Chapter 06:Optical Amplifiers

Ex6.1:pg-249

In [2]:
# Example 6.1
# Calculation of the gain
# Page no 249

import math

#Given data

n=1.5;                    # Refractive ondex of air
lambdaa=1550*10**-9;        # Wavelength 
c=3*10**8;                 # Velocity of light
p=5.73*10**-17;            # Power spectral density
h=6.63*10**-34             # Planck constant


# Gain
f=c/lambdaa;

G=(p/(2*n*h*f))+1;

#Displaying results in the command window            
print "\n Gain G =",round(G)
 Gain G = 150.0

Ex6.2:pg-255

In [1]:
# Example 6.2
# Calculation of the variance of the signal–ASE beat noise
# Page no 255

import math



#Given data

a=1.3*10**-16;                 # PSD of an amplifier
f=7*10**9;                    # Cut off frequency
Pi=10*10**-6;                 # Input power
R=0.8;                      # Responsivity
G=20;                       # Gain of an amplifier

# The variance of the signal–ASE beat noise
G=10**(G/10.0);
P=G*Pi;

r=4*R**2*P*a*f;
r=r*10**9;




#Displaying results in the command window            
print "\n The variance of the signal–ASE beat noise current is = ",round(r,2)," x 10**-9  A**2"

# The answers vary due to round off error
 The variance of the signal–ASE beat noise current is =  2.33  x 10**-9  A**2

Ex6.3:pg-257

In [3]:
# Example 6.3
# Calculation of the (a) the variance of the signal–ASE beat noise current, (b) the variance of the ASE–ASE beat noise current, and (c) the total variance.
# Page no 257


import math


#Given data

G=30;       # Gain

nsp=5;
R=0.8;
f1=16*10**9;
fe=9*10**9;
#Pi1=-60;
c=3*10**8;             # Velocity of light
h=6.63*10**-34        # Planck constant
lambdaa=1530*10**-9;   # Wavelegth
Pi1=-27;            # Input power
Pi2=-60;
f0=16*10**9;


#(a) The variance of the signal–ASE beat noise current for Pin=-27dBm
Po=G+Pi1;
Po=10**(Po/10.0);
Po=Po*10**-3;
G=10**(G/10.0);
f=c/lambdaa;
r=nsp*h*f*(G-1);
B=8*10**9;
#B=min(f/2,fe);
r0=4*R**2*Po*r*B;
#r0=r0*10**12;

#(b) The variance of the ASE–ASE beat noise current

r1=R**2*r**2*((2*f0)-fe)*fe;

#r1=r1*10**11;
# (c) The total variance.

rt=r0+r1;

# Displaying results in the command window            
print "\n (a) The variance of the signal–ASE beat noise current for Pin=-27dBm"

print "\n The variance of the signal–ASE beat noise current = ",round(r0*10**8,2)," x 10**-8 A**2"
print "\n The variance of the ASE–ASE beat noise current = ",round(r1*10**11,2)," x 10**-11 A**2"

print "\n The total variance = ",round(rt*10**8,3)," x 10**-8 A**2"
# The answers vary due to round off error


#Given data

G=30;       # Gain
nsp=5;
R=0.8;       # 
f1=16*10**9;
fe=9*10**9;
#Pi1=-60;
c=3*10**8;    # Velocity of light
h=6.63*10**-34    # Planck constant
lambdaa=1530*10**-9;  # Wavelegth
Pi1=-27;              # Input power
Pi2=-60;
f0=16*10**9;

#(b) The variance of the signal–ASE beat noise current for Pin=-60dBm
Po2=G+Pi2;
Po=10**(Po2/10);
Po=Po*10**-3;
G=10**(G/10);
f=c/lambdaa;
r=nsp*h*f*(G-1);
B=8*10**9;
#B=min(f/2,fe);
r0=4*R**2*Po*r*B;
#r0=r0*10**12;

#(b) The variance of the ASE–ASE beat noise current

r1=R**2*r**2*((2*f0)-fe)*fe;

#r1=r1*10**11;
# (c) The total variance.

rt=r0+r1;

# Displaying results in the command window            
print "\n \n (b) The variance of the signal–ASE beat noise current for Pin=-60dBm"

print "\n The variance of the signal–ASE beat noise current = ",round(r0*10**11,2)," x 10**-11 A**2"
print "\n The variance of the ASE–ASE beat noise current = ",round(r1*10**11,2)," x 10**-11 A**2"

print "\n The total variance = ",round(rt*10**11,2)," x 10**-11 A**2 "
 (a) The variance of the signal–ASE beat noise current for Pin=-27dBm

