Transcript System Design - UniMAP Portal

Chapter 8
Basic System Design
System factors for designing from scratch: Design Verification
Factor
Available choices
Type of fiber
Single mode, multimode, plastic
Dispersion
Repeaters, compensation
Fiber nonlinearities
Operating wavelength
(band)
Fiber characteristics, wavelengths
used, transmitter power
780, 850, 1310, 1550, 1625 nm
typical
Transmitter power
~0.1 to 20 mw typical; usually
expressed in dBm
Light source
LED, laser
Receiver characteristics Sensitivity, overload
Multiplexing scheme
None, CWDM, DWDM
System factors (continued)
Factor
Detector type
Modulation scheme
End-end bit error rate
Signal-to-noise ratio
Max number of
connectors
Max number of splices
Environmental
Mechanical
Available choices
PIN diode, APD, IDP
OOK, multilevel, coherent
<10-9 typical; may be much lower
Specified in dB for major stages
Loss increases with number of
connectors
Loss increases with number of
splices
Humidity, temperature, sunlight
exposure
Flammability, strength,
indoor/outdoor/submarine
Optical link loss budget
• Key calculations in designing a simple fiber optic
link
• Objective is to determine launch power and
receiver sensitivity
• Variables
– Environmental and aging
– Connector losses
– Cable losses
– Splices
– Amplifier
– Other components

The basic system design verification can be done through:
1- Power budget: The Ratio of PT/PR expressed in dB is the amount
of acceptable loss that can be incurred.
2- Rise time budget: A rise-time budget analysis is a convenient
method to determine the dispersion limitation of an optical link.

The power budget involves the power level calculations from
the transmitter to the receiver.
1. Attenuation
5. SNR requirements
2. Coupled power
6. Minimum power at
3. Other losses
detector
4. Equalization penalty
7. BER
(DL)
8. Safety margin (Ma)
The system margin can be expressed as:
Ma= PT(dBm)-PR(dBm)- system loss.
A (+)positive system margin ensures proper operation of the circuit.
A (-) negative value indicates that insufficient power will be reach
the detector to achieve the required BER.
The optical power budget is then assembled taking into account
ALL these parameters.
Pi = (Po + CL + Ma + DL) dBm
where Pi = mean input power launched in the fiber
Po = mean optical power required at the receiver
CL = total channel loss
DL =dispersion-equalization or ISI penalty,
*The sensitivity of the detector is the minimum detectable power.
 Risetime
following:
budget
includes
the
1. Risetime of the source, TS
2. Risetime of the fiber (dispersion),
TF
3. Risetime of the amplifier, TA
4. Risetime of the detector, TD
The risetime budget is assembled as:
Tsyst = 1.1(TS2 + TF2 + TD2 + TA2)1/2
For non-return-to-zero (NRZ) data
Tsyst
0 .7

BT
For return-to zero (RZ) data
Tsyst
0.35

BT
Example 8.1
We need to design a digital link to connect two points 10-km apart.
The bit rate needed is 30Mb/s with BER = 10-12.
Determine whether the components listed are suitable for the link.
Source: LED 820nm GaAsAl; couples 12µW into 50µm
fiber; risetime 11ns
Fiber: Step Index fiber; 50µm core; NA = 0.24; 5.0
dB/km loss; dispersion 1ns/km; 4 connectors with
1.0dB loss per connector
Detector: PIN photodiode; R = 0.38A/W; Cj = 1.5pF,
Id = 10pA; risetime = 3.5ns; minimum mean optical
power = - 86dBm
Calculate also the SNR of the link if RL given is 5.3kΩ
Solution :
For this example, 3 factors need to be considered:
a)
Bandwidth
b)
Power levels
c)
Error rate (SNR)
Risetime Budget
We start with the risetime budget. Assume using NRZ
coding, the system risetime is given by:
Tsyst
0.7
0.7


