Fundamental fiber optic receiver operation

1. Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 1

2. Fundamental Receiver Operation
 The first receiver element is a pin or an avalanche photodiode, which
produces an electric current proportional to the received power level.
 Since this electric current typically is very weak, a front-end amplifier
boosts it to a level that can be used by the following electronics.
 After being amplified, the signal passes through a low-pass filter to
reduce the noise that is outside of the signal bandwidth.
 The also filter can reshape (equalize) the pulses that have become
distorted as they traveled through the fiber.
 Together with a clock (timing) recovery circuit, a decision circuit
decides whether a 1 or 0 pulse was received,
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 2

3. Error Sources
!
)(
)(
0
n
e
NnP
E
h
dttP
h
N
N
n
r
−
=
== ∫
τ
ν
η
ν
η
E is energy received in a time interval and is photon energy,
where is the probability that n electrons are emitted in an
interval .
torphotodetecinpairshole-
electronofnumberaveragetheis−N
efficiencyquantumdetectortheis−η
τ
νh
)(nPr
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 3

4. InterSymbol Interference (ISI)
Pulse spreading in an optical signal, after traversing along optical fiber, leads
to ISI. Some fraction of energy remaining in appropriate time slot is
designated by , so the rest is the fraction of energy that has spread into
adjacent time slots. γ
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 4

5. Receiver Configuration
The binary digital pulse train incident on the photodetector can be written
in the following form:
t.allforpositiveiswhichshapepulsereceivedtheis)(and
digitmessageththeofparameteramplitudeanisperiod,bitiswhere
)()(
th
nbT
nTthbtP
p
nb
n
bpn∑
+∞
−∞=
−=
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 5

6.  We assume the digital pulses with amplitude V represents bit 1 and 0
represents bit 0. Thus can take two values corresponding to each
binary data. By normalizing the input pulse to the photodiode to
have unit area
represents the energy in the nth pulse.
the mean output current from the photodiode at time t resulting from
pulse train given by (neglecting the DC components arising from dark
current noise):
nb
)(thp
∫
+∞
∞−
=1)( dtthp
nb
∑
+∞
−∞=
−ℜ==
n
bpno nTthbMtMP
h
q
ti )()()(
ν
η
sponsivityo Re−ℜ
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 6

7. Bit Error Rate (BER)
 Probability of Error= probability that the output voltage
is less than the threshold when a 1 is sent + probability that
the output voltage is more than the threshold when a 0 has
been sent.
 Typical error rates for optical fiber telecom systems range
from 10–9 to 10–12 (compared to 10-6 for wireless systems)
 The error rate depends on the signal-to-noise ratio at the
receiver (the ratio of signal power to noise power).
b
e
t
e
TB
Bt
N
N
N
t
t
/1
duringedtransmittpulsesof#total
intervalmecertain tiaovererrorof#
ErrorofyProbabilitBER
==
=
==
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 7

8. Probability distributions for received logical 0 and 1 signal pulses.
the different widths of the two distributions are caused by various signal
distortion effects.
thv
edtransmitt0if,exceedstageoutput volequalizerthat theprobablity)0|()(
edtransmitt1if,thanlessistageoutput volequalizerthat theprobablity)1|()(
0
1
vdyypvP
vdyypvP
v
v
∫
∫
∞
∞−
=
=
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 8

9.  Where are the probabilities that the transmitter sends 0
and 1 respectively.
 For an unbiased transmitter
∫∫
∞
∞−
+=
+=
th
th
v
v
ththe
dyypqdyypq
vPqvPqP
)1|()1|(
)()(
01
0011
[7-7]
01 and qq
5.010 == qq
10 1 qq −=
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 9

10. Gaussian Distribution
dv
bv
dyypvP
dv
bv
dyypvP
thth
thth
vv
th
v
on
v
th
∫∫
∫∫
∞∞
∞−∞−





 −
−==





 −
−==
2
off
2
off
off
0
2
on
2
on
1
2
)(
exp*
2
1
)0|()(
2
)(
exp*
2
1
)1|()(
σσπ
σσπ
mea
n
mea
n
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 10

11.  If we assume that the probabilities of 0 and 1 pulses are equally likely,
then using above equationn, BER becomes:
Q
Q
Q
dxxQP
Q
e
/2)exp(-
2
1
)
2
(erf1
2
1
)exp(
1
)(BER
2
2/
2
π
π
≈






−=−== ∫
∞
dyyx
vbbv
Q
x
thth
∫ −=
−
=
−
=
0
2
on
on
off
off
)exp(
2
)(erf
π
σσ
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 11

12. Approximation of error function
Variation of BER vs Q, .
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 12

13. Special Case
In special case when:
Vbb ==== onoffonoff ,0&σσσ
From eq [7-29], we have: 2/Vvth =
Eq [7-8] becomes:






−= )
22
(erf1
2
1
)(
σ
σ
V
Pe
[7-11]
ratio.noise-rms-to-signalpeakis
σ
V
Prepared by: Mr.P.Gunasekaran., AP/ECE14-Aug-19 13