ece477_7.ppt

Chapter 7
Photonic Transmission
Systems (Digital & Analog)

Content
• Digital Photonic transmission System
• Digital Photonic Receiver (BER, Quantum Limit)
• Analog Photonic Transmission System
• Photonic Digital Link Analysis & Design
- Link Loss budget, Link Power budget, Rise Time budget
• System Rise Time and Information Rate

Digital Transmission System (DTS)
• The design of optical receiver is much more complicated than that of optical transmitter
because the receiver must first detect weak, distorted signals and the n make decisions
on what type of data was sent.

Error Sources in DTS
!
)
(
)
(
0
n
e
N
n
P
E
h
dt
t
P
h
N
N
n
r



 





is the average number of electron-hole pairs in photodetector,
is the detector quantum efficiency and E is energy received in a time
interval and is photon energy, where is the probability
that n electrons are emitted in an interval .
N

 
h )
(n
Pr

[7-1]
[7-2]

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.


Receiver Configuration
The binary digital pulse train incident on the photodetector can be written in the
following form:
t.
all
for
positive
is
which
shape
pulse
received
the
is
)
(
and
digit
message
th
the
of
parameter
amplitude
an
is
period,
bit
is
where
)
(
)
(
t
h
n
b
T
nT
t
h
b
t
P
p
n
b
n
b
p
n





 [7-3]

• In writing down eq. [7-3], 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 in eq. [7-3] is (neglecting the DC components arising from dark
current noise):
n
b
)
(t
hp




 1
)
( dt
t
hp
n
b








n
b
p
n
o nT
t
h
b
M
t
MP
h
q
t
i )
(
)
(
)
(


[7-4]

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.
b
e
t
e
T
B
Bt
N
N
N
t
t
/
1
during
ed
transmitt
pulses
of
#
total
interval
me
certain ti
a
over
error
of
#
Error
of
y
Probabilit
BER





[7-5]

Probability distributions for received logical 0 and 1 signal pulses.
the different widths of the two distributions are caused by various signal
distortion effects.
th
v
ed
transmitt
0
if
,
exceeds
tage
output vol
equalizer
that the
probablity
)
0
|
(
)
(
ed
transmitt
1
if
,
than
less
is
tage
output vol
equalizer
that the
probablity
)
1
|
(
)
(
0
1
v
dy
y
p
v
P
v
dy
y
p
v
P
v
v







[7-6]

• Where are the probabilities that the transmitter sends 0 and 1
respectively.
• For an unbiased transmitter









th
th
v
v
th
th
e
dy
y
p
q
dy
y
p
q
v
P
q
v
P
q
P
)
1
|
(
)
1
|
(
)
(
)
(
0
1
0
0
1
1
[7-7]
0
1 and q
q
5
.
0
1
0 
 q
q
1
0 1 q
q 


Gaussian Distribution
dv
b
v
dy
y
p
v
P
dv
b
v
dy
y
p
v
P
th
th
th
th
v
v
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
|
(
)
(






mean
mean
[7-8]

• If we assume that the probabilities of 0 and 1 pulses are equally likely, then
using eq [7-7] and [7-8] , BER becomes:
Q
Q
Q
dx
x
Q
P
Q
e
/2)
exp(-
2
1
)
2
(
erf
1
2
1
)
exp(
1
)
(
BER
2
2
/
2













 

[7-9]
dy
y
x
v
b
b
v
Q
x
th
th
 





0
2
on
on
off
off
)
exp(
2
)
(
erf



[7-9]
[7-10]

Approximation of error function
Variation of BER vs Q,
according to eq [7-9].