 23.3ns
6
BT 30 x10
Also:
Tsyst = 1.1(TS2 + TF2 + TD2)1/2
Now we can assemble the total system risetime:
Total system risetime = 23.3 ns
Risetime of the source, TS = 11.0ns
Risetime of the fiber (dispersion), TF 10 x 1.0ns = 10.0ns
Allowance for the detector risetime, TD
2
 Tsys 
2
2
TD  
  TF  TS  15.09ns
 1.1 
Power Budget
Total power launched into fiber
= -19dBm
Losses: Fiber attenuation 5dB/km x 10 = 50dB
4 connectors 1dB x 4 = 4dB
Power available at detector =[( -19dBm – 50dB- 4dB)] = -73 dBm
Since power available at the detector is –73 dBm, the sensitivity of
the detector must be less than this.
The safety margin, Ma = -73-(-86) dB
= 13dB
The choice of components are suitable because;
a)
b)
TD calculated is greater than TD given
Total power available at the detector is greater than
the minimum power required by the detector i.e Ma is
positive.
Example 8.2
An optical link is to be designed to operate over an 8-km length
without repeater. The risetime of the chosen components are:
Source:
8 ns
Fiber: Intermodal
5 ns/km
Intramodal
1 ns/km
Detector
6ns
From the system risetime considerations estimate the maximum bit
rate that may be achieved on the link using NRZ code.
Solution:
Tsyst
= 1.1(TS2 + TF2 + TD2)
= 1.1 [82 + (8 x 5)2 + (8 x 1)2 + 62)1/2]
= 46.2 ns
Max bit rate = BT (max)
0.7

 15.2Mbps
Tsyst
Maximum bit rate = 15.2Mbps
Or 3 dB optical BW = 7.6MHz
Exercise 1
The following parameters were chosen for a long haul single mode
optical fiber system operating at 1.3µm.
Mean power launched from laser = 0 dBm
Cabled fiber loss = 0.4 dB/km
Splice loss = 0.1 dB/km
Connector loss at transmitter and receiver = 1 dB each
Mean power required at the APD
When operating at 35Mbps(BER = 10-9)
-65 dBm
When operating at 400Mbps(BER = 10-9) -54 dBm
Required safety margin = +7 dB
Estimate:
a)
b)
c)
maximum possible link length without repeaters
when operating at 35Mbps. It may be assumed
that there is no dispersion-equalization penalty at
this rate.
maximum possible link length without repeaters
when operating at 400Mbps.
the reduction in the maximum possible link length
without repeaters of (b) when there is dispersionequalization penalty of 1.5dB.
Solution
a)35Mbps
Pi – Po = [(Fiber cable loss + Splice losses ) x L + Connector loss + Ma ]dB
[-3dBm – (-55 dBm)] = (0.4 + 0.1)L + 2 + 7
0.5L = 52 –2-7
L = 86km
b) 400 Mbps
Pi – Po = [(Fiber cable loss + Splice losses ) x L + Connector loss + Ma ]dB
[-3dBm – (-44 dBm)] = (0.4 + 0.1)L + 2 + 7
0.5L = 41 –2-7
L = 64km
• Including dispersion-equalization penalty of 1.5dB
Pi – Po = [(Fiber cable loss + Splice losses ) x L + Connector loss + DL + Ma]dB
[-3dBm – (-44 dBm)] = (0.4 + 0.1)L + 2 + 1.5 + 7
0.5L = 41 –2 -1.5 – 7
L = 61km
Note: a reduction of 3 km in the maximum length without repeaters when DL is
taken to account.
Exercise 2
Calculate the flux density to construct an optical link of 15 km
and bandwidth of 100 Mb/s. Components are chosen with the
following characteristics: Receiver sensitivity -50 dBm (at 100
Mb/s), fiber loss 2 dB/km and transmitter launch power into the
fiber is 0 dBm, detector coupling loss is 1 dB. It is anticipated
that in addition, 10 splices each of loss 0.4 db are required.
Determine where the system operate with sufficient power
margin or not?.
Example 8.4
An optical link was designed to transmit data at a rate of 20 Mbps
using RZ coding. The length of the link is 7 km and uses an LED at
0.85µm. The channel used is a GRIN fiber with 50µm core and
attenuation of 2.6dB/km.
The cable requires splicing every kilometer with a loss of 0.5dB per
splice. The connector used at the receiver has a loss of 1.5dB. The
power launched into the fiber is 100µW. The minimum power
required at the receiver is –71dBm to give a BER of 10-10. It is also
predicted that a safety margin of 6dB will be required.
Show by suitable method that the choice of components is suitable
for the link.
Solution
The power launched into the fiber
Minimum power required at the receiver
Total system margin
Fiber loss
7 x 2.6
Splice loss
6 x 0.5
Connector loss
Safety margin
100µW = -10 dBm
- 71dBm
61 dB
18.2dB
3.0 dB
6.0 dB
28.7dB
Excess power margin = 61 dB - 28.7 dB = 32.3 dB
Based on the figure given, the system is stable and provides an excess
of 2.3 dB power margin. The system is suitable for the link and has
safety margin to support future splices if needed..
Example 8.5
An optical communication system is given with the following
specifications:
Laser:  = 1.55µm,  = 0.15nm, power = 5dBm, tr = 1.0ns
Detector: tD = 0.5ns, sensitivity = -40dBm
Pre-amp: t A = 1.3ns
Fiber: total dispersion (M+Mg) = 15.5 psnm-1km-1, length = 100km,
 = 0.25dB/km
Source coupling loss = 3dB
Connector (2) loss = 2dB
Splice (50) loss = 5dB
System: 400 Mbps, NRZ, 100km
Show by suitable method that the choice of components is suitable for
the link
Solution
For risetime budget
system budget, Tsyst =
source
fiber
detector
pre-amp
for receiver,
ts
tF
tD
tA
0.7
0.7

