chapter8 - Optical receiver operation

Example 8.2.1, page 8-6

In [5]:
from __future__ import division
from numpy import log, sqrt, pi, log10
P=10**-9     #probability of error
eta=1       #ideal detector
h=6.626*10**-34 #plank's constant
c=3*10**8      #speed of light
lamda=1*10**-6  #wavelength
B=10**7      #bit rate
Mn = -log(P) 
print "The quantum limit at the receiver to maintain bit error rate 10**-9 is (%.1f*h*f)/eta." %Mn
f=c/lamda
Popt= 0.5*Mn*h*f*B/eta      #computing optical power
Popt_dB = 10 * log10(Popt) + 30     #optical power in dbm
Popt=Popt*10**12 
print "Minimum incident optical power is %.1f W or %.1f dBm." %(Popt,Popt_dB) 

The quantum limit at the receiver to maintain bit error rate 10**-9 is (20.7*h*f)/eta.
Minimum incident optical power is 20.6 W or -76.9 dBm.

Example 8.2.2, page 8-8

In [7]:
SN_dB=60     #signal to noise ratio
h=6.626*10**-34 #plank's constant
c=3*10**8      #speed of light
lamda=1.3*10**-6  #wavelength
eta=1 
B=6.5*10**6      #Bandwidth
SN=10**(SN_dB/10) 
f=c/lamda
Popt= 2*SN*h*f*B/eta      #computing optical power
Popt_dB = 10 * log10(Popt) + 30     #optical power in dbm
Popt=Popt*10**6 
print "Incident power required to get an SNR of 60 dB at the receiver is %.4f microWatt or %.3f dBm" %(Popt,Popt_dB)
#Calculation error in the book.They have take SN as 10**5 while calculating, which has lead to an error in final answer
#answer in the book 198.1nW and -37.71 dBm

Incident power required to get an SNR of 60 dB at the receiver is 1.9878 microWatt or -27.016 dBm

Example 8.3.1, page 8-11

In [9]:
lamda=0.85*10**-6 
h=6.626*10**-34     #plank's constant
c=3*10**8       #speed of light
q=1.6*10**-19   #charge of electron
eta=65/100  #quantum efficiency
P0=300*10**-9   #optical power
Id=3.5     #dark current
B=6.5*10**6      #bandwidth
K=1.39*10**-23  #Boltzman constant
T=293       #temperature
R=5*10**3      #load resister
Ip= 10**9*eta*P0*q*lamda/(h*c) 
Its=10**9*(2*q*B*(Ip+Id)) 
Its=sqrt(Its) 
print "rms shot noise current is %.2f nA." %(Its) 
It= 4*K*T*B/R 
It=sqrt(It) 
It=It*10**9 
print "Thermal noise is %.2f nA." %(It) 
#answer given in book for Thermal noise is wrong.

rms shot noise current is 0.53 nA.
Thermal noise is 4.60 nA.

Example 8.3.2, page 8-12

In [13]:
lamda=0.85*10**-6 
h=6.626*10**-34     #plank's constant
c=3*10**8       #speed of light
q=1.6*10**-19   #charge of electron
eta=65/100  #quantum efficiency
P0=300*10**-9   #optical power
Id=3.5     #dark current
B=6.5*10**6      #bandwidth
K=1.39*10**-23  #Boltzman constant
T=293       #temperature
R=5*10**3      #load resister
F_dB=3      #noise figure
F=10**(F_dB/10) 
Ip=10**9*eta*P0*q*lamda/(h*c) 
Its=10**9*(2*q*B*(Ip+Id)) 
It1= 4*K*T*B*F/R 
SN= Ip**2/(Its+It1) 
SN_dB=10*log10(SN) 
SN=SN
print "SNR is %.2e or %.2f dB." %(SN,SN_dB) 
#answer given in the book is wrong.

SNR is 6.25e+04 or 47.96 dB.

Example 8.4.1, page 8-14

In [14]:
Cd=7*10**-12 
B=9*10**6 
Ca=7*10**-12 
R=(2*3.14*Cd*B)**-1 
B1=(2*3.14*R*(Cd+Ca))**-1 
R=R/1000 
B1=B1/10**6 
print """Thus for 9MHz bandwidth maximum load resistance is %.2f Kohm
Now if we consider input capacitance of following amplifier Ca then Bandwidth is %.2fMHz
Maximum post detection bandwidth is half."""%(R,B1) 
#answer for resistance in the book is wrong

Thus for 9MHz bandwidth maximum load resistance is 2.53 Kohm
Now if we consider input capacitance of following amplifier Ca then Bandwidth is 4.50MHz
Maximum post detection bandwidth is half.

Question 7, page 8.44

In [16]:
w=25*10**-6     #width
v=3*10**4       #velocity
t=w/v       #computing drift time
BW=(2*pi*t)**-1         #computing bandwidth
rt=1/BW     #response time
rt=rt*10**9 
print "Maximum response time is %.2f ns." %(rt) 
#Answer in the book is wrong.

Maximum response time is 5.24 ns.