Solved problems 1

SOLVED PROBLEMS

1. Consider the full-duplex network shown in figure 1.1. A 3-dB directional couplers are used. If

the directional couplers each have 1.5 dB of excess loss and all connectors (one at each

directional coupler port, one at the transmitter, and one at the receiver) have 0.8 dB of loss. The

fiber has 4-dB loss. Compute the total loss from transmitter to receiver.

T T

Connector λ1 Connector

Directional coupler Directional coupler

R R

Figure 1.1 full-duplex communications system. T, transmitter; R, receiver.

The unused ports are also shown.

Solution:

= 3 + 1.5 = 4.5 dB.

The total loss = 2 (4.5) + 6 0.8 + 4 = 17.8 dB.

2. A four-port directional coupler has a 4:1 splitting ratio and an excess loss equal to 2 dB. The

coupler’s directionality is 40 dB.

(a) What fraction of the input power goes to each of the ports?

(b) Compute the throughput loss and tap loss.

(c) Compute the loss due to radiation, scattering, and absorption in the coupler.

Solved problems 2

l

Port 1 Port 2

P1 P2

Fibers or wave-guides

Port 4 Port 3

P4 (Coupling length) P3

Figure 1.2 a four ports directional coupler.

Solution

(a)

,

2 dB

0.631

= 0.631 and

= 0.631

For port2:

0.505

For port 3:

=

0.126

For port 4:

40 dB

= 0.0001

(b)

2.967 dB

2.967 +2 = 4.967 dB.

8.996 dB.

8.996 +2 = 10.996 dB.

(c)

= 2 dB.

Solved problems 3

3. A five-terminal tee network is structured like one shown in figure 1.3. The tee couplers are like

the one shown in figure 1.4. Assume ideal 3-dB couplers, ideal fibers, and lossless connectors.

(a) Draw the entire network.

(b) Compute the transmission loss to each of the receivers when terminal 1 is the transmitter.

Terminals

2 3 4

Transmitter connector Connector Transmitter

Terminal1 splice Terminal 5

Receiver Receiver

Directional coupler Tee coupler Bus fiber (trunk fiber) Directional coupler

Figure 1.3 Tee network interconnecting N terminals.

1 2 Connector splice

Fiber bus

Directional coupler

4 3

2 3

Directional coupler

1 4 connector

T R

Transmitter Receiver

Figure 1.4 tee coupler using two directional couplers.

Solved problems 4

If we transmit from terminal 1 to the other terminals

( ) for terminal N

for 1

Coupler description (dB) (dB) Splitting ratio

3 dB 3 3 1:1

For terminal 2 : 2 2 dB.

For terminal 3: 3 3 dB.

For terminal 4: 4 dB.

For terminal 5: ( ) 15 dB.

4. Repeat problem 3 when the directional couplers each have 1.5 dB of excess loss, all connectors

have 0.8 dB of loss, the fiber loss is 35 dB/km, there is 100 m between terminals, and splices

produce a 0.2 dB loss. The number of connectors and/or splices is up to you to specify.

3+1.5 = 4.5 dB.

3+1.5 = 4.5 dB.

If we transmit from terminal 1 to the other terminals

( ) ( ) ( ) for terminal N

( ) ( )

for 1

For terminal 2 : 2 dB.

For terminal 3: 3 30.2dB.

For terminal 4: 4 dB.

For terminal 5: ( ) 45.3 dB.

5. A star network like that in figure 1.5. Connects five terminals. The excess loss of the star

coupler is 2 dB, connector losses are 0.8 dB, splice losses are 0.2 dB, and the fiber loss is 35

dB/km. Terminals 1, 2, 3 and 4 are 100 m from the star coupler. Terminal 5 is 20 m from the

star coupler. Include all the connectors and splices you think you need, and note them on your

sketch.

(a) Sketch the network.

(b) If terminal 1 transmits, compute the total transmission loss to each of the receivers.

