Ch-2 Optical Fibers & Fiber Cables

Example 2.1 Page no 31

In [67]:
from __future__ import division
from math import sqrt, pi
# Calculation of core diameter
# Given data
n1=1.5#               # Refractive index of core
n2=1.48#              # Refractive index of cladding
N=1000#               # No of modes
lamda=1.3#           # Light wavelength
V=sqrt(2*N)#          # Mode parameter

#core diameter
d=(lamda*V)/(2*pi*sqrt(n1**2-n2**2))#   

print"Mode parameter = %0.2f  "%(V)#
print"Core diameter = %0.0f micrometer "%(d)#

# Answer is wrong in the book.

Mode parameter = 44.72  
Core diameter = 38 micrometer

Example 2.2 Page no 35

In [68]:
from math import asin, sin, degrees
# Calculation of (a) critical angle (b) acceptance angle and oblique angle (c) numerical aperature 
# (d) percentage of light collected by the fiber and (e) diameter of fiber

# Given data
n1=1.5#                           # Refractive index of core
n2=1.45#                          # Refractive index of cladding
V=2.405#                          # Mode parameter
lamda=1.55                       # Wavelength of fiber

# (a) Critical angle of the
theatha=degrees(asin(n2/n1))#             

# (b) Oblique angle
oa=90-theatha#                       
 #Acceptance angle   
t1=n1*sin(oa*pi/180)#  
th1=degrees(asin(t1))# 

# (c) Numerical aperature                      
NA=sqrt(n1**2-n2**2)#                   

# (d) Percentage of light collected in fiber
P=(NA)**2*100#                        

#(e) Diameter of fiber
d=V*lamda/pi*(1/sqrt(n1**2-n2**2))#  


print"Critical angle = %0.0f degrees "%(theatha)#
print"Oblique angle = %0.0f degrees "%(oa)#
print"Acceptance angle = %0.0f degrees "%(th1)#
print"Numerical aperature = %0.4f  "%(NA)#
print"Percentage of light collected in fiber = %0.1f  "%(P)#
print"Diameter of fiber = %0.2f micrometer "%(d)#

Critical angle = 75 degrees 
Oblique angle = 15 degrees 
Acceptance angle = 23 degrees 
Numerical aperature = 0.3841  
Percentage of light collected in fiber = 14.7  
Diameter of fiber = 3.09 micrometer

Example 2.3 Page no 38

In [69]:
#  Calculation of (a) critical angle (b) numerical aperature and (c) acceptance angle

# Given data
n1=1.5#                           # Refractive index of core
n2=1.47#                          # Refractive index of cladding)

# (a) Critical angle
theatha=degrees(asin(n2/n1))#            
# (b) Numerical aperature
NA=sqrt(n1**2-n2**2)#               
# (c) Acceptance angle
theatha1=degrees(asin(NA))#                     


print"Critical angle = %0.1f degrees "%(theatha)#
print"Numerical aperature = %0.2f  "%(NA)#
print"Acceptance angle = %0.1f degrees "%(theatha1)#

Critical angle = 78.5 degrees 
Numerical aperature = 0.30  
Acceptance angle = 17.4 degrees

Example 2.4 Page no 46

In [70]:
# Calculation of output power

# Given data
Pi=1#                                       # Input power
A=0.5#                                     # Atteuation
L=15#                                     # Fiber link length

# Output Power
Po=Pi*10**((-A*L)/10)#                     


print"Output Power = %0.3f mW "%(Po)#

Output Power = 0.178 mW

Example 2.5 Page no 47

In [71]:
from math import log10
# Calculation of maximum transmission distance


# Given data
Pi=1*10**-3#                   # Input power
A=0.5#                        # Atteuation
Po=50*10**-6#                  # Output Power

