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Optical Communication Systems
SET 4524
Photonics Technology Centre, Photonics Technology Centre, Photonics Technology Centre, Photonics Technology Centre, Faculty Of Electrical EngineeringFaculty Of Electrical EngineeringFaculty Of Electrical EngineeringFaculty Of Electrical Engineering
Universiti Teknologi MalaysiaUniversiti Teknologi MalaysiaUniversiti Teknologi MalaysiaUniversiti Teknologi Malaysia
DR SEVIA MAHDALIZA IDRUSBEE (UTM), MEE (UTM), PhD (Warwick, UK)
P06-210
03-5535451/019-7755038
Final Tutorial
Tutorial 1
You are required to design a 100-kmdigital optical link to transmit data of400Mbps-NRZ. You are supplied withcomponents having a parameters asgiven below.
• You are given a photodetector for your design,which requires a sensitivity of -40 dBm.
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You are given a photodetector for your design,which requires a sensitivity of -40 dBm.
• The design requires 2 connector each having aloss of 1 dB, a source coupling loss of 3 dB andsplice losses of 5dB.
• Choose suitable components for your design.
• Verify your choice by using power budget andthe rise time budget.
Property LED LD 1 LD2
Spectral width (nm) 50 0.15 5
Rise time (ns) 400 1 75
Ouput power (mW) 1 3.2 2
Wavelength (nm) 1550 1550 1300
Costs Low High High
Description Loss (dB/km) Wavelength (nm)
Multimode
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Glass
SI 5 850
GRIN 5 850
PCS SI 8 800
Plastic ST 200 580
Single mode
Glass 4 850
Glass 0.5 1300
Glass 0.25 1550
Solution Tutorial 1
Given
L=100km
Data rate 400Mbps(NRZ)
Minimum sensitivity -40 dBm
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Minimum sensitivity -40 dBm
2 connectors; total loss=2dB
Source coupling loss=3dB
Splice loss=5dB
Solution T1
Tsystem=0.7/BT=0.7/400M=1.75ns
From the system risetime calculated it can be concluded that the only suitable source for this design is Laser Diode 1 with a risetime of 1 ns.
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Others will make it impossible to achieve the required system risetime as they have a risetime of 400ns and 75ns respectively.
Since LD 1 emits at a wavelength of 1550nm, the suitable fiber for it is the single mode glass fiber having an attenuation of 0.25 dB/km at 1550nm.
Assuming that material dispersion dominates,
the value for it can be approximated as;
Solution T1
nmkmpsD ZDm /)1(122 −=
λ
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nmkmps
nmkmps
nmkmpsDm
/68.19
/)1550
13001(122
/)1(122
=
−=
−=λ
Risetime Budget;
The risetime for the photodiode is not given. Therefore it is logical to calculate the limits for this value
T =1ns
2221.1 DFSSys TTTT ++=
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TS=1ns
TF=Dmx∆λxL=19.68x0.15nmx100km=295.2ps
Thus
Therefore, the photodiode must have a risetime of 1.2ns or lower.
( ) nsTTT
T FS
Sys
D 2.11.1
22 =−−
=
Input power, Pi=3.2mW=5.05dBm
Losses
=connectors + source coupling + splice loss
+cable loss
=2+3+5+(0.25x100)
=35dB
Minimum sensitivity, Po=-40dBm
Power Budget;
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Minimum sensitivity, Po=-40dBm
Pi-Po=Losses+ safety margin, Ma
Hence, Ma= 5.05-(-40)-35= 10.05dB
Since the safety margin is positive, the choice of components are SUITABLE with the condition that the photodiode has a risetime of 1.2ns or LOWER.
Tutorial 2
a) The advantages of single mode SIfiber over the multimode SI fiber.
[3]
b) Why multimode SI fiber are morefavorable compared to single mode SIfiber at lower bandwidth application.
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fiber at lower bandwidth application.[3]
c) Explain the advantages of amultimode GRIN index fiber comparedto a multimode SI fiber.
