Chapter 8

Basic System Design

The basic system design verification can be done through:

1. Power budget 2. Risetime budget.

The power budget involves the power level calculations from the transmitter to the receiver.

1. Attenuation2. Coupled power3. Other losses4. Equalization penalty

(DL)

5. SNR requirements6. Minimum power at

detector7. BER8. Safety margin (Ma)

The optical power budget is then assembled taking into account ALL these parameters.

Pi = (Po + CL + Ma + DL) dB

where Pi = mean input power launched in the fiberPo = mean optical power required at the receiverCL = total channel loss DL =dispersion-equalization or ISI penalty,

*The sensitivity of the detector is the minimum detectable power.

Risetime budget includes the following:

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 + TF

2 + TD2 + TA

2)1/2

For non-return-to-zero (NRZ) data

For return-to zero (RZ) data

Tsyst B

T7.0

Tsyst B

T35.0

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) Bandwidthb) Power levelsc) Error rate (SNR)

Risetime Budget We start with the risetime budget. Assume using NRZ

coding, the system risetime is given by:

Also:

Tsyst = 1.1(TS2 + TF

2 + TD2)1/2

nsxB

TT

syst 3.231030

7.07.06

Now we can assemble the total system risetime:

Total system risetime = 23.3 nsRisetime of the source, TS = 11.0nsRisetime of the fiber (dispersion), TF 10 x 1.0ns = 10.0ns

Allowance for the detector risetime, TD

nsTTTsys

T SFD 09.151.1

222

Power Budget

Total power launched into fiber = -19dBmLosses: Fiber attenuation 5dB/km x 10 = 50dB

4 connectors 1dB x 4 = 4dB

Power available at detector =[( -19dBm – 50dB- 4dB)] = -73 dBmSince 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) TD calculated is greater than TD given b) 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 nsFiber: Intermodal 5 ns/km

Intramodal 1 ns/kmDetector 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 + TF

2 + TD2)

= 1.1 [82 + (8 x 5)2 + (8 x 1)2 + 62)1/2]= 46.2 ns

Max bit rate =

Maximum bit rate = 15.2MbpsOr 3 dB optical BW = 7.6MHz

MbpsT

Bsyst

T 2.157.0

(max)

Example 8.3The following parameters were choosen for a long haul single mode optical fiber system operating at 1.3µm.

Mean power launched from laser = - 3dBmCabled fiber loss = 0.4 dB/kmSplice loss = 0.1 dB/kmConnector loss at transmitter and receiver = 1 dB eachMean power required at the APD

When operating at 35Mbps(BER = 10-9) -55 dBmWhen operating at 400Mbps(BER = 10-9) -44 dBm

Required safety margin = +7 dB

Estimate:a) maximum possible link length without repeaters when operating at 35Mbps. It may be assumed that there is no dispersion-equalization penalty at this rate.b) maximum possible link length without repeaters

when operating at 400Mbps. c) the reduction in the maximum possible link

length without repeaters of (b) when there is dispersion-equalization penalty of 1.5dB.

Solutiona)35MbpsPi – 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 MbpsPi – 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

c) Including dispersion-equalization penalty of 1.5dBPi – 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.

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 –41dBm 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.

SolutionThe power launched into the fiber 100µW = -10 dBmMinimum power required at the receiver - 41dBmTotal system margin - 31 dBm

Fiber loss 7 x 2.6 18.2dBSplice loss 6 x 0.5 3.0 dBConnector loss 6.0 dBSafety margin 28.7dB

Excess power margin = -31 dBm - 28.7 dB = 2.3 dBmBased 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 = 1.0nsDetector: tD = 0.5ns, sensitivity = -40dBmPre-amp: t A = 1.3nsFiber: total dispersion (M+Mg) = 15.5 psnm-1km-1, length = 100km, = 0.25dB/kmSource coupling loss = 3dBConnector (2) loss = 2dBSplice (50) loss = 5dBSystem: 400 Mbps, NRZ, 100km

SolutionFor risetime budgetsystem budget, Tsyst = = 1.75ns

source ts = 1.0ns …(1)fiber tF =

= 0.25ns …(2)detector tD = 0.5nspre-amp tA = 1.3nsfor receiver,

total = = 1.39ns …(3)

System risetime from (1),(2) and (3)= = 1.73ns

22 39.125.00.1

610400

7.07.0

TB

1005.15

2

AD tt

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 5 dBmSource coupling loss 3 dBConnector loss 2 dBSplice loss 5 dBAttenuation in the fiber 25 dBTotal loss 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,