# Maximum transmission distance
L=(10/A)*log10(Pi/Po)#         


print"Maximum transmission distance = %0.0f km "%(L)#

Maximum transmission distance = 26 km

Example 2.6 Page no 48

In [72]:
# Calculation of output power

# Given data
Pin=1*10**-3#                  # Input power
AL1=10#                       # Attenuation 1
AL2=20#                       # Attenuation 2          
#Output power 1 and 2
Po1=Pin/10#                      
Po2=Pin/20#                          
Po1=Po1*10**3#
Po2=Po2*10**6#

print"Output power = %0.1f mW"%(Po1)#
print"Output power = %0.0f uW"%(Po2)#

Output power = 0.1 mW
Output power = 50 uW

Example 2.7 Page no 50

In [73]:
from math import exp

# Calculation of attenuation and Rayleigh scattering coefficient for fiber


# Given data
n=1.46#                                      # Refractive index
p=0.286#                                     # Average photoelastic coefficient
B=7.25*10**-11#                               # Isothermal compressibility
k=1.38*10**-23#                               # Boltzmann's constant
T=1350#                                      # Fictive temperature
l1=1*10**-6#                                  # Wavelength 1
l2=1.3*10**-6#                                # Wavelength 2
L=10**3#                                      # Length

# Rayleigh scattering coefficient for length 1
y1=8*(pi)**3*(n)**8*(p)**2*B*k*T/(3*(l1)**4)#   
# Rayleigh scattering coefficient for length 2
y2=8*(pi)**3*(n)**8*(p)**2*B*k*T/(3*(l2)**4)#   
y1=y1#
y2=y2#
#Attenuation 1
T1=exp(-(y1*L))#                               
#Attenuation 2
T2=exp(-(y2*L))#                               


print"First Rayleigh scattering coefficient = %0.6f m**-1 "%(y1)#
print"Second Rayleigh scattering coefficient = %0.6f m**-1 "%(y2)#

print"Attenuation (@ Length 1) = %0.2f dB/km"%(T1)#
print"Attenuation (@ Length 2) = %0.2f dB/km"%(T2)#

First Rayleigh scattering coefficient = 0.000189 m**-1 
Second Rayleigh scattering coefficient = 0.000066 m**-1 
Attenuation (@ Length 1) = 0.83 dB/km
Attenuation (@ Length 2) = 0.94 dB/km

Example 2.8 Page no 52

In [74]:
# Calculation of threshold power of stimulated Brillouin scattering and Raman Scattering



# Given data
A=0.5#                       # Attenuation
d=5#                         # Core diameter
lamda=1.3#                  # Operating wavelength
v=0.7#                       # Bandwith of laser diode

# Threshold power of stimulated Brillouin scattering
Pb=4.4*10**-3*d**2*lamda**2*A*v#
Pb=Pb*10**3#

#Threshold power stimulated Raman Scattering
Pr=5.9*10**-2*d**2*lamda*A# 


print"Threshold power of stimulated Brillouin scattering = %0.2f mW"%(Pb)#
print"Threshold power stimulated Raman Scattering = %0.2f W"%(Pr)#

Threshold power of stimulated Brillouin scattering = 65.06 mW
Threshold power stimulated Raman Scattering = 0.96 W

Example 2.1 Page no. 479

In [75]:
# Compuatation of mode parameter



#Given data
n1=1.503# # refractive index of core
n2=1.50# # refractive index of cladding
a=4*10**-6# # core radius
lamda=1*10**-6# # light wavelength

# Mode parameter computation
V=(2*pi*a*sqrt(n1**2-n2**2))/(lamda)#

#Displaying the result in command window
print"Mode parameter is = %0.3f "%(V)#

# The answer vary due to round off error

Mode parameter is = 2.385

Example 2.2 Page no. 479

In [76]:
# Calculation of numerical aperature



#Given data
v=2.111#                  # Mode parameter
a=4.01*10**-6#             # Core radius in m
lamda=1.3*10**-6#         # Wavelength of laser light m