 The variance of the signal–ASE beat noise current =  2.65  x 10**-8 A**2

 The variance of the ASE–ASE beat noise current =  5.59  x 10**-11 A**2

 The total variance =  2.659  x 10**-8 A**2

 
 (b) The variance of the signal–ASE beat noise current for Pin=-60dBm

 The variance of the signal–ASE beat noise current =  1.33  x 10**-11 A**2

 The variance of the ASE–ASE beat noise current =  5.59  x 10**-11 A**2

 The total variance =  6.92  x 10**-11 A**2 

Ex6.4:pg-262

In [4]:
# Example 6.4
# Calculation of the amplifier gain
# Page no 262

import math

#Given data

Po=0;                   # Signal output of amplifier
#f=7*10**9;             # Cut off frequency
B=7.5*10**9;             # Bandwidth
R=0.9;                  # Responsivity
c=3*10**8;               # Velocity of light
lambdaa=1550*10**-9;      # Operating frequency
fn=4.5;                 # Noise figure
Ro=0.066*10**-3;         # Beat noise current
h=6.626*10**-34;         # Planck constant

# The amplifier gain
P=10**(Po/10)*10**-3;
r=Ro**2/(4*R**2*B*P);
fn=10**(fn/10);
f=c/lambdaa;
G=(1/fn)*(((2*r)/(h*f))+1);




#Displaying results in the command window            
print "\n The amplifier gain = ",round(G)


# The answers vary due to round off error
 The amplifier gain =  992.0

Ex6.5:pg-263

In [5]:
# Example 6.5
# Calculation of the OSNR in a bandwidth of 12.49 GHz.
# Page no 263



import math

#Given data

G=25;                 # Gain
c=3*10**8;             # Velocity of light
h=6.63*10**-34         # Planck constant
lambdaa=1545*10**-9;    # Wavelegth
Pi=-22;              # Input power
fn=6;      
B=12.49*10**9;

# The OSNR in a bandwidth of 12.49 GHz
Po=G+Pi;
Po=10**(Po/10.0);
Po=Po*10**-3;
fn=10**(fn/10.0);
G=10**(G/10.0);
f=c/lambdaa;
r=(G*fn-1)*(h*f/2.0);
O=Po/(2*r*B);
O=10*math.log10(O);

# Displaying results in the command window            
print "\n The OSNR in a bandwidth of 12.49 GHz = ",round(O,2)," dB"



# The answers vary due to round off error
 The OSNR in a bandwidth of 12.49 GHz =  29.94  dB

Ex6.6:pg-268

In [6]:
# Example 6.6
# Calculation of the single-pass gain and 3-dB bandwidth
# Page no 268



import math

#Given data

c1=3*10**8;                # Velocity of light
f=7*10**9;                # Cut off frequency
L=500*10**-6;             # Input power
Gp=15;                   # Peak gain
n=3.2;
Gs=2.52;
R=0.32;
a=0.1024;
b=-0.6546;
c=1;

# The single-pass gain

x1 =( -1*b+ math.sqrt ((b **2) -4*a*c)) /(2* a); # 1 s t r o o t
x2 =( -1*b- math.sqrt ((b **2) -4*a*c)) /(2* a); # 2nd r o o t

# The 3-dB bandwidth
G=10**(Gp/10.0);
x=(1-(R*x2))/(2*math.sqrt(R*x2));
f=(c1/(math.pi*L*n))*math.asin(x);
# f=f*10**-9; 

# Displaying results in the command window            

print "Single pass gain Gs= ",round(x1,2),"   or"
print "\nSingle pass gain Gs= ",round(x2,2)
print "\nThe the 3-dB bandwidth = ",round(f*10**-9,2)," GHz "


# The answers vary due to round off error
Single pass gain Gs=  3.87    or

Single pass gain Gs=  2.52

The the 3-dB bandwidth =  6.39  GHz 

Ex6.7:pg-273

In [7]:
# Example 6.7
# Calculation of (a) the saturation power and (b) the bias current I
# Page no 273

import math




#Given data

c=3*10**8;                # Velocity of light
lambdaa=1530*10**-9;      # Wavelength
t=0.3                   # Overlap factor
r=7.3*10**-20;           # Gain cross section
r0=1*10**-9;            # Carrier lifetime
q=1.609*10**-19;     
v=7.5*10**-16;         # Active volume
h=6.63*10**-34         # Planck constant
A=5*10**-6;            # Effective area
g=4.82*10**3;          # Small signal gain coeffifient
N=3.5*10**23;          # 