[4]
d) A multimode step index fiber with a corediameter of 50µm large enough to beconsidered by ray theory. It has a corerefractive index of 1.50 and a relative indexdifference of 1%. If the operating wavelengthis 850nm, estimate;
i. the critical angle at the core-cladding interface. [2]
ii. the numerical aperture for the fiber [2]
iii. the acceptance angle in air for the fiber [2]
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iii. the acceptance angle in air for the fiber [2]
iv. the total number of guided modes [3]
v. new core size to decreased the total number ofguided modes obtained above by 50%. [3]
vi. Determine the cut-off wavelength for the fiber toexhibit single mode operation corresponding to the
new core size. Is it practical? [3]
a) The advantages of SMF/SI over MMF/SI
The SMF/SI has an advantage of low intermodaldispersion (broadening of transmitted light pulses),the signal dispersion caused by the delay differencesbetween different modes in a MMF may be avoided.
[1.5]
The MMF/SI has higher intermodal dispersion due tothe differing group velocities of the propagatingmodes. This restricts the maximum bandwidth
Sol. Tutorial 2
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modes. This restricts the maximum bandwidthattainable with MMF/SI, when compared to the singlemode fibers. [1.5]
b) The advantages of MMF/SI over SMF/SI at lower bandwidthapplication.
The use of spatially incoherent optical sources (e.g.most light emitting diodes) which cannot beefficiently coupled to single mode fibers. [1]
Larger numerical apertures & core diameters,facilitating easier coupling to optical sources. [1]
Lower tolerance requirements on fiber connectors.[1]
c) The advantages of a MMF/GRIN compared toMMF/ SI. [4]
The advantage of the GRIN fiber compared toMMF/SI fiber is the considerable decrease inmodal dispersion.
The most common refractive index profile for agraded-index fiber is very nearly parabolic.The parabolic profile results in continualrefocusing of the rays in the core, andminimizes modal dispersion.
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minimizes modal dispersion.
A GRIN fibers do not have a constantrefractive index in the core but a decreasingcore index n(r) with a radial distance from theaxis the cladding.Thus, light rays followsinusoidal paths down the fiber.
d) Specification given;Type: MMF/SI; diameter, d = 50x10-6m; radius,a=25x10-6m; n1=1.5; ∆=0.01; λ=0.85x10-6;
i. the critical angle at the core-cladding interface
ii. the numerical aperture for the fiber
[2]
2
1
2
2
2
1
2n
nn −=∆ 48.15.1)01.0(25.12 222
1
2
12 =−=∆−= nnn
o
cn
n6.80
50.1
48.1sinsin 1
1
21 === −−θ
→
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ii. the numerical aperture for the fiber
iii. the acceptance angle in air for the fiber
21.0)01.02(5.1)2( 2
1
2
1
1 ==∆= xnNA [2]
o
a NA 24.12sin 1 == −θ [2]
iv. the total number of guided modesthe normalised frequency;
Total number of guided modes
v. new operating wavelength to decreased the totalnumber of guided modes obtained above by 50%.
[2]
[1]
81.381085.0
21.0)1025(2)(
26
6
===−
−
x
xxNA
aV
πλπ
7532
81.38
2
22
===V
M
DrSMI 15
v. new operating wavelength to decreased the totalnumber of guided modes obtained above by 50%.
New core size;
[2]
M2=377 ; 46.27)3772(V ==
nmxx
x
NA
Va 1771077.1
21.02
)1085.0(46.27)
2
56
==== −−
ππλ
[1]
vi. Determine the cut-off wavelength for the fiber toexhibit single mode operation corresponding to thenew core size.
For single mode, Vc=2.405
Not practical because the λc is in the outsidecommunication windows. [1]
[2].71.9405.2
21.0)1077.1(22 5
mxx
V
aNA
c
c µππ
λ ===−
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communication windows. [1]
Tutorial 3
a) Briefly describe how light is generatedin an injection laser diode (ILD).
[4 marks]
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b) Compare the various properties oflight generated by an ILD with thatgenerated by a light emitting diode(LED). [3 marks]
c) A 150-km digital optical link wasimplemented using a single mode fiber and asuitable light source. The light source has aspectral width of 2 nm and output power of 0dBm. By making reasonable assumption(s)answer the following questions.
i. Estimate the bit rate achievable by the
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i. Estimate the bit rate achievable by thefiber. [4 marks]
ii. Estimate the power emitted at the outputend of the fiber. [3 marks]
iii. List the assumptions made to solve (i) and(ii). [3 marks]
a) ILDs light generation [4]
recombination of the injected carriers by theprovision of an optical cavity in the crystalstructure for the feedback of photons; wherein general feedback process;
a photon colliding with an atom in theexcited energy state causes the
Sol. Tutorial 3
DrSMI 19
a photon colliding with an atom in theexcited energy state causes thestimulated emission of a second photonand then both these photons release twomore. Creates avalanche multiplication.