Special Case
In special case when:
V
b
b 


 on
off
on
off ,
0
&



From eq [7-29], we have: 2
/
V
vth 
Eq [7-8] becomes:







 )
2
2
(
erf
1
2
1
)
(


V
Pe
[7-11]
Study example 7-1 pp. 286 of the textbook.
ratio.
noise
-
rms
-
to
-
signal
peak
is

V

Quantum Limit
• Minimum received power required for a specific BER assuming that the
photodetector has a 100% quantum efficiency and zero dark current. For
such ideal photo-receiver,
• Where is the average number of electron-hole pairs, when the incident
optical pulse energy is E and given by eq [7-1] with 100% quantum
efficiency .
• Eq [7-12] can be derived from eq [7-2] where n=0.
• Note that, in practice the sensitivity of receivers is around 20 dB higher
than quantum limit because of various nonlinear distortions and noise
effects in the transmission link.
)
exp(
)
0
(
1 N
P
Pe 

 [7-12]
N
)
1
( 


Analog Transmission System
• In photonic analog transmission
system the performance of the system
is mainly determined by signal-to-
noise ratio at the output of the
receiver.
• In case of amplitude modulation the
transmitted optical power P(t) is in
the form of:
where m is modulation index, and s(t)
is analog modulation signal.
• The photocurrent at receiver can be
expressed as:
)]
(
1
[
)
( t
ms
P
t
P t 

B
I
I
m


)]
(
1
[
)
( 0 t
ms
MP
t
i r
s 

 [7-13]
Analog LED modulation

• By calculating mean square of the signal and mean square of the total
noise, which consists of quantum, dark and surface leakage noise currents
plus resistance thermal noise, the S/N can be written as:
t
eq
B
D
P
P
t
eq
B
D
r
r
N
s
F
R
TB
k
B
M
F
M
I
I
q
MmI
F
R
TB
k
B
M
F
M
I
P
q
MmP
i
i
N
S
)
/
4
(
)
(
)
(
2
)
)(
2
/
1
(
)
/
4
(
)
(
)
(
2
)
)(
2
/
1
(
2
2
2
0
2
0
2
2









power
optical
received
average
:
amplifier;
baseband
of
figure
noise
:
amplifier
and
load
tor
photodetec
of
resistance
equivalent
:
bandwidth;
noise
effective
:
factor
noise
photodiode
excess
:
)
(
current;
leakage
-
Surface
:
current;
dark
bulk
primary
:
;
nt
photocurre
primary
: 0
r
t
eq
x
L
D
r
P
P
F
R
B
M
M
F
I
I
P
I



[7-14]

pin Photodiode S/N
• For pin photodiode, M=1:
F
R
TB
k
P
m
F
R
TB
k
m
I
N
S
eq
B
r
t
eq
B
P
)
/
4
(
)
2
/
1
(
)
/
4
(
)
)(
2
/
1
(
2
2
0
2
2


 Low input signal level [7-15]
qB
P
m
N
S r
4
0
2

 Large signal level [7-16]

SNR vs. optical power for photodiodes

Photonic Digital Link Analysis & Design
• Point-to-Point Link Requirement:
- Data Rate
- BER
- Distance
- Cost & Complexity
• Analysis Methods:
- Link loss & S/N analysis (link power budget analysis and loss
allocation) for a prescribed BER
- Dispersion (rise-time) analysis (rise-time budget allocation)

System Design Choices:
Photodetector, Optical Source, Fiber
• Photodetectors: Compared to APD, PINs are less expensive
and more stablewith temperature. However PINs have lower
sensitivity.
• Optical Sources:
1- LEDs: 150 (Mb/s).km @ 800-900 nm and larger than 1.5
(Gb/s).km @ 1330 nm
2- InGaAsP lasers: 25 (Gb/s).km @ 1330 nm and ideally around
500 (Gb/s).km @ 1550 nm. 10-15 dB more power. However
more costly and more complex circuitry.
• Fiber:
1- Single-mode fibers are often used with lasers or edge-emitting
LEDs.
2- Multi-mode fibers are normally used with LEDs. NA and
should be optimized for any particular application.