Solved problems 5

Transmitters

1 2 5 3 4

Connector

Splice

Star coupler

Splice

Connector

1’ 2’ 5’ 3’ 4’

Receivers

Figure 1.5 star network

Solution

For 1 5,

+ 2 2L

+ 2 + 2 0.8+ 2 0.2 +2 0.1 35 = 18 dB.

For N,

+ 0.120

+ 2 + 2 0.8+ 0.2 + 0.120 35 = 15 dB.

Solved problems 6

6. This problem compares different simple add/drop multiplexer architectures.

(a) First consider the fiber Bragg grating-based add/drop element shown in figure 1.6. Suppose

a 5% tap is used to couple the added signal into the output, and the grating induces a loss of

0.5 dB for the transmitted signals and no loss for the reflected signal. Assume the circulator

has a loss of 1 dB per pass. Carefully compute the loss seen by a channel that is dropped, a

channel that is added, and a channel that is passed through the device. Suppose the input

power channel is 15 dBm. At what power should the add channel be transmitted so that

the powers on all the channels at the output are the same?

λ1 λ3 λ4

λ2

1 2 Coupler

Fiber Bragg grating

3

λ1 λ2 λ3 λ4 λ1 λ2 λ3 λ4

λ2 λ2

Drop Add

(b)

Figure 1.6 Optical add/drop elements based on fiber Bragg Gratings.

(b) Suppose you had to realize an add/drop multiplexer that drops and adds four wavelengths.

One possible way to do this is to cascade four add/drop elements of the type shown in figure

1.6 in series. In this case, compute the best-case and worst-case loss seen by the channel that

is dropped, a channel that is added, and a channel that is passed through the device.

(c) Another way to realize a four-channel add/drop multiplexer is shown in figure 1.7. Repeat

the preceding exercise for this architecture. Assume that the losses are as shown in the

figure. Which of the two would prefer from a loss perspective?

Solved problems 7

λ1 λ2 λ3 λ4 2 λ1 λ2 λ3 λ4 Coupler λ1 λ2 λ3 λ4

1 10%

Fiber Bragg gratings

3

Splitter 6 dB 6 dB Combiner

Filters 1 dB λ1 λ2 λ3 λ4

λ1 λ2 λ3 λ4

Figure 1.7 A four channel add/drop multiplexer architecture.

(d) Assume that fiber gratings cost $500 each, circulators $3000 each, filters $ 1000 each, and

splitters, combiners, and couplers $ 100 each. Which of the two preceding architectures

would you prefer from a cost point of view?

7. An alternative to the star couplers is to construct cascading 3 dB couplers to obtain a N N

device formed by using constructed 2 couplers as the like shown in figure 1.11, where N is

an even integer. Show that the total losses increase logarithmically with N.

The number of 3-dB couplers needed to construct an N N cascaded star coupler is

If the fraction of power traversing each 3-dB coupler element is FT, with FT 1, a fraction

(1 FT) of power is lost in each 2 coupler.

Solved problems 8

The excess loss

Total loss = splitting loss excess loss (

)

(

)

λ1 λ1 λ2… λ8

λ2

λ3

λ4

λ5

λ6

λ7

λ8

λ1 λ2… λ8

Figure 1.11 Example of an 8 8 star coupler formed by interconnecting twelve 2 couplers.

Example: consider a commercial available 32 single mode coupler made from a cascade of 3-dB

fused-fiber 2 couplers, where 5% of the power is lost in each element. Determine:

(i) The excess loss

(ii) The splitting loss

(iii) Total loss.

Directional coupler

The fused biconically tapered directional coupler, sketched in figure 1.12, has been designed to provide

low-loss couplers with a range of splitting ratios. The couplings to ports 2 and 3 are given by

( )

( )

Where is the coupling coefficient (given in radians per meter) between the two waveguides and L

is the length of the fiber over which interaction exists. All the power appears at port 3 when the length

of the interaction region is

( ) , where the coupling length is .

Solved problems 9

L

Cladding

1 Core 2

Cladding

4 Core 3

Figure 1.12 Fused bi-conically tapered directional coupler.