When the electromagnetic wavesassociated with these photons are inphase, amplified coherent emission isobtained
b) ILD vs LED sources [3]
Power
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BW; ILD> LEDLED emits spontaneous radiation, the speed ofmodulation is limited by the spontaneousrecombination time of the carriers and not verylarge (a few hundred megahertz). ILD has very fastmodulation (up to 10 GHz).
Spectra: ILDs have narrower spectra than LEDs.The spectrum of an LD remains more stable withtemperature than that of an LED.
ILDs advantages (will be considered).
High radiance due to the amplifying effect ofstimulated emission. Injection lasers will generallysupply milliwatts of optical output power.
Narrow linewidth of the order of 1 nm or less whichis useful in minimizing the effects of materialdispersion.
Modulation capabilities which at present extend upinto the gigahertz range and will undoubtedly beimproved upon.
DrSMI 21
improved upon.
Relative temporal coherence which is consideredessential to allow heterodyne (coherent) detection inhigh capacity systems, but at present is primarily ofuse in single mode systems.
Good spatial coherence permits efficient coupling ofthe optical output power into the fiber even for fiberswith low numerical aperture.
c) Calculation
i. Dispersion =
Since the fiber is SMF, there is no modal dispersion;Thus
Total Dispersion=
Using Gaussian approximation
nmkmpsDm /68.1955.1
3.11122 =
−=
nsxxxLxDm 904.5150268.19 ==∆λ [2]
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Using Gaussian approximation
Bit rate, BT=0.2/σ=0.2/5.904ns=33.87Mb/s [2]
ii. Power emitted at the fiber end
Total attenuation=0.154x150=23.1dB
kmdB /154.055.1
85.07.1
85.07.1
44
=
=
=λ
α
Assuming the odBm is the power coupled into thefiber, the power emitted at the output end of thefiber, PR is
PR=0dBm-23.1dB= –23.1dBm= 4.89µW [3]
iii. Assumptions made [3]
It is a long distance link; therefore we require thefiber to contribute the lowest attenuation. This isachieved by assuming λ=1550nm.
Assume the major attenuation mechanism is Raleigh
DrSMI 23
Assume the major attenuation mechanism is Raleighscattering and the value of the attenuation isapproximately given by
The major dispersion contributing factor is thematerial dispersion and can be approximated by
485.0
7.1
=λ
α
dB/km
−=
λλZD
mD 1122 ps/nmkm
Tutorial 4
a) Compare the signal to noise ratio (SNR)achievable between optical receivers using PINphotodiode and avalanche photodiode (APD).
[5 marks]
b) An optical link was implemented using a silicasingle mode fiber between Skudai and Yong Peng
DrSMI 24
single mode fiber between Skudai and Yong Peng100 km apart. The light source is a laser diodeemitting at a wavelength (λ) of 1550 nm. Thepower coupled into the fiber was found to be 1dBm. Calculate the signal to noise ratio (SNR) ofthe link for both cases below. You are required tostate any assumption(s) made
i. Using PIN photodiode as given in Table Q3(A)[6 marks]
Parameter Symbol Test Condition Min Typical Max Unit
Spectral
response range
λ -- 900 -- 1700 nm
Peak sensitivity
wavelength
λ P -- -- 1550 -- nm
Responsivity R λ = 1550 nm -- 0.95 -- A/W
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Responsivity R λ = 1550 nm
λ = 1300 nm
--
--
0.95
0.9
--
--
A/W
Cut-off
frequency
fc VR = 5V
RL = 50ohm
-- 2 -- GHz
Dark current Ip VR = 5V -- 0.02 0.4 nA
Capacitance Cj -- -- 1 1.5 pF
Table Q3(A) PIN photodiode I specifications
ii. Using avalanche photodiode (APD) as given in TableQ3(B) [6 marks]
Parameter Symbol Test Condition Min Typical Max Unit
Wavelength λ -- 1250 -- 1620 nm
APD breakdown
voltage
V BR I D =10µA 40 60 80 V
APD
responsivity
R APD λ =1550 nm, M=10
λ =1310 nm, M=10
8
7.5
8.5
8.5
--
--
A/W
DrSMI 26
responsivity λ =1310 nm, M=10 7.5 8.5 --
Bandwidth BW RL=50ohm, M=10 1650 1950 -- MHz
Sensitivity Prmin -- -- -34 -33 dBm
Table Q3(B) Avalanche photodiode (APD) specifications
a) SNR achieved by PD vs APD
SNR PIN [2.5]
Sol. Tutorial 4
( ) ( )L
ndp
p
amp
L
dp
p
R
KTBFIIeB
Ior
iR
KTBIIeB
I
N
S
42
42
2
2
2
+++++=
DrSMI 27
The SNR obtained by summing all the noisecontributions.
