# Chapter 9: Optical Communication Systems¶

## Example 9.1, Page Number 449¶

In [1]:
from math import sqrt

#Given P-I-N Diode
q=0.6 #Quantam Efficiency
l=1.3*(10**-6) #Wavelength in Meters
i=3*(10**-9) #Reverse Bias Leakage Current in Ampere
r=50 #Resistance in Ohm
b=500*(10**6) #Bandwidth in Hertz
P=10*(10**-6) #Optical Power in Watt
e=1.6*(10**-19) #Charge of a Electron
h=6.6*(10**-34) #Plancks Constant
c=3*(10**8)  #Speed Of Light
k=1.38*(10**-23) #Boltzmann Constant

ip=(q*P*e*l)/(h*c)  #Where i is the Photogenerated current

print "The Photogenereated Current is %.2e A"%(ip)

itotal=sqrt(2*(i+ip)*e*b) #Where itotal is the Total Shot Noise Current using equation 9.11

print "The total shot noise Current is %.2e A"%(itotal)

ij=sqrt(4*k*r*b*300)/r #Where ij is the Total Johnson Noise Current using Equation 9.13

print "The Total Johnson Noise Current is "+str(ij)+" A"

sn=(ip**2)/((itotal**2)+(ij**2)) #Where sn is the Signal to Noise Ratio in Decibel
sn=round(sn,1)
print "The Required Signal to noise ratio is "+str(sn)

rl=1/(2*3.14*c1*b) #Where rl is the optimum Load Resistance
rl=round(rl,1)
print "The optimum Load Resistance "+str(rl)+" ohm"

ij2=sqrt(4*k*rl*b*300)/rl

print "The Optimum Johnson Noise Current is %.2e A"%(ij2)

sn1=(ip**2)/((itotal**2)+(ij2**2))
sn1=round(sn1,1)
print "The New Signal to noise Ratio is "+str(sn1)

The Photogenereated Current is 6.30e-06 A
The total shot noise Current is 3.18e-08 A
The Total Johnson Noise Current is 4.06939798988e-07 A
The Required Signal to noise ratio is 238.5
The optimum Load Resistance 318.5 ohm
The Optimum Johnson Noise Current is 1.61e-07 A
The New Signal to noise Ratio is 1471.1


## Example 9.2, Page Number 462¶

In [2]:
print "The Transmitter Output is 0 Dbm"
print "The Receiver Sensitivity is -50 Dbm"
print "The Required Margin is 50 Dbm"
print "System Loss-"
print "Fiber Loss 2 Db/Km,15 Km                                        30 Db"
print "Detector Coupling Loss                                           1 Db"
print "Total Splicing Loss (0.5 DB x 10)                                5 Db"
print "Headroom for Temperature range,ageing effects & Future Splices   5 Db"
print "Total Attenuation                                               41 Db"
f=30
d=1
t=5
h=5
t=f+d+t+h  #The total power attenuation
p=50-t

print "The Excess Power Margin required is "+str(p)+" Db"

The Transmitter Output is 0 Dbm
The Receiver Sensitivity is -50 Dbm
The Required Margin is 50 Dbm
System Loss-
Fiber Loss 2 Db/Km,15 Km                                        30 Db
Detector Coupling Loss                                           1 Db
Total Splicing Loss (0.5 DB x 10)                                5 Db
Headroom for Temperature range,ageing effects & Future Splices   5 Db
Total Attenuation                                               41 Db
The Excess Power Margin required is 9 Db


## Example 9.3, Page Number 474¶

In [3]:
from math import sqrt
n1=2.286 #The Ordinary Refractive Index
d=6*(10**-3) #Refractive Index Change
n2=n1-d #The Difference of the Two

NA=sqrt((n1**2)-(n2**2))

first=1/(4*NA)
first=round(first,2)

second=3/(4*NA)
second=round(second,2)

print "The Requirement for Single Mode Behaviour becomes"
print str(first)+" <= d/lambda <= "+str(second)  #Where d=5*Lambda for suitable thickness design

#The Higher Region is Miscalculated in the Book

The Requirement for Single Mode Behaviour becomes
1.51 <= d/lambda <= 4.53