1. Unit - II
Transmission characteristic of Optical Fiber
EC 8751 Optical Communication
Dr. J. Martin Leo Manickam
Professor
St. Joseph's college of Engineering
Chennai
2. Signal Attenuation & Distortion in Optical Fibers
• What are the loss or signal attenuation mechanism in a fiber?
• Why & to what degree do optical signals get distorted as they
propagate down a fiber?
• Signal attenuation (fiber loss) largely determines the maximum
repeaterless separation between optical transmitter & receiver.
• Signal distortion cause that optical pulses to broaden as they travel
along a fiber, the overlap between neighboring pulses, creating
errors in the receiver output, resulting in the limitation of
information-carrying capacity of a fiber.
3. Attenuation (fiber loss)
• Power loss along a fiber:
The parameter is called fiber attenuation coefficient in a units of for
example [1/km] or [nepers/km]. A more common unit is [dB/km] that is
defined by:
Z=0
P(0) mW
Z= l l
p
e
P
l
P
)
0
(
)
( mw
z
p
e
P
z
P
)
0
(
)
( [3-1]
p
]
km
/
1
[
343
.
4
)
(
)
0
(
log
10
]
dB/km
[ p
l
P
P
l
[3-2]
4. Fiber loss in dB/km
Where [dBm] or dB milliwat is 10log(P [mW]).
z=0
Z=l
]
dBm
)[
0
(
P
]
km
[
]
dB/km
[
]
dBm
)[
0
(
]
dBm
)[
( l
P
l
P
[3-3]
Km
in
distance
-
z
dB,
in
n
attenuatio
P
P in(dbm)
out(dbm)
z
6. A fiber has an attenuation of 0.5 dB/Km at 1500 nm. If 0.5 mW of
optical power is initially launched into the fiber, estimate the
power level after 25 km. (Apr 19) (Nov 19)
Km
in
distance
-
z
dB,
in
n
attenuatio
P
P in(dbm)
out(dbm)
z
dBm
5
.
15
25
5
.
0
10
1
10
5
.
0
log
10
P 3
3
out(dbm)
7. When the mean optical power launched into a 8 Km fiber is 120 μw.
The mean optical power at the fiber output is 3 μw. Calculate the
overall attenuation in dB assuming there are no splices. (Apr 19)
2.0025
16.02/8
(dB/Km)
n
attenuatio
02
.
16
10
1
10
120
log
10
10
1
10
120
log
10
P
P
(dB)
n
attenuatio 6
6
6
6
out(dBm)
in(dbm)
8. When the mean optical power launched into an 8km length of fiber is 120μW, the mean optical power
at the fiber output is 3 μW. Determine (a) Overall signal attenuation in dB/km (b) The overall signal
attenuation for a 10 km optical link using the same fiber with splices at 1Km intervals, each giving an
attenuation of 1dB. (c) Numerical input/output power ratio.(Apr 18)
2.0025
16.02/8
(dB/Km)
n
attenuatio
02
.
16
10
1
10
120
log
10
10
1
10
120
log
10
P
P
(dB)
n
attenuatio 6
6
6
6
out(dBm)
in(dbm)
Overall Signal attenuation for a 10 KM optical link = 10 x 2 = 20 dB/km
The overall signal attenuation for a 10 km optical link using the same fiber with
splices at 1Km intervals, each giving an attenuation of 1dB = 20 + 9 = 29 dB /
Km
Numerical input/output power ratio = 3
.
794
1010
29
1
O
P
P
9. A continuous 40km long optical fiber link has a loss of 0.4dB/km i)What is the
minimum optical power level that must be launched into the fiber to maintain
an optical power level of 2 µw at the receiving end? ii)What is the required input
power if the fiber has a loss of 0.6 dB/km? (Nov 18)
A Continuous 12 km long optical fiber link has a loss of 1.5dB/km. Propose a
proper solution to find the minimum optical power that must be launched into
the fiber to maintain an optical power level of 0.3 µw at the receiving end (Apr 19)
10.
11. Absorption
Absorption is caused by three different mechanisms:
1- Absorbtion by atomic defects (ionic radiation)
2- Extrinsic absorbtion:Impurities in fiber material from
transition metal ions (must be in order of ppb) & particularly
from OH ions with absorption peaks at wavelengths 2700
nm, 400 nm, 950 nm & 725nm.
3- Intrinsic absorption (fundamental lower limit): electronic
absorption band (UV region) & atomic bond vibration band
(IR region) in basic SiO2.