Main sources are dark current noise and quantumnoise, both regarded as shot noise on thephotocurrent thermal noise from the detector loadresistor and active elements tends to dominatebecause the dark currents in well-designed siliconphotodiodes can be made very small.
SNR APD [2.5]
The random gain mechanism introduces excessnoise into the receiver in terms of increased shotnoise. The photocurrent is increased by a factor M,then the shot noise is also increased by an excess
( ) 2
2
42 −++
=M
R
KTBFMIIeB
I
N
S
L
nx
dp
p
DrSMI 28
then the shot noise is also increased by an excessnoise factor Mx
b)
Pin=1dBm L=100km λ=1550nm K=1.38x10-23 J/K
e=1.6 x10-19C h=6.66 x10-3 Js T=27oC =300K
Tx : P coupled =1dBmSMF : Rx (Using PIN or APD)
i. SNR using PIN photodiode [6]
From the data sheet Q3(A)
R=0.95A/W Ct=1.5pF (use max) RL=50Ω
( ) ( )L
n
dp
p
amp
L
dp
p
R
KTBFIIeB
Ior
iR
KTBIIeB
I
N
S
42
42
2
2
2
+++++=
( )L
dp
p
R
KTBIIeB
I
N
S
42
2
++=
DrSMI 29
Use
Estimated attenuation at 1550nm
Total attenuation=0.1537x 100 km =15.37 dB
L
GHzpCR
BtL
12.25.1502
1
2
1=
⋅⋅==
ππ
kmdB /1537.055.1
85.07.1
85.07.1
44
=
=
=λ
α
Power arriving at the detector is,
PR=1dB-15.37dB = -14.37dB=0.0366mW
Photocurrent;
Ip=RPR=0.0366mW x 0.95=0.0347mA =34. 7µA
Given the photodiode current, Id=0.02-0.4nA (takethe max. value of 0.4nA)
Assume in ideal case with the noise figure, F=1
Thus the SNR are given by2IS
DrSMI 30
( )
( ) [ ][ ] [ ]
( )1314
2
2319
192
2
1002.71034.2
7.34
50
112.23001038.14)4.07.34(12.2106.12
)4.07.34(12.2106.127.34
42
−−
−−
−
+=
+++
++=
++=
xx
A
GxxxxxnGxxx
nGxxxA
R
KTBFIIeB
I
N
S
L
ndp
p
µ
µ
µµ
=1641 = 32.2dB
ii. SNR using APD [6]
Photocurrent, Ip=PRR=0.0366mW x 8.6=0.3111mA
Assume negligible excess noise, i.e x=0 and F=1
From the datasheet: M=10, B=1.950GHz
The dark current Id not available thus normallyIp>>>Id
Thus the SNR
( ) 2
2
42 −++
=M
R
KTBFMIIeB
I
N
S
L
nx
dp
p
DrSMI 31
Thus the SNR
( )
( )
[ ] [ ]
( )1513
2
2
2319
2
2
2
1045.61094.1
3111.0
1050
1950.13001038.14)3111.0(950.1106.12
3111.0
412
−−
−−
+=
+
=
++=
xx
m
x
GxxxxxmGxxx
m
MR
KTBFIIeB
I
N
S
L
n
dp
p
= 482830 = 56.83dB
If x=1
( )
( )
[ ] [ ]
( )1512
2
2
2319
2
2
2
1045.61094.1
3111.0
1050
1950.13001038.1410)3111.0(950.1106.12
3111.0
42
−−
−−
+=
+
=
+=
xx
m
x
GxxxxxmGxxx
m
MR
KTBFMIeB
I
N
S
L
np
p
DrSMI 32
= 49723 = 46.97dB
Tutorial 5a) Describe the techniques normally carried-out in
verifying a particular design of optical fibercommunication link or network. [4 marks]
b) You are given specifications of various different opticalcomponents that can be used to implement an opticalcommunication link. They include optical fiber cableas given in Table Q4(A), laser diode as given in TableQ4(B), PIN photodiode II as given in Table Q4(C) and
DrSMI 33
Q4(B), PIN photodiode II as given in Table Q4(C) andAvalanche photodiode (APD) as given in Table Q4(D).Study the specification carefully.
