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Optical Receivers 229

is called the intermediate frequency. To obtain Eq. (5.88), we have used the formula
∗
Re (X)=(X + X )∕2. (5.90)
2
When the LO power P LO is much larger than the signal power A , the first term in Eq. (5.88) can be neglected.
r
Since the LO output is cw, A 2 is a constant and it leads to a d.c. component in the photocurrent which can be
LO
removed by capacitive coupling from the photodetector to the front end of the electrical amplifier. Therefore,
the signal that goes to the front end can be written as
I (t)= RA A Re {s(t) exp [−i( t + − )]}. (5.91)
d r LO IF c LO
With
2
P = A , (5.92)
r r
P = A 2 , (5.93)
LO LO
Eq. (5.91) may be rewritten as
√
I (t)= R P P Re {s(t) exp [−i( t + − )]}. (5.94)
d r LO IF c LO

5.6.1.1 Homodyne Single-Branch Receiver
If = 0, such a receiver is known as a homodyne receiver.If < 2B , where B is the symbol rate, it
IF IF s s
is sometimes referred to as intradyne. Otherwise, it is called a heterodyne receiver. For a homodyne receiver,
the phase of the received carrier should be exactly the same as the phase of the local oscillator. This can
c
be achieved using an optical phase-locked loop or it can be post-corrected using digital phase estimation
techniques (see Chapter 11). When the phases are exactly aligned ( = LO ), Eq. (5.94) can be written as
c
I (t)= RP Re [s(t)], (5.95)
d
0
√
where P = P P . If the transmitted signal is real such as that corresponding to binary OOK or PSK, the
0 r LO
real part of s(t) has all the information required to retrieve the transmitted data. If the transmitted signal is
complex, as in the case of amplitude and phase-modulated signals, in-phase and quadrature (IQ) receivers are
required to estimate the transmitted information, which will be discussed in Sections 5.6.3 and 5.6.4. Note that
√
in Eq. (5.95), the responsivity R is multiplied by P P . If we choose a very large local oscillator power,
r LO
P LO , the effective responsivity, RP , could be increased and, therefore, the sensitivity of the coherent receiver
0
was significantly larger than that of the direct detection receiver. This was one of the reasons for pursuing
coherent receivers in the 1980s. If the phase of the local oscillator is not fully aligned with the phase of the
carrier, Eq. (5.94) can be written as
I (t)= RP Re [s(t) exp (−iΔ)], (5.96)
d
0
where Δ ≡ − LO is the phase error. If Δ is ∕2 and s(t) is 1 within a bit interval, I (t)= 0 and, therefore,
d
c
the phase error leads to bit errors.
Example 5.6
A BPSK-NRZ signal is transmitted over a fiber of length 100 km. The peak power of the signal at the transmit-
ter is 12 dBm. The fiber loss is 0.2 dB/km. Assuming that the receiver is a homodyne single-branch receiver,
find the peak current if (a) LO power = 10 dBm, (b) LO power =−10 dBm. Assume R = 0.9A/W.