Link Power/Loss Analysis
Margin
System
]
[
]
/
[
]
[
2
]
[
]
[
]
[






km
L
km
dB
dB
l
P
dBm
P
dBm
P
dB
P
f
c
T
R
s
T

Total Power Loss

Receiver Sensitivities vs. Bit Rate
The Si PIN & APD and InGaAsP PIN plots for BER= . The InGaAs APD plot is for
BER= .
9
10
11
10

Link Loss Budget [Example 8.1]

Link Power Budget Table [Example 8.2]
• Example: [SONET
OC-48 (2.5 Gb/s)
link]
Transmitter: 3dBm
@ 1550 nm;
Receiver: InGaAs
APD with -32 dBm
sensitivity @ 2.5
Gb/s;
Fiber: 60 km long
with o.3 dB/km
attenuation; jumper
cable loss 3 dB each,
connector loss of 1
dB each.
Component/loss
parameter
Output/sensitivity
/loss
Power margin
(dB)
Laser output 3 dBm
APD Sensitivity
@ 2.5 Gb/s
-32 dBm
Allowed loss 3-(-32) dBm 35
Source connector
loss
1 dB 34
Jumper+Connect
or loss
3+1 dB 30
Cable attenuation 18 dB 12
Jumper+Connect
or loss
3+1 dB 8
Receiver
Connector loss
1 dB 7(final margin)

Dispersion Analysis (Rise-Time Budget)
2
/
1
2
2
2
2
2
0
2
2
/
1
2
2
2
mod
2
350
440
]
[
































rx
q
tx
rx
GVD
tx
sys
B
L
D
B
L
t
t
t
t
t
t


source
the
of
width
Spectral
:
[nm]
Dispersion
:
)]
.
/(
[
dispersion
velocity
group
to
due
time
-
rise
:
[ns]
7
.
0
fiber;
the
of
km
1
the
of
:
]
[
fiber
the
of
Length
:
]
[
BW
Electrical
3dB
:
]
[
dispersion
modal
:
]
[
time
rise
receiver
:
]
[
time
rise
er
transmitt
:
]
[
0
mod


nm
km
ns
D
t
q
BW
MHz
B
km
L
MHz
B
n
t
ns
t
ns
tx
t
GVD
rx
rx


Two-level Binary Channel Codes

System rise-Time & Information Rate
• In digital transmission system, the system rise-time limits the
bit rate of the system according to the following criteria:
period
bit
RZ
of
%
35
period
bit
NRZ
of
%
70


sys
sys
t
t

Example
• Laser Tx has a rise-time of 25 ps at 1550 nm and spectral
width of 0.1 nm. Length of fiber is 60 km with dispersion 2
ps/(nm.km). The InGaAs APD has a 2.5 GHz BW. The rise-
time budget (required) of the system for NRZ signaling is 0.28
ns whereas the total rise-time due to components is 0.14 ns.
(The system is designed for 20 Mb/s).

Example: Transmission Distance for MM-Fiber
• NRZ signaling, source/detector: 800-900 nm LED/pin or AlGaAs
laser/APD combinations. ; LED output=-13 dBm;fiber loss=3.5
dB/km;fiber bandwidth 800 MHz.km; q=0.7; 1-dB connector/coupling loss
at each end; 6 dB system margin, material dispersion ins 0.07 ns/(km.nm);
spectral width for LED=50 nm. Laser ar 850 nm spectral width=1 nm; laser
ouput=0 dBm, Laser system margin=8 dB;
9
10
BER 


Example:Transmission Distance for a SM Fiber
• Communication at 1550 nm, no modal dispersion, Source:Laser;
Receiver:InGaAs-APD (11.5 log B -71.0 dBm) and PIN (11.5log B-60.5
dBm); Fiber loss =0.3 dB/km; D=2.5 ps/(km.nm): laser spectral width 1
and 3.5 nm; laser output 0 dBm,laser system margin=8 dB;

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