i. Choose any suitable components to implement apoint-to-point optical link and draw the blockdiagram of your optical link clearly naming thechosen component in the diagram.[3marks]
ii. By using suitable method(s) determine the specificationsof the link. You are required to state any assumption(s)
made. [10 marks]
Parameter Condition Typical Unit
Attenuation λ = 1310 nmλ = 1550 nmλ = 1625 nm
0.350.200.23
dB/km
Cladding diameter -- 125±0.3 µm
Mode field diameter λ = 1310 nmλ = 1550 nm
9.2±0.410.4±0.5
µm
Dispersion λ = 1550 nmλ = 1625 nm
18.022.0
ps/nm-km
DrSMI 34
λ = 1625 nm 22.0
Table Q4(A) Optical fiber specifications
POWER OPTION Unit
0.5 1 mW
Wavelength range 1520-1580 1520-1580 nm
Spectral width 3 3 nm
Risetime 0.5 0.5 ns
Threshold current 40 40 mA
Maximum forward voltage 2 2 V
Table Q4(B) Laser diode specifications
Parameter Symbol Test Condition Min Typical Max Unit
Spectral response range
λ -- 1250 -- 1620 nm
Peak sensitivity wavelength
λ P -- -- 1550 -- nm
Responsivity R λ = 1550 nmλ = 1310 nm
0.750.70
0.800.75
----
A/W
Bandwidth BW RL=100ohm -- 875 -- MHz
Sensitivity Prmin -- -- -27 -- dBm
Table Q4(C) PIN photodiode II specifications
DrSMI 35
Parameter Symbol Test Condition Min Typical Max Unit
Wavelength λ -- 1250 -- 1620 nm
APD breakdown voltage
V BR I D =10µA 40 60 80 V
APD Responsivity R APD λ=1550 nm, M=10λ=1310 nm, M=10
87.5
8.58.5
----
A/W
Bandwidth BW RL=50ohm, M=10 1650 1950 -- MHz
Sensitivity Prmin -- -- -34 -33 dBm
Table Q4(D) Avalanche photodiode (APD) specifications
a) The basic system design verification can be donethrough 2 methods:
i. Power budget [2]
involves the power level calculations from thetransmitter to the receiver. Which the parametersthat considered are Attenuation, Coupled power,Other losses (splices, reflection losses, connectors
Sol. Tutorial 5
DrSMI 36
Other losses (splices, reflection losses, connectorsetc), Equalization penalty (DL), SNR requirements,Minimum power at detector, BER, Safety margin(Ma).
The optical power budget calculation;
Pi = (Po + CL + Ma + DL) dB
ii. Risetime budget [2]
the calculations involve the BW, which the followingparameters involves.
Risetime of the source, TS
Risetime of the fiber (dispersion), TF
Risetime of the amplifier, TA
Risetime of the detector, TD
The risetime budget is assembled as:
Tsyst= 1.1(TS2 + TF
2 + TD2 + TA
2) 1/2
DrSMI 37
b) System design with the given datasheets [3]
Laser diode Fiber PIN Diode
λ=1550nm SMF, 0.20dB/km λp=1550nm
Po=1mW D=18 ps/nmkm Sensitivity=-27dBm
∆λ=3nm L=1km R=0.80A/W
Transmitter Receiver
i. Rise time budget [5]
Tsys(NRZ)=0.7/BT or Tsys(RZ)=0.35/BT
Risetime of the source, laser: TS= 0.5ns=500ps
Risetime of the fiber,
TF=18x ∆λxL=18x3nx1km=54ps
Risetime of the detector PIN,
Tr=2.19RLCD
Bandwidth, BW=1/2π RLCD = 875MHz=1/π 50CD
Hence C =1.82pF
DrSMI 38
Hence CD=1.82pF
Thus, Tr=2.19x100x1.82pF=398ps
Total system risetime, Tsys=1.1(TS2+ TF
2+ TD2) 1/2
= 1.1(542+5002+3982) 1/2 = 705.5ps
Bit rate (NRZ)=0.7/705.5=992.2Mbps
Bit rate (RZ)=0.35/705.5=496.1Mbps
If we use T=0.35/BT=and assume T=Tpin
Then Tpin=0.35/875M=400ps
If we use NRZ, then T=0.7/BT
But BT=2xBW=1750MHz
Thus T=0.7/1750=400ps
Which is same rise time.