Assuming that the coupler is lossless, the expression for the power P3 coupled from the first fiber to the

second fiber over axial distanced z is

P3 P1 ( )

Where is the coupling coefficient describing the interaction between the fields in the two fibers. By

conservation of power, for identical-core fibers we have

P2 P1 P3 P1 [1 ( ) ] P1 ( ).

This shows that the phase of the driven fiber always lags behind the phase of the driving fiber.

1.0

0.8 P2/ P1

0.6

0.4 ⁄

0.2 P3/P1

0

Coupler draws length (mm)

Figure 1.13 normalized coupled powers P2/ P1 and P3/P1 as function of coupler draw length for a 1300

nm power level P1 launched into fiber 1.

Norm

alized p

ow

er

Solved problems 10

Thus, when the power is launched into fiber 1, at z 0 the phase in fiber 2 lags behind that in fiber

1. This lagging phase relationship continues for increasing z, until at a distance that satisfies

,

all of the power has been transferred from fiber 1 to fiber 2. Now fiber 2 becomes the driving fiber, so

that for

the phase in fiber 1 lags behind that in fiber 2, and so on. As a result of this

phase relation, the 2 coupler is a directional coupler. That is no energy can be coupled into a wave

traveling backward in the negative-z direction in the driven waveguide.

1.0

0.8 P3/ P1

0.6

0.4

0.2 P2/P1

0

1200 1300 1400 1500 1600

Wavelength (nm)

Figure 1.14 illustrating the dependence on wavelength of coupled powers in the completed 15 mm

long coupler.

8. Consider the unidirectional device for de-multiplexing and multiplexing shown in figure 1.15.

Suppose that a 10% tap is used to couple the added signal into the output, and the grating

induces a loss of 0.5 dB for the transmitted signals and no loss for the reflected signal. Assume

the circulator has a loss of 1 dB per pass. Carefully compute the loss seen by:

(i) channels that are dropped

for λ2

for λ3

Norm

alized p

ow

er

Solved problems 11

λ1 λ2 λ3 λ4 Circulator λ1 λ2 λ3 λ4

λ2 λ3

1 2 Coupler

Fiber Bragg gratings

3

λ1 λ2 λ3 λ4 λ1 λ2 λ3 λ4

Splitter 6 dB 6 dB Combiner

λ3 λ2

Filters 1 dB

λ2 λ3

Figure 1.15 Optical add/drop elements based on fiber Bragg Gratings and circulator.

(ii) channels that are added

for λ2, λ3

(iii) Channels that is passed through the device.

For λ1, λ4

( ) ( )

(iv) Suppose the input power channel is 20 dBm. At what power should the add channels be

transmitted so that the powers on all the channels at the output are the same?

PO Pin 20 dBm Pin PO

Padd PO Padd

Pin PO

Solved problems 12

Important notes:

λ4 λ3

λ1 λ2 λ3 λ4 λ1 λ2 λ4

Fiber Bragg gratings 3-ports circulator

λ1 λ2 λ4

λ1 λ2 λ4

(a) This design does not work for wavelengths λ1, λ2, and λ4.

λ1 λ2 λ4

λ3 λ3

λ1 λ2 λ3 λ4 λ1 λ2 λ4

Fiber Bragg gratings 3-ports circulator

λ1 λ2 λ4

(b) This design will work will for all wavelengths.

Figure 1.16 illustrates the OADM system diagram.

Solved problems 13

OPTICAL ADD/DROP MULTIPLEXERS

Optical add/drop multiplexers (OADMs) provide a cost-effective means for handling pass-

through traffic in both metro and long-haul networks. Figure 1.17 illustrates a WDM network.

Node A Node B Node C

OADM

λ4

λ3

λ2

λ1

OLT OLT transponder

Add/drop

(a)

Node A Node B Node C

Add/drop

(b)

Figure 1.17 an example of a three-node linear network that illustrates the role of optical add/drop

multiplexers. Three wavelengths are needed between nodes A and C, and one wavelength each

between nodes A and B and between nodes B and C. (a) A solution using point-to-point WDM

systems. (b) A solution using an optical add/drop multiplexer at node B.

Why the transponders are needed in the solution of figure 1.17 (a) to handle the pass-through

traffic?

