# Chapter20 Fibre Optic Communication¶

## Example 20.2.1,Pg.no.753¶

In [12]:
import math
from math import sqrt
n1=1.55  #RI of glass
#NA of the fibe is given as
NA=n1*sqrt(2*(n1-n2)/n1)
NA=round(NA,2)
print 'The numerical aperture is',NA
#Acceptance angle is given as
acc_angle=math.asin(NA)
acc_angle=round(acc_angle,2)
print 'The acceptance angle is',acc_angle

The numerical aperture is 0.35
The acceptance angle is 0.36


## Example 20.2.2,Pg.no.761¶

In [13]:
import math
from math import pi
d=50*10**-6
wav=0.8*10**-6
NA=0.352
#Determination of V number
V=(pi)*d*NA/wav
V=round(V,2)
print 'The V number is',V
#Determination of approximate number of modes
N=(V**2)/2
N=round(N,1)
print 'the approximate no . of modes are',N

The V number is 69.12
the approximate no . of modes are 2388.8


## Example 20.2.3,Pg.no.762¶

In [14]:
import math
from math import pi
d=5*10**-6
wave=1.3*10**-6
NA=0.35
#Determination of V number
V=(pi)*d*NA/wave
V=round(V,4)
print 'The v number is',V
print 'From the table it is seen that 6 modes have cut off v less than 4.23'

The v number is 4.2291
From the table it is seen that 6 modes have cut off v less than 4.23


## Example 20.2.4,Pg.no.763¶

In [32]:
import math
V=69.1  #normalized cutoff frequency
N=2390  #number of modes supported as a step index fiber
#Determination of no . of modes supported by graded index fiber
N_a=(N*a)/(a+2)
print 'No . of modes supported by graded index fiber=',N_a

No . of modes supported by graded index fiber= 1195


## Example 20.2.5,Pg.no.763¶

In [30]:
import math
from math import pi
d=10*10**-6
wav=1.3*10**-6
n1=1.55
V_max=2.405
NA=(V_max*wav)/(pi*d)
#a) Determination of maximum normailized index difference
del1=(0.5)*((NA/n1)**2)
del1=round(del1,3)
print 'a) the normilized index difference is',del1
#b) Determination of r effective index of claddin glass
n2=n1*(1-del1)
n2=round(n2,2)
print 'b) cladding index required is',n2
#Determination of the fiber acceptance angle
theta_max=math.asin(NA)
theta_max=round(theta_max,3)
print'The max acceptance angle is',theta_max

a) the normilized index difference is 0.002
b) cladding index required is 1.55
The max acceptance angle is 0.1


## Example 20.3.1,Pg.no.766¶

In [29]:
import math
A_max=25
A1=2
A2=0.3
#a) Determination of repeater dist at 0.9um wavelength
z1=A_max/A1
print 'a)The repeater dist for 0.9um wavelength is',z1,'km'
#b) Determination of repeater dist at 1.5um wavelength
z2=A_max/A2
z2=round(z2,2)
print 'b)The repeater dist for 1.5um wavelength is',z2,'km'

a)The repeater dist for 0.9um wavelength is 12 km
b)The repeater dist for 1.5um wavelength is 83.33 km


## Example 20.4.1,Pg.no.772¶

In [28]:
import math
#given
n1=1.55
del1=0.0258
l=12.5
z=1000
c=3*10**8    #velocity of light
#a) Determination of intermodal dispersion
del_per_km=(n1*z*del1)/((1-del1)*c)*10**7
del_per_km=round(del_per_km,2)*10**-7
print 'The intermodal dispersion is',del_per_km ,'s/km'
#b) Determination of intermodal dispersion for l =12.5
del_l=del_per_km*l/1000*10**9
del_l=round(del_l,2)*10**-9
print 'The intermodal dispertion for l=12.5 is',del_l,'s'

The intermodal dispersion is 1.37e-07 s/km
The intermodal dispertion for l=12.5 is 1.71e-09 s


## Example 20.4.2,Pg.no.773¶

In [19]:
import math
n1=1.55
del1=(258.0)*(10**-2)
z=1000
c=3*10**8
z_disp=12.5
#Determination of intermodal dispersion
del_total=round(del_total,2)*10**-5
print 'The intermodal dispersion is',del_total,'sec'