12. Absorbtion by atomic defects
Imperfactions in the atomic structure of the fiber material
Example: Missing molecules, high density clusters of atom groups or oxygen defects
Negligible but significant when exposed to ionizing radiations
14. Optical Fiber Attenuation
0.5 dB/km (1310 nm)
0.3 dB/km (1550 nm)
Nominal attenuations:
Attenuation peak
P(x) = P0e-ax
= (10/x) log P0/P(x)
OFS AllWave fiber: example of a “low-water-peak” or “full spectrum” fiber. Prior to 2000 the fiber transmission bands were
referred to as “windows.” 14
Optical fibers operated in E
band (near 1440 nm) is
called Low water peak or
Full spectrum fibers
15. Intrinisic absorbtion by the basic constitutent
atoms of the fiber material
Electronic absorbtion bands (band gaps) in the ultra violet region and from
atomic vibration bands in the near infrared region
UV edge of the electron absorbtion bands follow Urbach's rule
0
E
E
UV
Ce
C and E0 are Empirical constants
E is the photon Energy
UV absorbtion decreases exponentially with
increasing wavelength
Near IR absorbtion is determined by the
presence of OH- ions and the inherent IR
aborbtion of the constituent
16. Absorption & scattering losses in fibers
Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000
63
.
4
exp
10
60
6
.
46
2
.
154 2
x
x
UV
48
.
48
exp
10
81
.
7 11
IR
UV absorbtion in db/km
Infrared absorbtion in db/km
x - mole fraction of GeO2
Loss is low at 0.148 dB/Km at 1.57 μm in a single mode fiber
17. Scattering Loss
• Microscopic variations in the material density
• Compositional fluctuations
• structural inhomogenities or defects during fiber manufacture
Refractive index variation leads to Rayleigh type scattering
Scattering loss in dB/Km is given by T
f
B
scat
T
k
p
n
2
8
4
3
3
8
n-refractive index
p- photoelastic coefficient
kB-Boltzmann's constant
βT-isothermal compressibility of the material
Tf- temperature at which density fluctuations are frozen in the material
18. Typical spectral absorption & scattering attenuations for a single
mode-fiber
Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000
For wavelengths below 1 μm,
Rayleigh scattering dominates.
As a result, attenuation decreases
with increasing wavelength.
At longer wavelengths, IR
absorbtion dominates
21. Bending Losses in Fibers
• Optical power escapes from tightly bent fibers
• Bending loss increases at longer wavelengths
– Typical losses in 3 loops of standard 9-m single-mode fiber (from:
Lightwave; Feb 2001; p. 156):
– 2.6 dB at 1310 nm and 23.6 dB at 1550 nm for R = 1.15 cm
– 0.1 dB at 1310 nm and 2.60 dB at 1550 nm for R = 1.80 cm
• Progressively tighter bends produce higher losses
• Bend-loss insensitive fibers have been developed and now are
recommended
• Improper routing of fibers and incorrect storage of slack fiber
can result in violations of bend radius rules
R
Test setup for
checking bend loss:
N fiber loops on a
rod of radius R
21
22. Bending Losses in Fibers (2)
The total number of modes that can be supported by
a curved fiber is less than in a straight fiber.
22
23. Core and cladding loss
Core and cladding have different indices of refraction leading to
different attenuation coefficients
Let α1 and α2 be attenuation coefficients of core and cladding,
then the loss for a mode of order (v,m) is given by
P
P
P
P clad
core
vm 2
1
P
P
clad
vm
)
( 1
2
1
25. Intermodal delay or dispersion
occurs in multimode fibers
modal delay is a result of each mode having a different value of the group
velocity at a single frequency
Phase Velocity versus Group Velocity: Wave Dispersion - Bing video
26. Consider an axial ray making an angle 0 degrees with the axis and extreme
meridional ray makes an angle θ with the axis
Let L be the fiber length
The time taken by the axial ray to travel a distance of L is given by
C
n
L
n
C
L
Taxial
1
1
/
where n1 be refractive index of core and C be the velocity of light
------------(1)
Velocity = distance / time
27. The time taken by the extreme meridional ray which makes an angle θ
to travel a distance of L is given by
cos
/
)
cos
/
( 1
1
C
Ln
n
C
L
Tmer
cos
)
90
sin(
sin
1
2
n
n
C
------------(2)
------------(3)
From the figure,
Sub (3) in (2), we get
2
2
1
n
C
n
L
Tmer
------------(4)
28. Time delay between the axial ray and extreme meridional ray
1
2
1
1
1
2
2
1
n
n
C
n
L
C
n
L
n
C
n
L
T
T
T axial
mer
1
2
1
n
n
As
------------(5)
, then equation (5) becomes
C
n
L
T
1
29. Bit rate distance product
C
n
n
BL 2
1
2
rms impulse response σs due to intermodal dispersion SI MM fiber
C
n
NA
L
C
Ln
S
1
2
1
3
4
)
(
3
2
rms impulse response σs due to intermodal dispersion GI fiber (α=2)
C
Ln
S
3
20
2
1
30. Consider a 1 - Km long MM SI fiber in which n1=1.480 and ∆=0.01.
What is the modal delay per length in this fiber?
km
ns
C
n
L
T
/
50
10
3
01
.