The power level of a signal coming into node B from node A might be also so low that it

cannot be passed through for another hop to node C.

Solved problems 14

The power of the signals added at a node must ideally be equal to the power of the signals

passing through.

OADMs are inflexible, static elements and do not allow in-service selection under software

control of what channels are dropped and passed through.

OADM ARCHITECTURES

Several architectures have been proposed for building OADMs. These architectures typically use one

or more of the multiplexers/filters. Most practical OADMs use either fiber Bragg gratings, dielectric

thin-film filter, or array wave-guide gratings.

Different OADM architectures that are commercially implemented

Parallel architecture

An arbitrary subset of channels can be dropped and the remaining passed through. The loss

through the OADM is fixed, independent of how many channels are dropped and added.

Since all channels are de-multiplexed and multiplexed at all the OADMs, each light path passes

through many filters before reaching its destination.

λw λw

λ1, λ2, ..…, λw λ1, λ2, …, λw

λ1

λ2 λ2

λ1 λ1

Figure 1.18 Parallel architecture, where all the wavelengths are separated and multiplexed back.

WDM networks provide circuit-switched end-to-end optical channels, or light paths, between

network nodes to their users, or clients. A light path consists of an optical channel, or

wavelength, between two network nodes that is routed through multiple intermediate nodes.

Intermediate nodes may switch and convert wavelengths. These networks may be thought of as

wavelength-routing networks.

Solved problems 15

Advances In Reconfigurable Add Drop Multiplexer ROADM Technologies

Large amounts of information traveling on multiple wavelengths around an optical network need to be

switched at the network nodes. Information arriving at a node is forwarded to its final destination via

the best possible path, which is determined by such factors as distance, cost, and the reliability of

specific routes. The conventional way to switch the information is to convert the input fiber optical

signal to an electrical signal, perform the switching in the electrical domain, then convert the electrical

signal back to an optical signal that goes down the desired output fiber. This optical-electrical-optical

O-E-O conversion uses systems that are expensive, bulky, and are bit-rate/protocol dependent. The

different optical component technologies that have been developed for use in ROADM subsystems

include MEMS, liquid crystals (liquid crystal devices LCD and liquid crystal on silicon LCoS

technologies), and monolithic and hybrid planar light wave circuits PLC based on silica on silicon and

polymer on silicon platforms.

ROADMs allow avoiding the unnecessary O-E-O conversion, enabling O-O-O systems that use optical

switching, which has significant advantages for carriers and service providers. ROADMs route the

optical signals directly, and are bit-rate/protocol transparent, so future upgrades of bit-rate or protocol

can be accommodated without the need to upgrade the switch.

A ROADM network element typically includes:

Transponder

ROADM subsystem

Optical service channel

Optical power monitoring

Amplifiers(Pre-amp &Post-amp)

Dispersion compensation module

Type II ROADM

A type II ROADM offers colorless Add/Drop ports. Three generations of subsystems are based on the

wavelength blocker WB and small switch array SSA.

Wavelength blocker WB

The WB-based subsystem has a broadcast and select architecture and is based on free-space

optics, which can use MEMS or LCD actuation. It is mostly used in long-haul networks, and

typically has 80 channels with a channel spacing of 50 GHz. The ports are made colorless

through the use of tunable filters at the drop ports and tunable lasers at the add ports, without

having an impact on the through path. Figure 1.19 shows generation-I WB.

Solved problems 16

Wavelength blocker

Splitter Splitter OCM

Drop Add

1 or 1 N or M

Splitter Combiner

Tunable filters

Tunable lasers

Receivers

Figure 1.19 architectures of Gen 1 type II ROADM subsystem based on WB.

OCM: optical channel monitor.

Small Switch Array SSA.

Figure 1.20 shows generation-I type II ROADM subsystem based on SSA.

1 DCE OCM 1 DCE OCM

Demux Mux

Drop Add

N M OXC M N OXC

Receivers Transmitters

Figure 1.20 architectures of Gen 1 type II ROADM subsystem based on SSA.

