Chapter 4: Signal Degradation in Optical Fibers¶

Example 4.1, Page number 100¶

In [3]:
'''Calculate the loss of fiber'''

from sympy import *
import math

#Variable declaration
L = 400./1000.   #distance(km)
Pi = Symbol("Pi")
Po = 0.25*Pi

#Calculations
Loss = (10*math.log10(Pi/Po))/L

#Result
print "Loss =",round(Loss),"dB/km"

Loss = 15.0 dB/km


Example 4.2, Page number 101¶

In [4]:
'''Calculate power level'''

#Variable declaration
L = 3            #distance(km)
a = 0.5          #attenuation loss(dB/km)
lamda = 0.82     #wavelength(um)
Pi = 1           #input power(mW)

#Calculation
Po = Pi*10**((-a*L)/10)

#Result
print "Power level =",round(Po,2),"mW"

Power level = 0.71 mW


Example 4.3, Page number 101¶

In [5]:
'''Determine maximum possible link length'''

#Variable declaration
a = 0.5                  #attenuation loss(dB/km)
Pi = 1.5*10**-3          #input power(uW)
Po = 2.*10**-6           #output power(uW)

#Calculation
L = (10*math.log10(Pi/Po))/a

#Result
print "Maximum possible link length =",round(L,2),"km"

Maximum possible link length = 57.5 km


Example 4.4, Page number 102¶

In [6]:
'''Determine pulse dispersion per unit length'''

#Variable declaration
d = 30            #distance

#Calculation
D = pd/d

#Result
print "Pulse dispersion per unit length =",D,"ns/m"

Pulse dispersion per unit length = 4e-05 ns/m


Example 4.5, Page number 102¶

In [7]:
'''Determine the optical output power'''

#Variable declaration
L = 30           #distance(km)
a = 0.8          #attenuation loss(dB/km)
lamda = 1300     #wavelength(nm)
Pi = 200         #input power(uW)

#Calculation
Po = Pi*10**((-a*L)/10)

#Result
print "Optical output power =",round(Po,3),"uW"

Optical output power = 0.796 uW


Example 4.6, Page number 103¶

In [8]:
'''Determine
a)Overall signal attenuation
b)signal attenuation per km
d)input output power ratio'''

import math

#Variable declaration
L = 8            #distance(km)
Pi = 120*10**-6  #input power(W)
Po = 3*10**-6    #output power(W)

#Calculations
#Part a
a = 10*math.log10(Pi/Po)

#Part b

#Part c
l = 10
x = adb*l       #loss occured along 10km of fiber
y = 9           #loss due to splices(dB)
dB = x+y

#Part d
pi_by_po = 10**(round(dB)/10)

#Results
print "a) Overall signal attenuation =",round(a),"dB"
print "b) Signal attenuation per km =",round(adb),"dB/km"
print "c) Signal attenuation for the link =",round(dB),"dB"
print "d) Input output power ratio =",round(pi_by_po,1)

a) Overall signal attenuation = 16.0 dB
b) Signal attenuation per km = 2.0 dB/km
c) Signal attenuation for the link = 29.0 dB
d) Input output power ratio = 794.3


Example 4.7, Page number 104¶

In [6]:
'''Determine theoretical attenuation for two given wavelengths'''

import math

#Variable declaration
Tf = 1400           #fictive temperature(K)
p = 0.286           #photoelastic coeeficient for silica
n = 1.46            #refractive index
Bo = 7*10**-11      #isothermal compressibility
K = 1.381*10**-23   #Boltzman's constant(J/k)
L = 10**3

#Calculation
#For lamda = 0.63 um
lamda1 = 0.63*10**-6 #wavelength(m)
Yr1 = (8*math.pi**3*n**8*p**2*Bo*K*Tf)/(3*lamda1**4)
E1 = math.exp(-Yr1*L)
a1 = 10.*math.log10(1/E1)

#For lamda = 1.30 um
lamda2 = 1.30*10**-6 #wavelength(m)
Yr2 = (8*math.pi**3*n**8*p**2*Bo*K*Tf)/(3*lamda2**4)
E2 = math.exp(-Yr2*L)
a2 = 10.*math.log10(1/E2)

#Result
print "(solutions of example 4.8 and 4.7 are interchanged in the textbook)"
print "Theoretical attenuation =",round(a1,3),"dB/km"
print "Theoretical attenuation =",round(a2,3),"dB/km"

(solutions of example 4.8 and 4.7 are interchanged in the textbook)
Theoretical attenuation = 5.21 dB/km
Theoretical attenuation = 0.287 dB/km


