Signal attenuation in optical fibers is caused by several mechanisms: 1. Material absorption from defects in the glass fiber material and impurities, which accounts for the majority of losses. 2. Scattering from irregularities in the fiber structure that cause light to change direction. 3. Bending losses, where tight bends of the fiber cause light to radiate out of the fiber, with losses increasing sharply for bends below a critical radius. Signal attenuation limits the maximum distance between optical transmitters and receivers in a fiber system before the signal needs to be regenerated.

1. Signal Attenuation

2. Signal Attenuation & Distortion in
Optical Fibers
• What are the loss or signal attenuation mechanisms in a fiber?
• Why & to what degree do optical signals get distorted as they
propagate down a fiber?
• Signal attenuation (power loss) largely determines the
maximum repeaterless separation between optical transmitter
& receiver.
• Signal distortion cause 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 (power loss)

4. Attenuation (power loss)

5. 






)
(
)
0
(
log
10
]
dB/km
[
z
P
P
z

Power 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 

 

6. Example 1

7. Solution

8. Computing Power Level in Decibel (dB)
P1 P2
Relative Power
dB = Decibels
1
2
Ratio
Power
P
P

1
2
10 log
dB
P
P
 
 
 
 
 P2 < P1 , dB < 0 Losses
P2 > P1 , dB > 0 Gain  
3 2
10 log
dB 
1
3
2
10 log
dB
 
 
 
 
2 1
10
10
dB
P P
 
 
 
 

1 3 2 1
3
4 4 2
P
P P P
P P P P
  
1 3 2 1
3
4 4 2
10 log 10 log
P
P P P
dB
P P P P
 
 
 
  
   
   
 

3 2 1
3
4 2
10 log 10 log 10 log
P
P P
dB
P P P
 
   
 
   
   
 
   
 
 
 4 3 2
dB dB dB dB



System
P1 P4
P2
P3
System System System

9. P1 P4
P2
P3
Example 2: Suppose that three elements in the figure have losses of 11, 6 and 3 dB respectively. Find
the total loss of the commination system. Find the output power if the input power is 5 mW.
Solution
 
2
10log
m
dB P

11 6 3 20
loss
T dB
     
10
10
dB
out in
P P
 
 
 
 

Example 3: A Light Emitting Diode radiates 2 mW. Compute the dBm value of this radiated power.
This power travels through a group of components having a combined loss of 23 dB. Compute the
output power.
Absolute Power Assume P1 = 1 mW
 
2
10 log
m
dB P
   3
10 log 2
m m
dB dB


20
10 10
0.01
10 10
dB
out mW
P
   
   
   
   

  
The output Power 3 23 20
m m
dB dB dB
 
 
10
10
dB
out in
P P
 
 
 
 

or
Solution
1
2
10 log
dB
P
P
 
 
 
 

20
10
0.01 0.05
5 10 5 mW
 
 
 
 


   
23
10
0.01
2 10 mW
 
 
 
 

  

10. Fiber attenuation (losses) mechanisms
• Material absorption
• Scattering losses
• Bending losses

11. Material absorption losses

12. Fiber Attenuation
A- Material Absorption
1- Intrinsic Absorption by atomic defects in the glass
Crystallography:
an imperfection in
a crystal caused by
the presence of an
extra atom in an
otherwise
complete lattice.
Vacancies
Composition (i.e. missing
molecules, high density clusters of
atoms, group or oxygen defects in
the glass structure
Usually absorption from these defects are small and negligible compared to other sources
2- Extrinsic Absorption by Impurities atoms
Substitutional Impurities
Atomic defects act like impurities:
they can also contribute to light
attenuation

13. Intrinsic absorption

14. Fiber attenuation characteristics

15. Extrinsic absorption

16. OH absorption peak

17. Scattering losses

18. Scattering losses

19. Example: Optical Fiber Attenuation
Wavelength SMF28 62.5/125
850 nm 1.8 dB/km 2.72 dB/km
1300 nm 0.35 dB/km 0.52 dB/km
1380 nm 0.50 dB/km 0.92 dB/km
1550 nm 0.19 dB/km 0.29 dB/km

20. Bending Losses
• Macrobending Loss: The
curvature of the bend is much
larger than fiber diameter.
Lightwave suffers severe loss
due to radiation of the evanescent
field in the cladding region. As
the radius of the curvature
decreases, the loss increases
exponentially until it reaches at a
certain critical radius. For any
radius a bit smaller than this
point, the losses suddenly
becomes extremely large. Higher
order modes radiate away faster
than lower order modes.

21. Macrobending Loss- Example
For SMF-28 fiber a loss of 0.05dB will be introduced with 100
turns around a mandrel of diameter of 75mm.
Loss will increase rapidly for a smaller bending radius.
Typically, a bending radius should be more than 150 times the
cladding diameter of the fiber for long term applications and
more than 100 times the cladding diameter for short term
applications. For example, for fiber with 125mm cladding
diameter, the bending radius should be larger than 19mm.

22. Bending Losses
• Microbending Loss:
Microscopic bends of the fiber
axis that can arise when the
fibers are incorporated into
cables. The power is dissipated
through the microbended fiber,
because of the repetitive
coupling of energy between
guided modes & the leaky or
radiation modes in the fiber.
Small-scale fluctuations in the radius of curvature of the fiber axis leads to
microbending losses. Microbends can shed higher-order modes and can
cause power from low-order modes to couple to higher-order modes

23. Bending Losses
Keeping bend radii greater than 10 cm makes bend loss negligible
It is common to wrap optical fiber in communication systems