If we use T =2.19R C and BW=1/2π R C
DrSMI 39
If we use Tr=2.19RLCD and BW=1/2π RLCD
Then CD=1.82pF= 1/2π 100X 875M
And Tr=400ps=Tpin
Then Tsys=1.1(TS2+ TF
2+ TD2) 1/2
= 1.1(542+5002+4002) 1/2 = 706.8ps
– Bit rate (NRZ)=0.7/706.8 =990.4Mbps
– Bit rate (RZ)=0.7/706.8 =495.2Mbps
ii. Power budget [5]
We can also determint te maximum length of the linkwith power budget
Assume power into the fiber = P emitted into thefiber,
Pcouple=1mW=0dBm
And given, attenuation=0.2dB/km and
sensitivity=-27dBm
Allowable attenuation is 27dB
DrSMI 40
Allowable attenuation is 27dB
Maximum distance considering power budget only is27/0.2=135km.
Remember, with the length, the bit rate achievablewill be less than 990Mbps because TF will then be18x3x135=7290ps and BT=87.5Mbps.
Tutorial 6
a) Fiber links are limited in path length byattenuation and if this is the major problem,the link is said to be power limited. Anumber of mechanisms are responsible forthe signal attenuation within optical fibers.These mechanisms are influenced by thematerial composition, the preparation andpurification technique, and the waveguide
DrSMI 41
material composition, the preparation andpurification technique, and the waveguidestructure. Please discuss the followingmatters;
i. Two major types of linear scattering losses thatcontribute to the fiber attenuation. [4 marks]
ii. Four effective methods to reduce the linearscattering losses [2 marks]
b) Fiber bandwidth is determined by an effectcalled dispersion. Dispersion causes distortionof digital and analog signals, and was a majorcharacteristic to be considered when choosinga fiber. Thus, please briefly explain;
i. an effective intramodal dispersion for asingle mode fiber at longer wavelength.
[3 marks]
DrSMI 42
[3 marks]
ii. the main dispersion type that contribute topulse spreading for a multimode fiber.
[3 marks]
iii. Two methods to reduce pulse broadeningdue to intramodal dispersion.
[2 marks]
c) A 10 km fiber optic link of MMF/SI fiber with acore refractive index of 1.5 and a relativerefractive index difference of 1.5%.Calculate;
i. The delay difference between the fundamental
mode to the highest mode at the fiber output
[3 marks]
ii. The rms pulse broadening due to
DrSMI 43
ii. The rms pulse broadening due tointermodal dispersion. [3 marks]
iii. The maximum bit rate using calculated rms by (ii).
[3 marks]
iv. The bandwidth-length product corresponding to
(iii). [2 marks]
a) Atthenuationi. Two major types of linear scattering losses
Rayleigh Scattering: the dominant intrinsic loss mechanism inthe low absorption window between the ultraviolet and infraredabsorption tails. It results from inhomogeneities of a randomnature occurring on a small scale compared with the wavelengthof the light. These inhomogeneities manifest themselves asrefractive index fluctuations and arise from density andcompositional variations which are frozen into the glass latticeon cooling. The scattering due to the density fluctuations, which
Sol. Tutorial 6
DrSMI 44
on cooling. The scattering due to the density fluctuations, whichis in almost all directions, produces an attenuation proportionalto l/λ4 following the Rayleigh scattering formula. [2]
Mie Scattering: Occur at inhomogeneities which arecomparable in size to the guided wavelength. These result fromthe nonperfect cylindrical structure of the waveguide and maybe caused by fiber imperfections such as irregularities in thecore-cladding interface, core-cladding refractive indexdifferences along the fiber length, diameter fluctuations, strainsand bubbles. The scattering created by such inhomogeneities ismainly in the forward direction. [2]
ii. Any four effective methods to reduce the linear
scattering losses [2]
Rayleigh scattering reduced by operating at
the longest possible wavelength. The
theoretical attenuation due to Rayleigh
scattering in silica at wavelengths of 0.63, 1.00
and 1.30 µm.