#Numerical aperture computation
NA=(v*lamda)/(2*pi*a)# 

#Displaying the result in command window
print"Numerical aperature = %0.2f"%(NA)#

Numerical aperature = 0.11

Example 2.3 Page no 480

In [77]:
from math import log
# Calculation of potential difference



# Given data
na=10**24#                     # Accepter impurity level
nd=10**22#                     # Donor impurity level
ni=2.4*10**19#                 # Intrinsic electron 
T=290#                        # Room temperature
e=1.602*10**-19#               # Electric charge
K=1.38*10**-23#                # Boltzmann constant


#Potential difference
V=(K*T)/e*(log(na*nd/(ni)**2))#


print"Potential difference = %0.2f V "%(V)#
# The potential difference varies with the variation of Na, Nd and ni

Potential difference = 0.42 V

Example 2.4 Page no 480

In [78]:
# Calculation of (a) Numerical aperature and (b) critical angle



# Given data
n1=1.5#                   # Refractive index of core
n2=1.47#                  # Refractive index of cladding

# (a) Numerical aperature
NA= sqrt(n1**2-n2**2)#                

# (b) Critical angle
theatha=degrees(asin(n2/n1))#               


print"Numerical aperature = %0.2f "%(NA)#
print"Critical angle = %0.1f degrees "%(theatha)#

Numerical aperature = 0.30 
Critical angle = 78.5 degrees

Example 2.5 Page no 480

In [79]:
import math
# Computation of (a) normalized frequency and (b) no. of guided modes



#Given data
lamda=0.85*10**-6#      # wavelength of fiber
a=40*10**-6#             # core diameter of fiber
delta=0.015#            # relative refractive index 
n1=1.48#                # refractive index of core

# (a) Normalized frequency 
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda# 
#(b) Number of guided modes
m=v**2/2#       
m=math.ceil(m)#
#Displaying results in the command window            
print"Normalized frequency is = %0.1f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#

Normalized frequency is = 75.8  
Number of guided modes = 2873

Example 2.6 Page no 480

In [80]:
# Computation of normalized frequency and no of guided modes


#Given data
lamda=1.30*10**-6#        # Wavelength of fiber
a=25*10**-6#               # Core diameter of fiber
delta=0.01#               # Relative refractive index 
n1=1.50#                  # Refractive index of core

# (a) Normalized frequency 
v=((2*pi*a*n1)/(lamda))*((2*delta)**(1/2))# 
#(b) Number of guided modes
m=v**2/2#                    
#m=ceil(m)#

#Displaying results in the command window            
print"Normalized frequency = %0.2f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#

#Answer varies due to round off error

Normalized frequency = 25.63  
Number of guided modes = 329

Example 2.7 Page no 481

In [81]:
# Calculation of normalized frequency and no of guided modes



#Given data
lamda=1.55*10**-6#      # Wavelength of fiber
a=30*10**-6#             # Core diameter of fiber
delta=0.015#            # Relative refractive index 
n1=1.48#                # Refractive index of core

# (a) Normalized frequency 
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda# 
#(b) Number of guided modes
m=v**2/2#                               

#Displaying results in the command window            
print"Normalized frequency = %0.2f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#
# The answers vary due to round off error

Normalized frequency = 31.17  
Number of guided modes = 486

Example 2.8 Page no 481

In [82]:
# Calculation of normalized frequency and no of guided modes


#Given data
lamda=1.55*10**-6#         # Wavelength of fiber
a=4*10**-6#                 # Core diameter of fiber
delta=0.01#                # Relative refractive index 
n1=1.48#                   # Refractive index of core
# (a) Normalized frequency 
v=(2*pi*a*n1*(2*delta)**(1/2))/lamda# 
#(b) Number of guided modes
m=v**2/2#                               

#Displaying results in the command window            
print"Normalized frequency = %0.3f  "%(v)#
print"Number of guided modes = %0.0f  "%(m)#
# The answers vary due to round off error