# (a) the saturation power and 


f=c/lambdaa;
Ps=(h*f*A)/(t*r*r0);
Ps=Ps*10**-3;

# (b) the bias current I

I=(g/(r*r0)+N/r0)*q*v;
I=I*10**3;
# Displaying results in the command window            
print "\n The saturation power Psat = ",round(Ps,3)," mW "

print "\n The bias current I = ",round(I,3)," mA "


# The answers vary due to round off error
 The saturation power Psat =  29.68  mW 

 The bias current I =  50.204  mA 

Ex6.9:pg-290

In [10]:
# Example 6.9
# Calculation of the variance of the signal–ASE beat noise current.
# Page no 290


import math

#Given data

si=30.0;                # Electrical SNRs at the amplifier input
so=25.0;               # Electrical SNRs at the amplifier output
p=0;                 # Signal power at output 
r=-126.0;             # Signal power at input 
R=0.9;              # Planck constant
f=195*10.0**12;       # Frequency
b=20*10.0**9;         # Bandwidth

# The variance of the signal–ASE beat noise current
p1=10**(p/10)*10**-3;
rn=10**(r/10)*10**-3;
r1=rn*b;
r0=2*R**2*p1*r1;


#Displaying results in the command window            
print "\n The variance of the signal–ASE beat noise current = ",round(r0*10**9,2)," x 10**-9 A**2  W/Hz"


# The value of noise power given in example as -126 but for calculation it is taken as -128 in book. Therefore answer is varying.
 The variance of the signal–ASE beat noise current =  8.14  x 10**-9 A**2  W/Hz

Ex6.12:pg-296

In [11]:
# Example 6.12
# Calculation of the ASE power spectral density per polarization.
# Page no 296


import math
#Given data

si=30.0;                # Electrical SNRs at the amplifier input
so=25.0;               # Electrical SNRs at the amplifier output
po=2;                # Signal power at output 
pi=-13;             # Signal power at input 
h=6.626*10**-34;     # Planck constant
f=195*10**12;

# The ASE power spectral density per polarization
fn=si-so;
fn=10**(fn/10);
G=po-pi;
G=10**(G/10.0);
r=(h*f*(G*fn-1))/2;
r=r*10**18;

#Displaying results in the command window            
print "\n The ASE power spectral density per polarization = ",round(r,3)," x 10**-18 W/Hz "
 The ASE power spectral density per polarization =  6.396  x 10**-18 W/Hz 

Ex6.13:pg-296

In [14]:
# Example 6.13
# Calculation of the geometric mean of the facet reflectivity R
# Page no 296

import math

#Given data
Gm=20.0;
G1=5.0;

# The geometric mean of the facet reflectivity R
Gmax=10**(Gm/10);                    # Peak Gain
Gs=10**(G1/10);                      # Single pass gain
R=(math.sqrt(Gs)-10)/(math.sqrt(Gs)-Gs*10);




#Displaying results in the command window            
print "\n The geometric mean of the facet reflectivity R = ",round(R,3)
 The geometric mean of the facet reflectivity R =  0.275

Ex6.14:pg-297

In [15]:
# Example 6.14
# Calculation of the upper bound on the single-pass gain
# Page no 297
import math

#Given data

n=3.5;                    # Refractive index
c1=3*10**8;                # Velocity of light
L=200*10**-6;             # Amplifier length
a=0.09;
b=-(1.2*0.1805**2+0.6);
c=1;

# The geometric mean of the facet reflectivity R
f=c1/(2*n*L);

x1 =( -1*b+ math.sqrt ((b **2) -4*a*c)) /(2* a); # 1 s t r o o t
x2 =( -1*b- math.sqrt ((b **2) -4*a*c)) /(2* a); # 2nd r o o t



#Displaying results in the command window            
print "\n The geometric mean of the facet reflectivity R = ",round(f*10**-9,2)," GHz "
print "\n The upper bound on the single-pass gain Gs = ",round(x1,2)," or  "
print "\n The upper bound on the single-pass gain Gs = ",round(x2,2)


# The answers vary due to round off error
 The geometric mean of the facet reflectivity R =  214.29  GHz 

 The upper bound on the single-pass gain Gs =  4.77  or  

 The upper bound on the single-pass gain Gs =  2.33