Improving fabrication process i.e cooling and
DrSMI 45
Improving fabrication process i.e cooling and
heating
removing imperfections due to the glass
manufacturing process
carefully controlled extrusion and coating of
the fiber
increasing the fiber guidance by increasing the
relative refractive index difference.
b) Dispersion
i. an effective dispersion for a single mode fiber at longer
wavelength.
Waveguide dispersion: an intramodal dispersion. This
results from the variation in group velocity with
wavelength for a particular mode. Pulses of same mode
but different wavelengths need to travel at different angle
therefore have different velocities. Waveguide dispersion
occurs because the effective refractive index, neff for any
one mode varies with λ. D depends on the V parameter
DrSMI 46
one mode varies with λ. DW depends on the V parameter
of the fiber. For a SMF whose propagation constant is β
the fiber exhibits waveguide dispersion when d2β/dλ2 ≠ 0.
for MMF, where the majority of modes propagate far from
cutoff, are almost free of waveguide dispersion and it is
generally negligible compared with material dispersion.
However, with SMF where the effects of the different
dispersion mechanisms are not easy to separate,
waveguide dispersion may be significant. [3]
ii. the main dispersion for a multimode fiber.
in MMF all three dispersion mechanism exist
simultaneously that is material dispersion, waveguide
dispersion, and multimode dispersion. However
intermodal dispersion is unique to MMF.
Intermodal dispersion occurs due to different modes
propagating through the fiber will have different net
velocities and will arrive at different time at the output.
DrSMI 47
This causes the waveform to spread. Depend on ∆λ.
Therefore even if the source has ∆λ = 0, then DM and DW
will be zero, but it will still suffer multimode dispersion.
The amount of modal dispersion or spreading is easily
developed by the difference in travel time between mode
propagating at the steepest angle with respect to the axis.
[3]
iii. methods to reduce intramodal dispersion. [2]
Dispersion shifted fiber: The waveguide dispersion is
exploited to interact with the material dispersion to shift
the zero dispersion wavelength to a value which will have
the lowest attenuation.
Dispersion flattened fiber: The fiber is modified to
achieve low dispersion window over the low loss
wavelength region between 1.3 µm and 1.6 µm.
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Depressed cladding fiber : The fiber is made so that the
core is surrounded by a thin inner cladding whose index
is low and an outer cladding whose index is slightly
higher.
c) A 10 km fiber optic link of MMF/SI fiber with a core refractive
index of 1.5 and a relative refractive index difference of 1.5%.
i. The delay difference
ii. The rms pulse broadening due to intermodal dispersion.
nsx
xxx
c
LnT 750
10998.2
015.05.110108
3
1 ==∆
=δ [3]
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ii. The rms pulse broadening due to intermodal dispersion.
nsnsT
c
Lns 5.216
32
750
3232
1 ===∆
=δ
σ [3]
iii. The maximum bit rate using calculated rms by (ii).
iv. The bandwidth-length product corresponding to (iii).
[2]
[3]1
(max) 92.05.216
2.02.0 −=== Mbitsns
Bs
T σ
MHzkmkmMHzxxLB 2.91092.0 ==
DrSMI 50
[2]MHzkmkmMHzxxLBopt 2.91092.0 ==
Tutorial 7You are required to verify the capability of an optical link
designed by your colleague. Table XX shows the components
used in the implementation of the link. The link is to be used
for distribution of high-speed data from a data terminal to three
remote terminals each of them 80-km away from the data
terminal. The splitting of the optical signal is done through a 1:
3 optical splitter. You are required to suggest the most suitable 3 optical splitter. You are required to suggest the most suitable
optical fiber for this application.
Draw the block diagram of the link clearly labelling the various
components.
Estimate the maximum bit rate achievable for this link.
Determine whether the components chosen are suitable.
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Parameter Symbol Test Condition Min Typical Max UnitAvalanche photodiode
Wavelength λ -- 1250 -- 1620 nm
APD Responsivity RAPD λ =1550 nm
λ =1310 nm
--
--
10
8.5
--
--
A/W
Sensitivity Prmin -- -- -34 -- dBm
Rise time Tdet -- 0.1 ns
Dark current I -- 20 nA
Table XX
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Dark current Idark -- 20 nA
Laser diode
Output power Pout -- 0.5 mW
Wavelength range λ -- 1520-
1580
nm
Spectral width ∆λ -- 3 nm
Risetime tsource -- 0.5 ns
Threshold current Ith -- 40 mA
Laser Drive Circuit
OPTICAL
SPLITTER
data terminal
Solution T7
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APD and receiver circuit
INPUT
0.5 mW or –3.01 dBm
LASER
Node 1
Node 2
Node 3Optical Splitter
Splits optical input equally to
the output ports.