0
480
.
1
8
1
This means that a pulse broadens by 50 ns after
travelling a distance of 1 Km ithis type of fiber
31. Consider the following two MM fibers: (a) a step index fiber with a core index
n1=1.458 and a core cladding index difference ∆=0.01 (b) a parabolic profile GI
fiber with the same values of n1 and ∆. Compare the rms pulse broadening per km
for these two fibers
km
ns
C
n
L
S
/
14
10
3
3
2
)
01
.
0
(
458
.
1
3
2 8
1
km
ps
C
n
L
S
/
14
10
3
3
20
)
01
.
0
(
458
.
1
3
20 8
2
2
1
32. Factors Contributing to Dispersion
• The wave propagation constant β is a slowly varying function of the angular
frequency ω.
• One can see where various dispersion effects arise by expanding β in a Taylor
series about a central frequency ω0.
1. The 1st term β0z describes a phase shift of the optical wave.
2. The factor β1(ω0)z produces a group delay τg = z/Vg, where Vg is the group velocity
Propagation constants in x and y directions are β1x and β1y for a particular mode.
Corresponding group delays are τgx = zβ1x and τgy = zβ1y at a distance z
The difference in the propagation times of thjse two modes is called POLARIZATION
MODE DISPERSION.
∆τPMD = z|β1x - β1y|
32
33. Factors Contributing to Dispersion
3. The factor β2 shows that the group velocity (https://youtu.be/EIqKG5TiSYs) of a
monochromatic wave depends on the wave frequency.
• Thus the different group velocities of the frequency components of a pulse cause
it to broaden with distance.
• The group velocity spreading is the GVD
• The dispersion D is related to β2 by
Unit: [ps/(nm.km)]
D is the sum of material and waveguide dispersion
4. The factor β3 is the third-order dispersion.
– The third-order dispersion can be related to the dispersion D and the dispersion slope S0 = ∂D/∂λ
(the variation in the dispersion D with wavelength)
33
34. Chromatic Dispersion (CD)
• It also is known as intramodal dispersion
(https://youtu.be/SAEQND4NyoM)
• This pulse spreading is known as the group velocity dispersion (GVD)
– It is pulse spreading that occurs within a single mode, because the GVD is a
function of wavelength
– Since CD depends on wavelength, its effect on signal distortion increases with
source spectral width
• The dispersion is given in units of ps/(nm × km)
Laser
LED
34
35. Material Dispersion
Material dispersion occurs because the index of
refraction varies as a function of λ
•Because the group velocity of a mode is a function
of the refractive index, the various spectral
components of a given mode will travel at different
speeds, depending on λ.
•The pulse spread σmat for a source of spectral width
σλ is
where Dmat is the material dispersion
35
36. Problem: A manufacturer's data sheet lists the material dispersion Dmat of
Ge02 doped fiber to be 110 ps/(nm.km) at a wavelength of 860 nm. Find
the rms pulse broadening per km due to material dispersion if the optical
source is GaAlAs LED that has aspectral width σλ of 40 nm at an output
wavelength of 860 nm
k m
ns
k m
nm
ps
nm
D
L
ma t
ma t
/
4
.
4
)]
.
/(
110
[
)
40
(
40. Polarization-Mode Dispersion
• A varying birefringence along its length will cause each polarization mode to travel
at a slightly different velocity. (https://youtu.be/DKCHYUxXYXo)
• The difference in propagation times ΔτPMD between the two orthogonal
polarization modes will result in pulse spreading.
• This is the polarization-mode dispersion (PMD)
• PMD is a random, time-varying function in a fiber, so it needs to be measured
statistically.
• If DPMD, which is measured in ps/(km)0.5, is the average PMD parameter, then
Typical values of DPMD range
from 0.05 to 1.0 ps /(km)0.5. 40
41. Dispersion Calculation
• The broadening σ of an optical pulse over a fiber of length L
is given by σ = D(λ)σλ , where σλ is the half-power spectral
width of the optical source.
• For a non-dispersion shifted fiber
For a dispersion shifted fiber: D(λ) = (λ – λ0)S0.
S0 is the dispersion slope S(λ) = dD/dλ at λ0, given in ps/(nm2•km).
41
42. A multimode graded index fiber exhibits total pulse broadening of 0.1μs over a distance of 15 km. Estimate
(i)The maximum possible bandwidth on the link assuming no intersymbol interference
(ii)The pulse dispersion per unit length
(iii)The bandwidth length product for the fiber