Example 4.8, Page number 104¶

In [7]:
'''Calculating the refractive index of glass'''

import math

#Variable declaration
lamda = 1*10**-6    #wavelength(m)
p = 0.245           #photoelastic coefficient
Bc = 8.4*10**-11    #isothermal compressebility(m^2/N)
Tf = 758            #fictive temperature(K)
K = 1.381*10**-23   #Boltzman's constant(J/k)
alpha = 0.46        #attenuation(Db/Km)
L = 1.*10**3        #distance(km)

#Calculations
Yr = (8*math.pi**3*p**2*Bc*K*Tf)/(3*lamda**4)
n= (alpha/(4.34*Yr*10**3))**(1./8.)

#Result
print "Refractive index =",round(n,2)

Refractive index = 1.49


Example 4.9, Page nuber 106¶

In [11]:
'''Calculating the operating wavelength of laser and attenuation of fiber'''

#Variable declaration
Pb = 150.*10**-3     #threshold optical power for brillouin(W)
Pr = 1.5             #threshold optical power for raman(W)
d = 8.0               #core diameter(um)
v = 1                #bandwidth(GHz)

#Calculations
'''
Pb is given by,
Pb = 4.4*10**-3*d^2*lamda^2*alpha*v ----(1)

Pr is given by,
Pr = 5.9*10**-2*d^2*lamda*alpha  ----(2)

Dividing (1) by (2), we get,
'''

lamda = (Pb*5.9*10**-2*d**2)/(Pr*4.4*10**-3*d**2)

alpha = Pr/(5.9*10**-2*d**2*lamda)

#Results
print "Operating wavelength =",round(lamda,2),"um"
print "Attenuation =",round(alpha,3),"dB/km"

Operating wavelength = 1.34 um
Attenuation = 0.296 dB/km


Example 4.10, Page number 107¶

In [12]:
'''Estimating the threshold optical powers for Brillowin and raman scattering'''

#Variable declaration
d = 6          #core diameter(um)
v = 0.8        #bandwidth(GHz)
lamda = 1.5    #wavelength(um)
alpha = 0.5    #attenuation(dB/Km)

#Calculations
Pb = 4.4*10**-3*d**2*lamda**2*alpha*v

Pr = 5.9*10**-3*d**2*lamda*alpha

#Results
print "Threshold optical power for SBS =",round((Pb/1E-3),2),"mW"
print "Threshold optical power for SRS =",round((Pr/1E-1),2),"W"

Threshold optical power for SBS = 142.56 mW
Threshold optical power for SRS = 1.59 W


Example 4.11, Page number 107¶

In [13]:
'''Determing the critical radius of curvature'''

import math

#Variable declaration
delta = 3./100      #relative refracive index difference
n1 = 1.5            #refractive index of core
lamda = 0.82*10**-6 #operating wavelength(m)

#Calculations
n2 = math.sqrt(n1**2-2*delta*n1**2)

Rc = (3*n1**2*lamda)/(4*math.pi*(n1**2-n2**2)**0.5)

#Results
print "The critical radius of curvature is",round((Rc/1E-6),3),"um(Calculation mistake in textbook)"

The critical radius of curvature is 1.199 um(Calculation mistake in textbook)


Example 4.12, Page number 108¶

In [9]:
'''Calculating the crtitical radius of curvature'''

#Variable declaration
delta = 3./100      #relative refracive index difference
n1 = 2.0            #refractive index of core
lamda = 1.55*10**-6 #operating wavelength(m)
d = 8               #core diameter(um)

#Calculations
a = d/2
n2 = math.sqrt(n1**2-2*delta*n1**2)
lamda_c = (2*math.pi*a*10**-6*n1*((2*delta)**0.5))/2.405
Rc = (20*lamda*(2.748-0.996*(lamda/lamda_c))**-3)/((n1**2-n2**2)**0.5)

#Result
print "The critical radius of curvature is",round((Rc/1E-6),3),"um"

#note There is a calculation mistake in calculating 'n' in the textbook. Hence, the difference in results

The critical radius of curvature is 4.322 um


Example 4.13, Page number 109¶

In [15]:
'''Determine-
a)Maximum possible bandwidth
b)pulse dispersion per unit length
c)bandwidth length product'''

#Variable declaration
L = 10            #distance(km)

#Calculatons
#Part a
Bt = 1/(2*t)