The intermodal dispersion is 5.37e-05 sec


## Example 20.4.3,Pg.no.774¶

In [20]:
import math
#given
wav_0=0.8*10**-6
Dm=-0.15
wav_3=1.5
z=12.5
del_t=Dm*wav_3
#Determination of total material dispersion
del_md=del_t*z
print 'The total material dispersion is',del_md,'ns'

The total material dispersion is -2.8125 ns


## Example 20.4.4,Pg.no.775¶

In [21]:
import math
#given
Dm=6.6
z=12.5
del_3=6
del_wg=Dm*z*del_3
print'Expected waveguide dispersion is',del_wg,'ps'

Expected waveguide dispersion is 495.0 ps


## Example 20.4.5,Pg.no.776¶

In [27]:
import math
#given
del_imd=0
del_md=2.81
del_wgd=0.495
t_w=2.5
del_tot=((del_imd**2)+(del_md**2)+(del_wgd**2))**(1/2)
print 'The total dispersion is',del_tot,'ns'
t_r=((t_w**2)+(del_tot**2))**(1/2)
#Determination of max allowed bit rate
B=(1000/(2*t_r))
print 'The max allowed bit rate is',B,'Mbps'

The total dispersion is 1.0 ns
The max allowed bit rate is 500.0 Mbps


## Example 20.4.6,Pg.no.778¶

In [23]:
import math
#given
del_t=4.0
B=10.0
#a) Determination of BW distance product
BDP=1/(2*del_t)
print'a)The BW distance product for fiber is',BDP,'Mbps−km'
#b) Determiation of dispersion limited length
z_max_disp=BDP/(B*10**-3)
print'b)The disp limited length for a fiber is',z_max_disp,'km'

a)The BW distance product for fiber is 0.125 Mbps−km
b)The disp limited length for a fiber is 12.5 km


## Example 20.5.1,Pg.no.780¶

In [24]:
import math
E1=1.9
E2=1.46
E3=0.954
eV=1.9       #All in eV
c=3*10**8    #speed of light
#a) Determination of wavelength and freq for E1=1.9
wav1=1.241/E1
f1=c/(wav1)
wav1=round(wav1,1)
f1=round(f1,1)
print 'a) i) the wavelength is',wav1,'um'
print 'a) ii) the freq is',f1,'MHz'
#b) Determination of wavelength and freq for E2=1.46
wav2=1.241/E2
f2=c/(wav2)
print 'b) i) the wavelength is',wav2,'um'
print 'b) ii) the freq is',f2,'MHz'
#c ) Determination of wavelength and freq for E3=0.945
wav3=1.241/E3
f3=c/(wav3)
wav3=round(wav3,1)
f3=round(f3,1)
print'c)i) the wavelength is',wav3,'um'
print'c)ii) the freq is',f3,'MHz'

a) i) the wavelength is 0.7 um
a) ii) the freq is 459307010.5 MHz
b) i) the wavelength is 0.85 um
b) ii) the freq is 352941176.471 MHz
c)i) the wavelength is 1.3 um
c)ii) the freq is 230620467.4 MHz


## Example 20.8.1,Pg.no.799¶

In [25]:
import math
#given
pt=0
pr=-57
Nc=2
BER=10**-9
N=5
Lpt=6
Lpr=6
Lc=1
Ls =0.5
Lf=2
M=5
del_t=0.505
B=35
Ns=5
#a) Determination of loss−limited fiber length
z=(pt-pr-M-(Nc*Lc)-(Ns*Ls)-Lpt-Lpr)/Lf
print 'a) the loss−limited fiber is',z,'km'
#b) Determination of max BW for loss−limited fiber length
B_max=1/(5*del_t*z)
B_max=round(B_max,3)
print 'b) the max BW for loss−limited length is',B_max,'Gbps'
#c ) Determination of dispersion−limited length
z_disp=1000/(5*del_t*B)
z_disp=round(z_disp,2)
print 'the dispertion limited length is',z_disp,'km'

a) the loss−limited fiber is 17.75 km
b) the max BW for loss−limited length is 0.022 Gbps
the dispertion limited length is 11.32 km