141)
⎢ ⎥
11
⎢ ⎥
⎣−ii ⎦
∘
Let the output of the 2 × 490 hybrid be
1
⎡q ⎤
out
⎢ ⎥
2
out
q out = ⎢q ⎥ . (5.142)
⎢ 3 ⎥
q
⎢ out⎥
q
⎢ 4 ⎥
⎣ out⎦
∘
Now, the input–output relationship of the 2 × 490 hybrid can be written as
out in
q = Tq . (5.143)
Example 5.8
Repeat Example 5.7(a) with the BPSK signal replaced by the QPSK signal. Let s(t) be 1∠3∕4. Find the
in-phase and quadrature components of the current of a balanced IQ receiver.