#Part b
D = t/L

#Part c
Bl = Bt*L

#Results
print "a)Maximum possible bandwidth =",Bt/1E+6,"MHz"
print "b)Pulse dispersion per unit length =",D/1E-6,"us/km"
print "c)Bandwidth length product =",Bl/1E+6,"Mhz km"

a)Maximum possible bandwidth = 5.0 MHz
b)Pulse dispersion per unit length = 0.01 us/km
c)Bandwidth length product = 50.0 Mhz km


Example 4.14, Page number 109¶

In [12]:
'''Estimating the wavelength of transmitted light'''

#Variable declaration
Rc = 84*10**-6 #radius of curvature(m)
n1 = 1.46     #core refractive index
n2 = 1.45     #cladding refractive index

#Calculations
lamda = (Rc*4*math.pi*(n1**2-n2**2)**0.5)/(3*n1**2)

#Result
print "Wavelength =",round(lamda/1e-6,2),"um(Calculation mistake in textbook)"

Wavelength = 28.16 um(Calculation mistake in textbook)


Example 4.15, Page number 110¶

In [17]:
'''Calculating the pulse broadening due to dispersion'''

#Variable declaration
lamda = 1.5       #wavelength(um)
M = 20            #dispersion parameter(ps n/m-km)
L = 30            #length of fiber(km)
s_l = 2           #spectral width(nm)

#Calculations
s_m = s_l*L*M

#Result

Pulse broadening = 1200 ps


Example 4.16, Page number 110¶

In [17]:
'''Calculating material dispersion parameter and rms pulse broadening'''

#Variable declaration
lamda = 850       #wavelength(nm)
s_l = 20          #spectral width(nm)
Dh = 0.025        #material dispersion
c = 2.998*10**5   #speed of light(km/s)
L = 1             #length of fiber(km)

#Calculations
M = Dh/(c*lamda)

s_m = s_l*L*M

#Results
print "Material dispersion parameter =",round((M/1E-12),2),"ps n/m km^-1"

Material dispersion parameter = 98.1 ps n/m km^-1
RMS pulse broadening = 1.96 ns/km


Example 4.17, Pge number 111¶

In [21]:
'''Determing the pulse boradening per km'''

#Variable declaration
lamda = 0.9       #wavelength(um)
s_l = 45          #spectral width(um)
Dh = 4*10**-2     #material dispersion
c = 3*10**8       #speed of light(m/s)
L = 1             #length of fiber(km)

#Calculations
s_m = (s_l*L*Dh*lamda)/c

#Result

Pulse boradening = 5.4 ns/km


Example 4.18, Page number 111¶

In [22]:
'''Determing the pulse boradening per km'''

#Variable declaration
lamda = 0.85       #wavelength(nm)
M = 95             #material dispersion parameter(ps n/m km^-1)

#Calculations
s_l = 0.0012*lamda
s_m = s_l*L*M

#Result

Pulse boradening = 0.0969 ns/km


Example 4.19, Page number 112¶

In [18]:
'''Calculating the pulse broadening'''

#Variable declaration
NA = 0.275       #numerical aperture
n1 = 1.48        #core refractive index
L = 5            #length of fiber(km)
c = 3.*10**5     #speed of light(km/sec)

#Calculations
STs = (L*NA**2)/(2*n1*c)

#Result
print "Pulse broadening =",round((STs/1E-9),2),"ns(Calculation mistake in textbook)"

Pulse broadening = 425.82 ns(Calculation mistake in textbook)


Example 4.20, Page number 112¶

In [24]:
'''Calculate -
a)delay difference between modes
c)maximum bit rate
d)bandwidth length product'''

from math import sqrt

#Variable declaration
L = 3*10**3   #length of link(m)
n1 = 1.5      #refractive index of core
D = 2./100     #refractive index difference
c = 2.998*10**8 #speed of light(m/s)

#Calculations
#Part a
Sts = (L*n1*D)/c

#Part b
s_s = (L*n1*D)/(2*sqrt(3)*c)

#Part c
Bt = 1/(2*Sts)

#Part d
Bl = Bt*L

#Results
print "Delay difference is",round(Sts/1E-9),"ns"
print "rms pulse broadening due to intermodal dispersion is",round((s_s/1E-9),1),"ns"
print "Maximum bit rate is",round((Bt/1E+6),1),"Mbits/sec"
print "Bandwidth length product is",round((Bl/1E+9),1),"MHz km"

Delay difference is 300.0 ns
rms pulse broadening due to intermodal dispersion is 86.7 ns
Maximum bit rate is 1.7 Mbits/sec
Bandwidth length product is 5.0 MHz km


Example 4.21, Page number 113¶

In [25]:
'''Calculating the rms pulse broadening per km for multimode step index fiber and graded index fiber'''