BT
400  10 6
= 1.0ns
= 15.5   100
= 0.25ns
= 0.5ns
= 1.3ns
= tD  t
= 1.39ns
System risetime from (1),(2) and (3)
11111111
1.0  0.252  1.392
=
= 1.73ns
total
2
A
= 1.75ns
…(1)
…(2)
…(3)
Since the calculated Tsyst is less than the available Tsyst the
components is suitable to support the 400 Mbps signal.
For the power budget:
Laser power output
Source coupling loss
Connector loss
Splice loss
Attenuation in the fiber
Total loss
5 dBm
3 dB
2 dB
5 dB
25 dB
35 dB
Power available at the receiver = (5 dBm -35 dB) = -30 dBm
The detector’s sensitivity is -40 dBm which is 10 dB less. Therefore
the chosen components will allow sufficient power to arrive at the
detector. Safety margin is +10 dB,
Exercise: An analog optical link of length 2km employs an LED
which launches mean optical power of 10 dBm into a multimode
optical fiber. The fiber cable exhibits loss of 3.5 dB/km with splice
losses calculated at 0.7 dB/km . In addition there is a connector loss
at the receiver of 1.6 dB. The PIN photodiode receiver has a
sensitivity of -25dBm for an SNR of 50 dB and with a modulation
index of 0.5. it is estimated that a safety margin of 4 dB is required.
Assume threre is no dispersion –equalization penalty:
a) Perform an optical power budget for the system operating under
the above conditions and ascertain its viability.
b) Estimate any possible increase in link length which may be
achieved using an injection laser source which launches mean
optical power of 0 dBm into the fiber cable. In this case the
fafety margin must be increased to 7 dB.
Exercise :
A single TV channel is transmitted over an analog optical link
using direct intensity modulation. The video which has a
bandwidth of 5 MHz and a ratio of luminance to composite video
of 0.7 is transmitted with a modulation index of 0.8. The receiver
contains p-i-n photodetector with a responsitivity of 0.5 A/W and
a preamplifier with an effective input impedance of 1 M ohms
together with a noise figure of 1.5dB. Assuming the receiver is
operating at a temperature of 20 C and neglecting the dark
current in the photodiode, determine the average incident optical
power required at the receiver (i.e, receiver sensitivity) in order to
maintain a peak to peak signal power to rms noise power ratio of
55 dB.