SPLITTER
80 km
ii.
Output at each nodes of splitter is 0.5/3 mW =
0.1667mW or -7.781 dBm. Alternatively, the
splitting loss is 10 log(1/3)=-4.771 dB.
Therefore the output at each nodes of the
splitter is –3.01 dBm – 4.771 = -7.781 dBm.
The best wavelength is at 1550 nm to The best wavelength is at 1550 nm to
achieve lowest attenuation.
Assuming silica singlemode fiber, the
attenuation coefficient can be estimated as
dB/km. This will then be 0.154 dB/km.
The total attenuation for each link will then be
0.154 x 80-km =12.32 dB.DrSMI 54
The estimated power arriving at each APD will then be
–7.781 dBm – 12.32 dB = -20.101 dBm. Since the
sensitivity of the APD is given as –34 dBm, the
components chosen has no problem with the power
budget.
It is not mentioned in the description of the system on
the data rate required, therefore verification could be
done by estimating the maximum bit rate achievable
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by the system.
Dispersion can be estimated by assuming that the
only dispersion present is material dispersion since
modal dispersion is zero for singlemode fiber and
waveguide dispersion is negligible.
Therefore the material dispersion is
DM=122(1-lZD/l) = 122(1-1.276/1.550) = 21.57 ps/nmkm.
For the link, total dispersion is Tfiber = 21.57 ps/nmkm
x 3 nm x 80 km = 5.177 ns.
Tsys = 1.1(0.5 ns2 + 5.177 ns2 + 0.1 ns2)1/2= 5.722 ns
For non-return-to-zero (NRZ) data:
21
2det
221.1
++= TfiberTsourceTsysT
sysTTB
7.0=
For return-to-zero (RZ) data:
DrSMI 56
sys
sysTTB
35.0=
Therefore, Bit rate,
BT = 0.7/5.722 ns = 122.4 Mb/s for a NRZ data format.
BT = 61.2 Mb/s for RZ data format.
The estimated power arriving at each APD
will then be –7.781 dBm – 12.32 dB = -20.101
dBm.
Since the sensitivity of the APD is given as –
34 dBm, the components chosen has no
problem with the power budget. problem with the power budget.
It is not mentioned in the description of the
system on the data rate required, therefore
verification could be done by estimating the
maximum bit rate achievable by the system.
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(a)Discuss the two main reasons why optical fiber is a favored communication channel for long-haul communications compared to copper channels. (4marks)
(b) Draw a typical block diagram for a high-speed optical communication link indicating clearly the various components. (3 marks)
(c) You are required to prepare a proposal for the selection of a suitable optical fiber for a particular application. Describe in detail what are the
TUTORIAL 8
DrSMI 58
optical fiber for a particular application. Describe in detail what are the factors that you will consider in making your decision. (10 marks
(d) A 20-km optical link was implemented utilizing a light emitting diode (LED) with an optical output of 0 dBm at 1300 nm. By making reasonable assumption(s), estimate the power at the output of the fiber for these two cases: (8 marks)
i. The fiber used is a multimode graded index fiber an NA of 0.20. ii. The fiber used is a multimode step index fiber with an NA of
0.32.
Answer T8
a) Refer Note
b) Refer Note
c) Similar to answer T5-(b) but limits to
-rise time budget
-power budget
Transmitter Receiver
d) Refer tutorial and note.
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Transmitter Receiver
Laser diode Fiber PIN Diode
λ=1550nm SMF, 0.20dB/km λp=1550nm
Po=1mW D=18 ps/nmkm Sensitivity=-27dBm
∆λ=3nm L=1km R=0.80A/W
Presentation 26 March 2009
1. World Undersea Fiber Optic System2. Long Haul Optical Fiber Communication
Systems3. Technology Review on The Optical SourceTechnology Review on The Optical Source4. Photonics in Medical Application5. Photonics Sensor 6. Optical Power Budget Analysis
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