代写EEEE3086 H63EEJ Optical Networks代写留学生Matlab语言

日期：2024-07-18 05:18

H63EEJ Optical Networks

09 January 2022

Coursework:

“Design of the Local Optical Fibre Network”

25% course work 12.5 hours student time. Up to 3000 words CWK on passive Optical Network components.

Report Deadline: 4^th May 2023

AIM

Your aim is to design local optical fibre network that consists of a minimum 3 passive and 1 active optical communication component. In your design, you will need to calculate losses in the optical network and describe your approach. You will also need to calculate mean down time of the design network by providing appropriate values for Failure Rate and Mean Time to Repair of each of the used components.

In your max 3000 words report you will need to:

1. Demonstrate a basic understanding of the principle of design of the local optical fibre networks.

2. Demonstrate a basic understanding of the principle of operations of the used in your design passive and active optical network components.

3. List main functionalities of the used in your design passive and active optical network components.

4. Sketch the designed network clearly labelling all components used.

5. Calculate losses in the optical network.

6. Describe and discuss your findings in terms of losses, risks and what can be done to minimise those losses and mitigate the risks.

7. Calculate the Mean Down Time of the design network.

Theory

Passive/access optical network: 

Access network connects the exchanges (that were interconnected by the metro network) to the ultimate user (home or business) by a variety of techniques.

When fibre is used the optical network units/terminations (ONUs/ONTs) may be connected point to point or in a tree/branch configuration (often called a ‘passive optical network’ (PON)) to the OLT in the nearest exchange. Access rings are also sometimes used.

Both access networks and metro/regional networks can improve their survivability (ability to function in the event of loss of node or link) by dual homing, which basically means two connections into thPhysically the PON connects an OLT (with one or more wavelengths) to optical network units (ONUs) which may be located in homes (fibre-to-the-home: FTTH), in buildings (fibre-to-the-building: FTTB or fibre-to-the-premises: FTTP), to the kerb (FTTK: fibre-to-the-kerb) etc. The network topology is usually tree/branch. In fact the name PON is typically synonymous with that topology.

Passive optical components:

Coupler/splitter

Isolator

Circulator

Filter/multiplexer

Array waveguide grating

Mach-Zehnder interferometer (MZI)

Variable optical attenuator

Active optical components:

Transmitters

Amplifiers

Receivers

Mean down time calculations:

1. In the following simple PON network calculate the mean down time (in min/year) experienced by ONUs.

Assume the following data:

The length of fibre between ONU and nearest splitter is 0.5 km, between the splitters is 1.5 km, and the feeder is 8 km long.

Failure rate (in FITs – failures in 10⁹ hours): LT = 3000, Splitter = 250, NU = 800, fibre (per km) = 2500 (this is due to a cable cut rather than an actual failure of the fibre/cable material). [You will need to calculate the Mean Time Between Failures (MTBF) from these values]

Mean Time to Repair (MTTR): LT = 4 hours, Feeder fibre = 5 hours, Splitter = 10 hours, Distribution fibre = 10 hours, NU = 36 hours

Show diagrammatically, and with the appropriate calculation how we might (economically) improve the MDT to less than 53 min/year.

We are trying to find the mean down time (MDT) experienced by the users of the network. So we want the probability of unavailability.

For the OLT have p_dOLT=MTTR_OLT/(MTTR_OLT+MTBF_OLT)

But MTBF_OLT (in hours)=1/failure rate per hour = 10⁹/FITrate  =10⁹/3000

(FITrate is failures per 10⁹ hours)

So p_dOLT=MTTR_OLT/(MTTR_OLT+MTBF_OLT)=4/(4+10⁹/3000)=1.20x10^-5

MDT_OLT=p_dOLTx365x24x60=6.31 min/year

For the ONU:

p_dONU=MTTR_ONU/(MTTR_ONU+MTBF_ONU)=36/(36+10⁹/800)=2.88x10^-5

MDT_ONU=p_dONUx365x24x60=15.14 min/year

For the feeder:

p_dfeeder=MTTR_feeder/(MTTR_feeder+MTBF_feeder)=5/(5+10⁹/(2500x8))=1.00x10^-4

MDT_feeder=p_dfeederx365x24x60=52.55 min/year

For the distribution fibre:

p_ddistrib=MTTR_distrib/(MTTR_distrib+MTBF_distrib)=10/(10+10⁹/(2500x(1.5+0.5)))=5.00x10^-5

MDT_distrib=p_ddistribx365x24x60=26.28 min/year

For the splitters (both the same):

p_dsplitter=MTTR_splitter/(MTTR_splitter+MTBF_splitter)=10/(10+10⁹/250)=2.49x10^-6

MDT_splitter=p_dsplitterx365x24x60=1.31 min/year

As all the downtime probabilities are small we can neglect the probability of multiple failures occurring at the same time and write that:

MDT_system= MDT_OLT+ MDT_feeder+ MDT_splitterx2+ MDT_distrib+ MDT_ONU

    =6.31+52.55+2.62+26.28+15.14

    =102.9 min/year

This is, to a good approximation, what we might expect an ONU/user to experience on average.

We can (without too much extra expense) improve the MDT to less than 53 min/year by having a redundant diversely routed protection feeder.

This protection approximately reduces the system MDT to 102.9-52.55=50.35 min/yr

More accurately use p_{dfeedercombination}=(p_dfeeder)²  =10^-8

Which we see is negligible! (even compared to splitter!)