Solved problems 17

The SSA-based PLC solution is smaller and more cost-effective. It has M colorless Add/Drop

ports (M can be up to N, the total number of channels), which is achieved through the use of

M N switches. Today, N is typically 8, 16, 32, or 40 channels, and M is typically 4 or 8 ports.

This subsystem occupies 1 slot. This solution is typically used in metro networks, and has a

channel spacing of 100 GHz.

Wavelength Selective Switch WSS

A 1 N WSS can be used either for degree-N connectivity as in ring-to-ring inter-connect, or

for adding/dropping channels as s ROADM with 1Express port and N Drop ports. A WSS

allows any number of channels to exit any port. When used as a ROADM, the Add

functionality is implemented separately, typically through the use of a Multiplexer and a tap.

For a 1 N WSS with n channels, n 1 N switches are needed. A connectivity node of degree-4

where two fiber-pair rings interconnect, requiring one 1 WSS per fiber, and a total of 4 1

WSSs for the node. A fiber-pair ring Add/Drop node needing N Drop ports requires one

1 N WSSs per fiber, so two 1 N WSSs for the node. Figure 1.21 shows schematically a

1 WSS with 8 Drop ports.

Mux

Any number of λ’s Out

n 1 N switches

λ1 Any number of λ’s

Any number of λ’s

In

λ1, λ2,…, λn Any number of λ’s

Any number of λ’s

λn

Any number of λ’s

Any number of λ’s

Any number of λ’s

Any number of λ’s

Mux

Figure 1.21 functional diagram of n-channel 1 WSS used at an Add/Drop node, providing one

Express port and 8 Drop ports.

Solved problems 18

Next Generation ROADM Network (www.enablence.com)

The ROADMs are typically defined by the number of degrees, or directions, of switching they can

perform. The first generation of ROADMs could only switch wavelengths in two degrees, typically

called East and West as illustrates in figure 1.22. The second generation of ROADMs could switch in

four degrees, North, South, East, and West as illustrated in figure 1.23.

However, with these first two generations of ROADM, typically based on wavelength selective

switching WSS, automated switching in the optical layer can only take place at the intermediate nodes

across a network link.

The selected pass through Wavelengths

AWG AWG AWG AWG

RX TX TX RX

Figure 1.22 illustrates the architecture diagram of two degrees ROADM.

Wavelengths remain in the optical layer while passing through intermediate nodes on the network,

operators do not have to deploy transponders or convert between optical and electrical signals. These

ROADMs are also more elegant than previous architectures, which used fixed optical add/drop

multiplexers with external optical patch panels and cabling.

The next generation ROADM allows for the traffic to be remotely switched at the wavelength level.

The physical characteristics of this ROADM are defined as:

Colorless

Directionless

Contention less

WS

S W

SS

S

pli

tter

Splitter

Solved problems 19

WS

S W

SS

S

plitter

Spli

tter

RX TX

Splitter WSS

TX AWG AWG TX

AWG AWG

AWG AWG

RX RX

Splitter WSS

RX AWG AWG TX

Figure 1.23 the typical architecture of a four-degree ROADM. With this design, the add/drop points at

the end points of the network must still be physically assigned or reassigned by technicians.

Solved problems 20

Optical path switch

λ1

λ1

λ2

λ2

I/O customer optical signal

Pass- through traffic add/drop traffic,

Only signal exists at east direction

Optical path switch Link failure

λ1

λ1

λ2

λ2

I/O customer optical signal

Pass- through traffic add/drop traffic,

Only signal exists at west direction

Figure 1.16 illustrating the operation of the ROADM with protection. (a) Working light path in

operating mode. (b) Switching the working light path to the protected route, protected

mode.

Solved problems 21

Optical path switch

λ1

λ1

λ2

λ2

Pass- through traffic I/O low-priority I/O working

signal signal

Optical path switch

λ1

λ1

λ2

λ2

Pass- through traffic No signal I/O protected

signal

Figure 1.17 illustrating the operation of the ROADM with protection. (a) Working light path in

operating mode. (b) Switching the working light path to the protected route, protected mode.