#Variable declaration
L = 6           #length of link(km)
n1 = 1.5        #core refractive index
delta = 1./100  #refractive index difference
c = 2.998*10**8 #speed of light(m/s)

#Calculations
s_m = s_s/L

L1 = 10**3
s_g = (L1*n1*delta**2)/(20*sqrt(3)*c)

#Result
print "The rms pulse broadening per km for multimode step index fiber is",s_m,"ns/km"
print "The rms pulse broadening per km for graded index fiber is",round((s_g/1E-12),2),"ps/km"

The rms pulse broadening per km for multimode step index fiber is 14.45 ns/km
The rms pulse broadening per km for graded index fiber is 14.44 ps/km


Example 4.22, Page number 114¶

In [26]:
'''Estimating the total pulse broadening per km '''

#Variable declaration
NA = 0.4         #numerical aperture
n1 = 1.48        #core refractive index
n2 = 1.47        #cladding refractive index
M = 30           #material dispersion parameter(ps n/m /km)
s_l = 25         #spectral width(ns)
L = 1            #length of fiber(km)
delta = 1./100    #refractive index difference

#Calculations
s_m = M*L*s_l
s_s = (L1*n1*delta**2)/(20*sqrt(3)*c)
s_t = math.sqrt(s_m**2+s_s**2)

#Results
print "Total pulse broadening per km is",s_t,"ps"

Total pulse broadening per km is 750.0 ps


Example 4.23, Page number 115¶

In [39]:
'''Calculate-
b)bandwidth length product'''

import math

#Variable declaration
NA = 0.3         #numerical aperture
n1 = 1.45        #core refractive index
M = 250          #material dispersion parameter(ps n/m /km)
s_l = 50         #spectral width(ns)
L = 1            #length of fiber(km)
c = 2.998*10**8  #speed of light(m/s)

#Calculations
#Part a
s_m = (M*L*s_l)*10**-3                           #ns/km
s_s = ((L*10**3*NA**2)/(4*math.sqrt(3)*n1*c*10**3))*10**12  #ns/km
s_t = math.sqrt(s_m**2+s_s**2)

#Part b
Bl = 0.2/s_t

#Results
print "b)bandwidth length product =",round((Bl/1E-3),1),"Mhz km"

a)total pulse broadening = 32.39 ns/km
b)bandwidth length product = 6.2 Mhz km


Example 4.24, Page number 116¶

In [28]:
'''Finding the beat length'''

#Variable declaration
lamda = 1.32       #wavelength(um)
Lbc = 100          #length of fiber(km)
Sf = 1.5           #spectral width(nm)

#Calculations
Bf = lamda**2/(Lbc*Sf)
Lb = lamda/Bf

#Result
print "Beat length =",round(Lb,1),"km"

Beat length = 113.6 km


Example 4.25, Page number 116¶

In [29]:
'''Calculate modal birefringence'''

#Variable declaration
lamda = 1.3*10**-6  #wavelength(um)
Lb1 = 0.5*10**-3    #length of fiber(mm)
Lb2 = 60            #(m)

#Calculations
#for beat length of 0.5mm,
Bf1 = lamda/Lb1

#for beat length of 60m,
Bf2 = lamda/Lb2

#Result
print "The modal birefringences are ",Bf1/1E-3, "*10^-3 and",round((Bf2/1E-8),2),"*10^-4"

The modal birefringences are  2.6 *10^-3 and 2.17 *10^-4


Example 4.26, Page number 117¶

In [30]:
'''Calculate the modal birefringence, coherence length and difference between propagation constants'''

#Variable declaration
lamda = 0.5*10**-6  #wavelength(um)
Lb = 5*10**-2       #length of fiber(km)
S_l = 1*10**-9      #spectral width(nm)

#Calculations
Bf = lamda/Lb

Lbc = lamda**2/(Bf*S_l)

B = (2*math.pi)/Lb      #Bx-By

#Results
print "Modal birefringence =",Bf
print "Coherence length =",Lbc,"m"
print "Difference between propagation constants =",round(B,1)

Modal birefringence = 1e-05
Coherence length = 25.0 m
Difference between propagation constants = 125.7


Example 4.27, Page number 117¶

In [31]:
'''Calculating the maximum bit rate'''

#Variable declaration
L = 10*10**3        #length of fiber(m)
St2 = 600*10**-12   #mode dispersion(s/km)

#Calculations
Bt = 0.9/(0.55*St2*L)

#Result
print "Maximum bit rate =",round(Bt/1E+3),"Kbps"

Maximum bit rate = 273.0 Kbps