Normalized frequency = 3.394  
Number of guided modes = 6

Example 2.9 Page no 481

In [83]:
# Calculation of Core radius

#Given data
lamda=0.85*10**-6#           # Wavelength of fiber
delta=0.015#                 # Relative refractive index 
n1=1.48#                     # Refractive index of core
v=2.403#                     # Normalized frequency for single mode fiber
# Computation of core radius
a=v*lamda/(2*pi*n1*sqrt(2*delta))#
a=a*10**6#

print"Radius of core = %0.1f micrometer "%(a)#

Radius of core = 1.3 micrometer

Example 2.10 Page no 482

In [84]:
# Calculation of Cut off wavelength

# Given data
V=2.403#              # Normalized frequency
delta=0.25#           # Refractive index of core
n1=1.46#              # Relative refractive index
a=4.5*10**-6#          # Radius of core

# Cut off wavelenth
lamda=(2*pi*a*n1*(sqrt(2*delta)))/V#       


print"Cut off wavelenth = %0.0f nm "%(lamda*10**8)#

# The answers vary due to round off error

Cut off wavelenth = 1215 nm

Example 2.11 Page no 482

In [85]:
#Calculation of (a) reflection and (b) loss of light signal at joint areas.



# Given data
n1=1.5#                  # Refractive index of core
n=1#                     # Refractive index of air

# (a) Reflection at the fiber air interface 
R=((n1-n)/(n1+n))**2#     

# (b) Light loss due to fiber air interface
l= -10*log10(1-R)#


print"Reflection at the fiber air interface = %0.2f  "%(R)#
print"Light loss due to fiber air interface = %0.2f dB "%(l)#

Reflection at the fiber air interface = 0.04  
Light loss due to fiber air interface = 0.18 dB

Example 2.12 Page no 482

In [86]:
# Computation of (a) numerical aperature and (b) maximum angle of entrance

#Given data
n1=1.48#             # Refractive index of core
n2=1.46#             # Refractive index of cladding

# (a) Numerical Aperture 
NA=sqrt(n1**2-n2**2)#

#(b) Maximum angle of entrance
theata=degrees(asin(NA))

#Displaying result in the command window            
print"Numerical Aperture  = %0.3f  "%(NA)#
print"Maximum angle of entrance = %0.0f degress "%(theata)#

# Final answer in the book is wrong. Please refer example 2.11 of 
# Fiber Optic Communication by Gerd Keiser book.

 

Numerical Aperture  = 0.242  
Maximum angle of entrance = 14 degress

Example 2.13 Page no 483

In [87]:
# Calculation of (a) core radius and (b) maximum value of angle of acceptance of the fiber

#Given data
lamda=1320*10**-9#          # Wavelength of fiber
delta=0.077#                # Relative refractive index 
n1=1.48#                    # Refractive index of core
n2=1.478#                   # Refractive index of cladding
v=2.403#                    # Normalized frequency

# (a) Core radius
a=v*lamda/(2*pi*delta)#
a=a*10**6#

#Numerical Aperture
NA=sqrt(n1**2-n2**2)#

# (b) Angle of acceptance
theata = degrees(asin(NA))


print"Radius of core = %0.1f micrometer "%(a)#
print"Numerical aperture = %0.5f  "%(NA)#
print"Angle of acceptance = %0.0f degrees "%(theata)#

# The answers vary due to round off error

Radius of core = 6.6 micrometer 
Numerical aperture = 0.07692  
Angle of acceptance = 4 degrees

Example 2.14 Page no 483

In [88]:
# Calculation of critical wavelength



#Given data
a=3*10**-6#            # Core diameter of fiber
delta=0.15#           # Relative refractive index 
v=2.405#              # Normalized frequency

# Critical wavelength
lamda=(2*pi*a*delta)/v#
lamda*=10**9#


print"Critical wavelength = %0.0f nm "%(lamda)#

# The answers vary due to round off error

Critical wavelength = 1176 nm